Let's solve the problem step by step: ## Given Data - **Solar flux, \( G \) = 700 W/m²** - **Collector area, \( A \) = 3 m²** - **Fraction absorbed = 90% (0.9)** - **Fraction reflected = 10% (0.1)** - **Cover glass temperature, \( T_{glass} \) = 30°C = 303 K** - **Emissivity of cover glass, \( \epsilon \) = 0.94** - **Sky temperature, \( T_{sky} \) = -30°C = 243 K** - **Ambient air temperature, \( T_{a} \) = 25°C = 298 K** - **Convection coefficient, \( h \) = 10 W/m²·K** - **Water flow rate, \( \dot{m} \) = 0.01 kg/s** - **Specific heat of water, \( c_p \) = 4179 J/kg·K** --- ## (a) **Overall Energy Balance & Useful Heat Collected (\( \dot{q}_u \))** ### Energy Absorbed by the Plate: \[ \dot{Q}_{in} = (\text{absorbed fraction}) \times (\text{solar flux}) \times (\text{area}) \] \[ = 0.9 \times 700 \times 3 = 1890 \text{ W} \] ### Losses from the Collector (per unit area): #### 1. **Convection to Ambient Air:** \[ \dot{q}_{\text{conv}} = h (T_{glass} - T_a) \] \[ = 10 \times (303 - 298) = 10 \times 5 = 50 \text{ W/m}^2 \] #### 2. **Radiation to Sky:** \[ \dot{q}_{\text{rad}} = \epsilon \sigma \left( T_{glass}^4 - T_{sky}^4 \right) \] Where \( \sigma = 5.67 \times 10^{-8} \) W/m²·K⁴. \[ T_{glass} = 303 \text{ K}, \quad T_{sky} = 243 \text{ K} \] \[ \dot{q}_{\text{rad}} = 0.94 \times 5.67 \times 10^{-8} \left(303^4 - 243^4\right) \] Calculate \( 303^4 \) and \( 243^4 \): - \( 303^4 = 8.467 \times 10^9 \) - \( 243^4 = 3.487 \times 10^9 \) - \( 303^4 - 243^4 = 8.467 \times 10^9 - 3.487 \times 10^9 = 4.98 \times 10^9 \) \[ \dot{q}_{\text{rad}} = 0.94 \times 5.67 \times 10^{-8} \times 4.98 \times 10^9 \] \[ = 0.94 \times (5.67 \times 4.98 \times 10^{1}) \] \[ = 0.94 \times (28.27) \] \[ = 26.58 \text{ W/m}^2 \] #### 3. **Total losses per unit area:** \[ \dot{q}_{\text{loss}} = \dot{q}_{\text{conv}} + \dot{q}_{\text{rad}} = 50 + 26.58 = 76.58 \text{ W/m}^2 \] #### 4. **Useful heat collected per unit area (\( \dot{q}_u \)):** Absorbed energy per unit area: \[ \dot{q}_{\text{abs}} = 0.9 \times 700 = 630 \text{ W/m}^2 \] \[ \dot{q}_u = \dot{q}_{\text{abs}} - \dot{q}_{\text{loss}} = 630 - 76.58 = \boxed{553.4~\text{W/m}^2} \] --- ## (b) **Temperature Rise of Water (\( T_2 - T_1 \))** **Total Useful Heat Collected:** \[ \dot{Q}_u = \dot{q}_u \times A = 553.4 \times 3 = 1660.2 \text{ W} \] **Energy balance for water:** \[ \dot{Q}_u = \dot{m} c_p (T_2 - T_1) \] \[ T_2 - T_1 = \frac{\dot{Q}_u}{\dot{m} c_p} = \frac{1660.2}{0.01 \times 4179} = \frac{1660.2}{41.79} = 39.74~\text{K} \] --- ## (c) **Collector Efficiency (\( \eta \))** \[ \eta = \frac{\dot{Q}_u}{\dot{Q}_{in}} = \frac{1660.2}{1890} = 0.879 = \boxed{87.9\%} \] --- ## **Summary of Answers** ### (a) **Useful heat collected per unit area:** \[ \boxed{553.4~\text{W/m}^2} \] ### (b) **Temperature rise of the water:** \[ \boxed{39.7~\text{K}} \] ### (c) **Collector efficiency:** \[ \boxed{87.9\%} \]