# Logic applied to making isothermal assumption

## Homework Statement

A thin electrical heater is inserted between a long circular rod and a concentric tube with inner and outer radii of 20 and 40 mm. The rod (A) has a thermal conductivity of kA = 0.15 W/(m*K), while the tube (B) has a thermal conductivity of kB = 1.5 W/(m*K) and its outer surface is subjected to convection with a fluid of temperature T∞ = -15°C and heat transfer coefficient 50 W/(m2 *K). The thermal contact resistance between the cylinder surfaces and the heater is negligible.
(a) Determine the electrical power per unit length of the cylinders (W/m) that is required to maintain the outer surface of cylinder B at 15°C.
(b) What is the temperature at the center of cylinder A?

## Homework Equations

Rconv = 1/(hA)
Rconduction = ln(r2/r1)/(2piLk)
Q=delT/R

## The Attempt at a Solution

a) Q is constant throughout cylinder.
L=length of cylinder
Rconv = 1/(50*2pi*.04*L)=.07958/L
Ts=outer surface temp
Q/L = (Ts-Tinfinity)/(L*Rconv) = 376.98 W/m
b)
T1 is temperature of the inner surface
Q/L=(T1-Ts)/(Rconduction*L)
Rconduction = ln(.04/.02)/(2*pi*L*1.5)=.07355/L
T1=(Q/L)*(Rconduction*L)+Ts = 42.73C

The answer for temperature at the center is 42.73 which is also the temperature of the outer radius of the rod.

What leads one to an isothermal assumption that allows for correct calculation?

$$\frac{1}{r}\frac{d}{dr}\left(r\frac{dT}{dr}\right)=0$$
If we integrate this once, we obtain: $$r\frac{dT}{dr}=C$$where C is a constant. Integrating again gives:
$$T=C\ln(r)+D$$where D is another constant. If the temperature is finite at r = 0, then C must be equal to zero. Therefore, T = D = const.