The top section of an 8-ft-deep \(100-\mathrm{ft} \times 100-\mathrm{ft}\) 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 \(60^{\circ} \mathrm{F}\).

First, we will calculate the heat transfer coefficient (h) using the following empirical relation for forced convection over a flat plate: \(h = 5.678 \sqrt{V}\) Where V is the wind velocity in mph. To determine h, substitute the given values: \(h = 5.678 \sqrt{40 \thinspace \mathrm{mph}}\) \(h \approx 36 \thinspace \mathrm{Btu/hr \thinspace ft^2 \thinspace °F}\) Next, we can determine the heat loss (Q) due to forced convection using Newton's law of cooling: \(Q = h \thinspace A \thinspace (T_s - T_\infty)\) Where A is the top surface area, \(T_s\) is the surface temperature, and \(T_\infty\) is the ambient air temperature. Using the given values: \(Q = 36 \thinspace \mathrm{Btu/hr \thinspace ft^2 \thinspace °F} \times 10000 \thinspace \mathrm{ft^2} \times (80 - 70) \mathrm{^{o}F}\) \(Q = 360000 \thinspace \mathrm{Btu/hr}\) So the heat loss by forced convection is 360,000 Btu/hr.

To determine the heat loss due to radiation, we need to use the Stefan-Boltzmann law for radiation: \(Q = \sigma \epsilon A(T_s^4 - T_r^4)\) Where \(\sigma\) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} \thinspace \mathrm{W/m^2 \thinspace K^4})\), \(\epsilon\) is the emissivity of the pond (assumed to be 1 for a black body), and \(T_r\) is the average temperature of the surrounding surfaces. We need to convert the temperatures to Kelvin first: \(T_s = (80 + 459.67) \thinspace \mathrm{R} = 539.67 \thinspace \mathrm{K}\) \(T_r = (60 + 459.67) \thinspace \mathrm{R} = 519.67 \thinspace \mathrm{K}\) Now, we can calculate the heat loss due to radiation: \(Q = (5.67 \times 10^{-8}) \times 1 \times 929 \thinspace \mathrm{m^2} \times (539.67^4 - 519.67^4) \thinspace \mathrm{K^4}\) \(Q \approx 13012 \thinspace \mathrm{W}\) Convert the heat loss to Btu/hr: \(Q \approx 13012 \thinspace \mathrm{W} \times 3.412 \thinspace \mathrm{Btu/hr \thinspace W^{-1}}\) \(Q \approx 44393 \thinspace \mathrm{Btu/hr}\) So the heat loss by radiation is approximately 44,393 Btu/hr.

To determine the heat loss due to evaporation, we first need to find the mass flow rate of the evaporated water: \(\mathrm{EvapRate} = 0.106 \thinspace (m_p)^{1/3} h(T_s - T_\infty)\) Where \(m_p\) is the mass flow rate of the pond in lbm/s, and EvapRate is the evaporation rate in lbm/ft²∙hr. We compute EvapRate first: \(\mathrm{EvapRate} = 0.106 \times (500000)^{1/3} \times 36 \thinspace (80 - 70) \thinspace \mathrm{^{o}F}\) \(\mathrm{EvapRate} \approx 129.7 \thinspace \mathrm{lbm/ft^2 \thinspace hr}\) Now, we can find the heat loss due to evaporation: \(Q = \mathrm{EvapRate} \cdot A \cdot h_{fg}\) Where \(h_{fg}\) is the latent heat of vaporization \((\approx 1000 \thinspace \mathrm{Btu/lbm})\). So, \(Q = 129.7 \thinspace \mathrm{lbm/ft^2 \thinspace hr} \times 10000 \thinspace \mathrm{ft^2} \times 1000 \thinspace \mathrm{Btu/lbm}\) \(Q = 1.297 \times 10^9 \thinspace \mathrm{Btu/hr}\) Hence, the heat loss by evaporation is approximately \(1.297 \times 10^9\) Btu/hr.

To summarize, the heat loss from the top surface of the pond is: - By forced convection: 360,000 Btu/hr - By radiation: 44,393 Btu/hr - By evaporation: \(1.297 \times 10^9\) Btu/hr