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Chapter 6: Problem 171

A window air conditioner that consumes \(1 \mathrm{kW}\) of electricity when running and has a coefficient of performance of 3 is placed in the middle of a room, and is plugged in. The rate of cooling or heating this air conditioner will provide to the air in the room when running is \((a) 3 \mathrm{kJ} / \mathrm{s},\) cooling \((b) 1 \mathrm{kJ} / \mathrm{s},\) cooling \((c) 0.33 \mathrm{kJ} / \mathrm{s},\) heating \((d) 1 \mathrm{kJ} / \mathrm{s},\) heating \((e) 3 \mathrm{kJ} / \mathrm{s},\) heating

Short Answer

Expert verified
a. 3 kJ/s b. 2 kJ/s c. 1 kJ/s d. 4 kJ/s Answer: a. 3 kJ/s

Step by step solution

01

Calculate the heat energy provided or removed by air conditioner

To find the rate of heating or cooling provided by the air conditioner, multiply the power consumption with the coefficient of performance. Rate of energy = Power consumption × Coefficient of performance Rate_of_energy = 1 kW × 3 = 3 kJ/s Since, it's a window air conditioner, it will be used for cooling purposes. Rate of cooling = 3 kJ/s Now, the answer can be matched with the options provided. The rate of cooling provided by the air conditioner is 3 kJ/s which corresponds to the answer choice (a).

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Most popular questions from this chapter

A heat pump with refrigerant-134a as the working fluid is used to keep a space at \(25^{\circ} \mathrm{C}\) by absorbing heat from geothermal water that enters the evaporator at \(60^{\circ} \mathrm{C}\) at a rate of \(0.065 \mathrm{kg} / \mathrm{s}\) and leaves at \(40^{\circ} \mathrm{C}\). Refrigerant enters the evaporator at \(12^{\circ} \mathrm{C}\) with a quality of 15 percent and leaves at the same pressure as saturated vapor. If the compressor consumes \(1.6 \mathrm{kW}\) of power, determine \((a)\) the mass flow rate of the refrigerant, \((b)\) the rate of heat supply, \((c)\) the \(\mathrm{COP}\), and \((d)\) the minimum power input to the compressor for the same rate of heat supply.

A well-established way of power generation involves the utilization of geothermal energy-the energy of hot water that exists naturally underground-as the heat source. If a supply of hot water at \(140^{\circ} \mathrm{C}\) is discovered at a location where the environmental temperature is \(20^{\circ} \mathrm{C},\) determine the maximum thermal efficiency a geothermal power plant built at that location can have. Answer: 29.1 percent

Refrigerant-134a enters the condenser of a residential heat pump at \(800 \mathrm{kPa}\) and \(35^{\circ} \mathrm{C}\) at a rate of \(0.018 \mathrm{kg} / \mathrm{s}\) and leaves at \(800 \mathrm{kPa}\) as a saturated liquid. If the compressor consumes \(1.2 \mathrm{kW}\) of power, determine \((a)\) the COP of the heat pump and \((b)\) the rate of heat absorption from the outside air.

A typical new household refrigerator consumes about \(680 \mathrm{kWh}\) of electricity per year and has a coefficient of performance of \(1.4 .\) The amount of heat removed by this refrigerator from the refrigerated space per year is \((a) 952 \mathrm{MJ} / \mathrm{yr}\) (b) 1749 MJ/yr \((c) 2448\) MJ/yr \((d) 3427 \mathrm{MJ} / \mathrm{yr}\) \((e) 4048 \mathrm{MJ} / \mathrm{yr}\)

A heat pump with a COP of 3.2 is used to heat a perfectly sealed house (no air leaks). The entire mass within the house (air, furniture, etc.) is equivalent to \(1200 \mathrm{kg}\) of air. When running, the heat pump consumes electric power at a rate of \(5 \mathrm{kW}\). The temperature of the house was \(7^{\circ} \mathrm{C}\) when the heat pump was turned on. If heat transfer through the envelope of the house (walls, roof, etc.) is negligible, the length of time the heat pump must run to raise the temperature of the entire contents of the house to \(22^{\circ} \mathrm{C}\) is (a) \(13.5 \mathrm{min}\) (b) \(43.1 \mathrm{min}\) \((c) 138 \min\) \((d) 18.8 \mathrm{min}\) \((e) 808 \mathrm{min}\)
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