Consider a rod of uniform thermal conductivity whose one end (x = 0) is insulated and the other end (x = L) is exposed to flow of air at temperature T∞ with convective heat transfer coefficient h. The cylindrical surface of the rod is insulated so that the heat transfer is strictly along the axis of the rod. The rate of internal heat generation per unit volume inside the rod is given as

$\dot{\mathrm{q}}=\cos \frac{2\pi \mathrm{x}}{\mathrm{L}}$

The steady-state temperature at the mid-location of the rod is given as TA. What will be the temperature at the same location, if the convective heat transfer coefficient increases to 2h?

The correct answer is T_A.

1. Physical Analysis and Boundary Conditions
We are given a 1D steady-state heat conduction problem along a rod of length $L$ with uniform thermal conductivity $k$. The cylindrical surface is insulated, restricting heat flow solely along the axis of the rod ($x$-direction).

The boundary conditions are:
• At $x=0$, the surface is insulated:
${\left.\frac{dT}{dx}\right|}_{x=0}=0$
• At $x=L$, the surface is exposed to convection with fluid at temperature $T_{\infty }$ and convective heat transfer coefficient $h$:
$-k{\left.\frac{dT}{dx}\right|}_{x=L}=h\left(T\left(L\right)-T_{\infty }\right)$

2. Total Heat Generation
The rate of internal heat generation per unit volume inside the rod is:
$\dot{q}\left(x\right)=\cos \left(\frac{2\pi x}{L}\right)$

Let us calculate the net heat generation rate within the entire volume of the rod of cross-sectional area $A$:
$Q_{\text{gen}}={\int }_0^L\dot{q}\left(x\right)Adx=A{\int }_0^L\cos \left(\frac{2\pi x}{L}\right)dx=A{\left[\frac{L}{2\pi }\sin \left(\frac{2\pi x}{L}\right)\right]}_0^L=0$

Because the net heat generated in the rod is exactly zero, the net heat transfer out of the rod at the boundaries under steady-state conditions must also be zero:
$Q_{\text{out}}=Q_{\text{gen}}=0$

Since the boundary at $x=0$ is insulated, all net heat exchange must happen at $x=L$:
$-k{\left.\frac{dT}{dx}\right|}_{x=L}=0\Rightarrow h\left(T\left(L\right)-T_{\infty }\right)=0$

Assuming $h>0$, this simplifies to:
$T\left(L\right)=T_{\infty }$

This tells us that the boundary temperature at $x=L$ is strictly locked to the ambient fluid temperature $T_{\infty }$, independent of the value of the convective heat transfer coefficient $h$.

3. Steady-State Temperature Distribution
The one-dimensional steady-state heat conduction equation with internal heat source is:
$\frac{d^{2}T}{d{x}^2}+\frac{\dot{q}\left(x\right)}{k}=0\Rightarrow \frac{d^{2}T}{d{x}^2}=-\frac{1}{k}\cos \left(\frac{2\pi x}{L}\right)$

Integrating once with respect to $x$:
$\frac{dT}{dx}=-\frac{L}{2\pi k}\sin \left(\frac{2\pi x}{L}\right)+C_1$

Applying the insulated boundary condition at $x=0$:
$0=0+C_1\Rightarrow C_1=0$

Integrating a second time to find the temperature profile $T\left(x\right)$:
$T\left(x\right)=\frac{L^2}{4{\pi }^{2}k}\cos \left(\frac{2\pi x}{L}\right)+C_2$

Applying the boundary condition $T\left(L\right)=T_{\infty }$ to find $C_2$:
$T_{\infty }=\frac{L^2}{4{\pi }^{2}k}\cos \left(2\pi \right)+C_2=\frac{L^2}{4{\pi }^{2}k}+C_2\Rightarrow C_2=T_{\infty }-\frac{L^2}{4{\pi }^{2}k}$

Substituting $C_2$ back gives the complete temperature distribution:
$T\left(x\right)=T_{\infty }+\frac{L^2}{4{\pi }^{2}k}\left[\cos \left(\frac{2\pi x}{L}\right)-1\right]$

4. Temperature at the Mid-Location
At the mid-location of the rod, $x=\frac{L}{2}$:
$T_A=T\left(\frac{L}{2}\right)=T_{\infty }+\frac{L^2}{4{\pi }^{2}k}\left[\cos \left(\pi \right)-1\right]=T_{\infty }-\frac{L^2}{2{\pi }^{2}k}$

As derived, the temperature distribution $T\left(x\right)$ is completely independent of the convective heat transfer coefficient $h$. Therefore, increasing the convective heat transfer coefficient to $2h$ will have no effect on the temperature at any location, and the temperature at the mid-location remains T_A.