Question

Consider a flat-plate solar collector placed horizontally on the flat roof of a house. The collector is \(5 \mathrm{ft}\) wide and \(15 \mathrm{ft}\) long, and the average temperature of the exposed surface of the collector is \(100^{\circ} \mathrm{F}\). The emissivity of the exposed surface of the collector is \(0.9\). Determine the rate of heat loss from the collector by convection and radiation during a calm day when the ambient air temperature is \(70^{\circ} \mathrm{F}\) and the effective sky temperature for radiation exchange is \(50^{\circ} \mathrm{F}\). Take the convection heat transfer coefficient on the exposed surface to be $2.5 \mathrm{Btu} / \mathrm{h} . \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}$.

Short answer

Answer: The rate of heat loss from the solar collector by convection and radiation during a calm day is approximately 5931.31 Btu/h.

Step-by-step solution

Calculate the area of the collector

First, we need to find the area of the solar collector. The given dimensions are 5 ft wide and 15 ft long. Multiply the width by the length to find the total area: A = width × length = (5 ft) × (15 ft) = 75 ft^2.

Calculate the heat transfer rate due to convection

We are given the convection heat transfer coefficient, h, the surface temperature, Ts, and the ambient air temperature, T_inf. We can use the following formula to calculate the heat transfer rate due to convection, Q_conv: Q_conv = h × A × (Ts - T_inf) Plug in the known values: Q_conv = (2.5 Btu/h.ft^2.°F) × (75 ft^2) × (100°F - 70°F) = 2.5 × 75 × 30 = 5625 Btu/h

Calculate the heat transfer rate due to radiation

Now we need to find the heat transfer rate due to radiation, Q_rad. We are given the emissivity, ε, and the surface temperature, Ts, as well as the effective sky temperature, T_sky. We will use the Stefan-Boltzmann constant, σ, which is equal to 0.1714×10^{-8} Btu/h.ft^2.°R^4. Remember to convert temperatures to absolute temperature (Rankine) for radiation calculations. We can use the following formula to calculate Q_rad: Q_rad = ε × σ × A × (Ts^4 - T_sky^4) Convert temperatures to Rankine: Ts_R = 100°F + 459.67 = 559.67°R T_sky_R = 50°F + 459.67 = 509.67°R Plug in the known values: Q_rad = 0.9 × (0.1714×10^{-8} Btu/h.ft^2.°R^4) × (75 ft^2) × (559.67^4 - 509.67^4) ≈ 306.31 Btu/h

Calculate the total heat loss rate

Now, we sum the heat transfer rates due to convection and radiation to find the total heat loss rate, Q_total: Q_total = Q_conv + Q_rad = 5625 Btu/h + 306.31 Btu/h ≈ 5931.31 Btu/h The rate of heat loss from the solar collector by convection and radiation during a calm day is approximately 5931.31 Btu/h.