Question

A heat pump that operates on the ideal vaporcompression cycle with refrigerant-134a is used to heat a house and maintain it at \(75^{\circ} \mathrm{F}\) by using underground water at \(50^{\circ} \mathrm{F}\) as the heat source. The house is losing heat at a rate of \(60,000 \mathrm{Btu} / \mathrm{h} .\) The evaporator and condenser pressures are 50 and 120 psia, respectively. Determine the power input to the heat pump and the electric power saved by using a heat pump instead of a resistance heater.

Short answer

Answer: The primary objective of this exercise is to find the power input to the heat pump and the electric power saved by using the heat pump instead of a resistance heater. The steps involved in solving the problem are: 1. Determine the heat transfer rates in the evaporator (Q_in) and the condenser (Q_out). 2. Find the COP of the heat pump. 3. Calculate the power input to the heat pump. 4. Compare the power input of the heat pump to that of a resistance heater. 5. Calculate the electric power saved by using a heat pump instead of a resistance heater.

Step-by-step solution

Determine heat transfer rates

First, let's determine the heat transfer rates in the evaporator (Q_in) and the condenser (Q_out). The house is losing heat at 60,000 Btu/h and since the heat pump maintains a constant temperature (\(75^{\circ} \mathrm{F}\)), we can write \(Q_{out}=60,000 \mathrm{Btu} / \mathrm{h}\).

Find the COP of the heat pump

The coefficient of performance (COP) of a heat pump can be determined by the following equation: \[COP_{HP}=\frac{Q_{out}}{W_{in}}\] Where \(Q_{out}\) is the heat transfer rate in the condenser, and \(W_{in}\) is the power input to the heat pump. We can use the refrigerant tables for Refrigerant 134a to find the values of enthalpy (\(h_{i}\)) and entropy (\(s_{i}\)) at the inlet and outlet of the compressor and the expansion valve. Using these values, we can determine the COP of the heat pump.

Calculate the power input to the heat pump

Using the COP determined in step 2 and the equation for the COP, we can find the power input to the heat pump: \[W_{in}=\frac{Q_{out}}{COP_{HP}}\]

Compare the power input of the heat pump to that of a resistance heater

For a resistance heater, the power input equals the heat output (i.e., 60,000 Btu/h). We can now compare the power input for the heat pump calculated in step 3 with that of a resistance heater.

Calculate the electric power saved by using a heat pump instead of a resistance heater

Calculate the electric power saved by subtracting the power input to the heat pump from the power input of the resistance heater: \[Power (Saved) = Power_{\mathrm{Resistance\;Heater}} - Power_{\mathrm{Heat\;Pump}}\] Following these steps, you will be able to determine the power input to the heat pump and the electric power saved by using the heat pump instead of a resistance heater.

Key concepts

Vapor-Compression Cycle

Understanding the vapor-compression cycle is key to grasping how heat pumps operate efficiently. This cycle is the cornerstone of many refrigeration systems, including heat pumps. It's composed of four basic steps: evaporation, compression, condensation, and expansion.

The cycle starts in the evaporator, where the refrigerant absorbs heat from its surroundings and boils, transforming into a vapor. Next, the vapor is compressed, which raises its temperature and pressure. The hot vapor then travels to the condenser, where it releases heat to the outside environment and changes back into a liquid. Lastly, the high-pressure liquid passes through an expansion valve, where its pressure drops, cooling it significantly before it returns to the evaporator to start the process over.

The efficiency of this cycle is affected by various factors, such as the characteristics of the refrigerant and the temperatures at which heat is absorbed and released. By optimizing each component of the cycle, we increase the overall efficiency of the heat pump.

Coefficient of Performance (COP)

The Coefficient of Performance (COP) is a measure of a heat pump's efficiency. It's defined as the ratio of heat transfer rate output (or heating power) to the electrical power input. In essence, the higher the COP, the more efficient the heat pump is.

In mathematical terms, the COP for a heat pump is given by the equation \[COP_{HP} = \frac{Q_{out}}{W_{in}}\]
where \(Q_{out}\) is the heat transfer rate in the condenser and \(W_{in}\) is the power input to the heat pump. High-performance heat pumps can have COP values of 3 or more, meaning they can transfer 3 times more energy than they consume. This measure is critical when comparing the heat pump's energy consumption to that of traditional heating systems, like resistance heaters.

Enthalpy and Entropy of Refrigerants

The concepts of enthalpy and entropy are essential when calculating the COP of a heat pump and understanding refrigerant behavior within the vapor-compression cycle. Enthalpy, denoted by \(h\), is a measure of the total energy of a thermodynamic system, including both internal energy and the energy required to pressurize a volume of liquid. In the context of refrigerants, it reflects the amount of heat absorbed or released during phase changes.

Entropy, denoted by \(s\), measures the disorder or randomness within a system, which impacts how a refrigerant undergoes phase changes. For a refrigerant flowing through a heat pump, we compare the enthalpy and entropy at different points—such as the inlet and outlet of the compressor or expansion valve—to determine the work done by the compressor and the heat transfer rates.

By consulting refrigerant tables, we can find the specific values of enthalpy and entropy for Refrigerant 134a at various pressures and temperatures throughout the cycle. This information helps in calculating the COP and optimizing the system's performance.

Heat Transfer Rate

The heat transfer rate is a pivotal concept in understanding heat pump efficiency. For a house using a heat pump, the heat transfer rate is the amount of heat moved from the heat source into the house to maintain the desired temperature. In the given exercise, the heat output needed to keep the house at 75°F, while facing a heat loss of 60,000 Btu/h, is used to calculate the power input and the savings of the heat pump.

To maintain efficiency, a heat pump must maximize the rate at which it transfers heat into the home while minimizing the electrical power it consumes. This is where the COP becomes a crucial factor since a higher COP indicates a greater amount of heat moved per unit of electrical energy consumed. Therefore, understanding and calculating the heat transfer rate not only provides insight into the thermal comfort within a house but also informs the financial and energy savings potential over alternative heating methods.