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Главная Журналы RETURN ENTRY OUTPUT IFdTER.EQ. 15)THEN KSOLVE(1)=0 KSOLVE(2)=1 COND=CP*AMU/PR ENDIF ASUM=0. WSUM=0. TSUM=0. DO 200 J=2,M2 DO 200 1=2,L2 AR=XCVd) *YCV (J) ASUM=ASUM+AR WSUM=WSUM+W(I,J)*AR TSUM=TSUM+W(I,J)*T(I,J)*AR 200 CONTINUE WBAR=WSUM/ASUM TB=TSUM/(WSUM+SMALL) WP=XL+YL DH=4.*ASUM/WP FRIC=-2.*DPDZ*DH/(DEN*WBAR**2 + SMALL) HP=YL DTDZ=QW*HP/(RHOCP*WSUM+SMALL) TWAV=0. DO 210 J=2,M2 TWAV=TWAV+T(1,J)*YCV(J) 210 CONTINUE TWAV=TWAV/YL ANU=QW*DH/(COND*(TWAV-TB)+SMALL) IFdTER.GE.l) AMU=DH*WBAR*DEN/RE DO 220 IUNIT=IU1,IU2 IFdTER.EQ. 0) WRITE (lUNIT, 230) 230 FORMAT(IX, ITER,3X, W(8,8) ,4X, T(8,8) 1 ,3X, AMU,7X, F,9X,NU) WRITE (lUNIT, 24 0)ITER,W(8,8),T(8,8) ,AMU,FRIC,ANU 240 FORMAT(1X,I3,1X,1P5E10.2) 220 CONTINUE IF(ITER.EQ.LAST) THEN DO 250 1=1,Ll DO 250 J=1,M1 W(I, J)=W(I,J)/WBAR T(I,J)=(T(I,J)-TWAV)/(TB-TWAV) 250 CONTINUE CALL PRINT CALL PLOT ENDIF RETURN ENTRY PHI IF(NF.EQ.l) THEN REL=1.-REGAM DO 300 J=2,M2 DO 300 1=2,L2 IFdTER.EQ.0) THEN GAM(I,J)=AMU ELSE TAUW=ABS(FLUXIl(J,1)) XPLUS(I,J)=X(I)*SQRT(TAUW*DEN)/AMU TAUW=ABS(FLUXJl(1,1)) YPLUS(I,J)=Y(J)"SQRT(TAUW*DEN)/AMU RLX=DFACT(XPLUS(I,J))*CAPL(XL,X(I)) RLY=DFACT(YPLUS(I, J))*CAPL(YL,Y(J)) RL=(RLX**(-20)+RLY**(-20) )* * (-0.05) FP=1 . IF(I.NE.L2) FP=XCV(I)/(XCV(I)+XCV(I+l)) WP=FP*W(I+1,J)+(1.-FP)*W(I,J) FM=1. IF(I.NE.2) FM=XCV(I)/(XCV(I)+XCV(I-1)) WM=FM*W(I-l,J)+(1.-FM)*W(I,J) DWDX=(WP-WM)/XCV(I) FP=1 . IF(J.NE.M2) FP=YCV(J)/(YCV(J)+YCV(J+1)) WP=FP*W(I,J+1)+(1.-FP)*W(I,J) FM=1 . IF(J.NE.2) FM=YCV(J)/(YCV(J)+YCV(J-1)) WM=FM*W(I,J-1)+(1.-FM)*W(I,J) DWDY=(WP-WM)/YCV(J) GAMT=DEN*RL**2*SQRT(DWDX**2+DWDY*2) IF(ITER.EQ.l) AMUT(I,J)=GAMT AMUT(I,J)=REGAM*GAMT+REL*AMUT(I,J) GAM(I,J)=AMU+AMUT(I, J) ENDIF 300 CONTINUE COME HERE TO ADJUST THE PRESSURE GRADIENT IFdTER.EQ. 1) THEN GAMAV=0. DO 305 J=2,M2 DO 305 1=2,L2 GAMAV=GAMAV+GAM(I, J)* XCV(I)* YCV(J) 305 CONTINUE GAMAV=GAMAV/ASUM DPDZ=DPDZ*GAMAV/GAMAVP GAMAVP=GAMAV ENDIF DO 307 J=2,M2 DO 307 1=2,L2 SC(I,J)=-DPDZ 307 CONTINUE ENDIF IF(NF.EQ.2) THEN DO 310 J=2,M2 DO 310 1=2,L2 GAM(I,J)=COND+CP*AMUT(I,J)/PRT SC(I,J)=-RHOCP*W(I,J)DTDZ 310 CONTINUE ENDIF COME HERE TO SPECIFY BOUNDARY CONDITIONS DO 320 1=2,L2 KBCMl(I)=2 320 CONTINUE DO 330 J=2,M2 KBCLl(J)=2 330 CONTINUE С IF(NF.EQ.2) THEN DO 340 1=2,L2 KBCJl(I)=2 340 CONTINUE DO 350 J=2,M2 KBCIl(J)=2 FLXCIl(J)=QW 350 CONTINUE ENDIF RETURN END С CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC 11.3.5. Результаты расчетов Results of conduct for cartesian coordinate system ************************************************** turbulent flow in a square duct
x= 0.ooe+oo 2.50e-04 2.25e-03 8.75e-03 2.28e-02 4 ,73e-02 8.52e-02 x= 1.40e-0i 2.14e-01 3.10e-01 4.32e-01 5.ooe-01 j= 1 2 3 4 5 6 7 y= o.ooe+00 2.50e-04 2.25e-03 8.75e-03 2.28e-02 4.73e-02 8.52e-02 j= 8 9 10 11 12 y= 1.40e-01 2.i4e-01 3.10e-0i 4.32e-01 5.ooe-01 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 [ 82 ] 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 |
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