Question:
A solid slab of thickness \( H1 \) is initially at a uniform temperature \( T0 \). At time \( t = 0 \), the temperature of the top surface at \( y = H1 \) is increased to \( T1 \), while the bottom surface at \( y = 0 \) is maintained at \( T0 \) 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 \( y = \frac{H1}{2} \) to reach 99% of its final steady value is \( T1 \). If the thickness of the slab is doubled to \( H2 = 2H1 \), and the time required for the temperature at \( y = \frac{H2}{2} \) to reach 99% of its final steady value is \( T2 \), then \( \frac{T2}{T1} \) is .
A solid slab of thickness \( H1 \) is initially at a uniform temperature \( T0 \). At time \( t = 0 \), the temperature of the top surface at \( y = H1 \) is increased to \( T1 \), while the bottom surface at \( y = 0 \) is maintained at \( T0 \) 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 \( y = \frac{H1}{2} \) to reach 99% of its final steady value is \( T1 \). If the thickness of the slab is doubled to \( H2 = 2H1 \), and the time required for the temperature at \( y = \frac{H2}{2} \) to reach 99% of its final steady value is \( T2 \), then \( \frac{T2}{T1} \) is .
Updated On: Nov 19, 2024
- 2
- \(\frac14\)
- 4
- \(\frac12\)
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The Correct Option is C
Solution and Explanation
The correct option is (C) :4
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