Let's solve each part step-by-step: --- ## **Given Data** - **Panel area**: \( A = 4\, \text{m} \times 2\, \text{m} = 8\, \text{m}^2 \) - **Solar Constant**: \( G_s = 1,\, \text{W/m}^2 \) - **Reflectivity**: \( \rho_{\text{solar}} = .10 \) - **Absorptivity**: \( \alpha = 1 - \rho_{\text{solar}} = .90 \) - **Conversion Efficiency**: \( \eta_{\text{panel}} = .20 \) - **Emissivity**: \( \epsilon = .8 \) (both sides) - **Temperature of space**: \( T_{\text{space}} = \, \text{K} \) --- ## **a. Write the energy balance equation for the panel** ### **Energy Input** - **Absorbed Solar Power:** \[ Q_{\text{in}} = \alpha G_s A \] ### **Energy Output** - **Electrical Power Drawn:** \( P_{\text{power}} \) - **Radiative Losses:** Both sides radiate to space, \[ Q_{\text{rad}} = 2 \cdot \epsilon \sigma A T^4 \] Where \( \sigma = 5.67 \times 10^{-8}\, \text{W/m}^2\text{K}^4 \) is the Stefan-Boltzmann constant. ### **Energy Balance Equation** \[ \boxed{ \alpha G_s A = P_{\text{power}} + 2 \epsilon \sigma A T^4 } \] --- ## **b. How much convective heat loss will the panel experience in outer space?** - **Answer:** **None.** There is no air in outer space, so **no convective heat loss** occurs. --- ## **c. What is the electrical power output of this panel?** \[ P_{\text{power}} = \eta_{\text{panel}} \times (\alpha G_s A) \] - \( \eta_{\text{panel}} = .20 \) - \( \alpha = .90 \) - \( G_s = 1365\, \text{W/m}^2 \) - \( A = 8\, \text{m}^2 \) ### **Calculation:** 1. **Absorbed Power:** \[ \alpha G_s A = .90 \times 1365 \times 8 = .90 \times 10,920 = 9,828 \text{ W} \] 2. **Electrical Power:** \[ P_{\text{power}} = .20 \times 9,828 = 1,965.6 \text{ W} \] **Final Answer:** \[ \boxed{P_{\text{power}} = 1,966 \ \text{W} } \] --- ## **d. What is the steady-state temperature of the solar panel?** At steady-state, energy in = energy out: \[ \alpha G_s A = P_{\text{power}} + 2 \epsilon \sigma A T^4 \] Plug in the numbers: - \( \alpha G_s A = 9,828 \) W - \( P_{\text{power}} = 1,966 \) W - \( \epsilon = .8 \) - \( \sigma = 5.67 \times 10^{-8} \) - \( A = 8 \) \[ 9,828 = 1,966 + 2 \times .8 \times 5.67 \times 10^{-8} \times 8 \times T^4 \] \[ 9,828 - 1,966 = 2 \times .8 \times 5.67 \times 10^{-8} \times 8 \times T^4 \] \[ 7,862 = 2 \times .8 \times 5.67 \times 10^{-8} \times 8 \times T^4 \] \[ 7,862 = 9.072 \times 10^{-7} \times T^4 \] \[ T^4 = \frac{7,862}{9.072 \times 10^{-7}} \] \[ T^4 = 8.667 \times 10^9 \] \[ T = (8.667 \times 10^9)^{1/4} \] \[ T \approx 303.5 \text{ K} \] **Final Answer:** \[ \boxed{T = 304\, \text{K}} \] --- ## **e. What is the steady-state temperature if we stop drawing electrical power?** Set \( P_{\text{power}} = \): \[ \alpha G_s A = 2 \epsilon \sigma A T^4 \] \[ 9,828 = 9.072 \times 10^{-7} \times T^4 \] \[ T^4 = \frac{9,828}{9.072 \times 10^{-7}} \] \[ T^4 = 1.0837 \times 10^{10} \] \[ T = (1.0837 \times 10^{10})^{1/4} \] \[ T \approx 323.1 \text{ K} \] **Final Answer:** \[ \boxed{T = 323\, \text{K}} \] --- ## **Summary Table** | Part | Answer | |------|------------------------------| | (a) | \( \alpha G_s A = P_{\text{power}} + 2\epsilon \sigma A T^4 \) | | (b) | Zero (no convection in space) | | (c) | \( 1,966\, \text{W} \) | | (d) | \( 304\, \text{K} \) | | (e) | \( 323\, \text{K} \) | Let me know if you need any step explained further!