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EGR 511 NUMERICAL METHODS _______________________ LAST NAME, FIRST Problem set #5 1. Determine the stiffness ratio for the following set of equations dy1 dy 2 = -600y1 + 550y2; = 550y1 – 600y2 dx dx SR = __________ Ans: SR = 23 2. Express the third-order equation '' ' y''' + ty - ty - 2y = t, y(0) = y''(0) = 0, y'(0) = 1, as a set of first-order equations and solve at t = 0.2, 0.4, 0.6 by the Fourth-order Runge-Kutta method (h = 0.2). Ans: t = 0.6, y = 0.62097, y’ = 1.1379, y” = 0.67326 3. Use the Runge-Kutta method with h = 0.2 for to approximate the solutions of the following systems of first-order differential equations and compare the results to the actual solutions. u1’ = 3u1 + 2u2 – (2t2 + 1)e2t, for 0 t 1 with u1(0) = 1; u2’ = 4u1 + u2 + (t2 + 2t - 4)e2t, for 0 t 1 with u2(0) = 1; 1 1 1 2 actual solutions u1(t) = e5t - e-t + e2t and u2(t) = e5t + e-t + t2e2t 3 3 3 3 Fourth-order Runge-Kutta method (h = 0.2) u1p='3*u1+2*u2-(2*t*t+1)*exp(2*t)'; u2p='4*u1+u2+(t*t+2*t-4)*exp(2*t)'; h=0.2;t=0;u1=1;u2=1;hs=h/2; for i=1:5 u1s=u1;u2s=u2; k11=eval(u1p);k12=eval(u2p); t=t+hs;u1=u1s+hs*k11;u2=u2s+hs*k12; k21=eval(u1p);k22=eval(u2p); u1=u1s+hs*k21;u2=u2s+hs*k22; k31=eval(u1p);k32=eval(u2p); t=t+hs;u1=u1s+h*k31;u2=u2s+h*k32; k41=eval(u1p);k42=eval(u2p); u1=u1s+h*(k11+2*k21+2*k31+k41)/6; u2=u2s+h*(k12+2*k22+2*k32+k42)/6; end fprintf('Calculated u1(1) = %8.6f , u2(1) = %8.6f\n',u1,u2) u1c=exp(5)/3-exp(-1)/3+exp(2); u2c=exp(5)/3+2*exp(-1)+exp(2); fprintf('Actual u1(1) = %8.6f , u2(1) = %8.6f\n',u1c,u2c) >> s5p3 Calculated u1(1) = 55.661181 , u2(1) = 56.030503 Actual u1(1) = 56.737483 , u2(1) = 57.595868 4. a) Solve the following stiff initial-value problems using Euler’s method and compare the results with the actual solution. i) y’ = - 9y, for 0 t 1 with y(0) = e and h = 0.1; actual solution y(t) = e1-9t. 1 ii) y’ = - 20(y – t2) + 2t, for 0 t 1 with y(0) = and h = 0.1; 3 1 actual solution y(t) = t2 + e-20t. 3 b) Repeat Exercise (a) using the Runge-Kutta Fourth-Order method. y1p='-9*y1'; y2p='-20*(y2-t*t)+2*t'; h=.1;t=0;y1=exp(1);y2=1/3; for i=1:10 y1=y1+h*eval(y1p);y2=y2+h*eval(y2p); t=t+h; end disp('Euler method') fprintf('Calculated y1(1) = %8.6f , y2(1) = %8.6f\n',y1,y2) t=0;y1=exp(1);y2=1/3;hs=h/2; for i=1:10 y1s=y1;y2s=y2; k11=eval(y1p);k12=eval(y2p); t=t+hs;y1=y1s+hs*k11;y2=y2s+hs*k12; k21=eval(y1p);k22=eval(y2p); y1=y1s+hs*k21;y2=y2s+hs*k22; k31=eval(y1p);k32=eval(y2p); t=t+hs;y1=y1s+h*k31;y2=y2s+h*k32; k41=eval(y1p);k42=eval(y2p); y1=y1s+h*(k11+2*k21+2*k31+k41)/6; y2=y2s+h*(k12+2*k22+2*k32+k42)/6; end disp('Runge-Kutta 4th order') fprintf('Calculated y1(1) = %8.6f , y2(1) = %8.6f\n',y1,y2) y1c=exp(-8); y2c=t*t+exp(-20)/3; fprintf('Actual y1(1) = %8.6f , y2(1) = %8.6f\n',y1c,y2c) >> s5p4 Euler method Calculated y1(1) = 0.000000 , y2(1) = 1.333333 Runge-Kutta 4th order Calculated y1(1) = 0.000372 , y2(1) = 1.002506 Actual y1(1) = 0.000335 , y2(1) = 1.000000 5. The temperatures (K) at the nodal points of a two dimensional system are shown. The left-side surface is held at a uniform temperature of 500oK, while the surface A is subjected to a convection boundary condition with a fluid temperature of 500oK and a heat transfer coefficient of 10 W/m2oK. Calculate the heat rate leaving the surface A per unit thickness normal to the page. Estimate the thermal conductivity of the material. Ans: Heat rate 349 W/m, thermal conductivity 0.77 W/mK 6. The steady state temperature (oC) associated with selected nodal points of a two-dimensional system having a thermal conductivity of 1.5 W/moK are shown. The isothermal boundary is at 200oC. a) Determine the temperature at nodes 1, 2, and 3. b) Calculate the heat transfer rate per unit thickness normal to the page from the system to the fluid. Ans: a) Node 1: 160.7 oC, node 2: 95.6 oC, node 3: 48.73 oC b) 743 W/m 7. a) Show that the Implicit Trapezoidal method, yn+1 = yn + h*(f(xn, yn) + f(xn+1, yn+1))/2, is A-stable (Stable with any h). b) Show that the Backward Euler, yn+1 = yn + h* f(xn+1, yn+1) , method is A-stable. 8. Determine the stiffness ratio at x = 1, y1 = 1, y2 = -1 for the following set of equations dy2 dy1 = - 90xy1 – 100y2; = 200y1y2 – 90xy2 dx dx Ans: SR = 1.123 9. (P. Chapra 17.6) Use least-squares regression to fit a straight line, y = a + bx, to x y 2 9 3 6 4 5 7 10 8 9 9 11 5 2 5 3 (a) Along with the slope and intercept, compute the standard error of the estimate and the correlation coefficient. Plot the data and the straight line. Assess the fit. (b) Recompute (a), but use polynomial to fit a parabola, y = a + bx + cx2, to the data. Compare the results with those of (a). Solution % Problem 5.9 x=[2 3 4 7 8 9 5 5]; y=[9 6 5 10 9 11 2 3]; n=length(x); coef=polyfit(x,y,1); ycal=polyval(coef,x); S=sum((ycal-y).^2); yave=mean(y); Sdev=sum((y-yave).^2); stde=sqrt(S/(n-2)); cor=sqrt(1-S/Sdev); coef1=coef; fprintf('y = a+bx, a = %8.5f, b = %8.5f\n',coef(2),coef(1)) fprintf('Standard error = %8.4f\n',stde) fprintf('Correlation coefficient = %8.4f\n',cor) coef=polyfit(x,y,2); ycal=polyval(coef,x); S=sum((ycal-y).^2); yave=mean(y); Sdev=sum((y-yave).^2); stde=sqrt(S/(n-2)); cor=sqrt(1-S/Sdev); fprintf('y = a+bx+cx^2, a = %8.5f, b = %8.5f, c = %8.5f\n',coef(3),coef(2),coef(1)) fprintf('Standard error = %8.4f\n',stde) fprintf('Correlation coefficient = %8.4f\n',cor) xmax=max(x);xmin=min(x); x1=[xmin xmax]; y1=polyval(coef1,x1); dx=(xmax-xmin)/50; x2=xmin:dx:xmax; y2=polyval(coef,x2); plot(x,y,'o',x1,y1,x2,y2) xlabel('x');ylabel('y') >> s5d9 y = a+bx, a = 3.48955, b = 0.62985 Standard error = 3.2214 Correlation coefficient = 0.4589 y = a+bx+cx^2, a = 16.02696, b = -4.80692, c = 0.48894 Standard error = 1.9254 Correlation coefficient = 0.8474 14 12 y 10 8 6 4 2 2 3 4 5 6 x 7 8 9