Questions: Isentropic Efficiency of Turbines, Compressors, and Pumps

Question 1

A compressor has an isentropic efficiency of 0.80. Compared to an ideal isentropic compressor performing the same pressure rise, the actual compressor requires...

• A. 80% of the work an ideal compressor would require — it is more efficient than the ideal
• B. The same work as the ideal compressor — isentropic efficiency only affects heat transfer
• C. More work than the ideal compressor — specifically, actual work = isentropic work / 0.80
• D. 20% less work than the ideal compressor, since 80% efficiency means 20% is saved

Answer: (C)

Compressor isentropic efficiency is defined as η_c = w_isentropic / w_actual. Rearranging: w_actual = w_isentropic / η_c = w_isentropic / 0.80. Since 0.80 < 1, w_actual > w_isentropic — the real compressor requires MORE work than the ideal. Irreversibilities (friction, flow separation) convert some of the input work to heat within the fluid rather than raising pressure, so you must supply extra work to achieve the same pressure rise. Options A and D reverse this logic — they reflect the mistaken intuition that higher efficiency always means using less of something, forgetting that for a compressor, the 'something' being minimized is work input.

Question 2

Why is the isentropic efficiency formula for a compressor the inverse of the formula for a turbine (η_c = w_s/w_actual vs. η_t = w_actual/w_s)?

• A. Because compressors use a different thermodynamic cycle than turbines
• B. In both formulas, the smaller quantity is in the numerator and the larger is in the denominator, so that efficiency is always less than 1
• C. Because turbine efficiency accounts for heat transfer while compressor efficiency does not
• D. Because the isentropic process produces more work in a compressor than in a turbine

Answer: (B)

The formulas are constructed to keep efficiency below 1 for both devices. For a turbine: actual work output < isentropic (ideal) work output, so putting actual in the numerator gives a ratio < 1. For a compressor: actual work input > isentropic (ideal) work input, so putting isentropic in the numerator gives a ratio < 1. In both cases, efficiency = (what you actually get or ideally need) / (the larger quantity). The inversion is not arbitrary — it reflects the different direction of energy flow (out vs. in) while maintaining the logical meaning that η = 1 is the unattainable ideal.

Question 3

For a real turbine operating between fixed inlet and exit pressures, the actual exit enthalpy is higher than the isentropic exit enthalpy.

• T. True
• F. False

Answer: True

In a turbine, the isentropic exit state (2s) represents maximum work extraction — the fluid loses as much enthalpy as thermodynamically possible while entropy stays constant. Real irreversibilities (friction, turbulence) convert some of that potential work to internal heat within the fluid, leaving the fluid with more enthalpy at the exit than it would have after a perfect isentropic expansion. On an h-s diagram, the actual exit state (2a) lies to the right and above the isentropic exit state (2s). Since actual work = h1 − h2a and h2a > h2s, actual work < isentropic work — confirming η_t < 1.

Question 4

An isentropic efficiency of 0.90 for a turbine means that 90% of the kinetic energy entering the turbine is converted to shaft work.

• T. True
• F. False

Answer: False

Isentropic efficiency compares actual work output to the maximum work that could be extracted in a hypothetical isentropic process between the same inlet state and exit pressure — not to the total energy content of the incoming fluid. The fluid still retains significant enthalpy at the exit even in the ideal case (it doesn't become zero-energy). Isentropic efficiency = (actual work) / (maximum possible work from an isentropic expansion). This is a relative, not absolute, efficiency measure. Stating it as '90% of total energy converted' would require knowing absolute inlet enthalpy relative to a reference state, which is not how the definition works.

Question 5

A steam turbine inlet is at state 1 with enthalpy h1 and entropy s1. The isentropic exit state (2s) and actual exit state (2a) are both at the same exit pressure. On an h-s diagram, which state has higher enthalpy — 2s or 2a — and what does this tell you about actual versus ideal turbine work output?

Model answer: State 2a (actual) has higher enthalpy than state 2s (isentropic). Actual turbine work = h1 − h2a; isentropic work = h1 − h2s. Because h2a > h2s, the actual work output is smaller than the isentropic work — the turbine delivers less shaft work than the ideal. The 'missing' work went into increasing the fluid's entropy through irreversibilities (friction, heat transfer, turbulence), which appears as extra enthalpy in the exit steam.

On the h-s (Mollier) diagram, both exit states sit on the same isobar (same exit pressure). The isentropic path is a vertical line (constant s), while the actual path curves to the right as entropy increases. Since enthalpy generally increases with entropy at constant pressure in the two-phase and superheated regions, the actual exit state 2a is above and to the right of 2s. This diagram-reading skill is essential for turbine and compressor problem-solving.