Question

The top section of an 8-ft-deep, 100-ft \(\times 100-\mathrm{ft}\) heated solar pond is maintained at a constant temperature of \(80^{\circ} \mathrm{F}\) at a location where the atmospheric pressure is \(1 \mathrm{~atm}\). If the ambient air is at \(70^{\circ} \mathrm{F}\) and 100 percent relative humidity and wind is blowing at an average velocity of \(40 \mathrm{mph}\), determine the rate of heat loss from the top surface of the pond by \((a)\) forced convection, \((b)\) radiation, and (c) evaporation. Take the average temperature of the surrounding surfaces to be \(60^{\circ} \mathrm{F}\).

Short answer

Answer: The heat loss due to forced convection from the top surface of the heated solar pond is approximately 153,932 W.

Step-by-step solution

(a) Forced Convection Heat Loss

To calculate the rate of heat loss due to forced convection, we have to use the following equation for convective heat transfer: $$q=h \times A \times (T_{1} - T_{2})$$ Where: - \(q\) is the rate of heat transfer - \(h\) is the convective heat transfer coefficient - \(A\) is the area of the surface (in this case, the pond) - \(T_1\) is the temperature of the surface (in this case, the pond) - \(T_2\) is the temperature of the ambient air We're given the surface temperature \(T_1\) (80°F) and the ambient air temperature \(T_2\) (70°F), along with the dimensions of the pond (100 ft × 100 ft), so we can determine the area. The convective heat transfer coefficient \(h\) can be calculated using the empirical relation: $$h=10.45 - v + 10 \sqrt{v}$$ where \(v\) is the wind velocity in m/s. First, let's convert the wind velocity from mph to m/s: $$40 \,\mathrm{mph} \times \frac{1609.34 \, \mathrm{m}}{3600 \,\mathrm{s} \times 1 \, \mathrm{mile}} \approx 17.88 \, \mathrm{m/s}$$ Now we can calculate the convective heat transfer coefficient: $$h \approx 10.45 - 17.88 + 10 \sqrt{17.88} \approx 30.23 \, \mathrm{W/m^2K}$$ Note that we'll need to convert the temperature difference to Kelvin for the heat transfer calculation. The temperature difference in Fahrenheit is: $$T_{1} - T_{2} = 80 - 70 = 10^{\circ} \mathrm{F}$$ Converting to Kelvin: $$\Delta T_{K} = \frac{10^{\circ} \mathrm{F}}{1.8} \approx 5.56 \, \mathrm{K}$$ Now we can find the surface area of the pond (converting to m²): $$A = 100 \,\mathrm{ft} \times 100\, \mathrm{ft} \times \left(\frac{0.3048\, \mathrm{m}}{1\, \mathrm{ft}}\right)^{2} \approx 929.03 \, \mathrm{m^2}$$ Finally, we can calculate the heat loss due to forced convection: $$q = h \times A \times \Delta T_K \approx 30.23\, \mathrm{W/m^2K} \times 929.03\, \mathrm{m^2} \times 5.56\, \mathrm{K} \approx 153,\! 932 \, \mathrm{W}$$ The heat loss due to forced convection is approximately 153,932 W.