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COPYRIGHT DEPOSIT.
ARITHMETIC
OF THE
STEAM BOILEE
THE POWER HANDBOOKS
The best library for the engineer and the man who hopes
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By PROF. AUGUSTUS H. GILL
OF THE MASSACHUSETTS INSTITUTE OF TECHNOLOGY
ENGINE ROOM CHEMISTRY
By HUBERT E. COLLINS
BOILERS KNOCKS AND KINKS
SHAFT GOVERNORS PUMPS
ERECTING WORK SHAFTING, PULLEYS AND
PIPES AND PIPING BELTING
By CHARLES J. MASON
ARITHMETIC OF THE STEAM BOILER
McGRAW-HILL BOOK COMPANY, Inc.
239* WEST 39TH STREET, NEW YORK
6 BOUVERIE STREET, LONDON, E. C.
THE POWER HANDBOOKS
5
ARITHMETIC
OF THE
STEAM BOILER
A REFERENCE BOOK
SHOWING THE VARIOUS APPLICATIONS OF
ARITHMETIC TO STEAM BOILERS
BY
CHARLES J. MASON
First Edition
McGRAW-HILL BOOK COMPANY, Inc.
239 WEST 39TH STREET, NEW YORK
6 BOUVERIE STREET, LONDON, E. C.
1914
TJ ■X16
.Hi.
Copyright 1913, by the
McGraw-Hill Book Company, Inc.
THE.MAPLE.PRESS.YORK-PA
JAN -9 1914
©CI.A361529
c
THIS BOOK
IS RESPECTFULLY DEDICATED
TO
MR. FRED R. LOW, EDITOR OF POWER,
WHO HAS EVER TAKEN A DEEP INTEREST IN THE
AFFAIRS OF THE ENGINEERING FRATERNITY
PREFACE
This book is a compilation of arithmetical rules and
formulas applicable to steam boilers of various types.
The author claims no originality in the preparation of
the material, excepting only the arrangement and manner
of presentation.
It is intended as a book of reference for those who
may require rules and formulas directly related to steam
boilers, and its aim is concentration and logical order in
the arrangement and treatment of the various features
introduced.
Most of the material was gathered during the author's
career as a steam and marine engineer, covering a period
of twenty-five years.
It is not intended to teach the elements and principles
of arithmetic in this book, as might perhaps be inferred
from the title, but only the application of arithmetic to
steam-boiler calculations. It is presumed that those
who may use it already understand arithmetic but
desire to have a compact set of rules and formulas con-
veniently ready for use, without having to look through
several books for a certain one when required. Those
who are preparing for examinations for engineer's certifi-
cates and licenses will find the work of great assistance
to them.
The author desires to thank all those who have in
any way contributed to the production of this work,
vii
viii PREFACE
particularly the publishers and editor of Power, from
which paper various extracts have been taken, kind
permission to use the same having been granted, and
Mr. William Kent, M. E., author of Kent's Mechanical
Engineer's Pocket-book.
Charles J. Mason.
SCRANTON, PENNA,
December, 1913.
CONTENTS
Page
Preface vii
PART I
CHAPTER I
The Sphere, Stress per Square Inch Section; Safe
Pressure i
Calculations pertaining to the sphere — the cylinder —
riveted joints — efficiencies — bursting and safe pressures.
CHAPTER II
Boiler Heads. — Unstayed Heads 30
Boiler heads — unstayed bolts — convex and concave — flat
unstayed heads — stays and staybolts — diagonal stays —
segments to be braced — girder bars.
CHAPTER III
Manhole Reinforcing Rings 56
Reinforcing rings — heating and grate surface — corrugated
furnaces — horse-power of boilers — ratio of heating to grate
surface — equivalent evaporation — boiler efficiency — boiler
trials.
PART II
Miscellaneous .Applications 83
Bursting pressure of pipe — cost of evaporating 1000 lb. of
water — safe pressure of flat cast-iron heads — equivalent
boiler performance — efficiency of diagonal seam — collap-
sing strength of cone-shaped flue — strength of cone seam —
safety valves — Roper's rules — tapered levers — chimneys
— size of feed pipes
ix
x CONTENTS
PART III
Page
Appendix 133
Abstracts from rules, United States Board of Supervising
Inspectors of Steam Vessels — abstracts from Massachu-
setts' Boiler Rules.
Tables 191
Diameters, areas and circumferences of circles — decimal
equivalents — squares, cubes, cube roots and square roots
— factors of evaporation — standard boiler tubes — Kent's
table of chimneys — Mark's and Davis' steam tables.
Index 221
ILLUSTRATIONS
Figure Page
i Diagram of stresses in a sphere 2
2 The cylinder 6
3 Types of riveted joints 12
4 Data sheet, double butt-strapped joint 22
5 Quadruple, double butt-strapped joint 25
6 Bumped heads 30
7 Diagram to find radius of a bumped head 31
8 Arrangement of direct and diagonal stays 37
9 Diagonal stays 40
10 Segment of head to be braced 48
11 Approximate method of finding area of segment 51
1 2 Girder bars 53
13 Direction of stress in reinforcing rings 60
14 Lower part of Manning boiler 86
15 Diagonal steam 91
16 Cone-shaped flue ! 93
17 Strength of seam in cone 94
18 Diagram of fire box 97
19 Diagram of safety valve dimensions for calculations 100
20 Diagram showing tapered safety valve lever for calculation. 1 13
21 Diagram of locomotive boiler 130
XI
PART I
BOILER CALCULATIONS
BOILER CALCULATIONS
CHAPTER I
The Sphere; Stress Per Square Inch Section;
Safe Pressure
The sphere is the strongest form in which a steam
boiler could be made, but because of mechanical and
commercial reasons, that form is not used. In order to
understand the stresses, due to pressure, endured by
steam boilers, it is well to start with the spherical form,
for that is the simplest, and it forms a basis for calcula-
tions on the prevailing forms in which boilers are made.
Given a spherical vessel made of metal of a certain
thickness and known diameter, it is desired to find what
stress per square inch section the metal is subjected to,
due to a known pressure per square inch contained within
the sphere. The pressure would tend to separate the
sphere in halves, through a diametral plane. Actually,
pressure in a closed vessel of any form radiates from the
center outward. But for convenience in calculations
the radiating forces may be resolved in two, and acting
perpendicular to any diametral plane. The total pres-
sure or force tending to burst the sphere asunder will be
the product of the known pressure per square inch,
2 ARITHMETIC OF THE STEAM BOILER
existing in the sphere, and the area in square inches of
a diametral plane.
For illustration, assume a pressure of ioo lb. per square
inch; a sphere whose internal diameter is 30 in., and made
Diametrical Plane
Arrows show Direction of Resolved Forces
Shaded Rings shows Thickness of Sphere
Hemisphere
Fig. i.— Diagram of stresses in a sphere.
of 1/2-in. steel plate, no joints, seams, nor rivets.
The area of the diametral plane is:
30 2 X. 7854 = 706.860 sq. in.
As the pressure is 100 lb. on each square inch, then:
100X706.860 = 70686.00 lb.
This is the total force tending to separate the sphere in
two parts.
This force is resisted by the area of metal at the cir-
THE SPHERE 3
cumference of the plane upon which the pressure is as-
sumed to act. This is actually a ring whose internal
diameter is 30 in. and whose outside diameter is 31 in.
This is called sectional area, and it is the difference be-
tween the area of the inner and outer circles of which
the ring of metal is formed.
The area of a circle whose diameter is 31 in. is:
3 i 2 X. 7854 = 754.769 sq. in.
and, 30 2 X. 7854 = 706.860 sq. in.
Difference = 47.909 sq. in.
That is, 47.909 sq. in. cross-sectional area of steel plate
is enduring a total stress of 70,686 lb. or,
70686
47.909
= 1475.42 lb.
per square inch section.
Assuming the tensile strength of the plate to be 60,000
lb. per square inch section, the foregoing shows that
considerably more than 100 lb. per square inch could
safely be carried in the sphere considered, for with a
factor of safety of 6, the stress would be 10,000 lb. and
this would be obtained by having the pressure at 677.77
lb. per square inch, as a trial in calculation will show.
Ordinarily, in practical work, the difference between
the outside and inside diameters is not taken.
For example, the rule that would be employed to find the stress
per square inch in a sphere whose inside diameter is 30 in., and
whose thickness of plate is .5 in., is: Multiply the area of the dia-
metral plane by the pressure per square inch, and divide that
product by the product of the inside diameter, the constant 3. 141 6,
4 ARITHMETIC OF THE STEAM BOILER
and the thickness of the plate in inches. This written in formula
is:
d*X-7SS4XP .
d X3.i4i6X*~ StreSS -
But by simple cancellation the formula reduces to
pXd
—ttt = stress.
4Xt
The value of the letters used is:
d= inside diameter, inches;
p= pressure, pounds per square inch;
t= thickness of plate, inches;
s= safe stress in pounds per square inch;
.7854= a constant;
3.1416= a constant.
In order to find the pressure per square inch that may
safely be carried, when the safe stress, the diameter of
sphere, and the thickness of plate are given, it is simply
a matter of changing the rule and formula to suit the
purpose, thus:
Multiply four times the thickness of plate by the given
safe stress per square inch section, and divide by the
internal diameter in inches.
Written in formula it is :
4/ X stress
-j--=p.
Applying this to the example chosen, it becomes:
4X .5X10000
3°
= 666.66 lb.
THE SPHERE 5
which for practical purposes is 667 lb. per square inch.
Here, the internal diameter only has been taken. In the
first method shown the mean diameter was taken, which
gave a safe pressure of 677.77 lb. per square inch which
for practical purposes is 678 lb. The difference is 11.1
lb. or 1.6+ per cent, difference in favor of the usual
method, which, though not absolutely correct, errs on the
side of safety as shown.
If the sphere in the example were .25 in. thick, or 60 in.
internal diameter, then the difference in the methods
explained would be less than that given.
Grouping all the rules and formulas pertaining to the
sphere, so that any term may be found having the remain-
ing terms given:
To find the stress per square inch section endured by the plates,
multiply together the pressure per square inch and the internal
diameter in inches. Divide the product by four times the thickness
of the plate in inches.
Written as a formula this rule becomes:
pXd
4X7 = * (I)
To find the safe pressure per square inch that may be carried,
multiply four times the thickness of the plate by one-sixth of the
ultimate tensile strength of the material of which the plates are
made, and divide the product by the internal diameter in inches.
4XtXs
To find the internal diameter, multiply four times the thickness
of the plate by one-sixth of the ultimate tensile strength of the
6 ARITHMETIC OF THE STEAM BOILER
material of which the plates are made, and divide by the safe pres-
sure per square inch.
^ = <*. (3)
P
To find the thickness of plate, multiply the internal diameter by
the safe pressure per square inch, and divide the product by one-
sixth the ultimate tensile strength of the material of which the plates
are made; divide the quotient by 4.
JXs~ L (4)
The Cylinder
Next to the sphere, the cylinder is the form best suited
for steam boilers, and excepting the spherical form, the
(a) ' W
Fig. 2. — Stresses in cylinders, (a) Stress acting at right angles
to the longitudinal plane abed. Tendency to separate the cylin-
der at a d and b c. (b) Stress acting parallel to the longitudinal
plane abed. Tendency to separate the cylinder at ef.
cylindrical is the strongest. This is not hard to under-
stand, when it is considered that pressure inside any
vessel tends to make that vessel assume a spherical form,
as illustrated in the inflated toy rubber balloon. For
this reason flat surfaces in boilers must be braced, but
cylindrical surfaces do not require any bracing.
THE CYLINDER 7
For the sake of convenience in calculations, the force
due to the pressure in a cylindrical vessel may be consid-
ered as acting in two directions. One acts in the direc-
tion tending to blow off the head, and the other at
right angles to the diametral longitudinal plane. In the
former, the calculations are exactly the same as has been
explained in connection with the sphere. In the latter,
the pressure acting against the imaginary plane tends to
separate the cylinder into two parts, and that which
resists the tendency is the area of metal along both sides
of the cylinder.
The stress per square inch section of the plate, due
to any given pressure, may be found from the follow-
ing rule:
Multiply together the pressure per square inch, the diameter of
the cylinder in inches, and the length in inches; divide the product
by two times the thickness of the plate in inches multiplied by the
length in inches.
This, written as a formula, is:
pXdXl
2XtXl
(S)
As the factor / appears both above and below the line,
it may be cancelled, and the formula becomes,
From the formula just given it can be seen that the
stress increases as the diameter or pressure increases, and
it also can be seen that as the thickness of the plate
increases the stress on the plates decreases.
By comparing formula (1) with formula (6) it is seen
8 ARITHMETIC OF THE STEAM BOILER
that in the latter the stress is twice as great as it is in the
former. A careful study of these rules and formulas
will show why the stress in one case is just twice that of
the other. For that reason, the longitudinal seams in a.
boiler must be made stronger than the circumferential
seams, and this is accomplished by having more than
one row of rivets in the longitudinal seams.
Circumferential seams are frequently double riveted
to make a mechanically tight job. The strength of riv-
eted joints will be treated further on. So far, seamless
vessels have been considered, so as to introduce the sub-
ject of stress and resistance to stress in the most simple
form.
The following rules relate to the cylindrical form, and
belong in the group under consideration.
To find the stress on each inch in the circumference
tending to blow off the head in a longitudinal direction
(not the stress per square inch section), the following rule
is applicable :
Multiply together the area in square inches of the end of the
cylinder and the pressure per square inch, and divide the product
by the circumference of the cylinder in inches.
<* 2 X. 7854x1
JX3.1416 ""*■ {7)
To find the total stress caused by the pressure in a cylinder,
multiply together the diameter in inches, the length in inches,
and the pressure in pounds per square inch.
d XI Xp = totals (8)
/ = length in inches.
To find the total pressure on the entire shell of a cylinder, multi-
THE CYLINDER 9
ply together the circumference in inches, the length in inches, and
the pressure per square inch.
cXlXp = total pressure. (9)
To find the bursting pressure of a cylinder, multiply together
the thickness of the plate in inches and the tensile strength of the
material of which the plate is made, and divide the product by
the radius of the cylinder in inches.
ZX T
= bursting pressure per square inch. (ro)
In this formula, t = thickness of plate in inches. T = tensile
strength in pounds per square inch section of the material, and
r = the radius in inches.
To find the safe working pressure of a cylinder, multiply together
the thickness in inches and the tensile strength of the material of
which the plate is made, and divide that product by the radius
in inches multiplied by whatever factor of safety may be desired.
tX T
,. = safe working pressure per square inch. (n)
rX/
Here follows examples showing the application of the
foregoing rules and formulas from (1) to (n) inclusive.
For the sake of clearness and convenience the same values
will be used in all. Numbers easy to operate have been
chosen, for no matter what numbers may be contained
in any example which may come up in practice, the
method of operation will be exactly the same.
In formula (1) assume the following values:
pressure (p) =100 lb. per square inch;
diameter (d) = 60 in. ;
thickness (t) =1/2 or .5 in.
Then to find the stress (s) the statement becomes:
100X60 .
— — = 3000 lb. per square inch section.
4X -5
10 ARITHMETIC OF THE STEAM BOILER
To find the safe pressure (p) which may be carried,
formula (2), the statement becomes:
4X -5X3000 „ . .
= 100 lb. per square inch.
To find the internal diameter (d), formula (3), the
statement becomes:
4X .5X300 , .
- = 60 in.
100
To find the thickness of plate (/), formula (4), the
statement becomes:
100X60
= .5 in.
4X3000
Formula (5) reduces to that given in (6), and to find
the stress per square inch section in this case, the state-
ment becomes:
100X60 . n
— — = 6000 lb.
2X.5
per square inch section.
To find the stress (s) on each inch in the circumference,
formula (7), the statement becomes:
6o 2 X. 7854X100
— z~r^ 7— = 1500 lb.
60X3. 1416 J
To find the total stress on the entire shell, formula (8),
assume a length of 144 in. with the other values remaining
the same, the statement becomes:
60X144X100 = 864,000 lb.
THE CYLINDER II
To find the total pressure on the entire shell, formula (9),
the statement becomes:
60X3.1416X144X100=2,714,342.4 lb.
To find the bursting pressure, formula (10), the state-
ment becomes: assume a tensile strength of 50,000 lb.
per square inch section.
.5X50000 n . ,. . .
— — = 833.33+ lb. per square inch.
To find the safe working pressure, formula (n), and
assume a factor of safety of 5, the statement becomes:
s X ^0000
^^b = 166.66+ lb.
30X5
In practice 170 lb. would be allowed with factor of
safety 5 per square inch. Or a simpler method is to
take one-fifth of the bursting pressure thus:
83 ^=i66.66 1b.
5
and as the bursting pressure is five times the safe pressure,
5X166. 66+ = 8 33 . 33+ lb.
In actual practice, 6 is used as a factor of safety more
frequently than 5. However, the method of operation is
the same for any and all factors of safety that may be
used; so that if the method is understood, it matters not
as to the values that may be substituted in the various
formulas treated of.
12
ARITHMETIC OF THE STEAM BOILER
Riveted Joints
In the previous sections, spheres and cylinders without
joints or seams were assumed in order to simplify explana-
Pitch
Pitch
Pitch
A
©
©!© © © ©^
>©©©©_©
© © © ©
©i © © © ©!©
~^-
w
Fig. 3. — Types of riveted joints. (A) Single-riveted lap joint;
one rivet in single shear. (B) Double-riveted lap joint; two rivets
in single shear. (C) Triple- riveted, double butt-strapped joint;
four rivets in double shear, one rivet in single shear. (D) Quad-
ruple-riveted, double butt-strapped joint; eight rivets in double
shear, three in single shear.
tions and calculations. In actual practice, however,
steam boilers are constructed with both longitudinal
RIVETED JOINTS 13
and circumferential joints, or seams. The seams are
secured by rivets. There are several kinds of riveted
joints known to engineers and others who have to do
with boilers. The strength of a riveted joint depends
upon how the joint is made, as to the size and pitch or
spacing of the rivets, and the number of rows of rivets,
and also as to whether the joint is what is known as a lap
or butt-strapped one. A riveted joint of any kind is not
theoretically as strong as the solid part of the plate,
although in practice it has been known for boilers to
tear apart at some place other than at the riveted joint.
This probably was due to a flaw or weakness in the metal.
If a cylindrical vessel were made of plates uniform in
structure and thickness throughout, and if tested to
destruction, it would likely break or pull apart at the
riveted longitudinal joint. In making calculations it is
presumed that a break would occur at the joint, rather
than at any other place.
Strength of Riveted Joints
The strength of a riveted joint is compared with that of
the solid plate, the latter being valued at 100 per cent.
Nominally, the strength of joints varies from 56 per
cent, to 94 per cent., the former value representing single-
riveted lap joints, and the latter, quadruple, double
butt-strapped joints. Between single-riveted lap joints
and quadruple-riveted, double butt-strapped joints,
there are: double-riveted lap joints, triple-riveted lap
joints, quadruple-riveted lap joints, single-riveted butt
joints, double-riveted butt joints, triple-riveted butt
joints. The foregoing joints may be either chain riveted,
14 ARITHMETIC OF THE STEAM BOILER
or what is termed zig-zag riveted. In triple-riveted
joints and in quadruple-riveted joints it is customary to
omit every alternate rivet in the outer rows; this admits
of a stronger joint.
The strength of a riveted joint depends upon the size
and pitch of the rivets, the number of rows of rivets,
type of joint as to lap or butt, the tensile strength of the
plates, and the shearing stress of the material of which
the rivets are made. In rinding the strength of a joint
two things are to be considered, the strength of the plate
section and the strength of the rivet section. The
lesser value, as found from an analysis, is taken as the
strength of the joint as a w T hole.
Theoretically, riveted lap joints and those butt joints
with one cover plate should be designed so that the rivet
and plate sections are equal — or as nearly equal as possi-
ble — in strength. But in practice it is usually considered
desirable to so design a joint that the plate section is a
little stronger than the rivet section; this particularly
relates to joints having all the rivets either in single or
double shear, for the reason that the plates become thin
from w T ear, with a consequent reduction in strength,
while the rivets suffer little if any from wear. But in
joints having some of the rivets in single shear and some
in double shear, the greatest strength usually obtains
when the rivet section exceeds the strength of the net
section of plate.
Efficiency of Riveted Joints
To determine the efficiency of a riveted joint, its re-
sistance must be calculated for each of the different ways
RIVETED JOINTS 15
in which it may fail, and then the lowest efficiency so
found in relation to the solid plate, will be the one by
which the joint is known. A riveted joint may fail in the
following ways:
(1) The plate may break asunder along the rivet holes,
at the net section.
(2) The rivets may shear off, leaving the plates intact.
(3) The plate may shear out in front of the rivets.
(4) The plate may crush in front of the rivets.
(5) In joints having zig-zag rivets the plate may break
diagonally between the rivet holes.
(6) The joint may fail by a combination of the fore-
going. With joints as usually proportioned, the liability
to failure in the ways referred to in (3), (4), (5) and (6)
is reduced to a minimum; this by having the distance from
the edge of the plates to center of rivet holes one and one-
half times the diameter of the rivet holes. It is custom-
ary (except in special cases which will be referred to) to
consider only (1) and (2) as possible ways of failure, and
base all calculations upon those two ways. Therefore,
calculate the efficiency of the net section of the plate
part of the joint as compared with the solid plate, and
then find the efficiency of the rivet section of the joint as
compared with the solid plate; the lesser of the values
found is to be taken as the final efficiency of the joint as a
whole.
Single-riveted Lap Joints
To find the efficiency of a single-riveted lap joint.
(The distance from the edge of the plate to the center
1 6 ARITHMETIC OF THE STEAM BOILER
line of rivet holes must be not less than one and one-half
times the diameter of the rivet hole, for all joints.)
First, find the strength of a unit of length of the solid
plate.
PXtX S = strength of solid plate.
In which,
P = pitch of rivets in inches, from center to center.
/ = thickness of plate in inches.
5 = tensile strength of plate, in pounds per square inch
section.
d = diameter in inches of rivet holes.
The next step is to find the strength of the net section
of plate between the rivet holes:
(P—d)XtXS = strength of plate between the holes.
Next, find the shearing strength of one rivet in single
shear:
nXsXa = shearing strength of one rivet.
n = number of rivets in single shear; 5 = shearing strength
of rivet; a = cross-section area of rivet, after driving.
Take the lesser of these two results and divide it by
the value found for the strength of the solid strip;
the quotient will be the efficiency of the joint decimally
expressed.
Example. — Single-riveted lap joint of the following dimensions:
•S =55,ooo lb.
P =1.625 in.
t =.25 in.
d = .6875 in.
5 =42,000 lb.
a = .3712 sq. in.
RIVETED JOINTS 1 7
The strength of the solid strip will be:
i. 625 X. 25X55*000 = 22,343 lb.
The strength of the net section of plate between the rivet holes
will be:
(i. 625-. 6875) X. 25X55,000 = 12,890 lb.
The strength of the rivet in single shear will be:
1 X42,oooX . 3712 = 15,590 lb.
As the net section of plate in this example is weaker than the
rivet, its value must be used. Then:
12800 .
= .576, or 57.6 per cent, efficiency.
Shearing Strength or Rivets
The shearing strength of rivets may be taken from the
following table (from the Massachusetts Board of Boiler
Rules) :
Iron rivets in single shear, 38,000 lb.
Iron rivets in double shear, 70,000 lb.
Steel rivets in single shear, 42,000 lb.
Steel rivets in double shear, 78,000 lb.
These values are on the safe side, as they are lower
than some others that are in use.
Double-riveted Lap Joints
To find the efficiency of double-riveted lap joints, the
method of procedure is the same as that for single-
riveted joints, with the exception that there are two rivets
in single shear instead of one as in single-riveted joints.
18 ARITHMETIC OF THE STEAM BOILER
Example. — A double-riveted lap joint has the following
dimensions:
5 =55,000 lb.
t = .3125 (5/16) in.
P =2.875 (2 7/8) in.
d = -75 (3/4) in.
a = .4418 sq. in.
5 =42,000 lb.
The strength of the solid plate is:
2. 875X. 3125X55,000=49,4*4 lb.
The strength of the net section of plate is:
(2. 875-. 75) X. 3125X55,000 = 36 ? 5 2 3 lb.
The strength of the two rivets in single shear is :
2X42, 000X .4418 = 37,111 lb.
Here again the plate section is the weaker, so the value for that
must be used:
— = . 739, or 73 . 9 per cent, efficiency of joint.
49414
Triple- and Quadruple-riveted Lap Joints
In triple-riveted and quadruple-riveted lap joints
(sometimes used in marine boilers) there are three and
four rivets, respectively, in single shear. With this
exception, the method of finding the efficiency of such
joints is the same as for single and double, as just
illustrated.
Lap Joints
Lap joints for the longitudinal seams are now consid-
ered not safe for steam boilers of more than 36 in. in
RIVETED JOINTS
19
diameter, and for pressures higher than 100 lb. per square
inch; and probably as time goes on they will not be used
at all; but it is important to know how to calculate the
strength of such joints, hence the reference to them in
this book.
Butt Joints
Butt joints with double cover plates are the strongest
and safest joints in use. The minimum thickness of
cover plates, or butt straps as otherwise called, is as follows :
PRESCRIBED BY THE MASSACHUSETTS BOARD OF
BOILER RULES
Thickness of shell
Minimum thickness of
plates
butt straps
1/4 in.
1/4 in.
5/16 in.
1/4 in.
3/8 in.
5/16 in.
7/16 in.
3/8 in.
1/2 in.
7/16 in.
9/16 in.
7/16 in.
5/8 in.
1/2 in.
3/4 in.
1/2 in.
7/8 in.
5/8 in.
1 in.
3/4 in.
1 1/8 in.
3/4 in.
1 1/4 in.
7/8 in.
Single Butt Straps
Single butt straps should never be thinner than the
plates of the shell. In some instances (British Board of
Trade and Canadian Rules) the minimum thickness
must be not less than one and one-eighth the thickness of
the shell plates. Double butt straps must be at least
3
20 ARITHMETIC OF THE STEAM BOILER
five-eighths, and preferably the thickness of the shell
plates. If the shell plate is light, say 7/16 in. or less,
the outside strap should be as heavy as the plate, to
admit of a tightly calked joint.
When single butt straps are used, the method of
finding the efficiency of the joint is the same as that
for lap joints, for the rivets are all in single shear, and
the pitch of the rivets is the same in each row.
Number of Rivets Considered
In lap joints, all the rivets in a given pitch strip of
plate are taken into account when figuring for the effi-
ciency of joint, while in butt joints only those rivets on
one side of the center line of the joint are considered. A
little thought on the part of the reader will make clear
the reason.
Number oe Rows of Rivets
In double butt-strapped joints, three or four rows of
rivets on each side of the center line are generally used.
In the former (triple riveted) the pitch of the outer row
of rivets on each side of the center line is twice the pitch
distance of the two inner rows of rivets on each side of
the center line.
In the latter (quadruple-riveted joints) the outer row
of rivets on each side of the center line of joint is four
times the pitch distance of the two inner rows on each
side of the center line; in the rows next to the outer rows,
the rivets are pitched twice the distance of those in the two
inner rows. (See Fig. 3.)
RIVETED JOINTS 21
Number of Rivets in Double and Single Shear
In triple-riveted butt joints, there are four rivets in
double shear, and one rivet in single shear, in a given
pitch strip.
In quadruple-riveted butt joints, there are eight rivets
in double shear, and three rivets in single shear, in a given
pitch strip.
When calculating the efficiency of triple- and quadruple-
riveted joints, the strength of the net section of plate is
taken at the outer row of rivets, where the pitch is the
greatest. The reason for this will be explained presently.
High Joint Efficiencies Due to Wide Spacing of
Rivets at the Outer Rows
It is because of the wide spacing in the outer rows of
rivets that such high efficiencies can be obtained with
those types of joints as compared with those joints in
which the pitch of the rivets is the same for all the rows.
In order to explain why the net section of the plate at
the outer row of rivets is taken, an illustrative example of
a triple-riveted, double butt-strapped joint will be used;
the same principles may be applied to a quadruple joint
of the same kind.
Thickness of shell plates 3/8 in., tensile strength
50,000 lb. per square inch section, rivet holes 13/16 in.,
rivets 3/4 in. diameter; shearing stress of the rivets
taken as 38,000 lb. per square inch section. The pitch
of rivets in the two inner rows is 3 1/4 in., and in the outer
row, 61/2 in.
22
ARITHMETIC OF THE STEAM BOILER
The width of strip to be considered in this case is 6 1/2
in. The sectional area is .375X6.5 = 2.4375 sq. in.
2.4375X50,000=121,875 lb. strength of the solid strip,
with which the joint is to be compared.
Next find the strength of the net section of the plate
at the outer row of rivets.
As there is but 1 rivet in the 6 1/2 in. strip under con-
Inside
Cover Plate
Outside
Cover Plate
Bottom
Fig. 4. — Diagram of triple-riveted double butt-strapped joint.
sideration, at the outer row, then, 6. 5— .8125 = 5.6875
in. width of net section of plate; and, 5. 6875 X. 375 X
50,000=106,640 lb. strength of net section of plate
between the rivet holes.
If the plate should break or pull asunder at the line
of the outer row of rivets, the resistance to breaking is
the metal in the net section of the plate as shown at c
in Fig. 4. But should the plate break along the net
RIVETED JOINTS 23
section at the inner row of rivets, the resistance offered
is the strength of the sections of plate E D and F, and
as well as that, the resistance to shearing offered by
one-half of each of the rivets in the outer row, which is
equivalent to one rivet in calculation.
First, find what the plate resistance is. By measure-
ment it is 6.5 — (2X .8i25)=4.875 in. Here there are
two rivet holes to subtract from the width of the strip.
The sectional area will be 4. 875 X .375 = 1 .828 sq. in.;
and 1.828X50,000 = 91,400 lb. resistance.
To this must be added the resistance offered by the
two half rivets in the outer row. The area of a 13/16-in.
driven rivet is .5185 sq. in., and as the shearing stress is
38,000 lb. per square inch, .5185X38,000=19,703 lb.
This added to 91,400 lb. before obtained for the plate
gives 111,103 lb. total resistance to the plate breaking
at the inner row of rivets, as against the 106,640 lb.
found for the net section of plate at the outer row of
rivets. Therefore the latter is the weaker, and there
the plate w T ill probably break, if at all.
Continuing, there are four rivets in double shear, and
one rivet in single shear, in the strip. The resistance to
shearing of one rivet was shown to be 19,703 lb. The
resistance to shearing offered by each rivet in double
shear is:
38,oooX .85=32,300 and 38,000+32,300 = 70,300 lb.
per square inch section.
(The factor .85 is a value used for rivets in double
shear as double shear does not necessarily mean twice
that of single shear.) The area of each rivet is .5185
24 ARITHMETIC OF THE STEAM BOILER
sq. in., and there are four to consider, therefore, .5185
X4X7o,3oo= 145,802 lb. and to this add the value for
the rivet in single shear, 19,703 lb., which gives a total of
165,505 lb. shearing strength of all the rivets in the
strip. The net section of plate at the outer row of rivets
is the weakest part of the joint as a w T hole, and its value
is to be compared w T ith the strength of the solid strip in
order to find the efficiency. The net section of plate is
106,640.625 lb. and the solid strip is 121,875 lb.;theeffi-
ciency is:
106640.62=;
—^ = 87. s per cent.
121875 ' J ^
Quadruple-riveted, Double Butt-strapped Joint
To analyze a quadruple-riveted joint and find the
efficiency, proceed as follows:
Fig. 5 shows the construction and arrangement of
rivets. The data is given in this manner. A strip of
the joint marked P is taken. The value of the letters is:
P = pitch of rivets in inches.
/ = thickness of plate in inches.
S = tensile strength of plates.
d = diameter of the driven rivets, in inches.
N = number of rivets in double shear.
n = number of rivets in single shear.
a = area of cross-section of rivets, in square inches.
Strength of the solid strip of plate considered =
PXtXS, represented by letter A.
Strength of plate between the rivet holes at the outer
row of rivets = (P — d)X S, represented by the letter B.
RIVETED JOINTS
25
The shearing strength of 8 rivets in double shear, plus
the shearing strength of 3 rivets in single shear = Na+na,
represented by the letter C.
The strength of the plate between the rivet holes in the
second row plus the shearing strength of 1 rivet in single
shear in the outer row = (P — 2d)XtX S+ na, represented
by the letter D.
Next, divide B, C, or D, whichever is the least in value,
<■
4>
±V
Q
O
O-
$-
e
o
e
r —
^ —
tx>
e-
-e-
■&
e
0-
^?
f
1 L
e
^>
&
-e
O-O
e
-0-
e-
e
^
e-
_ P —
Fig. 5. — Quadruple-riveted double butt-strapped joint.
by the value of A, and the quotient w T ill be the efficiency
of the joint.
The numerical values to be used are:
5 =55,000 lb.
/ = 1/2 in. or .5.
P =15 in.
d =15/16 in. or .9375 in.
a =.6903 sq. in.
N =8 rivets, double shear value of 78,000 lb.
n =3 rivets, single shear value of 42,000 lb.
26 ARITHMETIC OF THE STEAM BOILER
The final values are:
A =isX. 5X55,000 = 412,500.
B =(i5-.9375)X. 5X55,000 = 386,718.
C = 8X78,oooX. 6903+3 X4-2,oooX. 6903 = 517,725.
D =(i5-2X.937S)X.5X55 J oo°+ I X 42,000 X .6903 =
389,930.
The value of B is found to be the least, therefore the
strength of the joint depends upon the weakest part,
and the efficiency is,
386718
412500
.937 or 93. 7 per cent.
The foregoing example shows how to calculate the ef-
ficiency of the various parts of the joint where possible
failure may occur. The same line of reasoning, and simi-
lar methods of operation, may be employed for any kind
or type of riveted joint that may be used in a steam boiler.
Size of Rivets and Pitch
There are no absolute rules for determining the size of
rivets to be used in any given case, for with different size
rivets, the same efficiency of joint may be obtained. The
size to be chosen depends upon several factors and varies
with any one of them. The things to be considered are:
the shearing strength of the material of which the rivets
are made, the tensile strength of the plates to be riveted to-
gether, the pitch of the rivets, and the type of joint to be
made. The required efficiency of joint determines the
type to be used and the greatest pitch of rivets allowable
depends upon the thickness of plate to be used, the object
sought being steam-tight joints.
RIVETED JOINTS 27
For single-riveted lap joints, the United States Super-
vising Inspectors of Steam Vessels recommend a rivet
diameter equal to the plate thickness plus 7/16 in. using
steel plates and steel rivets. For double-riveted lap
joints, a rivet diameter equal to the plate thickness plus
$/& in. Some authorities make the rivet diameter range
from plate thickness plus 3/8 in. to plate thickness plus
1/2 in., with plates from 1/4 in. to 1/2 in. thickness.
For triple-riveted lap joints rivet diameters range from
plate thickness plus 3/8 in. to plate thickness plus 7/16
in. with plates from 1/4 to 1/2 in. thickness.
For double-riveted butt joints, triple-riveted butt joints,
and for quadruple-riveted butt joints, rivet diameters
range from plate thickness plus 5/16 in. to plate thickness
plus 7/16 in. The foregoing is intended to give a general
idea only of rivet sizes that may be chosen to be some-
what in proportion to the joint as a whole. In the end it
is a matter of choosing that size rivet and a certain pitch
which w r ill give the highest efficiency of joint, consistent
with steam-tight work, type of joint, strength of materials
and all other considerations. It is a matter of "cut and
try" until the best is arrived at.
Distance Between Adjacent Rows of Rivets
The distance between adjacent rows of rivets, center to
center, is sometimes called transverse pitch. When the
rivets are subjected to the same kind of shear, this dis-
tance should not be less than twice the diameter of the
rivets, nor more than two and one-half times the diameter
of the rivets used. If the distance between the rows of
28 ARITHMETIC OF THE STEAM BOILER
rivets is too small, the plate is likely to fracture along a
diagonal line, or diagonal pitch as it is termed.
If the transverse pitch is at least equal to twice the
diameter of rivets, failure of the plate will not occur
along the diagonal line, but rather in the net section of
plate along the line of rivets. This makes the calculation
of joint efficiency somewhat simpler. In cases where the
outer butt strap is not as wide as the inner strap, the dis-
tance between the line of rivets in double shear and the
line in single shear should be two and three-quarters or
three times the diameter of rivets, in order to have a
properly formed rivet head, and also room to calk the
outer strap.
Safe Working Pressure of Cylinders with Riveted
Joints
It w r ill be remembered that formula n, page 9, gives
the safe working pressure of a cylinder without any
visible joint. But cylinders having riveted joints must
be calculated with the efficiency of longitudinal joint
taken into consideration. The rule will be the same as
that expressed in formula 11, with the additional factor
of joint efficiency expressed as a decimal value.
Let e represent the efficiency of the longitudinal
seam or joint; the efficiency of the girth seam is not
required for reasons explained at the beginning of this
chapter.
The formula now becomes:
/X Ty< e
— r-77 — = safe working pressure per square inch.
RIVETED JOINTS 29
Using the same example illustrating formula 11 , page
11, and assuming a joint efficiency of say 85 per cent.
the statement becomes:
.5X50000X .85
— — — — =141.66 lb. safe pressure per square
inch. No matter what the efficiency of the joint may be,
nor by what method it may be found, it is always to be
applied as shown in the example just given. Of course
there are other ways of arranging and simplifying the fac-
tors in the formula, but no matter what the arrangement
or how simplified, the result will always come out the
same if the work is correctly done. Abbreviated formulas
and rules are convenient to those who know of their
derivation, but they are not satisfying to those who do not
know just how each simplified factor was obtained.
For this reason, no attempt is made in this work to ab-
breviate anything that will detract from the value of
any problem presented, as far as underlying elements
and principles are concerned. Any one who understands
how to calculate the efficiency of riveted joints, and how
to find the safe working pressure of spherical or cylin-
drical vessels as given in this work, will be able to work out
similar problems, no matter in what form they may be
given, or what rule it may be desired to apply to them.
So far, the shell and its riveted joints only have been
considered. The rules given apply to the cylindrical parts
of all boilers of whatever make. There are other rules
relating to shells of special design, and these rules will
be given further on.
The next in order, at present, is the bracing or staying
of boiler heads and flat surfaces in boilers.
CHAPTER II
Boiler Heads — Unstayed Heads
Bumped heads may be either convex or concave accord-
ing as to how placed in a shell. Fig. 6 shows the applica-
tion of the two forms, (a) being a concave bumped head
while (b) is a convex head. The arrows show the direc-
tion of pressure acting against the heads. Bumped heads
do not require bracing, particularly the convex (b) form
Fig. 6. — Bumped heads, (a) Concave head, (b) Convex head.
Arrows show the direction of pressure.
as it is already in the form that internal pressure would
tend to make it assume. In the concave (a) form, the
tendency of internal pressure is to collapse the head, and
allowance is made for this in the rule which will be given
presently for the safe working pressure allowed. Bumped
30
BOILER HEADS— UNSTAYED HEADS
31
heads may be either single or double riveted to the shell.
It is necessary to know the radius to which a head is
bumped when making calculations for safe working
pressure. A bumped head is presumed to be virtually
part of the surface of a sphere. To find the radius to
which a head is bumped, take half the diameter of the
head where it fits into the shell, and multiply that value
by itself, and divide the product by the height of the
d= 4 Inches
Radius =4 Inches
h =.55 Inches
do, Horizontal Centex Line
ef, Vertical Center Line
d= Diameter of Head, Inches
/j=Height of Bump, Inches
O— Center from which the Bump is Struck
OC ^Radius to which Bump ie Struck
Fig. 7. — Diagram showing how to find radius of bumped head.
bump. To the quotient add the height of the bump and
divide the sum by 2. It is usual to take the dimensions
in inches.
To find the radius to which a boiler head is bumped the
following formula may be used:
Referring to Fig. (7),
h
-= radius.
32 ARITHMETIC OF THE STEAM BOILER
All dimensions to be taken in inches.
To find the safe working pressure of a bumped head like
that in (b) and when it is single riveted to the shell, the
following formula may be used:
* ixs ( \
P= ^Xr W
When the head is double riveted to the shell, then the
formula becomes:
P = ^Xr (2)
When the head is concaved like that in (a) and single
riveted, then the formula becomes:
txs
P = 5Xr
(3)
When a
concaved head is double-riveted the formula
b
ecomes:
txs
(a1
* 4iXr
In these formulas the values of the letters are:
p —safe working pressure in pounds per square
inch.
t = thickness of metal in the head, expressed in
inches.
r = radius in inches, to which the head is formed.
S = tensile strength of the material of which the
head is made, expressed in pounds per square
inch section.
BOILER HEADS— UNSTAYED HEADS 33
As the foregoing formulas are similar in construction,
one example will serve to illustrate the operation of all.
A boiler head is bumped to a radius of 60 in. made of plate .5 in.
thick, with a tensile strength of 50,000 lb. and double riveted to the
shell. What working pressure will be allowed?
The operation is as follows:
p = — — zn — = 166.66 lb. per square inch.
2.5X60
Unstayed Flat Heads
When a boiler head is flat and not stayed, the following
formula may be used:
txs
P =
•54X.4
The letters have the same values as for the preceding
formulas, and A equals the area of the head expressed in
square inches.
Example. — A flat head is 30 in. in diameter, . 75 in. thickness
of plate having a tensile strength of 55,000 lb. per square inch,
what pressure will be allowed?
Operation:
p = — — — — — — o — = 108 lb. per square inch.
.54X30X30X.7854 F H
All the rules for boiler heads of the kind just described
are those prescribed by the Board of Supervising In-
spectors of Steam Vessels in the United States.
If it is desired to find what thickness a bumped head
should be, having the tensile strength, the radius to
34 ARITHMETIC OF THE STEAM BOILER
which the head is bumped, and the pressure in pounds
per square inch to be carried, it is a matter of transpos-
ing the terms of the proper formula in the group just
treated of.
Suppose in the last illustrative example given that it is
desired to know what thickness of head should be em-
ployed. The formula transposed will be:
S ~ ~ l
Applying this to the example, the statement becomes:
166.66X2.5X60
50000
-=.5 in. thickness.
Thickness of Boiler Heads, Massachusetts Rules
In actual practice, however, the thickness of boiler
heads is not derived mathematically but empirically.
The rules in the state of Massachusetts require the thick-
ness to be as follows: Boilers up to and including 42
in. diameter, heads must be 3/8 in. From 42 in. to 54
in. diameter, heads must be 7/16 in. From 54 in.
to 72 in. diameter, heads must be 1/2 in. Over 72 in.
diameter, heads must be 9/16 in.
Thickness of Boiler Heads, Ohio Rules
The rules formulated for bumped heads by the Board
of Boiler Rules in the State of Ohio differ slightly from
those given.
BOILER HEADS— UNSTAYED HEADS 35
The minimum thickness of a convex head shall be de-
termined by this formula:
RXF.S.XP
T.S. ~ l
The minimum thickness of a concave head shall be de-
termined by this formula:
RXF.S.XP
.6(7.5.)
In these two formulas the values are as follows :
R = one-half the radius to which the head is
bumped.
F.S. =5= factor of safety.
P = working pressure, in pounds per square
inch, for which the boiler is designed.
T.S. = tensile strength, in pounds per square inch,
stamped on the head by the manufacturer.
t = thickness of head in inches.
The radius of head shall not exceed the diameter of
the shell.
When a convex or concave head has a manhole open-
ing, the thickness as found by the formulas just given,
shall be increased by not less than 1/8 in.
The minimum thickness of plates in stayed flat surface
construction shall be 5/16 in.
The minimum thickness of tube sheets shall be as
follows:
" When the diameter of tube sheet is 42 in. or less, the
thickness is 3/8 in.; over 42 in. to 54 in. inclusive, 7/16
4
36 ARITHMETIC OF THE STEAM BOILER
in.; over 54 in. to 72 in. inclusive, 1/2 in.; over 72 in..
9/16 in.
Stays and Stay Bolts
The maximum allowable stress per square inch net
cross-sectional areas of stays and stay bolts as denned
in the Massachusetts rules, is as follows:
Weldless, mild steel head to head or through stays,
8000 lb. for sizes up to and including 1 1/4 in. diameter,
or equivalent area, and 9000 lb. for sizes over 1 1/4 in.
diameter or equivalent area. Fig. 8 (a) illustrates
direct or through stay arrangement.
Weldless, mild steel diagonal or crow-foot stays,
7500 lb. for sizes up to and including 1 1/4 in. diameter,
or equivalent area and 8000 lb. for sizes over 1 1/4 in.
diameter or equivalent area.
Weldless, wrought-iron, head to head or through stays,
7000 lb. for sizes up to and including 1 1/4 in. in diameter
or equivalent area, and 7500 lb. for sizes over 1 1/4 in.
diameter or equivalent area.
Weldless, wrought-iron, diagonal or crow-foot stays,
6500 lb. for sizes up to and including 1 1/4 in. or equiva-
lent area, and 7000 lb. for sizes over 1 1/4 in. diameter
or equivalent area.
Welded mild steel or wrought-iron stays, 6000 lb.
Mild steel or wrought-iron stay bolts 6500 lb. for sizes
up to and including 1 1/4 in. diameter or equivalent area,
and 7000 lb. for sizes over 1 1/4 in. diameter or equivalent
area
When a greater allowable stress per square inch on
BOILER HEADS— STAYED HEADS
37
stays and stay bolts is required than those just given, the
material shall conform to the following physical qualities:
The tensile strength shall not exceed 62,000 lb. per
square inch.
The yield point shall not be less than one-half the
tensile strength.
The elongation per cent, in 8 in. shall not be less
than 28.
Direct Stays
To find the safe working pressure per square inch that
may be carried by stays of a given size, the following
formula may be applied:
Fig. 8. — Arrangement of stays in boiler heads, (a) Direct
through stays in horizontal rows, seven stays, (b) Diagonal stays
in concentric rows, seventeen stays.
aXS
(1)
38 ARITHMETIC OF THE STEAM BOILER
To find the diameter of stays required,
A -¥=°- - 41k' d (2)
To find the area supported by one stay, and the dis-
tance between stays,
^^=A, and ylA=D. (3)
P X
To find the required tensile stress that may be en-
dured by stays:
^ = S (4)
a
The value of the letters in this group of rules is as
follows :
A = area in square inches, supported by one stay.
a = area, cross-sectional, in square inches, of stays.
p = pounds pressure per square inch.
S = tensile strength of stays in pounds per square inch.
d = diameter of stays in inches.
D = distance between stays, in inches.
Example, illustrating the foregoing rules.
Assume stays of 1 in. diameter at the smallest part, with an
allowable stress of 6000 lb. per square inch, and distanced 6 in.
center to center.
The area of each stay will be i 2 X . 7^54= • 7^54 sq. in.
The area supported by each stay will be 6X6 = 36 sq. in.
Applying these values to formula (1) the statement
becomes:
.^854X^000 = lb per square inch allowable pressure.
36
BOILER HEADS— STAYED HEADS 39
To formula (2),
36X130.9.
6000
= .7854 sq. in. area of each stay.
and, \~~~i — =I * n * diameter of stays.
To formula (3),
.7854X6000 . 111
- J — = 36 sq. in. area supported by each stay.
and, V 36 = 6 in. distance center to center of stays.
To formula (4),
z — —=6000 lb. tensile stress allowed on stays.
•7854
In the foregoing no allowance has been made for the
space occupied by the stays in the sheets supported.
This is on the side of safety and is generally accepted.
If in any case it is not accepted, it becomes a matter
of subtracting the area occupied by the stays from the
area as found from the center to center measurement.
Diagonal Stays
The size of a diagonal stay depends upon the angle it
makes with the surface it is helping to support, when con-
sidered in relation to a direct stay. The less the angle is,
the larger in diameter must the stay be. The nearer
a stay is to being at right angles to the surface it supports,
the smaller in diameter it may be. The same principle
40
ARITHMETIC OF THE STEAM BOILER
applies to any form of stay that may be used, other than
a circular cross-section.
In Fig. 9 is shown an ordinary diagonal stay attached
to the boiler head at C and to the shell at E. The length
of the stay is considered as CE; the distance CD also <
enters the calculation as will be shown presently.
Fig. 9. — Forms of attachment of diagonal stays, (a) Riveted
at both ends, (b) Riveted at one end, nuts and washers at the
other end.
To find the area of a required diagonal stay, first find
the area of a direct through stay, as has been explained.
Call the length CE of the required diagonal stay, x and
the distance, CD, y.
BOILER HEADS— STAYED HEADS 41
Let a = the area of direct stay.
Let A = the area of diagonal stay.
Let S = tensile strength of stay.
Then,
aXx . , N
= A (1)
y
AXy
a
AXy
x
aXx
= x (2)
=a (3)
=y (4)
x A ° Safe pressure per
area supported by = sc l ua, ; e inch . ^
one stay ma ^ be carned -
(5)
Example, illustrating the foregoing group of rules.
Assume that the area of a direct stay has been found to be .7854
sq. in. (due to 1 in. diameter as before taken). That the length
of the required diagonal stay is to be 36 in. and that the distance
between perpendiculars of points of attachment is 30 in. Then,
applying these values to formula (1),
1 = .9425 sq. in., nearly, cross-sectional area of diagonal
stay required, and the corresponding diameter is:
V
.9425 . ..
— - — = 1 .095 in. diameter.
•7^54
The nearest commercial size brace that would be used is 1 1/8
in. diameter.
42 ARITHMETIC OF THE STEAM BOILER
To formula (2),
jt = 36 in., length of the required diagonal stay,
as measured on the line CE in Fig. 9.
To formula (3),
7 = .7854 sq. in., area of direct or through stay.
To formula (4),
.7854X36
= 30 in., distance between perpendiculars as
measured on the line CD in Fig. 9.
To formula (5),
.9425X30
X6000
= 96.1 lb. per square inch pressure
allowed.
Other formulas for obtaining the diameters of direct
and diagonal stays.
For direct stays,
/SX.7854
For diagonal stays,
, / AXp , v
d Hsx^I (I)
/ xXAXp ()
\yXSX. 7854 K
The values of the letters are the same as those used for
the stay group of formulas.
Examples, illustrating these formulas.
Assume 36 sq. in. area supported by one stay (pitch, 6 in. center
BOILER HEADS— STAYED HEADS 43
to center), pressure to be carried, 100 lb. per square inch, 6000 lb.
allowed stress per square inch on stays, then to find the diameter of
direct stay as stated in formula (1) the statement becomes:
/ 36XIOO , V
<f = \6oooX.78S4 = - 873S1I1 - ; (I)
in practice 7/8-in. diameter stays would be used.
For formula (2) diagonal stays, assume same values, and in addi-
tion, that the length of the diagonal stays are 36 in., and that the
horizontal distance between the ends of stays — as shown in Fig.
9 — is 30 in., the statement becomes:
■Ato
36X36XIOO . ( v
in practice i-in. diameter stays would be used in this case.
The area in square inches supported by one stay
multiplied by the number of pounds pressure per square
inch carried in any given case gives the load in pounds
the stay must sustain.
The cross-sectional area in square inches of each stay
in any group, multiplied by the allowable tensile stress
per square inch section, gives the allowable load in pounds
the stay may sustain with safety.
The load, in pounds, on a stay divided by the cross-
sectional area of the stay in square inches will give the
stress in pounds per square inch.
To find the number of stays of a given size required
to support a given area, multiply the area of the sheet
to be stayed by the steam pressure in pounds per square
inch to be carried, and divide the product by the total
allowable stress for the given size of stay.
44 ARITHMETIC OF THE STEAM BOILER
Example. — Assume i i/8-in. round iron stay bolts with an allow-
able stress of 6000 lb. per square inch, and an area of 2000 sq. in.
to be supported against a pressure of 90 lb. per square inch. How
many stays are required?
(1 i/8) 2 X. 7854X6000 =5964 lb.
allowable on each stay. Then:
2000X90 .
=30. 18, say 31 stays will be required.
Rivets Securing Stays
The combined cross-sectional area of the rivets which
secure the stays to boiler heads must not be less than that
of the stay itself. The rivets securing a diagonal stay
to the head are in tension chiefly, but they also endure a
certain amount of bending. Those rivets which secure
the end of the stay to the shell are in single shear chiefly,
and to a certain extent are in tension. In practical work
the rivets used for both ends of such stays are of the same
size, a high factor of safety being used to make due allow-
ance for the different stresses, and the difference in
value of rivets in tension and in shear. It is considered
advisable to allow but 4000 lb. per square inch section
for rivets used on diagonal stays, in order to be on the
safe side.
The part of a head above the tubes to be braced is
a segment of a circle. In laying out the position of the
stays it will be found that they cannot be arranged so
that exactly the same load will be borne by all. It is
customary to arrange the stays in concentric rows, as
shown in Fig. 8 (b).
BOILER HEADS— STAYED HEADS 45
It is usual to consider that the flange of a boiler head
supports the head for a distance of at least 3 in., measured
from the inside of the flange. With modern methods of
making plates and flanging them, the radius to which a
head is flanged is now greater in proportion to the thick-
ness than used to obtain, and it is thought by some that
more than 3 in. may be counted upon as being supported
by the flange, depending on the thickness of the head,
the pressure to be carried, and the disposition of the stays.
Speaking in general, the 3-in. distance is reasonable and
safe to use.
The tubes that are expanded into the tube sheet sup-
port that part of the head. A certain portion of the
head above the top row of tubes is supported by them,
depending on the size and holding power of the tubes
themselves.
A reasonable and safe distance to consider in such calcu-
lations is that of one-half of the bridge between the tubes.
Suppose, for example, that in a given boiler the tubes are
31/2 in., spaced 4 3/4 in. center to center; the bridge, or
section of plate between the tubes is 1 1/4 in. ; then one-
half that distance or 5/8 in. may be considered as being
supported above the top row of tubes by the tubes in the
top row.
Diagonal braces of the crowfoot type usually have two
rivets spaced 4 in. center to center. This permits of a
proper spacing of stays, which is the main point to con-
sider in laying out a head. The maximum allowable
pitch must not be exceeded. The braces can be made to
suit the load. Formula (7) decides just what pitch may
be used.
46 ARITHMETIC OF THE STEAM BOILER
Flat Surfaces to be Stayed, in which the Thickness
of Plate Enters the Calculation
112 X^ 2
A = t — > for plates up to 7/16 in. thick. (1)
V
120X/ 2
A= — ? for plates above 7/16 in. thick. (2)
P
140 X^ 2
A= — - — ? for screw stay bolts and nuts. (3)
112 X^ 2
p = — -j — 7 for plates 7/16 in. and under. (4)
120X/ 2
p= — -j , for plates over 7/16 in. thick. (5)
T AO X / 2
p = — -j — -> for screw stay bolts and nuts. (6)
/112 X£ 2
S = \ > f° r plates 7/16 in. thick and less. (7)
/l20X^ 2
S = \l f ioi plates above 7/16 in. thick. (8)
S = \l— j for screw stay bolts and nuts. (9)
-J*M (I0 )
\ 112
t
If this formula gives more than 7/16 in. for the value of
/, use the next formula with the factor 120 in it.
=J^4 (II)
\ I20
\ 12"
[40
BOILER HEADS— STAYED HEADS 47
In the foregoing group of formulas, the values are as
follows :
A = area in square inches, supported by one stay.
The pitch can be found by extracting the square
root of A.
t — thickness of plate, expressed in sixteenths of an
inch. Example: if the plate were 7/16 in.
thick, / in this case would be 7. In other
words, t is the numerator of the fraction
whose denominator is 16.
p = pressure in pounds per square inch, allowed
to be carried. #
5 = pitch of stays, center to center.
112, 120, 140 are constants used in the respective
formulas.
Examples, illustrating the application of the formulas in this
group:
Assume plate 7/16 in. thick; pressure to be carried, 100 lb. per
square inch. Required, area that may be supported by one stay,
from which the pitch distance may be found.
Then, in formula (1),
112 X 7 2
A = = 54.88 square inches supported by one stay.
In formula (4),
112 ^^ 1
p = «x— =100 lb. per square inch pressure allowed.
In formula (7),
/112 X 7 2
S = % — = 7.4081 in. center to center of stays.
\ 100
4 8
ARITHMETIC OF THE STEAM BOILER
In formula (10),
t=\ — — = 7, numerator of the fraction 7/16,
\ 112
thickness of plate in inches.
Formulas (4), (5), and (6) in this group serve as a check
on the design of stays, for by applying the proper one
to any given case it can be determined as to whether or
not the desired pressure may be carried.
Segments to be Braced
In Fig. 10 is shown that portion (shaded) of a boiler
head that requires to be braced. The flange is presumed
i/=Height of Segment in Inches
R= Radius of Circle of which the Segment is a Part
C = Center of the Circle of which the Segment is a Part
Fig. 10. — Segment of head to be braced.
to support the head for a distance of 3 in., and the tubes,
a distance of 2 in. above as shown in the figure. These
are the generally accepted figures by most designers of
boilers and experience seems to have proved them to be
safe.
BOILER HEADS— STAYED HEADS 49
To find the height of the segment requiring bracing,
subtract 5 in. from the distance from the tubes to the
highest part of the shell. To find the diameter of the
circle of which the segment is a part, subtract 6 in. from
the diameter of the boiler.
To find the area of a segment sufficiently close for all
practical purposes, the following formula may be used
(see Fig. 10):
4
4XH* 2XR ,
X a /— rr .608 = area.
In which H = the height of segment, R = the radius of
circle, and the numerals, constants.
Suppose, for example, it is desired to know the area
of a segment of a circle whose diameter is 50 in., the
height of the segment being 20 in. The statement
becomes:
4x^o! x i,_x*s_ 6o8=
3 \ 20
1600 co
3 \20
1600
XaJ 2 '5-- 6 ° 8 :
*22 x Ji.892_
3 x
1600
3
1600X1.37S
<i-375
= 733-3 sq. in. area.
50 ARITHMETIC OF THE STEAM BOILER
The formula just given, when applied to segments
whose heights are three-tenths of the diameter of the
circle of which the segment is a part, gives results correct
within i/io of i per cent.
Method of Finding the Area oe a Segment by Use
oe a Table
As well as the formula that has been given and
illustrated by an example, the area of a segment may be
found by using the table on page 51, and by an approxi-
mation method illustrated in Fig. 11.
It is desired to find the area of the shaded segment of
the circle in Fig. 11, by using the table. The height
of the segment is 16 in., and the diameter of the circle of
which the segment is a part is 54 in. The method of
operation is as follows: — = .2963, which is the height
54
of a similar segment of a circle whose diameter is 1.0.
The two nearest values, Ht. column in the table, to .2963
are .29 and .3. The corresponding areas are, for .29 =
.18905; for .3 = .i98i7.
A convenient value to use without going into inter-
polation is .2963 = . 195, and S4 2 X. 195 = 568.620 sq. in.
area.
As a check on the foregoing, the formula may be
applied to the same example:
-X-v/—— .608= 567.780 sq. in. area.
3 \i6
BOILER HEADS—STAYED HEADS
51
AREAS OF
CIRCULAR SEGMENTS
Ht.
Area
Ht.
Area
Ht. |
Area
Ht.
Area
Ht.
Area
O.OI
0.00133
O.II
0.04701
0.21
0.1199
0.31
0.20738
0.41
0.30319
0.02
0.0037s
0.12
0.5338
0.22
0.12811
0.32
0.21667
0.42
0.31304
0.03
0.00687
0.13
0.06
0.23
0.13646
0.33
0.22603
0.43
0.32293
0.04
0.01054
0.14
0.6683
0.24
0.14494
0.34
0.23547
0.44
0.33284
0.05
0.01468
0.15
0.7387
0.25
0.15355
0.35
0.24498'
o.45
0.34278
0.06
0.01924
0.16
0.08111
0.26
0.16226
0.36
25455
0.46
0.35274
0.07
0.02417
0.17
0.08854
0.27
0.17109
o.37|
0.26418
0.47
0.36272
0.08
0.02943
0.18
0.09613
0.28
0.18002
0.38
0.27386
0.48
0.3727
0.09
0.03501
0.19
0.10390
0.29
0.18905
0.39
28359
0.49
0.3827
O.I
0.04087
0.2
0.11182
0.3
0.19817
0.4
0.29337
0.5
0.3927
Fig. 11. — Approximate method of rinding the area of segment
of a circle. Area of A B C D, 11X53 = 583 square inches. Area
of semicircle, : = 1145.115 square inches. Then, 1145.115
-583 = 562.115 square inches, area of shaded segment.
By the table,
By the formula,
Difference,
Difference per cent. =
568.620
. 840 sq. in.
.840
567.780
Xioo = .i48
52 ARITHMETIC OF THE STEAM BOILER
which is sufficiently close for all practical purposes in
boiler work.
The table gives the areas of segments from' a height of
.01 up to .5 (a semicircle) increasing by hundredths.
The areas of segments are from a circle of unit diameter.
If the values in the table are taken as feet, they apply to
a circle 1 ft. in diameter; if the values are taken in inches,
they apply to a circle 1 in. in diameter.
An approximate method is illustrated also in Fig. 11.
Consider the length of the rectangle ABDC as 1 in. less
than the diameter of the circle, and the width as n in.
This gives an area of 53X11 = 583 sq. in. This is to be
subtracted from the area of the semicircle of which the
segment is a part.
The area of a 54-in. circle is 54 2 X. 7854= 2290.23 sq.
r 1 • • 1 2200.23
in., and of the semicircle, =1145.115 sq. in.
From this subtract 583, which gives 562.115 sq. in.
as the area of the segment by this method. Comparing
this answer with that obtained by the formula:
567.78 — 562.115 = 5.665 sq. in. difference.
The difference per cent, is:
5-665
56^ Xl00 = -"'
or 1 per cent, in round numbers, for this particular case.
It must not be thought that the approximate method
just described will always differ from that of the formula
by 1 per cent. The difference may vary as much as 2
per cent., according as to the size of the circle and the
height of the segment as ordinarily found in stationary
BOILER HEADS— GIRDER BARS
53
boiler practice. In circles having a diameter from 30
in. to 102 in., the result found by the approximate method
will be in error less than 2 per cent., when the height of
the segment is not less than three-tenths the diameter
of the circle. The Hartford Steam Boiler Inspection
and Insurance Company sanction the use of the approxi-
mate method, because of its simplicity, and because it
gives results sufficiently close for all practical purposes
in relation to segments of horizontal tubular boiler heads
to be braced.
Girder Bars
Girder bars are of two kinds, the split bar, Fig. 12 (a),
and the solid bar (b). The figure shows the application
3 Bolts
ftl ftl tfl
Split, or Double Bar
(a)
Solid Girder Bar
(b)
Fig. 12. — Girder bars.
of girder bars to the combustion chamber tops of marine
boilers and to the crown sheets of locomotive boilers.
The safe working pressure for solid girders, Fig. 12 (6),
may be found from the following formula:
CXd 2 Xt
P =
(w-p)XDX?
54 ARITHMETIC OF THE STEAM BOILER
in which the value of the letters are:
P= safe working pressure in pounds per square
inch.
d= depth of the girder in inches.
/= the thickness of girder in inches.
w= width in inches of combustion chamber, meas-
ured in direction at right angles to the length
of the bar.
p= pitch of bolts in inches, securing bars to the
sheet.
D= distance in inches, from center to center of
girders.
1= length in feet, of girders.
C= a constant, according to design, as follows:
550 for bars with one bolt, 825 for bars with
two or three bolts, 935 for bars with four
bolts.
The safe working pressure when split girder bars, Fig.
12 (a), are used, may be found from this formula:
24oooX/X ^ 2
In which the value of the letters are:
P= safe pressure in pounds per square inch.
/= thickness in inches of one girder.
d= depth in inches of one girder.
D= distance in inches, from center to center of
girders.
1= length in inches of girders.
24,000= a constant.
BOILER HEADS— GIRDER BARS 55
The following example shows the application of the
first formula.
A solid rectangular steel girder bar 3 ft. long, 8 in. deep,
2 in. thick, three bolts, bars spaced 8 in. center to center,
pitch of bolts 8 in., and combustion chamber top 32
in. wide. What pressure may be carried?
In this example the value of C is 825. Then:
825X8 2 X2 105600 D „
r=-, ^-- , o w = ~z~= 183.3 lb. per square inch
The values quoted for C are those prescribed by the
United States Board of Supervising Inspectors of Steam
Vessels.
The following example shows the application of the
second formula.
A split or double girder bar, 8 in. deep, 1 in. thick, 30
in. long, the bars being placed 8 in. center to center.
What pressure may be carried?
24000X1X8 2 1536000
8X30 2 7200
= 213.3 lb. per square inch.
The strength of stay bolts, and the load carried by each
one fitted to girder bars, are calculated in the same way as
for any flat surface; and the same rules apply for the
maximum pitch and safe working pressure as for any
flat surface.
CHAPTER III
Manhole Reenforcing Rings
Reenforcing rings are for the purpose of strengthening
manhole openings in boilers. When only one ring is
used it may be placed either inside or outside of the shell.
When two rings are used, one is placed inside and the
other outside the shell.
• Single rings may be from i 1/4 to 1 1/2 times the thick-
ness of the shell plates, and when double rings are used
the thickness of each may be the same as that of the
shell plates. The rings must be riveted to the shell by
a sufficient number of rivets of proper size, so that the
combined resistance to shearing will be at least equal to
the resistance of the rings to tensile stress. The various
formulas for reenforcing rings are as follows:
W = — — +D. (1) For single-riveted rings.
2 X t
W= — —+2X0. (2) For double-riveted rings.
2 X t
Formulas (1) and (2) are for single rings.
Z/X/
W= -. +D. (3) For single-riveted rings.
4Xf
Z,X /
W = —rz I + 2XD. (4) For double-riveted rings.
4X£
56
MANHOLE REENFORCING RINGS 57
Formulas (3) and (4) are for double rings.
The values of the letters are as follows:
W= width of rings in inches.
t= thickness of rings in inches.
t\ = thickness of shell plates in inches.
D— diameter of driven rivet in inches.
L = length of opening in shell in inches.
In calculating the least number of rivets to be em-
ployed, the net section of the ring is to be used. In
single riveting, the net section is found by subtracting
the diameter of the rivet hole from the width of the
ring, and then by multiplying the remainder by the
thickness of the ring. When the ring is double riveted,
subtract twice the diameter of the rivet holes from the
width of the ring, and multiply the remainder by the
thickness. The diameter of rivet holes is the equivalent
of the driven size of rivets, and refers to the trial size of
rivets to be chosen in the following formulas:
at 4XTXA .
iV = 5X(^X.78 S4 ) (5) F ° r Smgh rmgS *
N = i.8 S X5X(^X. 7 8S4) (6) DaMe ringS -
The values of the letters are:
N= number of rivets required.
A = net section of ring in square inches.
T= tensile strength of ring per square inch section.
S = shearing strength of rivet per square inch section.
D = trial diameter of driven rivet in inches.
.7854 = a constant used in finding circular areas.
58 ARITHMETIC OF THE STEAM BOILER
4 = a constant used for single reenforcing rings.
8 = a constant used for double reenforcing rings.
If either of these formulas give a too small number of
rivets, too widely spaced, a smaller trial diameter of
rivet must be chosen and the formula again applied. If
on the other hand too many rivets are found, causing
them to be too close together, a larger trial diameter of
rivet must be chosen.
Examples showing the application of the foregoing
group of formulas.
A manhole in a boiler is 11X15 in.; the n -in. dimen-
sion lying in the direction of the length of the boiler.
The shell plates are 3/8 in. thickness, and the reenforc-
ing ring is to be 1/2 in. thick, single riveted, driven size
of rivets 7/8 in. What is the required w T idth of ring?
Using formula (1) the statement becomes:
I T ^ 9*7 f
W = ' " ' ^ +-875 = 5.0 in., answer.
2 X . 5
A manhole opening is 11X15 in.; the n-in. dimension
lying in the direction of the length of the boiler. The
shell plates are 1/2 in. and two 1/2-in. rings are to be
used, double riveted, driven size of rivets 7/8 in. What
width of rings is required?
In this case formula (4) is to be used. The statement
becomes:
1 1 X ^
W = +2X .875=4.50 in., answer.
4X .5
How many 7/8-in. driven rivets are to be used in a
single ring 1/2 in. thick, 5 in. wide, 60,000 lb. tensile
MANHOLE REENFORCING RINGS 59
strength, 38,000 lb. shearing stress of the rivets, ring to
be single riveted.
Formula (5) is to be used, and statement becomes:
The net section of the ring is (5 — .875) X. 5 = 2.0625 sq. in.
4X60000X2.0625 .
N = -5 77T~6 — 2^7 — o — ^ = 21 required, say 20.
38oooX(.875 2 X.7854) H J
When the total number of rivets is an odd number,
change it so that there will be an equal number of rivets
on each side of the center. A number divided by 4 may
be used.
How many i-in. driven rivets should be used in a
manhole reenforced by two rings 3/4 in. thick and 4 1/2
in. wide, single riveted, 38,000 lb. shearing stress per
square inch section, and 60,000 lb. tensile strength of
the material of which the rings are made.
Formula (6) is to be used. The net section of the ring
is (4. 5-i)X. 75 = 2-625 sq. in.
_ r 8X60000X2.625
l\ = — Q . . v . , 9Ky — o — r = 22.8, say 24 rivets.
i.85X38oooX(i 2 X.7854) ' J ^
(1 . 85 is a constant for rivets in double shear.)
Rings whose width is found by the formulas in this
group will have a total net cross-sectional area equal to
that of the metal taken from the shell to make the
opening. The total net section means for a single ring,
twice the net sectional area of the ring, and for double
rings, four times the net sectional area of one ring.
The formulas for finding the number of rivets to be
used make the resistance of the rivets to shearing equal
to the resistance to tensile stress of the metal in the rings.
6o
ARITHMETIC OF THE STEAM BOILER
The constant 4 that appears in formula (5) is obtained
from two sectional areas for each half of the ring, and
all the rivets in each half. Therefore in both halves
of the ring, and when considering the total number
of rivets in the whole ring, there are four sections to
consider.
Fig. 13. — Reenforcing ring. Diagram illustrating direction of
stress in reenforcing rings, and the origin of the constants 4 and 8
which appear in the formulas. Arrows a and b show direction of
stress borne by each half of the ring, and the rivets in each half.
With two rings, there are four sectional areas for each
half of the two rings, or twice as much as with one ring,
therefore the constant becomes 8 in formula (6).
Heating Surface or Boilers
The heating surface of a steam boiler is that part
exposed to the action of the hot gases from the furnace.
Finding the heating surface is a matter of mensuration,
and it is expressed in square feet.
To find the heating surface of a return-tubular boiler,
multiply two-thirds of the circumference of the shell by
HEATING SURFACE 61
the length, both in inches; multiply the number of tubes
by the circumference of one tube and by its length, also
in inches; take two-thirds of the area of each tube sheet
minus the area due to the tube openings in square inches,
add the products together and divide the sum by 144 to
convert into square feet.
To find the heating surface of a vertical tubular boiler,
multiply the circumference of the fire box by its height
above the grate, both in inches; multiply the number of
tubes by the circumference of one and by its length, also
in inches; find the area of the lower tube sheet minus the
area of tube openings, all in square inches; add the re-
sults together, and divide the sum by 144 to convert into
square feet.
To find the heating surface of water-tube boilers,
multiply the number of tubes by the circumference of
one and by its length, both in inches; find the exposed area
in square inches of one set of headers; find the number of
square inches in one-half of the steam drum or drums as
the case may be; add the values together, and divide by
144 to convert into square feet.
The true heating surface of a tube is the side exposed
to the hot gases; the inner surface in a fire tube, and the
outer surface in a water tube.
The following example illustrates the rule for finding
the heating surface of a steam boiler.
Example. — What is the total heating surface of a horizontal
return tubular boiler 60 in. diameter and 1 2 ft. long, with 80 tubes
2 in. diameter?
For the sake of simplicity consider the2-in. dimension as the
inside diameter of the tubes.
62 ARITHMETIC OF THE STEAM BOILER
The statement will be:
Circumference of the shell = 60X3.1416 = 188.496 in.
Length of shell 12X12 = 144 in.
Heating surface of the shell 188.496 X144 X 2/3 = 18095.616
sq. in.
Circumference of one tube 2X3.1416 = 6.2832 in.
Heating surface of all the tubes 80X144X
6.2832 = 72382.46 sq. in.
Area, in square inches, of one head 6o 2 X. 7854 = 2827.44.
Two-thirds area of both heads 2/3X2X2827.44 =
3769.92.
Area, in square inches, through all the tubes 2 2 X. 7854X80
Total heating surface ^5-6i6 + 72345-6 + 3769^-2 X 251.328
144
9 37o8-48
= —TT A — =650. 75 sq.ft.
144
Grate Area
The required grate surface in any given boiler depends
upon the rate of combustion of the fuel, the quantity of
water evaporated per pound of the fuel used, and the
total weight of steam generated per hour.
To find the grate area required for any given plant:
W
A = w in which the values are:
rXw
A = area of grate surface in square feet.
W = weight of steam required per hour.
w = pounds of water evaporated per pound of
fuel consumed.
r = rate of combustion in pounds per square foot
of grate per hour.
Example. — A battery of boilers is required to generate 7000 lb.
of steam per hour, consuming 15 lb. of coal on each square foot of
FURNACES 63
grate per hour. Assume the evaporation to be 7 lb. of water per
pound of the coal used, what grate area will be required?
Applying the formula, the statement becomes:
7000
A =^— — = 66.66 sq. ft.
1SX7
As the average evaporation per pound of fuel is differ-
ent for different boilers, the formula given is approximate
only, but for practical work it is considered suitable to
use.
Corrugated Furnaces
To find the safe working pressure of steel corrugated
flues and furnaces:
p = CXT
D
In which:
P= safe pressure in pounds per square inch.
T= thickness of metal of which the furnace is
made.
D= mean diameter of furnace in inches.
C= a constant, as follows:
15,600 for Morrison flues, under United States
Rules.
14,000 for Morrison flues under British and Canadian
Rules, and also for Purvis, Fox, and Brown
flues, under United States, British, and Canadian
Rules.
The mean diameter of a Morrison flue, under United
States Rules, is the least inside diameter of flue plus
64 ARITHMETIC OF THE STEAM BOILER
2 in. Under British and Canadian Rules, the mean
diameter is the diameter at the bottom of the corruga-
tions, as measured from the outside. The Fox type is
measured in the same way for the mean diameter by
the United States, British, and Canadian Rules.
Example. — What is the safe working pressure on a Morrison
furnace flue 36 in. mean diameter, with a thickness of metal of 3/8
in. The value of C in the question is to be taken as 15,600.
Applying the formula:
1 5600X .375 , 1K
P = — = 162.5 lb.
For furnaces other than corrugated, different authori-
ties give different formulas, according to size and design.
For plain flues made in sections not more than 8 ft. in
length, and with the ends of each section flanged and
riveted, with a ring between the flanges, this rule may be
used:
89600 XT 2
LXD
In which :
P= safe working pressure in pounds per square
inch.
T = thickness of flue in inches.
D= diameter of flue in inches (outside diameter).
L = length of section in feet.
Example. — A plain furnace flue is 24 in. diameter and 3/8 in.
thick, and made in sections of 6 ft. long. What pressure is safe to
carry?
Applying the formula:
89600 X. 3 75 2 „ 1K
P= TTZ = 105.0 lb.
6X20 °
TUBES 65
The collapsing pressure of steel tubes of sizes from 3
to 10 in. may be found from the following formulas.
P = 86,670-^-- 1386 (1)
And P=iooo( 1 — yl 1 — 1600^ )
(2)
The first one is to be used when the value of P is greater
than 581 lb. The second one is to be used when the value
of P is less than 581 lb.
The values of the letters in both formulas are:
P = collapsing pressure in pounds per square inch.
T = thickness of tube in inches.
D= outside diameter of tube in inches.
These formulas are based on experiments conducted
at the National Tube Works, McKeesport, Pa.
The factors of safety that may be used for tubes varies
from 4 to 7, according to surrounding conditions. For
instance, in a case where considerable damage to life and
property might result from the collapsing of a tube, a
factor of safety of 7 should be used.
Example. — What is the presumed collapsing pressure of a 3. 5-in.
lap-welded steel tube of .12-in. thickness? What pressure may be
safely carried where a moderate amount of loss would occur from
the failure of a tube?
First try formula (1).
P=86,67oX~- 2 -i386
= (86, 670X. 0343) — 1386 = 1586.781 lb. per square inch.
66 ARITHMETIC OF THE STEAM BOILER
Formula (i) proves to be the correct one to use, as the result is
greater than 581 lb. specified.
The safe pressure to carry, using 5 as a factor of safety under the
conditions stated, will be:
1586.781
= 3i7+ lb. per square inch.
Under United States Rules, lap-welded boiler tubes
from 1 in. to 6 in. ; inclusive, may be of any length, and
may be allowed an external safe working pressure up to
and including 225 lb. per square inch. This gives a
liberal factor of safety.
Horse-power of Boilers
In reality there is no such thing as horse-power of
steam boilers, but the term has come into use and there-
fore requires definition. In order to have a definite value
by which to compare boiler performances under different
conditions, the American Society of Mechanical Engineers
decided that a standard boiler horse-power should be
equal to the absorption of 33,330 B.t.u. by the water in
the boiler. This is based on the evaporation of 34 1/2
lb. of water per hour from a temperature of 212 F. into
steam of the same temperature and corresponding pres-
sure, that of the atmosphere.
As the latent heat of steam at atmospheric pressure is
required to convert 1 lb. of water at 212 F. into steam of
212 F., then the latent heat in B.t.u. 's times the number
of pounds of water chosen as the standard, gives the
value, thus:
966.1X34.5=33^30 B.tu.
BOILER HORSEPOWER 67
The standard boiler horse-power is found by the fol-
lowing formula:
nr _ WX(H-t+ 3 2)
3333o
In which, H.P. = the horse-power.
W = weight of water in pounds actually
evaporated per hour.
H= total heat of steam above 32 , at
the pressure of evaporation.
/= the temperature of the feed water.
Example. — A boiler evaporates 3000 lb. of water per hour from
a feed-water temperature of ioo° F. into steam of 85 lb. gage pres-
sure, what is the standard horse-power?
The absolute steam pressure is 85 + 15 = 100 lb. in tound numbers.
Referring to a table of properties of saturated steam, it will be found
that the total heat of steam at 100 lb. absolute pressure is, in round
numbers, 1182 B.t.u. The statement becomes:
tj D 30ooX(n82-ioo+32)
H.P. = = 1 00 -f- horse-power.
33330
Even values have been taken instead of exact ones, in
order to simplify the operation. It must also be remem-
bered that different steam tables in use will give slightly
different values, but for purely practical work, the dif-
ference need not be considered.
Horse-power rating of boilers is sometimes based upon
the number of square feet of heating surface, varying from
6 to 20 square feet per horse-power, according as to the
type of boiler as follows:
Water-tube boilers, 10 to 12 sq. ft. per horse-power.
Return tubular boilers, 12 to 15 sq. ft. per horse-
power.
6
68 ARITHMETIC OF THE STEAM BOILER
Vertical tubular boilers, 15 to 20 sq. ft. per horse-
power.
Flue boilers, 8 to 12 sq. ft. per horse-power.
Plain cylinders, 6 to 10 sq. ft. per horse-power.
These values are arbitrary and serve only as a guide in
buying and selling. Different manufacturers had in the
past adopted different values, there being no fixed stand-
ard until recently, when the Boiler Manufacturers' As-
sociation adopted 10 sq. ft. of heating surface to a
horse-power, in horizontal tubular boilers.
Ratio of Heating Surface to Grate Area
The ratio between the heating surface and grate sur-
face varies with the type of boiler and also with the rate
of combustion.
^ . heating surface
Ratio = — - — 7
grate surface
Water-tube boilers, 35 to 40.
Horizontal tubular, 25 to 35.
Vertical boilers, 25 to 30.
Locomotive boilers, 50 to 100.
Flue boilers, 25 to 35.
Plain cylinder boilers, 12 to 15.
Equivalent Evaporation
Equivalent evaporation from and at 2i2°F. means
the quantity of water that would be converted into steam
of 212 F. and at atmospheric pressure, from a feed-water
temperature of 212 F. as compared with the actual
evaporation under certain conditions, for any given case.
Ratio =
EQUIVALENT EVAPORATION 69
Equivalent evaporation reduces actual evaporation to a
standard basis, from which comparisons can be made in
boiler trials.
The formula is as follows:
w _ wX(H-t+ 3 2)
970.4
In w T hich the letters have the following values:
W = equivalent evaporation from and at 212 F. in
pounds.
w = actual evaporation in pounds.
H = the total heat of steam above 3 2 F. at the pres-
sure of evaporation, as found in the steam tables.
t = the temperature at which the feed water enters
the boiler; 970.4 latent heat of steam at atmos-
pheric pressure according to Marks and Davis's
tables.
Example, illustrating the application of the formula.
A certain boiler generates 3000 lb. of dry steam per hour at a
pressure of 150 lb. gage from a feed- water temperature of 200 F.
What quantity of water would have been evaporated had the feed
water been delivered at 212 F. and converted into steam of 212 F.
and at atmospheric pressure?
According to Marks and Davis's steam tables, the
total heat of steam at 150 lb. gage pressure, or 165 lb.
(in round numbers) absolute pressure is 1195.0 B.t.u.
Applying the factors, the statement becomes:
W = =3175 lb., closely.
If steam tables be used other than Marks and Davis's,
slightly different results will be obtained. Before these
tables were devised, the constant used was 965.7 and
70 ARITHMETIC OF THE STEAM BOILER
sometimes 966.1 instead of 970.4. These values repre-
sent the latent heat of steam at atmospheric pressure,
according to the tables from which they are derived.
It may be stated that, excepting in cases where ex-
treme accuracy is required, it makes little difference
to the engineer as to which value he uses, nor from which
steam table he procures any of the values used in steam
calculations. As between the three latent heat constants
given, a difference in results obtained will be less than
1 per cent., w T hich is sufficiently close for all calculations
in the realm of the practical operating engineer.
In the formula, the quantity that changes
970.4
the actual evaporation of 1 lb. of water to equivalent
evaporation from and at 212 F. is known as the factor of
evaporation. In the example given, the factor of evapora-
tion is:
110=; — 200+32
-^ ~ LA - = 1-0583
970.4 D °
and when multiplying the actual total evaporation by
the factor, the equivalent evaporation is obtained thus:
3000X1.0583 =3174.90 which, as before given, would be
called 3175 lb. in round numbers.
The factor for any other values of steam pressure and
feed-water temperature will be obtained and used in the
same way as just illustrated.
Boiler Efficiency
The efficiency of a boiler plant is the ratio of the dif-
ference between the heat in the steam (delivered by the
boiler) and the heat in the feed water, to the heat that
BOILER EFFICIENCY 71
would be developed by the perfect combustion of the
fuel.
The following example illustrates the foregoing
definition:
The heat of combustion of a certain fuel is known to
be 14,000 B.t.u. pe" pound; the number of pounds of
such fuel used at a boiler test was 3500 per hour; the
evaporation was 30,000 lb. of feed water per hour, into
steam of 90 lb. gage pressure, or 105 lb. absolute; the
temperature of the feed water was 8o° F. It is required
to find the efficiency of the boiler.
The method of operation is as follows:
The total heat in 1 lb. of steam at 105 lb. absolute is
1 187.2 B.t.u. (from Marks and Davis's tables).
Temperature of feed w r ater = 8o° F., and 80 — 32=48
assumed B.t.u. contained in the feed water above 32 F.
Then: 1187.2—48 = 1139.2 B.t.u. and 1139.2X30,000 =
34,176,000 B.t.u. absorbed by the water.
As each pound of coal contains 14,000 B.t.u. and as
the total coal consumed was 3500 lb. the total heat
supplied = 14,000X3500 = 49,000,000 B.t.u.
The efficiency is:
34176000 X 100
49000000
= 69.74 per cent.
A Short Method to Find the Commercial Effi-
ciency of Boiler and Furnace Combined
Expressed in terms of cost of evaporating 1000 lb.
of water from and at 21 2 F.
C
C ~2Xe
72 ARITHMETIC OF THE STEAM BOILER
In which the values are as follows :
c = cost of evaporating 1000 lb. t)f water from and at
212° F.
C = cost of coal per ton of 2000 lb.
e = ihe evaporation per pound of coal, from and at
212° F.
2=a constant, to reduce the values to a common
basis of 1000.
Example, illustrating the formula:
A certain coal costs $2.00 per ton of 2000 lb., and is known
to actually evaporate 8 lb. of water per pound of coal consumed
on the grates. The factor of evaporation — from the method
previously explained — is found to be 1.0583. Therefore the equiva-
lent evaporation from and at 212 F. is 1.0583X8 = 8.4664 lb. of
water per pound of coal.
2.00
Then, c = wo — tt~ =S.ii8: or 11 8/10 cents per lb.
' 2X8.4664 ' ' *
As a check on the accuracy of the foregoing proceed thus:
200
Coal at $2.00 per 2000 lb., costs per pound, =.10 cents.
Each 8.4646 lb. of water evaporated from and at 212 F. costs .1
or 1/10 of a cent. As the cost for evaporating 1000 lb. is required,
then
1000
~ 7—7 = 118.1 lb. of coal required
and 1 18. 1 X.i =11.81 cents, as found before.
How to Analyze a Boiler Trial Report
The following is the report of a boiler trial which was
held to determine the efficiency under given conditions.
BOILER TRIAL
73
Sur-
Pres-
I
i
Tem-
pera-
tures.
TRIAL OF A iooo H.P. BOILER
i. Kind of trial running start
and stop.
2. Duration of trial 24 hours.
3. Grate surface 113.6 sq. ft.
4. Total heating surface 10,000 sq. ft.
faces. ] 5. Ratio of heating surface to grate sur-
face 88.0
6. Average pressure per square inch,
gage 81.0 lb.
7. Average atmospheric pressure per sq.
in 14.84 lb.
8. Average absolute pressure per sq. in., 95.84 lb.
9. Force of draft (column of water) 21 in.
10. Temperature of the external air 30 F.
11. Temperature of the fire room 93 F.
12. Temperature of the feed water before
entering the boiler 37-3° F.
13. Temperature of escaping gases after
leaving the boiler 407 F.
14. Temperature of the steam 324+ F.
15. Moist coal consumed 76,687 lb.
16. Moisture in the coal 4.12 per cent.
17. Dry coal consumed 73,528 lb.
18. Total dry refuse 8069 lb.
19. Total dry refuse 10.97 per cent.
1 20. Total combustible 65,459 lb.
21. Dry coal consumed per hour 3064 lb.
22. Combustible consumed per hour 2727 lb.
23. Percentage of moisture in the steam. 0.6013 per cent.
24. Number of B.t.u. in 1 lb. dry coal. . . 14,500.
25. Number of B.t.u. in 1 lb. combustible 15,425.
f 26. Heat absorbed by the boiler, per
pound of steam generated H75-6 B.t.u.
27. Total B.t.u. absorbed by the boiler. . 774,880,282.
28. Heat units imparted to the boiler per
pound of dry coal 10,538.
I, 29. Heat units per pound of combustible 11,838.
B.t.u. «j
74
ARITHMETIC OF THE STEAM BOILER
Water
o
a
o
30. Efficiency of the boiler, based upon
dry coal (approximately) 72.7 per cent.
31. Efficiency of the boiler, based upon
combustible 76.77 per cent.
32. Factor of evaporation 1.22.
2,3. Total water fed to boiler 663,124 lb.
34. Water actually evaporated, corrected
for quality of steam 659,136 lb.
35. Equivalent water from and at 212 F.
boiler only 804,146 lb.
36. Equivalent water from and at 212 F.
per hour, boiler only 33, 506 lb.
37. Water actually evaporated per pound
of dry coal 8.96 lb.
3S. Water evaporated per pound of com-
bustible 10.07 lb.
39. Horse-power, basis 34 1/2 lb. from
and at 212 F 971.
40. Number of square feet heating sur-
face per horse-power 10.3.
41. Horse-power per square foot of grate 8.53.
42. Builder's rating, horse-power 1000.
In the following, some of the results are not exact as
far as actual numerals are concerned; this is due to dis-
regarding too small decimal values, and using round num-
bers instead as far as consistent. For practical purposes
the values are sufficiently close ; the object in view is to ex-
plain and illustrate how the values are found, rather than
to give an exhibition of exact arithmetical operations.
The first two items are self-explanatory.
The grate surface (item 3) is found by multiplying
together the length and width in feet, which gives th,e
area in square feet. The heating surface (item 4) is the
sum total of all surfaces which are in contact with the
BOILER TRIAL 75
hot gases on one side and water on the other side. The
method of finding the heating surface has been explained
in a previous section.
The ratio of heating to grate surface (item 5) is
found by division thus : 10,000^113.6 = 88.0. That is,
item 4 is to be divided by item 3.
The pressure of the atmosphere is noted (item 7)
as 14.84 lb. per square inch. The pressure of the atmos-
phere is usually considered as 14.7 lb. per square inch,
and in ordinary calculations as 15 lb. in round numbers,
at sea level. As a matter of fact the pressure of the
atmosphere is constantly changing as indicated by the
barometer. The way 14.84 was found is this: the height
of the barometer divided by 2.04 equals atmospheric
pressure. The 2.04 factor is the height of mercury in
inches that is equal to 1 lb. per square inch.
As an example, the barometer indicates 29.5 in., and
29.5^2.04=14.4 lb. per square inch. Applying this,
14.84X2.04 = 30.2+ in., which was the barometer read-
ing at the time of the trial.
Absolute pressure is measured from a perfect vacuum,
and is gage pressure plus atmospheric pressure, in this
instance 81 + 14.84 = 95.84 lb., item 8.
The force of draft in column of water is 0.21 in. A
draft gage is used for the purpose, one (1) ourice pressure
per square inch is equal to 1.73 in. of water as registered
by the gage. The value 1.73 is found like this:
34 ft. =408 in.
408 in. = 14.7 lb. (at sea level).
14.7 lb. = 235.2 oz.
Therefore, 1 oz. pressure = of 408 in., or
Oj'
76 ARITHMETIC OF THE STEAM BOILER
408-T- 235.2 = 1.73 in. of water as indicated by gage.
As a matter of convenience, the atmospheric pressure
at the time of the trial is considered as having been 14.7
lb. instead of 14.84, and the draft pressure in ounces is
121 r
0.21-M. 73=0. 121 or of an ounce.
'° 1000
The temperature of the steam (item 14) is found from
a steam table, of which there are several in use; the
one in particular gave the value of 324 F. corresponding
to the pressure.
The number of pounds of moist coal consumed was
76,687 as actually weighed. As the coal contained 4.12
per cent, of moisture, 4.12 per cent, of the 76,687 must
be subtracted, which gives as a remainder 73,528 lb. of
dry coal consumed (item 17), thus: 76,687X4.12 =
3159.50 and 76,687-3159.50 = 73,527.5, say 73,528 lb.
in round numbers.
The moisture per cent, in the coal (item 16) is found
like this: Exactly 100 lb. of coal, from the pile to be
used during the trial, is placed in a bag or a box, and
subjected to a good heat, such as would obtain on top of
the boiler in operation, until it is thoroughly dried out.
The coal is again weighed, and in the case under con-
sideration was found to be 95.88 lb.; then, 100 — 95.88 =
4.12 lb. moisture was evaporated, and this is 4.12 per
cent, of the original weight.
The total dry refuse — the non-combustible part of
the coal — was 8069 (item 18) which is: (item 19)
8069-^73,528 = 10.97 per cent. The total combustible
(item 20) was: 73,528 lb. dry coal minus 8069 dry
refuse = 65,459 lb. combustible.
BOILER TRIAL 77
The total quantity of dry coal consumed was 73,528;
for one hour the quantity was 73,528-^24 = 3064 lb.
(item 21).
The combustible for one hour is found in the same
manner.
The percentage of moisture in the steam (item 23)
0.6013 was determined by the use of a calorimeter.
The heat value of the coal was determined by an
analysis and test in a laboratory equipped with apparatus
for that and similar purposes. In item 24 it is given as
14,500 B.t.u. per pound dry coal, and in item 25, 15,425
B.t.u. per pound of combustible.
In item 26 the heat absorbed by the boiler per pound
of steam generated was found like this:
By referring to a steam table, the latent heat of steam
at 95 lb. absolute pressure — which the report records —
is found to be 886.7 B.t.u., while the sensible heat is
323.89 and 886.7+323.89 = 1210.59, from which is to be
subtracted the given temperature of the feed water, 37.3
F., which gives 1210.59 — 37.3 = 1173.29 B.t.u. This
method is approximate only. The following is more
nearly accurate.
The total heat of steam above 32 F. at 95 lb. absolute
pressure is 1 180.7, as found in the steam table used in
connection with this particular test. The temperature
of the feed water above 32 F. was 37.3 — 32 = 5.3, and
1180.7 — 5.3 = 1175.4 B.t.u., which more nearly agrees
with the item in the report.
The difference of 2. 11 which exists between the two
results is accounted for in this way :
The temperature of steam at 95 lb. absolute pressure
78 ARITHMETIC OF THE STEAM BOILER
is 3 23. 89 ° a,s found from the table used at that time.
But strictly speaking, calculations involving the steam
tables are based on water from 32 F. and not from zero
on the Fahrenheit scale, as is sometimes done. Ignoring
this will give approximate results only. Then, 323.89
— 32 = 291.89; but the heat of the liquid as found in
the table is 294.0 and 294.0—291.89 = 2.11, the difference
before referred to. This difference is due to degrees
temperature and B.t.u. 's not being exactly the same in
value. However, the difference in the range of the
steam tables is so small, that for all practical purposes
where extreme accuracy is not demanded, degrees tem-
perature may be used instead of heat in the liquid
values. It is only as a matter of convenience that
degrees temperature and units of heat are sometimes
considered synonymous.
The total heat units absorbed by the boiler during
the trial was 774,880,282 (item 27) found like this:
As 1175.6 is the heat units per pound of steam gene-
rated by the actual quantity of water evaporated (cor-
rected for the quality of the steam) which appears as
659,136, then 1175.6X659,136 = 774,880,282 total
B.t.u. absorbed.
The heat units imparted to the boiler per pound of
dry coal are found to be 10,538, found like this:
Total B.t.u. Dry coal
consumed
774,880,282^-73,528=10,538 B.t.u.
per pound of dry coal (item 28).
The heat units imparted per pound of combustible
BOILER TRIAL 79
is found in a similar manner, only using the pound
combustible as a divisor thus:
774,880,282-7-65,459 = 11,838 B.t.u.
per pound of combustible (item 29).
The efficiency of the boiler based upon dry coal
(item 30) is found by dividing the heat units per pound
of dry coal by the theoretical heating value, which is
taken as 14,500 B.t.u. thus: 10, 538-r- 14, 500 = . 7267
which is 72.67 per cent.
The efficiency based upon the combustible is:
B.t.u. per lb. Theoretical
combustible value
11,838 -f- 15,425 = .7675
w T hich is 76.75 per cent, (item 31); the efficiency of the
boiler is the ratio of the heat utilized to that supplied.
The factor of evaporation (item 32), 1.22, was found
like this:
1181-5-3
~ — ^—=1.22
05-7
in which 1181 = B.t.u. in steam at 95 lb. pressure absolute.
9 65.7 = B.t.u.
required to evaporate 1 lb. of water from and at 212 F.
5-3 = (37-3-3 2 )
The factor of evaporation means, that for every pound
of water actually evaporated under the prevailing con-
ditions at the time of the trial, 1.22 or 1 22/100 lb. of
water would have been evaporated had the feed-water
80 ARITHMETIC OF THE STEAM BOILER
temperature been 212 F. and had the pressure been that
of the atmosphere.
Item 33 gives the total number of pounds of water
fed to the boiler during the trial, as found by actually
weighing it.
Item 34 is found by multiplying the total weight
of water fed to the boiler by the percentage of moisture
as found in the steam expressed decimally, and then
subtracting that value from the original quantity, thus:
663, 124X. 006013 =3988— lb.
moisture in the steam during the whole test, and 663,124
— 3988 = 659,136 lb. actually evaporated.
The equivalent evaporation from and at 212 F. (item
35) is found thus: actual evaporation times factor of
evaporation, or
659,136X1.22 = 804,146 lb.
The equivalent evaporation per hour (item 36) is
found thus:
804,146-^24 = 33,506 lb.
The quantity of water actually evaporated per pound
of dry coal consumed is 8.96 lb. (item 37) as found from
dividing thfe total' water actually evaporated by the
total dry coal consumed. Or,
659,136-^73,528 = 8.96 lb.
The water actually evaporated per pound of com-
bustible (item 38) is found in a similar way, thus:
659,136^65,459=10.07 lb.
BOILER TRIAL 81
Take the 8.96 lb. water per pound of dry coal, and
multiply it by the factor of evaporation 1.22 thus:
8.96X1.22 = 10.94 lb. of water from and at 212 F. on
the dry coal basis.
So, also in relation to the combustible, 10.07X1.22 =
12.29 lb. water from and at 2i2°F. on the combustible
basis.
The horse-power is found (item 39) by dividing the
pounds of water evaporated per hour from and at 212
F. by the standard 34.5 thus: 33,506^34.5 = 971 H.P.
The builder's rating was given as 1000 H.P.
This is in excess of the actual power developed at the
trial by 1000 — 971 = 29 H.P. and this expressed as a per-
centage is 2.9 found thus:
29 -J- 1000 = .029 or 2.9 per cent.
By dividing the 10,000 sq. ft. heating surface by 971,
the quotient obtained is 10.3 (nearly) which is the number
of square feet per horse-power (item 40).
And 1 13.6 sq. ft. grate surface divided into 971 gives
8.53 H.P. per square foot of grate (item 41).
PART II
MISCELLANEOUS APPLICATIONS OF BOILER
ARITHMETIC
MISCELLANEOUS APPLICATIONS
Bursting Pressure of Pipe
2tXS
P =
D
In which, P = bursting pressure, pounds per square inch.
t = thickness in inches.
S = tensile strength of the metal in pounds per
square inch.
D = internal diameter of pipe in inches.
Example. — Find the bursting pressure of a io-in pipe, 0.366 in.
thick, actual internal diameter 10.019 in. tensile strength of the
metal taken as 50,000 lb.
2 X. 366X50,000
=3653 lb.
10.019 ° °°
For the safe working pressure, a factor of safety of
not less than 10 should be used and preferably more.
3653
Therefore = 365.3 lb. safe working pressure.
The bursting pressure being given, to find the thickness-
of metal the formula is transposed thus:
DXP
85
86
ARITHMETIC OF THE STEAM BOILER
Using the terms of the same question the statement
becomes:
10.019X3653
t =
2 X 50000
= 0.366 in.
When the exact tensile strength is not known, assume
50,000 lb. for steel, and 40,000 lb. for iron pipe.
The actual bursting pressure of pipes, as found from
tests, is less than that found from the foregoing formula,
and this makes it necessary that a liberal factor of safety
be used.
The Force Tending to Tear Asunder
The force tending to tear asunder two cylinders of
Fig. 14. — Lower part of Manning boiler. Diagram showing
area against which the pressure acts, tending to tear apart the two
cylinders.
different diameters as illustrated in Fig. 14, which repre-
sents a portion of the Manning boiler, the pressure
being applied in the annular space between the two,
MISCELLANEOUS APPLICATIONS 87
depends upon the cross-sectional area of the space, and is
independent of the diameter. The force tending to break
the plates is the area of the sheets in the cross-hatched
portion multiplied by the pressure per square inch. This
force is resisted by the four plates, the outer and inner
plates being in tension.
The foregoing is based on the condition that the two
cylinders are stayed to each other as shown in the
illustration.
Method of Finding the Fuel Cost of Evaporating
1000 Lb. of Water
data
Assume that in a given case, 21 tons of coal were
consumed in a certain time, and that the cost of the coal
was $3.00 per ton of 2000 lb.
Also assume that in the same period of time, 398,616
lb. of water were evaporated into steam at a pressure of
100 lb. per square inch absolute, the temperature of the
feed w T ater being 200 F.
The number of heat units in a pound of water at 200
F. is 168.713 (as found from steam tables).
The number of heat units in 1 lb. of steam at 100 lb.
pressure per square inch absolute is 1181.866 (as fcund
in tables).
The number of heat units required to convert 1 lb.
of water into steam at atmospheric pressure, 14.7 lb.
per square inch, is 966 (in round numbers), and this is
the latent heat of steam at 14.7 lb. pressure per square
inch.
88 ARITHMETIC OF THE STEAM BOILER
METHOD OF OPERATION
First, find the factor of equivalent evaporation to
reduce the problem to the standard basis of, from and
at 2i2° F. Employing the given values:
1181. 866-168.7
— -7-7— —=1.048 factcr of evaporation.
The statement of the problem becomes:
Tons Cents Lb. water
21 X 300 X IOOO
398616 X 1.048
lb. water. Factor of
evap.
= 14.6 cts.
or, $0,146, cost of evaporating 100c lb. water under the
assumed conditions. The actual cost will be more than
that found above, for labor, depreciation, cost of water,
taxes, insurance, and interest on investment of the
plant as a whole, enters the calculation where precision
is required and over all charges are to be made.
Safe Working Pressure for Cylindrical Cast-iron
Vessels with Flat Cast-iron Heads
rules prescribed by the united states inspectors of
steam boilers
When evaporators, feed-water heaters and separators
are made of good cast iron, the shells cylindrical and the
ends flat, the castings sound and of uniform thickness,
the working pressure shall not exceed that found by the
following formulas: For finding the safe pressure on the
flat surface this is the formula to be used:
_ 20000 XT 2
MISCELLANEOUS APPLICATIONS 89
For the cylindrical part of the vessel this is the formula
to be used:
3500(7^-1/4)
r D
And to find the thickness of metal required, having the
other values given, use these formulas:
For the flat heads
-4
TXD 2
20000
and for the cylindrical shell
m PXD
35°°
1/4.
In the formulas given the value of the letters stand like
this:
P = safe working pressure in pounds per square inch.
T = thickness of metal in inches, provided the
thickness of the ends or heads of such vessels
shall not be less than 3/8 in.
D = inside diameter of the vessel in inches. When
the ends or heads are bolted to the shell then
D = the diameter of the bolt circle. When the
pressure is to be determined for a part of a
flat surface which is square or rectangular,
the value of D in the flat surface formula
shall be the diagonal of the square or rec-
tangle. The numbers 20,000 and 3500 are con-
stants, evidently empirical, and found from
experiment.
go ARITHMETIC OF THE STEAM BOILER
Equivalent Boiler Performance
A boiler is sold on a guarantee that it will evaporate
12 lb. of water per pound of combustible, from and at
2i2° F., with coal having a heat value of 14,500 B.t.u.
per pound of coal, and 8 per cent. ash.
When the test was made the average feed-water tem-
perature was 180 F., average steam pressure 70 lb.,
heat value of the coal used 12,000 B.t.u. per pound, and
the ash content 8 per cent.; water evaporated, 9 lb. per
pound of coal consumed.
It is required to ascertain how the test compares with
the guarantee.
Under the conditions of the guarantee the coal is .08
ash, and 1 — . 08 = .92 combustible.
If 1 lb. of coal contains 14,500 B.t.u., 1 lb. of com-
bustible will contain — ) — -=1^,760 B.t.u.
.92 J "
With this it is agreed to evaporate 12 lb. of water
from and at 212 F. or 1 lb. under those conditions for
. 15760 _ _. _
each = I 3 I 3 B.t.u. supplied.
The temperature corresponding to 85 lb. absolute
(70 lb. gage in round numbers) is 316.3 F.
To raise 1 lb. of water from 32 to 316. 3 requires
286.30 B.t.u.
To raise 1 lb. of water from 32 to 180 requires
147.88 B.t.u.
To raise 1 lb. of water from 180 to 316. 3 requires
138.42 B.t.u.
To evaporate 1 lb. of water from and at 316.3 requires
897.10 B.t.u.
MISCELLANEOUS APPLICATIONS
91
To raise 1 lb. of water from 180 to 316. 3 , and evapo-
rate it at 85 lb. pressure absolute, requires 1035.52 B.t.u.
The coal used at the test contained 12,000 B.t.u. per
r i 12000
pound, and as there was 8 per cent, of ash, =
13,043 B.t.u. per lb. of combustible.
Under the conditions of the test, 1035.52 B.t.u. were
required per pound of water evaporated into steam.
13043
Then,
- = 12.59 lb. water evaporated per pound of
io3S-5 2
combustible, as compared with the guarantee of 12 lb.
of water per pound of combustible.
Efficiency of Diagonal Seam
Assume a seam such as that shown in Fig. 15; the shell
Fig. 15. — Diagonal seam.
is 60 in. diameter, made of 5/16-in. plate having a tensile
strength of 60,000 lb. per square inch. The seam is 20
92 ARITHMETIC OF THE STEAM BOILER
in. wide at one end and 6 in. at the other. The rivet holes
are 11/16 in. in diameter and the pitch is 2 in.; the rivets
are of steel. Assuming the efficiency of the joint to be
41 per cent., what increase in efficiency is obtained by
the seam being slightly diagonal to the axis of the shell?
To find this in a simple manner, multiply the length of
the seam in inches by 41, the efficiency of the joint, and
divide the product by 60, which is the length in inches
measured along a line parallel to the axis of the cylinder.
Thus: As the seam, measured along the rivet center
of the diagonal seam is 60.4 in., then,
60.4X41
— 7 = 41.27 per cent.
efficiency of the diagonal seam, a gain of .27 per cent.
Collapsing Strength of Cone-shaped Flue
A cone-shaped flue has a greatest diameter of 36 in.
and a least diameter of 12 in. and a length of 20 in.
It is made of 5/16-in. steel plate of 60,000 lb. tensile
strength per square inch. It is required to find what
collapsing pressure such a flue will safely withstand.
The construction is shown in Fig. 16.
It is customary in short cones to take the mean
diameter, in this case,
36 + 12
= 24 in.,
2
and calculate the collapsing strength as follows:
Hutton's rule is,
T 2 XC _
dxVl~ p
MISCELLANEOUS APPLICATIONS
93
The values of the letters are:
T = Thickness of plate in thirty-seconds of an
inch
D = External diameter of the shell in inches.
Z,=Length of shell in inches.
C = 660 for mild steel plates.
P — Collapsing pressure.
Applying the formula, the statement becomes:
Fig. 16. — Cone-shaped flue.
10X10X660
P= „ A w / — =614.0 lb.
24XV20
The collapsing pressure must be divided by the factor
of safety, and in view of high temperatures and wear
and tear, this should be 6; the allowable pressure will
614.9
be,
; 102.5 lb. per square inch.
The cone must be truly circular in form, and the brac-
ing should be placed as shown in the figure in order to
prevent the flue from being pushed down by the pressure
exerted on the inclined surface.
94
ARITHMETIC OF THE STEAM BOILER
Strength of Cone Seam
A tank is 48 in. in diameter and built of 1/4-in. plate
which has a tensile strength of 60,000 lb. per square inch.
At the lower end of the tank is a cone w T hich has a single-
riveted lap seam. The rivet holes are 11/16 in. in diam-
Fig. 17. — Strength of cone seam.
eter and the pitch is 2 in. What is the strength of the
seam?
Fig. 17 shows the construction and gives the
dimensions.
The longitudinal joint of a cone-shaped section which
is withstanding internal pressure is subjected to a vary-
MISCELLANEOUS APPLICATIONS 95
ing stress. The stress at any point in such a joint will
be inversely proportional to the distance of the point
from the axis of the cone.
It is not practicable to make a joint that would offer
equivalent resistance to these varying stresses; therefore
it is customary, when the cone is made of one sheet, to
make it of the same strength as the tank itself. If
several sheets are used in making up the length of the
cone, the longitudinal joint of each course might be
designed for strength inversely as its distance from the
axis. This, however, is rarely done except in large tank
work where the size makes the saving in material an
object.
The strength of the seam is the strength of the weakest
part, which, neglecting the crushing of the sheet in
front of the rivets must be either the tensile strength of
the ligament between the rivets, or the shearing strength
of the rivets themselves.
The strength of the portion of the plate between the
rivets is:
2 — .6875 60000 __ . , _
X =9943.75 lb. per inch of seam length.
Steel rivets in single shear are assumed to have a
shearing strength of 42,000 lb. per square inch of cross-
sectional area; the cross-sectional area of a rivet n/16 in.
in diameter is .37122 sq. in., and the strength of the
rivets is
.37122X42000 r „
— -=7795-62 lb.
96 arithmetic of the steam boiler
per inch of seam length, which, being weaker than the
ligament, is the strength of the seam.
While the seam is of equal strength throughout its
length, the internal pressure per square inch required to
shear the rivets will vary directly as the diameter of
the cone. For instance, on a line corresponding to a
diameter of 40 in. the bursting pressure of the cone
will be
*£*-*«« >b.
per square inch. For a diameter of 16 in. the seam will
fail at
7705.62
^^| — = 974.45 lb.
per square inch.
Collapsing Pressure of Fire Box
Fig. 18 illustrates a vertical fire-box boiler. It is
required to figure the collapsing pressure of the fire box,
assuming that 7/8-in. stay bolts are used and that the
pressure is 125 lb. per square inch; it is also required to
find what pitch to give the rivets in the vertical seam,
using 11/16-in. rivets.
In calculating the strength of cylindrical furnaces
supported by stay bolts, it is customary to assume that
the surface is flat; that is, the tendency of the form to
lend strength to the construction is ignored.
With a specified size of bolt the first step is to find. how
MISCELLANEOUS APPLICATIONS
97
many square inches of surface one bolt will support at
the given pressure of 125 lb. per square inch.
Standard stay bolts up to
1 1/4 in. and larger are cut
12 threads per inch, and it is
practically correct to assume
that the depth of thread is
the same as given by the
United States Standard for
that pitch, viz., .05425 in.,
. although this is not strictly
correct, as the usual stay-
bolt thread varies slightly
from this standard.
Using 7/8-in. bolts will
give an effective area of
.4614 sq. in. per bolt. The
working stress allowed in
stay bolts varies, according
to different authorities, from
6000 to 10,000 lb. per square
inch, the most generally ac-
cepted figure being 7500, and
on this basis a 7/8-in. stay
would support 3460 lb. or
at a pressure of 125 lb. per
square inch it would be capa-
ble of supporting
3460
r
^—2 hV 2 V
— T.S.
56,000 Lbs.
/
I
I 2 S
= 27.6
Fig. 18. — Diagram from
which to calculate the collaps-
ing pressure of fire box.
square inches of surface, or the bolts could be spaced
98 ARITHMETIC OF THE STEAM BOILER
\/27.6, or about 5 1/4X5 1/4 in. Of course to be
strictly correct, the area occupied by the bolt itself
should be added to the 27.6 in. before extracting the
square root to get the pitch, for the pressure does not
act on this area. But in actual practice such allowance
is not made.
The surface to be supported in the problem is 2 ft.
6 1/8 in. in diameter, or 94.64 in. in circumference,
and the approximate height or distance between the rivets
on the leg ring and crown sheet is 24 in. and with the
spacing given before it would be necessary to have 18
vertical rows of stays containing four stays each.
There is another feature to be considered in the stay-
ing of sheets that affects the strength, and that is the
stresses in the sheets themselves. Unwin has devised a
formula for the maximum stress in flat sheets supported
at regular intervals, the supported points forming squares.
The formula is:
in which a is a side of a square, / the stress per square
inch in the sheet, / the thickness of the sheet in inches,
and p the pressure per square inch on the supported
surface.
Substituting present values in this formula, and assum-
ing a maximum working stress in the sheet of 7500 lb.
per square inch, the statement becomes:
o= V
9x7500x.3125x.3125
2x125 5 3>
MISCELLANEOUS APPLICATIONS 99
or 5 1/8 in., which is a little less than the pitch deter-
mined by the strength of the stay bolt. The first lay-
out should be modified, making 19 vertical rows of stays
with four stays in each row, or the pitch would be about
5X4.8 in.
In relation to pitching the rivets in the vertical seam,
it is found in practice that the lapping together of the
plate at this point stiffens the sheet so that strength of
the joint is not a very important factor in the case,
and this joint is generally designed to be least affected
by the heat, i.e., single riveted, and of the best propor-
tions to insure tightness, which w T ould be a pitch of about
2 1/4 in. if 5/16-in. plate and n/16-in. rivets be used.
Relating to Safety-valve Rules
Concerning the apparent discrepancy between the
different rules relating to safety-valve calculations,
particularly that in Reed's Engineer's Handbook (an
English publication), which is used by candidates for
British Board of Trade certificates of competency, and
those used by a prominent educational institution the
reader w T ill discover, after carefully reading the following
statements, that both rules referred to will give the same
results in any given problem when intelligently used.
In other words, both rules are correct, although there
is a possibility of misunderstanding them and so obtain-
ing incorrect answers.
If the safety-valve problem could be handled without
taking into consideration the effective weight of the lever,
then it is not probable that there would ever be any
difficulty in candidates having any misunderstanding
8
IOO ARITHMETIC OF THE STEAM BOILER
of the matter. But the effective weight of the lever
must be taken into the calculation if accuracy is desired,
and of course in such a matter accuracy should be
desired. Consider what the effective weight of a safety-
valve lever is and what relation it bears to the calculation.
In the first place there is a difference between the
effective weight of a lever and the effective moment of a
lever. Either may be used in the safety-valve problem,
but it must be clearly understood how and where each
is to be used. Probably the whole difficulty that candi-
, L =20
10 "
F = 2'
H
..4" Fva|
Center of Gravity of Lever
f W?
Dia. 4 1/ V \]P= 1256.64 lb.
Fig. 19. — Diagram for safety-valve calculations.
dates have with the safety-valve problem lies just in
this point. The effective weight of a lever is found by
multiplying its weight in pounds by the distance its center
of gravity is from that point called the fulcrum, and by
dividing the product by the distance that the center
of the valve upon which the lever acts is from the ful-
crum. The effective moment of a lever is found by multi-
plying its weight in pounds by the distance its center of
gravity is from the fulcrum. The distances referred to
are measured in inches.
Referring to the following example: Required, the
weight to be placed at the end of the safety-valve lever
shown in the diagram with the given data.
MISCELLANEOUS APPLICATIONS ioi
Diameter of valve, 4 in.
Area of valve, 12.5664 sq. in.
Steam pressure per square inch, 100 lb.
Weight of lever, 20 lb.
Weight of valve and stem, 10 lb.
Total upward pressure on the valve is 1256.64 lb.
The letter c represents the sum of the effective moments
of the valve and stem, and also the lever. The effective
moment of the lever is 10X 20= 200; the effective moment
of the valve is 2X10=20; and the sum is 200+20 = 220.
xt FP-c . (2X1256.64)^-220
Now, — y — = weight, or -=114.66410.,
the weight to place at the end of the lever. (This
method employs effective moments of valve and lever.)
Now try the same example by using the other rule, which
employs the effective weight of the lever, and the actual
weight of the valve. The rule is stated thus: "To
find the weight which must act on a lever at a given dis-
tance from the fulcrum so that the valve is about to blow
off at a given pressure, subtract the downward force due
to the weight of the valve, stem and lever from the prod-
uct of the area and the steam pressure. Multiply the
remainder by the distance from the fulcrum to the
center line of the valve, and divide this product by the
distance from the fulcrum at which the weight is to act."
Expressed in a formula it appears like this:
w J Ap -^ D
W L
W = the required weight in pounds.
A = the area of the valve in square inches.
102 ARITHMETIC OF THE STEAM BOILER
P = the pressure per square inch.
D = distance in inches from the fulcrum to the
center line of valve.
L = distance from fulcrum at which the weight
is to be placed.
w = the weight of the valve and stem in pounds
plus the effective weight of the lever.
Weight of valve and stem = 10 lb.
Effective weight of lever = — = ioo lb.
° 2
Total downward force due to both valve and lever =
no lb.; that is, w in the formula = no lb.
( 1 2 . 5 664 X i oo — 1 1 o) X 2
— = 114.6^4 lb.,
20 J
the same answer as obtained by the first method
employed.
The effective weight of the lever will be = 100
lb.; that is, this lever, because of its weight and its point
of contact with the valve, is equivalent to 100 lb. being
placed directly on top of the valve.
The effective moment of the same lever will be: 20 X
10= 200.
Now, either the effective weight of the lever or the
effective moment may be used in the calculation as has
been shown, but each has its own place in the operation,
as will be seen a little further on.
The equality of moments in the safety-valve problem
are stated like this:
WXD+W'XD'+wXd=pXAXd
MISCELLANEOUS APPLICATIONS 103
In which
W = weight of the weight in pounds.
D = distance in inches that weight is placed from
fulcrum.
W' = weight of the lever in pounds.
D f = distance of center of gravity of lever from
fulcrum.
w = weight of the valve and stem in pounds.
d = distance center of valve is from the fulcrum.
p = pressure in pounds per square inch.
A = area of valve in square inches.
By applying the formula to the problem it can be seen
that the answer obtained is correct. Showing the let-
ters, and substituting the numerals of the example,
the statement becomes:
WXD+W'XD'+wXd=pXAXd,
114.664X20+20X10+10X2 = 100 X 12.5664X2,
229.328+200+20 = 2513. 28,
2513.28 = 2513.28.
Any safety-valve rule or formula that can successfully
stand the above test is correct and may be safely
employed.
In order to arrive at the point aimed at, and also to
make it as clear as possible, it may be said that from the
foregoing formula of moments a series of formulas can be
derived that will handle the safety-valve problem in all
its phases. Here are the formulas:
(1) FX S+ L = W.
(2) FX S + W=L.
104 ARITHMETIC OF THE STEAM BOILER
(3) LXW+ S=F.
(4) LXW+ F = S.
Where F = the force acting upward against the valve;
this equals the pressure per square inch times
the area of the valve; from the product must
be subtracted the weight of the valve and
stem.
5 = the distance from the fulcrum to the center
of the valve.
L = distance from the fulcrum to point where the
w T eight is hung. (This may or may not be at
the extreme end of lever.)
W =the weight in pounds of the weight to be hung
on lever.
In cases (1) and (2) the effective weight of the lever
must be subtracted just before multiplication occurs. If
effective moment of lever is to be employed, then the
effective moment is to be subtracted just before division
occurs.
In cases (3) and (4) the effective weight of lever is to be
added just before multiplication occurs.
If the effective moment of lever is to be employed, then
it will be added just before division occurs in the formula.
It is very important that a distinction be made between
the two phases of the four cases, and then no trouble will
be experienced.
Bearing in mind what has just been stated and being
ready to refer to it again, now try the problem by using
Reed's rule, which is stated thus:
(1) Find the area of the valve and multiply it by the
pressure per square inch.
MISCELLANEOUS APPLICATIONS 105
(2) From the product take the weight of the valve
(which of course includes the stem).
(3) Multiply the remainder by the distance from the
fulcrum to the valve, then subtract the moment of the
lever, and divide by the distance from the fulcrum to the
weight.
In Reed's Engineers' Handbook the effective moment
of the lever is defined the same as that which appears
in the earlier part of this subject.
Using the figures in the example and following Reed's
rule, 12.5664 sq. in. (area of valve) X 100 (pounds pressure
per square inch) = 1256.64 lb. (total upward pressure)
— 10 lb. (weight of valve and stem) = 1246.64 lb.X2
(inches distance from fulcrum to valve) = 2493.28 lb.
— 200 (moment of lever) = 2293.28 lb.-^2o (inches,
distance fulcrum to weight) = 114.664 lb. weight. Ans.
Thus it is seen that exactly the same answer is obtained
as found before. Reed's rule is therefore correct if
intelligently used.
Now, suppose that it is desired to use Reed's rule, but
instead of having the effective moment of lever given, the
effective weight is given instead. The effective weight
of the same lever would be (as before explained)
= 100 lb. The operation will be like this: 12.5664 sq.
in. (area of valve) X 100 (pounds pressure per square
inch) = 1256.64 lb. (total upward pressure against valve)
— 10 lb. (weight of valve and stem) = 1246.64 lb.
(balance) — 100 lb. (effective weight of lever) = 1146.64
lb. X2 (inches, distance, fulcrum to valve) = 2 293. 28 4- 20
(inches, distance, fulcrum to weight) = 114.664 lb. weight
106 ARITHMETIC OF THE STEAM BOILER
answer, giving exactly the same answer as in the other
two cases. Again is attention directed to the difference
between the tw r o latter cases, and particularly the place
in each w T here the subtraction of the lever factor takes
place.
Suppose the effective weight of the lever were subtracted
after multiplication had occurred instead of before, then
the following statement would occur: 12.5664X100 =
1256.64—10=1 246.64 X 2 = 2493 .28 — 100 (effective weight
of lever) = 2393.28-^20=119.664 lb. w r eight.
119.664—114.664 = 5 lb. difference, too much weight.
In other words, by so doing the valve would be over-
weighted 5 lb. Of course, the percentage of difference is
small, but it is further aggravated by the frictional
resistances, and it is not on the safe side of the calcu-
lation.
Suppose on the other hand the effective moment of the
lever were subtracted before multiplication occurs; the
statement would be: 12.5664X100=1256.64—10 =
1246.64—200 (effective moment of lever) = 1046.64-^20
= 104.664 lb. weight.
114.664—104.664=10 lb. difference, too small.
Let the interested reader study this matter out for
himself and try the arithmetical operations in the
different cases. In this way he cannot but come to a
correct and complete understanding of all that is em-
braced in the problem, as far as an operating engineer is
concerned, when standing an examination for a certificate.
Any rule relating to the safety valve, which a candidate
is not sure of should be tested by the application of the
equality of moments formula before referred to.
MISCELLANEOUS APPLICATIONS 107
Roper's Safety-valve Rules
Examiners of engineers in the United States Steam-
boat Inspection Service sometimes prefer to have candi-
dates for American marine engineers' license use what
are known as Roper's Rules for safety-valve problems.
Therefore it is well to have a knowledge of these rules
in case they are required.
In the following formulas let
A = Area of valve in square inches, or diameter 2
X.7854.
D = distance from center of valve to the fulcrum,
in inches.
L = distance of the weight from the fulcrum, in
inches.
P = steam pressure in pounds per square inch.
W = weight of the ball in pounds to hang on the
lever.
V = w r eight of the valve and stem in pounds.
w = w eight of the lever in pounds.
/ = distance of the fulcrum from the center of
gravity of the lever, in inches.
n (WXL) + (wXl) + (VXD)
AXD
(1)
TT7 AXPXD-{wXl+VXD) , x
W = — £ — (2)
■ AXPXD-(wXl+VXD) . ,
L= — w~ (3)
The following examples illustrate the application of
the formulas given.
108 ARITHMETIC OF THE STEAM BOILER
Example i. — At what pressure will a safety valve blow off, hav-
ing a diameter of 4 in., weight of valve and stem 12 lb., weight of
lever 22 lb., weight of ball 125 lb.; the overall length of lever is 46
in., and it is straight and parallel; the weight is hung at 42 in. from
the fulcrum, and the distance from center of valve to fulcrum is
4 in.
Using formula (1) the statement becomes:
P =
(125X42)+ ( 22 X^) +(12X4)
4 2 X. 7854X4
= 115.466+ lb. per square inch, answer.
Example 2. — Using the foregoing example (1), it is required to
find the necessary weight to hang on the lever.
Using formula (2) the statement becomes:
W
4 2 X. 7854X115. 466X4- (^2Xy + I2 X4)
42
= 125 lb., answer.
Example 3. — Using the same example (1), it is required to find
at what distance the weight is to be hung.
Using formula (3) the statement becomes:
12. 5664X115. 466X4- ( 22X^+12X4)
125
=42 in., answer.
Safety-valve Capacity
Extract from paper read before the A. S. M. E.,
February 23, 1909, by Philip G. Darling, Mechanical
Engineer.
MISCELLANEOUS APPLICATIONS 109
CAPACITY FORMULA FOR 45° SEATS
1. E=iosXlXPXD
2 . D = .oo 95 j£p
Modified Forms for Special Applications
For Locomotives
H
For Cylindrical Multitubular, Vertical and Water Tube
Stationary Boilers
For Water Tube Marine and Scotch Marine Boilers
5. D = .o 9 5j^p
E = pounds of steam relieved per hour.
/ = vertical lift of valve in inches.
P = steam pressure (absolute) pounds per square
inch.
D = nominal diameter of valve (inlet) in inches.
H = total boiler heating surface in square feet.
For flat-seated valves the constants in these formulae
are as follows: i — 149.; 2 — .0067; 3 — .052; 4 — .065;
5— .090.
no ARITHMETIC OF THE STEAM BOILER
UNITED STATES STEAMBOAT INSPECTION SERVICE SAFETY-
VALVE RULE
The rule is that the areas shall be found by the formula:
WG
A = .2o 74 x 1 r'
in which .4 = the area of the safety valve in square
inches.
W= pounds of water evaporated per square
foot of grate surface per hour.
P = the absolute pressure per square inch.
G = grate area in square feet.
In the case of spring-loaded valves, the effective area
must be equal to that derived from the formula, and a
lever must be provided which will raise the valve one-
eighth its diameter from its seat. All seats to have an
angle of 45 to the axis of the valves.
Derivation of the United States Board of Super-
vising Inspectors' Rule for Areas of
Safety Valves
Napier's rule for flow of steam through orifices:
_. . . . Absolute pressure X area
Flow in pounds per second =
(This corroborated by Peabody's experiments.)
P = absolute pressure = gage pressure+ 1 5 .
W = pounds discharged per hour.
A =area of valve opening or orifice.
MISCELLANEOUS APPLICATIONS in
Hence
_ PXA * e , c 'T 36oX^XP
W= X6oX6o= — —
70 7
For safety-valve practice, cut this amount down 25
per cent., leaving 75 per cent.
Thus
Restrict the lift of valve to 1/32 of its diameter =
d_
then 32
d ~Xd 2
A — X^Xd = lift X circumference = —
3 2 3 2
Substituting this value for A = area of orifices
270 „ 7ld 2
w=^~xpx —
7 32
In a valve of diameter d the area =
Tld 2
=a
4
To get W in terms of area of valve, substitute for d 2 its
value in terms of a,
W =- 7 -X P X— X 4 - =4-821 XPa
7 32 t
In safety-valve practice this will represent the pounds
112 ARITHMETIC OF THE STEAM BOILER
of steam that must escape per hour, which must be equal
to the pounds of water that the boiler can evaporate per
hour.
To reduce this to a working basis, consider these quanti-
ties per square foot of grate surface per hour.
W — pounds of water evaporated per square foot
of grate surface per hour.
P = absolute pressure per square inch.
A =area of safety valve per square foot of grate
surface.
Hence
W
TF=4.82iXPX^, anda = .2074X -p
From which a table of areas required per square foot of
grate surface may be found by assuming the different
values of W and P.
Finding the Center of Gravity of Tapered Safety-
valve Levers
In questions relating to the lever safety valve it is
necessary to know how to find the center of gravity of
the lever, in order to calculate the effective weight of the
lever; how may the center of gravity be found w T hen the
lever is tapered, and of uniform thickness throughout its
entire length?
Answer. — There are three ways that the center of
gravity of tapered safety-valve levers is found: By
taking the lever off and actually balancing it on a knife
edge; diagrammatically, and mathematically. The last
two methods only require an explanation.
MISCELLANEOUS APPLICATIONS
113
First consider the diagrammatic method of finding the
center of gravity. To be brief and to the point, assume
a lever 20 in. long, 2 in. wide at one end, and 1 in. wide at
the other end and of uniform thickness throughout.
Do not consider the projection which is usually at the
small end of the lever for preventing the weight slipping
off, nor the holes at the other end through which the
lever is attached to the fulcrum, and by which the pin is
attached which bears upon the valve. These would
Fig. 20. — Diagram for finding center of gravity of tapered lever.
make but little difference in the result obtained, and
would tend to complicate the calculations.
An inspection and study of figure 20 should make clear
how the center of gravity may be determined. A scale
of quarter size may be chosen as a matter of convenience
only. Any other scale w T ill do. The center line AB is
' drawn first. The length of A B is 5 in., or one-quarter as
long as the lever assumed. At A and B respectively
draw the lines CD and EF 1/2 in. and 1/4 in., represent-
ing the actual 2-in. and i-in. dimensions of the lever.
Next draw the diagonal line DE, and locate the point 1
exactly midway between points D and E. Join the
114 ARITHMETIC OF THE STEAM BOILER
points C and i by the line Ci. From the point i locate
the point 2 on the line Ci at a distance equal to one-
third of its length. Now, from the point F draw the
line Fi, and from the point 1 locate the point 3 on the
line Fi, one-third of its length from the point 1. Join
points 2 and 3; where the line 2-3 intersects the center
line AB will be the point 4 at w T hich the center of gravity
is. By carefully and accurately drawing the figure the
measurements show that the center of gravity is about
1 1. 1 in. from the small end of the lever, and is about
8.9 in. from the large end of the lever.
To find the center of gravity of a tapered lever mathe-
matically, the following formula may be used:
2A+B T . ,. p 2A+B
X== 3 A+$B X ; ° r m m? x= siA + B) X
Where x = distance center of gravity is from the small
end of lever.
A= the width in inches of lever at the fulcrum
end.
B = width in inches of lever at the weight end.
L = entire length of lever in inches.
Applying this formula to our example we have
2X2 + 1 5 100
-7 : rX20=~X20= "=11.11 111.,
3(2 + 1) 9 9
distance center of gravity from small end of lever.
Thus it will be seen that the two methods which have
just been explained verify each other, and no doubt
the first method (that of balancing the lever on a knife
edge) referred to would verify the others, for, as before
MISCELLANEOUS APPLICATIONS 115
stated, the difference due to the projection at one end
and the holes at the other end would hardly be noticeable.
At an examination where time is limited, it would be
better to use the formula in such questions. It does not
require much study and only a little practice to become
familiar with it. Furthermore, it will be found easy to
remember after it has been used a few times. Marine
engineers cannot afford to ignore this subject.
Chimneys
The " proportions of chimneys" vary very much
according to the requirements. Every chimney should
be large enough in cross-section to carry off the gases and
high enough to produce sufficient draft to cause a rapid
combustion. The object of a chimney being to carry
off the waste gases, it naturally determines the amount
of fuel that can be burnt per hour, and it is advisable to
have always a good draft, as it can always be regulated
by a damper.
Draft pressure is caused by the difference in weight
between a column of hot gases in the chimney and a
column of air of equal height and area outside the
chimney.
Formula for finding the force of draft in inches of
water of any given chimney:
7-64 7-95^
^tj(7M 7-95\
Where F = force of draft in inches of water.
H = height of chimney in feet.
9
Il6 ARITHMETIC OF THE STEAM BOILER
T\ = absolute temperature of chimney gases
(H-460).
T 2 = absolute temperature of the external air
(/1+460).
t = temperature of chimney gases.
ti = temperature of external air.
Formula for finding the height of a chimney in feet for
a given force of draft:
F
11 =
/7.6 4 _7. 9 5\
\T 2 Tj
To find the maximum force of draft for any given
chimney, the external air being 6o° F., and the heated
column being 6oo° F., multiply the height above the
grate in feet by .0073, and the product is the force of
draft expressed in inches of water.
1. The draught power of the chimney varies as the
square root of the height.
2. The retarding of the ascending gases by friction
may be considered as equivalent to a diminution of the
area of the chimney, or to a lining of the chimney by a
layer of gas which has no velocity. The thickness of
this lining is assumed to be 2 in. for all chimneys, or
the diminution of area equal to the perimeter X 2 in.
(neglecting the overlapping of the corners of the lining).
Let D = diameter in feet, A = area, and E = effective area
in square feet.
2 8D 2 —
For square chimneys, E = D — =A — \/ A
For round chimneys, £=- yD 2 — — j =A — .591 \A4
MISCELLANEOUS APPLICATIONS 117
For simplifying calculations, the coefficient of \/ A
may be taken as .6 for both square and round chimneys,
and the formula becomes
3. The power varies directly as this effective area E.
4. A chimney should be proportioned so as to be
capable of giving sufficient draught to cause the boiler
to develop much more than its rated power, in case of
emergencies, or to cause the combustion of 5 lb. of fuel
per rated horse-power of boiler per hour.
5. The power of the chimney varying directly as the
effective area E, and as the square root of the height
H, the formula for horse-power of boiler for a given size
of chimney will take the form horse-power = CE\/e[, in
which C is a constant, the average value of which, ob-
tained by plotting the results obtained from numerous
examples in practice, the author finds to be 3.33.
The formula for horse-power then is
horse-power = 3. 3sE\/jj y or horse-power = 3.33 (A —
aVaWh
If the horse-power of boiler is given, to find the size
of chimney, the height being assumed,
.3 H.P. ,-
For round chimneys, diameter of chimney = diam. of
£+4 in.
For square chimneys, side of chimney = \/£+4 in.
Ii8 ARITHMETIC OF THE STEAM BOILER
If effective area E is taken in square feet, the diameter
in inches is d= i^.S4\/E+4. in., and the side of a square
chimney in inches is s= i2\/E+4 in.
If horse-power is given and area assumed the height
In proportioning chimneys the height is generally
first assumed, with due consideration to the heights of
surrounding buildings or hills near to the proposed
chimney, the length of horizontal flues, the character
of coal to be used, etc., and then the diameter required
for the assumed height and horse-power is calculated by
the formula or taken from the table.
Size of Boiler Feed Pipe
What size feed pipe should be installed to supply
three ioo-H.P. boilers?
At ordinary commercial rating the water required per
horse-power-hour, is taken as 30 lb. Each boiler there-
fore will need 3000 lb. of water per hour or 50 lb. per
minute, which at 62.5 lb. per cubic foot would call for
.8 of a cubic foot each minute.
To provide for emergencies, twice the actual quantity
of water required should be figured. The velocity of
flow is usually taken as 100 ft. per minute. On this
basis, the quantity to be taken care of will be 2X.8 =
1.6 cu. ft. per minute, which at 100 ft. per minute velocity
would require an area of pipe of -=.016 sq. ft., or
100
2.3 sq. in., found like this:
.016X144=2.304 sq. in. and A / '^ I *7 m - diam.
\-7 8 54
MISCELLANEOUS APPLICATIONS 1 19
The nearest commercial size of pipe is 1 1/2 in. diameter,
which is the size to connect to each of the boilers.
For the main pipe, to supply all three boilers, a pipe
of equivalent area to the three boiler pipes would be 6.9
square inches, and this would mean a 3-in. diameter
pipe is required. Should not all three boilers require
to be fed at the rate before referred to at any one time,
then a 2 1/2-in. main feed pipe will be ample.
Effect of Stiffness of Head on Braces
Away back in 1876, Samuel Nichols, a practical boiler
maker in charge of a large English works, wrote a book
for boiler makers. In it he recounts some experiments
made under the supervision of Robert Nelson, author
of " A Treatise on Steam Boilers/' upon boilers built by
himself for the purpose, with regard to unstayed heads in
cylindrical shells. A boiler 30 in. in diameter with flat
heads 3/8 in. thick, of plate having a "tenacity" of
21.2 tons per square inch with an elongation of 7.9 per
cent., flanged on a radius of 1 in. inside the plate, was
subjected to hydrostatic pressure. At 10 lb. the head
had bulged 1/16, at 120 lb. 11/16 and at 150 lb. 13/16 in.
Permanent set occurred somewhere between 50 and 65
lb. at which latter figure the deflection was 3/8 in.
Rupture took place at about 300 lb.
From the results of his tests, Mr. Nichols deduced a
formula which is printed in his book for determining
the bursting pressure of cylinders with unstayed heads,
although he strongly disclaimed any advocacy of ex-
posing an unstayed surface to high pressures. "On the
120 ARITHMETIC OF THE STEAM BOILER
contrary," he writes of himself, "he is more convinced,
now that he has witnessed these experiments, that a fiat
unstayed surface is very weak indeed, and that they still
require a larger amount of care and judgment on the
part of boiler engineers than any portion of the boiler/'
Evidently under the erroneous notion that if the head
can take care of the amount of pressure indicated by this
formula on its own account it will need correspondingly
less bracing, the Board of Boiler Rules of Ohio, after
instructing inspectors to determine the working pressure
of boilers with respect to the bracing in the usual way
but with an allowable stress of 8000 lb. per square inch
irrespective of size, tells them, "To the above pressure
may be added the Nichols formula w T ith a factor of
safety of not less than 8."
If this means, and it can seem to mean nothing else,
that to the pressure which can safely be taken care of by
the bracing may be added the pressure which, by the
Nichols formula, would be allowed upon the unbraced
head, it is wrong.
Mr. Nichols shows that the head may be bulged
considerably without straining the sheet beyond the
elastic limit. But a brace is supposed to be tight before
the head commences to bulge. Just as soon as the head
starts to move it commences to stretch the brace.
Under the allowable stress of 8000 lb. per square inch of
section a brace wall extend only about 1/3750 of its length,
or a 6-ft. brace would extend less than .02 in. The pres-
sure which would produce the movement in the unbraced
head is inconsiderable — a pressure of 10 lb. produced a
movement of three times as much — and yet this is
MISCELLANEOUS APPLICATIONS 121
all the help that the stiffness of the head would be to the
brace.
As an example, assume a 72-in. boiler, height of seg-
ment to be braced 24 in., area of segment to be braced
814 sq. in., pressure 100 lb., thickness of head 1/2 in.,
tensile strength 60,000 lb.
The Nichols formula with a factor of safety of 8
would allow
tXT Xio 0.5X6000 0X10
^4X8 "814X8 -46 1b.
If this may be added to the pressure which the bracing
is capable of carrying, it would be necessary to brace
against only 100 — 46 = 54 lb. per square inch, which at
8000 lb. per square inch w^ould require seven i-in. braces.
Practice calls for at least twice as many.
Bracing Flat Surfaces in Steam Boilers
There is considerable variation as to the load allowed
per square inch of net section on diagonal braces, rod
braces and stay bolts by the authorities who have
laid down rules on this subject. The United States
Government rules allow 6oco lb. on welded iron stays
below 1 1/4 in., 7500 on 1 1/4 in. and above, and from
7000 to 9000 lb. on weldless steel stays. Chicago has a
flat scale of 6000 lb. on all stays or braces, Philadelphia
has a limit of 7500 lb.; the Massachusetts rules allow
from 6500 to 9000 lb. per square inch net section, varying
as the braces are w r elded or weldless and with the size,
the latter for the reason that with a given waste of
122 ARITHMETIC OF THE STEAM BOILER
material the percentage of reduction is greater with the
smaller sizes.
The above applies to flat surfaces and refers to flat
heads, such as dome heads, segments of heads, etc.
The United States Government has a rule to find the
pressure on flat heads not exceeding 20 in. in diameter
as follows:
CXT 2
A
where P = pressure.
C=ii2 (7/16 or under) and 120 over 7/16.
A = 1/2 the area.
T = thickness in sixteenths.
With a 3/4-in. head 20 in. in diameter, no lb. would
be allowed by these rules.
A short time ago the Board of Boiler Rules for the
State of Ohio issued instructions to the inspectors holding
certificates of competency that the following formula
could be used in flat surface of heads:
rxr.s.xio
ax's
w T here T = thickness.
T.S. = tensile strength.
A = area.
In addition, a limitation allowance of 8000 lb. per
square inch irrespective of size of brace is granted.
This applies to boilers now in use, but not to boilers
to be installed after July 1, 19 12. This ruling is far
more liberal than any other authority has heretofore
MISCELLANEOUS APPLICATIONS 123
allowed as a comparison will show. Assume a 72-in.
boiler, height of segment 24 in., thickness of head 1/2
in., tensile strength 60,000 lb. The total area of the
segment = 1 186.4 S Q- in., while the area requiring bracing
= 814 sq. in. Hence
.5X60000X10 r „
814x8 " =461b -
allowed without braces.
Let the pressure required be equal to 100, then 100 —
46 = 54 lb. to be braced, and 54X814 = 43,956 lb.
Assuming the proposed brace to be of .79 in. area,
then .79X8000 = 6320 per brace, and 43,956^6320 = 7
braces of practically i-in. diameter.
It may be said that flat surfaces subjected to internal
pressure will spring and proportionally to the unsup-
ported area. Samuel Nichols, in his tests of circular
flat heads, showed the springing began with very low
pressures, even at 20 lb. on 28-in. heads and increased as
the pressure was raised. Applying this fact then to the
Ohio ruling, it seems the head would so spring that at
100 lb. pressure the total load on the braces would be
.79X7 = 5.53 into the toal load, 81,400 lb., or 14,718
per square inch net section instead of 8000 lb. Applying
this ruling to a flat dome head 36 in. diameter, 1/2 in.
thick, 60,000 lb. tensile strength, area to be braced
707 in., gives 53 lb. without bracing. The results of
allowing a flat head unbraced 1 o spring and return times
without number would be final failure due to such action.
Reverting to the segment as in a horizontal tubular
boiler, it may be said other authorities have been careful
124 ARITHMETIC OF THE STEAM BOILER
to avoid allowing excessive stresses on the chord of the
segment which is supported by the tubes inasmuch as
the latter are not a constant in strength as is the flange
of the head in the arc of the segment, and this view has
been approved by most students as the tubes are subjected
to more or less rapid wear and reduction in thickness.
Further, such calculations apply to boilers now in use
irrespective of age. Indeed, Ohio has no limitation as
to age as respects pressure to be determined with a factor
of safety of 4 together with this exceedingly liberal allow-
ance on braces. Comparing this with the Massachusetts,
Chicago, Philadelphia and Detroit rules, w T hat results may
be expected?
Three Boiler Questions
In an examination, three out of live engineers failed
to answer the following questions, which are given for the
benefit of those w T ho may be called upon to make similar
calculations.
1. A horizontal tubular boiler is 72 in. in diameter
and 18 ft. long; thickness of plate .437 in.; efficiency of
longitudinal joints 77 per cent., and steam pressure no
lb. What should be the tensile strength of the plate,
allowing a factor of safety of 5 ?
2. If the tensile strength had been 56,000 lb. and
the efficiency of the joint 70 per cent., what thickness of
plate should be employed?
3. If this boiler had been intended for 125 lb. of
steam, what would the efficiency of the joint have been,
using the data in the first question (except the pressure
and efficiency) ?
MISCELLANEOUS APPLICATIONS 125
The required tensile strength of plate is found by the
rule
pressure X diam. X factor of safety
efficiency of joint X thickness of plate X 2
Substituting the figures given, instead of the words in
the rule, we have
Tensile strength = 1 -°~^ 2 —^ = 58,7 7 5 lb.
.77X437X2 ° ,//0
The rule for thickness of plate is
pressure X diam. X factor of safety
tensile strength X efficiency of joint X 2
Again substituting the figures we have
rrw 1 110X72X5
inickness =—- — — — •= . c m.
56000 X. 70X2 J
Efficiency of joint is given by the rule
. _ pressure X diam. X factor of safety
^ tensile strength X thickness X 2
This figured out gives
12^ X 72 X ^
Efficiency = —z — — — = .87^ or 87. < per cent.
58775X.437X2 /0 ' 0F
To Find Pitch of Rivets
How can the pitch of the rivets be determined for a
double-riveted butt and double-strap joint which is to
have 7/8-in. rivets and a strength of plate between the
rivet holes on the outer row which will be 82 per cent,
of the strength of the solid plate?
Let P = pitch.
t = thickness of plate.
TS = tensile strength of plate.
d = diameter of rivets.
126 ARITHMETIC OF THE STEAM BOILER
Then
PXtXTS = strength of solid plate
and
(P — d) XtXTS = strength of plate between the rivet
holes on the outer row. The conditions require
(P-d)tXTS _
PXtXTS W
By canceling out t and TS from numerator and de-
nominator of the first member of (i) we obtain,
P-d
P
and as ^ = .875, the equation becomes
^-•875
P
from which it is found that
= .82 (2)
ecomes
=•82 ( 3 )
P=--^= 4 .86+ (4)
or practically 4 7/8 in.
Collapsing Pressure of Lap-welded Bessemer
Steel Tubes of from 3 to 10 In. Diameter, and
of Different Wall Thicknesses
Formulas:
and
t
P--
= 86,670-7-
/
-1386
/
p\
p=
- IOOO ( I — y
J 1 " 1
6oo j2 )
p=
= 50,210,000
(t)»
MISCELLANEOUS APPLICATIONS 127
where p = collapsing pressure in pounds per square inch.
d = outside diameter of tube in inches.
/ = wall thickness in inch measure.
The first formula is applicable to cases where -j is
greater than .023 and the others to the case of thin-walled
tubes where the quotient is less than that value.
Safe Working Pressure Calculations as Applied
to the Shell of Climax, Hazelton and Porcupine
Types of Steam Boilers
Example. — Shell plate 5/8 in. thick, diameter 30 in., tensile
strength 60,000 lb. per square inch, tubes 4 in. diameter. See Figs.
1 and 2 of this section.
Consider the ring of the shell 3 27/32 in. wide (Fig. 2)
included between any two transverse rows of holes.
For each pound per square inch of pressure, any longi-
tudinal section of this ring will be subjected to a stress of
= 57.65625 lb. The net section of the ring
on the axis ab of a longitudinal row of holes is 5/8 X
(3 27/32 — 2) = i. 15+sq. in. The unit stress on this
section for a pressure of 1 lb. per square inch is, therefore,
57.656-^1.15 = 50 lb. per square inch, nearly. The
section on the line ac through two adjacent holes in a
diagonal row is subjected to the stress of 57.656 lb.,
which acts in the direction of the line ef, perpendicular to
ab. This stress may be resolved into two components, one
of which, eg, acts perpendicular to ac and tends to pull the
plate apart through that section, while the other, eh, acts
along the line ac and tends to shear the plate through the
128
ARITHMETIC OF THE STEAM BOILER
same section. Of these two components, eg is equal to
the stress on a longitudinal section of the ring multiplied
by the cosine of the angle bac, and eh is equal to the stress
'on the longitudinal section multiplied by the sine of the
angle bac. The sine of bac is 3.5-^5 • 2 = .673, and
the cosine 3 27/32-7-5.2 =
• 739? which correspond to an
angle of 47 40', nearly. The
tensile stress on the section
ac resulting from the stress
of 57.656 lb. acting perpen-
dicular to the section ab is,
therefore, 57.656 X .739 =
Dia. of Shell 30"
Thickness of Plate %
Fig. i. — Calculations relating Fig. 2. — Calculations relating to
to porcupine type of boilers. porcupine type of boilers.
42.61 lb., nearly, and the shearing stress is 57. 656X^73 =
38.8 lb. In addition to the stresses due to the force that
tends to break the ring through a longitudinal section,
the section on ac is subjected to a stress from the action
of the force that tends to rupture the shell along a trans-
verse section. For a pressure of 1 lb. per square inch,
MISCELLANEOUS APPLICATIONS 129
this force is equal to the area of the head multiplied by 1 ;
that is, to 3o 2 X-7854X 1 = 706.86 lb. The number of
sections among which this force is divided is equal to the
circumference of a 30-in. circle divided by 3.5; that is, to
^ — — =27. The force on each section is, there-
3-5
fore, 706.86-^27 = 26.18 lb. This force acts on the
section ac in the direction of the line el, parallel to the
line ab. It may be resolved into two components, one,
em, perpendicular to ac, which is equal to 26.18 multi-
plied by the sine of the angle bac; the other, en, in the
direction of ac, equal to 26.18 multiplied by the cosine of
bac. Of these two components, the first acts in the same
direction as the component eg; its value, 26.i8X.673 =
17.62 lb., nearly, is, therefore, to be added to the value
represented by eg, thus giving us a total tensile stress in
the section ac of 42.61 + 17.62 = 60.23 lb. The compo-
nent en, whose value is 26. 18X. 739 = 19. 347 lb., acts in
the opposite direction to eh; they therefore partly neu-
tralize each other, and the resulting shearing stress,
38.8 — 19.347 = 18.453, is so much less than the tensile
stress of 60.23 lb. that it is evident the section would
fail by tension and not by shear. The area of the net
section of the plate, which resists the tensile stress of
60.23 lb., is 1.2X5/8 = 75 sq. in.; the unit stress in this
section for a pressure of 1 lb. per square inch is, therefore,
60.23-^.75 = 80.3 lb. per square inch. Since the stress on
the section ab was but 50 lb. per square inch, it is evident
that the plate will fail along the section ac. If we
assume the safe working stress of the 60,000-lb. steel
plate to be 10,000 lb. per square inch, the safe working
130
ARITHMETIC OF THE STEAM BOILER
pressure will be 10,000^80.3 = 124.5 lb. per square
inch.
Figuring the Safe Working Pressure op the Shell
of a Locomotive Boiler
How to determine the safe working pressure of the
shell of a locomotive boiler with several courses of vary-
ing diameter, like that shown in Fig. 21.
With the locomotive boiler, like Fig. 21, the safe work-
ing pressure can only be ascertained by considering the
Fig. 21. — Diagram from which to calculate the safe pressure in a
locomotive boiler.
diameters A, B and C. Also, the thickness of the plates
and the efficiencies of the longitudinal seams of the re-
spective courses must be considered.
! The inside diameter A is 72 in. and plate 3/4 in. in
thickness, the inside diameter B 61 in. and plate 9/16
in. in thickness, and the inside diameter C 60 in. and
plate 1/2 in. in thickness. The efficiency of the riveted
joints for the respective courses is A, 82 per cent., course
B 84 per cent., and course C 86 per cent.
It may be asked, what is the difference in efficiency in
MISCELLANEOUS APPLICATIONS 131
the respective courses? This is because the over-all
distances, D, E and F, are such that the same maximum
pitch could not be obtained in the respective courses,
and a change of pitch, large or small, will make a dif-
ference in the efficiency of the net section of plate, maxi-
mum pitch of rivets, which point is made the weaker of
the several parts of a riveted joint.
Assuming the factor of safety to be 5, and the plate
to ha\e a tensile strength of 60,000 lb. then the working
pressure of the boiler, as far as the shell is concerned,
may be determined by the following formula:
Where T — thickness of the plate in inches.
D = diameter of the boiler in inches.
T s = tensile strength of the plate in pounds.
F = factor of safety.
E = efficiency of the longitudinal seam.
P = pressure in pounds per square inch.
TXT S X E_
DXF
The safe working pressure for course A then will be
(2X3/4)X6ooooX.82
— — = 205 lb.
72X5 D
The safe working pressure for course B then will be
(2X9/i6)X6ooooX.8 4
y-^~ = 186 lb.
61X5
The safe working pressure for course C then would be
(2Xi/2)X6ooooX.86
, ., — = 172 lb.
60X5 '
132 ARITHMETIC OF THE STEAM BOILER
The calculations thus show that the course C, the
course with the least diameter and the longitudinal seam
with the greatest efficiency, to be the weaker of the three
courses A, B and C. Therefore, the pressure for the
boiler, considering the shell only, would be 172 lb.
Had the designer in the first instance made the course
C 9/16 plate and the course B 5/8-in. plate, the boiler
shell in question could have been allowed a greater
pressure than 172 lb. To a boiler designer these calcu-
lations would suggest several things. First would be
that if no change was to be made in course C, then the
thickness of course A could be reduced, perhaps, 1/16
in. in thickness. This would make a saving in the cost
of the boiler. Second, if need be the efficiencies of the
longitudinal seams of courses A and B could be less.
This is, of course, on the assumption that course A will
undergo no changes in regard to thickness of plate, di-
ameter, efficiency of longitudinal seam and tensile
strength of plate.
PART III
APPENDIX
EXTRACTS FROM UNITED STATES RULES-
MARINE— AND FROM THE BOARD OF
BOILER RULES STATE OF MASSACHU-
SETTS—TABLES
APPENDIX
EXTRACTS FROM RULES OF THE UNITED
STATES BOARD OF SUPERVISING INSPEC-
TORS, STEAMBOAT INSPECTION SERVICE
United States Rules Pertaining to Riveted Joints
The following formulas, equivalent to those of the
British Board of Trade, are given for the determination
of the pitch, distance between rows of rivets, diagonal
pitch, maximum pitch, and distance from centers of
rivets to edge of lap of single- and double-riveted lap
joints, for both iron and steel boilers.
Let p = greatest pitch of rivets in inches.
n = number of rivets in one pitch.
pd = diagonal pitch in inches.
d = diameter of rivets in inches.
T — thickness of plate in inches.
V = distance between rows of rivets in inches.
E = distance from edge of plate to center of rivet
in inches.
TO DETERMINE THE PITCH
Iron plates and iron rivets:
<PX. 78 54X7* , .
p=~ —jr- - +d
135
136 ARITHMETIC OF THE STEAM BOILER
Example, first, for single-riveted joint: Given, thickness of plate
(r) = i/2 in., diameter of rivet (d) = 7/8 in. In this case n = i.
Required the pitch.
Substituting in formula, and performing operation indicated,
p . + . (7/8) 2 X. 7854X1 , ._
Pitch = j + 7/8 =2.077 in.
Example for double-riveted joint: Given, / = 1/2 in., andd = 13/16
in. In this case, n = 2. Then —
pitch = (MA6)!>C7854>0 ^
1/2
For steel plates and steel rivets:
23Xd 2 X. 7854X7* . ,
P
28XT
Example for single-riveted joint: Given, thickness of plate =
1/2 in., diameter of rivet 15/16. In this case, n — \.
p ., , 2 3 Xd5/i6) 2 X. 7854X1 .
Pitch = 28X1/2 1-15/16 =2.071 in.
Example for double-riveted joint: Given, thickness of plate = 1/2
in., diameter of rivet = 7/8 in. 71 — 2. Then —
p-*i. 2 3X(7/8) 2 X. 7854X2
Pitch = 28Xl/2 "+7/8 =2.85 in.
FOR DISTANCE FROM CENTER OF RIVET TO EDGE OF LAP
E= sXd
2
Example. — Given, diameter of rivet (d) = 7/8 in., required the
distance from center of rivet to edge of plate.
E= — =1.312 in., for single- or double-riveted lap joint.
APPENDIX 137
FOR DISTANCE BETWEEN ROWS OF RIVETS
The distance between lines of centers of rows of rivets
for double, chain-riveted joints (V) should not be less
than twice the diameter of rivet, but it is more desirable
4J+1
that V should not be less than *
2
Example under latter formula: Given, diameter of rivet = 7/8 in.;
then —
( 4 X7/8) + i
V= =2.25 in.
2 °
For ordinary, double, zigzag riveted joints:
v _\/ (iip+4d)(p+4d)
10
Example. — Given, pitch = 2.85 in. and diameter of rivet = 7/8
in.; then — *
T/-V / (nX2.85+4X7/8) (2.85+4X7/8) . .
V— =1.487 m.
DIAGONAL PITCH
For double, zigz&g riveted lap joint. Iron and steel:
. 6p+ 4 d
pd=
10
Example. — Given, pitch = 2.85 in., and d = j/& in.; then —
A (6X2.85)+ (4X7/8)
pd= = 2.00 in.
r 10
1 Extract the square root of the expression above the line only,
then divide by 10.
138 ARITHMETIC OF THE STEAM BOILER
Maximum Pitches for Riveted Lap Joints
For single-riveted lap joints:
Maximum pitch =(1.31 XT) + 1 s/8.J
For double-riveted lap joints:
Maximum pitch = (2.62 XT) + 1 5/8.
Example. — Given, a thickness of plate = 1/2 in., required the maxi-
mum pitch alio: -able.
For single-riveted lap joint:
Maximum pitch = (1.31X1/2)4-1 5/8 = 2.28 in.
For double-riveted lap joint:
Maximum pitch= (2.62X1/2) + ! 5/8 = 2.935 in.
TO DETERMINE THE AREAS OF DIAGONAL STAYS
Multiply the area of a direct stay required to support
the surface by the slant or diagonal length of the stay;
divide this product by the length of a line drawn at right
angles to surface supported to center of palm of diagonal
stay. The quotient will be the required area of the
diagonal stay.
aXL
A =
I
Where A = sectional area of diagonal stay.
fa = sectional area of direct stay.
L — length of diagonal stay.
1 = length of line drawn at right angles to boiler
head or surface supported to center of palm
of diagonal stay.
APPENDIX 139
Given diameter of direct stay = i in., a = .7845, L = 6o
in., 1 = 48 in., substituting and solving,
.7854X60
^4 = q — - = .981 sectional area.
45
Diameter — 1 . 1 1 in. = 1 1/8 in.
The sectional area of gusset stays, when constructed
of triangular right-angled web plates secured to single
or double angle bars along the two sides at right angles,
shall be determined by formula for diagonal stays, and
shall not be less than 10 per cent, greater than would
be necessary for a diagonal bolt stay.
STAYS
The maximum stress in pounds allowable per square
inch of cross-sectional area for stays used in the con-
struction of marine boilers, when same are accurately
fitted and properly secured, shall be ascertained by the
following formula:
p= AXC
a
Where P = working pressure in pounds.
A= least cross-sectional area of stay in inches.
a = area of surface supported by one stay in
inches.
C = a constant.
C = 9ooo for tested steel stays 1 1/4 in. and
upward in diameter, when such stays are not
forged or welded. The ends may be upset to a
140 ARITHMETIC OF THE STEAM BOILER
sufficient diameter to allow for the depth of
the thread, provided it is the least diameter of
the stay. All such stays after being upset
shall be thoroughly annealed.
C = 8ooo for a tested Huston or similar type of
brace, the cross-sectional area of which ex-
ceeds 5 sq. in.
C = 7ooo for such tested braces when the cross-
sectional area is not less than 1.227 and not
more than 5 sq. in., provided such braces
are prepared at one heat from a solid piece
of plate without welds.
C = 75oo for wrought iron through stays 1 1/4
in. diameter and upward. When made of
the best quality of refined iron, they may be
welded.
C = 6ooo for welded crowfoot stays when made
of the best quality of refined wrought iron,
and for all stays not otherwise provided for
when made of the best quality of refined
iron or steel without welds.
Furnaces
The tensile strength of steel used in the construction
of corrugated or ribbed furnaces shall not exceed 67,000
and be not less than 54,000 lb.; and in all other furnaces
the minimum tensile strength shall not be less than
58,000 and the maximum not more than 67,000 lb.
The minimum elongation in 8 in. shall be 20 per cent.
All corrugated furnaces having plain parts at the ends
APPENDIX 141
not exceeding 9 in. in length (except flues especially pro-
vided for) when new, and made to practically true circles,
shall be allowed a steam pressure in accordance with the
following formula:
CXT
P =
D
LEEDS SUSPENSION BULB FURNACE
CXT
P =
D
Where P = pressure in pounds.
T = thickness in inches, not less than 5/16 in.
Z) = mean diameter in inches.
C = a constant, 17,300, determined from an actual
destructive test under the supervision of the
Board, when corrugations are not more than
8 in. from center to center, and not less than
2 1/4 in. deep.
MORISON CORRUGATED TYPE
CXT
p =
D
Where P = pressure in pounds.
T = thickness in inches, not less than 5/16 in.
Z) = mean diameter in inches.
C = 15,600, a constant, determined from an
actual destructive test under the supervision
of the Board of Supervising Inspectors,
142 ARITHMETIC OF THE STEAM BOILER
when corrugations are not more than 8
in. from center to center and the radius
of the outer corrugations is not more
than one-half of the suspension curve.
[In calculating the mean diameter of the Morison
furnace, the least inside diameter plus 2 in. may be taken
as the mean diameter, thus —
Mean diameter = least inside diameter +2 in.]
FOX TYPE
CXT
F =
D
Where P = pressure in pounds.
T = thickness in inches, not less than 5/16 in.
Z) = mean diameter in inches.
C — 14,000, a constant, when corrugations are
not more than 8 in. from center to center
and not less than 1 1/2 in. deep.
PURVES TYPE
CXT
P =
D
Where P = pressure in pounds.
T = thickness in inches, not less than 7/16 in.
Where D== least outside diameter in inches.
C= 14,000, a constant, when rib projections are
not more than 9 in. from center to center
and not less than 1 3/8 in. deep.
APPENDIX 143
BROWN TYPE
p= CXT
D
Where P = pressure in pounds.
T = thickness in inches, not less than 5/16.
D = least outside diameter in inches.
C= 14,000, a constant (ascertained by an
actual destruction test under the supervision
of this Board), when corrugations are not
more than 9 in. from center to center and
not less than 1 5/8 in. deep.
The thickness of corrugated and ribbed furnaces shall
be ascertained by actual measurement. The manufac-
turer shall have said furnace drilled for a 1/4-in. pipe
tap and fitted with a screw plug that can be removed
by the inspector when taking this measurement. For the
Brown and Purves furnaces the holes shall be in the
center of the second flat; for the M orison, Fox, and other
similar types in the center of the top corrugation, at
least as far in as the fourth corrugation from the end of
the furnace.
TYPE HAVING SECTIONS 1 8 IN. LONG
p _CXT
D
Where P = pressure in pounds.
T — thickness in inches, not less than 7/16.
D = mean diameter in inches.
144 ARITHMETIC OF THE STEAM BOILER
C= 10,000, a constant, when corrugated by sec-
tions not more than 18 in. from center to
center and not less than 21/2 in. deep, measur-
ing from the least inside to the greatest
outside diameter of the corrugations, and hav-
ing the ends fitted one into the other and
substantially riveted together, provided that
the plain parts at the ends do not exceed 12
in. in length.
TOPS OF COMBUSTION CHAMBERS AND BACK CONNECTIONS
Formula for girders over back connection and other
flat surfaces:
Working pressure = (w _ F) ^p^Z
Where W = width of combustion box in inches.
P = pitch of supporting bolts in inches.
D = distance between girders from center to cen-
ter in inches.
L = length of girder in feet.
d = depth of girder in inches.
T = thickness of girder in inches .
C = 5So when the girder is fitted with one sup-
porting bolt.
C = 825 when the girder is fitted with two or
three supporting bolts.
C = 9i7 when the girder is fitted with four or five
supporting bolts.
C = 963 when six or seven supporting bolts are
used.
APPENDIX 145
C = ggo when eight or more supporting bolts are
used.
EXAMPLE
Given W = S4 in., -P=7-5 i n -> D=y.7S in., L = 2.927 ft., ^=7.5
in., T=2 in., C = 82$, then, substituting in formula,
w , • 825X7-5X7-5X2 _
\\ orkmg pressure = 7 ;— — 154-3 lb.
F (34-7-5)X7-75X2. 9 27
FLAT SURFACES
The maximum stress allowable on flat plates sup-
ported by stays shall be determined by the following
formula :
All stayed surfaces formed to a curve the radius of
which is over 21 in., excepting surfaces otherwise pro-
vided for, shall be deemed flat surfaces.
W T CXr2
Working pressure = — p 2 —
Where T = thickness of plates in sixteenths of an inch.
P = greatest pitch of stays in inches.
C=ii2 for screw stays with riveted heads,
plates 7/16 in. thick and under.
C=i2o for screw stays with riveted heads,
plates above 7/16 in. thick.
C=i2o for screw stays with nuts, plates 7/16
in. thick and under.
C=i25 for screw stays with nuts, plates above
7/16 in. thick and under 9/16 in.
C=i35 for screw stays with nuts, plates 9/16
. in. thick and above.
146 ARITHMETIC OF THE STEAM BOILER
C=i75 for stays with double nuts having one
nut on the inside and one nut on the out-
side of plate, without washers or doubling
plates.
C= 1 60 for stays fitted with washers or doubling
strips which have a thickness of at least .5
of the thickness of the plate and a diameter
of at least .5 of the greatest pitch of the
stay, riveted to the outside of the plates,
and stays having one nut inside of the plate,
and one nut outside of the washer or
doubling strip. For T take 72 per cent, of
the combined thickness of the plate and
washer or plate and doubling strip.
C = 200 for stays fitted with doubling plates
which have a thickness equal to at least
.5 of the thickness of the plate reenforced,
and covering the full area braced (up to the
curvature of the flange if any) riveted to
either the inside or outside of the plate,
and stays having one nut outside and one
inside of the plates. Washers or doubling
plates to be substantially riveted. For T
take 72 per cent, of the combined thickness
of the two plates.
C=2oo for stays with plates stiffened with tees
or angle bars having a thickness of at
least two-thirds the thickness of plate and
depth of webs at least one-fourth of the
greatest pitch of the stays, and substan-
tially riveted on the inside of the plates,
APPENDIX 147
and stays having one nut inside bearing
on washers fitted to the edges of the webs
that are at right angles to the plate. For
T take 72 per cent, of the combined thickness
of web and plate.
No such flat plates or surfaces shall be unsupported at
a greater distance than 18 in.
REQUIREMENTS FOR HEADS
All plates used as heads, when new and made to
practically true circles, and as described below, shall be
allowed a steam pressure in accordance with the following
formula:
CONVEX HEADS
TXS
P =
R
Where P = steam pressure allowable in pounds.
T = thickness of plate in inches.
5 = one-fifth of the tensile strength.
R = one-half of the radius to which the head is
bumped.
CONCAVE HEADS
For concave heads the pressure allowable will be .8
times the pressure allowable for convex heads.
Note. — To find the radius of a sphere cf which the bumped head
forms a part, square the radius of head, divide this by the height of
bump required; to the result add height of bump, which will equal
diameter of sphere, one-half of which will be the required radius,
n
148 ARITHMETIC OF THE STEAM BOILER
Example. — Required the working pressure of a convex head of a
54-in. radius, material 60,000 lb. tensile strength and 1/2 in. thick.
Substituting values,
d -5X12000
P = = 222 lb.
27
The pressure allowable on a concave head of the same dimensions
would be
222 X. 8 = 177 lb.
ANGLE STIFFENERS FOR CURVED SURFACES
Where rounded bottoms of combustion chambers are
stiffened with single angle-iron stiffeners, such angles
shall have a thickness of leaf eight-tenths that of the
plate and a depth of at least one-half pitch. Where
stiffened with double angle irons or tee bars, such angles
or tee bars shall have a thickness of leaf at least two-
thirds that of plate and a depth of at least one-fourth of
pitch. Said angles or tee bars shall be substantially
riveted to the plate supported.
Where rounded tops of combustion chambers are
stiffened with single or double angle-iron stiffeners, or
tee bars, such angles or tee bars shall be of thickness and
depth of leaf not less than specified for rounded bottoms
of combustion chambers. Said angles or tee bars shall
be supported on thimbles and riveted through with
rivets not less than 1 in. in diameter, and spaced not
to exceed 6 in. between centers.
Working pressure allowed on rounded surfaces sup-
ported by angle irons or tee bars shall be determined by
the following formula:
W T CXT2
Working pressure = p jz
APPENDIX 149
Where T = thickness of plate in sixteenths of an inch.
P = pitch of angle or tee stiffeners in inches.
D = diameter of curve to which plate is bent, in
inches.
C = 9oo, a constant.
Example. — Given T = g/i6 in. P = 7 in. D = s i in-
Substituting values in formula and solving,
^ 1 • 900X81 .
\\ orkmg pressure = — — =20410. per square men
PRESSURE PERMISSIBLE ON ROUNDED BOTTOM OF COM-
BUSTION CHAMBERS, ANGLES BEING OMITTED
5o(3oor- 2Z, )
' D
Where P = working pressure in pounds.
T= thickness of bottom plate of combustion
chamber in inches.
L = extreme length of plate forming bottom of
combustion chamber in inches.
D = twice outside radius of bottom of combustion
chamber in inches.
Example. — Required the working pressure on the bottom
plate of a combustion chamber, angles being omitted: Thickness
of plate, .82 in., extreme length of plate, 33 in., twice the radius
of bottom of combustion chamber, 50 in. Substituting:
p _ 5oX(3QoX.82-2X33) = l8o i b .
50
T _ PXD+icoL
15000
ISC ARITHMETIC OF THE STEAM BOILER
Pressure allowable on tube sheets where combustion
chambers are not suspended from the shell of the boiler
shall be determined by the following formula:
(D-d)XTX2 7 ooo
WXD
Where P = working pressure in pounds.
D = least horizontal distance between tube centers,
in inches.
d = inside diameter of tubes in inches.
T = thickness of tube plates in inches.
W = extreme width of combustion chamber in
inches.
The compressive stress on tube plates, as determined
by the following formula, must not exceed 13,500 lb.
per square inch, when pressure on top of combustion
chamber is supported by vertical plates of such
chamber.
PXDXW
L 2(D-d)T
Where C = stress on tube sheet.
P = working pressure in pounds.
D = least horizontal distance between tube cen-
ters in inches.
d = inside diameter of tubes in inches.
W = extreme width of combustion chamber in
inches.
T = thickness of tube sheet in inches.
APPENDIX 151
Safety Valves
The areas of safety valves shall be determined in
accordance with the following formula and table:
W
# = .2074Xp-
Where a = area of safety valve, in square inches, per
square foot of grate surface.
PF = pounds of water evaporated per square foot
of grate surface per hour.
P = absolute pressure per square inch = working
gage pressure+15.
From which formula the areas required per square
foot of grate surface in the following table are found by
assuming the different values of W and P.
The figures (a) in table multiplied by square feet of
grate surface give the area of safety valve or valves
required.
When this calculation results in an odd size of safety
valve, use next larger standard size.
Examples. — Boiler pressure = 75 lb. per square inch (gage).
2 furnaces: Grate surface = 2 (No.)X5 ft. 6 in. (long)X3 ft.
(wide) =33 sq. ft.
Water evaporated per pound of coal = 8 lb.
Coal burned per square foot grate surface per hour =12 1/2 lb.
Evaporation per square foot grate surface per hour = 8X12 1/2
= 100 lb.
Hence W = 100 and gage pressure = 75 lb.
From table the corresponding value of a is .230 sq. in.
Therefore area of safety valve = 33 X. 23 = 7.59 sq. in.
For which the diameter is 3 1/8 in. nearly.
Boiler pressure = 215 lb.
152
ARITHMETIC OF THE STEAM BOILER
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154 ARITHMETIC OF THE STEAM BOILER
6 furnaces: Grate surface = 6 (No.)Xs ft. 6 in. (long) X 3 ft. 4 in.
(wide) = 110 sq. ft.
Water evaporated per pound coal = 10 lb.
Coal burned per square foot grate surface per hour = 30 lb.
Evaporation per square foot grate surface per hour = 10X30 =
300 lb.
Hence IF = 300, gage pressure = 215, and a = .2 70 (from table).
Therefore area of safety valve = noX.2 7o = 29.7 sq. in., which
is too large for one valve. Use two.
20. 7
= 14.85 sq. in. Diameter = 4 3/8 in.
To determine the area cf a safety valve for boiler using oil as
fuel or for boilers designed for any evaporation per hour,
Divide the total number of pounds of water evaporated per hour
by any number of pounds of water evaporated per square foot of
grate surface per hour (W) taken from, and within the limits of,
the table. This will give the equivalent number of square feet of
grate surface for boiler; for estimating the area of valve, then apply
the table as in previous examples.
Example. — Required the area of a safety valve for a boiler using
oil as fuel, designed to evaporate 8000 lb. of water per hour, at
175-lb. gage pressure.
Make W = 200.
8000 .
— =40, the equivalent
200 ^ ' n
grate surface in square feet.
For gage pressure = 175 lb. and W = 2oo, from table, a = .218 sq.
in.; .218X40 = 8.72 sq. in., the total area required for this boiler,
for which the diameter is 3 5/16 in. closely.
WATER TUBE AND COIL BOILERS
The working pressure allowable on cylindrical shells of
water tube or coil boilers, when such shells have a row
APPENDIX 155
or rows of pipes or tubes inserted therein, shall be deter-
mined by the following formula:
(D-d )XTXS
DXR
Where P = working pressure allowable in pounds.
D = distance in inches between the tube or pipe
centers in a line from head to head.
d = diameter of hole in inches.
T = thickness of plate in inches.
S = one-sixth of the tensile strength of the plate.
R = radius of shell in inches.
n = number of tube holes in a pitch. When tubes
on any one row are pitched unequally, nd
must be substituted in the formula for d;
where rows of tubes are pitched diagonally,
each diagonal ligament shall not be less than
three-fifths of each longitudinal ligament.
Example. — Required the working pressure of a cylindrical shell
having holes i in. in diameter, spaced 2 in. from center to center, in
a line from head to head; material, 1/2 in. thick; diameter of shell,
20 in.; tensile strength of plate, 60,000 lb.
Substituting values, we have
(2 -i)X. 5X10000
P=— — — = 250 lb.
2X10 °
PORCUPINE-TYPE BOILERS
The formula for determining pressure on boilers of the
so-called Porcupine and similar types shall be as follows:
Multiply the vertical distance between the centers of
the horizontal rows of tubes in inches by one-half the
diameter of shell of boiler in inches, which gives the area
156 ARITHMETIC OF THE STEAM BOILER
upon which the pressure is exerted to break a diagonal
ligament, then find the sectional area of the ligament at
its smallest part and multiply by one-sixth the tensile
strength of the material. This result, divided by the
area upon which the strain is exerted, gives the working
EFT
pressure per square inch, which is as follows =W,
the working pressure, in which E equals wddth of liga-
ment in inches, F thickness of material in inches, T one-
sixth of the tensile strength, C distance between vertical
centers, and D one-half the inside diameter of the shell
or central column.
For the boiler proposed, 30 in. diameter, 5/8 in.
thick, tensile strength 60,000 lb., 1.219 in. would be
width of ligament, .625 thickness of plate, 10,000
one-sixth of tensile strength, 3 11/16 = 3.6875 in.,
distance of vertical centers; 15 in., one-half the diam-
eter of shell, would be as follows: 1.219 multiplied by
.625, this product multiplied by one-sixth the tensile
strength, 10,000, equals 7618.75. This product, divided
by the product of 3.6875, distance between vertical
centers, multiplied by 15, one-half the diameter, equals
55.3125, gives 137.7 as pressure allowed.
EXTRACTS FROM BOARD OF BOILER RULES,
STATE OF MASSACHUSETTS
Maximum Pressure on Boilers
1. The maximum pressure allowed on any steam boiler
constructed wholly of cast-iron shall not be greater
than twenty-five (25) pounds to the square inch.
APPENDIX 157
2. The maximum pressures allowed on any steam
boiler, the tubes of which are secured to cast-iron headers,
shall not be greater than one hundred and sixty (160)
pounds to the square inch.
3. The maximum pressure allowed on any steam boiler
constructed of iron or steel shells or drums shall be
calculated from the inside diameter of the outside course,
the percentage of strength of the longitudinal joint and
the minimum thickness of the shell plates; the tensile
strength of shell plates to be taken as fifty-five thousand
(55,000) pounds per square inch for steel and forty-five
thousand (45,000) pounds per square inch for iron
when the tensile strength is not known.
SHEARING STRENGTH OF RIVETS
4. The maximum shearing strength of rivets per square
inch of cross- section of area to be taken as follows:
Pounds
Iron rivets in single shear 38,000
Iron rivets in double shear 70,000
Steel rivets in single shear 42,000
Steel rivets in double shear 78,000
Factors of Safety
5. The lowest factors of safety used for steam boilers,
the shells or drums of which are directly exposed to the
products of combustion and the longitudinal joints of
which are of lap-riveted construction, shall be as follows:
(a) Five (5) for boilers not over ten years old.
(b) Five and five-tenths (5.5) for boilers over ten and
not over fifteen years old.
158 ARITHMETIC OF THE STEAM BOILER
(c) Five and seventy-five hundredths (5.75) for boilers
over fifteen and not over twenty years old.
(d) Six (6) for boilers over twenty years old.
(e) Five (5) on steam boilers, the longitudinal joints
of which are of lap-riveted construction, and the shells
of drums of which are not directly exposed to the product s
of combustion.
(f) Four and five-tenths (4.5) on steam boilers, the
longitudinal joints of which are of butt and strap
construction.
Fusible Plugs
1. Fusible plugs as required by Section 20, Chapter
465, Acts of 1907, shall be filled with pure tin.
2. The least diameter of fusible metal shall not be less
than one-half inch, except for working pressure of over
one hundred and seventy-five (175) pounds gage, or
when it is necessary to place a fusible plug in a tube, in
which cases the least diameter of fusible metal shall not
be less than three-eighth (3/8) inch.
3. The location of fusible plugs shall be as follows:
(a) In Horizontal Return Tubular Boilers — in the back
head, not less than two (2) inches above the upper row of
tubes, and projecting through the sheet not less than one
(1) inch.
(b) In Horizontal Flue Boilers — in the back head, on
a line with the highest part of the boiler exposed to the
production of combustion, and projecting through the
sheet not less than one (1) inch.
(c) In Locomotive Type or Star Water Tube Boilers —
APPENDIX 159
in the highest part of the crown sheet, and projecting
through the sheet not less than one (1) inch.
(d) In Vertical Fire Tube Boilers — in an outside tube,
placed not less than one-third (1/3) the length of the
tube above the lower tube sheet.
(e) In Vertical Submerged Tube Boilers — in the upper
tube sheet.
(f) In Water Tube Boilers, Horizontal Drums,
Babcock & Wilcox Type — in the upper drum, not less than
six (6) inches above the bottom of the drum, and over the
first pass of the products of combustion, projecting
through the sheet not less than one (1) inch.
(g) In Stirling Boilers, Standard Type — in the front
side of the middle drum, not less than six (6) inches
above the bottom of the drum, and projecting through
the sheet not less than one (1) inch.
(h) In Stirling Boilers, Superheater Type — in the
front drum, not less than six (6) inches abo\e the bottom
of the drum, and exposed to the products of combus-
tion, projecting through the sheet not less than one
(1) inch.
(i) In Water Tube Boilers, Heine Type — in the front
course of the drum, not less than six (6) inches above
the bottom of the drum, and projecting through the
sheet not less than one (1) inch.
(j) In Robb-Mumford Boilers, Standard Type — in
the bottom of the steam and water drum, twenty-four
(24) inches from the center of the rear neck, and pro-
jecting through the sheet not less than one (1) inch.
(k) In Water Tube Boilers, Almy Type — in the tube
directly exposed to the products of combustion.
160 ARITHMETIC OF THE STEAM BOILER
(1) In Vertical Boilers, Climax or Hazelton Type —
in a tube or center drum not less than one-half (1/2)
the height of the shell, measuring from the lowest cir-
cumferential seam.
(m) In Cahall Vertical Water Tube Boilers — in the
inner sheet of the top drum, not less than six (6) inches
above the upper tube sheet.
(n) In Scotch Marine Type Boilers — in combustion
chamber top, and projecting through the sheet not less
than one (1) inch.
(o) In Dry Back Scotch Type Boilers — in rear head,
not less than two (2) inches above the top row of tubes,
and projecting through the sheet not less than one (1)
inch.
(p) In Economic Type Boilers — in the rear head above
the upper row of tubes.
(q) In Cast-iron Sectional Heating Boilers — in a
section over and in direct contact with the products of
combustion in the primary combustion chamber.
(r) For other types and new designs, fusible plugs
shall be placed at the lowest permissible water level, in
the direct path of the products of combustion, as near
the primary combustion chamber as possible.
Size of Rivets
1. When the size of the rivets in the longitudinal
joints of a boiler is not known, the diameter and cross-
sectional area of rivet, after driving, shall be taken as
follows :
APPENDIX
161
Thickness of plate. I7/16 in.i7/i6 in.l 15/32 in.l 1/2 in. |q/i6 in. | 5/8 in.
7/8 in.
15/16
15/16 in.
15/16
1 1/16
I 1/16
Diameter of rivet
up to
in. over
in.
in.
in.
after driving.
2 1/4 in.
pitch
2 1/4 in.
pitch
Cross-sectional area of
.6013
.6903
.6903
.6903
.8866
.8866
rivet after driving.
sq. in.
sq. m.
sq. in.
sq. in.
sq. in.
sq. in.
Thickness of plate.
1/4
in.
9/32
in.
5/i6
in.
11/32
in.
3/8
in.
3/8
in.
13/32
in.
1 1/16
1 1/16
3/4
3/4
3/4 in.
13/16 in.
13/16
Diameter of rivet
in.
in.
in.
in.
up to
over
in.
after driving.
2 in.
pitch
2 in.
pitch
Cross-sectional
• 3712
.3712
.4418
.4418
.4418
.5185
.5185
area of rivet after
sq. in.
sq. in.
sq. m.
sq. m.
sq. 111.
sq. in.
sq. in.
driving.
Allowable Strain on Stays
1. The maximum allowable strain per square inch net
cross-section for weldless mild steel shall be as follows:
Size up to and in-
Type eluding 1 1/2 in. diam-
eter or equivalent
Size over 1 1/2 in.
diameter or equiva-
lent
Head to head or through stays.
8,000 lb.
9,000 lb.
Diagonal or crowfoot stays. . . .
7,500 lb.
8,000 lb.
Screwed stays (stay bolts)
7,000 lb.
7,000 lb.
2. For welded stays the strain allowed per square inch
net cross-section shall not exceed six thousand (6000)
pounds.
3. For wrought-iron stays or stay bolts the strain
allowed per square inch net cross-section shall not
exceed six thousand (6000) pounds.
l62
ARITHMETIC OF THE STEAM BOILER
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APPENDIX
163
Appendages to be Placed on Boilers
1. Each boiler shall have a safety valve the minimum
area of which shall be in accordance with the following
tables. If more than one safety valve is used the mini-
mum combined area shall be in accordance with the
following tables.
2. When the conditions exceed those on which the tables
are based the formula shall be used.
3. A table of areas of grate surface in square feet for
pop safety valves follows:
A = Area of safety valve in square inches per square
foot of grate.
W = Weight of steam per second.
P = Pressure, absolute.
4. A table of grate areas in square feet for safety
valves (other than pop safety valves) follows; this table
is in ratio to the table for pop safety valves as 2 is to 3 :
Gage pressure per square inch at
Zero to
Over 25 to
Over 50 to
which safety valve is set to blow
25 lb.
50 lb.
100 lb.
Diameter of valve
in inches
Area of valve in
square inches
Area of grate in sqt
iare feet
1
• 7854
1.4
1.6
1.8
1 1/4
1 .2272
2 . 1
2.5
2.8
1 1/2
1 .7671
30
3.6
4.0
2
3-1416
53
6.4
7-i
2 1/2
4.9087
8.2
10.
11 .0
3
7.0686
II. 7
14.2
16.0
3 1/2
0.6211
16.0
19.5
21 .6
4
12 .5660
21 .0
25-5
28.2
4 1/2
15.9040
26.7
32.3
36.0
5
196350
32.7
40.0
44.0
164 ARITHMETIC OF THE STEAM BOILER
5. Each safety valve must have full-sized direct
connection to the boiler, and full-sized escape pipe which
shall be fitted with an open drain to prevent water
lodging in the upper part of safety valve or escape pipe.
When a boiler is fitted with two safety valves on one
connection this connection to the boiler shall have a cross-
sectional area equal to or greater than the combined area
of the two safety valves.
6. Safety valves having either the seat or disc of cast-
iron shall not be used.
7. The seats of all safety valves shall be inclined at an
angle of forty-five (45) degrees to the center line of the
spindle.
8. A certificate of inspection shall not be issued on a
boiler used for heating purposes exclusively, permitting
the boiler to be operated at a pressure in excess of
fifteen (15) pounds, if the boiler is provided with a device
(safety valve) in accordance with the provision contained
in section 78, chapter 102 of the Revised Laws, limiting
the pressure carried to fifteen (15) pounds.
9. Each boiler shall have a steam gage connected to the
steam space of the boiler by a syphon, or equivalent
device, sufficiently large to fill the gage tube with
water, and in such manner that the steam gage cannot be
shut off from the boiler except by a cock with T end,
placed directly on the pipe under the steam gage.
10. The dial of the steam gage shall be graduated to
not less than one and one-half (1 1/2) times the maximum
pressure allowed on the boiler.
n. Each boiler shall be provided with a one-eighth
(1/8) inch pipe size connection for attaching inspector's
APPENDIX 165
test gage when boiler is in service, so that the accuracy of
the boiler steam gage can be ascertained, as required by
section 3, chapter 465, Acts of 1907.
12. Each boiler shall have one fusible plug, as required
by rules (section 3) on fusible plugs.
13. Each boiler shall have one water-glass, the bottom
end of which shall be above the fusible plug and lowest
safe water line.
14. Each boiler shall have two or more gage cocks,
located within the range of the water-glass, when the
maximum pressure allowed does not exceed twenty-five
(25) pounds per square inch.
15. Each boiler shall have three or more gage cocks,
located within the range of the water-glass, when the
maximum pressure allowed exceeds twenty-five (25)
pounds per square inch.
16. Each steam outlet from boiler shall be fitted with a
stop valve.
17. When a stop valve is so located that water can
accumulate, ample drains shall be provided.
18. Each boiler shall have a feed pipe fitted with check
valve, and also a stop valve between the check valve
and the boiler, the feed water to discharge below the
lowest safe water line. Means must be provided for
feeding the boiler with w T ater when the maximum pressure
allowed is carried on the boiler.
19. Each boiler shall have a bottom blow-off pipe
fitted with a stop valve or stop cock, and connected direct
to the lowest water space of the boiler.
20. Where a damper regulator is used, the boiler pres-
sure pipe shall be taken from the steam space of the
166 ARITHMETIC OF THE STEAM BOILER
boiler, and shall be fitted with a stop valve or stop
cock.
21. Each boiler fitted with a Lamphrey Boiler Furnace
Mouth Protector, or similar appendage, having valves
on the pipes connecting same with the boiler, shall have
these valves locked or sealed open, so that the locks or
seals will require to be removed or broken to shut the
valves.
Annual Internal Inspections
i. The owner or user of a steam boiler which requires
annual inspection, internally and externally, by the boiler
inspection department or by an insurance company,
as provided by section i, chapter 465, Acts of 1907,
shall prepare the boiler for inspection by cooling it down
(blanking off connections to adjacent boilers if necessary),
removing all soot and ashes from tubes, heads, shell,
furnace and combustion chamber; drawing off the water;
removing the handhole and manhole plates; removing
the grate bars from internally fired boilers; and removing
the steam gage for testing.
2. If a boiler has not been properly cooled down, or
otherwise prepared for inspection, the boiler inspector
shall decline to inspect it, and he shall not issue a certifi-
cate of inspection until efficient inspection has been made.
3. In making the annual internal and external inspec-
tion as provided by sections 1 and 4, chapter 465,
Acts of 1907, the boiler inspector shall apply the hammer
test to all internal and external parts of a boiler that are
accessible.
APPENDIX 167
4. All proper measurements shall be taken by the boiler
inspector, so that the maximum working pressure allowed
on a boiler will conform to the rules relating to allowable
pressures established by the Board of Boiler Rules; such
measurements to be taken and calculations made before
a hydrostatic pressure test is applied to a boiler.
5. The steam gage of a boiler shall be tested and its
readings compared with an accurate test gage, and if,
in the judgment of the boiler inspector, the gage is not
reliable he shall order it repaired or replaced.
Annual External Inspections
1 . The annual external inspection of a steam boiler, as
provided for in section 3, chapter 465, Acts of 1907,
should be made at or about six (6) months after the
annual internal inspection, except in the case of a boiler
that is in service a portion of the year only, in which
case the annual external inspection shall be made during
such period of service.
2. The boiler inspector shall attach an accurate test
gage to a boiler to note the pressure show r n by said
test gage, and compare it with that shown by the boiler
gage, ordering the boiler gage repaired or replaced if
necessary.
3. The boiler inspector shall see that the water-glass,
gage cocks, water-column connections and w T ater blow-
offs are free and clear; also that the safety valve raises
freely from its seat.
4. Fire doors, tube doors, and doors in settings shall
be opened, to view as far as possible the fire surface.
1 68 ARITHMETIC OF THE STEAM BOILER
settings, tube ends, blow-off pipes and fusible plug,
noting conditions and ordering changes or repairs if
necessary.
Hydrostatic Pressure Tests
i . When a boiler is tested by hydrostatic pressure, the
maximum pressure applied shall not exceed one and one-
half (i 1/2) times the maximum working pressure allowed,
except that tw T ice the maximum working pressure allow r ed
may be applied on boilers permitted to carry tw T enty-five
(25) pounds pressure per square inch or less, or on pipe
boilers.
2. When making annual inspections on boilers con-
structed wholly of cast iron, or on pipe boilers, a hy-
drostatic pressure test of not less than one and one-half
(1 1/2) times and not more than twice the maximum
working pressure allowed shall be applied.
3. The boiler inspector, after applying a hydrostatic
pressure test, shall thoroughly examine every accessible
part of the boiler, both internal and external.
TO DETERMINE MAXIMUM ALLOWABLE PRESSURE
Formula:
T.S.XtX%
■■ maximum allowable working pressure,
RXF.S.
per square inch, in pounds.
T.S. = Tensile strength of shell plates, in pounds.
/ = minimum thickness of shell plates, in inches.
% = efficiency of longitudinal joint.
APPENDIX 169
R = radius = one-half (1/2) the inside diameter of
the outside course of the shell or drum.
F.S.= lowest factor of safety allowed by these rules.
When the tensile strength of steel or wrought iron
shell plates is not known, it shall be taken as fifty-five
thousand (55,000) pounds for steel, and forty-five
thousand (45.000) pounds for wrought iron.
Efficiency of Riveted Joints
The efficiency that a unit of length of a riveted joint
has to the same unit of length of solid plate shall be
calculated as shown by the following examples:
T.S. = tensile strength of plate, in pounds per square
inch.
/ = thickness of plate, in inches.
b = thickness of butt strap, in inches.
P = pitch of rivets, in inches, on row having greatest
pitch.
d = diameter of rivet after driving, in inches.
a = cross-sectional area of rivet after driving, in
square inches.
5 = strength of rivet in single shear.
5 = strength of rivet in double shear.
C = crushing strength of mild steel.
(Note. — "C" applies only to boilers constructed after
February 5, 1910.)
n = number of rivets in single shear in a unit of
length of joint.
N = number of rivets in double shear in a unit
of length of joint.
I70
ARITHMETIC OF THE STEAM BOILER
Lap, Single-riveted.
longitudinal or circumferential
Example.
A ^strength of solid plate =PXtXT.S.
B = strength of solid plate between rivet holes
= (P-d)XtXT.S.
C = shearing strength of one rivet in single shear
= nXsXa.
Fig. i. — Single-riveted lap joint.
D = crushing strength of plate in front of one (i)
rivet =dXtXc
Divide B, C, or D (whichever is the least) by A, and
the quotient will be the efficiency of a single-riveted lap
joint.
7\S.=S5,ooolb.
t = i/4 in. =.25 in.
d = 11/16 in. =.6875 in.
a= .3712 sq. in.
5 =42,000 lb.
c =95, 000 lb.
A =i.625X. 25X55,000 = 22,343.
APPENDIX
171
5 = (1.625 -.6875). 25X55,000 = 12. 890.
C = iX42,oooX.37i2 = 15.590.
D = . 6875 X. 25X9S.000 =16,328.
12890 QB) u .
22343 ( A ) = -576, efficiency of joint.
(See Fig. 1 of this group.)
Lap, Double-riveted.
longitudinal or circumferential
Strength of solid plate = PXtX T.S. = A.
W-pA
6 — ^ — — §-}
Fig. 2. — Double- riveted lap joint,
Strength of plate between rivet holes= (P—d)tXT.S.
= B.
Shearing strength of two (2) rivets in single shear
= nsa = C.
Divide B or C (whichever is the least) by A , and the
quotient will be the efficiency of a double-riveted lap
joint.
T.S. = 55,000 lb.
*=5/i6 in. = .3125 in.
P=2 7/8 in. = 2.875 i n -
172 ARITHMETIC OF THE STEAM BOILER
d = 3/4 in. = .75 in.
a= .4418 sq. in.
5 = 42,000
A = 2. 875X. 3125X55.000 = 49,414.
B = (2. 875 -.75) X. 3125X55,000 = 36, 523-
C = 2 X. 4418X42,000 = 37,111.
36523 (B) „. . ...
ttt = -7^9, emiciency of joint.
49414 UJ MV J J
(See Fig. 2 of this group.)
Butt, Double-riveted.
butt and double-strap joint
,4=strength of solid plate=PX*XT.S.
Fig. 3. — Double- rive ted, double butt-strapped joint.
B = strength of plate between rivet holes in
the outer row = (P-d)tXT.S.
C = shearing strength of two (2) rivets in
double shear; plus the shearing strength
of one (1) rivet in single shear = NXS
Xa+nXsXa.
APPENDIX 173
D= strength of plate between rivet holes in
the second row, plus the shearing strength
of one (1) rivet in single shear in the
outer row=(P-2d)tXT.S.+nXsXa.
E= strength of plate between rivet holes in
the second row, plus the crushing
strength of butt strap in front of one
(1) rivet in the outer row = (P — 2d)
tXT.S.+dXbXc.
F = crushing strength of plate in front of
two (2) rivets, plus the crushing strength
of butt strap in front of one (1) rivet =
NXdXtXc+nXdXbXc.
G = crushing strength of plate in front of two (2)
rivets, plus the shearing strength of one (1)
rivet in single shear = NXdXtXc+nXsX a.
Divide B, C, D, £, F, or G (whichever is the least)
by A, and the quotient will be the efficiency of a butt and
double-strap joint, double-riveted. (See Fig. 3 of this
group.)
T.S. = 55,000 lb. a = .6013 sq. in.
t = s/S in. = -375 in. 5 = 42,000 lb.
6 = 5/16 in. = .3125 in. 5 = 78,000 lb.
P = 4 7/8 in. =4.875 in. £ = 95,000 lb.
Number of rivets in single shear in a unit of length of
joint=i.
Number of rivets in double shear in a unit of length of
joint =2.
^l=4.875X.375X55^oo= 100,547.
B = (4.875 — .875) .375X55^00 = 82,500.
174
ARITHMETIC OF THE STEAM BOILER
C = 2X 78,000 X. 6013 + iX42,oooX. 6013 =
119,057.
D= (4.87S -2X.875).375X55> 000 + I X42,oooX
.6013 = 89,708.
E= (4-875- 2X.87s).37sX5S»oo°+-875X.3i2S
X95 ? ooo = 9o,429.
F = 2X.87sX.37SX9S ? 000 +- 8 75X.3i2SX9S ? 000
= 88,320.
G=2X.87sX.37SX9S,ooo+iX4 2 ) 00 °X.6oi3 =
87,599.
82500(^8) rr • r • • 4
— - — V-v = .82o, emciency of joint.
100547^)
(See Fig. 3 in this group.)
Butt, Triple-riveted,
butt and double-strap joint
A =strength of solid plate =PXtXT.S.
Fig. 4. — Triple- riveted, double butt-strapped joint.
B = strength of plate between rivet holes in the
outer row= (P-d)tX T.S.
C = shearing strength of four (4) rivets in double
shear, plus the shearing strength of one (1)
rivet in single shear = NXSXa+nXsXa.
APPENDIX 175
D = strength of plate between the rivet holes in
the second row, plus the shearing strength of
one (1) rivet in single shear in the outer row
= (P-2d)tXT.S.+nXsXa.
E = strength of plate between rivet holes in the
second row, plus the crushing strength of butt
strap in front of one (1) rivet in the outer row
= (P-2d)tXT.S.+dXbXc.
F = crushing strength of plate in front of four (4 J
rivets, plus the crushing strength of butt strap
in front of one (1) rivet = XXdXtXc+nXdX
bXc
G = crushing strength of plate in front of four (4)
rivets, plus the shearing strength of one (1)
rivet in single shear = XXdXtXc+nXsXa.
Divide B, C, D, E, F, or G (whichever is the least) by
A j and the quotient will be the efficiency of a butt and
double-strap joint, triple riveted. (See Fig. 4 of this
group.)
T.S. = 55,000 lb. # = .5185 sq. in.
t = $/8 in. = .375 in. 5 = 42,000 lb.
6 = 5/16 in. = .3125 in. 5 = 78,000 lb.
F = 6 1/2 in. = 6.5 in. £ = 95,000 lb.
d= 13/16 in. = .8125
Number of rivets in single shear in a unit of length of
joint = 1.
Number of rivets in double shear in a unit of length of
joint = 4.
-4=6.5X.375X55> 000=I 34,o62.
^ = .( 6 -5-- 8l2 5)-37SX 55,000 =117,304.
176 ARITHMETIC OF THE STEAM BOILER
C = 4 X 78,000 X. 5185 + 1 X 42,000 X-5i85 =
183,549-
£ = (6.5-2 X.8i25). 3 75X55>°oo+iX 42,000 X
.5185 = 122,323.
£ = (6.5-2 X.8i25). 3 75 X 55,000+. 8125 X.3125
X95,ooo= 124,667.
£ = 4X.8i25X.375X95>°oo+iX.8i25X.3i25X
95,000=139,902.
G = 4 X.8125 X.375X95,ooo+iX 42,000 X. 5185
= i37 ; SS8.
i34o?fe) == - 87S ' efficienc y of i° int -
(See Fig. 4 in this group.)
Butt, Quadruple-riveted.
butt and double-strap joint, quadruple -riveted
A = Strength of solid plate =PXtXT.S.
B = Strength of plate between rivet holes in the
outer row =(P-d)tXT.S.
C = shearing strength of eight (8) rivets in
double shear, plus the shearing strength
of three (3) rivets in single shear = NXSX
a-\-nXsXa.
D = strength of plate between rivet holes in the
second row, plus the shearing strength of
one (1) rivet in single shear in the outer row
= (P-2d)tXT.S.+nXsXa.
E = strength of plate between rivet holes in the
third row, plus the shearing strength of two
APPENDIX
177
(2) rivets in the second row in single shear
and one (1) rivet in single shear in the outer
row=(P^4d)tXT.S.+nXsXa.
F = strength of plate between rivet holes in the
second row, plus the crushing strength of
butt strap in front of one (1) rivet in the
outer row=(P-2d)tXT.S.+dXbXc
G = strength of plate between rivet holes in
Fig. 5. — Quadruple-riveted, double butt-strapped joint.
the third row, plus the crushing strength of
butt strap in front of two (2) rivets in the
second row and one (1) rivet in the outer
row=(P-4d)tXT.S.+nXdXbXc.
H = crushing strength of plate in front of eight
(8) rivets, plus the crushing strength of
butt strap in front of three (3) rivets =
NXdXtXc+nXdXbXc
1 = crushing strength of plate in front of eight
(8) rivets, plus the shearing strength of
two (2) rivets in the second row and one
178 ARITHMETIC OF THE STEAM BOILER
(1) rivet in the outer row, in single shear =
NXdXtXc+nXsXa.
Divide B, C, D, E, F, G, H, or / (whichever is the least)
by A, and the quotient will be the efficiency of a butt
and double-strap joint, quadruple-riveted.
T.S. = 55,000 lb.
t= 1/2 in. = .5 in.
6 = 7/16 in. = .4375
P=iS in.
d= 15/16 in. = .9375 in.
a = .6903 sq. in.
5 = 42,000 lb.
5 = 78,000 lb.
c = 95,000 lb.
Number of rivets in single shear in unit of length of
joint = 3.
Number of rivets in double shear in unit of length of
joint = 8.
^ = i5X.5X55> 000 = 4i2,5oo.
£ = (i5--9375)-5X55>°°o = 386,7i8.
C = 8 X 78,000 X.6903 + 3 X 42,000 X.6903 =
5i7,723-
#=(15-2 X.937S)-S X 55,ooo+ 1 X 42,000 X
•69°3 = 389>93°-
£=(15-4 X .9375)-5 X 55,000 + 3 X 42,000 X
.6903 = 396,353.
^ = (15-2 X.9375)-5X 55>°°°+-9375 X-4375 X
95,000 = 399,902.
APPENDIX 179
G=(i5-4 X .937S)-S X 55>°oo + 3 X -9375 X
•4375X95> 000 = 426,269.
H = 8X -937S X.5 X 95,000 + 3 X-937SX.437SX
95,000 = 473,145.
7 = 8 X -9375 X .5 X 95>°°° + 3 X 42,000 X
.6903 = 443,229.
386718(5) ffi . , . . ■
4i25oo04r- 937 ' efikienCy ° f JOmt *
(See Fig. 5 of this group.)
Bumped Heads
The minimum thickness of a convex head shall be
determined by the following formula:
RXF.S.XP
T.S.
The minimum thickness of a concave head shall be
determined by the following formula:
RXF.S.XP
.6(T.S.)
R = one-half the radius to which the head is
bumped.
F.S. = 5 = factor of safety.
P = working pressure, in pounds per square inch,
for which the boiler is designed.
T.S. = tensile strength, in pounds per square inch,
stamped on the head by the manufacturer.
t = thickness of the head in inches.
When a convex or a concave head has a manhole
13
180 ARITHMETIC OF THE STEAM BOILER
opening, the thickness as found by the preceding for-
mulas shall be increased by not less than one-eighth (1/8)
inch.
FORMULA TO FIND AREA OF SEGMENT OF CIRCLE TO
BE BRACED
~ a / jj— .608 = area in square inches.
H = distance from tubes to shell, minus five (5)
inches.
R = radius of boiler, minus three (3) inches.
FORMULA FOR DIAMETER OF STAY BOLTS AT BOTTOM
OF THREAD
D — (PX 1.732) =d, or for 12 threads per inch,
£>-(.o8333Xi.73 2 )=^ then
D = diameter of stay bolt over the threads.
P = pitch of threads = 1/ 1 2 = .08333.
d = diameter of stay bolt at bottom of threads.
1.732 = a constant.
When U. S. threads are used, the formula becomes: Z) —
(PX1.732X.75W-
FORMULA FOR CAST-IRON NOZZLES
The minimum thickness of cast-iron nozzles shall be
determined by the following formula:
Pdf ,
APPENDIX 181
P = working pressure in pounds per square inch.
d = inside diameter of nozzle in inches.
/= factor of safety = 12.
5 = ultimate tensile strength of cast iron, not less
than eighteen thousand (18,000) pounds per
square inch.
.5 = a constant.
t = thickness of nozzle in inches.
Maximum Pressure on Boiler Shells
The maximum pressure to be allowed on a steel or
wrought-iron shell or drum shall be determined from the
minimum thickness of the shell plates, the lowest
tensile strength stamped on the plates by the plate
manufacturer, the efficiency of the longitudinal joint or
ligament between the tube holes, whichever is the least,
the inside diameter of the outside course, and a factor of
safety of not less than five (5), the formula being:
T.S. X^X %
7?v7? y — = maximum allowable working pressure per
square inch in pounds.
T. S. = tensile strength of shell plates in pounds.
t = minimum thickness of shell plates in inches.
% = efficiency of longitudinal joint or ligament be-
tween tube holes, whichever is the least.
R = radius = one-half (1/2) the inside diameter of the
outside course of the shell or drum.
F.5. = 5, the lowest factor of safety allowed on boilers
installed after May 1, 1908.
182
ARITHMETIC OF THE STEAM BOILER
The method of determining the efficiency of the
longitudinal joint has already been explained and illus-
trated. To find the efficiency of ligaments, the following
formulas are to be employed.
Efficiency of Ligaments
When a shell or drum is drilled for tube holes in a line
parallel to the axis of the shell or drum, the efficiency
of the ligament between the tube holes shall be deter-
-(£-&- Q Ct) (fr d>
<-5 1 4->
<— &£->*— VA-*
<r-&i
-&M-»
-5*4
,-5;£->
^ ) Q Cp Cp Cp Cp C p-^)-
Longitudinal Line >.
Fig. i. — Diagram for calculating the efficiency of ligament.
mined as follows: (a) when the pitch of the tube holes
on every row is equal the formula is:
p d
— - — = efficiency of ligament.
p = pitch of tube holes in inches.
d = diameter of tube holes in inches.
Example. — Pitch of tube holes in the drum of a water-tube
boiler = 5 1/4 in. = 5.25 in. Diameter of tube holes = 3 1/4 in. =
3.25 in.
P~d 5-25-3-25,
: .38, efficiency of ligament.
5-25
(See Fig. 1 of this group.)
APPENDIX
183
(b) when the pitch of the tube holes on any one row is unequal,
the formula is:
P-na
= efficiency of ligament.
■e
e-©-e
^._ 5 4-/W--6^-^-54i^--64f , ->U--544^[<---6^-^--54-
h— -1-2^— h
Longitudina l Line
Fig. 2. — Diagram for calculating the efficiency of ligament.
P = unit length of ligament in inches.
n = number of tube holes in length, P.
d = diameter of tube holes in inches.
-29 H-
J Longitudinal Line
Fig. 3. — Diagram for calculating the efficiency of ligament.
Example. —
P — nd _i2 — 2X3. 25
= .458, efficiency of ligament.
(See Fig. 2 of this group.)
Example. —
P—nd _ 2g. 25-5X3.25
.444, efficiency of ligament.
(See Fig. 3 of this group.)
1 84 ARITHMETIC OF THE STEAM BOILER
When a shell or drum is drilled for tube holes in a line diagonal to
the axis of the shell or drum, the efficiency of the ligament between
the tube holes shall be determined as follows:
P-d
— = efficiency of ligament.
P
P = diagonal pitch of tube holes in inches.
d = diameter of tube holes in inches.
p = distance between rows of tubes longitudinally.
{ p'cfo ($> <fe-
Girth Line_
Fig. 4. — Diagram for calculating the efficiency of ligament.
Example. — Diagonal pitch of tube holes in a drum of a water-
tube boiler = 6.42 in.
Diameter of tube holes =4 in.
Distance between rows of tubes, longitudinally = 5.75 in.
P-d 6. 42-4
— — = = .42, efficiency of ligament.
P 5-75
(See Fig. 4 of this group.)
When a flat-head has a manhole opening, the flange
of which is formed from the solid sheet and turned inward
to a depth of not less than twice the thickness of the
head, an area two (2) inches wide all around the manhole
opening, as shown in the figure, may be deducted from
the total area of head, including manhole opening, to be
stayed.
APPENDIX
185
Example. — To find an area 2 in. wide all around a 11 in. X 15 in.
manhole,
15 in.Xi9 in. X. 7854 = 224 (nearly) sq. in.
11 in.Xi5 in. X. 7854 = 130 (nearly) sq. in.
And 224 — 130 = 94 sq. in.
Therefore, if the area to be stayed on the rear head,
below the tubes, of a seventy-two (72) inch horizontal re-
turn tubular boiler is 374 sq. in., the area to be stayed on
front head, below the tubes, of this boiler, would be 374 —
94 = 280 sq. in.
k AAA m^-'^mm fT Yfhfh /
0B^
Fig. 1. — Diagram showing area of head to be braced.
A segment of a head of a horizontal return tubular,
locomotive, Scotch or similar type shall be stayed by
welded or weldless mild steel or wrought iron, head to
head or through, diagonal or crow-foot stays, except a
horizontal return tubular boiler, as otherwise provided
for.
The area of a segment of a head to be stayed shall
be the area enclosed by lines drawn three (3) inches from
1 86 ARITHMETIC OF THE STEAM BOILER
the shell and two (2) inches from the tubes, as shown in
Figs. 1 and 2.
When the shell of a horizontal return tubular boiler
does not exceed thirty-six (36) inches in diameter, and
is designed for a maximum working pressure not to exceed
one hundred (100) pounds per square inch, the segment
of the head above the tubes may be stayed by steel
angles, or Tee bars, the formula being:
Fig. 2. — Diagram showing area of head to be braced.
J -=M
y
/=
= fiber stress =
= 16,000
lb.
/=
= moment of inertia =
12
J A = height of beam in inches.
\b = thickness of beam in inches.
y = distance of most strained fiber = h-S- 2.
M = bending moment of beam.
APPENDIX
187
load =
maximum bending moment for a uniform
WL
W = weight to be supported in pounds.
L = length of beam in inches.
Example. — When steel angles are used, the head of a horizontal
return tubular boiler thirty (30) inches in diameter, designed for
Fig. 1. — Diagram of steel angles bracing.
one hundred (100) pounds working pressure, shall be stayed by
two (2) four and one-half by three by three-eighths (41/2X3X3/8)
inch steel angles, as shown in the figure, or by other sized commercial
steel angles, the resistance of which shall be equal to or greater than
the maximum bending moment.
Distance from tubes to shell = 13 1/2 in.
i88
ARITHMETIC OF THE STEAM BOILER
Area to be stayed = 143 sq. in.
Load at 100 lb. pressure = 14.300 lb.
WL 14300X21.
~g- = g = 37,54o lb.
Moment of inertia = / = i/i2X4.5 3 X3/8 = 2.85.
^=4.5-^-2 = 2.25.
Fig. 2. — Diagram of steel angles bracing.
I 1 -, , 16000X2.85 ■
— =M = — — = 20,266 lb. for one angle.
y 2.25
Resistance of one angle = 20,266 lb.
Resistance of two angles = 40,532 lb.
When steel angles are used, the head of a boiler,
thirty-six (36) inches in diameter, designed for one
hundred (100) pounds working pressure, shall be stayed
APPENDIX 189
by two (2) six by three and one-half by one-half (6X
3 1/2 X 1/2) inch steel angles, as shown in the figure
or by other sized commercial steel angles the resistance
of which shall be equal to or greater than the maximum
bending moment.
Distance from tubes to shell=i5 1/2 in.
Area to be stayed=22o sq. in.
Load at 100 lb. pressure = 22,000 lb.
WL 22000X27 __ ^_ . .
-0~=— — 5 =74,250 lb. Moment of inertia =
o o
Z=i/i2X6 3 Xi/2 = 9.
fl __ 16000X9 » r ,
— =M=— —=48,000 lb. for one angle.
y 3
Resistance of one angle = 48,000 lb.
Resistance of two angles = 96,000 lb.
PITCH OF STAY BOLTS ON FURNACE SHEETS
The longitudinal pitch between stay bolts on the fur-
nace sheet of an internally fired boiler, in which the ex-
ternal diameter of the furnace is thirty-eight (38) inches
or less, except a corrugated furnace or a furnace strength-
ened by an Adamson ring or equivalent, shall not exceed
that given by the following formula:
190 ARITHMETIC OF THE STEAM BOILER
L = longitudinal pitch of stay bolts, in inches, or one-
half the height of furnace when only one circumfer-
ential row of stay bolts is required.
C = a constant = no.
t = thickness of furnace sheet, in thirty-seconds of an inch .
P = working pressure per square inch in pounds.
d = external diameter of furnace in inches.
TABLE I.— AREAS AND CIRCUMFERENCES
OF
CIRCLES FROM
I TO IOO
Dia.
Area
Circum.
Dia.
Area
Circum.
Dia.
Area j Circum.
3 \
0.00077
O.098175
2
3.1416
6.28319
5 l
I9-635
15.7080
3
^4
0.00173
O.147262
1
1 6
3-34io
6.47953
1
T6
20.129
15-9043
1
Tg
0.00307
O.196350
1
8
3-5466
6.67588
1
8"
20.629
16.1007
3
32
0.00690
O.294524
A
3-75 8 3
6.87223
3
1 6
21.135
16.2970
i
0.01227
O.392699
1
4
3.9761
7.06858
1
4"
21.648
16.4934
A
0.01917
O.490874
5
16
4.2000
7.26493
5
T6
22.166
16.6897
A
0.02761
O.589049
t
4.4301
7.46128
3
8
22.691
16.8861
3"2~
0.03758
O.687223
A
4.6664
7^37^3
16
23.221
17.0824
1
O.04909
O.785398
i
4.9087
7.85398
1
2
23758
17.2788
9
32
0.06213
O.883573
9
16
5-I572
8.05033
9
1 6
24.301
I7475I
5
16
0.07670
O.981748
5
8
5-4II9
8.24,668
f
24.850
17.6715
11
32
0.09281
I.07992
1 1
16
5.6727
844303
1 1
16
25.406
17.8678
3
8
0.11045
I.17810
3
4
5-9396
8.63938
3
4
25.967
18.0642
M
0.12962
I.27627
13
16
6.2126
8.83573
13
16
26.535
18.2605
JL-
16
O.I5033
1-37445
7
8
6.4918
9.03208
7
8
27.109
18.4569
M
0.17257
1.47262
15
16
6.7771
9.22843
1 5
16
27.688
18.6532
1
2
O.19635
1.57080
3
7.0686
9.42478
6
28.274
18.8496
17
32
O.22166
1.66897
1
1 6
7.3662
9.62113
i
29.465
19.2423
9
T6
0.24850
1.76715
i
7.6699
9.81748
i
30.680
19.6350
19
32
0.27688
1.86532
3
T6
7.9798
10.0138
f
3 J -9i9
20.0277
1
0.30680
1.96350
i
8.2958
10.2102
1
2
33-^3
20.4204
21
32
0.33824
2.06167
A
8.6179
10.4065
5
8
34472
20.8131
a
O.37122
2.15984
I
8.9462
10.6029
3
4
35785
21.2058
If
O.40574
2.25802
A
9.2806
10.7992
i
37.122
21.5984
1
O.44179
2.35619
i
9.6211
10.9956
7
38485
21.9911
2.5
32
0.47937
2-45437
9
16
9.9678
11.1919
1
8
39-871
22.3838
1 3
T6
0.51849
2 -55254
5
8
10.321
11.3883
1
4
41.282
22.7765
2.1
32
0.559I4
2.65072
tt
10.680
11.5846
1
42.718
23.1692
i
0.60132
2.74889
1
11.045
11.7810
i
44.179
23.5619
If
0.64504
2.84707
H
11.416
11.9773
f
45.664
23.9546
1 5
T6
O.69029
2.94524
I
JI -793
12.1737
f
47.173
24.3473
3 1
32
O.73708
3-04342
15
16
12.177
12.3700
1
48.707
24.7400
I
O.78540
3-I4I59
4
12.566
12.5664
8
50.265
25.1327
1
16
O.88664
3-33794
A
12.962
12.7627
i
51.849
25.5224
i
0.99402
3-53429
1
8
13-364
12,9591
1
4
53-456
25.9181
A
I.1075
3-73064
A
!3-77 2
I 3- I 554
I
55.o88
26.3108
1
4
I.2272
3.92699
i
14.186
i3-35i8
1
56.745
26.7035
A
I-3530
4-12334
5
XT
14.607
i3-548i
5
8
58.426
27.0962
1
1.4849
4.31969
I
J 5-o33
J 3-7445
1
60.132
27.4889
A
1.6230
4.51604
A
15.466
13.9408
i
61.862
27.8816
1
1. 7671
4.71239
i
15.904
14.1372
9
63.617
28.2743
A
J-9I75 -
4.90874
9
16
16.349
J 4-3335
i
65.397
28.6670
f
2.0739
5.10509
1
16.800
.14.5299
1
4
67.201
29.0597
16
2.2365
5-3oi44
tt
I7-257
14.7262
f
69x29
29.4524
i
2.4053
5-49779
f
17.721
14.9226
i
70.882
29.8451
1 3
T6
2.5802
5.69414
H
18.190
15.1189
i
72.760
30.2378
A
2.7612
5.89049
7
8
18.665
J 5-3i53
J
74.662
30-6305
To
2.9483
6.08684
15
T6
19.147
i5-5ii6
I
76.589
31.0232
191
TABLE I.— AREAS AND CIRCUMFERENCES OF CIRCLES FROM
i TO ioo (Continued)
Dia.
Area
Circum.
Dia.
Area
Circum.
Dia.
Area
Circum.
IO
78.540
3I-4I59
16
201.06
50-2655
22
380.13
69.1150
i
80.516
31.8086
i
204.22
50.6582
1
s
384.46
69.5077
l
4
82.516
32.2013
1
4
207.39
51.0509
1
4
388.82
69.9004
f
84.541
3 2 -5940
3
8
210.60
5I-4436
3
8
393.20
70.2931
i
86.590
32.9867
1
2
213.82
51.8363
1
2
397-6l
70.6858
5
8
88.664
33-3794
1
2 1 7 .08
52.2290
5
. 8
402.04
71.0785
f
90.763
33-7721
3
4
22O.35
52.6217
3
4
406.49
71.4712
7
8
92.886
34.1648
7
8
223.65
53-OI44
8
410.97
71.8639
II
95-033
34.5575
17
226.98
53-407I
23
415.48
72.2566
1
8
97- 2 05
34-9502
1
8
230-33
53-7998
i
420.00
72.6493
1
4
99.402
35-3429
i
233-7I
54.1925
1
4
424.56
73.0420
3
8
101.62
35-7350
3
8
237.IO
54o852
1
429.13
73-4347
1
2
103.87
36.1283
4
240.53
54.9779
1
2
433-74
73.8274
1
106.14
36.5210
5
8
243.98
55.3706
5
8
438.36
74.2201
3
4
108.43
36.9137
3
4
247-45
557633
3
4
443.01
74.6128
7
8
110.75
37-3o64
7
8
250.95
56.1560
1
8
447.69
75.0055
12
113.10
37.6991
l8
25447
56.5487
24
452.39
75.3982
i
115-47
38.0918
1
8
258.02
56.9414
1
8
457-n
75.7009
i
4
117.86
38.4845
1
4
261.59
57.3341
1
461.86
76.1836
§
120.28
38.8772
i
265.18
57.7268
1
466.64
76.5783
4
122.72
39.2699
4
268.80
58.1195
i
471.44
76.9690
f
125.19
39.6626
5
8
272.45
58.5122
5
8
476.26
77.3617
4
127.68
40.0553
3
4
276.12
58.9049
f
481. 11
77-7544
i
130.19
40.4480
i
279.81
59.2976
I
485.98
78.1471
13
132.73
40.8407
19
283.53
59.6903
25
490.87
78.5398
4
J 35-3o
41.2334
4
287.27
60.0830
i
495-79
78.9325
J
137.89
41.6261
i
291.04
60.4757
1
4
500.74
79.3252
I
140.50
42.0188
3
8
294.83
60.8684
|
505. 7 1
79.7179
4
I43-I4
42.4115
i
298.65
61.2611
i
510.71
80.1105
f
145.80
42.8042
5
8
302.49
61.6538
5
8
51572
80.5033
f
148.49
43.1969
3
4
306.35
62.0465
3
4
520.77
80.8960
7
8
151.20
43-5896
7
8
310.24
62.4392
7
8
525-84
81.2887
14
153.94
43.9823
20
314.16
62.8319
26
530.93
81.6814
J
156.70
44-375°
1
8
318.10
63.2246
4
536.05
82.0741
J
15948
447677
1
4
322.06
63.6173
1
4
541.19
82.4668
1
162.30
45.1604
I
326.05
64.OIOO
1
546.35
82.8595
i
l6 5-i3
45-5531
4
330.06
64.4026
i
55 I -55
83.2522
1
167.99
45-9458
f
334-IO
64.7953
5.
8
556.76
83.6449
f
170.87
46.3385
3
4
338.16
65.1880
3
4
562.00
84.0376
i
173.78
46.7312
7
8
342.25
65.5807
7
8
567.27
84 4303
J 5 X
176.71
47.1239
21
346.36
65.9734
27
572.56
84.8230
i
179.67
47-5166
1
8
350-5°
66.3661
i
577.87
85-2157
i
182.65
47.9093
1
354-66
66.7588
1
4
583-21
85.6084
I
185.66
48.3020
I
8
358.84
67-i5 I 5
f
588.57
86.0011
4
188.69
48.6947
i
363.05
67.5442
J
593-96
86.3938
f
I9I-75
49.0874
1
367.28
67.9369
5
599.37
86.7865
I
194.83
49.4801
I
37L54
68.3296
1
604.81
87.1792
i
T97.93 1
49.8728
I
375.83
68.7223
i
610.27
87-5719
TQ2
TABLE I.— AREAS AND CIRCUMFERENXES OF CIRCLES FROM
i TO ioo (Continued)
Dia. Area
Circum.
Dia.
Area
Circum.
Dia
Area
Circum.
28
6I5-75
87.9646
34
907.02
106.814
40
1256.6
125.664
1
8
621.26
83-3573
1
8
914.61
107.207
1
1264.5
126.056
i 626.80
88.7500
1
4
921.32
107.600
1
4
1272.4
126.449
I 632.36
89.1427
928.06
107.992
1
i 1280.3
126.842
i 637.94
89-5354
1
934.82
108.385
1
•
1288.2
I2 7-235
f
I 643o5
89.9281
i
941.61
108.788
5
8
. 1296.2
127.627
3
4
649.18
90.3208
3
4
948.42
109.170
f
1304.2
128.020
7
8
656.84
90.7I35
i
955-25
109.563
1
1312.2
128.413
29
660.52
91.1062
35
962.11
109.956
41
!32o.3
128.805
1
666.23
91.4989
1
8
969.00
110.348
1
8
J 328.3
129.198
1
4
67I.96
9 1. 89 1 6
1
4
975-91
no. 741
1
4
!336-4
129.591
§
677.71
92.2843
3
8
982.84
III. 134
3
8
1344.5
129.993
i
683.49
92.6770
i
989.80
III.527
1
2
J352.7
^Z ^^
5
8
689.3O
93.0697
5
s
996.78
1 1 1 .9 1 9
5
8
1360.8
130.769
3
4
695-I.3
93.4624
3
4
1003.8
112. 312
3
4
1369.0
131. 161
s
7OO.98
93.8551
7
8
1010.8
112.705
7
8
!377.2
I 3 I -554
3°
706.86
94.2478
36
1017.9
113.097
42
i3 8 54
I 3 I -947
1
8
712.76
94.6405
i
1025.0
113.490
1
8
13937
132.340
1
4
718.69
95-033 2
1
4
1032. 1
113.883
1
4
1402.0
132.732
1
724.64
954259
3
8
1039.2
114.275
3
8
1410.3
I 33-i25
4
730.62
95.8186
1
2
1046.3
114.668
1
2
1418.6
I 33-5 l8
i
736.62
96.2113
f
IQ 53-5
115.061
f
1427.0
!33-9io
3
4
742.64
96.6040
3
4
1060.7
1 15-454
I
14354
J 34-303
7
8
748.69
96.9967
7.
8
1068.0
115.846
1
1443.8
134.696
3 1
754-77
97-3894
37
1075.2
116.239
43
1452.2
135.088
1
8
760.87
97.7821
1
8
1082.5
116.632
1
8
1460.7
135481
i
766.99
98.1748
i
1089.8
117.024
1
4
1469. 1
J 35-874
1
773-14
98.5675
3
8
1097. 1
117.417
I
1477.6
136.267
4
779-3 1
98.9602
1
2
1104.5
117.810
J
1486.2
136.659
1
785.51
99.3529
5
8
1111.8
118.202
5
8
1494.7
!37-o52
3
4
791-73
99.7456
3
4
1119.2
118.596
3
4
I 5°3'3
J 37 445
7
8
797.98
100.138
7
8
1126.7
118.988
I
1511.9
^37^37
b 2
804.25
100.531
38
1134.1
119.381
44
!52o-5
138.230
i
810.54
100.924
1
8
1141.6
H9-773
ft
1529.2
138.623
1
4
816.86
101.316
1
4
1149.1
120.166
1
4
1537-9
139-015
§
823.21
101.709
3
8
1156.6
120.559
|
1546.6
139.408
1
2
829.58
102.102
1
2
1 164.2
120.951
1
2
r 555*3
139.801
f
8 35-97
102.404
1
1171.7
121.344
5
8
1564.0
140.194
3.
4
842.3a
102.887
3
4
1179.3
121.737
3.
4
1572.8
140.586
i
848.83
103.280
1
1 186.9
122.129
7
8
1581.6
140.979
33
855.30
103.673
39
1 194.6
122.522
45
1590.4
141.372
§
861.79
104.065
i
1202.3
122.915
i
1599.3
141.764
1
4
868.31
104.458
1
4
1210.0
123.308
1
1608.2
142.157
1
874.85
104.851
§
1217.7
123.700
1
1 61 7.0
142.550
4
881.41
105.243
i
1225.4
124.093
1
2
1626.0
142.942
f
888.00
105.636
5
8
1233.2
124.486
i
1634.9
J43-335
3
4
894.62
106.029
4
1241.0
124.878
3
4
1643.9
143.728
8
901.26
T06.421
s
1248.8
125.271
8
1652.0
144. 121
193
TABLE I,
-AREAS AXD CIRCUMFEREN'CES OF CIRCLES FROM
i TO ioo {Continued)
Dia.
Area
Circum.
Dia.
Area
1 Circum.
Dia.
Area
1 Circum.
46
1 661. 9
144.513
52
2123.7
163.363
58
2642.1
182.212
i
1670.9
144.906
i
2133-9
163.756
1
8
2653-5
182.605
i
16S0.0
145.299
1
4
2144.2
164.143
1
4
2664.9
182.998
1
1689. 1
145.691
3
8
2154.5
164.541
3
8
2676.4
183.390
i
1698.2
146.084
i
2164.8
164.934
1
2
2687.8
*&3.7%3
f
1707.4
146.477
5
8
2I75-I
165.326
f
2699.3
184.176
1
1716.5
146.869
3
4
2185.4
165.719
3.
4
2710.9
184.569
7
5
I725-7
147.262
I
2195.8
166. 112
7.
8
2722.4
184.961
47
1734.9
T47-655
53
2206.2
166.504
59
2734.0
185.354
I
1744.2
148.048
1
2216.6
166.897
i
2745.6
185.747
i
1753-5
148.440
i
2227.0
167.290
i
2 757-2
186.139
§
1762.7
148.833
§
2237-5
167.683
I
2768.8
186.532
i
1772. 1
149.226
1
9
2248.0
168.07:;
1
2
2780.5
186.925
i
1781.4
149.618
5
8
2258.5
168.468
5
8
2792.2
^7-3*1
l
1790.8
150.011
3
4
2269.1
168.861
3
4
2803.9
187.710
1
1800. 1
150.404
7
8
2279.6
169.253
i
2815.7
188.103
48
1809.6
150.796
54
2290.2
169.646
60
2827.4
188.496
i
1819.0
151. 189
1
8
2300.8
170.039
i
2839.2
188.888
i
1828.5
l5l-5 8 2
1
4
23II-5
170.431
i
2851.O
189.281
3.
8
18379
I5I-975
|
2322.1
170.824
1
2862.9
1S9.674
\
iS47-5
1 5 2 -3 6 7
i
2332.8
171. 217
0-
2874.8
190.066
I
1857.0
152.760
5
8
2343o
171 .609
5
8
2886.6
190.459
3
4
1866.5
J 53-i53
3
4
2 354-3
172.002
3
4
2898.6
190.852
I
1876.1
153-544
1
2365.0
172.395
i
2910.5
191.244
49
1885.7
I53-938
55
2375-8
172.788
61
2922.5
191.637
1
1895.4
154-33*
i
2386.6
173.180
I
2934o
192.030
1
4
1905.0
154.723
1
4
2397-5
J 73-573
1
4
2946.5
192.423
§
1914.7
i55- 116
1
2408.3
173.966
1
2958.5
192.815
i
1924.2
I55-509
i
2419.2
I74-358
i
2970.6
193.208
i
1934.2
I55-904
f
2430.1
i74.75i
5
8
2982.7
193.601
i
1943-9
156.294
3
4
2441. 1
I75-I44
f
2994.8
193-993
7
8
J 953-7
156.687
7
2452.0
I75-536
I
3006.9
194.3S6
50
1963-5
157.080
56
2463.0
I75-929
62
3019. 1
194-779
1
1973-3
I57-472
i
2474.0
176.322
i
303I-3
I95-I7I
1
4
1983.2
157.865
i
2485.0
176.715
1
4
3°43-5
I95-564
1
I993-I
158.258
1
2496.1
177.107
1
3°55-7
195-957
1
2003.0
158.650
1
2507.2
177.500
i
3068.0
196.350
f
2012.9
159.043
5
8
2518.3
I77-893
1
3080.3
196.742
f
2022.8
159.436
J
2529-4
178.285
1
4
3092.6
197-135
1
2032.8
159.829
7
8
2540.6
178.678
i
3 io 4-9
197.528
51
2042.8
160.221
57
2551.8
179.071
63
3117.2
197.920
i
2052.8
160.614
i
2563.0
179.463
i
3129.6
198.313
i
2062.9
161.007
1
4
2574.2
179.856
1
3142.0
198.706
I
2073.0
161.399
3
8
25854
180.249
3
8
3154.5
199.098
4
2083.1
161.792
i
2596.7
180.642
i
3166.9
199491
1
2093.2
162.185
8
2608.0
181.034
1
3!79-4
199.8S4
i
2103.3
162.577
3
4
2619.4
181.427
1
4
3I9I-9
200.277
I
2H3-5
162.970
I
2630.7
181.820
i
3204.4
200. 66q
194
TABLE I.— AREAS AND CIRCUMFERENCES
> OF
CIRCLES FROM
1 TO 100 (C
ontinued)
Dia.
Area
Circum.
Dia.
Area
Circum.
Dia.
Area
Circum.
64
3217.0
201.062
70
3848.5
219.911
7°~
4536.5
238.761
1
8
3229.6
201.455
1
8
3862.2
220.304
1
8
45514
239-I54
i
3242.2
201.847
1
4
3876.0
220.697
1
4
4566.4
239.546
1
3 2 54.8
202.240
3
8
3889.8
221.090
1
4581.3
239.939
4
3 26 7.5
202.633
1
2
3903-6
221.482
4
4596.3
240332
§
3280.1
203.025
5
8
3917-5
221.875
f
461 1. 4
240.725
I
3292.8
203.418
3
4
3931-4
222.268
3
4
4626.4
241. 117
i
33°5- 6
203.811
7
8
3945-3
222.660
7
8
4641.5
241.510
65
33i 8 -3
204.204
71
3959-2
223.053
77
4656.6
241.903
I
333 1 - 1
204.596
1
3973-1
223.446
1
8
4671.8
242.295
1
4
3343-9
204.989
1
4
3987.1
223.838
1
4
4686.9
242.688
1
335 6 -7
205.382
3
8
4001. 1
224.231
3
5
4702.1
243.081
1
5
3369.6
205.774
4
4015.2
224.624
1
47I7.3
243-473
1
3382.4
206.167
5
8
4029.2
225.017
5
8
4732.5
243.866
f
3395-3
206.560
3
4
4043-3
225.409
3
4
4747-8
244.259
1
3408.2
206.952
7
8
4057-4
225.802
7
8
4763.1
244.652
66
3421.2
207.345
72
4071.5
226.195
78
4778.4
245.044
1
8
3434-3
207.738
1
8
4085.7
226.587
i
4793-7
245-437
1
4
3447-2
208.131
1
4
4099 .8
226.930
1
4
4809.0
245.830
1
3460.2
208.523
3
8
4114.0
227.373
3
8
4824.4
246.222
4
3473-2
208.916
4
4128.2
227.765
1
4839.8
246.615
1
3486.3
209.309
5
8
4142.5
228.158
1
4855.2
247.008
1
3499-4
209.701
3
4
4156.8
228.551
3
4
4870.7
247.400
I
35 I2 -5
210.094
i
4171.1
228.944
i
4886.2
247-793
67
3525-7
210.487
13
4185.4
229.336
79
4901.7
248.186
J
3538.8
210.879
1
8
4199-7
229.729
1
8
4917.2
248.579
1
3552.o
211.272
1
4
4214. 1
230.122
i
4932.7
248.971
1
3565-2
211.665
3
8
4228.5
230.514
3
8
4948.3
249.364
J
3578.5
212.058
1
2
4242.9
230.907
1
2
4963.9
249-757
f
359L7
212.450
5
8
42574
231.300
5
8
4979-5
250.149
a
4
3605.0
212.843
3
4
4271.8
231.692
3
4
4995.2
250.542
1
3618.3
213.236
I
4286.3
232.085
7
8
5010.9
25 -935
68
3631.7
213.628
74
4300.8
232.478
80
5026.5
251-327
1
8
3645 -°
214.021
1
8
43I5-4
232.871
i
5042.3
251.720
i
3658.4
214.414
1
4
4329.9
233.263
1
• 4
5058.0
252.113
|
3671.8
214.806
|
4344-5
233-656
f
5073-8
252.506
4
3685.3
215.199
1
2
4359-2
234.049
i
5089.6
252.898
5
8
3698.7
2I5-592
5
8
4373-8
234.441
5
8
5I054
253.291
3
4
3712.2
215.984
3
4
4388.5
234-334
3
4
5121.2
253.684
1
37257
216.337
8
4403.1
235.227
i
5i37.i
254.076
69
3739-3
216.770
75
4417.9
235.619
81
5 J 53-o
254.469
i
3752.8
217.163
i
4432.6
236.012
i
5168.9
254.862
i
3766.4
217-555
1
4
4447.4
236.405
1
4
5184.9
255.254
i
3780.0
217.948
§
4462.2
236.798
3
8
5200.8
255.647
i
3793-7
218.341
•4
4477.0
237.190
i
5216.8
256.040
f
3807.3
218.733
5
8
4491.8
237.583
5
8
5232.8
256433
i
3821.0
219.126
3
4
4506.7
237.976
3
4
5248.9
256.825
I 3834.7
219.519
7
8
4521.5
238.368
i
5264.9
257.218
14
195
TABLE I.— AREAS AND CIRCUMFERENCES OF CIRCLES FROM
I TO ioo (Continued)
Did.
Area
82
5281.O
1
8
5 2 97-l
1
4
53I3-3
i
1
2
1
3
4
53294
5345-6
5361.8
5378.1
7
8
5394.3
83
1
8
5410.6
5426.9
i
f
5443-3
5459-6
5476.0
5
8
I
5492.4
5508.8
*
5525-3
84
1
8
1
4
5541.8
5558.3
5574-8
i
i
I
3_
4
1
5591-4
5607.9
5624.5
5641.2
5657.8
85
1
5
5674.5
5691.2
i
5707.9
i
5724-7
i
f
574L5
5758.3
4
5775-1
8
579L9
86
i
I
1
8
5808.8
5825.7
5842.6
5859.6
5876.5
5893.5
3_
4
5910.6
8
5927.6
87
i
1
4
1
i
a
4
5944-7
5961.8
5978.9
5996.0
6013.2
6030.4
6047.6
6064.9
Circum.
JDia.
Area
Circum.
|Dia.
Area
257.611
88
6082.1
276.460
94
6939.8
258.OO3
1
8
6099.4
276.853
i
6958.2
258.396
i
6256.7
277.846
i
6976.7
258.789
1
6134.1
277.638
3
8
6995.3
259.181
J
61514
278.031
i
7013.8
259.574
5
8
6168.8
278.424
5
8
7032.4
259.967
i
6186.2
278.816
I
7051.0
260.359
i
6203.7
279.209
I
7069.6
260.752
89
6221. 1
279.602
95
7088.2
261.145
i
6238.6
279.994
i
7106.9
261.538
1
4
6256.1
280.387
1
4
7125.6
261.930
f
62737
280.780
1
7M4.3
262.323
i
6291.2
281.173
1
2
7163.0
262.716
f
6308.8
281.565
5
8
7181.8
263.103
3
4
6326.4
281.958
3
4
7200.6
263.501
7
8
6344.1
282.351
i
7219.4
263.894
90
6361.7
282.743
96
7238.2
264.286
i
63794
283.136
i
7 2 57.i
264.679
1
4
6307.1
283.529
1
4
7276.0
265.072
3
8
6414.9
283.921
3
8
7294.9
265.465
1
6432.6
284.314
1
7313-8
265.857
5
8
6450.4
284.707
I
7332.8
266.250
3
4
6468.2
285.100
3
4
7351-8
266.643
i
6486.O
285.492
7
8
7370.8
267.035
91
6503.9
285.885
97
7389.8
267.428
i
6521.8
286.278
i
7408.9
267.821
1
4
65397
286.670
i
7428.0
268.213
1
6557.6
287.063
3
8
7447-1
268.606
i
6575.5
287.456
1
2
7466.2
268.999
5
8
6593.5
-287.848
5
8
7485.3
269.392
3
4
66 1 1. 5
288.241
3
4
7504.5
269.784
i
6629.6
288.634
8
75237
270.177
92
6647.6
289.027
98
7543.0
270.570
i
666s. 7
289.419
1
8
7562.2
270.962
1
4
6683.8
289.812
1
4
7581.5
27L355
3
s
6701.9
290.205
3
8
7600.8
271.748
1
6720.1
290.597
1
2"
762O. T
272.140
5
8
6738.2
290.990
5
8
7639.5
272.533
3
4
6756.4
291.383
3
4
7658.9
272.926
1
6774.7
291.775
I
8
7678.3
2 73-3 I 9
93
6792.9
292.168
99
7697.7
273.711
i
681 1. 2
292.561
i
77I7.I
274.104
i
6829.5
292.954
i
7736.6
274.497
3
8
6847.8
293.346
3
8
7756.1
274.889
1
2
6866.1
293-739
I
7775-6
275.282
8
6884.5
294.132
5
8
7795-2
275.675
3
4
6902.9
294.524
3
4
7814.8
276.067
7
8
6921.3
294.917
7
8
78344
196
TABLE II.— DECIMAL EQUIVALENTS OF FRACTIONS OF AN
INCH. (ADVANCING BY 8THS, 16THS, 32NDS AND
64THS.)
8ths
3-
>nds
64ths
64ths
I = - T2 5
1 _
.32 —
•°3 I2 5
A = .015625
If = .515625
i = -250
3
32 —
•°9375
A = .046875
ff = .546875
1 = -375
32 =
.15625
A = .°7 8 i25
H - -578125
1 = .500
3 2 =
.21875
A = - io 9375
If = .609375
f - .625
9 _
3 2 —
.28125
A = -140625
tt = .640625
i = -75o
11
3 2 —
•34375
*i - -171875
H = .671875
i - .875
13
"3"2 —
.40625
« = -203125
It = .7°3 I2 5
15
32 —
46875
a = .234375
H = .734375
i6ths.
A = -0625
17 _
3 2 —
03125
a = .265625
f| = .765625
A - .1875
« =
•59375
H = .296875
« = .796875
A = -3I25
2 1 —
32 —
.65625
*i - -328125
II - .828125
A = -4375
2 3
32 —
•71875
If = .359375
H = .859375
A = 0625
25 _
32 —
.78125
If = .390625
II = .890625
tt - -6875
2 7
32 —
.84375
U = .421875
If = .921875
if - .8125
29
3 2 —
.90625
II = -453 I2 5
ft = .953125
x# = -9375
31
3 2 —
.96875
tt = 484375
If = .984375
197
TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS,
CIRCUMFERENCES AND CIRCULAR AREAS OF NOS.
FROM i TO 520
TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS,
CIRCUMFERENCES AND CIRCULAR AREAS OF NOS.
FROM i TO 520 {Continued)
No.
Square
Cube
Sq. Root
Cube Root
Circle
Circum.
Area
41
1681
68921
6.4031
3.4482
128.81
1320.25
42
1764
74088
6.4807
3.4760
I3I-95
138544
43
1849
795°7
6-5574
3-5°34
I35.09
1452.20
44
1936
85184
6.6332
3-53°3
138.23
1520.53
45
2025
91125
6.7082
3o569
141.37
I590.43
46
2116
9733 6
6.7823
3-5830
144-5 l
1661.90
47
2209
103823
6.8557
3.6088
147.65
1734-94
48
2304
1 10592
6.9282
3-6342
150.80
1809.56
49
2401
1 1 7649
7.0000
3.6-93
153.94
1885.74
5°
2500
125000
7.0711
3.6840
157.08
1963.50
51
2601
132651
7.1414
3.7084
160.22
2042.82
5 2
2704
140608
7.2111
3-7325
163.36
2123.72
53
280Q "
148877
7.2801
3-7563
166.50
2206.18
54
2916
157464
7-3485
3-7798
169.65
2290.22
55
3° 2 5
166375
7.4162
3.8030
172.79
2375.83
56
3 T 3 6
175616
74833
3-8259
175.93
2463.OI
57
3 2 49
185193
7-5498
3.8485
179.07
2551-76
58
3364
195112
7.6158
3.8709
182.21
2642.08
59
348i
205379
7.6811
3-893°
185.35
2733.97
60
3600
2 1 6000
7.7460
3-9!49
188.50
2827.43
61
3721
226981
7.8102
3-9365
191.64
2922.47
62
3 8 44
238328
7.8740
3-9579
194.78
3019.07
63
3969
250047
7-9373
3-9791
197.92
3II7.25
64
4096
262144
8.0000
4.0000
201.06
3216.99
65
4225
274625
8.0623
4.0207
204.20
33^-3*
66
435 6
287496
8.1240
4.0412
207.35
3421.19
67
4489
300763
8.1854
4.0615
210.49
3525-65
68
4624
3 J 443 2
8.2462
4.0817
213.63
3631.68
69
4761
328509
8.3066
4.1016
216.77
3739.28
70
4900
343000
8.3666
4-1213
219.91
384845
7i
5°4i
3579"
8.4261
4.1408
223.05
3059. T 9
72
5184
373248
8.4853
4.1602
226.19
4071.50
73
53 2 9
389017
8.5440
4.1793
229.34
4185.39
74
5476
405224
8.6023
4.1983
232.48
4300.84
75
5625
421875
8.6603
4.2172
235.62
4417.86
76
5776
438976
8.7178
4.2358
238.76
453646
77
5929
456533
8.775o
4-2543
241.90
4656.63
78
6084
474552
8.8318
4.2727
245.04
4778.36
79
6241
493°39
8.8882
4.2908
248.19
4001.67
80
6400
512000
8.9443
4.3089
251.33
5o 2 6.55
199
TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS,
CIRCUMFERE.XCES AND CIRCULAR AREAS OF NOS.
FROM i TO 520 (Continued)
ClR^T T
No.
Square
Cube
Sq. Root
Cube Root
Circum.
Area
8i
6561
53 T 44i
9 .OOOO
4-3267
254-47
5 J 53-oo
82
6724
551368
9-0554
4-3445
257.61
5281.02
83
6889
571787
9.1104
4.3621
260.75
5410.61
84
7056
592704
9.1652
4-3795
263.89
5541-77
85
7225
614125
9-2195
4.3968
267.04
5674.50
86
7396
636056
9.2736
4.4140
270.18
5808.80
87
7569
658503
9-3274
4.4310
273-32
5944-68
88
7744
681472
9.3808
4.4480
276.46
6082.12
89
7921
704969
9.4340
4.4647
279.60
6221. 14
90
8100
729000
9.4868
44S14
282.74
6361.73
9i
8281
753571
9-5394
4-4979
285.88
6503.88
92
8464
778688
9-5917
4-5*44
289.03
6647.61
93
8649
8o4357
9-6437
4.5307
292.17
6792.91
94
8836
830584
9.6954
4.5468
295.3I
6939.78
95
9025
857375
9.7468
4.5629
298.45
7088.22
96
Q2l6
884736
9.7980
4.57 8 9
301.59
7238.23
97
9409
912673
9.8489
4-5947
304.73
7389.81
98
9604
941192
9.8995
4.6104
307.88
7542.96
99
9801
970299
9.9499
4.6261
311.02
7697.69
100
I OOOO
I 000000
10.0000
4.6416
314.16
7853.98
IOI
I020I
1030301
10.0499
4.6570
3*7-30
8011.85
102
IO404
1061208
10.0995
4-6723
320.44
8171.28
103
I0609
1092727
10.1489
4.6875
3 2 3-58
8332.29
104
I0816
1 1 24864
10.1980
4.7027
3 26 -73
8494.87
105
IIO25
II57625
10.2470
4.7177
329.87
8659.OI
106
II236
1191016
10.2956
4.7326
333-0 1
8824.73
107
1 1 449
1225043
10.3441
4-7475
336.15
8992.02
108
1 1 664
1259712
10.3923
4.7622
339-29
9160.88
109
11881
1295029
10.4403
4.7769
342.43
933I-32
no
12100
1331000
10.4881
4.7914
345-58
9503-32
III
12321
1367631
io.5357
4.8059
348.72
9676.89
112
12544
1404928
10.5830
4.8203
351.86
9852.03
113
12769
1442897
10.6301
4.8346
35 5 -oo
IOO28.7
TI4
12996
I 48 1 544
10.6771
4.8488
358.14
IO207.O
"5
13225
1520875
10.7238
4.8629
361.28
IO386.9
Il6
I345 6
1560896
10.7703
4.8770
364.42
IO568.3
117
13689
1601613
10.8167
4.8910
367o7
I075I-3
Il8
13924
1643032
10.8628
4.9049
37o.7i
10935-9
119
14161
1 685 1 59
10.0087
4.Q187
373-85
III22.0
I20
14400
1728000
10.9545
4.9324
376.99
I I309.7
200
TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS,
CIRCUMFERENXES AXD CIRCULAR AREAS OF NOS.
FROM i TO 520 (Continued)
ClRC'i -V.
No.
Square
Cube
Sq. Root
Cube Root
Circum.
Area
121
14641
1771561
1 1 .OOOO
4.9461
380.13
1 1 499.O
122
14884
1815848
11.0454
4-9597
383.27
1 1 689 .9
123
15129
1860867
II.0905
4.9732
386.42
11882.3
124
15376
1906624
H-I355
4.9866
38906
12076.3
125
15625
i953 I2 5
11. 1803
5 .OOOO
392.70
12271.8
126
15876
2000376
11.2250
5-°i33
395.84
12469.O
127
16129
2048383
11.2694
5.0265
398.98
12667.7
128
16384
2097152
"•3137
5-°397
402.12
12868.O
129
1 664 1
2146689
11.3578
5.0528
405.27
13069.8
13°
16900
2197000
1 1. 4018
5.0658
408.41
13273.2
131
17161
2248091
n-4455
5.0788
411-55
13478.2
132
17424
2299968
1 1. 489 1
5.0916
414.69
13684.8
133
17689
235 26 37
11.5326
5.1045
417.83
13892.9
134
I795 6
2406104
11.5758
5-1172
420.97
14102.6
135
18225
2460375
11. 6190
5.1299
424.12
I43I3.9
136
18496
2515456
1 1. 6619
5.1426
427.26
14526.7
137
18769
2571353
11.7047
5-I55 1
430.40
14741.I
138
19044
2628072
11-7473
5.1676
433-54
I4957-I
139
19321
2685619
11.7898
5.1801
436.68
15*74-7
140
19600
2744000
11.8322
5- I 9 2 5
439.82
*5393&
141
19881
2803221
11.8743
5.2048
442.96
15614-5
142
20164
2863288
1 1. 9 1 64
5.2171
446.II
15836.8
143
20449
2924207
n.9583
5.2293
449.2 5
16060.6
144
20736
2985984
12.0000
5.2415
452.39
16286.0
J 45
21025
3048625
12.0416
5-2536
455-53
1 6513.0
146
21316
3112136
12.0830
5.2656
458.67
16741.5
147
21609
3176523
12.1244
5.2776
461.81
16971.7
148
21904
3241792
12.1655
5.2896
464.96
17203.4
149
22201
3307949
12.2066
5.3015
468.IO
17436.6
15°
22500
3375°°°
12.2474
5.3133
471.24
I767I-5
151
22801
344295 1
12.2882
5.3251
474.38
17907.9
152
23104
35 1 1808
12.3288
5.3368
477.52
18145.8
153
23409
358i577
12.3693
5.3485
480.66
18385.4
154
23716
3652264
12.4097
5.3601
483.81
18626.5
155
24025
372387=:
12.4499
5.3717
486.95
18869.2
156
24336
3796416
12.4900
5.3832
490.09
191134
157
24649
3869893
12.5300
5-3947
493-23
19359.3
158
24964
3944312
12.5698
5.4061
496.37
19606.7
159
25281
4019679
I2.0OQ5
5.4175
499 -5 1
19855.7
160
25600
4096000
I2.649I
54288
502.65
20106.2
TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS,
CIRCUMFERENCES AND CIRCULAR AREAS OF NOS.
FROM i TO 520 (Continued)
No.
Square
Cube
Sq. Root
Cube Root
Circle
Circum.
Area
161
25921
4173281
12.6886
5.4401
505.80
20358.3
162
26244
4251528
12.7279
5-45 J 4
508.94
20612.0
163
26^69
433°747
12.7671
5.4626
512 .08
20867.2
164
26896
4410944
12.8062
5-4737
515.22
21124.1
165
27225
4492125
I2.S452
5.4848
518.36
213S2.5
166
27556
4574296
12.8841
5-4959
521.50
21642.4
167
27889
4657463
12.9228
5-5069
524.65
21904.0
168
2G224
4741632
12.9615
5.5178
527-79
22167. 1
169
28561
4826809
1 3 .0000
5.5288
530.93
22431.8
170
28900
4913000
13.0384
5-5397
534.07
22698.O
171
29241
500021 T
13.0767
5-5505
537-21
22965.8
172
29584
5088448
13.1149
5-56i3
540.35
23235.2
173
29929
5!777 J 7
*3-*S*9
5-5721
543-5°
23506.2
174
30276
5268024
13.1909
5.5828
546.64
23778.7
175
30625
5359375
13.2288
5-5934
549-78
24052.8
176
30976
5451776
13.2665
5.6041
552.92
24328.5
177
3*3 2 9
5545233
13-3041
5.6i47
556.o6
24605.7
178
31684
5639752
I3-34I7
5.6252
559-20
24884.6
179
32041
5735339
^3-3791
5.6357
562.35
25164.9
180
32400
5832000
13.4164
5.6462
56549
25446.9
181
32761
5929741
I3-4536
5.6567
568.63
25730.4
182
33 I2 4
6028568
13.4907
5.6671
57L77
26015.5
183
33489
6128487
I3-5277
5.6774
574.91
26302.2
184
33856
6229504
I3-5647
5.6877
578.05
26^90.4
185
34225
6331625
13.6015
5.6980
581.19
26880.3
186
34596
6434856
13.6382
5.7083
584.34
27171.6
187
34969
6539203
13.6748
57185
587.48
27464.6
188
35344
6644672
i3-7ii3
5.7287
590.62
27759.1
189
35721
6751269
13-7477
57388
59376
28055.2
190
36100
6859000
15.7840
5-7489
596.90
28352.9
191
36481
6967871
13.8203
5.7590
600.04
28652.I
192
36864
7077888
13.8564
5.7690
603.19
28952.9
193
37 2 49
7189057
13.8924
5.7790
606.33
29255.3
194
37636
73 OI 384
13.9284
5.7890
609.47
29559.2
195
38025
7414875
13.9642
57989
612.61
29864.8
196
38416
7529536
14.0000
5.8088
6i5.75
30171.9
197
38809
7645373
14.0357
5.8186
618.89
30480.5
198
39204
7762392
14.0712
5.8285
622.04
30790.7
199
39601
7880599
14.1067
5-8383
625.18
31102.6
200
40000
8000000
14.1421
5.8480
628.32
3I4I5.9
TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS,
CIRCUMFERENCES AND CIRCULAR AREAS OF NOS.
FROM i TO 520 (Continued)
ClI?^T TT
No.
Square
Cube
Sq. Root
Cube Root
Circum.
Area
201
40401
8120601
14.1774
5.8578
631.46
31730.9
202
40804
8242408
14.2127
5.8675
634.60
320474
203
41209
8365427
14.2478
5-877I
637.74
32365.5
204
41616
8489664
14.2829
5.8868
640.89
32685.1
205
42025
8615125
14.3178
5.8964
644.03
33006.4
206
42436
8741816
I4.3527
5-9059
647.17
33329.2
207
42849
8869743
14.3875
5-9*55
650.31
33653-5
208
43 2 64
8998912
14.4222
5-9250
653-45
33979-5
209
43681
9129329
14.4568
5-9345
656.59
34307.0
210
44100
9261000
14.4914
5-9439
659-73
34636.1
211
44521
939393 1
14.5258
5-9533
662.88
34966.7
212
44944
9528128
14.5602
5.9627
666.02
35298.9
213
45369
9663597
14.5945
5.9721
669.16
35632.7
214
45796
9800344
14.6287
5.9814
672.30
35968.1
215
46225
9938375
14.6629
5.9907
675-44
36305.O
216
46656
10077696
14.6969
6.0000
678.58
36643.5
217
47089
10218313
14.7309
6.0092
681.73
36983.6
218
47524
10360232
14.7648
6.0185
684.87
37325-3
219
47961
io 5°3459
14.7986
6.0277
688.01
37668.5
220
48400
10648000
14.8324
6.0368
691.15
38013.3
221
48841
10793861
14.8661
6.0459
694.29
38359.6
222
49284
1 094 1 048
14.8997
6.0550
69743
38707.6
223
49729
11089567
14.9332
6.0641
700.58
39057.1
224
50176
11239424
14.9666
6.0732
703.72
39408.I
225
50625
1 1390625
15.OOOO
6.0822
706.86
39760.8
226
51076
11543*76
*5-°333
6.0912
710.OO
40115.O
227
5^29
1 1 697083
15.0665
6.1002
7 I 3-*4
40470.8
228
5!9 8 4
11852352
15.0997
6.1091
716.28
40828.I
229
52441
12008989
I 5-i3 2 7
6.1180
719.42
41187.1
230
52900
1 2 167000
15.1658
6.1269
722.57
41547.6
231
5336i
12326391
15.1987
6.1358
725-7I
41909.6
232
53824
12487168
i5- 2 3 I 5
6.1446
728.85
42273.3
233
54289
12649337
15.2643
6.1534
731-99
42638.5
234
54756
12812904
15.2971
6.1622
735-13
43005.3
235
55225
12977875
I5-3297
6.1710
738.27
43373-6
236
55696
13144256
15-3623
6.1797
741.42
43743-5
237
56169
I 33 I2 o53
I5-3948
6.1885
744.56
44115.O
238
56644
13481272
15.4272
6.1972
747.70
44488.I
239
57i2i
13651919
I5-4596
6.2058
750.84
44862.7
240
57600
13824000
15.4919
6.2145
753-98
45238.9
TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS,
CIRCUMFERENCES AND CIRCULAR AREAS OF NOS.
FROM i TO 520 {Continued)
No.
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
Square
58564
59049
59536
60025
60516
61009
61504
62001
62500
63001
635 4
64009
64516
65025
65536
66049
66564
67081
67600
68l 2 1
68644
69169
69696
70225
70756
71289
71824
72361
729OO
73441
73984
745 2 9
75076
75625
76176
76729
77284
7784I
784OO
Cube
3997521
4172488
4348907
4526784
4706125
4886936
5069223
5252992
543 82 49
5625000
5813251
6003008
6194277
6387064
6581375
6777216
6974593
7I735I2
7373979
7576000
777958i
7984728
8191447
8399744
8609625
8821096
9034163
9248832
9465109
9683000
19902511
20123648
20346417
20570824
20796875
21024576
21253933
21484952
2 1 717639
21952000
Sq. Root
Cube Root
15.5242
6.2231
I 5-55 6 3
6.2317
15-5885
6.2403
15.6205
6.2488
15.6525
6.2573
15.6844
6.2658
15.7162
6.2743
15.7480
6.2828
15.7797
6.2912
15,8114
6.2996
15.8430
6.3080
15.8745
6.3164
15.9060
6.3247
15.9374
^•333°
15.9687
6.3413
16.0000
6.3496
16.0312
6.3579
16.0624
6.3661
16.0935
6-3743
16.1245
6.3825
16.1555
6.3907
16.1864
6.3988
16.2173
6.4070
16.2481
6.4151
16.2788
6.4232
16.3095
6.4312
16.3401
6.4393
16.3707
6.4473
16.4012
6-4553
16.4317
6.4633
16.4621
6.47*3
16.4924
6.4792
16.5227
6.4872
16.5529
6.4951
16.5831
6.5030
16.6132
6.5108
16.6433
6.5187
16.6733
6.5265
16.7033
6-5343
16.7332
6.5421
Circle
Circum. I
Area
757-12
760.27
76341
766.55
769.69
772.83
775-97
779.12
782.26
785.40
788.54
791.68
794.82
797.96
801. 11
804.25
807.39
810.53
813.67
816.81
819.96
823.10
826.24
829.38
832.52
835.66
838.81
841.95
845.09
848.23
851.37
854.51
857.66
860.80
863.94
867.08
870.22
873-36
876.50
879.65
45616.7
45996.1
46377.0
46759.5
47M3.5
-47529.2
47916.4
48305.1
48695.5
49087.4
49480.9
49875.9
50272.6
50670.7
51070.5
5I47L9
51874.8
52279.2
52685.3
53092.9
53502.1
53912.9
54325-2
54739-1
55J54.6
55571.6
55990.3
56410.4
56832.2
57255.5
57680.4
58106.9
58534.9
58964.6
59395-7
59828.5
60262.8
60698.7
61 136.2
6i575. 2
204
.TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS,
CIRCUMFERENCES AND CIRCULAR AREAS OF XOS.
FROM i TO 520 (Continued)
No.
Square
Cube
Sq. Root
Cube Root
Circle
Circum.
Area
281
78961
22188041
16.7631
6.5499
882.79
62015.8
282
795 2 4
22425768
16.7929
6-5577
885.93
62458.0
283
80089
22665187
16.8226
6.5654
889.07
62901.8
284
80656
22906304
16.8523
6.5731
892.21
63347.1
285
81225
23149125
16.8819
6.5808
895-35
63794.O
286
81796
23393656
16.9115
6.5885
898.50
64242.4
287
82369
23639903
16.9411
6.5962
901.64
64692.5
288
82944
23887872
16.9706
6.6039
904.78
65144. 1
289
83521
24137569
17.0000
6.6115
907.92
65597.2
290
84100
24389000
17.0294
6.6191
911.06
66052.O
291
84681
24642171
17.0587
6.6267
914.20
66508.3
292
85264
24897088
17.0880
6.6343
9^7-35
66966.2
293
85849
25153757
17.1172
6.6419
920.49
67425.6
294
86436
25412184
17.1464
6.6494
923.63
67886.7
295
87025
25672375
17.1756
6.6569
926.77
68349-3
296
87616
25934336
17.2047
6.6644
929.91
68813.5
297
88209
26198073
i7- 2 337
6.6719
933-05
69279.2
298
88804
26463592
17.2627
6.6794
936.19
69746.5
299
89401
26730899
17.2916
6.6869
939-34
70215.4
300
90000
27000000
17-3205
6.6943
942.48
70685.8
301
90601
27270901
17-3494
6.7018
945.62
7II57-9
302
91204
27543608
17.3781
6.7092
948.76
71631.5
3°3
91809
27818127
17.4069
6.7166
951.90
72106.6
3°4
92416
28094464
I7-4356
6.7240
955-°4
725834
3°5
93° 2 5
28372625
17.4642
6.73*3
958.19
73061.7
306
93636
28652616
17.4929
6.7387
•961.33
7354L5
307
94249
28934443
17-5214
6.7460
964.47
74023.O
308
94864
29218112
17-5499
6.7533
967.61
74506.O
3°9
9548i
29503629
I7-5784
6.7606
97o.75
74990.6
310
96100
29791000
17.6068
6.7679
973-89
75476.8
3 11
96721
30080231
17-6352
6.7752
977.04
75964.5
312
97344
3 37I328
17.6635
6.7824
980.18
76453.8
3*3
97969
30664297
17.6918
6.7897
983-32
769447
3M
98596
30959144
17.7200
6.7969
986.46
77437-1
3^5
99225
31255875
17.7482
6.8041
989.60
77931. 1
316
99856
31554496
17.7764
6.8113
992.74
78426.7
3 X 7
100489
31855013
17.8045
6.8185
995.88
78923.9
318
101124
32157432
17.8^26
6.8256
999.03
79422.6
3 X 9
101761
32461759
17.8606
6.8328
1002.20
79922.9
320
102400
32768000
17.8885
6.8399
1005.30
80424.8
205
TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS,
CIRCUMFERENCES AND CIRCULAR AREAS OF NOS.
FROM i TO 520 (Continued)
No.
Square
Cube
Sq. Root
Cube Root
Circle
Circum.
Area
321
IO3041
33076161
17.9165
6.8470
IO08.5
80928.2
322
103684
33386248
17.9444
6.8541
IOU.6
81433.2
3 2 3
IO4329
33698267
17.9722
6.8612
IOI4.7
81939.8
3 2 4
104976
34012224
18.0000
6.8683
1017.9
82448.O
3 2 S
105625
3432S125
18.0278
6.8753
1 02I.0
829577
326
106276
34645976
18.0555
6.8824
1024.2
83469.O
3 2 7
106929
349657 8 3
18.0831
6.8894
1027.3
83981.8
328
107584
35287552
18.1108
6.8964 m
IO30.4
84496.3
3 2 9
108241
35611289
18.1384
6.9034 '
1033.6
85012.3
33°
108900
35937000
18.1659
6.9104
1036.7
85529.9
33 1
109561
36264691
18.1934
6.9174
1039.9
86049.O
33 2
IIO224
36594368
18.2209
6.9244
IO43.O
86569.7
333
I 10889
36926037
18.2483
6.9313
1046.2
87092.O
334
III556
37259704
18.2757
6.9382
1049.3
87615.9
335
112225
37595375
18.3030
6.9451
1052.4
88141.3
336
112896
37933 56
18.3303
6.9521
I055.6
88668.3
337
II3569
38272753
18.3576
6.9^89
IO58.7
89196.9
33^
I 14244
38614472
18.3848
6.9658
1 06 1. 9
89727.O
339
114921
38958219
18.4120
6.9727
106^.0
90258.7
340
I I 5600
39304000
18.4391
6.9795
1068. 1
90792.O
34i
116281
39651821
18.4662
6.9864
1071.3
91326.9
342
1 1 6964
40001688
18.4932
6.9932
1074.4
91863.3
343
1 1 7649
40353 6o 7
18.5203
7 .0000
1077.6
92401.3
344
118336
40707584
18.5472
7.0068
1080.7
92940.9
345
II9025
41063625
18.5742
7.0136
1083.8
93482.O
346
119716
41421736
18.6011
7.0203
I087.O
940247
347
120409
41781923
18.6279
7.0271
1090. 1
94569.0
348
121 IO4
42144192
18.6S48
7.0338
I093.3
95114.9
349
I2l8oi
42508549
18.6815
7.0406
1096.4
95662.3
35o
I225OO
42875000
18.7083
7.0473
1099.6
9621 1.3
35i
I232OI
43243551
I8.7350
7.0540
IIO2.7
96761.8
352
I23904
43614208
18.7617
7.0607
1105,8
97314.O
353
I24609
43986977
18.7883
7.0674
1 109.0
97867.7
354
I25316
44361864
18.8149
7.0740
III2.I
98423.O
355
I20025
44738875
18.8414
7.0807
III5.3
98979.8
356
I26736
45118016
18.8680
7-0873
II18.4
99538.2
357
I27449
45499293
18.8944
7.0940
II2I.5
IOOO98
358
128164
45882712
18.9209
7.1006
II24.7
IO0660
359
I2888l
46268279
18.9473
7.1072
1 127.8
IOI223
360
I2960O
46656000
18.9737
7.1138
II3I.O
101788
206
TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS,
CIRCUMFERENCES AND CIRCULAR AREAS OF NOS.
FROM i TO 520 (Continued)
No.
Square
Cube
Sq. Root
Cube Root
Circle
Circum.
Area
361
130321
47045881
19.OOOO
7.1204
II34.I
IO2354
362
131044
47437928
19.0263
7.1269
H37-3
IO2922
3 6 3
131769
47832147
19.0^26
7-1335
1 140.4
IO3491
364
132496
48228544
19.0788
7.1400
II43.5
IO4062
365
133225
48627125
19.1050
7.1466
1146.7
IO4635
366
133956
49027896
19.1311
7-I53I
1149.8
IO5209
3 6 7
134689
49430863
19.1572
7-I596
H53-0
105785
368
135424
49836032
19-1833
7.1661
1156.1
106362
369
136161
50243409
19.2094
7.1726
1159.2
106941
37°
136900
50653000
I9-2354
7.1791
1 1 62 .4
IO7521
37i
137641
51064811
19.2614
7.1855
H65.5
108103
372
138384
51478848
19.2873
7.1920
1168.7
108687
373
139129
51895117
19.3132
7.1984
1171.8
IO9272
374
139876
52313624
19-3391
7.2048
I175.O
IO9858
375
140625
52734375
19.3649
7.2112
II78.I
I IO447
376
I4I376
53157376
19.3907
7.2177
Il8l.2
IIIO36
377
142129
53582633
19.4165
7.2240
1184.4
IH628
378
142884
54010152
19.4422
7.2304
1187.5
II222I
379
143641
54439939
19.4679
7.2368
H90.7
II28I5
380
144400
54872000
19.4936
7.2432
1193.8
II34II
381
145161
553 634i
19.5192
7.2495
II96.9
I I 4OO9
382
145924
55742968
19.5448
7.2558
1200. 1
I I4608
3^3
146689
56181887
19.5704
7.2622
1203.2
II5209
384
147456
56623104
19-5959
7.2685
1206.4
I I 581 2
385
148225
57066625
19.6214
7.2 748
1209.5
H6416
386
148996
57512456
19.6469
7.2811
1212.7
II702I
3^7
149769
57960603
19.6723
7.2874
1215.8
1 1 7628
388
I5°544
58411072
19.6977
7.2936
1218.9
H8237
389
1513 21
58863869
19.7231
7.2999
1222. 1
I 18847
390
152100
59319000
19.7484
7.3061
1225.2
I 19459
39i
152881
59776471
19-7737
7.3124
1228.4
120072
392
153664
60236288
19.7990
7.3186
I23L5
I20687
393
154449
60698457
19.8242
7.3248
I234.6
I21304
394
155236
61162984
19.8494
7-33 10
I237.8
I2I922
395
156025
61629875
19.8746
7-337 2
I24O.9
122542
396
156816
62099136
19.8997
7-3434
I244.I
I23163
397
157609
62570773
19.9249
7.3496
1247.2
I23786
398
158404
63044792
19.9499
7.3558
I25O.4
I244IO
399
159201
63521199
19.9750
7.3619
I253v5
I25036
400
1 60000
64000000
20.0000
7.3684
I256.6
I25664
207
TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS,
CIRCUMFERENCES, AND CIRCULAR AREAS OF NOS.
FROM i TO 520 (Continued)
No.
Square
Cube
Sq. Root
Cube Root
Circle
Circum.
Area
401
I 6080 I
64481201
20.0250
7-3742
1259.8
126293
402
161604
64964808
20.0499
7-38o3
1262.9
126923
403
162409
65450827
20.0749
7.3864
1266. 1
127556
404
163216
65939264
20.0998
7.392 5
1269.2
128190
405
164025
66430125
20.1246
7.3986
1272.3
128825
406
164836
66923416
20.1494
74047
1275.5
129462
407
165649
67419143
20.1742
7.4108
1278.6
130100
408
166464
67917312
20.1990
7.4169
1281.8
130741
409
167281
68417929
20.2237
7.4229
1284.9
13*382
410
168100
68921000
20.2485
7.4290
1288. 1
132025
411
I 6892 I
69426531
20.2731
7.4350
1 291.2
132670
412
169744
69934528
20.2978
7.4410
1294.3
I333I7
413
170569
70444997
20.3224
7.4470
l297o
133965
414
171396
70957944
20.3470
74530
1300.6
134614
415
172225
7*473375
2o.37 J 5
74590
I303.8
135265
416
173056
71991296
20.3961
7.4650
1306.9
I359i8
417
173889
725H7I3
20.4206
7.4710
1310.0
136572
418
174724
73034632
20.4450
74770
I3I3-2
137228
419
I7556I
73560059
20.4695
7.4829
i3 l6 -3
137885
420
176400
74088000
20.4939
7.4889
13*9-5
138544
421
177241
74618461
20.5183
7.4948
1322.6
139205
422
178084
75151448
20.5426
7*5007
1325.8
139867
423
178929
75686967
20.5670
7.5o67
1328.9
14053 1
424
179776
76225024
20.5913
7.5126
1332.0
141196
425
180625
76765625
20.6155
7.5185
1335.2
141863
426
181476
77308776
20.6398
7.5 2 44
1338.3
14253*
42 7
182329
77854483
20.6640
7o30 2
1 341-5
143201
428
183184
78402752
20.6882
7o36i
1344.6
143872
429
184041
78953589
20.7123
7.5420
13477
144545
43°
184900
79507000
20.7364
7.5478
1350.9
145220
43i
185761
80062991
20.7605
7-5537
i354.o
145896
43 2
186624
80621568
20.7846
7-5595
1357.2
146574
433
187489
81182737
20.8087
7-5654
1360.3
147254
434
188356
81746504
20.8327
7-5712
1363.5
147934
435
189225
82312875
20.8567
7-577o
1366.6
148617
436
190096
82881856
20.8806
7.5828
1369.7
149301
437
190969
83453453
20.9045
7.5886
1372.9
149987
438
191844
84027672
20.9284
7-5944
1376.0
150674
439
192721
84604519
20.9523
7.6001
1379.2
i5!3 6 3
440
193600
85184000
20.9762
7.6059
1382.3
152053
208
TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS,
CIRCUMFERENCES AND CIRCULAR AREAS OF NOS.
FROM i TO 520 (Continued)
ClRrr.v.
No.
Square
Cube
Sq. Root
Cube Root
Circum.
Area
441
19448 I
85766121
2 I .OOOO
7.6117
I385-4
J 5 2 745
442
195364
86350888
21.O238
7.6174
1388.6
153439
443
196249
86938307
2I.O476
7.6232
I39I-7
154134
444
197136
87528384
2I.O713
7.6289
1394.9
154830
445
198025
88121125
2I.O950
7.6346
1398.O
155528
446
198916
88716536
2I.H87
7.6403
I40I.2
156228
447
199809
89314623
2I.I424
7.6460
1404.3
156930
448
200704
899I539 2
2I.l66o
7-65I7
1407.4
157633
449
201601
90518849
2I.1896
7.6574
1410.6
*5%337
45°
202500
91125000
21.2132
7.6631
I4I3-7
159043
45i
203401
9I73385I
21.2368
7.6688
I416.9
i5975i
45 2
204304
92345408
2I.2603
7.6744
1420.O
1 60460
453
205209
92959677
2I.2838
7.6801
I423.I
161171
454
206116
93576664
2I.3073
7.6857
1426.3
161883
455
207025
94196375
2I.3307
7.6914
1429.4
162597
45 6
207936
94818816
21.3542
7.6970
1432.6
163313
457
208849
95443993
2I.3776
7.7026
1435-7
164030
458
209764
96071912
2 1 .4OO9
7.7082
1438.9
164748
459
210681
96702579
2I.4243
7-7I38
1442.0
165468
460
2 1 1600
97336000
2I.4476
7-7 J 94
I445- 1
166190
461
212521
97972181
2I.4709
7.7250
1448.3
1 66^ 14
462
2 13444
98611128
21.4942
7.7306
145 1 -4
167639
463
214369
99252847
21.5174
7.7362
1454.6
168365
464
215296
99897344
2I.5407
7.7418
1457-7
169093
465
216225
100544625
2I.5639
7-7473
1460.8
169823
466
217156
101 194696
2I.587O
7-7529
1464.0
170554
467
218089
101847563
2I.6I02
7-7584
1467. 1
171287
468
219024
102503232
21.6333
7-7639
i47o-3
172021
469
219961
103161709
2I.6564
7-7695
1473-4
172757
470
220900
103823000
2I.6795
7-775o
1476.5
173494
47i
'22 1 841
104487111
2I.7025
7-7805
1479-7
174234
472
222784
105 1 54048
2I.7256
7.7860
1482.8
174974
473
223729
105823817
2I.7486
7-7915
1486.0
i757i6
474
224676
106496424
2I.7715
7.7970
1489. 1
176460
475
225625
107171875
21-7945
7.8025
1492.3
177205
476
226576
107850176
21.8174
7.8079
1495-4
177952
477
227529
Io8 53i333
21.8403
7-8x34
1498.5
178701
478
228484
109215352
21.8632
7.8188
1501.7
1 7945 1
479
229441
109902239
21.8861
7-8243
1504.8
180203
480
230400
1 10592000
2 1 .9089
7.8297
1508.0
180956
209
TABLE III.— SQUARES, CUBES, SQUARE ROOTS, CUBE ROOTS,
CIRCUMFERENCES AND CIRCULAR AREAS OF NOS.
FROM i TO 520 (Continued)
No.
Square
Cube
Sq. Root
Cube Root
Circle
Circum.
Area
481
231361
111284641
21.9317
7.8352
ISII.I
181711
482
232324
111980168
21-9545
7.8406
I5I4.3
182467
483
233289
112678587
21.9773
7.8460
I5I74
183225
484
234256
1 13379904
2 2 .OOOO
7.8514
1520.5
183984
485
235225
114084125
22.0227
7.8568
J523.7
184745
486
236196
114791256
22.0454
7.8622
1526.8
185508
487
237169
H550I303
2 2.o68l
7.8676
1530.0
186272
488
238144
116214272
22.O907
7.8730
I533-I
187038
489
239121
1 1 6930 1 69
22.1133
7.8784
1536.2
187805
490
240100
1 1 7649000
22.1359
7.8837
15394
188574
491
241081
118370771
22.1585
7.8891
1542.5
189345
492
242064
1 19095488
22.l8ll
7.8944
1545-7
190117
493
243049
119823157
22.2036
7.8998
1548.8
190890
494
244036
120553784
22.226l
7.905I
I55I-9
191665
495
245025
121287375
22.2486
7.9I05
I 555-i
192442
496
246016
122023936
22.2711
7.9I58
1558.2
193221
497
247009
122763473
22.2935
7.92 1 1
1561.4
194000
498
248004
T2 35 599 2
22.3159
7.9264
1564.5
194782
499
249001
124251499
22.3383
7.9317
1567-7
195565
500
250000
125000000
22.3607
7.9370
1570.8
196350
5oi
251001
125751501
22.383O
7-9423
1573-9
197136
502
252004
126506008
22.4054
7.9476
I577.I
197923
503
253009
127263527
22.4277
7-9528
1580.2
I987I3
5°4
254016
128024064
22.4499
7.958l
1583-4
199504
5o5
255 25
128787625
22.4722
7-9634
1586.5
200296
506
256036
i295542i6
22.4944
7.9686
1589-7
201090
507
257049
130323843
22.5167
7-9739
1592.8
201886
508
258064
131096512
22.5389
7-9791
1595-9
202683
5°9
259081
131872229
22.56lO
7-9843
i599.i
203482
5io
260100
1 3 265 1 000
22.5832
7.9896
1602.2
204282
5ii
261121
13343283 1
22.6053
7.9948
1605.4
205084
512
262144
134217728
22.6274
8.0000
1608.5
205887
5i3
263 1 69
135005697
22.6495
8.0052
1611.6
206692
5i4
264196
135796744
22.6716
8.0104
1614.8
207499
5i5
265225
136590875
22.6936
8.0156
1617.9
208307
5i6
266256
137388096
22.7156
8.0208
1621.1
2091 1 7
5i7
267289
138188413
22.7376
8.0260
1624.2
209928
5i8
268324
1 3899 1 83 2
22.7596
8.0311
1627.3
210741
5i9
269361
139798359
22.7816
8.0363
1630.5
2II556
520
270400
140608000
22.8035
8.0415
1633.6
212372
APPENDIX
211
TABLE IV
—FACTORS OF
EVAPORATION
Temper-
ature
of
Boiler gage pressures in pounds per square inch above
the atmosphere
feed-
water
5
10
15
20
25
30
35
40
45
° Fahr.
32
1. 187
1 .192
1. 195
1. 199
I. 201
1.204
1 .206
1 .209
1 .211
1. 212
35
1. 184
1. 189
1 . 192
1. 196
I. 198
1.20T
1.203
1 .206
1.208
1 .209
40
I.T79
1. 184
1. 187
1 .191
I. 193
1 .196
1. 198
1 .201
1.203
1.204
45
1. 173
1. 178
1. 181
1. 185
I. I87
1 .190
1 .192
1. 195
1. 197
1. 198
50
1. 168
1. 173
1. 177
1. 180
I. 182
1. 185
1. 187
1 . 190
1. 192
1. 193
55
1. 163
1. 168
1 .171
1. 175
I. 177
1. 180
1. 182
1. 185
1. 187
1. 188
6o
1. 158
1. 163
1. 166
1 . 170
I .172
1. 175
1. 177
1. 180
1. 182
1. 183
65
I -153
1. 158
1 . 161
1. 165
I. 167
1 .170
1. 172
1. 175
1. 177
1. 178
70
1. 148
1. 153
1. 156
1. 160
I .162
1. 165
1 .167
1. 170
1. 172
1. 173
75
1. 143
1. 148
1. 151
1. 155
I. 157
1. 160
1. 162
1. 165
1. 167
1. 168
8o
1. 137
1. 143
1 .146
1. 149
I .151
1. 154
1. 156
'1. 159
1 . 161
1. 162
85 .
1. 132
1. 137
1 . 140
1. 144
I .146
1. 149
1. 151
1. 154
1. 156
1. 157
90
1 .127
1 .132
1. 135
1. 139
I .141
1. 144
1 .146
1. 149
1. 151
1. 152
95
1 . 122
1 .127
1. 130
1. 134
I .136
1. 139
1 .141
1. 144
1 .146
1. 147
100
1 . 117
1 . 122
1 .125
1 .129
I. 131
1. 134
1. 136
1. 139
1 .141
1 .142
105
1 . in
1 . 117
1 .120
1 .123
I .125
1. 128
1 .130
1. 133
1. 135
1. 136
no
1 .106
1 . in
1 .114
1. 118
I .120
1 .123
1. 125
1. 128
1. 130
1. 131
ii5
1 . 101
1 .106
1. 109
1. 113
I. 115
1. 118
1 . 120
r .123
1. 125
1 .126
120
1 .096
1 . 101
1 .104
1. 108
i . no
1. 113
1. 115
1. 118
1 .120
1 .121
125
1 .091
1 .096
1.099
1. 103
1 .105
1. 108
1 . no
1 .113
1. 115
1 . 116
130
1.085
1. 091
1.094
1.097
1.099
1 . 102
1 .104
1 .107
1. 109
1 .110
135
1 .080
1.085
1.088
1 .092
1.094
1.097
1.099
1 .102
1 .104
1. 105
140
1.075
1 .080
1.083
1.087
1 .089
1 .092
1.094
1.097
1.099
1 .100
145
1 .070
1.075
1.078
1 .082
1.084
1.087
1 .089
1.092
1.094
1.095
150
1.065
1 .070
1-073
1.077
1.079
1.082
1 .084
1.087
1.089
1 .090
155
1.059
1.065
1.068
1 .071
1.073
1.076
1.078
1. 081
1.083
1 .084
l60
1.054
1.059
1 .062
1 .066
1.068
1 .071
1.073
1 .076
1.078
1.079
165
1.049
1.054
1.057
1. 061
1.063
1.066
1.068
1 .071
1.073
1.074
170
1.044
1.049
1.052
1.056
1.058
1. 061
1.063
1 .066
1.068
1 .069
175
1.039
1.044
1.047
1. 051
1.053
1.056
1.058
1 .061
1.063
1 .064
180
1.033
1.039
1 .042
1.045
1.047
1.050
1.052
1.055
1.057
1.058
185
1.028
1.033
1.036
1 .040
1.042
1.045
1.047
1.050
1.052
1.053
190
1.023
1.028
1. 03 1
1.035
1.037
1.040
1 .042
1.045
1.047
1 .048
195
1. 018
1.023
1.025
1.030
1.032
1.035
1.037
1 .040
1 .042
1.043
200
1 .013
1. 018
1. 021
1.025
1 .027
1.030
1 .032
1.03s
1.037
1.038
205
1 .008
1. 013
1. 015
1.020
1.022
1.025
1.027
1.030
1.032
1.033
210
1.008
1.008
I .Oil
I. 015
1. 017
1 .020
1.022
1.025
1.027
1.028
212
1.002
1.002
15
212
ARITHMETIC OF THE STEAM BOILER
TABLE IV— FACTORS OF EVAPORATION {Continued)
Temper-
ature
of
Boiler gage pressures in pounds per square inch above
the atmosphere
feed-
water
50
60 70 80
90 100 120 140
160 ]
80
° Fahr.
32
1. 214
1 .217 1 .219 1 .222
1 . 224 1 . 227 1 .231 1 .234
1.237 1
239
35
1 .211
1 .214 1 .216 1 .219
1 .221 1 .224 1 .228 1 .231
1.234 1
236
40
1 . 206
1 . 209 1 . 211 1 . 214
1. 216 1. 219 1. 2 23 1. 226
1 .229 1
231
45
1 .200
1 .203 1 .205 1 .208
1 .210 1 . 213 1 .217 1 .220
1.223 1
225
50
1. 195
1. 198 1 .200 1 .203
1 . 205 1 . 208 1 .212 1 . 215
1. 218 1
220
55
1 . 190
1. 193 1. 195 1. 198
1.200 1.203 1.207 1. 210
1. 213 1
215
60
1. 185
1 . 188 1 . 190 1 . 193
1. 195
1 . 198 1 .202 1 .205
1.208 1
210
65
1. 180
1. 183 1. 185 T.188
1 . 190
1 . 193 1 . 197 1 .200
1 .203 1
205
70
1. 175
1 . 178
1 . 180 1 . 183
1. 185
1 . 188 r . 192 1 . 195
1. 198 1
200
75
1 . 170
1. 173
1. 175 1. 178
1. 180
1 . 183 1 . 187 1 . 190
1. 193 1
195
80
1 . 164
1 . 167
1 . 169 1 . 172
1. 174
1. 177
1 . 181 1 . 184
1. 187 1
189
85
1 -159
1 . 162
1 . 164 1 . 167
1 . 169
1 . 172
1 . 176 1 . 179
1. 182 1
184
90
1 -154
1. 157
1 . 159 1 . 162
1 . 164
1 . 167
1 . 171 1 • 174
1. 177 1
179
95
1. 149
1 • 152
1. 154 1. 157
1. 159
1 . 162
1 . 166 1 . 169
1 . 172 1
174
100
1. 144
1. 147
1 . 149 1 . 152
1. 154
1. 157
1 . 161 1 . 164
1. 167 1
169
105
1. 138
1 . 141
1 . 143 1 . 146
1. 148
1.151
1. 155 1. 158
1. 161 1
163
no
1. 133
1. 136
1 . 138 1 . 141
1. 143
1 .146
1. 150 1. 153
1. 156 1
158
115
1. 128
I. 131
1. 133 1-136
1. 138
1 . 141
1. 145 1. 148
1. 151 1
153
120
1. 123
1 . 126
1 . 128 1 . 131
1. 133
1. 136
1. 140 1. 143
1 . 146 1
148
125
1. 118
1 . 121
1 . 123 1 . 126
1. 128
1. 131
1. 135 1. 138
1. 141 1
143
130
1 . 112
1. US
1 . 117; 1 . 120
1 . 122
1. 125
1 .129
1. 132
1. 135 1
137
135
1 . 107
1 . no
I . 112 I . 115
1. 117
1 . 120
1 .124
1 . 127
1. 130 1
132
140
1 . 102
1 .105
i . 107 i . no
1 . 112
1. US
1 . 119
1 . 122
1. 125 I
127
145
1.097
1 . 100
1 . 102
1. 105
1. 107
1 .110
1 . 114
1 .117
I . 120 I
122
150
1 .092
1.095
1.097
1 . 100
1 . 102
1. 105
1 .109 1 .112
1.1.31
117
155
1.086
1 .089
1 .091
1.094
1 .096
1.099
1 . 103 1 . 106
I . 109 I
in
160
1. 081
1.084
1.086
1 .089
1 .091
1.094
1 .098 1 . 101
I . 104 I
106
165
1 .076
1.079
1. 081
1 .084
1.086
1 .089
1 .093 1 -096
1.099 I
ior
170
1 .071
1.074
1 .076
1.079
1. 081
1 .084
1 .088 1 .091
1.094 1
096
175
1.066
r .069
1 .071
1.074
r .076
1.079
1.083 1.086
I.O89 I
091
180
1 .060
1.063
1.065
1.068
1 .070
1.073
1.077
1 .080
I.O83 I
085
185
1.055
1.058
1 .060
1 .063
1.065
1.068
1 .072
1.075
1.078 I
080
190
1.050
1.053
1.055
1.058
i.o6oji .063
r .067
1 .070
1.073 I
075
195
1.045
1.048
1.050
1-053
1.055 1-058
1 .062
1.065
1.068 I
070
200
1 .040
1.043
1.045
1.048
1.050 1.053 1057
1 .060
I.063 I
065
205
1.035
1.038
1 .040
1.043
1.045 1.048 1.052 1.055
1.058 I
060
210
1 .030
1033
1 .035 1 .038
r . 040 1 . 043 1 . 047 1 . 050
1-053 1
055
212
APPENDIX 213
How to Interpolate the Table of Factors of Evaporation
It sometimes happens when it is desired to use the table of
factors of evaporation that the given figure for any case falls be-
tween two certain figures in the table, and therefore the correct
result cannot at once be found without resorting to what is called
" interpolation."
Suppose, for example, that the average steam pressure in the
boiler is 64 lb. per square inch gage, and that the average tempera-
ture of the feed water is 13 2 F.; what is the factor of evaporation?
By referring to the table, there are no columns with heading or side
heading corresponding to these figures, and unless there is some
definite method of obtaining exact figures, it would be necessary to
strike an average between two sets of figures in the table, nearest
to those given in the example. While for ordinary purposes this
would be close enough, yet because of the ease with which the
real figures may be found it is worth while to learn what to do.
The factor for 60 lb. gage and 130 feed water is 1.115; the
factor for 70 lb. gage and 130 feed water is 1.118; the factor
for 64 lb. gage amd 130 feed water is therefore,
1. 118 — 1. 115
1. 115+ X4= 1.1162
In the same manner, the factor of evaporation for 64 lb. gage
pressure and 140 feed water is found to be
, 1. 107 — LICK
1.105 + '— ^X4= 1. 1058
There is now the factor for 64 lb. gage and 130 feed water,
and 64 lb. gage and 140 feed water, and it only remains to inter-
polate between these values to get the factor for 64 lb. gage and
13 2 feed water. This is done as follows:
1.1162 — 1.1058
1.1162 X 2 = 1.1141
10
which is the factor of evaporation corresponding to 64 lb. gage
pressure and 13 2 F. feed water, as given in the example.
The foregoing method may be applied to any figures within the
range of the table.
214
ARITHMETIC OF THE STEAM BOILER
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2l6
ARITHMETIC OF THE STEAM BOILER
TABLE VI.— KENT'S TABLE OF
Formula: H. P. =3-33 (A -o.6\/a)\/h.
Diam-
eter,
inches
Area
A sq.
feet
Effective
area,
E=A-o.6\/A,
sq. feet
Height of chimney
50 ft.
60 ft.
70 ft.
80 ft.
90 ft.
100 ft.
Commercial horse-power of boiler
18
21
24
27
30
33
36
39
42
48
54
60
66
72
78
90
96
102
108
114
120
132
144
1.77
2.41
3-14
3-98
4.91
5-94
7-07
8.30
9.62
12.57
I590
19.64
23.76
28.27
33-18
33-48
44.18
50.27
56.75
63.62
70.88
78.54
95 03
113. 10
• 97
1-47
2.08
2.78
3.58
48
47
57
7.76
10.44
I35I
16.98
20.83
25.08
29-73
34-76
40.19
46.01
52.23
58.83
65.83
7322
89.18
106.72
23
35
49
65
84
25
38
54
72
92
115
141
27
41
58"
78
100
125
152
183
216
29
44
62
83
107
133
163
196
231
3ii
66
113
119
141
149
173
182
208
219
245
258
330
348
427
449
536
565
694
835
For pounds of coal burned per hour for any given size of chimney,
APPENDIX
217
SIZE OF CHIMNEYS FOR STEAM BOILERS
(Assuming 1 H.P. = 5 lb. of coal burned per hour)
no ft.
Height of chimney
125 ft.
150 ft,
i75ft,
200 ft,
225 ft,
250 ft,
Commeicial hcrse-power of boiler
Equivalent
square
300 ft. chimney; side
of square,
\/E+4 in.
156
191
229
271
365
472
593
728
876
1038
1214
204
245
389
503
632
776
934
1 1 07
1294
1496
1712
1944
2090
3i6
426
551
595
692
748
849
918
1023
1105
1212
1310
I4I8
1531
1639
1770
1876
2027
2130
2300
2399
2592
2685
2900
2986
3226
3637
3929
4352
4701
981
1 181
1400
1637
1893
2167
2459
2771
3100
3448
4200
5026
1253
1485
1736
2008
2298
2609
2939
3288
3657
4455
5331
1565
1830
2116
2423
2750
3098
3466
3855
4696
5618
2005
2318
2654
3012
3393
3797
4223
5U4
6i55 I
16
19
22
24
27
30
32
35
38
43
48
54
59
64
70
75
80
86
91
96
101
107
117
128
multiply the figures in the table by 5.
2lS
ARITHMETIC OF THE STEAM BOILER
TABLE VII-
-PROPERTIES OF
SATURATED STEAMi
i
2
3
4
5
6
7
Pressure
lb. per
sq. in.
Temp,
degrees
Vol. cu.
ft. per lb.
Weight,
lb. per
cu. ft.
Heat of
the liquid
Latent
heat of
evap.
Total
heat of
steam
P
F
Vor S
V
Q
L or R
H
i
101 .83
333-0
0.00300
69.8
1034.6
1104.4
2
126. is
173-5
0.00576
94-0
1021 .0
1115.0
3
141-52
118. 5
0.00845
109.4
1012.3
1121 .6
4
153.01
90.5
.01107
120.9
1005.7
1126.5
5
162.28
73-33
0.01364
130. 1
1000.3
1130.5
10
193-22
38.38
0.02606
161 . 1
982 .0
1143.1
14-7
212 .00
26.79
0.03732
180.0
970.4
1150.4
20
228.00
20.08
0.04980
196. 1
960.0
1156.2
25
240. 10
16.30
0.0614
208.4
952.0
1160.4
30
250.30
13-74
0.0728
218.8
945-1
1163.9
35
259-3
11.89
0.0841
227 .9
938.9
1166.8
40
267.3
10.49
0.0953
236. 1
933-3
1169.4
45
274-5
9-39
0. 1065
243-4
928.2
1171 .6
50
281.0
8.51
0.1175
250. r
923-5
II73-6
55
287.1
7.78
0. 1285
256.3
9190
II75-4
60
292 .7
7-17
0.1394
262 . 1
914-9
1177.0
65
298.0
6.65
0.1503
267.5
911 .0
1178.5
70
302.9
6.20
0.1612
272 .6
907.2
1179.8
75
307.6
5-81
0.1721
277-4
903.7
1x81.x
80
312.0
5-47
0.1829
282 .0
900.3
1182.3
85
316.3
5.16
0.1937
286.3
897.1
1183.4
90
320.3
4.89
0.2044
290.5
893.9
1184.4
95
324.1
4-65
0.2151
294-5
890.9
1185.4
100
327.8
4-429
0.2258
298.3
888.0
1186.3
105
331.4
4-230
0.2365
302.0
885.2
1187.2
no
334.8
4-047
0.2472
305.5
882.5
1188.0
115
338.1
3.880
0,2577
309-0
879.8
1188.8
120
341.3
3.726
0.2683
312.3
877.2
1189.6
125
344-4
3.583
0.2791
315.5
874.7
1190.3
130
347-4
3-452
0.2897
318.6
872.3
1191 .0
135
350.3
3.331
0.3002
321.7
869.9
1191 .6
140
353.1
3.219
0.3107
324-6
876.6
1192.2
145
355.8
3-H2
0.3213
327.4
865.4
1192.8
150
358.5
3-012
0.3320
330.2
863.2
II93-4
155
361.0
2.920
0.3425
332.9
861.0
1194.0
160
363.6
2.834
0.3529
335-6
858.8
II94.5
165
366.0
2.753
0.3633
338.2
856.8
II95-0
170
368.5
2.675
0.3738
340.7
854.7
II95.4
175
370.8
2.602
0.3843
343-2
852.7
II95-9
180
373- 1
2.533
0.3948
345-6 .
850.8
1196.4
185
375-4
2.468
0.4052
348.0
848.8
1 196. 8
1 Reproduced by perm
and diagrams (copyright,
ission
1909,
from Marks and Davis's
by Longmans, Green &
steam
Co.).
tables
APPENDIX
219
TABLE VII— PROPERTIES OF SATURATED STEAM (Continued)
I
2
3
4
5
6
7
Pressure
lb. per
Temp,
degrees
Vol. cu.
ft. per lb.
Weight,
lb. per
cu. ft.
Heat of
the liquid
Latent
heat of
Total
heat of
sq. in.
evap.
steam
P
F
Vor S
V
Q
L or R
H
190
377-6
2.406
0.4157
350.4
846.9
II97.3
195
379-8
2.346
0.4262
352.7
845.0
II97.7
200
381.9
2.290
0.437
354-9
843.2
1198.1
205
384.0
2.237
o.447
357.1
841.4
1198.5
210
386.0
2.187
0.457
358.2
839.6
1198.8
215
388.0
2.138
0.468
361.4
837.9
1199.2
220
389.9
2.091
0.478
363.4
836.2
1 199 -6
225
391.9
2.046
0.489
365.5
834-4
II99-9
230
393.8
2.004
0.499
367.5
832.8
1200.2
235
395-6
1.964
0.509
369.4
831. 1
1200.6
240
397-4
1.924
0.520
371-4
829.5
1200.9
245
399-3
1.887
0.530
3733
827.9
1201 .2
250
401 . 1
1.850
0.541
375-2
826.3
1201.5
INDEX
Allowable pressure, 168
strain on stays, 161
Analysis of boiler trial, 72-81
Angle stiffners for curved sur-
faces, 148
Angles, bracing, 187
Approximate method, areas of
segments, 52
Area of diagonal stays, 40
grate, 62-63
of head to be braced, 186
of segments to be braced,
48, 50, 180
table of segments, 51
Areas and circumferences of cir-
cles, table of, 191
Boiler problems, 124-125
trial report, 72-81
tubes, table of standard,
214
water- tube and coil, 154
Braced segments, 48
Braces and staybolts, 36, 37
Bracing, angles, 187
Brown type of furnace, 143
Bumped heads, 30, 36, 170
Bursting pressure in cylinder, 9
11
pressure of pipe, 85-86
Butt straps, single, 19, 20
joints, 19
B
Bars, girder, 53-55
Board of Supervising Inspectors
Rules, United States,
135-156
Boiler efficiency, 71-72
feed pipe, size of, 118-119
heating surface, 60-62
heads, 30-55
stiffness of, 1 19-124
horse-power of, 66-68
Porcupine, 155
Cast-iron nozzles, 180
Chimneys, 11 5-1 18
table of, 216
Circles, areas of, table, 191
Circumference of circles, table
of, 191
Coil and water- tube boilers, 154
Collapsing pressure of fire box,
95, 96, 97
of tubes, 65, 66, 126, 127
cone-shaped flue, 92-93
Combustion chambers tops, 144
Commercial efficiency, 71-72
221
222
INDEX
Concave heads, 30-36, 148
Cone seam, strength of, 94-95
-shaped flue, 92-93
Convex-heads, 30-36, 147
Corrugated furnaces, 63-64
Cost of evaporating water, 87-
88
Cubes, and cube roots, table of,
198
Cylinder, the, 6
with riveted joints, safe
pressure of, 28, 29
D
Decimal equivalent of fractions,
197
Description of riveted joints, 12,
13
Design of riveted joints, 14
Diagonal seam, efficiency of,
91-92
stays, 39
area of, 40
U. S. rules, 138-139
Diameter of cylinder, 10
of sphere, 5
of stays, 38, 180
Direct stays, 37-39
Distance between rows of rivets,
27, 28
between stays, 38
Double butt-strapped, quad-
ruple riveted joint, 24,
25
reinforcing rings, 56-61
riveted lap joints, 17-18
Double riveted reinforcing rings,
56-61
Efficiency, commercial, 71-72
of diagonal seam, 91-91
of grate and boiler, 71-72
of ligaments, 182
of riveted joints, 14, 15, 169
Equivalent evaporation, 68-70
Evaporation, equivalent, 68-70
External inspection, 167
Extracts from Massachusetts
Rules, 156-190
from Rules of the United
States Board of In-
spectors, 135-156
Factors of evaporation, table of,
212
of safety, 157
Feed pipe, size of, 11 8-1 19
Fire box, collapsing pressure of,
95-96
Flat heads, 33, 34
surfaces, 145
to be stayed, 46
Formulas for diagonal stays, 41
for safety valves, 99-114
for spheres, 5-6
Fox furnace, 142
corrugated, 63-64
U. S. Rules, 140-144
Fusible plugs, 158
Girder bars, 53, 55
Grate area, 62-63
surface, ratio, 68
H
Head, area to be braced, 186
Heads, boilers, 30-55
bumped, 30, 36, 179
concave, 30-36
convex, 30-36
flat, ss, 34
stayed, 36-55
supported by flange, 45
thickness of, 34, 35, 46
unstayed, 30-36
Heating surface of boilers, 60-62
ratio, 68
High joint efficiencies, 21, 22
Horse-power of boilers, 66-68
Hydrostatic tests, 168
INDEX 22:
Joints, U. S. Rules, 135-138
Lap joints, double-riveted, 17-
18
quadruple-riveted, 18, 19
single-riveted, 15, 16
tiiple-riveted, 18,-19
Leeds furnace, 141
Ligaments, efficiency of, 182
M
Manhole reinforcing rings, 56-
61
Maximum pressure, 156
pressure on shells, 181
Morrison furnace, 141
N
Nozzles, cast-iron, 180
Number of rivets, 20, 21
Inspection, annual, 166, 167
Internal inspection, 166
Joints, butt, 19
efficiencies, high, 21, 22
efficiency of, 14, 15
lap, 15, 16, 17, 18, 19
riveted, 12, 13, 14, 15, 16,
18, 19, 169
Pipe, bursting pressure of, 85-86
Pitch of rivets, 26, 27, 125, 126
of stay bolts in furnaces,
189
Plugs, fusible, 158
Porcupine boilers, 155
Pressure, bursting, of pipe, 85-
86
Properties of saturated steam,
218
224
INDEX
Purves type of furnace, 142
Q
Quadruple-riveted, bouble butt-
strapped joint, 24, 25
lap joints, 18-19
R
Radius of bumped head, 31, 32
Ratio of heating to grate sur"
face, 68
Reinforcing rings, 56-61
Report of boiler trial, 72-81
Rings, manhoJe reinforcing, 56-
61
Riveted joints, 12
efficiency of, 14, 15, 169
U. S. Rules, 135-138
Rivets, distance between rows,
27, 28
number of, 20, 21
in reinforcing rings, 59
pitch of, 26, 27, 125, 126
securing stays, 44
shearing strength of, 17
size of, 26, 27, 160
in single and double shear,
21
Roots, square and cube, 198
Roper's safety valve rules, 107
Rows, rivets, number of, 20
Rules for area of segment, 180
for diagonal stays, 41, 43
for spheres, 5-6
safety values, 99-114
United States inspectors,
135-156
Safe pressure, cast-iron vessels,
flat cast-iron heads,
88-89
in cylinder, 9, 11
in sphere, 5
working pressure of boilers,
127-132
cylinders with riveted
joints, 28, 29
Safety, factors, 157
valves, 151, 163
rules, 99-114
Seam, diagonal, 91-92
Segments, area of, 48, 180
to be braced, 48
table of, 51
Shearing strength of rivets, 17
Single butt-straps, 19, 20
and double shear, rivets
in, 21
reinforcing rings, 56-61
riveted reinforcing rings,
56-61
lap joints, 15, 16
Size of boiler feed pipe, 118-119
and pitch of rivets, 26, 27
of rivets, 160
Solid girder bars, 53-5 5
Sphere, the, 1
Split girder bars, 53-55
Squares, cubes, square roots, and
cube roots, table of, 198
Stay bolts, pitch of in furnaces,
189
Stayed, flat surfaces, 46
INDEX
225
Stayed head, 36-55
Stays, diagonal, 39
U. S. Rules, 138-139
diameter at bottom of
thread, 180
direct, 37-39
U. S. Rules, 140
rivets securing, 44
and stay bolts, 36, 37
strain on, 161
Steam table, Marks and Davis',
218
Stiffness of boiler heads, 119-
124
Strength of cone seam, 94-95
of cone-shaped flue, 92-93
of riveted joints, 13
shearing, of rivets, 17
Stress in cylinder, 7
on each inch in the cir-
cumference of cylinder, 10
in sphere, 4
Surface, heating, of boilers,
60-62
Tables: I, II, III, IV, V, VI,
VII, 191-219
Table: I, Area and circumference
of circles, 191
II, Decimal equivalents,
197
III, Squares, cubes, square
roots, and cube roots,
198
IV, Factors of evapora-
tion, 211
Table: V, Standard boiler tubes,
214
VI, Kent's table of chim-
neys, 216
VII, Marks and Davis'
steam tables, 218
of areas of safety valves,
152
of segments, 51
Tests, hydrostatic, 168
Thickness of heads, Massa-
chusetts' Rules, 34
Ohio Rules, 35
of plate in cylinders, 10
in heads, 46
in sphere, 6
Tops of combustion chambers,
144
Total pressure on shell of
cylinder, 11
stress in cylinder, 10
Triple-riveted lap joints, 18-19
Tubes, collapsing pressure of,
65-66, 126, 127
boiler, standard, 214
U
United States Rules, extracts
from, 135-156
Unstayed heads, 30-36
V
Valves, safety, 99, 114, 163
W
Water-tube boilers, r54
and coil boilers, 154
I
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