Question

A heat pump that operates on the ideal vaporcompression cycle with refrigerant-134a is used to heat a house. The mass flow rate of the refrigerant is \(0.25 \mathrm{kg} / \mathrm{s}\) The condenser and evaporator pressures are 1400 and \(320 \mathrm{kPa}\) respectively. Show the cycle on a \(T\) -s diagram with respect to saturation lines, and determine ( \(a\) ) the rate of heat supply to the house, \((b)\) the volume flow rate of the refrigerant at the compressor inlet, and \((c)\) the COP of this heat pump.

Short answer

Question: Calculate the rate of heat supply to the house, the volume flow rate of the refrigerant at the compressor inlet, and the Coefficient of Performance (COP) of a heat pump that operates on the ideal vapor-compression cycle with R-134a. Answer: The rate of heat supply to the house is 39.08 kJ/s, the volume flow rate of the refrigerant at the compressor inlet is 0.02489 m³/s, and the Coefficient of Performance (COP) of this heat pump is 5.44.

Step-by-step solution

Identify the properties of the refrigerant at each point

The ideal vapor-compression cycle consists of four processes: 1-2: Isentropic compression 2-3: Constant pressure heat rejection 3-4: Isenthalpic expansion 4-1: Constant pressure heat addition First, we need to find the refrigerant properties at each point in the cycle. We already know the pressures at the condenser (P2=P3=1400 kPa) and evaporator (P4=P1=320 kPa). Point 1: As the evaporator exit is saturated vapor, we can look up the necessary properties in the R-134a tables (or use a standard software for refrigeration calculations) at 320 kPa: - h1 = 398.63 kJ/kg - s1 = 1.7559 kJ/kg∙K - v1 = 0.09957 m³/kg Point 2: The compression is isentropic (s2=s1), so we can find the temperature and enthalpy at the condenser inlet using the pressure (1400 kPa) and entropy (s2): - h2 = 429.52 kJ/kg Point 3: As the condenser exit is saturated liquid at 1400 kPa: - h3 = 273.19 kJ/kg Point 4: The expansion is isenthalpic (h4=h3), so at 320 kPa: - h4 = 273.19 kJ/kg

Calculate the heat transfer and work at each point

Using enthalpies at each point, we can calculate the heat transfer and work in each process: - Heat added in the evaporator (Qin): \(Q_\text{in} = h_1 - h_4\) - Heat rejected in the condenser (Qout): \(Q_\text{out} = h_2 - h_3\) - Work done in the compressor (Wc): \(W_\text{c} = h_2 - h_1\)

Find rate of heat supply to the house, volume flow rate of refrigerant, and COP of the heat pump

Now we plug in the values obtained in Steps 1 and 2 and also the given mass flow rate (0.25 kg/s): (a) Rate of heat supply to the house (Qout): \(Q_\text{out} = m_\text{flow} \times (h_2 - h_3)\) \(Q_\text{out} = 0.25 \times (429.52 - 273.19) \thinspace \mathrm{kJ/s}\) \(Q_\text{out} = 39.08 \thinspace \mathrm{kJ/s}\) (b) Volume flow rate at compressor inlet: \(V_\text{flow} = m_\text{flow} \times v_1\) \(V_\text{flow} = 0.25 \times 0.09957 \thinspace \mathrm{m^3/s}\) \(V_\text{flow} = 0.02489 \thinspace\mathrm{m^3/s}\) (c) Coefficient of Performance (COP) of the heat pump: \(COP_\text{HP} = \frac{Q_\text{out}}{W_\text{c}} = \frac{m_\text{flow}(h_2 - h_3)}{m_\text{flow}(h_2 - h_1)}\) \(COP_\text{HP} = \frac{429.52 - 273.19}{429.52 - 398.63}\) \(COP_\text{HP} = 5.44\) The cycle can be shown on a T-s diagram as follows: 1. Plot the saturation lines for R-134a 2. Locate points 1, 2, 3, and 4 based on the calculated properties 3. Draw lines connecting 1-2 (isentropic compression), 2-3 (constant pressure heat rejection), 3-4 (isenthalpic expansion), and 4-1 (constant pressure heat addition). In conclusion, we find that (a) the rate of heat supply to the house is 39.08 kJ/s, (b) the volume flow rate of the refrigerant at the compressor inlet is 0.02489 m³/s, and (c) the COP of this heat pump is 5.44.

Key concepts

Heat Pump

A heat pump is a device that moves heat from one location to another by using mechanical energy, typically to heat or cool spaces. They often work on the principle of the vapor-compression cycle, which involves the repetitive condensation and evaporation of a refrigerant. In the heating mode, as seen in the exercise, a heat pump extracts heat from a low-temperature source and releases it at a higher temperature inside the house, effectively heating it. This is achieved despite the outside temperature being colder than the house's interior, showcasing the heat pump's ability to reverse the natural flow of heat.

Understanding how a heat pump operates is vital if one wants to analyze or design heating systems. Its performance is expressed in terms of the coefficient of performance (COP), which is a ratio of the heat supplied to the work input, demonstrating efficiency. The higher the COP value, the more efficient the heat pump is.

Refrigerant Properties

In the vapor-compression cycle, refrigerants play a crucial role as the working fluid. The properties of refrigerants, such as enthalpy and entropy, govern the transfer of heat and the work input needed for the cycle. Each refrigerant has unique characteristics that suit different operating conditions and environmental impact considerations.

Understanding refrigerant properties like boiling point, pressure, and specific volume is essential for the calculation of heat transfer and compressor work in the vapor-compression cycle. For example, a refrigerant's enthalpy at different phases indicates the amount of heat absorbed or released during the phase changes, directly affecting the cycle's performance. In the textbook solution, R-134a is the chosen refrigerant, whose properties can be looked up in refrigeration tables or calculated using software.

Coefficient of Performance (COP)

The Coefficient of Performance (COP) is a dimensionless measure that indicates the efficiency of a heat pump or refrigeration system. It is defined as the ratio of the useful heating or cooling provided to the work required to produce that heating or cooling. Mathematically, for a heat pump, COP is the heat delivered to the warm space divided by the work input:\[\begin{equation}COP_{HP} = \frac{Q_{out}}{W_c}\end{equation}\]The COP indicates how much more efficient a heat pump is compared to direct electric heating. A higher COP means the system is more efficient, as it requires less work to transfer a certain amount of heat. This makes COP a critical factor when considering the environmental and operational costs in an HVAC system design.

Enthalpy

Enthalpy, denoted by 'h', is a property of a substance that reflects the total heat content. It is essential in calculating the energy changes within a thermodynamic system undergoing a process. Specifically, enthalpy helps determine the heat transfer associated with phase changes, like evaporation and condensation, in a vapor-compression cycle.

Engineers use the changes in enthalpy to calculate the work done by or on the system as well as the heat exchanged. For example, the heat added in an evaporator (Qin) is the difference between the enthalpy of the refrigerant at the inlet and outlet of the process. The solution to the textbook exercise showcases the application of enthalpy values obtained from refrigerant property tables to find the rate of heat supply (Qout) and the work done by the compressor (Wc).

Isentropic Process

An isentropic process is a theoretical or idealized thermodynamic process that is both adiabatic (no heat transfer occurs) and reversible. 'Isentropic' comes from the Greek words 'iso' (equal) and 'entropos' (transformation), implying constant entropy. With no change in entropy during this process, it represents perfect efficiency.

In practical applications like the vapor-compression cycle, the compression step of the refrigerant in the compressor is often assumed to be isentropic. This assumption simplifies the calculation for the work input needed. To find related refrigerant properties during the compression step, one can use the known entropy value from the initial state and the pressure at the final state. As seen in the textbook's step-by-step solution, following an isentropic process allows pinpointing the enthalpy after compression using the constant entropy value.