Juan C. Ramirez, PhD, P.E.

What is the Difference Between Isentropic and Mechanical Efficiency?

​No energy conversion process is “perfect”. The compressor is no exception. Not all the energy consumed by the motor driving the compressor is used to raise the pressure of the gas; some of that energy is "wasted". This is an unavoidable consequence of the second law of thermodynamics. Also, auxiliary components such as bearings, belts, seals, and gear trains will always consume energy.
 
The concept of “efficiency” helps us quantify the energy losses associated with a compression process. The mechanical efficiency compares the energy delivered to the compressor with the energy the compressor delivers to the gas. Let’s look at an example for clarification. Remember that the power delivered by the motor to the compressor is called brake power, or sometimes, shaft power.
 
Consider an ammonia compressor with saturated vapor at 20°F at the inlet. A pressure gauge and a temperature probe at the compressor discharge indicate 155 psig and 220°F, respectively. If the ammonia flow rate is 4 pounds per minute, and the brake power is 10 hp, what is the mechanical efficiency?
 
To answer the question, we need to compare the brake power to the power delivered to the gas. In this process, the power delivered to the gas is the mass flow rate multiplied times the enthalpy change across the compressor. For the conditions shown, we can consult a pressure-enthalpy diagram for ammonia and determine that the enthalpy at the compressor inlet is about 617 Btu/lbm. At the compressor discharge, the enthalpy is about 718 Btu/lb. Therefore, the power delivered to the gas is (4 lb/min)(718 – 627)Btu/lbm = 364 Btu/min, which is equivalent to 8.6 hp. Therefore, the mechanical efficiency is 8.6/10 = 86%.
 
But, what about the isentropic efficiency? This concept compares the actual process experienced by the gas to the process that would occur if entropy were kept constant. The isentropic compressor efficiency is the ratio of enthalpy rise for the actual process to the enthalpy rise for the isentropic process between the same pressure extremes. To do this comparison, we draw both processes on a pressure-enthalpy diagram:

Pressure–enthalpy diagram illustrating the comparison between actual (blue) and isentropic (red) compression processes, highlighting enthalpy changes and the 170 psia discharge pressure.

In the figure above, the isentropic process is drawn in red. It follows a line of constant entropy from saturated vapor at 20°F up to the 170 psia isobar. The final enthalpy for the isentropic process is roughly 695 Btu/lb. Therefore, the enthalpy rise for the isentropic process is (695 – 627) = 68 Btu/lbm. The real process is shown in blue, and for this process we have already determined the enthalpy rise is (718 – 627) = 91 Btu/lbm. The isentropic efficiency is therefore 68/91 ≈ 75%.

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Now, here is a challenge for you. Give this problem a try and contact us ([email protected]) if you need help or to check your answer:
 
In an industrial facility, an R-134a compressor operates between pressures of 33 and 100 psia, with no superheat at the suction. The brake power is 8 hp and the flow rate is 15 lb/min. If the mechanical efficiency is 80%, the expected compressor discharge temperature (°F) is most nearly:
(A) 85
(B) 110
(C) 120
(D) 135

I have carefully crafted a course that covers all these concepts and much more, to help you walk into the exam with confidence and the peace of mind of knowing all your bases are covered.

Note: The correct answer is (C).