Question

A solid slab of thickness \( H_1 \) is initially at a uniform temperature \( T_0 \). At time \( t = 0 \), the temperature of the top surface at \( y = H_1 \) is increased to \( T_1 \), while the bottom surface at \( y = 0 \) is maintained at \( T_0 \) for \( t \geq 0 \). Assume heat transfer occurs only in the \( y \)-direction, and all thermal properties of the slab are constant. The time required for the temperature at \( y = H_1/2 \) to reach 99\% of its final steady value is \( \tau_1 \). If the thickness of the slab is doubled to \( H_2 = 2H_1 \), and the time required for the temperature at \( y = H_2/2 \) to reach 99\% of its final steady value is \( \tau_2 \), then \( \tau_2 / \tau_1 \) is:

• A. 2
• B. \( \frac{1}{4} \)
• C. 4
• D. \( \frac{1}{2} \)

Hint:

In transient heat conduction problems, the time required for temperature changes is proportional to the square of the slab thickness. Doubling the thickness increases the time by a factor of 4.

Answer 1

The Correct Option is C

Step 1: Understand heat conduction in a slab. The time required for a temperature change to propagate through a slab is proportional to the square of the slab's thickness. For transient heat conduction, the characteristic time is given by: \[ \tau \propto \frac{H^2}{\alpha}, \] where: - \( H \) is the thickness of the slab, - \( \alpha \) is the thermal diffusivity of the material (constant). Step 2: Relate the characteristic times. For the first case, the slab thickness is \( H_1 \), and the characteristic time is \( \tau_1 \). For the second case, the slab thickness is doubled to \( H_2 = 2H_1 \). The new characteristic time \( \tau_2 \) is proportional to \( H_2^2 \): \[ \tau_2 = \frac{(H_2)^2}{(H_1)^2} \cdot \tau_1 = \frac{(2H_1)^2}{H_1^2} \cdot \tau_1 = 4 \cdot \tau_1. \] Step 3: Conclusion. The ratio of the times is: \[ \frac{\tau_2}{\tau_1} = 4. \]