Главная Журналы 11.5.4. Листинг подпрограммы ADAPT с сссссссссссссссссссссссссссссссссссссссссссссссссссссссс SUBROUTINE ADAPT с--EXAMPLE 15 - WATER SEEPAGE UNDER A DAM С---- $INCLUDE:COMMON DIMENSION P(NI,NJ),U(NI,NJ),V(NI,NJ) EQUIVALENCE (F(1,1,1),P(1,1)),(F(1,1/2),U(1,1)), 1 (F(l,l,3) ,V(1,1) ) ENTRY GRID HEADER=WATER SEEPAGE UNDER A DAM PRINTF=PRINT15 PL0TF=PL0T15 CALL INTA4(NZX, 3 , NCVX(1) ,5,NCVX(2) ,2,NCVX(3),5) CALL DATA5(XZONE(1),3.,XZONE(2),.5,XZONE(3),3., 1 POWRX(1),-1.2,POWRX(3),1.2) CALL INTA3(NZY,2,NCVY(1),6,NCVY(2) ,3) CALL DATA2(YZONE(1),3.,YZONE(2) , 1.) CALL ZGRID RETURN ENTRY BEGIN TITLE(1)= PRESSURE TITLE(2)= и VELOCITY TITLE(3)= V VELOCITY CALL INTA8(KSOLVE(1),1,KPRINT(1),1,KPLOT(1),1,KPRINT(2), 1 1, KPL0T(2) , 1,KPRINT (3) , 1, KPLOT (3) , 1, LAST, 3) DC=1. DO 100 J=1,M1 DO 100 1=1,Ll P(I,J)=0. 100 CONTINUE DO 110 1=2,L2 IF(X(I).LT.3.) P(I,M1)=100. 110 CONTINUE RETURN ENTRY OUTPUT CALCULATE U AND V VELOCITIES AT THEIR STAGGERED LOCATIONS DO 200 J=3,M2 DO 200 1=2,L2 V(I,J)=-DC*(P(I,J)-P(I,J-1))/(Y(J)-Y(J-l)) 200 CONTINUE DO 210 1=2,L2 V(I,M1)=-FLUXM1(I,1) 210 CONTINUE DO 220 J=2,M2 DO 220 1=3,L2 U(I, J)=-DC*(P(I,J)-P(I-1,J) )/(X(I)-X(I-l) ) 220 CONTINUE DO 230 J=2,M2 DO 230 1=2,L2 IF(X(I).GT.3..AND.X(I).LT.3.5.AND.Y(J).GT.3.) THEN U(I,J)=0. и(I+l,J)=0. V(I,J)=0. V(I, J+1)=0. ENDIF 230 CONTINUE DO 240 IUNIT=IU1,IU2 IFdTER.EQ. 0) WRITE (lUNIT, 250) 250 FORMATdX, 4TER,2X, P(3,5) , 5X, U(3,5) , 5X, 1 V(3,5) ЧбХ, P(5,5) ,5X, U(5,5) ,5X, •V(5,5) ) WRITE (lUNIT,2 60) ITER,P(3,5),U(3,5),V(3,5), 1 P(5,5),U(5,5),V(5,5) 260 FORMAT(IX,12,1P6E11.2) 240 CONTINUE IFdTER.EQ.LAST) THEN CALL PRINT COME HERE TO FILL IBLOCK(I,J) BEFORE CALLING PLOT DO 270 J=2,M2 DO 270 1=2,L2 IF(X(I).GT.3..AND.X(I).LT.3.5.AND.Y(J).GT.3.) 1 IBL0CK(I,J)=1 270 CONTINUE CALL PLOT ENDIF RETURN ENTRY PHI DO 300 J=2,M2 DO 300 1=2,L2 GAM(I,J)=DC IF(X(I).GT.3..AND.X(I).LT.3.5.AND.Y(J).GT.3.) 1 GAM(I,J)=0. 300 CONTINUE COME HERE TO SPECIFY BOUNDARY CONDITIONS DO 310 1=2,L2 KBCJl(I)=2 310 CONTINUE DO 320 J=2,M2 KBCIl(J)=2 KBCLl(J)=2 320 CONTINUE RETURN END 11.5.5. Результаты расчетов RESULTS OF CONDUCT FOR CARTESIAN COORDINATE SYSTEM WATER SEEPAGE UNDER A DAM V(3,5) O.OOE+00 ITER P(3,5) U(3,5) 0 0.OOE+OO 0.OOE+OO 1 7.33E+01 2.92E+00 -6.14E+00 2 7.33E+01 2.92E+00 -6.15E+00 3 7.33E+01 2.92E+00 -6.15E+00 1=1 2 3 X = O.OOE+OO 3.52E-01 1.04E+00 1 1=8 9 10 X = 3.38E+00 3.72E+00 4.22E+00 4 P(5,5) U(5,5) V(5,5) O.OOE+OO O.OOE+OO O.OOE+OO 6.37E+01 9.50E+00 -4.94E+00 6.37E+01 9.49E+00 -4.95E+00 6.37E+01 9.49E+00 -4.95E+00 4 5 6 7 69E+00 2.28E+00 2.78E+00 3.13E+00 11 12 13 14 81E+00 5.46E+00 6.15E+00 6.50E+00 J=l 23 4 56 7 Y = O.OOE+OO 2.50E-01 7.50E-01 1.25E+00 1.75E+00 2.25E+00 2.75E+00 J = 8 9 10 11 Y = З.ПЕ+00 3.50E+00 3.83E+00 4.OOE+OO PRESSURE ******
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