"L I B R.AFLY OF THE UNIVERSITY Of ILLINOIS 510.84 It6r no. 349-354 cop. Z Digitized by the Internet Archive in 2013 http://archive.org/details/analysisofsomeeu352flow 5 /£'■* (1+) and Si puSi pSu d ,,^l ^,/^U\2 , c x cH oSc 3x " 3x ^ 5x' xix where V = ^-Ax PU more closely than n Bu pudu dp at ox ox „ Si pudi pdu _ md P St + olT 3i=° Some of the terms on the right hand side of equations (h) and (5) are proportional to the velocity gradient and can be related to the physical concept of viscosity. These additional terms arise as a result of the mass transport mechanism and are responsible for producing the dissipa- tive effects that give stability to the method. Thus, PIC and its related methods are a class of codes whose dissipative force is referred 12 to as effective viscosity or implied viscosity or smoothing. These terms allow PIC to give accurate approximations to high velocity flow. The V terms have their principal effectiveness in regions where the velocity gradient is large or where fluctuations occur in high speed flow. At low fluid speeds or in regions of near stagnation, the veloc- ity gradients are small and hence the effective viscosity terms are not large enough to make the dissipative force effective so that oscillations develop. These oscillations will grow until they become comparable to the local sound speed when the effective viscosity terms become large enough to prevent any further growth of the instabilities (as in test problem 6b). Thus, the instabilities are bounded. These difficulties can be alleviated by the use of artificial viscous pressure terms in the equations for momentum and energy (see section 2.k). However, these terms should be added only where the fluid speed is small com- pared to the local sound speed. If applied in high velocity regions, the artificial viscosity will smear out many features of interest. Some authors question whether the smoothing scheme in PIC f7 81 can be directly related to physical concepts. ' They have tried to show that doing so results in many contradictions. For example, the values of the additional terms far exceed the values of the anal- ogous physical terms. 13 3. DERIVATION OF THE DIFFERENCE EQUATIONS The region occupied by the fluid is considered to be discretized into a mesh of equal sized cells fixed in space. Each cell is of width Ax and is assigned some index j. Half integer indices are used to denote cell boundaries. Ax Ax Ax Figure 3. The Mesh of Cells The fluid is then described at any instant of time by specifying the velocity, density and pressure for each cell. These values are considered to be known at the center of each cell. The calculation advances explicitly in time steps i.e. cell quantities at time (t + At) are calculated in terms of values of those quantities at t. The system of equations to be solved is do 5 (pU) £ 5t = o dU UcVQ + idP _ cH dx Pdx (6) (7) lU || o * + i a (ro) . (8) at ox p ax v y and P = f ( P ,I) which represent the conservation of mass, momentum and energy and the equation of state, respectively, where the symbols have their usual meanings, As in PIC, the solution is accomplished by performing first a Lagrangian update, then rezoning to the original Eulerian mesh. During the Lagrangian phase, the operator ;— + °r is replaced by -rr, the total derivative. Equations (7) and (8) then become dt p ax 2| + I |Mi . (10) dt p ox Equation (10) can be rewritten as at N*^ % i <™> ■ ° ai p au J TTf aU 1 aP-, n or =7- + — r— + UL^r + — c— J = at p ax at p ax Now, equation (9) implies that j£ + £jffl . (11) at p ax The intermediate or tilde values of velocity and specific internal energy at the end of the Lagrangian advance are given by the following equations. 13 3- DERIVATION OF THE DIFFERENCE EQUATIONS The region occupied by the fluid is considered to be discretized into a mesh of equal sized cells fixed in space. Each cell is of width Ax and is assigned some index j. Half integer indices are used to denote cell boundaries. Ax Ax Ax -it- Figure 3. The Mesh of Cells The fluid is then described at any instant of time by specifying the velocity, density and pressure for each cell. These values are considered to be known at the center of each cell. The calculation advances explicitly in time steps i.e. cell quantities at time (t + At) are calculated in terms of values of those quantities at t. The system of equations to be solved is So o- (pU) 5t = (6) I + UcUJ + idp = dx pdx 5t (7) lU || + U^ + i-i(HT) . (8) dt dx p dx v ' and P = f ( P ,I) which represent the conservation of mass, momentum and energy and the equation of state, respectively, where the symbols have their usual meanings, As in PIC, the solution is accomplished by performing first a Lagrangian update, then rezoning to the original Eulerian mesh. During the Lagrangian phase, the operator :-— ■ + 4- is replaced by -rr-, the total derivative. Equations (7) and (8) then become dt p dx M t 1 |IM _ (10) dt p dx Equation (10) can be rewritten as §1 N^ ♦ ? §1 < ro > ■ ° or |i t £ |H + V M , i |E] . dt p dx dt p dx Now, equation (9) implies that dt p dx The intermediate or tilde values of velocity and specific internal energy at the end of the Lagrangian advance are given by the following equations. 15 u n = u. n - (p. n n - p. n J -^ < 12 > V = 1° - Cf",- p",) P .i" At (13) ° ' ° 3 2p. n * where U. = — [U. + U. ] J 2 j j These intermediate values correspond to values at time (n+l) At at the point in the fluid that was at the center of cell j at time n At. To first order, the tilde quantities are values at time (n+l) At at the Eulerian point x. + U. At. The use of time-centered values of velocity — n ~ n U. in the equation for I. results m greater stability and better behavior of fluid entropy than if the simple value U. were used. Equation (l^) J is used to calculate the intermediate value of specific energy. E. = I. + | (U.) 2 (11+) 3 3 2 j' The distorted Lagrangian mesh is now rezoned back to the original Eulerian mesh. In effect, this is a mass transport across cell boundaries. The mass flow across the interface between cell j and cell (j f l) is defined by n n "*n AM. 1 = p. 1 U. 1 At J+2 J+2 0+2 *n The quantity U. 1 is known as the weighted velocity. J+2 Three techniques for calculating U. 1 define the three differencing schemes used in this report which are called continuous rezone, OIL and donor. The interface values of density and velocity cannot be calculated by simply 16 averaging adjacent cell values, because the method obtained is highly- unstable and virtually useless. The density weighting and velocity weighting, as they are called, must be done with greater care. The equations for the density weighting and velocity weighting are simply stated in this section but are derived at length in section 3.1. The direction of the flow at the interface between cell j and cell (j+l) is defined by the sign of U 1 where U 1 = 5- [U + U ]. «J p then J + 2 n -> P. 1 = J+2 n P. If U. n l = then J+2 J+2 |[p." + If U. n i < then ^2 P A Velocity Weighting (a) continuous rezone If U, n i >0 i+5 - *n then U. 1 J+2 n+1-, p j ] u. n i J+2 i * C^i - Vi>-4i 2A x ^ PI U 1 kL .„ ,, *n, J+ 2 If U. 1 < then U. 1 Jf 2 J+ 2 , ,~ n « ik At 1 +■ (U " - U/ ) j+2 j '2A x (b) OIL ^_n 1 *n. ^ + 2 U. 1 J f o i j. fu n ~ n N At 2 1 ♦ (u » - U -) J+l j ' ^2Sc 19 density for the other two schemes gives fairly good results. The three velocity weighting schemes used in this report are described below. 3.1.1 Continuous Rezone Assume that the cell in the diagram is moving from left to right. i. n i j Figure k. Continuous Rezone The undashed rectangle in Figure h represents the position of cell j at time n. The dashed rectangle represents the position of the same rectangle after the initial Lagrangian flow. The density, specific internal energy and momentum are assumed constant throughout the dashed cell. The velocity and specific internal energy at the center of the dashed cell are given by equations (12) and (13) respectively. The quantity of mass which is in the dashed cell and has crossed over into Eulerian cell (j-t-l) during the rezone is calculated from the fraction c ' O _ J+2 J J+ 2 ■X-n ~ n U. 1 = if U. 1 = J+2 J+ 2 U. 1 = U." if U. 1 < n 0+2 J +1 J + 2 3.1.U Rich This method is included for the sake of completeness although it was not applied to the test problems due to its poor stability properties. The weighted values of density and velocity are obtained from the first two terms of a Taylor series expansion about the donor cell. Assuming that the flow is from left to right (U. n l X)), and that •v~ ^ 2 ~ n ~ ~ ~ n ~ n . Ax oU l n ~ n , U . , ., - U "2 U. 1 = U. + =~ s- . = U. + .1 + 1 J-1 > J+O J 2 dx [ J J -d r « then AM. n l = [U. n + U j + l" ,j-l ] [p . n + P t j+1 " p j-l ] At J + 2 ^ 1+ ^ 1+ Similarly, if U. n l < and if 3 2 -\~ ~ n ~ n U^l = U. n n - %^- | n , = U* - U j+2 - U j <0 j+5- j+1 2dx'j+l j+1 -" j-— • L - ~ n ~ n n n then AM. 1 = [U. . - j+2 j J [p. - ,]+2 K j J At j+2 J+1 — JT~^ J — T~^ Otherwise AM. 1 = J+2 23 3.2 Conservation of Mass, Momentum and Energy The change in the mass of cell j between the start and the end of each iteration is ... n r n+-l n-, . £M. = Lp . - p. Ax 3 3 3 = m. n i - m n i 3-2 J + 2 where the half integer subscripts indicate quantities moving across cell boundaries. Therefore, the change in mass of the entire system is J m n = Z , [ m n i - £M. n i] 3=1 3 "2 J + 2 n n Note that £M. 1 =AM // . 1 v 1 N because the mass transported across the right hand side of cell j equals the mass tranported across the left hand side of cell (j+l). Hence /M n = ®£ - ZM_ n l J. J+„ 2 * Therefore, the only change in mass of the total system occurs by mass crossing the boundaries of the system and hence, mass is conserved. Redistribution of energy and momentum occurs during the calculation of tilde values and also during the rezoning from the Lagrangian mesh to the Eulerian mesh. During the calculation of tilde values, the changes in energy and momentum are the following J n r ~n _n n A ^ = jLi P J [ j " V * J /Mom = Z n p n [U 1 ? - U 1 ?] Ax J=l 3 3 3 2*+ Therefore ~n AM = Z - [P. n l - P*l] At J-o J ' 2 As before, all terms cancel the first and the last, leaving AM n = (*rf - P T n i) At om 1 J+— ' 2 Also ^ = Z, k-?* At (U* - U. n J + P n Ax (U* + U n ) (U* - U*)] t ,i=i 2 ,i .i+i .i-i y .i .i y .1 .r J' = Z, % [-p? (u* -u") -u n (p.* -p. n J] .1=1 2 ,i ,1+1 ,i-l y ,i ,i+l .i-l J J J J At r n 1 O 1 O J J+l ,1 J+l = At p? if 1 + 5? P n - p n if 1 - P A u" 1 o 1 o o o J+^ J+l 2 2 - u T n i p* + p" u* J+P J+l J+l J+l Therefore, the only changes in momentum and energy of the system during the calculation of tilde values occurs by momentum and energy crossing the boundaries of the system. Hence, momentum and energy are conserved during this part of the calculation. The equations for the rezoning phase are directly derived from the equations for conservation of momentum and energy and hence momentum and energy must be conserved during the rezone. The energy and momentum changes of the system occurring during the rezone are 25 n ...n ~n A .„n n ~n omR 1 L J+2 R 2 d Thus, the mass, momentum and energy of the system are conserved identically despite finite difference approximations. However, corrections must be made at transmittive boundaries and pistons. It is interesting that the derivations of the conservation of mass, momentum and energy are independent of the velocity weighting scheme used. This is because the conservation laws are concerned with cell centered quantities but the velocity weighting schemes are defined by interface quantities. The only discrepancies arising between the theoretical energy of the system and the numerically approximated values are due to round- off errors. The values of system energy obtained for test problem 1A (see section 5.2) which are typical of the values obtained for the rest of the problems are: Initial system energy I.1363XIO gms. System energy at .1 seconds minus work done by piston (a) continuous rezone I.I36OXIO gms. (b) OIL 1.1359X10 9 gms. (c) donor 1.1353X10 9 gms. 26 3.3 Order of the Truncation Errors By applying Taylor's series expansions to the finite difference equations for continuous rezone, OIL and donor, and comparison with the equations of hydrodynamics, the order of the errors involved in the approxi- mations can be found. The methods are of at least the order calculated and might be higher. The derivations are carried out assuming that the flow is from left to right. Combinations of left to right and right to left flow have not been examined. The order of the truncation errors for all three velocity weighting schemes for each of the three equations of hydrodynamics are simply stated in this section and are derived in detail in APPENDIX B. 2 (a) continuous rezone is (a x) + (At) +■ (■— ) for each of the three conservation equations 2 (b) OIL is (Ax) + (At) + (j- ) for each of the three conservation equations (c) donor is (Ax) +0 (At) for the conservation of mass At 2 and momentum, and is (At) + (Ax) •*- (— ) for the ZjX conservation of energy. Thus, the set of finite difference equations representing each of the three schemes continuous rezone, OIL and donor is of order (A x) + o (At) + (g ) 27 k. TIME STEP CONTROL AND TREATMENT OF BOUNDARIES k.l Time Step Control Three criteria were used for calculating the time step. The first was the familiar Courant-Friedrichs-Lewy stability criterion which states that at any time step n, no signal travelling at the local sound speed C can propagate through more than one cell in one time step. Hence, for each n At n 1 -^ c n 3 for all j in the region of interest .,n „ . r A x-, or At = F min L J i c." < where < F - 1. F is defined as the Courant-Friedrichs-Lewy number. The second limitation which applies to supersonic flow problems only, prohibits the transport of mass across more than one cell in one time interval. The fluid model provides no mechanism for transfer of mass between non-adjacent cells. Hence At n 1 -& u. n for all j in the region of interest The third limitation is strictly applicable to the OIL method but was applied to the other schemes as well, enabling all three techniques 28 to use the same Courant-Friedrichs-Lewy number. Thus, continuous rezone and donor should he able to run faster than OIL. Suppose At is calculated from the Courant-Friedrichs-Lewy criterion. Then ., n < Ax , _ 7 /max . TT n . _ n N At — — - where V = max ( . U , C. ) V 3 ' 3 3 FA x < or At = — - — where F — 1 Assuming that the flow is from left to right, suppose that some U. n = V. Then 3 U. n At = FAx (28) 3 However, the fluid model for OIL prevents the left hand boundary of the cell whose center is at (j + p) from crossing into cell (j + l). Hence tt n a*. < A x U 5 At - — (29) By comparing equations (28) and (29) it can be seen that F — _- is a necessary condition in order to remain consistent with the fluid model. It may also ~ n v. 1 happen that some U. > V, in which case F < _- for stability. It is necessary to show that the condition U. > V can exist to support the hypothesis that J F V. 29 In the test problems, F was chosen to be equal to .h. This is equivalent to assuming that no U\ will increase by more than 25% over V in one time step. Some authors have experienced difficulties with negative energies and negative volumes and have incorporated additional time step controls to handle them. However, this was not found necessary in this report. k.2 Treatment of Boundaries A transmittive boundary at the interface (j + -|) is defined by the following equations : OX ' 0+2 t — • 1 = u ox j+i £ I .1 = ox j++ s'jiK For a left hand boundary at J = ■§-, the above equations are satisfied by using a dummy cell with index and letting U = U , P = p , P P and I = I . A right hand boundary is handled in a similar fashion. 30 It is necessary to calculate the mass and energy transported across a transmittive boundary in order to check the conservation laws. The conditions at a reflective boundary are the same as above except U = -U . Moving piston boundaries are complicated by the work terms done on the gas by the piston. 31 5. DESCRIPTIONS, NUMERICAL VALUES AND DISCUSSIONS OF TEST PROBLEMS 5.1 Introduction The test problems are intended to exhibit typical behavior of the hydrocodes in a wide variety of situations and are not difficult in r-5] themselves. In the report from which the test problems are taken . there are seven test problems with a few variations of each, which are designated A, B, C etc. The difference between these variations is in the initial numerical values of one or more of the fluid parameters. The test problems in [3] which were not run with continuous rezone, OIL and donor are described in the next paragraph. The format for the descrip- tion of each test problem will consist of a brief description, initial numerical values, boundary conditions and plot times together with initial and final plots of pressure, density, specific internal energy and velocity. Sometimes, initial plots are not given if one or more parameters is constant throughout the region of interest. For further details about the test problems, the reader should consult reference [2], Also, the PUFF, Lax Wendrof f and analytic solutions may be found in reference [3 ]. None of the last three parts of problem two and none of problem four was run due to the presence of a vacuum. When a vacuum arises, the density equals zero and equations using density as a divisor cannot be used. The last part of problem seven was not run, since it is just a scaling down of the first part by a factor of 100 and no new features are introduced. Also, problem three was not run mainly because it is a pure compression problem and in this respect would yield little informa- tion not already given by problem one. 32 The chief difference between the test problems in Hick's reports and the ones in this report are that the latter may be translated in the positive x direction to avoid negative or zero subscripts. This will be clearly stated where appropriate. Also, there might be a slight difference between the total number of cells in the mesh used in this report and in reference [3]. Neither of these differences will change the calculated error norms because the cell sizes in each report are the same and the discrepancy in the total number of cells occurs at the boundaries of the problem where the exact and approximate solutions are equal. Errors The analytic solution to the mathematical model of the physical problem is known as the exact solution. The difference between the exact solution and an approximate or hydrocode solution is referred to as the error. For purposes of comparison, the calculated error norms are the same as those used by Hicks and are defined below. For example, take a variable V whose exact and approximate values in cell j are V (j) and V (j) respectively. Let V be the absolute maximum of V (j) where j ranges over the region of interest. For V, make the E following definitions: Sum. Abs. Error = h~ ' V A^ " V E^ M ° Sum. Sqr. Error = -~ ? ( V d) " V j)) V J M Maximum Error = max I V A (J) - V E (J)1 sign (V.(j ) - V (j )) V M 33 where J,, is the index of . ' A E v For each test problem and each of the three schemes, error tables are given containing sum abs. error, sum sqr. error and maximum error. Large values of the maximum error norm can occur at the shock front due to the numerical approximations of the shock being spread over two or three cells. Thus, the maximum error norm is an unreliable norm with which to compare these methods. For all test problems, a polytropic equation of state is used as in equation (27). P = (7 - 1) P I (27) For all test problems, 7= l.k. 5.2 Test Problem 1 (a) Description In this problem, a piston moves into a fluid at constant velocity from left to right. The fluid is consequently compressed. At the start of the problem, a shock moving from left to right exists 50 meters ahead of the piston. The exact solution is two constant states separated by the shock discontinuity as shown in Figure 6. 3^ v T v r Figure 6. Test Problem 1 V V V V X V sound speed to the left of the shock sound speed to the right of the shock pressure to the left of the shock pressure to the right of the shock density to the left of the shock density to the right of the shock velocity to the left of the shock velocity to the right of the shock piston velocity shock velosity initial position of shock MC where M is a positive constant r c X = initial position of piston 35 (b) Numerical Values for 1A M = 1 Ax = 1 meter h / 2 P = 10 dynes/ cm (■ .3 p =10 gm/cm V = r V, a: 1.18 x 10 5 cm/sec V ~ 2.08 x 10 5 cm/sec s ' k 2 P» ~ 3.^7 x 10 dynes/cm Q o - 2.30 x 10" gm/cm^ X = 51 meters in this report and 50 meters in [3] Right boundary at 300 meters X =1 meter in this report and meters in [3] P (c) Boundary Conditions On the right, a transmittive "boundary and on the left, a piston starting at 1 meter and moving to the right with velocity V . (d) Plot Times 0. and .1 sec. (e) Numerical Values for IB The same as 1A except M = 100 V» ~ 1.18 x 10' cm/sec 8 2 P» ~ 1.68 x 10 dynes/cm C» ~ 6.26 x 10 cm/sec V ~ 1.42 x 10^ cm/sec s ' Pa 2: 5.99 x 10 gm/sec (f) Plot Times 0. and 10" 3 sec. 36 Computer Time = 100 Sec. Sum Abs. Error Table 1. Errors for Test Problem 1A Problem Time = 10 J Sec. Continuous Rezone Iteration Number = ^6k Pressure Density Energy Velocity .885 .655 .1+65 .737 Computer Time = 105 Sec. Sum Abs. Error Pressure Density Energy Velocity .835 .638 .1+87 .709 Computer Time = 105 Sec. Sum Abs. Error Pressure Density Energy Velocity 1.031 .886 .658 I.0I+5 OIL Sum Sqr. Maximum Position of Maximum Error Error Error .101 .285 current shock position .056 .201 current shock position .019 .122 current shock position .1^3 .326 current shock position Iteration Number = 36^ Sum Sqr. Maximum Position of Maximum Error ' Error Error .076 .2^0 current shock position .OUl .15*+ current shock position .018 .109 current shock position .133 .285 current shock position Iteration Number = 36^ Sum Sqr. Maximum Position of Maximum Error Error Error . LOU -.236 current shock position .078 -.238 current shock position .027 -.115 current shock position .236 -.390 current shock position Donor 37 Computer Time = 99 Sec, Sum Abs. Error Table 2. Errors for Test Problem IB Problem Time = 10 J Sec. Continuous Rezone Iteration Number = 300 Pressure Density Energy Velocity .9^5 .963 .980 .828 Computer Time = 98 Sec, Sum Abs. Error Pressure Density Energy Velocity 1.113 I.165 1.229 1.005 Computer Time = 98 Sec, Sum Abs. Error Pressure 1.569 Density 1.752 Energy 1.715 Velocity 1.370 Sum Sqr. Error .119 .100 .358 .326 OIL Sum Sqr. Maximum Error Error Maximum Position of Maximum Error Error -.25^ current shock position -.266 current shock position -.587 current shock position -.556 current shock position Iteration Number = 299 Position of Maximum Error .171 -.356 current shock position .157 -.366 current shock position .U63 -.658 current shock position .U^3 -.6Ul current shock position Iteration Number = 300 Position of Maximum Error .312 -.^95 current shock position .320 -.50^ current shock position .658 -.720 current shock position .6537 -.719 current shock position Donor Sum Sqr. Maximum Error Error 38 (g) Discussion of 1A For each of the three schemes run, there are spikes at the shock front and slight waves behind the shock front as shown in Figure 7. Although the waves travelling back from the shock front are larger in the donor method, the donor spike obtained at the shock front is smaller than the spike for OIL or continuous rezone. PUFF has very good shock front definition and slight rarefaction dips behind the shock. Lax Wendroff and PUFF both have spikes in density and internal energy at the initial position relative to the piston, i.e. 50 meters in front of the piston. In addition, Lax Wendroff also has spikes at the shock front for all parameters. J\ Figure 7* Typical Wareform for Problem 1 (h) Discussion of IB Continuous rezone, OIL and donor show the same basic behavior as in 1A. Again, waves travel back from the shock front as in 1A. The largest waves are also obtained in the donor scheme. However, the small waves may be preferable to the very large spikes obtained in PUFF and Lax 39 Wendroff . PUFF has better shock front definition than any of the three schemes tried here but at the expense of large spikes 50 meters in front of the piston. Lax Wendroff has oscillations both at the piston face and at the shock front. 5.3 Test Problem 2 (a) Description In this problem, a piston moves with constant velocity from right to left thus evacuating a pipe which initially contains constant state gas at rest as in Figure 8. The motion of the piston creates a rarefaction wave moving to the right. P p r r Figure 8. Test Problem 2 (b) Numerical Values for 2A h / 2 P = 10 dynes/ cm -6 ? 10 gm/cm C = 1.4 x 10 cm /sec r ' V = - C /(y + 1) p r ' Ax = 100 cm Uo X (o) = 1300 meters in this report and 100 meters in [3] . X = 1500 meters in this report and 300 meters in [3] . H (c) Boundary Conditions On the right, a transmittive boundary and on the left, a piston moving to the left away from the gas with velocity V . (d) Plot Times .1 seconds (e) Numerical Values for 2B the same as 2A except V = r 7+1 Computer Time = 92 Sec. Table 3. Errors for Test Problem 2A Problem Time = .1 Sec. Continuous Rezone Iteration Number = 296 Sec. Sum Abs. Sum Sqr. Maximum Posi tion of Maximum Error Error Error Error Pressure 1.071 .016 -.031 X K Density .821 .011 -.026 X R Energy .456 .003 -.014 X E Velocity 2.1+7 .111 -.088 x „ Computer Time = 93 Sec. Sum Abs. Error Pressure 1.096 OIL Iteration Number = 296 Sum Sqr. Maximum Position of Maximum Error Error Error .017 -.032 X_ 4l Density- .848 .011 -.026 Energy .470 .003 -.014 Velocity 2.536 .115 Donor -.Q89 Computer Time = 90 Sec, L E X T X R Iteration Number = 296 Sum Abs. Sum Sqr. Maximum Position of Maximum Error Error Error Error Pressure 3.102 .218 .135 rarefaction wave Density 2.726 .151 .120 rarefaction wave Energy 1.088 .019 .042 rarefaction wave Velocity 5.586 .680 .269 rarefaction wave Table 4. Errors for Test Problem 2B Problem Time = .1 Sec. Continuous Rezone Computer Time = 95 Sec. Iteration Number = 296 Sum Abs. Sum Sqr. Maximum Position of Maximum Error Error Error Error Pressure 1.226 .014 -.023 2600 metres Density 1.052 .011 -.023 2600 metres Energy .924 .007 -.017 2600 metres Velocity 2.034 .055 -.057 2600 metres OIL Computer Time = 101 Sec. Sum Abs. Error Sum Sqr. Error Maximum Error Iteration Number = 296 Position of Maximum Error Pressure 1.201 .013 .023 2600 metres 1+2 Density 1.021+ .011 -.023 2600 metres Energy .906 .006 -.017 2600 metres Velocity 1.998 .053 Donor -.055 2600 metres Computer Time = 96 Sec. Iteration Number = 296 Sum Abs. Sum Sqr. Maximum Position of Maximum Error Error Error Error Pressure 2.818 .172 .135 2786 metres Density 2.528 .120 .122 2786 metres Energy 1.101+ .017 .01+2 2788 metres Velocity 3.072 .164 .133 2778 metres (f ) Discussion of 2A The results obtained for continuous rezone and OIL are very good. Donor displays the start of strong instabilities in the rarefaction wave. In continuous rezone, OIL and donor, there is a slight rise "in density and slight drop in specific internal energy just in front of the piston as in Figure 9 which seems to have little or no effect on the rare- faction region. This effect is slightly worse in donor than the other two. The reason for this effect might be connected with always choosing the donor value of density in the density weighting. Also, OIL, continuous rezone and donor underround at X_ and at X . PUFF and Lax Wendroff show L K undershoots to the left of X . Lax Wendroff even oscillates slightly at K the left of X_. K **3 Xr Xc Figure 9- Internal Energy for OIL and Continuous Rezone (g) Discussion of 2B The "basic behavior of 2B is similar to that of 2A. Again, donor breaks down in the rarefaction wave. PUFF and Lax Wendroff , due to the undershoots obtained to the left of X^, seem inferior to continuous rezone K and OIL for this problem. 5.U Test Problem 5 (a) Description This problem is a special case of the Riemann problem and is known as either the shock tube problem or the ruptured membrane problem. The problem is to find the hydrodynamic variables after the removal of a membrane at time t = separating two constant unequal states at rest as in Figure 10. We adopt the convention P > P > and investigate the three cases pp>p >p = p ,p,

P • The second part examines the collision of two symmetrical shocks. p l p r p r 1 ■ ■■ V r X s^(t) X Sr(t) Figure 12. Test Problem 6 (b) Numerical Values for 6A Ax = 1 meter X -(0) = 75 meters X (0) = 125 meters sr v ' P = 10 dynes /cm 51 p =10 gm/cm V cm/s ec Q o P = 10 dynes /cm 7 2 P =10 dynes/cm r ' Left boundary at 70 metres in this report and metres in [3]. Right boundary at 180 metres in this report and 200 metres in [3]. (c) Boundary Conditions Transmittive boundary conditions are applied at both the left hand and right hand boundaries. -4 Plot Times 0., and 7 x 10 seconds (d) Numerical Values for 6B The same as 6A except O p P = P =10 dynes/cm r ' Table 8. Errors for Test Problem 6A Problem Time = .0007 Sec. Continuous Rezone Computer Time = 75 Sec Iteration Number = . Sum Ab£ Error • Sum Sqr. Error Maximum Error Posit ion of Maximum Error Pressure 1.477 .431 • 515 -* X SR Density 1.914 • 373 .412 -* X SR Energy 1.924 • 334 .336 x 1 col Velocity 1.104 .148 -.229 X S1 52 OIL Computer Time = 75 Sec, Iteration Number = 163 Sum Abs. Sum Sqr. Maximum Position of Maximum Error Error Error Error Pressure 1.298 .309 •398 X SR Density 1.784 •293 •307 -* X SR Energy 1.866 • 314 • 330 x 1 col Velocity 1.064 .120 Donor -.226 * X S1 Computer Time = 79 Sec. Iteration Number = ! Sum Abs. Sum Sqr. Maximum Position of Maximum Error Error Error Error Pressure 1.315 .369 .465 * X S1 Density 2.166 .449 -.397 * X SR Energy 1.951 .325 •297 x 1 col Velocity 1.375 .305 -.443 X QT3 (f ) Discussion of 6A Continuous rezone and OIL both have small overshoots at X„ in pressure, density and velocity as shown in Figure 13 while donor has almost no overshoot. Also, there is a general smearing of fine detail as has been noticed in previous problems. The specific internal energy plots for continuous rezone and OIL are slightly better than donor. OIL and continuous rezone are practically identical. PUFF has large spikes in the density and specific internal energy. The Lax-Wendroff solution is barely recognizable due to large oscillatory components. 53 XsR Figure 13- Pressure for Problem 6A Table 9. Errors for Test Problem 6B Problem Time = .0007 Sec. Continuous Rezone Computer Time = 73 Sec. Iteration Number = . Sum Abs. Sum Sqr. Maximum Po£ sition of Maximum Error Error Error Error Pressure 5.51+1 1.336 .1+93 83 metres Density 1+.322 .808 .31+7 83 metres Energy 2.571 .1+29 .382 83 metres Velocity 30.831 26.560 -.990 76 metres 51+ OIL Computer Time = 75 Sec Iteration Number = 162 Sum Abs. Sum Sqr. Maximum Position of Maximum Error Error Error Error Pressure 1+-923 1.163 .1+63 83 metres Density 3.91+8 .69I+ .312 83 metres Energy 2.308 .381+ .368 83 metres Velocity 31.001 26.I+9I+ Donor -.990. 76 metres Donor -will not run to .0007 sec. due to instabilities. -J Figure ll+. Pressure for Continuous Rezone in Problem 6B (g) Discussion of 6B More difficulty was obtained with this problem than any other. The donor scheme for 6B would not run to 7 X 10~ seconds. The reasons for this difficulty are not, as yet, understood. It can- not be blamed on rarefaction regions, since none exist in this problem. The donor scheme runs up to and just beyond the collision but then 'blows up'. Both OIL and continuous rezone show serious oscillations 55 soon after collision as in Figure 1^. However, no asymmetries were observed. Several variations of the basic continuous rezone scheme were tried not using artificial viscosity, but none of them showed fewer oscillations than the standard method. PUFF shows some large spikes in the density and specific internal energy but the pressure and velocity are very good. The Lax Wendroff solution has instabilities comparable to OIL and continuous rezone. In summary, the collision of two asymmetric shocks works reasonably well but a lot of difficulty is obtained with the collision of two symmetrical shocks. From the plots, it can be seen that the instabilities for continuous rezone and OIL arise in the region where the velocity is supposed to be zero. Hence, in this region the effective viscosity will be close to zero and there will be almost no damping effect for any slight perturbations. 5.6 Test Problem 7 (a) Description This problem is concerned with the overtaking of one shock wave by another as in Figure 15- When two shock waves are travelling in the same direction, the one behind will always overtake the one in front. This is a consequence of the fact that the front side of a shock is supersonic relative to the shock front and the backside is subsonic. When the overtaking has occurred, a rarefaction wave travels to the left and a stronger shock continues moving to the right. 56 V sH V sr p i V i P p r r V X s^> X (t) sr v ' Figure 15- Test Problem 7 (b) Numerical Values for 7 A Ax = 1 meter h , 2 P = 10 dynes/ cm p =10 gm/cm V = r -,^8 / 2 P. = 10 dynes/ cm 12 2 P = 10 dynes/cm X (0) = 25 meters X (0) = 100 meters sr Left boundary at 2 meters in this report and meters in [3]. Right boundary at 200 meters . (c) Boundary Conditions Transmittive boundary conditions are applied at both the left hand and right hand boundaries. -5 (d) Plot Times 0. and 3 x 10 seconds 57 Table 10. Errors for Test Problem Problem Time = 3. X 10 J Sec. Continuous Rezone 7A Computer Time = 98 Sec. Ite ration Number = Sum Abs. Sum Sqr. Maximum Pos ition of Maximum Error Error Error Error Pressure 2.667 .167 .142 185 meters Density 3.230 .208 -.135 172 meters Energy 5.146 1.229 .U65 172 meters Velocity 1.581 .181 -.340 186 meters OIL Computer Time = 99 Sec. Iteration Number = ' Sum Abs. Sum Sqr. Maximum Position of Maximum Error Error Error Error Pressure 2.772 .169 .139 185 meters Density 3.260 .207 -.138 172 meters Energy 5.376 1.316 .474 172 meters Velocity I.687 .214 Donor -.374 186 meters Computer Time = 93 Sec. Iteration Numbe] Sum Abs. Sum Sqr. Maximum Position of Maximum Error Error Error Error Pressure 3.024 .177 .155 185 meters Density 3.330 .202 -.140 172 meters Energy 5.73^ 1.430 .485 172 meters Velocity 1.886 .260 -.397 186 meters = 331 58 (e) Discussion of 7 A _5 At 3 X 10 seconds, there is a slight dip to the left of the rarefaction wave on all plots for each of donor, continuous rezone and OIL at ninety meters as in figure l6. One does not know if these dips are obtained with PUFF and Lax Wendroff since in Hicks report, the plots for this problem start at 115 meters. A general smearing of shock corners is evidenced as in other problems. Continuous rezone has tiny overshoots in the pressure and density at X which do not arise in OIL or donor. The s plots obtained for velocity and specific internal energy for continuous rezone, OIL and donor, are practically identical. Also, there is very little difference between continuous rezone and OIL in general. PUFF has good overall definition but there are two very large spikes on the density and specific internal energy. 90 METERS Xs Figure 16. Pressure for Continuous Rezone in 7A 59 6. GENERAL DISCUSSION A comparison of the three differencing schemes described [3] in this report with PUFF and Lax Wendroff seems to refute the often asserted statement that Lagrangian codes are overwhelmingly superior [12] to Eulerian codes in one dimension. In general, donor preserves quite good definition at the shock front but is very unstable in rarefaction regions. Donor also shows waves behind the shock front (on the rarefaction side) which are evidence of the start of instabilities. Continuous rezone and OIL are stable in rarefaction but often have larger overshoots than donor at the shock. For many of the test problems, continuous rezone and OIL are almost indistinguisable. There is a general loss of fine detail in rarefaction regions in all three schemes as is evidenced by problems 5 and 6A. The mechanism for breakdown of donor in rarefaction is not, as yet, ascertained. In problem 5, instabilities seem to set in at the intersection of two curves of differing gradients, whereas in OIL and continuous rezone there is one continuous curve. For a general purpose hydrocode, donor, due to its rarefac- tion problems is less preferable than either OIL or continuous rezone. A hybrid scheme which uses donor at the shock front and one of the other two codes in the rest of the problem is probably best. 6o APPENDIX A PLOTS FROM THE TEST PROBLEMS 6i tt?*Q s-OIX) f£-*UJ3/UJD) JL1ISN30 wo 02'0 9\'o arc. — i 1 1 h- 80 '£ o 4- m CI Q CU- o r-5 o CO r\i | — o .-in X CD o o o o f sOIX) (2 00*052 DO'DOa 00 00 '00! 62 Dirosi ( eOIX) G3SAi:]UlI3Q~l3A OG 'SSI -t— 00*001 1— 00 DO OS h- 0Q'S2 h- 00 * 6G "6Z.fi -h- o o o o d m o o d o o CD d d 1 ^ ru ( — UJ 2bJ o O d o o o 3-0 DO'OftTi ( eOIX) OO'OOh 00'09£ (WD/gdh^) ;ohto oo'oze oo'oea ■33d£ oo-ofif 3 63 ( q-QIX) rt--UJ3/klO) WISN3U o eo'ct; 00 h DO'OfaE D0"00£ ( 2 0IX) DO "IIS, 013*002 (2**h3/g3nxq) imisj 00'OSt oo'oor 6k { E OIX) D3S/OUI3Q13A oo h~ 521 — h- 00 '001 h- 00 -I— 00"S2 h- o OO'Dd zr ■K GG v.LV OD'OfiFj ( e 0IX) k- oo'oo?^ l/S -*- +- o d o o a? o o d rj po oil GC LL_ r\j ) — ° h- °;z: o o OJ o O CO ru 00'09£ (N3/CGH^) jDH^G 00'0?£ IbN^lNl 00 00'OfiZ" DJdS 65 WO h— - ( s-0 IX) tL«*U3/H3) U1SNJG VO 112*0 ST'O h , o +■ fcircb- CDQ i->JG -I +Q -QJ 'OQt +- 00 *Gh? + 3 o 4- o o 4- LO 0> rj 4-crCr CvJ I — r j LjJ O O -i- o cm O o -- CD o o eOtX) (2**W0/S3NJia) IMlSS^bld CO *0SI OJ OO'OB o CO e •H EH a; a o N QJ K co 3 O G •H -P CI o o OJ 0) HO •H 66 I t OlX) mS/O UI3Q13A ■os I — h- 021 — b~ nnTi( OO'Oh 1— OQ'Oh-c-; =t o 4- =T o o o o o o CJ o I J d LiJ DC o zr CD o o d o OJ o o -- CD O d t-OJ cr. 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OCTSl oo-o- 1 1 h CD OO'SI-Q -4- oj o o o CO O o o zr o o o o o 02 I o H CO II EH CI o — o sen g cdCT. m LU o I - +* .o ° ft 3, •H O O d CO o o -- o o o 00'0fi2 00* 00c 00 "091 00 '0ZI 00'08 (siOIX) (W3/S9H3) J.0d3N3 1UNH31NI 00*0fi "33dS 00'CP lUi ( T,-DIX) (E**W3/W9) UISN3Q I— Ofi'O H- fuTO 91 '0 eo*o o 00'Qj 4- CM o 4- =r ru o CD Q_ a o O o a? nj o o o en GC •» uj o — ' u bD •H P-4 o o d CD o o +-0 rr o o iQ "021 00 '00 1 00 "08 00 '09 OO'Ofi 00 "OS (oioix) (z**wo/s3NJim 3ynss3yd 00*0° 142 OG' r :.L f— ( Z.OIX) G3S/W0) XlIG0n3A 00*03 OO'Sh OO'OE 00*51 OO'O- 1 h- 1 1 h o -+ cm CO o o o CD o o CD =r Oi]"0h[ o a) CO I o H X 00 '091 00-021 OO'OB (sioix) (HD/scy^) ji0cj3N3 idNy^iNi 00 -Oh °33dS 00*0° H O 1*6 Ofi'O J— U-OIX) 1 f£**N3/N0) JULISN3CI f\'d'0 91 "0 80'0 °4 CJ c? 4- OJ Co o o o CO OJ CD o o zr cu • o :3NJia) 3yn9S3yd Ikk Q.QSL 00 '09 ( zOIX) G3S/W3)UI3013A oo 00 *0£ OO'ST 1— oo-o- H o 00'SI-d CO o o o cc cu o o o I c o n CO'OftS < c c Hi •H CO '002 00 "091 00*021 oo *oe 00'Ofi 00*0 (siOIX) M3/SDU3U9U3N3 lUNy3J.NI "33dS l*+5 APPENDIX B DERIVATIONS OF ORDERS OF TRUNCATION ERRORS Conservation of Mass (a) Continuous Rezone It is necessary to find expressions for U. 1 X ~2 From equation (12) u n = u n - (p. n n - p. n J -&. 1 i i-fl i-l y _ n Therefore U n = U? l l 2p. Ax l f x | , . o(ax 2 )] & • (IT) p i Hence U. n l = U. n l - — |£ I . + O(AxAt) (l8) i noting that r-= ~ 1 + E if E «1 JL —ill Also t)A = UA . * || | + O(AxAt) (19) Therefore !J.*1 - ffA = g- | . ax + 0(ax 3 ) + O(AXAt) (20) n n *n In general AM, 1 = p. 1 U. 1 At 2 1+ 2 1+ 2 Ik6 Hence AM. 1 1+ 2 11 TT n T A 4- p. tf 1+ l At 1 + (u. n i -u. n i) ^ 1+2 i~2 Ax p n U A At 1 1+ 2 1 + |2 I . At + 0(At 2 ) + 0(Ax 2 At) OX ' 1 from equation (20) Therefore A ,, n n n ., AM . , 1 = p . At 1+ 2 U A - At 1+ 2 _1 dP n Sx" au . + U* i i ox + (AxAt) + 0(At ) (21) from equation (19) Also AM. n l X "2 11 AX P i-1 ^ U. n i - At 1 "2 n ox ' i l-l ox i-1 + O(AxAt) + 0(At ) The equation of continuity is M n+1 - M* = AM. n i - AM. n l i+: Therefore At i = P. , U. 1 - p. U. 1 + 0(At ) + O(AxAt) K i-1 1-2 i 1+2 lU7 = P^! U^! " P? lI ? + ^A* 2 ) + °( At2 ) + O(AxAt) Therefore P n+1 - P n ( P u) n - ( P u) n - 1 1 __ 1 1-1 At Ax - It + ^ + °C*0 ♦ °(a*) + < ) (b) OIL AMA 1+ 2 p n At U. n i 1+ 2 ! t (u* -U n ) ^ v l+l i y Ax (22) p n At U n l 1+2 1 + ^7 I niA t + O(A^At) + 0( A t 2 ) o x 1+ ?T Therefore AM. n l = p n At i+2 x U. n i - At 1+ 2 1_ dP n d~x P^ du l ■ l dx + O(AxAt) + 0(At ) (23) Since equation (23) is identical to equation (2l) ? the rest of the calculation for OIL is the same as for continuous rezone. (c) Donor AM. n l = p." U." At 1-7T i-l i-l Also p n ~n n n P - P. A . U. = p. U - i+1 i-l l l l l L -J At 2Ax from equation (12) 11+8 Therefore P^ 1 -P n P n u n -P» U* ,2 i i , 1 1 1-1 l-l _ , d p | aT" + S At ^2 'i-i Therefore n+1 n n TT n n TT n p. -P. p. IT. - p. . U. . ^ >, TT Conservation of Momentum (a) donor The form of the conservation of momentum equation which will be used is dpU dpU 2 dP _ 3T + 37" + Sx" = ° The finite difference equation used is M? +1 tf" 1 = M^U* + AM. n l U. , -AM. n l U. (2k) i-p i-1 i+o i Therefore M? +1 U? +1 - I^U" ' „ „ „ l_i. p." Tj. 2. p n u. 2 - (P." - P.M i-1 l-l i i l+l i-l At Therefore n+1 TT n+-l n TT n _ n _, n „ ~ 2 ~ 2 P . u . - p . u. P. , , - P. - p . t U. , - p .U. / oc -\ i j l l l+l i-l _ i-l i-l l l (25) At 2 Ax ~ Ax li+9 The right hand side equals p i-l l-l i 2 r i f^P I n^A 2 \\ £b n tt 11 A p I ft ^2^At Ax from equation (17) n /._ n s2 n /T7 n\2 p i-l (U i-l> - p i (U 1 } Ax + O(At) Therefore > n+i u n+l - P n u 2 ? P n d^) 2 - P . n n (u n y p.* - p. n 1 1 k i 1 1 v 1 1-1 v 1-1 1+1 1- At Ax 2Ax |pu + Ipu 2 + |p + Q(Ax) + Q(At) dt dx 3x (t>) continuous rezone U n AM. 1 = p n At 1 1+2 1 u. n i n dx I 1 1 ox I 1 p i _+ O(AxAt) + 0(At ) dP . rt / A 2v\ At ^ii + 0(Ax n- p i 1 from equations (17) and (2l) 150 i i At n AM. n i ff. n n - AM. 1 9? , 1-7T 1-1 1+7T 1 i i i+l i-1 2 2 p n + l ^+1 _ p n yn p _^ _ p ^ 2Ax AxAt from equation (25) The right hand side equals 'i-1 Ax u. n i u. n n i-^ i-i - At U n I -i- H'i-i 1 dP , „n SU ox ' 1 i-i ox ' 1 -'} - U i^(ii I i i + 0(Ax2) ) ^r + °(^ t) + 0(At2) 2 v. J p J n Ax U. n l U" - Atl^ri-^l. + U?|H i+7T i i ( n ox ' l l ox l p i ) " U i^ff I i + °(^ 2 ))^ + O(AxAt) + 0(At 2 ) 1 j n TT n n TT n n TT n. TT n\ — p. t U. 1 U. . - p . U. 1 u. ) Ax \ i-l i~2 1 " 1 x 1+ 2 V r TT n j oP , n TT n oU | \ 1-13 ox ' i i-l i-l ox ' i-l/ + 0(At) + 0(|| ) At Ax 1 \ OX 1 1 1 ox '} ;«4(lii..-»'- 2 i}-»4(£'.*°to 8 | 151 ^ + O(Ak) + O(At) v Ax Hence, equation (25) for continuous rezone becomes n+1 TT n+l n n n / TT ns2 n , _ n ,2 _ n _ n p. U. - p. IT. p. (U.) - p. . (U. , ) P. ,_ - P. . 1 i 1 1 1 i 'l-l v i-l / 14-1 1-1 At Ax + 2"Ax dpU , SpU 2 , dP , ^/ v _/ Aj v ^/At 2 at - ^ + §£■ 4- 0(ax) + O(At) + 0(^ ) dx ox V AX (c) OIL Since equation (23) is identical to equation (21) and the U. (i an integer) for continuous rezone and OIL are equal, the calculation for OIL is the same as for continuous rezone. Conservation of Energy The form of the conservation of energy equation which will be used is SfiE + SpJU + aPU = Q ~dt dx o\x It can be seen that 152 E n l = E p. At 1 P. Ax h U n At [u. n , + u. n n - u. n , - u. n J i+j. l+l 1-1 i-l J _ u At , n n s d , n — ^i+1 i-l/ At ^ ^ li i+l 2p*Ax 8(p^) 2 Ax 2 ^) 2 The conservation of energy equation in finite difference form is T,/r n+ l -n n+ l r/P- T=i n a*, n n ^ n *», ^, ~n M. E. = K. E. + AM. 1 E. , - AM. IE. li li i~2 i" 1 1+ 2 .,n+l -n+1 -Ji _n M. E. - M. E. or _i l ii At -F n r TT n ~ n T . n ~ n -, i [U. , , + U. , - U. , - U. , 1 -j7 i+1 l+l i-l i-l -U i (P n - P n ) -| V± i+1 H-l ; n N 2 ..n-t-1 _n+l „n n M. E. - M. E. Hence i l l l At +_^_ (p^ - p^) 8p n Ax n *n, ~ n n *n., ^n + p . _ U. 1 E. , - p. U. ,1 E. M i-1 i~2 i-l M i 1+2 x -dPU , n . J n *n., ~ n •s — . Ax + p. . U. IE. , OX ' 1 'l-l i-^- 1-1 n *n n <^n - p . U. 1 E. M i 1+2 i + 0(ax ) + O(AxAt) (26) (a) Donor U* n l = u" and U* n l = U n i"2 i- 1 1+2 x n *n, ~ n n *n.. ~n Hence p . _ U. 1 E. , - P . U. 1 E. = l-l i~2 i-l i 1+2 i 153 i-l i-l -,1 » n ^ n P." At E i-1 " -i=i (U ? + u n - u. n - u. n Q ) p.^Ax^ X X X " 2 X " 2 U. n , At _ -1=1 — (P n - p.*) +■ o(At 2 ) n A v l i-2 / 2p * n TT n p i U i T-.n _n A , E. - P. At /.. n ~ n 1 l (U. , + U. ., - v l+l i+l p n Ax 1| -u" -u. n J i-l i-l U? At " — (P* - P. n J + 0(At 2 ) _ n A v i+l i-l' v ' 2p . Ax l = p. n _ U. n , E. n . - p n U n E n + 0(At 2 ) + O(AxAt) i-l i-l i-l ill Hence equation (26) "becomes n+l „n+i n _n , _ T vn , _, T \n P. E. - p. E. «v_ T (pEU). - (pEU). . li li oPU in 4- v ' i x / i-. At dx 'i Ax = O(At) + O(^) 15k Therefore ,n+l (pE)f L - (pE)J ( pEU)* - (PEU)^ (PU)J - (PU)^ At A x Ax dpE dpEU dPU dt chc dx * O(At) + O(Ax) f 0(^-) (t>) continuous rezone n "*n ~n p. U. 1 E. i 1+2 1 n 1+ 2 \,2. P J 31 i+ 2 1+ 2 5x-|i| E^ - p? At ... n ~ n 1 -i ( Vl + U i + 1 p n Ax U l U n At l n ~ n -U." - U." ) l-l l-l ^l" P i-l) + °^> 2 Pi Ax Hence, the right hand side of equation (26) becomes . |2L |J ax + (Ax 2 ) + O(AxAt) U.n + |U | Ax _ At l-l ox 'i-l p 1 dP - v- I • 1 + U. 1 n n dU i-p- §x ' i-li + O(AxAt) + 0(At 2 ) + 0(Ax 2 ) X 155 P * At p n n _ _i-l [xj n + ^n _ ,n _ g n -, l-l l-l 4Ax l l 1-2 i-2 U • (P n - P. n o ) At + 0(At 2 ) 2Ax v i i-2' TT n dU I Ax . . f 1 dP I ' TT n. ^U p U . + -T- . -7T - At / T- ,1+ U. It- 4 i ox i 2 < 2 n dx i+^ i+^ dx i + O(AxAt) + 0(At 2 ) + 0(Ax 2 ) X P n At p n E n._i ^n ~n n . jjn j l l 4Ax l+l l+l l-l l-l - u (p n - p n ) i k l+l i-l ; At n / A+ 2^ "IS + 0(At ) Hence, it can be seen that M. E. - M. E. ^___ l l l l oPU At + d~F~ ' i . Ax + P n E n U* ill P* E* U* l-l l-l l-l = 0(At 2 ) + 0(Ax 2 ) + O(AtAx) 156 Therefore (pE)* +± - (p E )* (pEtl)* - (pEU)* (FU)* - (HI) * -— — — — + + — — — - At Ax Ax dpE SpEU oTU _/ A v «/ A .\ A /At N (c) OIL As stated previously, the mass transport terms for continuous ~n / \ rezone and OIL are equal. Also, the E. (i an integer) for continuous rezone and OIL are equal and hence the calculations of the order of the error for OIL and continuous rezone are the same. 157 LIST OF REFERENCES [1] Evans, M. W. , and Harlow, F. H. , "The Particle-in-Cell Method for Hydrodynamic Calculations", Los Alamos Scientific Laboratory- Report No. LA -2139 (1957). [2] Hicks, Darrell, "Hydrocode Test Problems ", Air Force Weapons Labora- tory Report No. AFWL-TR-67-127 (1968). [3] Hicks, Darrell and Pelzl, Robert, "Comparison between a Von Neumann- Richtmyer Hydrocode (AFWL's PUFF) and a Lax-Wendroff Hydrocode", Air Force Weapons Laboratory Report No. AFWL-TR-68-112 (1968). [k] Von Neumann, J., and Richtmyer, R., "A Method for the Numerical Cal- culation of Hydrodynamic Shocks", J. Appl. Phys., Vol. 21 (1950), P. 232. [5] Hicks, Darrell, "The Convergence of Numerical Solutions of Hydrodynamic Shock Problems", Air Force Weapons Laboratory Report No. AFWL-TR- 69-20 (1969), P. 32. [6] Landshoff, R. , "A Numerical Method for Treating Fluid Flow in the Presence of Shocks", Los Alamos Scientific Laboratory Report No. LA-1930 (1955). [73 Bjork, R., Brooks, N. , and Papetti, R., "A Numerical Technique for Solution of Multidimensional Hydrodynamic Problems", Rand Cor- poration Report No. RM-2628-PR (1963). [8] Kaplan, M. A., and Papetti, R. A., "An Analysis of the Two Dimensional Particle-in-Cell Method", Rand Corporation Report No. RM-U876-PR (1966). [9] Amsden, Anthony A. , "The Particle-in-Cell Method for the Calculation of the Dynamics of Compressible Fluids", Los Alamos Scientific Laboratory Report No. LA-3^6 (1966), p. 25. [10] Johnson, W. E. , "OIL: A Continuous Two -Dimensional Eulerian Hydro- dynamic Code", General Dynamics Corporation, General Atomic Division Report No. GAMD 5580 (1965), p. 23= [11] Rich, Marvin, "A Method for Eulerian Fluid Dynamics", Los Alamos Scientific Laboratory Report No. LAMS-2826 (1962). [12] Schulz, William D. , "Two -Dimensional Lagrangian Hydrodynamic Difference Equations", Meth. Comp. Phys., Vol. 3 (196*0, P- !• UNCLASSIFIED Security Classification DOCUMENT CCMTROL DATA -R&D (Security claealllcatlon ot tltlm, body of aba tree t and Inde Ing annotation null he tnxared mhrm the overall report la clmamlllmd ORIGINATING ACTIVITY (Corpcra la author) apartment of Computer Science liversity of Illinois at Urbana-Champaign -bana, Illinois 61801 2». NE^OIIT SECURITY C I A * II F ' C A T IOf UNCLASSIFIED 2b. OIIOUP REPORT TITLE ¥ ANALYSIS OF SOME EULERIAN METHODS FOR ONE DIMENSIONAL, WISCID, COMPRESSIBLE HYDRODYNAMICS descriptive NOTES (Type ol report and tr.clualve datum) 'search Report AUTHOR(S) (Flrmt nan*, middle Initial, leal nama) ,anley J. Flower dew REPORT DATE ?ptember 22, I969 70, TOTAL NO. OF PACES 163 76. NO. OF REFS 12 CONTRACT OR GRANT NO. 5-86-15-305 PROJECT NO. 5 af 30(602)>ay+ •a. ORIGINATOR'S REPORT NUMBER!*) DCS Report No. 352 •b. OTHER REPORT NO(S) (Any other number* that may be aaalgnad this report) DISTRIBUTION STATEMENT >.ialified requesters may obtain copies of this report from DCS. SUPPLEMENTARY NOTES DIE 12. SPONSORING MILITARY ACTIVITY Rome Air Development Center Griff iss Air Force Base Rome, New York 13440 I ABSTRACT The description and comparison of several explicit differencing schemes for the solution of the inviscid, non-thermal-conducting, single- material, one-dimensional equations of compressible hydrodynamics is in- cluded in this report. The comparison was based on the plots obtained from a series of one dimensional test problems which are included in APPENDIX A. FORM ..1473 IIWCLASSIPIED Security Classification UNCLASSIFIED Security Classification KEY WORDS explicit differencing schemes one-dimensional, inviscid, compressible hydrodynamics Eulerian methods particle-in-cell method LINK A MOLK «»L8 *T TTOflliAfiBTglED Security Classification