ME 353 Heat Transfer 1

M.M. Yovanovich

P2MT93.MS

Problem 2 of Midterm Exam, October 1993.

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Steel rod of diameter D = 30 mm, length L = 600 mm is welded

to a furnace wall passes through 200 mm of insulation before

emerging into surrounding air as shown in attached figure.

The furnace wall temperature is at 300 deg C and the air

temperature is at 20 deg C.

The Biot number Bi = h D/k = 0.0156. The parameter h is the

heat transfer coefficient over the exposed portion of the rod.

Steady-state, constant properties, radiation heat transfer is

negligible.

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> restart:

1. Show the thermal circuit of this system. Clearly label all nodes and resistances.

The system consists of two resistance in series: Rrod and Rfin.

The temperature nodes are T0, T1 and Tf where T0 is the temperature

of the furnace, T1 is the temperature of the rod at the point where it

emerges from the insulation and it is exposed to the air. Tf is the sink

temperature.

The length of the rod in the insulation is denoted L1, and the

length of the rod exposed to air is denoted L2.

2. The resistances with related parameters are given below.

> Rsys:= Rrod + Rfin;

> Rrod:= L1/(k*A);

> Rfin:= 1/((sqrt(h*P*k*A))*tanh(m*Lc));

> m:= sqrt((h*P)/(k*A)); #Fin parameter.

> Lc:= L2 + A/P; #Corrected length to account for end cooling.

> A:= Pi*Dia^2/4; P:= Pi*Dia;

>

3. Estimate the temperature of the end of the rod if the heat transfer

coefficient between the rod and the air is h = 13 W/m^2 K.

> krod:= h*Dia/Bi;

> krod:= evalf(subs(Dia = 30/1000, h = 13, Bi = 0.0156, krod), 4);

>

From relations given in Table 3.4 of Incropera and DeWitt

the temperature in a fin with perfect contact at the base and

end cooling is

> restart:

> theta_fin:=

> thetab*

> (cosh(m*(L2 - x)) + (h/(m*k))*sinh(m*(L2 - x)))/

> ((cosh(m*L2) + (h/(m*k))*sinh(m*L2)));

> theta_fin_end:= simplify(subs(x = L2, theta_fin));

> m:= sqrt((h*P)/(k*A));

> A:= Pi*Dia^2/4; P:= Pi*Dia;

> finpar:= (Dia = 30/1000, L2 = 40/1000, h = 13, k = 25,

> T0 = 300, Tf = 20);

> theta_fin_end:= evalf(subs(finpar, theta_fin_end), 5);

The temperature excess at the fin base thetab can be determined from the

system analysis which requires the heat transfer rate through the rod.

> Qsys:= (T0 - Tf)/Rsys;

> Rsys:= Rrod + Rfin:

> Rrod:= L1/(k*A): L1:= 20/1000:

> Rfin:= 1/((sqrt(h*P*k*A))*tanh(m*Lc)):

> m:= sqrt((h*P)/(k*A)): #Fin parameter.

> Lc:= L2 + A/P: #Corrected length to account for end cooling.

>

4. Heat transfer rate through the system, insulated portion and the exposed

portion are equal.

> Qsys:= evalf(subs(finpar, Qsys), 6);

> `T1 - Tf`:= evalf(subs(finpar, Qsys*Rrod), 6);

> thetab:= %;

> T_fin_end:= 20 + evalf(theta_fin_end, 5);