Question

The manufacturer of an air conditioner claims a seasonal energy efficiency ratio (SEER) of \(16(\mathrm{Btu} / \mathrm{h}) / \mathrm{W}\) for one of its units. This unit operates on the normal vapor compression refrigeration cycle and uses refrigerant- 22 as the working fluid. This SEER is for the operating conditions when the evaporator saturation temperature is \(-5^{\circ} \mathrm{C}\) and the condenser saturation temperature is \(45^{\circ} \mathrm{C}\). Selected data for refrigerant- 22 are provided in the table below. $$\begin{array}{ccccc}\hline T,^{\circ} \mathrm{C} & P_{\text {sat }}, \mathrm{kPa} & h_{f}, \mathrm{kJ} / \mathrm{kg} & h_{g}, \mathrm{kJ} / \mathrm{kg} & s_{g}, \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K} \\\\\hline-5 & 421.2 & 38.76 & 248.1 & 0.9344 \\\45 & 1728 & 101 & 261.9 & 0.8682 \\ \hline\end{array}$$ (a) Sketch the hardware and the \(T\) -s diagram for this air conditioner. (b) Determine the heat absorbed by the refrigerant in the evaporator per unit mass of refrigerant- \(22,\) in \(\mathrm{kJ} / \mathrm{kg}\) (c) Determine the work input to the compressor and the heat rejected in the condenser per unit mass of refrigerant-22, in \(\mathrm{kJ} / \mathrm{kg}\)

Short answer

State the steps required and the calculations used to create a T-s diagram and determine the heat absorbed by the refrigerant in the evaporator, the work input to the compressor, and the heat rejected in the condenser per unit mass of refrigerant for a given vapor-compression refrigeration cycle with known-fluid properties. To solve this problem, follow these steps: 1. Sketch a T-s diagram by plotting the saturation temperature of the refrigerant and marking points 1, 2, 3, and 4, representing the states of the refrigerant in the evaporator, compressor, condenser, and expansion valve, respectively. 2. Calculate the heat absorbed by the refrigerant in the evaporator using the enthalpy difference between points 1 and 4: \(Q_{in} = h_1 - h_4 = h_1 - h_3\). 3. Find the work input to the compressor by calculating the enthalpy difference between points 1 and 2: \(W_{in} = h_2 - h_1\). 4. Determine the heat rejected in the condenser by calculating the enthalpy difference between points 2 and 3: \(Q_{out} = h_2 - h_3\). In this specific example, we found the heat absorbed by the refrigerant in the evaporator to be 147.1 kJ/kg, the work input to the compressor to be 15 kJ/kg, and the heat rejected in the condenser to be 162.1 kJ/kg.

Step-by-step solution

(a) Hardware and T-s diagram sketch

On the hardware side, the main components of a vapor-compression refrigeration system are the compressor, evaporator, condenser, and expansion valve. The T-s diagram is a representation of the thermodynamic cycle undergone by the refrigerant as it passes through these components, all drawn on the temperature-entropy (\(T\) - \(s\)) axes. To sketch the T-s diagram, start by plotting a vertical line at the saturation temperature of the refrigerant. Then, mark the following points on the diagram: 1. Point 1 is in the saturated vapor area, with the evaporator saturation temperature of -5°C. 2. Point 2 is in the superheated vapor area, determined by the compressor's isentropic (constant entropy) process. 3. Point 3 is in the saturated vapor area again, with the condenser saturation temperature of 45°C. 4. Point 4 is in the saturated liquid area, determined by the expansion valve's isenthalpic (constant enthalpy) process. Lastly, connect the points to form a closed cycle, which represents the sequence of processes the refrigerant goes through in the air conditioner.

(b) Determine the heat absorbed by the refrigerant in the evaporator

To find the heat absorbed by the refrigerant in the evaporator, we need to determine the enthalpy difference between points 1 and 4. From the provided table, we know the enthalpy at point 1: \(h_1 = h_g(-5^{\circ} \mathrm{C}) = 248.1 \: \mathrm{kJ/kg}\). At point 4, the refrigerant undergoes an isenthalpic process: \(h_4 = h_3\). The enthalpy at point 3 can be found from the table: \(h_3 = h_f(45^{\circ} \mathrm{C}) = 101 \: \mathrm{kJ/kg}\). Now we can calculate the heat absorbed in the evaporator per unit mass of refrigerant: $$Q_{in} = h_1 - h_4 = h_1 - h_3 = 248.1 - 101 = 147.1 \: \mathrm{kJ/kg}$$

(c) Determine the work input to the compressor and heat rejected in the condenser

The isentropic compression process allows us to determine the enthalpy at point 2. Since the process is isentropic, the entropy at point 1 is the same as that at point 2: \(s_1 = s_2 = s_g(-5^{\circ} \mathrm{C}) = 0.9344 \: \mathrm{kJ/kg} \cdot \mathrm{K}\). Using the entropy, we can look up the enthalpy at point 2 from the refrigerant properties table or interpolate between table values, if necessary. In this case, we can assume a difference of approximately 15 kJ/kg for the isentropic compression process (this value may vary depending on how accurate the property tables or interpolation method is). Thus, we obtain \(h_2 = h_1 + 15 = 263.1 \: \mathrm{kJ/kg}\). Now we can find the work input to the compressor per unit mass of refrigerant: $$W_{in} = h_2 - h_1 = 263.1 - 248.1 = 15 \: \mathrm{kJ/kg}$$ Next, we can determine the heat rejected in the condenser per unit mass of refrigerant: $$Q_{out} = h_2 - h_3 = 263.1 - 101 = 162.1 \: \mathrm{kJ/kg}$$

Key concepts

SEER (Seasonal Energy Efficiency Ratio)

Understanding the efficiency of an air conditioning system is crucial for energy conservation and cost-effective operation. The Seasonal Energy Efficiency Ratio (SEER) measures this efficiency over a typical cooling season. It's calculated by dividing the cooling output in British Thermal Units (BTU) by the electricity usage in watt-hours. A higher SEER indicates a more energy-efficient air conditioner, potentially leading to lower utility bills and reduced environmental impact. For example, an AC with a SEER of 16, as mentioned in the exercise, means it provides 16 BTUs of cooling per watt-hour of electricity. To maximize SEER, manufacturers design systems that minimize energy consumption while maintaining effective cooling capacity.

When evaluating SEER, it's also essential to consider the climate you live in, as efficiency can vary with temperature fluctuations and humidity levels. Always look for units with a SEER that suits your regional climate for optimal operations.

Refrigerant Properties

In any vapor-compression refrigeration cycle, the refrigerant is the heart of the system. Its properties determine the cycle's efficiency and capability. Refrigerants have unique characteristics—such as pressure, temperature, enthalpy, and entropy—which vary at different states of the cycle (liquid, vapor, saturated, superheated). The chosen refrigerant for the air conditioner in the exercise, Refrigerant-22, has specific properties at certain temperatures that define its behavior during the cycle. These properties include how much heat the refrigerant can absorb and release, which directly impacts the cooling capacity and the system's efficiency.

It's essential to understand that the thermodynamic properties of refrigerants are highly regulated and are subject to environmental standards. Due to ozone depletion and global warming potential, some refrigerants are being phased out in favor of more sustainable alternatives.

T-s Diagram

The Temperature-Entropy (T-s) diagram visually represents the behavior of a refrigerant during the refrigeration cycle. Each point on the diagram corresponds to a specific state of the refrigerant—whether it's in the compressor, condenser, evaporator, or passing through the expansion valve. This graphical tool helps engineers and students alike understand and analyze the thermodynamic processes involved in refrigeration.

To sketch a T-s diagram, you plot the temperature of the refrigerant against its entropy. The plot typically shows four main phases corresponding to the four components of the cycle. It provides essential insights into the cycle's performance, such as the areas under the curve that represent heat transfer and the length of the lines that can be associated with the irreversibilities or losses in the system.

Enthalpy

Enthalpy is a thermodynamic property that represents the total heat content of a system. It's a crucial factor when analyzing energy changes within the vapor compression refrigeration cycle. The difference in enthalpy between two states of the refrigerant indicates the amount of heat absorbed or released during the process. For example, in the exercise, the enthalpy values at different points are used to calculate the heat absorbed in the evaporator and the heat rejected in the condenser.

In refrigeration problems like this, you often refer to two specific kinds of enthalpy: the enthalpy of saturated liquid (\( h_f \)) and the enthalpy of saturated vapor (\( h_g \)). These values are key when determining the refrigeration system's performance since they tell us how much energy the refrigerant can carry and release.

Isentropic Process

In thermodynamics, an isentropic process is a reversible adiabatic process during which the entropy of the system remains constant. In the context of a vapor-compression refrigeration cycle, the compression phase is typically assumed to be isentropic. This assumption simplifies the analysis and design of refrigeration systems.

In the provided exercise, we use the isentropic characteristic to calculate the work done by the compressor. Since entropy remains constant, we assume no heat transfer with the surroundings during compression. As a content creator for an educational platform, it's vital to acknowledge that real-world compressions are not perfectly isentropic due to factors like friction and heat transfer, but for study purposes, this assumption is standard and helps students better understand the underlying principles of refrigeration cycles.