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Chapter 1: Problem 109

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 \(2.5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\).

Short Answer

Expert verified
Answer: The total rate of heat loss from the solar collector during a calm day is 354 Btu/h.

Step by step solution

01

Compute the area of the solar collector

To calculate the heat loss, we first need to determine the area of the solar collector. The collector is given to be 5 ft wide and 15 ft long, so the area can be computed as: Area = width × length Area = 5 ft × 15 ft Area = 75 ft²
02

Calculate the heat loss due to convection

To calculate the heat loss due to convection, we use the convection heat transfer equation: Q_conv = h×A×(T_surface - T_ambient) Where: h = convection heat transfer coefficient = 2.5 Btu/h·ft²·°F A = area of the solar collector = 75 ft² T_surface = average temperature of the exposed surface = 100°F T_ambient = ambient air temperature = 70°F Plugging the values, we can compute the heat loss due to convection: Q_conv = 2.5 Btu/h·ft²·°F × 75 ft² × (100°F - 70°F) Q_conv = 187.5 Btu/h
03

Calculate the heat loss due to radiation

To calculate the heat loss due to radiation, we use the Stefan-Boltzmann law for the rate of radiation heat transfer: Q_rad = ε×σ×A×(T_surface^4 - T_sky^4) Where: ε = emissivity = 0.9 (dimensionless) σ = Stefan-Boltzmann constant = 5.67×10^{-8} W/m²·K⁴ = 1.714×10^{-9} Btu/h·ft²·°R⁴ (conversion 1 W = 3.412 Btu/h) A = area of the solar collector = 75 ft² T_surface and T_sky are the surface and sky temperatures, respectively, in °R: T_surface = 100°F + 460 = 560°R T_sky = 50°F + 460 = 510°R Plugging the values, we can compute the heat loss due to radiation: Q_rad = 0.9 × 1.714×10^{-9} Btu/h·ft²·°R⁴ × 75 ft² × (560^4°R - 510^4°R) Q_rad ≈ 166.5 Btu/h
04

Calculate the total heat loss

Finally, we can add the heat loss due to convection and radiation to find the total heat loss: Q_total = Q_conv + Q_rad Q_total = 187.5 Btu/h + 166.5 Btu/h Q_total = 354 Btu/h The total rate of heat loss from the solar collector during a calm day is 354 Btu/h.

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Most popular questions from this chapter

Two surfaces, one highly polished and the other heavily oxidized, are found to be emitting the same amount of energy per unit area. The highly polished surface has an emissivity of \(0.1\) at \(1070^{\circ} \mathrm{C}\), while the emissivity of the heavily oxidized surface is \(0.78\). Determine the temperature of the heavily oxidized surface.

On a still clear night, the sky appears to be a blackbody with an equivalent temperature of \(250 \mathrm{~K}\). What is the air temperature when a strawberry field cools to \(0^{\circ} \mathrm{C}\) and freezes if the heat transfer coefficient between the plants and air is \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) because of a light breeze and the plants have an emissivity of \(0.9\) ? (a) \(14^{\circ} \mathrm{C}\) (b) \(7^{\circ} \mathrm{C}\) (c) \(3^{\circ} \mathrm{C}\) (d) \(0^{\circ} \mathrm{C}\) (e) \(-3^{\circ} \mathrm{C}\)

An ice skating rink is located in a building where the air is at \(T_{\text {air }}=20^{\circ} \mathrm{C}\) and the walls are at \(T_{w}=25^{\circ} \mathrm{C}\). The convection heat transfer coefficient between the ice and the surrounding air is \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The emissivity of ice is \(\varepsilon=0.95\). The latent heat of fusion of ice is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\) and its density is \(920 \mathrm{~kg} / \mathrm{m}^{3}\). (a) Calculate the refrigeration load of the system necessary to maintain the ice at \(T_{s}=0^{\circ} \mathrm{C}\) for an ice rink of \(12 \mathrm{~m}\) by \(40 \mathrm{~m}\). (b) How long would it take to melt \(\delta=3 \mathrm{~mm}\) of ice from the surface of the rink if no cooling is supplied and the surface is considered insulated on the back side?

Consider a \(3-\mathrm{m} \times 3-\mathrm{m} \times 3-\mathrm{m}\) cubical furnace whose top and side surfaces closely approximate black surfaces at a temperature of \(1200 \mathrm{~K}\). The base surface has an emissivity of \(\varepsilon=0.4\), and is maintained at \(800 \mathrm{~K}\). Determine the net rate of radiation heat transfer to the base surface from the top and side surfaces. Answer: \(340 \mathrm{~kW}\)

\(80^{\circ} \mathrm{C}\). Also, determine the convection heat transfer coefficients at the beginning and at the end of the heating process. 1-133 It is well known that wind makes the cold air feel much colder as a result of the wind chill effect that is due to the increase in the convection heat transfer coefficient with increasing air velocity. The wind chill effect is usually expressed in terms of the wind chill temperature (WCT), which is the apparent temperature felt by exposed skin. For outdoor air temperature of \(0^{\circ} \mathrm{C}\), for example, the wind chill temperature is \(-5^{\circ} \mathrm{C}\) at \(20 \mathrm{~km} / \mathrm{h}\) winds and \(-9^{\circ} \mathrm{C}\) at \(60 \mathrm{~km} / \mathrm{h}\) winds. That is, a person exposed to \(0^{\circ} \mathrm{C}\) windy air at \(20 \mathrm{~km} / \mathrm{h}\) will feel as cold as a person exposed to \(-5^{\circ} \mathrm{C}\) calm air (air motion under \(5 \mathrm{~km} / \mathrm{h}\) ). For heat transfer purposes, a standing man can be modeled as a 30 -cm- diameter, 170-cm-long vertical cylinder with both the top and bottom surfaces insulated and with the side surface at an average temperature of \(34^{\circ} \mathrm{C}\). For a convection heat transfer coefficient of \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat loss from this man by convection in still air at \(20^{\circ} \mathrm{C}\). What would your answer be if the convection heat transfer coefficient is increased to \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) as a result of winds? What is the wind chill temperature in this case?
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