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Chapter 11: Problem 105

A heat pump water heater (HPWH) heats water by absorbing heat from the ambient air and transferring it to water. The heat pump has a COP of 3.4 and consumes \(6 \mathrm{kW}\) of electricity when running. Determine if this heat pump can be used to meet the cooling needs of a room most of the time for "free" by absorbing heat from the air in the room. The rate of heat gain of a room is usually less than \(45,000 \mathrm{kJ} / \mathrm{h}\).

Short Answer

Expert verified
Answer: Yes, the HPWH can meet the cooling needs of a room most of the time for "free," as it absorbs 20.4 kW of heat from the room, which is greater than the typical heat gain rate of 12.5 kW.

Step by step solution

01

Calculate the heat output by the HPWH

We know that the Coefficient of Performance (COP) is the ratio of heat output to the input electrical power. Using the given data, we can find the heat output by the HPWH. COP = \(\frac{\text{Heat Output}}{\text{Power Input}}\) Heat Output = COP × Power Input Plug in the given values: Heat Output = \(3.4 \times 6\,\text{kW}\) Heat Output = \(20.4\,\text{kW}\) The heat output by the HPWH is \(20.4\,\text{kW}\).
02

Calculate the heat absorbed by the HPWH

Since the HPWH absorbs heat from the ambient air and transfers it to the water, this means the heat output represents the absorbed heat. So, the heat absorbed by the HPWH is \(20.4\,\text{kW}\).
03

Convert the heat gain rate of a room to kW

The heat gain rate of a room is given in kJ/h. To compare this value with the heat absorbed by the HPWH, we need to convert this rate to kW. 1 kW = 1 kJ/s \(\frac{45000\,\text{kJ}}{1\,\text{h}}\) = \(\frac{45000}{3600}\,\text{kW}\) = \(12.5\,\text{kW}\) The heat gain rate of a room is \(12.5\,\text{kW}\).
04

Compare the absorbed heat rate to the room heat gain rate

The HPWH absorbs \(20.4\,\text{kW}\) of heat when running, while the typical rate of heat gain in a room is \(12.5\,\text{kW}\). As the absorbed heat rate is greater than the room heat gain rate, the HPWH can be used to meet the cooling needs of a room most of the time for "free" by absorbing heat from the air in the room.

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Most popular questions from this chapter

How does the COP of a cascade refrigeration system compare to the COP of a simple vapor-compression cycle operating between the same pressure limits?

Consider a regenerative gas refrigeration cycle using helium as the working fluid. Helium enters the compressor at \(100 \mathrm{kPa}\) and \(-10^{\circ} \mathrm{C}\) and is compressed to \(300 \mathrm{kPa}\). Helium is then cooled to \(20^{\circ} \mathrm{C}\) by water. It then enters the regenerator where it is cooled further before it enters the turbine. Helium leaves the refrigerated space at \(-25^{\circ} \mathrm{C}\) and enters the regenerator. Assuming both the turbine and the compressor to be isentropic, determine ( \(a\) ) the temperature of the helium at the turbine inlet, ( \(b\) ) the coefficient of performance of the cycle, and ( \(c\) ) the net power input required for a mass flow rate of \(0.45 \mathrm{kg} / \mathrm{s}\).

How is the second-law efficiency of a heat pump operating on the vapor- compression refrigeration cycle defined? Provide two alternative definitions and show that one can be derived from the other.

A gas refrigeration system using air as the working fluid has a pressure ratio of \(5 .\) Air enters the compressor at \(0^{\circ} \mathrm{C} .\) The high- pressure air is cooled to \(35^{\circ} \mathrm{C}\) by rejecting heat to the surroundings. The refrigerant leaves the turbine at \(-80^{\circ} \mathrm{C}\) and enters the refrigerated space where it absorbs heat before entering the regenerator. The mass flow rate of air is \(0.4 \mathrm{kg} / \mathrm{s}\). Assuming isentropic efficiencies of 80 percent for the compressor and 85 percent for the turbine and using variable specific heats, determine ( \(a\) ) the effectiveness of the regenerator, \((b)\) the rate of heat removal from the refrigerated space, and \((c)\) the \(\mathrm{COP}\) of the cycle. Also, determine \((d)\) the refrigeration load and the COP if this system operated on the simple gas refrigeration cycle. Use the same compressor inlet temperature as given, the same turbine inlet temperature as calculated, and the same compressor and turbine efficiencies.

Air enters the compressor of an ideal gas refrigeration cycle at \(7^{\circ} \mathrm{C}\) and \(35 \mathrm{kPa}\) and the turbine at \(37^{\circ} \mathrm{C}\) and \(160 \mathrm{kPa}\). The mass flow rate of air through the cycle is \(0.2 \mathrm{kg} / \mathrm{s}\). Assuming variable specific heats for air, determine ( \(a\) ) the rate of refrigeration, \((b)\) the net power input, and \((c)\) the coefficient of performance.
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