Question

Consider a laterally insulated rod of length L and constant thermal conductivity. Assuming one-dimensional heat conduction in the rod, which of the following steady-state temperature profile(s) can occur without internal heat generation? 

 

• A. A
• B. B
• C. C
• D. D

Hint:

For 1D steady-state conduction in a plane wall with no heat generation, the temperature profile is {always} linear. If you see a curved temperature profile, it implies either heat generation/consumption, non-constant thermal conductivity, or a non-planar geometry (like a cylinder or sphere).

Answer 1

The Correct Option is A, B

Step 1: Understanding the Concept:
The problem deals with the general heat conduction equation for a one-dimensional, steady-state system. We need to find the temperature profile \(T(x)\) that satisfies this equation under the condition of no internal heat generation.
Step 2: Key Formula or Approach:
The general one-dimensional heat conduction equation is: \[ \frac{\partial}{\partial x} \left( k \frac{\partial T}{\partial x} \right) + \dot{q} = \rho c \frac{\partial T}{\partial t} \] For the given conditions: - Steady-state: The temperature does not change with time, so \(\frac{\partial T}{\partial t} = 0\). - Constant thermal conductivity \(k\): \(k\) can be taken out of the derivative. - No internal heat generation: \(\dot{q} = 0\). The equation simplifies to: \[ k \frac{d^2T}{dx^2} = 0 \] Since \(k\) is a non-zero constant, this further simplifies to: \[ \frac{d^2T}{dx^2} = 0 \] Step 3: Detailed Explanation:
We need to find which of the given profiles satisfies the equation \(\frac{d^2T}{dx^2} = 0\).
- Integrating the equation once gives: \(\frac{dT}{dx} = C_1\), where \(C_1\) is a constant. This means the temperature gradient (slope) must be constant.
- Integrating a second time gives: \(T(x) = C_1x + C_2\), where \(C_2\) is another constant. This is the equation of a straight line.
Now, let's examine the plots:
- (A): This shows a temperature profile that is a straight line with a positive slope (\(C_1>0\)). This is a valid solution. It corresponds to heat flowing from right to left.
- (B): This shows a temperature profile that is a straight line with a negative slope (\(C_1<0\)). This is also a valid solution. It corresponds to heat flowing from left to right.
- (C): This shows a curved profile (parabolic). For such a profile, \(\frac{d^2T}{dx^2}\) is a non-zero constant (in this case, negative), which would imply the presence of uniform internal heat generation (\(\dot{q}>0\)). This is not a valid solution for \(\dot{q} = 0\).
- (D): This shows a wavy, non-linear profile. For this profile, \(\frac{d^2T}{dx^2}\) is not zero and is a function of \(x\). This would imply non-uniform internal heat generation. This is not a valid solution for \(\dot{q} = 0\).
Step 4: Final Answer:
Only the linear temperature profiles shown in (A) and (B) can occur without internal heat generation.
Step 5: Why This is Correct:
The governing differential equation for 1D, steady-state heat conduction with constant properties and no heat generation is \(\frac{d^2T}{dx^2} = 0\). The only mathematical functions that satisfy this are linear functions of the form \(T(x) = C_1x + C_2\). Both (A) and (B) represent such functions.