Question

Consider a cylindrical shell of length \(L\), inner radius \(r_{1}\), and outer radius \(r_{2}\) whose thermal conductivity varies in a specified temperature range as \(k(T)=k_{0}\left(1+\beta T^{2}\right)\) where \(k_{0}\) and \(\beta\) are two specified constants. The inner surface of the shell is maintained at a constant temperature of \(T_{1}\) while the outer surface is maintained at \(T_{2}\). Assuming steady onedimensional heat transfer, obtain a relation for the heat transfer rate through the shell.

Short answer

#Answer# The heat transfer rate through the cylindrical shell with varying thermal conductivity is given by the formula: $$ Q = k_0 \left[ (T_2-T_1) + \frac{\beta}{3}(T_2^3-T_1^3) \right] $$ Where \(k_0\) is the initial thermal conductivity, \(\beta\) is a constant related to the variation of conductivity with temperature, and \(T_1\) and \(T_2\) are the temperatures at the inner and outer surfaces of the shell, respectively.

Step-by-step solution

Write down the general heat transfer equation for a cylindrical shell

The general heat transfer equation for a cylindrical shell is: $$ \frac{d}{dr}\left( k(r) \frac{dT(r)}{dr} \right) = 0 $$ Where \(k(r)\) is the thermal conductivity, \(r\) is the radial distance, and \(T(r)\) is the temperature as a function of radial distance.

Express the thermal conductivity as a function of temperature

We are given that the thermal conductivity varies with temperature as \(k(T)=k_{0}\left(1+\beta T^{2}\right)\). We can substitute it in the above equation: $$ \frac{d}{dr}\left( k(T(r)) \frac{dT(r)}{dr} \right) = 0 $$

Solve the differential equation

Let's write the above equation as: $$ \frac{d}{dr}\left[ k(T) \frac{dT}{dr} \right] = \frac{dk(T)}{dT} \frac{dT}{dr} \frac{dT}{dr} = 0 $$ Now, integrate the equation $$ \int_{r_1}^{r_2} \frac{dk(T)}{dT} \frac{dT}{dr} \frac{dT}{dr} dr = 0 $$ Let's assume \(dT(r)=u\) and \(dr=v\). Then, $$ \int_{T_1}^{T_2} k(T) \frac{du}{dv} dv = 0 $$ Now separate and integrate again: $$ \int_{T_1}^{T_2} k(T) dT = \int_{r_1}^{r_2} \frac{du}{dv} dv $$

Integrate and apply boundary conditions to find the heat transfer rate

Integrate the left-side of the equation: $$ \int_{T_1}^{T_2} k_0(1+\beta T^{2}) dT = k_0\left[ \left. T\right\rvert_{T_1}^{T_2} + \frac{\beta}{3}\left. T^{3}\right\rvert_{T_1}^{T_2} \right] $$ Integrate the right-side of the equation: $$ \int_{r_1}^{r_2} \frac{du}{dv} dv = Q = \left. u\right\rvert_{r_1}^{r_2} $$ Where Q is the heat transfer rate through the shell. Now we have: $$ k_0 \left[ \left. (T_2-T_1) + \frac{\beta}{3}(T_2^3-T_1^3) \right]\right. = Q $$ So, the relation for the heat transfer rate through the shell is: $$ Q = k_0 \left[ (T_2-T_1) + \frac{\beta}{3}(T_2^3-T_1^3) \right] $$