ME 353 Heat Transfer 1

M.M. Yovanovich

P2MT96.MS

Problem 2 of Midterm Exam, October 1996.

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The system parameters are:

(a) solid rod radius: r1 = 20 mm and the cylinder radii:

r1 and r2 = 40 mm;

(b) thermal conductivities of the rod and cylinder:

k1 = 0.15 W/m K and k2 = 1.5 W/m K;

(c) heat transfer coefficient at outer surface:

h = 50 W/m^2 K and fluid temperature Tf = - 15 C;

(d) outer surface temperature must be maintained at Ts = 5 C;

(e) contact conductance is very large,

i.e. neglect contact resistance.

Calculate heater power per unit length of cylinder and

the temperature at the centerline of the rod.

Steady-state, constant properties, negligible radiation

heat transfer.

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> restart:

Define the system parameters.

> sys_par:=
(r1 = 20/1000, r2 = 40/1000, L = 1,
k1 = 0.15, k2 = 1.5, h = 50, Ts = 5, Tf = - 15);

a) Sketch of the system showing all components.

b) Show the equivalent thermal circuit with nodes and resistors,

and heat flow rate, all properly labeled.

Define the overall and component resistances of the system.

> Rsys:= Rcyl + Rf;

> Rcyl:= ln(r2/r1)/(2*Pi*k2*L);

> Rf:= 1/(h*2*Pi*r2*L);

Define the relationship between the heat flow rate through the system

and the temperature drop at the outer surface.

> Qsys:= (Ts - Tf)/Rf;

Define the relationship between the heat flow rate through the system

and the maximum temperature.

> Tmax:= Tf + Qsys*Rsys;

c) Calculate the heat transfer rate through the system and the maximum

temperature such that the outer surface temperature is 5 C.

> Q[sys]:= evalf(subs(sys_par, Qsys), 5);

> T[max]:= evalf(subs(sys_par, Tmax), 4)*C;

The electrical heater must deliver 251.3 W/m to maintain an outer surface

temperature of Ts = 5 C.

The maximum temperature is Tmax = 23.38 C which occurs at the interface

between the rod and the outer cylinder.

d) The temperature of the rod is isothermal at the maximum temperature because

there is no heat transfer into the rod.

The rod thermal conductivity does not enter into the solution.