ME Heat Transfer 1

M.M. Yovanovich

FEF97P4.MWS

Solution for Problem 4 of Final Examination, Fall 1997.

Assumptions:

1) Steady-state.

2) Constant properties.

3) Gray, diffuse isothermal surfaces.

4) Laminar natural convection inside the annular space.

> restart:

System parameters

> sys:= (Di = 70/1000, Do = 80/1000, epsilon1 = 0.9, epsilon2 = 0.1,
Ti = 400, To = 300, g = 9.81, beta = 1/(350), sigma = 5.67e-8);

Air properties at 350 K

> airprop:= (rho = 0.9950, cp = 1.009, mu = 208.2e-7, nu = 20.92e-6,
alpha = 29.9e-6, kf = 0.030, Pr = 0.700);

Convective correlation equation

> NuDi:= .74*(Pr/(0.861 + Pr))^(1/4)/
(1 + (Di/Do)^(7/5))^(5/4)*RaDi^(1/4);

> RaDi:= g*beta*(Ti - To)*Di^3/(alpha*nu);

Calculation of convective heat transfer rate.

> Qconv:= h*Ai*(Ti - To); Ai:= Pi*Di^2;

> RaDi1:= evalf(subs(sys, airprop, RaDi), 4);

> NuDi1:= evalf(subs(sys, airprop, NuDi), 4);

> h1:=
evalf(subs(sys, airprop, NuDi = NuDi1, kf*NuDi/Di), 4);

> Qconv1:= evalf(subs(sys, airprop, h = h1, Qconv), 4);

Calculation of radiative heat transfer rate

> Q12:= (Eb1 - Eb2)/(Rs1 + R12 + Rs2);
Eb1:= sigma*T1^4; Eb2:= sigma*T2^4;
Rs1:= (1 - epsilon1)/(A1*epsilon1);
Rs2:= (1 - epsilon2)/(A2*epsilon2);
R12:= 1/(A1*F12);
A1:= Pi*Di^2; A2:= Pi*Do^2; F12:= 1;

> Qrad1:= evalf(subs(T1 = Ti, T2 = To, sys, Q12), 4);

Conductive heat transfer rate

> Qcond:= k*S*(T1 - T2);

> S:= (4*Pi)/(1/(D1/2) - 1/(D2/2));

> Qcond1:=
evalf(subs(T1 = Ti, T2 = To, D1 = Di, D2 = Do, k = kf, sys, airprop, Qcond), 4);

Summary of input and output parameters.

> `system parameters` = sys;

> Rayleigh[Di] = RaDi1;
Nusselt[Di] = NuDi1;
h = h1*W/m^2/K;
Q[convection] = Qconv1*W;

> Q[radiation] = Qrad1*W;

> Q[conduction] = Qcond1*W;

Since the conduction heat transfer rate is larger than the convection heat transfer

rate, the heat transfer across the annular space is the sum of the conduction and the

radiation components. Therefore, Qtotal = 12.47 W.