ME 353 Heat Transfer 1

M.M. Yovanovich

FEF97P2.MS

Solution for Problem 2 of Final Examination, Fall 1997.

Assumptions:

1) Constant properties.

2) Radiative heat transfer is negligible.

3) Initial temperature is uniform.

> restart:

System parameters

> sys:= (L = 40/1000, k = 40, alpha = 8e-6, h = 200,
Ti = 440, Tf = 600, Tmin = 520);

Biot number

> Bi:= h*L/k; Bi_val:= evalf(subs(sys, Bi), 4);

Since the Biot number is not less than 0.2, the full solution is required.

The first term of the series solution will be used to find the heating time,

then the Fo number will be checked to see whether it is greater than

the critical Fo number, Foc = 0.2.

> phi1:= A1*exp(- delta1^2*Fo)*cos(delta1*zeta);

> phi:= (Tf - T)/(Tf - Ti);

> A1:= 2*sin(delta1)/(delta1 + sin(delta1)*cos(delta1));

> delta1:= Pi/2/(1 + (Pi/2/sqrt(Bi))^n)^(1/n);
n:= 2.139;

Calculate delta1, A1 and phi for zeta = 0.

> delta1_val:= evalf(subs(sys, delta1), 6);

> A1_val:= evalf(subs(sys, A1), 6);

> phi_val:= evalf(subs(T = Tmin, sys, phi), 6);

Substitute the above values into the expression for phi to get an equation in the

unknown Fo number. Solve for the Fourier number.

> eq1:= evalf(subs(sys, zeta = 0, phi1), 6) = phi_val;

> Fo_val:= evalf(solve(eq1, Fo), 4);

Since the calculated Fourier number is greater than the critical value,

the first term approximation is justified. Now we can find the heating time.

> Fo:= alpha*t/L^2;

> eq2:= subs(sys, Fo)= Fo_val;

> t_sec1:= evalf(solve(eq2, t), 5);

The heating time is approximately 770 seconds.

(c) Use the lumped capacitance model to estimate the heating time.

> phi_lump:= exp(- (h*As)/(rho*cp*V)*t);

> As:= 2*A; V:= 2*L*As;

> `rho cp` = k/alpha;

> phi_lump:= exp(- h*2*As*alpha*t/(2*As*L*k));

> eq3:= subs(sys, phi_lump) = phi_val;

> t_sec2:= evalf(solve(eq3, t), 5);

The lumped capacitance model underestimates the heating time.

(d) Determine the percent difference in the two heating times.

> percent_diff:= evalf((t_sec2 - t_sec1)/t_sec1*100, 4);

The lumped capacitance model underestimates the heating time by

approximately 10%.