Question

Refrigerant-134a at \(140 \mathrm{kPa}\) and \(-10^{\circ} \mathrm{C}\) is compressed by an adiabatic \(1.3-\mathrm{kW}\) compressor to an exit state of \(700 \mathrm{kPa}\) and \(60^{\circ} \mathrm{C}\). Neglecting the changes in kinetic and potential energies, determine ( \(a\) ) the isentropic efficiency of the compressor, \((b)\) the volume flow rate of the refrigerant at the compressor inlet, in \(\mathrm{L} / \mathrm{min}\), and \((c)\) the maximum volume flow rate at the inlet conditions that this adiabatic \(1.3-\mathrm{kW}\) compressor can handle without violating the second law.

Short answer

Answer: The isentropic efficiency of the compressor is 78%, the volume flow rate of the refrigerant at the compressor inlet is 666 L/min, and the maximum volume flow rate at the inlet conditions that this adiabatic 1.3-kW compressor can handle without violating the second law is 852 L/min.

Step-by-step solution

Isentropic efficiency

To find the isentropic efficiency, we can use the following relation between the actual work and the isentropic work: $$ \eta_c = \frac{h_{2s} - h_1}{h_2 - h_1} $$ Where \(\eta_c\) is the isentropic efficiency, \(h_1\) and \(h_2\) are specific enthalpies at the inlet and exit, and \(h_{2s}\) is the specific enthalpy for an isentropic exit state, which has the same entropy as the inlet state, \(s_1\) = \(s_{2s}\). We need to find these specific enthalpies. From the Refrigerant-134a tables, we can find \(h_1 = 249.55 \, \mathrm{kJ/kg} \,\) and \(s_1 = 0.9202 \, \mathrm{kJ/(kg.K)}\) at \(140 \, \mathrm{kPa}\), and \(-10^{\circ} \mathrm{C}\). At the exit state of \(700 \, \mathrm{kPa}\) and \(60^{\circ} \mathrm{C}\), we can find \(h_2 = 318.16 \, \mathrm{kJ/kg}\). For the isentropic exit state, we would have the same entropy as the inlet state. So, we can find \(h_{2s} = 297.6 \, \mathrm{kJ/kg}\) at \(700 \, \mathrm{kPa}\) and \(s_{2s} = s_1 = 0.9202 \, \mathrm{kJ/(kg.K)}\). Now we can find the isentropic efficiency as: $$ \eta_c = \frac{297.6 - 249.55}{318.16 - 249.55} = 0.780 $$

Volume flow rate at the compressor inlet

To find the volume flow rate at the compressor inlet, we will use the specific volume at the inlet condition and the mass flow rate, as follows: $$ \dot{V} = \dot{m} \cdot v_1 $$ Where \(\dot{V}\) is the volume flow rate, \(\dot{m}\) is the mass flow rate, and \(v_1\) is the specific volume at the inlet state. The actual work done by the compressor can be found as: $$ W_{actual} = \dot{m} (h_2 - h_1) $$ We are given the compressor power \(W_{actual} = 1.3 \, \mathrm{kW}\), so we can now find the mass flow rate as: $$ \dot{m} = \frac{1.3 \times 10^3 \, \mathrm{W}}{318.16 - 249.55} = 0.0512 \, \mathrm{kg/s} $$ From the Refrigerant-134a tables, we've determined that the specific volume at the inlet state \(v_1 = 0.2163 \, \mathrm{m^3/kg}\) Now, we can determine the volume flow rate at the compressor inlet as: $$ \dot{V} = 0.0512 \cdot 0.2163 = 0.0111 \, \mathrm{m^3/s} $$ Converting this to Liters per minute, we get \(0.0111 \times 1000 \times 60 = 666 \, \mathrm{L/min}\).

Maximum volume flow rate at the inlet conditions

To find the maximum volume flow rate at the inlet conditions without violating the second law, we set the compressor work equal to the isentropic work: $$ W_{max} = \dot{m}_{max} \cdot (h_{2s} - h_1) $$ Where \(\dot{m}_{max}\) is the maximum mass flow rate at the inlet conditions and \(W_{max} = 1.3 \, \mathrm{kW}\). We can now find the maximum mass flow rate at the inlet conditions as: $$ \dot{m}_{max} = \frac{1.3 \times 10^3 \, \mathrm{W}}{297.6 - 249.55} = 0.0657 \, \mathrm{kg/s} $$ The maximum volume flow rate at the inlet conditions can now be found as: $$ \dot{V}_{max} = 0.0657 \cdot 0.2163 = 0.0142 \, \mathrm{m^3/s} $$ Converting this to Liters per minute, we get \(0.0142 \times 1000 \times 60 = 852 \, \mathrm{L/min}\). So, the answers are (a) the isentropic efficiency of the compressor is \(78 \%\), (b) the volume flow rate of the refrigerant at the compressor inlet is \(666 \, \mathrm{L/min}\), and (c) the maximum volume flow rate at the inlet conditions that this adiabatic \(1.3-\mathrm{kW}\) compressor can handle without violating the second law is \(852 \, \mathrm{L/min}\).

Key concepts

Isentropic Efficiency Calculation

Understanding the isentropic efficiency of a compressor is crucial in evaluating its performance. Isentropic efficiency provides a measure of how close the compressor operates to the ideal, reversible process. An adiabatic, or isentropic, process is one that is both thermally insulated so that no heat exchange occurs with the surroundings, and occurs without friction or other dissipative effects. The isentropic efficiency calculation uses the specific enthalpies of the refrigerant at different states to compare the actual work input to the compressor with the work input if the process were isentropic.

The formula for calculating isentropic efficiency, \( \eta_c \) is as follows:
\[\eta_c = \frac{h_{2s} - h_1}{h_2 - h_1}\]
where:\

• \( h_1 \) is the specific enthalpy at the compressor inlet,
• \( h_2 \) is the specific enthalpy at the exit, and
• \( h_{2s} \) is the specific enthalpy at the exit for an isentropic process.

Calculating these values requires data from thermodynamic tables for the particular refrigerant being used. For an educational approach, prompts can be introduced to guide students through using these tables to find the required enthalpy values for given states.

Thermodynamic Properties of Refrigerant

The performance of a refrigeration system depends greatly on the thermodynamic properties of the refrigerant used. Refrigerants like R-134a have specific properties that determine their suitability for different applications. These properties include specific heat capacities, enthalpy, entropy, and specific volume, which vary with temperature and pressure.

For instance, the specific enthalpy (\(h\)), which represents the total heat content of the refrigerant per unit mass, is a key component in calculating work done in refrigeration cycles. Likewise, specific volume (\(v\)) influences the design and operation of compressors, as it impacts the volume flow rate through the compressor for a given mass flow rate of refrigerant. Understanding the relationship between pressure, temperature, and these thermodynamic properties is essential for accurate analysis and calculation of refrigeration cycles. Students can be encouraged to explore pressure-enthalpy diagrams, commonly known as P-H diagrams, or use refrigerant tables to understand how these properties interplay for various refrigerants.

Furthermore, it is essential to grasp that these properties, read from standard charts or calculated using equations of state, are vital to determining other performance parameters, such as isentropic efficiency and mass flow rate.

Second Law of Thermodynamics

The second law of thermodynamics is a fundamental principle that has far-reaching consequences in all areas of energy conversion and efficiency. In the context of an adiabatic compressor, it establishes limits on what is physically achievable. The law implies that for any spontaneous process, the entropy of the universe increases, or, in a more constrained system like a compressor, entropy does not decrease.

An important aspect of the second law for refrigeration cycles is that it dictates the maximum possible efficiency of the cycle; no process can be 100% efficient due to inevitable increases in entropy during real processes. This is why understanding the isentropic efficiency is critical because it gives us a metric by which we can compare the actual performance of a compressor to the theoretical limit imposed by the second law.

Additionally, the second law helps explain why practical compressors cannot handle an unlimited volume flow rate of refrigerant. The existence of limitations by the second law can be particularly enlightening for students who might consider only the energy perspective and neglect the implications of entropy. Exercises can integrate these concepts by involving calculations that respect the boundaries set by the second law, providing a realistic framework for students to appreciate the intricacies of thermodynamics in practical applications.