DoAssignment.ca

DoAssignment study guide

5.5 · Explain compressor efficiency and non-ideal cycle effects

Learn to explain compressor efficiency and non-ideal cycle effects through clear examples and targeted practice.

University of Alberta MEC E 340: Applied Thermodynamics

Refrigeration and Heat Pumps

MEC E 340 Applied Thermodynamics — Study topic 5.5

Start with a steady-flow compressor control volume and identify the working fluid. Label the inlet state 1, the actual outlet state 2, and the reference outlet state 2s. State 2s is an ideal comparison state, not a second physical outlet: it has the same pressure as state 2 and the same entropy as state 1. Unless a problem states otherwise, this lesson’s compressor examples assume adiabatic operation and negligible changes in kinetic and potential energy. With those assumptions, compressor work input per unit mass equals the fluid’s enthalpy rise. Use only property data or a gas model supplied or specified for the problem.

What you will learn

  • Define compressor isentropic efficiency and explain what its comparison means.
  • Use a steady-flow energy balance to calculate compressor work and outlet conditions.
  • Explain how compressor inefficiency can affect refrigeration-cycle performance.
  • Check assumptions, units, signs, and physical plausibility in compressor calculations.

1. Define the ideal comparison

For a one-inlet, one-outlet compressor operating steadily, mass conservation gives equal inlet and outlet mass flow rates. The reference state 2s is defined at the actual outlet pressure, with the same entropy as inlet state 1. It provides an ideal comparison for the specified compression; it does not describe another real flow path.
For the usual adiabatic comparison, actual compression requires more work than isentropic compression between the same inlet state and discharge pressure. Compressor isentropic efficiency compares the reference enthalpy rise with the actual enthalpy rise. An efficiency below one therefore means a larger actual enthalpy rise. Do not use the turbine efficiency definition for a compressor.
Use the property method given in the problem. For vapours or real gases, obtain enthalpies and entropy from supplied data or a named source. A problem may instead specify an ideal-gas model with constant heat capacities. That model can relate temperature change to enthalpy change, but it is an assumption, not a universal compressor rule.
ηc=h2s−h1h2−h1\eta_c=\frac{h_{2s}-h_1}{h_2-h_1}
  • State 2s has the actual outlet pressure and inlet entropy.
  • A lower efficiency means a larger actual enthalpy rise for the same reference rise.
  • A schematic plot shows process direction; it does not supply property values.

2. Use the energy balance to find work and outlet state

The steady-flow energy balance accounts for heat transfer, work transfer, and changes in enthalpy, kinetic energy, and potential energy. For an adiabatic compressor with negligible kinetic- and potential-energy changes, work supplied to the fluid per unit mass equals its enthalpy rise. If mass flow rate is known, multiply specific work by mass flow rate to obtain power.
For an ideal gas with constant specific heat, use the supplied specific heat to relate enthalpy change to temperature change. If the reference compression is also isentropic and the problem supplies or permits the ideal-gas model and heat-capacity ratio, use the corresponding temperature relation to find the reference outlet temperature. Apply that relation only under its stated assumptions.
Find the reference outlet first. Then use compressor efficiency to obtain the actual enthalpy rise, or actual outlet temperature when the stated model permits it. With property tables, use actual enthalpy and outlet pressure to identify state 2. In the stated ideal-gas, constant-specific-heat comparison, actual outlet temperature is higher than the reference temperature.
Take work input as positive. Under these assumptions, specific compressor work input is the outlet enthalpy minus the inlet enthalpy. Multiplying specific work by mass flow rate gives compressor power.
The governing equation for specific compressor work input under these assumptions is: w_{in}=h_2-h_1
  • Find state 2s before using efficiency to determine state 2.
  • Specific work can be expressed in kJ/kg; multiplying by kg/s gives kW.
  • A higher actual outlet temperature in this model does not mean heat was added.

3. Connect compressor losses to cycle performance

A compressor may operate as part of a power, refrigeration, or heat-pump cycle. For a fixed compressor inlet state and discharge pressure, lower isentropic efficiency raises compressor work relative to the isentropic reference. Other cycle states may also change if operating pressures or component conditions change, so distinguish a direct compressor comparison from a complete cycle recalculation.
In a refrigeration cycle, if the refrigerating effect is held fixed, more compressor work lowers refrigerator coefficient of performance. Refrigerator COP is heat removed from the cooled space divided by compressor work input. Heat removed and compressor work are different energy transfers and should not be confused.
Pressure losses in other parts of a cycle can change the pressures available at the compressor or other components. Those changes can alter state properties and cycle performance. The phrase “non-ideal effects” does not specify a numerical penalty: a numerical result requires supplied property data or an explicit model.
A cycle sketch can help keep state order and process direction clear. A pressure–specific-volume or temperature–entropy plot is schematic unless enough information is supplied to set exact coordinates. Do not infer enthalpies or pressure losses from the sketch.
The coefficient of performance is defined as the ratio of cooling output to work input: COPR=qLwin\mathrm{COP}_R=\frac{q_L}{w_{in}}.
  • Lower compressor efficiency raises work relative to the isentropic reference.
  • If cooling remains fixed, higher compressor work means lower refrigerator COP.
  • Quantify pressure-loss effects only when the required data or model is supplied.

Worked example

Ideal-gas compressor: actual outlet temperature and work

Air enters an adiabatic compressor at 300 K and 100 kPa and leaves at 600 kPa. Assume constant cp=1.005 kJ/(kg⋅K)c_p=1.005\ \mathrm{kJ/(kg\cdot K)}, k=1.40k=1.40, and compressor isentropic efficiency ηc=0.82\eta_c=0.82. Neglect kinetic- and potential-energy changes. Find the actual outlet temperature and specific work input.
Adiabatic compressor control volume
Adiabatic compressor control volumeCompressor1: 300 K, 100 kPa2: 600 kPa, actualWork inputControl-volume schematic

Air compressor at steady operation; adiabatic with negligible kinetic- and potential-energy changes. Device schematic, not a property plot.

  1. Set up states and assumptions
    The control volume is the compressor. State 1 is the inlet, state 2 is the actual outlet, and state 2s is the reference outlet at 600 kPa. Use the supplied ideal-gas, constant-property model and stated adiabatic assumption. Mass conservation gives equal inlet and outlet mass flow rates.
    m1=m2m_1=m_2
  2. Find the reference temperature
    The pressure ratio is 6. The supplied ideal-gas isentropic temperature relation gives the reference outlet temperature.
    T2s=300(6)(1.40−1)/1.40=500.6 KT_{2s}=300(6)^{(1.40-1)/1.40}=500.6\ \mathrm{K}
  3. Find the actual outlet temperature
    With constant specific heat, it cancels from the ratio of enthalpy rises. Use the efficiency definition to obtain the actual temperature rise.
    T2=300+500.6−3000.82=544.6 KT_2=300+\frac{500.6-300}{0.82}=544.6\ \mathrm{K}
  4. Calculate specific work input
    The adiabatic steady-flow energy balance makes specific work input equal to actual enthalpy rise. Apply the supplied constant specific heat to the actual temperature rise. w_{in}=1.005(544.6-300)=246\ kJ/kg
Answer: The actual outlet temperature is approximately 545 K, and specific work input is approximately 246 kJ/kg.
Check: The actual outlet temperature exceeds the 500.6 K reference temperature. Reference work is about 202 kJ/kg, so actual work is greater, as expected for an efficiency below one under the stated assumptions.

Worked example

Find efficiency from measured outlet temperature

A gas modelled as an ideal gas with constant k=1.40k=1.40 enters an adiabatic compressor at 300 K and 100 kPa. The discharge pressure is 400 kPa and the measured outlet temperature is 470 K. Use supplied constant cp=1.00 kJ/(kg⋅K)c_p=1.00\ \mathrm{kJ/(kg\cdot K)}. Neglect kinetic- and potential-energy changes. Find compressor isentropic efficiency and specific work input.
Measured compressor
Measured compressorCompressor1: 300 K, 100 kPa2: 470 K, 400 kPaWork inputControl-volume schematic

Gas treated with the supplied ideal-gas constant-property model; steady and adiabatic, with negligible kinetic- and potential-energy changes. Device schematic, not a property plot.

  1. Find reference outlet temperature
    The reference state uses the inlet temperature and actual discharge pressure. The pressure ratio is 4, and the supplied ideal-gas model applies.
    T2s=300(4)(1.40−1)/1.40=445.8 KT_{2s}=300(4)^{(1.40-1)/1.40}=445.8\ \mathrm{K}
  2. Calculate compressor efficiency
    With constant specific heat, it cancels from the ratio of enthalpy rises. Use the reference rise in the numerator and measured actual rise in the denominator.
    ηc=445.8−300470−300=0.8576\eta_c=\frac{445.8-300}{470-300}=0.8576
  3. Calculate specific work input
    For adiabatic steady flow with negligible kinetic- and potential-energy changes, work input equals enthalpy rise. Use the supplied specific heat and measured temperature rise. w_{in}=1.00(470-300)=170\ kJ/kg
Answer: The compressor isentropic efficiency is approximately 0.858, or 85.8%. Specific work input is 170 kJ/kg.
Check: The measured outlet temperature is above the 445.8 K reference, so the calculated efficiency is below one. The work unit is kJ/kg.

Worked example

Compressor efficiency and refrigeration COP

A simple vapour refrigeration cycle has supplied enthalpies h1=240 kJ/kgh_1=240\ \mathrm{kJ/kg}, h2s=270 kJ/kgh_{2s}=270\ \mathrm{kJ/kg}, and h3=h4=100 kJ/kgh_3=h_4=100\ \mathrm{kJ/kg}. The compressor is adiabatic and has ηc=0.75\eta_c=0.75. Assume refrigerating effect is h1−h4h_1-h_4. Find actual compressor outlet enthalpy and refrigerator COP.
Schematic vapour refrigeration cycle
Schematic vapour refrigeration cycleSpecific volume, vPressure, PCompressionHeat rejectionExpansionHeat absorption11223344Schematic · not to scale

Simple vapour refrigeration cycle using supplied enthalpies; schematic P–v plot, not to scale. State coordinates are not exact property values.

  1. Find actual outlet enthalpy
    The supplied state 2s enthalpy is the ideal reference at actual discharge pressure. Rearrange compressor efficiency to obtain actual outlet enthalpy.
    h2=240+270−2400.75=280 kJ/kgh_2=240+\frac{270-240}{0.75}=280\ \mathrm{kJ/kg}
  2. Find compressor work input
    For an adiabatic compressor, specific work input is the enthalpy rise from state 1 to state 2. w_{in}=280-240=40\ kJ/kg
  3. Calculate refrigerator COP
    Divide the supplied refrigerating effect by actual compressor work input. This compares cooling delivered with work required.
    COPR=240−10040=3.50\mathrm{COP}_R=\frac{240-100}{40}=3.50
Answer: The actual compressor outlet enthalpy is 280 kJ/kg, and refrigerator COP is 3.50.
Check: Reference compressor work is 30 kJ/kg, while actual work is 40 kJ/kg. With the supplied refrigerating effect fixed at 140 kJ/kg, greater actual work gives a lower COP than the reference comparison.

Common mistakes and how to avoid them

Defining the reference state using the actual outlet entropy.
Correction: State 2s has the inlet entropy and the actual outlet pressure.
Reversing the compressor efficiency ratio.
Correction: Use ideal-reference enthalpy rise divided by actual enthalpy rise.
Applying an ideal-gas temperature relation to every compressor fluid.
Correction: Use it only when the problem states or permits the ideal-gas and heat-capacity assumptions; otherwise use appropriate supplied property data.
Claiming an exact cycle penalty from compressor efficiency alone.
Correction: State what is held fixed and use only the supplied property and cycle data.

Lesson summary

  • Compressor isentropic efficiency compares ideal-reference and actual enthalpy rises for the same inlet state and discharge pressure.
  • For an adiabatic steady compressor with negligible kinetic- and potential-energy changes, work input per unit mass equals enthalpy rise.
  • Lower efficiency raises work input; if refrigerating effect is unchanged, refrigerator COP falls.
  • State assumptions, use an appropriate property method, carry SI units, and check that actual compression requires more work than the reference.

Check your understanding

Question 1

For the same inlet state and discharge pressure, what does compressor isentropic efficiency below one indicate in the usual adiabatic model?
  1. The actual enthalpy rise is greater than the reference enthalpy rise.
  2. The actual enthalpy rise is less than the reference enthalpy rise.
  3. The compressor delivers net work to the surroundings.
  4. The outlet pressure must be lower than the inlet pressure.
Show answer and explanation
The actual enthalpy rise is greater than the reference enthalpy rise.
Compressor efficiency is reference enthalpy rise divided by actual enthalpy rise. A value below one means the actual rise is greater.

Question 2

A refrigeration cycle has the same refrigerating effect but requires more compressor work per unit mass. What happens to its COP?
  1. It increases.
  2. It decreases.
  3. It stays unchanged.
  4. It becomes equal to compressor efficiency.
Show answer and explanation
It decreases.
Refrigerator COP is refrigerating effect divided by compressor work input. Increasing the denominator while holding the numerator fixed lowers COP.

Key terms

Isentropic reference state
A comparison outlet state at the actual discharge pressure and the same entropy as the compressor inlet.
Compressor isentropic efficiency
The reference enthalpy rise divided by the actual enthalpy rise for the specified compression.
Specific work input
Compressor work supplied per unit mass of working fluid, commonly expressed in kJ/kg.
Refrigerator COP
Heat removed from the cooled space per unit of compressor work input.

Continue through MEC E 340

View the complete MEC E 340 University of Alberta MEC E 340: Applied Thermodynamics curriculum and lessons

About this lesson and its review

Published by DoAssignment. This reviewed lesson follows University of Alberta MEC E 340: Applied Thermodynamics, study topic 5.5. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

Official curriculum reference

Report a correction or ask a question