/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 109 Consider a flat-plate solar coll... [FREE SOLUTION] | 91影视

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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 脳 length Area = 5 ft 脳 15 ft Area = 75 ft虏
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脳A脳(T_surface - T_ambient) Where: h = convection heat transfer coefficient = 2.5 Btu/h路ft虏路掳F A = area of the solar collector = 75 ft虏 T_surface = average temperature of the exposed surface = 100掳F T_ambient = ambient air temperature = 70掳F Plugging the values, we can compute the heat loss due to convection: Q_conv = 2.5 Btu/h路ft虏路掳F 脳 75 ft虏 脳 (100掳F - 70掳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 = 蔚脳蟽脳A脳(T_surface^4 - T_sky^4) Where: 蔚 = emissivity = 0.9 (dimensionless) 蟽 = Stefan-Boltzmann constant = 5.67脳10^{-8} W/m虏路K鈦 = 1.714脳10^{-9} Btu/h路ft虏路掳R鈦 (conversion 1 W = 3.412 Btu/h) A = area of the solar collector = 75 ft虏 T_surface and T_sky are the surface and sky temperatures, respectively, in 掳R: T_surface = 100掳F + 460 = 560掳R T_sky = 50掳F + 460 = 510掳R Plugging the values, we can compute the heat loss due to radiation: Q_rad = 0.9 脳 1.714脳10^{-9} Btu/h路ft虏路掳R鈦 脳 75 ft虏 脳 (560^4掳R - 510^4掳R) Q_rad 鈮 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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solar Collector
A solar collector is a device used to capture solar energy and convert it into usable heat. This heat can then be used for various purposes such as heating water, space heating, or even generating electricity. Solar collectors are typically installed on rooftops or open areas where they can receive maximum sunlight exposure.
The solar collector in this exercise is a flat-plate type, meaning it has a flat surface to absorb solar radiation. Its dimensions are 5 feet wide and 15 feet long, resulting in a total area of 75 square feet. This design is efficient for collecting heat over a large area. The absorbed heat is then either transferred to a heat transfer fluid or directly used by the application.
Flat-plate solar collectors are commonly used in residential and commercial solar thermal systems due to their simplicity and effectiveness. They work best in environments with consistent sunlight exposure and are relatively affordable, making them a popular choice for solar energy harnessing.
Convection
Convection is the process of heat transfer based on the movement of fluid or air. In the context of a solar collector, convection occurs when warm air near the surface of the collector is replaced by cooler air from the environment.
This exercise calculated heat loss through convection using the formula: \[ Q_{\text{conv}} = h \times A \times (T_{\text{surface}} - T_{\text{ambient}}) \]where:
  • \( h \) is the convection heat transfer coefficient.
  • \( A \) is the area of the collector.
  • \( T_{\text{surface}} \) and \( T_{\text{ambient}} \) are the temperatures of the collector's surface and the ambient air, respectively.
The coefficient \( h \) provided is 2.5 Btu/h路ft虏路掳F, indicating how effectively the heat is transferred from the surface to the air. The calculated convection heat loss is 187.5 Btu/h, representing the energy lost due to air movement around the collector.
Radiation
Radiation is another form of heat transfer that occurs through electromagnetic waves without the need for a medium, meaning it can even happen in a vacuum. For the solar collector, radiation involves the emission of heat from the collector's surface as infrared radiation.
The heat loss due to radiation in this exercise is calculated using the Stefan-Boltzmann law:\[ Q_{\text{rad}} = \varepsilon \times \sigma \times A \times (T_{\text{surface}}^4 - T_{\text{sky}}^4) \]where:
  • \( \varepsilon \) is the surface emissivity, a measure of how efficiently a surface emits thermal radiation.
  • \( \sigma \) is the Stefan-Boltzmann constant.
  • \( A \) is the area of the solar collector.
  • \( T_{\text{surface}} \) and \( T_{\text{sky}} \) are the surface and sky temperatures in Rankine.
This process resulted in a heat loss of approximately 166.5 Btu/h. Radiation mechanisms are significant when there are large temperature differences and clear sky conditions, influencing the overall heat balance of solar collectors.
Thermal Emissivity
Thermal emissivity is a property of surfaces that measures their efficiency in emitting thermal radiation. It is a dimensionless value between 0 and 1, representing the ratio of radiation emitted by a surface compared to that emitted by a perfect black body at the same temperature.
A higher emissivity means a surface is more effective at emitting radiative energy. For example, in this exercise, the solar collector's surface has an emissivity of 0.9, indicating it is highly efficient at radiating heat. This property significantly influences the calculation of radiative heat loss using the Stefan-Boltzmann law.
Understanding emissivity is crucial in solar thermal applications because it affects the efficiency and heat loss of solar collectors. Surfaces with higher emissivity may need to be optimized or balanced with other materials or coatings to better achieve desired thermal outcomes. In practice, selecting or designing materials with an appropriate emissivity is essential for improving the performance of solar thermal systems.

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Most popular questions from this chapter

Write down the expressions for the physical laws that govern each mode of heat transfer, and identify the variables involved in each relation.

A 300-ft-long section of a steam pipe whose outer diameter is 4 in passes through an open space at \(50^{\circ} \mathrm{F}\). The average temperature of the outer surface of the pipe is measured to be \(280^{\circ} \mathrm{F}\), and the average heat transfer coefficient on that surface is determined to be \(6 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\). Determine \((a)\) the rate of heat loss from the steam pipe and (b) the annual cost of this energy loss if steam is generated in a natural gas furnace having an efficiency of 86 percent, and the price of natural gas is $$\$ 1.10 /$$ therm ( 1 therm \(=100,000\) Btu).

A room is heated by a baseboard resistance heater. When the heat losses from the room on a winter day amount to \(9000 \mathrm{~kJ} / \mathrm{h}\), it is observed that the air temperature in the room remains constant even though the heater operates continuously. Determine the power rating of the heater, in \(\mathrm{kW}\).

Heat is lost steadily through a \(0.5-\mathrm{cm}\) thick \(2 \mathrm{~m} \times 3 \mathrm{~m}\) window glass whose thermal conductivity is \(0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The inner and outer surface temperatures of the glass are measured to be \(12^{\circ} \mathrm{C}\) to \(9^{\circ} \mathrm{C}\). The rate of heat loss by conduction through the glass is (a) \(420 \mathrm{~W}\) (b) \(5040 \mathrm{~W}\) (c) \(17,600 \mathrm{~W}\) (d) \(1256 \mathrm{~W}\) (e) \(2520 \mathrm{~W}\)

The deep human body temperature of a healthy person remains constant at \(37^{\circ} \mathrm{C}\) while the temperature and the humidity of the environment change with time. Discuss the heat transfer mechanisms between the human body and the environment both in summer and winter, and explain how a person can keep cooler in summer and warmer in winter.

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