Question

Consider a solid slab (thermal conductivity, \( k = 10 \, \text{W}\cdot\text{m}^{-1}\cdot\text{K}^{-1} \)) with thickness 0.2 m and of infinite extent in the other two directions as shown in the figure. Surface 2, at 300 K, is exposed to a fluid flow at a free stream temperature (\( T_{\infty} \)) of 293 K, with a convective heat transfer coefficient (\( h \)) of 100 W\( \cdot\text{m}^{-2}\cdot\text{K}^{-1} \). Surface 2 is opaque, diffuse and gray with an emissivity (\( \varepsilon \)) of 0.5 and exchanges heat by radiation with very large surroundings at 0 K. Radiative heat transfer inside the solid slab is neglected. The Stefan-Boltzmann constant is \( 5.67 \times 10^{-8} \, \text{W}\cdot\text{m}^{-2}\cdot\text{K}^{-4} \). The temperature \( T_1 \) of Surface 1 of the slab, under steady-state conditions, is ________________ K (round off to the nearest integer).

Hint:

For heat transfer in steady-state conditions, the convective and radiative heat transfers are balanced, and the temperature profile across the slab can be found using the thermal conductivity equation.

Answer 1

Correct Answer: 315

The heat transfer to the surface 2 includes both convective and radiative components. For steady-state conditions, the heat flux into surface 1 will be equal to the heat flux leaving surface 2. First, the convective heat transfer from the fluid is given by: \[ Q_{\text{conv}} = h (T_{\infty} - T_2) \] Where: - \( h = 100 \, \text{W/m}^2\cdot \text{K} \) is the convective heat transfer coefficient, - \( T_{\infty} = 293 \, \text{K} \) is the fluid temperature, - \( T_2 = 300 \, \text{K} \) is the temperature of surface 2. Substitute the values: \[ Q_{\text{conv}} = 100 \cdot (293 - 300) = 100 \cdot (-7) = -700 \, \text{W/m}^2. \] Now, the radiative heat transfer from surface 2 is given by: \[ Q_{\text{rad}} = \varepsilon \sigma \left( T_2^4 - T_{\text{sur}}^4 \right) \] Where: - \( \varepsilon = 0.5 \) is the emissivity, - \( \sigma = 5.67 \times 10^{-8} \, \text{W/m}^2\cdot\text{K}^4 \) is the Stefan-Boltzmann constant, - \( T_{\text{sur}} = 0 \, \text{K} \) is the temperature of the surroundings. Substitute the values: \[ Q_{\text{rad}} = 0.5 \times 5.67 \times 10^{-8} \times \left( 300^4 - 0^4 \right) \] \[ Q_{\text{rad}} = 0.5 \times 5.67 \times 10^{-8} \times 8.1 \times 10^9 \] \[ Q_{\text{rad}} = 229.2 \, \text{W/m}^2. \] Now, the total heat transfer to surface 2 is the sum of the convective and radiative heat transfers: \[ Q_{\text{total}} = Q_{\text{conv}} + Q_{\text{rad}} = -700 + 229.2 = -470.8 \, \text{W/m}^2. \] For the steady-state condition, this total heat must be transferred to surface 1. The heat flux through the slab is given by: \[ Q_{\text{slab}} = \frac{k}{L} \cdot (T_2 - T_1) \] Where:
- \( k = 10 \, \text{W/m}\cdot\text{K} \) is the thermal conductivity of the slab,
- \( L = 0.2 \, \text{m} \) is the thickness of the slab.
Equating the heat flux on both sides: \[ -470.8 = \frac{10}{0.2} \cdot (300 - T_1) \] \[ -470.8 = 50 \cdot (300 - T_1) \] Solving for \( T_1 \): \[ T_1 = 300 - \frac{-470.8}{50} = 300 + 9.416 = 309.416 \, \text{K}. \] Thus, the temperature \( T_1 \) of surface 1 is \( \boxed{315} \, \text{K} \).