I am in heat trasnfer using Priciples of Heat Transfer 8th edition by frank kreith. help me solve this problem "A horizontal, 1/8-inch-thick copper plate, 2 feet long and 2 feet wide is in air at 80 °F
and exposed to radiation from the sun on one side. If the total rate of incident radiation to the one side is
400 Btu/hr∙ft2 and the film coefficients for convection are 2 Btu/hr∙ft2∙°F for both the upper and lower surfaces,
determine the equilibrium temperature of the plate.

To solve for the equilibrium temperature of the copper plate, we can set up an energy balance at steady state, where the rate of energy absorbed by the plate equals the rate of energy dissipated by convection on both surfaces.

### Given Information:
1. **Dimensions of plate**: 2 ft by 2 ft
2. **Thickness of plate**: \( t = \frac{1}{8} \) in
3. **Incident radiation on one side**: \( G = 400 \, \text{Btu/hr·ft}^2 \)
4. **Film coefficient for convection (both sides)**: \( h = 2 \, \text{Btu/hr·ft}^2 \cdot \degree F \)
5. **Ambient air temperature**: \( T_\infty = 80^\circ \text{F} \)

### Assumptions:
- Steady-state conditions.
- Radiation is only incident on the top side of the plate.
- Convection occurs on both sides of the plate.
- The plate is thin, so temperature within the plate is uniform (lumped-capacitance assumption).
  
### Solution:

1. **Calculate Total Incident Radiation**:

   Since the plate’s area on the irradiated side is \( A = 2 \, \text{ft} \times 2 \, \text{ft} = 4 \, \text{ft}^2 \),
   \[
   \text{Total incident radiation} = G \times A = 400 \, \text{Btu/hr·ft}^2 \times 4 \, \text{ft}^2 = 1600 \, \text{Btu/hr}
   \]

2. **Set Up Energy Balance for Steady-State Condition**:

   At equilibrium, the absorbed solar radiation on the plate equals the convective heat loss from both sides.
   
   - **Heat Gain by Radiation**: \( Q_{\text{in}} = 1600 \, \text{Btu/hr} \)
   - **Heat Loss by Convection**: Convective losses happen on both the top and bottom surfaces, so:
     \[
     Q_{\text{out}} = 2 \times h \times A \times (T_{\text{plate}} - T_\infty)
     \]
   
   Setting \( Q_{\text{in}} = Q_{\text{out}} \):
   \[
   1600 = 2 \times 2 \times 4 \times (T_{\text{plate}} - 80)
   \]
   
3. **Solve for \( T_{\text{plate}} \)**:

   \[
   1600 = 16 \times (T_{\text{plate}} - 80)
   \]
   \[
   T_{\text{plate}} - 80 = \frac{1600}{16} = 100
   \]
   \[
   T_{\text{plate}} = 100 + 80 = 180^\circ \text{F}
   \]

### Answer:
The equilibrium temperature of the copper plate is \( T_{\text{plate}} = 180^\circ \text{F} \).

Would you like a detailed breakdown of any step?

### Relative Questions:
1. How would this solution change if the plate was exposed to radiation on both sides?
2. How does the thickness of the plate affect heat transfer calculations in general?
3. What would happen if the convection coefficients on the upper and lower surfaces were different?
4. How would the equilibrium temperature change if the incident radiation increased to 600 Btu/hr·ft²?
5. What additional factors would we need to consider if the plate were inclined instead of horizontal?

### Tip:
In heat transfer problems, always double-check units for consistency, especially when working with mixed units like Btu and °F.

I am in heat transfer using Principles of Heat Transfer 8th edition by Frank Kreith. Help me solve this problem: A horizontal, 1/8-inch-thick copper plate, 2 feet long and 2 feet wide is in air at 80 °F and exposed to radiation from the sun on one side. If the total rate of incident radiation to the one side is 400 Btu/hr∙ft2 and the film coefficients for convection are 2 Btu/hr∙ft2∙°F for both the upper and lower surfaces, determine the equilibrium temperature of the plate.

Solve for the equilibrium temperature of a 1/8-inch-thick copper plate, 2 feet by 2 feet, exposed to solar radiation at 400 Btu/hr·ft². Learn how to apply heat transfer principles, including energy balance, convection, and radiation. Suitable for undergraduate engineering students.

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