Air temperature on either side of a 10 cm thick brick wall are 25°C and 34°C. If the connective coefficients on each wall face is 1 W/m2 - °C and the wall conductivity is 0.10 W/m-°C, the wall face temperature are

To find the temperatures on the two faces of the brick wall, we need to apply the concept of steady-state heat conduction through the wall along with the heat transfer by convection on both sides.

### Given Data:
- Thickness of the brick wall, \( L = 10 \, \text{cm} = 0.10 \, \text{m} \)
- Thermal conductivity of the wall, \( k = 0.10 \, \text{W/m} \cdot \text{°C} \)
- Convective heat transfer coefficients on both sides of the wall, \( h_1 = h_2 = 1 \, \text{W/m}^2 \cdot \text{°C} \)
- Temperature on one side of the wall, \( T_\infty1 = 25 \, \text{°C} \)
- Temperature on the other side of the wall, \( T_\infty2 = 34 \, \text{°C} \)

### Approach:
1. **Assume the wall face temperatures**:
   - \( T_1 \) on the side facing the 25°C environment.
   - \( T_2 \) on the side facing the 34°C environment.

2. **Apply the heat balance**:
   - Heat conducted through the wall is equal to the heat convected from the wall surfaces.

3. **Heat conduction through the wall**:
   \[
   \text{Heat conduction rate } Q = \frac{k(T_2 - T_1)}{L}
   \]

4. **Heat convection from the surface to the environment**:
   - On the side at \( T_\infty1 = 25 \, \text{°C} \): 
     \[
     Q = h_1 (T_1 - T_\infty1)
     \]
   - On the side at \( T_\infty2 = 34 \, \text{°C} \):
     \[
     Q = h_2 (T_\infty2 - T_2)
     \]

### Forming the Equations:
From the heat balance at both faces:
\[
\frac{k(T_2 - T_1)}{L} = h_1 (T_1 - T_\infty1)
\]
\[
\frac{k(T_2 - T_1)}{L} = h_2 (T_\infty2 - T_2)
\]

### Substitute the given values:
\[
\frac{0.10(T_2 - T_1)}{0.10} = 1 (T_1 - 25)
\]
\[
T_2 - T_1 = T_1 - 25 \quad \text{(Equation 1)}
\]

\[
\frac{0.10(T_2 - T_1)}{0.10} = 1 (34 - T_2)
\]
\[
T_2 - T_1 = 34 - T_2 \quad \text{(Equation 2)}
\]

### Solving Equations:
From **Equation 1**:
\[
T_2 - T_1 = T_1 - 25
\]
\[
T_2 = 2T_1 - 25 \quad \text{(Equation 3)}
\]

From **Equation 2**:
\[
T_2 - T_1 = 34 - T_2
\]
\[
2T_2 = T_1 + 34 \quad \text{(Equation 4)}
\]

Substitute Equation 3 into Equation 4:
\[
2(2T_1 - 25) = T_1 + 34
\]
\[
4T_1 - 50 = T_1 + 34
\]
\[
3T_1 = 84
\]
\[
T_1 = 28 \, \text{°C}
\]

Substitute \( T_1 = 28 \, \text{°C} \) into Equation 3:
\[
T_2 = 2(28) - 25 = 56 - 25 = 31 \, \text{°C}
\]

### Results:
- The temperature on the side of the wall facing the 25°C environment, \( T_1 = 28 \, \text{°C} \).
- The temperature on the side of the wall facing the 34°C environment, \( T_2 = 31 \, \text{°C} \).

Do you want further details or have any questions?

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**Relative Questions:**

1. What is the general formula for heat conduction through a wall?
2. How does thermal conductivity affect the temperature distribution in a wall?
3. What are the units of convective heat transfer coefficients?
4. How would the wall face temperatures change if the convective coefficients were different?
5. What is the impact of wall thickness on heat transfer?
6. How do different materials affect the thermal conductivity of a wall?
7. What are the assumptions made in the steady-state heat conduction?
8. How does the temperature difference between the environments affect the heat transfer rate?

**Tip:** Always ensure unit consistency when applying formulas, especially in heat transfer problems where multiple types of units are involved.

Learn how to calculate the temperatures on both faces of a brick wall using heat conduction and convection. Given the wall's thickness, thermal conductivity, and convective coefficients on each side, find the temperatures based on the heat balance equations. Example problem solved step-by-step with detailed calculations.

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