\( 9-50 \) A \( 10-\mathrm{m} \)-long section of a \( 6-\mathrm{cm} \)-diameter horizontal hot- water pipe passes through a large room whose temperature is \( 27^{\circ} \mathrm{C} \). If the temperature and the emissivity of the outer surface of the pipe are \( 73^{\circ} \mathrm{C} \) and 0.8 , respectively, determine the rate of heat loss from the pipe by \( (a) \) natural convection and \( (b) \) radiation.

To determine the rate of heat loss from the pipe, we'll go through the calculations for both natural convection and radiation step by step. ### (a) Natural Convection The rate of heat loss by natural convection can be calculated using Newton's law of cooling: \[ Q_{\text{conv}} = h A (T_{\text{pipe}} - T_{\text{room}}) \] **Where:** - \( Q_{\text{conv}} \) = heat loss by convection (W) - \( h \) = convective heat transfer coefficient (W/m²·K) (typically for air around 5-25 W/m²·K) - \( A \) = surface area of the pipe (m²) - \( T_{\text{pipe}} \) = temperature of the pipe (K) - \( T_{\text{room}} \) = temperature of the room (K) 1. **Find the surface area \( A \) of the pipe:** \[ A = \pi D L \] Where: - \( D = 0.06 \) m (diameter of the pipe) - \( L = 10 \) m (length of the pipe) \[ A = \pi \times 0.06 \times 10 \approx 1.884 \ \text{m}^2 \] 2. **Temperature Conversion:** \[ T_{\text{pipe}} = 73 + 273.15 = 346.15 \ \text{K} \] \[ T_{\text{room}} = 27 + 273.15 = 300.15 \ \text{K} \] 3. **Assuming \( h \) = 10 W/m²·K (approximation):** \[ Q_{\text{conv}} = 10 \times 1.884 \times (346.15 - 300.15) \] \[ Q_{\text{conv}} \approx 10 \times 1.884 \times 46 = 868.32 \ \text{W} \] ### (b) Radiation The rate of heat loss by radiation can be calculated using the Stefan-Boltzmann law: \[ Q_{\text{rad}} = \varepsilon \sigma A (T_{\text{pipe}}^4 - T_{\text{room}}^4) \] **Where:** - \( Q_{\text{rad}} \) = heat loss by radiation (W) - \( \varepsilon \) = emissivity of the surface (unitless) - \( \sigma \) = Stefan-Boltzmann constant \((5.67 \times 10^{-8} \ \text{W/m}^2 \cdot \text{K}^4)\) - \( A \) = surface area of the pipe (m²) - Temperatures in Kelvin are used (already calculated above) 1. **Using the previously found area:** \[ Q_{\text{rad}} = 0.8 \times 5.67 \times 10^{-8} \times 1.884 \times \left( (346.15)^4 - (300.15)^4 \right) \] 2. **Calculating \( (346.15)^4 - (300.15)^4 \):** \[ (346.15)^4 \approx 1.515 \times 10^{11}, \ \ (300.15)^4 \approx 8.053 \times 10^{10} \] \[ (346.15)^4 - (300.15)^4 \approx 7.097 \times 10^{10} \ \text{K}^4 \] 3. **Final calculation:** \[ Q_{\text{rad}} = 0.8 \times 5.67 \times 10^{-8} \times 1.884 \times 7.097 \times 10^{10} \] \[ Q_{\text{rad}} \approx 0.8 \times 5.67 \times 1.884 \times 7.097 \times 10^{2} \approx 627.66 \ \text{W} \] ### Final Results The heat loss from the pipe by: - Natural Convection: **~868.32 W** - Radiation: **~627.66 W**