Question

An ideal vapor-compression heat pump cycle with Refrigerant \(134 \mathrm{a}\) as the working fluid provides heating at a rate of \(15 \mathrm{~kW}\) to maintain a building at \(20^{\circ} \mathrm{C}\) when the outside temperature is \(5^{\circ} \mathrm{C}\). Saturated vapor at \(2.4\) bar leaves the evaporator, and saturated liquid at 8 bar leaves the condenser. Calculate (a) the power input to the compressor, in \(\mathrm{kW}\). (b) the coefficient of performance. (c) the coefficient of performance of a Carnot heat pump cycle operating between thermal reservoirs at 20 and \(5^{\circ} \mathrm{C}\).

Short answer

Power input to the compressor is 2.01 kW, COP is 7.46, and Carnot COP is 19.54.

Step-by-step solution

- Determine properties from tables

Identify the relevant thermodynamic states for the refrigerant from the provided conditions: saturated vapor at 2.4 bar and saturated liquid at 8 bar. Use refrigerant tables to find the enthalpies at these states. Let's denote the state before the compressor as state 1 and after the compressor as state 2.

- Find enthalpies

From refrigerant tables: For saturated vapor at 2.4 bar (state 1): \(h_1 = 396 \text{ kJ/kg} \). For saturated liquid at 8 bar: \(h_3 = 248 \text{ kJ/kg} \).

- Use isentropic process

Assuming an isentropic process for the compressor, use the entropy values: For saturated vapor at 2.4 bar: \(s_1 = 1.250 \text{ kJ/kg·K} \). Then find the enthalpy after compression (state 2) using the isentropic relation: \(s_1 = s_2\), find out from the tables or diagrams: \(h_2 = 419 \text{ kJ/kg} \).

- Calculate the mass flow rate

Using the given heating rate (\(\text{Q} = 15 \text{ kW} \)) and the enthalpy change in the condenser: \( Q = \text{m} \times (h_2 - h_3) \). Solve for m: \( \text{m} = \frac{Q}{(h_2 - h_3)} = \frac{15}{(419 - 248)} = 0.0877 \text{ kg/s} \).

- Calculate power input to the compressor

Use the mass flow rate to find the power input to the compressor: \( \text{Power input} = \text{m} \times (h_2 - h_1) = 0.0877 \times (419 - 396) = 2.01 \text{ kW} \).

- Calculate the Coefficient of Performance (COP)

COP for the heat pump cycle is given by: \( \text{COP} = \frac{Q}{\text{Power input}} = \frac{15}{2.01} = 7.46 \).

- Calculate the COP of a Carnot heat pump cycle

For the Carnot heat pump, use temperatures in Kelvin: \( \text{T}_{\text{hot}} = 20 + 273.15 = 293.15 \text{ K} \) and \( \text{T}_{\text{cold}} = 5 + 273.15 = 278.15 \text{ K} \). \( \text{COP}_{\text{Carnot}} = \frac{T_{\text{hot}}}{T_{\text{hot}} - T_{\text{cold}}} = \frac{293.15}{293.15 - 278.15} = 19.54 \).

Key concepts

Isentropic Process

In the context of a vapor-compression heat pump cycle, an isentropic process is one where the entropy remains constant. Entropy is a measure of disorder or randomness in a system. For the compressor in our heat pump cycle, we assume it undergoes an isentropic process, meaning no entropy is generated within the compressor.

This isentropic assumption simplifies calculations and represents an ideal situation where the process is both adiabatic (no heat transfer) and reversible.

Key points in isentropic process:

• Entropy before and after the compressor remains the same: \[ s_1 = s_2 \]
• We utilize refrigerant tables or diagrams to obtain properties such as enthalpy and entropy for the refrigerant.
• This assumption allows us to use known values to find unknowns, making calculation straightforward and eliminating the need for more complex, real-world considerations like friction and heat losses.

The process, therefore, is critical in determining the enthalpy after compression, which is necessary for calculating things like power input to the compressor and ultimately the system's efficiency.

Enthalpy Calculation

Enthalpy is a measure of the total energy of a system, including both its internal energy and the product of its pressure and volume. In our vapor-compression heat pump cycle, calculating enthalpy at various points in the cycle is crucial for analyzing the system.

Steps for calculating enthalpy:

• Use refrigerant tables to find the enthalpy values corresponding to specific states of the refrigerant.
• For the given problem, state 1 involves saturated vapor at 2.4 bar with an enthalpy: \[ h_1 = 396 \text{ kJ/kg} \]
• State 3 concerns saturated liquid at 8 bar with an enthalpy: \[ h_3 = 248 \text{ kJ/kg} \]
• Using the isentropic assumption, we can find the enthalpy after compression (state 2) since \[ s_1 = s_2 \] leading to: \[ h_2 = 419 \text{ kJ/kg} \]
• These enthalpy values are then used to determine the mass flow rate and various power requirements.

Calculation steps can be summarized as:

• Obtain enthalpy values at different states from refrigerant tables.
• Use these values in energy balance equations to calculate quantities like mass flow rate and power input.

Enthalpy calculations link the thermodynamic properties of the refrigerant with the performance parameters of the heat pump system.

Coefficient of Performance

The Coefficient of Performance (COP) for a heat pump is a measure of its efficiency. It indicates how effectively the heat pump uses work input to transfer heat. A higher COP means higher efficiency.

Calculating COP:

• The general formula for COP in a heating mode: \[ \text{COP} = \frac{Q}{\text{Power Input}} \]
• In our problem, Q is the heating rate provided to the building, which is 15 kW, and the power input to the compressor is found to be 2.01 kW.
• Using these values, the COP is:\[ \text{COP} = \frac{15}{2.01} = 7.46 \]
• This means the heat pump delivers 7.46 units of heat for every unit of work.

Additionally, comparing this to the ideal Carnot heat pump cycle, which operates between specific temperatures (20°C and 5°C) and has a very high COP:

• Using the Carnot COP formula:\[ \text{COP}_{\text{Carnot}} = \frac{T_{\text{hot}}}{T_{\text{hot}} - T_{\text{cold}}} \]
• Convert temperatures to Kelvin (T_hot = 293.15 K and T_cold = 278.15 K):
• The Carnot COP is:\[ \text{COP}_{\text{Carnot}} = \frac{293.15}{293.15 - 278.15} = 19.54 \]

Real systems, like our vapor-compression heat pump, have lower COPs than the ideal Carnot cycle due to practical inefficiencies. Understanding COP helps in evaluating how well a heat pump performs and where improvements might be made.