Question

The left-hand side of a 20 cm thick wall is maintained at 25°C. The right-hand side of the wall is exposed to hot air at 50°C. There is no heat generation inside the wall and its thermal conductivity is 100 W/m·K. The convective heat transfer coefficient is 50 W/m²·K. Under steady state condition, the temperature (in °C) of the right-hand side surface of the wall is ........ (Rounded off to one decimal place).

Hint:

For heat conduction and convection problems, always start by setting up the energy balance between the heat entering and leaving the system. In this case, the conduction rate equals the convection rate.

Answer 1

To solve this problem, we will use the concept of heat conduction through a wall and convective heat transfer from the wall to the surrounding air.
1. Thermal Conductivity Equation (Heat Conduction): 
The steady-state heat conduction through the wall can be expressed as: \[ Q = \frac{kA(T_1 - T_2)}{L} \] Where:
- \( Q \) is the heat transfer rate,
- \( k \) is the thermal conductivity of the wall,
- \( A \) is the cross-sectional area of the wall,
- \( T_1 \) and \( T_2 \) are the temperatures at the two sides of the wall,
- \( L \) is the thickness of the wall.
2. Convective Heat Transfer Equation:
The heat lost from the wall to the air is governed by convection, which can be written as: \[ Q = hA(T_{{wall}} - T_{\infty}) \] Where:
- \( h \) is the convective heat transfer coefficient,
- \( A \) is the surface area,
- \( T_{{wall}} \) is the temperature of the wall surface,
- \( T_{\infty} \) is the temperature of the air.
3. Setting the Heat Transfer Rates Equal:
Since there is no heat generation inside the wall, the heat conducted through the wall must equal the heat lost to the surrounding air:
\[ \frac{kA(T_1 - T_2)}{L} = hA(T_{{wall}} - T_{\infty}) \] Simplifying, we get: \[ \frac{k(T_1 - T_2)}{L} = h(T_{{wall}} - T_{\infty}) \] 4. Substituting Known Values:
- \( k = 100 \, {W/m·K} \),
- \( T_1 = 25^\circ C \),
- \( T_2 = 50^\circ C \),
- \( L = 0.2 \, {m} \),
- \( h = 50 \, {W/m²·K} \),
- \( T_{\infty} = 50^\circ C \).
Substitute these values into the equation: \[ \frac{100(25 - T_{{wall}})}{0.2} = 50(T_{{wall}} - 50) \] 5. Solve for \( T_{{wall}} \): Expanding and solving the equation gives: \[ 500(25 - T_{{wall}}) = 50(T_{{wall}} - 50) \] \[ 12500 - 500T_{{wall}} = 50T_{{wall}} - 2500 \] \[ 15000 = 550T_{{wall}} \] \[ T_{{wall}} = \frac{15000}{550} = 27.27^\circ C \] Thus, the temperature of the right-hand side surface of the wall is approximately \( 27.3^\circ C \).