Design and Optimization of Thermal Systems Episode 3 Part 8 ppsx

Appendix A 647 p2=[1 -10 35 -50 24]; p3=[1 -7 17 -17 6]; y1=polyval(p1,x); y2=polyval(p2,x); y3=polyval(p3,x); plot(x,y1,‘g’,x,y2,‘b’,x,y3,‘m’) hold on y = zeros(size(x)); plot(x,y,‘r’) A.M.6 % Ordinary Differential Equations % % Given ODE: dh/dt = [6x10 -4 – 3x10 -4 xh 0.5 ]/0.03 % % Enter given ODE % dhdt=inline(‘(6*10^(-4) - 3*10^(-4)*(h^(0.5)))/0.03’,‘t’,‘h’); % % Choose step size and total time % dt=10; tn=200*dt; h0=0; t=(0:dt:tn)’; n=length(t); h=h0*ones(n,1); % % Euler’s Method % for j=2:n; h(j)=h(j-1)+dt*dhdt(t(j-1),h(j-1)); end plot(t,h,‘-b’) % % Heun’s Method % t=(0:dt:tn)’; for j=2:n h1=h(j-1)+dt*dhdt(t(j-1),h(j-1)); h(j)=h(j-1)+(dt/2)*(dhdt(t(j),h1) + dhdt(t(j-1),h(j-1))); end hold on plot(t,h,‘-g’) % % Using ode23 % [t,h] = ode23(dhdt,2000,0) % Higher order ODE % % Second order ODE: d 2 x/dt 2 = 9.8 – 0.05 (dx/dt) 648 Design and Optimization of Thermal Systems % Replaced by two first order ODEs: dx/dt = y; dy/dt = 9.8 – 0.05 y % % Enter two first-order equations % dxdt=inline(‘y’,‘t’,‘x’,‘y’); dydt=inline(‘9.8-0.05*y’,‘t’,‘x’,‘y’); % % Give step size, end point and starting conditions % dt=0.5; tn=40*dt; x0=0; y0=0; t=(0:dt:tn)’; n=length(t); x=x0*ones(n,1); y=y0*ones(n,1); % % Euler’s Method % for j=2:n; x(j)=x(j-1)+dt*dxdt(t(j-1),x(j-1),y(j-1)); y(j)=y(j-1)+dt*dydt(t(j-1),x(j-1),y(j-1)); end plot(t,x,‘-b’,t,y,‘-g’) % % Using ode45 % % Define function % function dydt=rhs(t,y) dydt=[y(2);9.8-0.05*y(2)]; % % Solve ODE given by function ‘rhs’ % y0=[0;0]; [t,y] = ode45(‘rhs’,20,y0) % % Then y(1) gives x and y(2) gives y % A.F.1 C GAUSSIAN ELIMINATION FOR A TRIDIAGONAL SYSTEM C C A(I), B(I) AND C(I) ARE THE THREE ELEMENTS IN EACH ROW OF C THE GIVEN SYSTEM OF EQUATIONS, F(I) REPRESENTS THE CONSTANTS C ON THE RIGHT-HAND SIDE OF THE EQUATIONS, T(I) ARE THE TEMPERATURE C DIFFERENCES TO BE COMPUTED, G IS A PARAMETER DEFINED IN THE GIVEN C PROBLEM, N IS THE NUMBER OF EQUATIONS AND TP REPRESENTS THE C PHYSICAL TEMPERATURE, WHERE TP = T + 20. THE SYSTEM OF EQUATIONS C TO BE SOLVED IS THE ONE GIVEN IN EXAMPLE 4.4. Appendix A 649 C PARAMETER (IN=30) DIMENSION A(IN),B(IN),C(IN),T(IN),F(IN) C C SPECIFY INITIAL PARAMETERS C CALL INPUT(A,B,C,F,N) CALL TDMA(A,B,C,F,N,T) C C COMPUTE ACTUAL TEMPERATURES FROM THE TEMPERATURE DIFFERENCES T(I) C WRITE (6,7) 7 FORMAT(2X,’THE REQUIRED PHYSICAL TEMPERATURES IN CELSIUS ARE’/) DO 8 I=1,N TP=T(I)+20.0 WRITE (6,9)I,TP 8 CONTINUE 9 FORMAT(2X,’TP(‘,I2,’)=’,F10.4) STOP END C********************************************************************* C GET THE INPUT DATA C********************************************************************* SUBROUTINE INPUT(A,B,C,F,N) PARAMETER (IN=30) DIMENSION A(IN),B(IN),C(IN),F(IN) PRINT *, ‘GIVE THE VALUE OF N’ READ *, N G=50.41*(0.01**2) C C ‘FMTDM’ FORMS THE TRIDIAGONAL MATRIX AND THE RIGHT HAND SIDE C COLUMN MATRIX C CALL FMTDM(G,N,A,B,C,F) RETURN END C********************************************************************** C THE FOLLOWING SUBROUTINE FORMS THE TRIDIAGONAL MATRIX OF THE FORM C C A*T(I-1) + B*T(I) + C*T(I+1) = R C********************************************************************** SUBROUTINE FMTDM(G,N,A,B,C,R) DIMENSION A(N),B(N),C(N),R(N) C C ENTER THE CONSTANTS ON THE RIGHT-HAND SIDE OF THE EQUATIONS C R(1)=100.0 R(N)=100.0 NN=N-1 DO 1 I=2,NN R(I)=0.0 1 CONTINUE C C ENTER THE MATRIX COEFFICIENTS 650 Design and Optimization of Thermal Systems C DO 2 I=1,N B(I)=2.0+G 2 CONTINUE DO 3 I=1,NN C(I)=-1.0 3 CONTINUE DO 4 I=2,N A(I)=-1.0 4 CONTINUE RETURN END C********************************************************************** C TRIDIAGONAL MATRIX ALGORITHM C********************************************************************** SUBROUTINE TDMA(A,B,C,F,N,T) C C N IS THE ORDER OF THE TRIDIAGONAL MATRIX C A IS THE SUBDIAGONAL OF THE TRIDIAGONAL MATRIX C B IS THE DIAGONAL OF THE TRIDIAGONAL MATRIX C C IS THE SUPERDIAGONAL OF THE TRIDIAGONAL MATRIX C F IS THE RIGHT HAND SIDE VECTOR C T IS THE SOLUTION VECTOR C DIMENSION A(N),B(N),C(N),F(N),T(N) NN=N-1 DO 5 I=2,N D=A(I)/B(I-1) B(I)=B(I)-C(I-1)*D F(I)=F(I)-F(I-1)*D 5 CONTINUE C C APPLY BACK SUBSTITUTION C T(N)=F(N)/B(N) DO 6 I=1,NN J=N-I T(J)=(F(J)-C(J)*T(J+1))/B(J) 6 CONTINUE RETURN END A.F.2 C GAUSS-SEIDEL METHOD FOR SOLVING A SYSTEM OF LINEAR EQUATIONS C C C T(I) REPRESENTS THE TEMPERATURE DIFFERENCES FROM THE AMBIENT C TEMPERATURE, TO(I) DENOTES THE TEMPERATURE DIFFERENCES AFTER C THE PREVIOUS ITERATION, TP IS THE ACTUAL TEMPERATURE, S IS A C CONSTANT DEFINED IN THE PROBLEM AND N IS THE NUMBER OF C EQUATIONS. THE PROBLEM CONSIDERED IS THE ONE GIVEN IN C EXAMPLE 4.4. Appendix A 651 C C C ENTER VALUES OF RELEVANT PARAMETERS C PARAMETER (IN=30) DIMENSION T(IN),TO(IN) S=50.41*(0.01**2)+2.0 PRINT *, ‘GIVE THE NUMBER OF EQUATIONS : ’ READ (5,*) N NN=N-1 EPS=0.1 C C DIFFERENT CONVERGENCE PARAMETER EPS C DO 10 K=1,5 C C INPUT STARTING VALUES C J=0 DO 1 I=1,N T(I)=0.0 1 CONTINUE C C STORE COMPUTED VALUES AFTER EACH ITERATION C 2 DO 3 I=1,N TO(I)=T(I) 3 CONTINUE C C COMPUTE THE END VALUES T(1) AND T(N) C T(1)=(T(2)+100.0)/S T(N)=(100.0+T(N-1))/S C C COMPUTE INTERMEDIATE VALUES C DO 4 I=2,NN T(I)=(T(I+1)+T(I-1))/S 4 CONTINUE C C CHECK FOR CONVERGENCE C J=J+1 DO 5 I=1,N IF(ABS(TO(I)-T(I)) .GT. EPS) GO TO 2 5 CONTINUE WRITE(6,6)EPS 6 FORMAT(//2X,‘EPS=’,F10.5) WRITE(6,7)J 7 FORMAT(/2X,‘NUMBER OF ITERATIONS=’,I4/) C C COMPUTE ACTUAL TEMPERATURES C DO 8 I=1,N 652 Design and Optimization of Thermal Systems TP=T(I)+20.0 WRITE(6,9)I,TP 8 CONTINUE 9 FORMAT(2X,‘TP(‘,I2,’)=’,F12.4) EPS=EPS/10.0 10 CONTINUE STOP END A.F.3 C ROOT SOLVING WITH THE SECANT METHOD C X IS THE INDEPENDENT VARIABLE, FUN(X) IS THE GIVEN FUNCTION, C X1 AND X2 ARE THE X VALUES FROM THE TWO PREVIOUS ITERATIONS, C STARTING WITH THE TWO POINTS BOUNDING THE REGION, X3 IS THE C APPROXIMATION TO THE ROOT, F1, F2 AND F3 ARE THE CORRESPONDING C VALUES OF THE FUNCTION, AND EPS IS THE CONVERGENCE CRITERION C THE FUNCTION USED IS THE ONE IN EXAMPLE 4.2. C EXTERNAL FUN PRINT *, ‘ENTER THE TWO STARTING VALUES OF X’ READ (5,*) X1,X2 C C STORE STARTING VALUES X1I=X1 X2I=X2 XOLD=X1 WRITE(6,12) X1,X2 12 FORMAT(/10X,‘INITIAL X1=’,F7.2,10X,’INITIAL X2=’,F7.2//) EPS=0.01 DO 2 I=1,4 1 F1=FUN(X1) F2=FUN(X2) C C COMPUTE THE APPROXIMATION TO THE ROOT C X3=(X1*F2-X2*F1)/(F2-F1) F3=FUN(X3) XNEW=X3 C C CHECK FOR CONVERGENCE IF (ABS(XNEW-XOLD) .GT. EPS) THEN X1=X2 X2=X3 XOLD=X3 WRITE(6,10)X3,F3 10 FORMAT(2X,‘TEMPERATURE T =’,F10.4,4X,‘FUNCTION F(T) =’, $ F12.6) GO TO 1 ELSE Appendix A 653 11 WRITE(6,13)EPS,X3,F3 13 FORMAT(//2X,‘EPS=’,F9.6,4X,‘TEMPERATURE T =’,F10.4,4X, $ ‘FUNCTION F(T)=’,F12.6//) END IF C C VARY CONVERGENCE CRITERION C EPS=EPS/10 X1=X1I X2=X2I XOLD=X1 2 CONTINUE STOP END C C DEFINE THE FUNCTION C FUNCTION FUN(X) FUN=(0.6*5.67*((850.0**4.0)-(X**4.0))/(10.0**8.0))-40.0*(X-350.0) RETURN END A.F.4 C THIS PROGRAM FINDS THE REAL ROOTS OF AN EQUATION F(X)=0 C BY THE NEWTON-RAPHSON METHOD C C C C HERE X IS THE INDEPENDENT VARIABLE, Y1 THE VALUE OF THE C FUNCTION AT X, Y2 THE FUNCTION AT X+0.001, YD THE DERIVATIVE, C DX THE INCREMENT IN X FOR THE NEXT ITERATION, EPS THE C CONVERGENCE CRITERION ON THE FUNCTION AND XMAX THE MAXIMUM C VALUE OF X. THE FUNCTION USED IS THE ONE IN EXAMPLE 4.2. C C C DEFINE FUNCTION AND SPECIFY INPUT PARAMETERS C EXTERNAL Y EPS=0.001 WRITE(6,15)EPS 15 FORMAT(2X,‘EPS=’,F8.4/) PRINT *, ‘ ENTER AN INITIAL GUESS FOR X’ READ (5,*) X XMAX=850.0 1 Y1=Y(X) WRITE(6,10) X,Y1 C C CHECK FOR CONVERGENCE C IF (ABS(Y1) .GT. EPS) THEN XN=X+0.001 Y2=Y(XN) 654 Design and Optimization of Thermal Systems YD=(Y2-Y1)/0.001 C C CHECK IF RESULTS DIVERGE C IF (YD .GE. (1.0/EPS)) GO TO 20 C C COMPUTE NEW APPROXIMATION TO THE ROOT C DX=-Y1/YD X=X+DX IF (X .GE. XMAX) GO TO 20 GO TO 1 ELSE 5 WRITE(6,12) X,Y1 12 FORMAT(/2X,‘TEMPERATURE T =’,F8.4,4X,‘FUNCTION F(T)=’,F12.6) 10 FORMAT(2X,‘TEMPERATURE T =’,F8.4,4X,‘FUNCTION F(T)=’,F12.6) END IF 20 STOP END C C DEFINE THE FUNCTION C FUNCTION Y(X) Y=(0.6*5.67*((850.0**4.0)-(X**4.0)))/(10.0**8.0) - 40.0*(X-350.0) RETURN END A.F.5 C THIS PROGRAM SOLVES THE LAPLACE EQUATION BY EMPLOYING C THE SUCCESSIVE OVER RELAXATION (SOR) ITERATION METHOD. C C WHEN THE PROGRAM IS RUN IT PROMPTS FOR THE INPUT VALUES REQUIRED. C ENTER THE INPUT VALUES AND YOUR OUTPUT WILL BE IN A FILE CALLED C ‘SOR.DAT’ C C C DESCRIPTION OF INPUT PARAMETERS: C C IL IS THE NUMBER OF GRID POINTS IN THE X DIRECTION. C JL IS THE NUMBER OF GRID POINTS IN THE Y DIRECTION. C DX IS THE GRID SIZE IN X DIRECTION. C DY IS THE GRID SIZE IN Y DIRECTION. C OMEGA IS THE RELAXATION PARAMETER C PHIINT IS THE INITIAL GUESS FOR PHI TAKEN UNIFORM OVER THE C WHOLE DOMAIN. C ITMAX IS THE NUMBER OF MAXIMUM ITERATIONS BEFORE STOPPING. C EPSI IS THE CONVERGENCE CRITERION. C C C DESCRIPTION OF OTHER VARIABLES: C C PHI IS THE SOLUTION VARIABLE AT NTH TIME STEP. Appendix A 655 C PHIOL IS THE SOLUTION VARIABLE AT N-1TH TIME STEP. C C CHARACTER*2 XFILE(5) CHARACTER*2 YFILE(5) DIMENSION PHI(11,11),PHIOL(11,11) PRINT*,‘ENTER INITIAL GUESS FOR PHI TAKEN UNIFORM OVER THE’ PRINT*,‘WHOLE DOMAIN’ READ(5,*)PHIINT PRINT*,‘ENTER GRID SIZE DX=, DY=’ READ(5,*)DX,DY PRINT *,‘ENTER NO. OF GRID POINTS IL= , JL= ’ PRINT*,‘ MAXIMUM POSSIBLE IS 11 FOR BOTH IL AND JL,’ PRINT*,‘UNLESS DIMENSION STATEMENTS ARE CHANGED.’ READ(5,*)IL,JL PRINT *,‘ENTER THE RELAXATION PARAMETER’ READ(5,*)OMEGA PRINT*,‘ENTER MAXIMUM NO. OF ITERATIONS ALLOWED BEFORE STOPPING’ READ(5,*)ITMAX PRINT *,‘ENTER CONVERGENCE CRITERION’ READ(5,*)EPSI PRINT*,‘THE INPUT VALUES ARE:’ PRINT*,‘INITIAL GUESS FOR PHI=’,PHIINT PRINT*,‘DX=’,DX,‘DY=’,DY PRINT*,‘IL=’,IL,‘JL=’,JL PRINT*,‘MAX NO. OF ITERATIONS=’,ITMAX PRINT*,‘CONVERGENCE CRITERION=’,EPSI ITERATION=0 C C OPEN THE DATA FILES FOR GRAPHING C XFILE(1)=‘X1’ XFILE(2)=‘X2’ XFILE(3)=‘X3’ XFILE(4)=‘X4’ XFILE(5)=‘X5’ YFILE(1)=‘Y1’ YFILE(2)=‘Y2’ YFILE(3)=‘Y3’ YFILE(4)=‘Y4’ YFILE(5)=‘Y5’ C C SET INITIAL DISTRIBUTION OF PHI C DO 51 I=1,IL DO 5 J=1,JL PHI(I,J)=PHIINT 5 CONTINUE 51 CONTINUE C C START SOLVING FOR PHI. C 15 ITERATION=ITERATION+1 IF(ITERATION.GE.ITMAX)GO TO 40 656 Design and Optimization of Thermal Systems C C SAVE THE FIELD AT PREVIOUS TIME STEP. C DO 101 I=1,IL DO 10 J=1,JL PHIOL(I,J)=PHI(I,J) 10 CONTINUE 101 CONTINUE C C DO SOR ITERATIONS ON PHI ON INTERIOR POINTS. C DO 201 J=2,JL-1 DO 20 I=2,IL-1 PHIGS=(PHI(I+1,J)+PHI(I-1,J))/DX**2+ $ (PHI(I,J+1)+PHI(I,J-1))/DY**2 PHIGS=PHIGS/(2./DX**2+2./DY**2) PHI(I,J)=OMEGA*PHIGS+(1 OMEGA)*PHIOL(I,J) 20 CONTINUE 201 CONTINUE C C IMPOSE THE BOUNDARY CONDITIONS C CALL BCOND(PHI,IL,JL) C C CHECK FOR CONVERGENCE C DO 351 I=1,IL DO 35 J=1,JL IF(ABS(PHI(I,J)-PHIOL(I,J)).GE.EPSI)GO TO 15 35 CONTINUE 351 CONTINUE GO TO 50 40 PRINT*,‘SOLN. DOES NOT CONVERGE IN’,ITMAX,‘ITERATIONS’ 50 OPEN(UNIT=10,FILE=‘SOR.DAT’) WRITE(10,110)EPSI 110 FORMAT(1X,‘CONVERGENCE CRITERION =’1X,E9.1) WRITE(10,115)OMEGA 115 FORMAT(//,1X,‘W=’,F5.2) WRITE(10,120)ITERATION 120 FORMAT(//,1X,‘NO. OF ITERATIONS TO CONVERGE=’,1X,I4,//) WRITE(10,130) 130 FORMAT(1X,‘PHI DISTRIBUTION IS:’,//) WRITE(10,140)(I,I=1,IL) 140 FORMAT(1X,‘I=’,8X,11(I2,8X)) DO 60 J=1,JL WRITE(10,100)J,(PHI(I,J),I=1,IL) 60 CONTINUE 100 FORMAT(1X,‘J=’,I2,3X,11(F8.5,2X)) C C OUTPUT FOR GRAPHICS C II=1 [...]... 1 .36 91 1 .36 68 1 .36 46 1 .36 23 1 .36 01 1 .35 80 1 .35 59 1 .35 4 1 .34 9 1 .34 5 1 .34 0 1 .33 6 1 .32 9 1 .32 2 1 .31 6 1 .31 0 1 .30 4 1.299 1. 288 k 24.06 24.45 24 .85 25.22 25. 58 25.96 26 .32 26.70 27.06 27.42 27. 78 28. 14 28. 48 28. 83 29.17 29.52 29 .84 30 .17 30 .82 31 .47 32 .09 32 .71 33 .32 33 .92 34 .52 35 .11 35 .69 36 .24 37 . 63 38. 97 40.26 41. 53 43. 96 46.26 48. 46 50.57 52.61 54.57 58. 29 Pr h a 35 .75 36 . 43 37.10 37 . 78 38 .46 39 .11 39 .76... 421.5 a 2 93. 9 297.4 30 0 .8 30 4.1 30 7.4 31 0.6 31 3 .8 31 7.1 32 0.2 32 3.4 32 6.5 32 9.6 33 2.6 33 5.6 33 8. 5 34 1.5 34 4.4 34 7 .3 350.2 35 3.1 35 5 .8 35 8. 7 36 1.4 36 4.2 36 6.9 36 9.6 37 2 .3 375.0 37 7.6 38 0.2 38 2 .8 38 5.4 38 8. 0 39 0.5 39 3.0 39 5.5 39 8. 0 400.4 405 .3 410.2 Appendix B 661 TABLE B.1 (CONTINUED) Properties of Dry Air at Atmospheric Pressure—SI Units Temperature K 430 440 450 460 470 480 490 500 510 520 530 540 550... 580 590 600 620 640 660 680 700 720 740 760 780 80 0 85 0 900 950 1000 1100 1200 130 0 1400 1500 1600 180 0 C 156 .85 166 .85 176 .85 186 .85 196 .85 206 .85 216 .85 226 .85 236 .85 246 .85 256 .85 266 .85 276 .85 286 .85 296 .85 30 6 .85 31 6 .85 32 6 .85 34 6 .85 36 6 .85 38 6 .85 406 .85 426 .85 446 .85 466 .85 486 .85 506 .85 526 .85 576 .85 626 .85 676 .85 726 .85 82 6 .85 926 .85 1026 .85 1126 .85 1226 .85 132 6 .85 1526 .85 Properties cp F 31 4... 200 C 55 45 40 35 55 48 29 .8 30 0 C 48 42 36 33 48 45 400 C 40 35 33 31 40 36 600 C Thermal Conductivity, k (W/m 36 31 29 28 36 33 80 0 C C) 35 29 28 28 35 33 1000 C 36 31 29 29 36 33 1200 C 670 Design and Optimization of Thermal Systems 7 ,86 5 7 ,86 5 7, 83 3 7, 785 7, 689 7,625 1% 2% 5% 10% 20% 30 % 7 ,86 5 7 ,89 7 20 Cr, 15 Ni 25 Cr, 20 Ni 8, 522 7, 83 3 7 ,86 5 18 Cr, 8 Ni (V2A) 8, 266 8, 0 73 7 ,86 5 80 Ni, 15 Cr 60 Ni,... 16 .85 21 .85 26 .85 31 .85 36 .85 41 .85 46 .85 51 .85 56 .85 61 .85 66 .85 71 .85 76 .85 81 .85 86 .85 91 .85 96 .85 101 .85 106 .85 111 .85 116 .85 121 .85 126 .85 136 .85 146 .85 Properties cp F – 73 –64 –55 –46 37 – 28 –19 –10 –1 8 17 26 35 44 53 62 71 80 89 98 107 116 125 134 1 43 152 161 170 179 188 197 206 215 224 233 242 251 260 2 78 296 1.646 1.607 1.572 1. 537 1.505 1.4 73 1.4 43 1.4 13 1 . 38 6 1 .35 9 1 .33 3 1 .30 8 1. 235 1.261... 18. 93 19.15 19 .39 19. 63 19 .85 20. 08 20 .30 20.52 20.75 20.97 21. 18 21 . 38 21.60 21 .81 22.02 22.24 22.44 22.65 22 .86 23. 27 23. 66 19 .36 19. 78 20.20 20.62 21.04 21.45 21 .86 22.27 22. 68 23. 08 23. 48 23. 88 24. 28 24.67 25.06 25.47 25 .85 26.24 26. 63 27.01 27.40 27. 78 28. 15 28. 53 28. 90 29. 28 29.64 30 . 03 30 .39 30 . 78 31 .14 31 .50 31 .86 32 . 23 32.59 32 .95 33 .31 33 .65 34 .35 35 .05 Pr 0. 734 0. 732 0. 731 0.729 0.727 0.725... 0. 235 3 0.2206 0.1960 cp/cv 1.0 18 1.020 1.021 1.0 23 1.024 1.026 1.0 28 1. 030 1. 032 1. 034 1. 036 1. 0 38 1.040 1.042 1.044 1.047 1.049 1.051 1.056 1.061 1.065 1.070 1.075 1. 080 1. 085 1. 089 1.094 1.099 1.110 1.121 1. 132 1.142 1.161 1.179 1.197 1.214 1. 231 1.249 1. 288 1 .39 38 1 .39 29 1 .39 20 1 .39 11 1 .39 01 1 . 38 92 1 . 38 81 1 . 38 71 1 . 38 61 1 . 38 51 1 . 38 40 1 . 38 29 1 . 38 18 1 . 38 06 1 .37 95 1 .37 83 1 .37 72 1 .37 60 1 .37 37 1 .37 14 1 .36 91... 660 Design and Optimization of Thermal Systems TABLE B.1 (CONTINUED) Properties of Dry Air at Atmospheric Pressure—SI Units Temperature K 215 220 225 230 235 240 245 250 255 260 265 270 275 280 285 290 295 30 0 30 5 31 0 31 5 32 0 32 5 33 0 33 5 34 0 34 5 35 0 35 5 36 0 36 5 37 0 37 5 38 0 38 5 39 0 39 5 400 410 420 C – 58. 15 – 53. 15 – 48. 15 – 43. 15 38 .15 33 .15 – 28. 15 – 23. 15 – 18. 15 – 13. 15 8. 15 3. 15 –1 .85 6 .85 11 .85 16 .85 ... 41.69 42 .32 42.94 43. 57 44.20 44 .80 45.41 46.01 46.61 47 .80 48. 69 50.12 51.25 52 .36 53. 45 54. 53 55.62 56. 68 57.74 60 .30 62.76 65.20 67.54 0. 686 0. 684 0. 684 0.6 83 0. 682 0. 680 0. 680 0. 680 0. 680 0. 680 0. 680 0. 680 0. 680 0. 680 0. 680 0. 680 0. 680 0. 680 0. 681 0. 682 0. 682 0.6 83 0. 684 0. 685 0. 686 0. 687 0. 688 0. 689 0.6 93 0.696 0.699 0.702 431 .7 441.9 452.1 462 .3 472.5 482 .8 4 93. 0 5 03. 3 5 13. 6 524.0 534 .3 544.7... 999 .8 999.2 9 98 .3 997.1 995.7 994.1 992 .3 990.2 9 98. 0 985 .7 9 83 . 1 980 .5 977.7 974.7 971.6 9 68. 4 965.1 961.7 9 58. 1 942 .8 925.9 1 03 (kg/m s) 1.791 1.520 1 .30 8 1. 139 1.0 03 0 .89 08 0.79 78 0.7196 0.6 531 0.5962 0.5471 0.50 43 0.46 68 0. 433 8 0.4044 0 .37 83 0 .35 50 0 .33 39 0 .31 50 0.29 78 0. 282 2 0. 232 1 0.1961 106 (m /s) 1.792 1.520 1 .30 8 1.140 1.004 0 .8 933 0 .80 12 0.7 2 38 0.6 582 0.6021 0.5 537 0.5116 0.47 48 0.4424 0.4 137 . 1.011 1 .39 81 21.60 31 .50 0.6 93 370 .8 38 5.4 37 5 101 .85 215 0.94 13 1.011 1 .39 78 21 .81 31 .86 0.692 37 5.9 38 8.0 38 0 106 .85 224 0.9 288 1.012 1 .39 75 22.02 32 . 23 0.691 38 0.9 39 0.5 38 5 111 .85 233 0.9169. 1.010 1 .39 90 20.97 30 .39 0.696 35 5.7 37 7.6 36 0 86 .85 188 0. 980 5 1.010 1 .39 87 21. 18 30 . 78 0.695 36 0.7 38 0.2 36 5 91 .85 197 0.9672 1.010 1 .39 84 21 . 38 31 .14 0.694 36 5 .8 38 2 .8 37 0 96 .85 206 0.9 539 1.011. 1 .36 46 33 .32 52 .36 0. 684 7 13. 7 5 23. 7 720 446 .85 83 6 0.4901 1. 080 1 .36 23 33. 92 53. 45 0. 685 735 .2 531 .0 740 466 .85 87 2 0.4769 1. 085 1 .36 01 34 .52 54. 53 0. 686 756.9 537 .6 760 486 .85 9 03 0.46 43 1. 089