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Compressor Isentropic Efficiency

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics
3 formulas
10 exam-style examples
~51 min
All Fluid Mechanics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

  • wa ≡ actual compressor work per unit mass
  • ws ≡ isentropic compressor work per unit mass
  • Tes ≡ isentropic exit temperature

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1
Blower / compressor fluid power — solve for shaft power — Compressor Isentropic Efficiency

a positive-displacement blower feeding an aeration basin Given volumetric flow (Q) = 11.5000 m^3/s; pressure rise (\Delta p) = 24,800 Pa; efficiency (\eta) = 0.7700, determine the shaft power (W) in W.

Given

  • volumetricflow(Q)=11.5000m3/svolumetric flow (Q) = 11.5000 m^3/s
  • pressurerise(Δp)=24,800Papressure rise (\Delta p) = 24,800 Pa
  • efficiency(η)=0.7700efficiency (\eta) = 0.7700

Find

shaft power (W), in W

Start with the thinking

  • The governing relation printed in this handbook section is Blower / compressor fluid power.
  • Everything except W is given, so isolate W symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • A blower raises the pressure of an air stream across an aeration basin diffuser system.

Step-by-step solution

  1. Step 1 — State the governing relation:

    W=QΔpηW = \dfrac{Q \Delta p}{\eta}
  2. Step 2 — Rearrange symbolically for W:

    W=QΔpηW = \dfrac{Q \Delta p}{\eta}
  3. Step 3 — List the givens: volumetric flow (Q) = 11.5000 m^3/s, pressure rise (\Delta p) = 24,800 Pa, efficiency (\eta) = 0.7700.

  4. Step 4 — Substitute the given values:

    W=11.5000248000.7700W = \dfrac{11.5000 24800}{0.7700}
  5. Step 5 — Evaluate:

    W=370390 WW = 370390\ \text{W}
  6. Step 6 — Check: returning W = 370,390 W to

    W=QΔpηW = \dfrac{Q \Delta p}{\eta}

    reproduces the given quantities, and both sides carry the same units.

Answer:
W=370390 WW = 370390\ \text{W}

Why the other options are there

  • 740,779 — kept a factor of two that cancels in the correct rearrangement.
  • 185,195 — dropped that same factor in the other direction.
  • 407,429 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fans, Blowers, and Compressors

Example 2
Blower / compressor fluid power — solve for volumetric flow — Compressor Isentropic Efficiency (2)

a centrifugal blower on a dust-collection duct Given shaft power (W) = 104,739 W; pressure rise (\Delta p) = 44,000 Pa; efficiency (\eta) = 0.6800, determine the volumetric flow (Q) in m^3/s.

Given

  • shaftpower(W)=104,739Wshaft power (W) = 104,739 W
  • pressurerise(Δp)=44,000Papressure rise (\Delta p) = 44,000 Pa
  • efficiency(η)=0.6800efficiency (\eta) = 0.6800

Find

volumetric flow (Q), in m^3/s

Start with the thinking

  • The governing relation printed in this handbook section is Blower / compressor fluid power.
  • Everything except Q is given, so isolate Q symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • A blower raises the pressure of an air stream across an aeration basin diffuser system.

Step-by-step solution

  1. Step 1 — State the governing relation:

    W=QΔpηW = \dfrac{Q \Delta p}{\eta}
  2. Step 2 — Rearrange symbolically for Q:

    Q=WηΔpQ = \dfrac{W \eta}{\Delta p}
  3. Step 3 — List the givens: shaft power (W) = 104,739 W, pressure rise (\Delta p) = 44,000 Pa, efficiency (\eta) = 0.6800.

  4. Step 4 — Substitute the given values:

    Q=1047390.680044000Q = \dfrac{104739 0.6800}{44000}
  5. Step 5 — Evaluate:

    Q = 1.6187\ \text{m^3/s}
  6. Step 6 — Check: returning Q = 1.6187 m^3/s to

    W=QΔpηW = \dfrac{Q \Delta p}{\eta}

    reproduces the given quantities, and both sides carry the same units.

Answer:
Q = 1.6187\ \text{m^3/s}

Why the other options are there

  • 3.2374 — kept a factor of two that cancels in the correct rearrangement.
  • 0.8093 — dropped that same factor in the other direction.
  • 1.7806 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fans, Blowers, and Compressors

Example 3
Blower / compressor fluid power — solve for efficiency — Compressor Isentropic Efficiency (3)

a compressor supplying pneumatic construction tools Given shaft power (W) = 83,942 W; volumetric flow (Q) = 0.4000 m^3/s; pressure rise (\Delta p) = 57,700 Pa, determine the efficiency (\eta).

Given

  • shaftpower(W)=83,942Wshaft power (W) = 83,942 W
  • volumetricflow(Q)=0.4000m3/svolumetric flow (Q) = 0.4000 m^3/s
  • pressurerise(Δp)=57,700Papressure rise (\Delta p) = 57,700 Pa

Find

efficiency (\eta)

Start with the thinking

  • The governing relation printed in this handbook section is Blower / compressor fluid power.
  • Everything except \eta is given, so isolate \eta symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • A blower raises the pressure of an air stream across an aeration basin diffuser system.

Step-by-step solution

  1. Step 1 — State the governing relation:

    W=QΔpηW = \dfrac{Q \Delta p}{\eta}
  2. Step 2 — Rearrange symbolically for \eta:

    η=QΔpW\eta = \dfrac{Q \Delta p}{W}
  3. Step 3 — List the givens: shaft power (W) = 83,942 W, volumetric flow (Q) = 0.4000 m^3/s, pressure rise (\Delta p) = 57,700 Pa.

  4. Step 4 — Substitute the given values:

    η=0.40005770083942\eta = \dfrac{0.4000 57700}{83942}
  5. Step 5 — Evaluate:

    η=0.2750\eta = 0.2750
  6. Step 6 — Check: returning \eta = 0.2750 to

    W=QΔpηW = \dfrac{Q \Delta p}{\eta}

    reproduces the given quantities, and both sides carry the same units.

Answer:
η=0.2750\eta = 0.2750

Why the other options are there

  • 0.5499 — kept a factor of two that cancels in the correct rearrangement.
  • 0.1375 — dropped that same factor in the other direction.
  • 0.3024 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fans, Blowers, and Compressors

Example 4
Blower / compressor fluid power — solve for shaft power (case 2) — Compressor Isentropic Efficiency (4)

a positive-displacement blower feeding an aeration basin Given volumetric flow (Q) = 8.1000 m^3/s; pressure rise (\Delta p) = 46,700 Pa; efficiency (\eta) = 0.5600, determine the shaft power (W) in W.

Given

  • volumetricflow(Q)=8.1000m3/svolumetric flow (Q) = 8.1000 m^3/s
  • pressurerise(Δp)=46,700Papressure rise (\Delta p) = 46,700 Pa
  • efficiency(η)=0.5600efficiency (\eta) = 0.5600

Find

shaft power (W), in W

Start with the thinking

  • The governing relation printed in this handbook section is Blower / compressor fluid power.
  • Everything except W is given, so isolate W symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • A blower raises the pressure of an air stream across an aeration basin diffuser system.

Step-by-step solution

  1. Step 1 — State the governing relation:

    W=QΔpηW = \dfrac{Q \Delta p}{\eta}
  2. Step 2 — Rearrange symbolically for W:

    W=QΔpηW = \dfrac{Q \Delta p}{\eta}
  3. Step 3 — List the givens: volumetric flow (Q) = 8.1000 m^3/s, pressure rise (\Delta p) = 46,700 Pa, efficiency (\eta) = 0.5600.

  4. Step 4 — Substitute the given values:

    W=8.1000467000.5600W = \dfrac{8.1000 46700}{0.5600}
  5. Step 5 — Evaluate:

    W=675482 WW = 675482\ \text{W}
  6. Step 6 — Check: returning W = 675,482 W to

    W=QΔpηW = \dfrac{Q \Delta p}{\eta}

    reproduces the given quantities, and both sides carry the same units.

Answer:
W=675482 WW = 675482\ \text{W}

Why the other options are there

  • 1,350,964 — kept a factor of two that cancels in the correct rearrangement.
  • 337,741 — dropped that same factor in the other direction.
  • 743,030 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fans, Blowers, and Compressors

Example 5
Blower / compressor fluid power — solve for volumetric flow (case 2) — Compressor Isentropic Efficiency (5)

a centrifugal blower on a dust-collection duct Given shaft power (W) = 29,393 W; pressure rise (\Delta p) = 16,300 Pa; efficiency (\eta) = 0.6300, determine the volumetric flow (Q) in m^3/s.

Given

  • shaftpower(W)=29,393Wshaft power (W) = 29,393 W
  • pressurerise(Δp)=16,300Papressure rise (\Delta p) = 16,300 Pa
  • efficiency(η)=0.6300efficiency (\eta) = 0.6300

Find

volumetric flow (Q), in m^3/s

Start with the thinking

  • The governing relation printed in this handbook section is Blower / compressor fluid power.
  • Everything except Q is given, so isolate Q symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • A blower raises the pressure of an air stream across an aeration basin diffuser system.

Step-by-step solution

  1. Step 1 — State the governing relation:

    W=QΔpηW = \dfrac{Q \Delta p}{\eta}
  2. Step 2 — Rearrange symbolically for Q:

    Q=WηΔpQ = \dfrac{W \eta}{\Delta p}
  3. Step 3 — List the givens: shaft power (W) = 29,393 W, pressure rise (\Delta p) = 16,300 Pa, efficiency (\eta) = 0.6300.

  4. Step 4 — Substitute the given values:

    Q=293930.630016300Q = \dfrac{29393 0.6300}{16300}
  5. Step 5 — Evaluate:

    Q = 1.1360\ \text{m^3/s}
  6. Step 6 — Check: returning Q = 1.1360 m^3/s to

    W=QΔpηW = \dfrac{Q \Delta p}{\eta}

    reproduces the given quantities, and both sides carry the same units.

Answer:
Q = 1.1360\ \text{m^3/s}

Why the other options are there

  • 2.2721 — kept a factor of two that cancels in the correct rearrangement.
  • 0.5680 — dropped that same factor in the other direction.
  • 1.2497 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fans, Blowers, and Compressors

Example 6
Blower / compressor fluid power — solve for efficiency (case 2) — Compressor Isentropic Efficiency (6)

a compressor supplying pneumatic construction tools Given shaft power (W) = 138,164 W; volumetric flow (Q) = 5.5000 m^3/s; pressure rise (\Delta p) = 34,000 Pa, determine the efficiency (\eta).

Given

  • shaftpower(W)=138,164Wshaft power (W) = 138,164 W
  • volumetricflow(Q)=5.5000m3/svolumetric flow (Q) = 5.5000 m^3/s
  • pressurerise(Δp)=34,000Papressure rise (\Delta p) = 34,000 Pa

Find

efficiency (\eta)

Start with the thinking

  • The governing relation printed in this handbook section is Blower / compressor fluid power.
  • Everything except \eta is given, so isolate \eta symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • A blower raises the pressure of an air stream across an aeration basin diffuser system.

Step-by-step solution

  1. Step 1 — State the governing relation:

    W=QΔpηW = \dfrac{Q \Delta p}{\eta}
  2. Step 2 — Rearrange symbolically for \eta:

    η=QΔpW\eta = \dfrac{Q \Delta p}{W}
  3. Step 3 — List the givens: shaft power (W) = 138,164 W, volumetric flow (Q) = 5.5000 m^3/s, pressure rise (\Delta p) = 34,000 Pa.

  4. Step 4 — Substitute the given values:

    η=5.500034000138164\eta = \dfrac{5.5000 34000}{138164}
  5. Step 5 — Evaluate:

    η=1.3535\eta = 1.3535
  6. Step 6 — Check: returning \eta = 1.3535 to

    W=QΔpηW = \dfrac{Q \Delta p}{\eta}

    reproduces the given quantities, and both sides carry the same units.

Answer:
η=1.3535\eta = 1.3535

Why the other options are there

  • 2.7069 — kept a factor of two that cancels in the correct rearrangement.
  • 0.6767 — dropped that same factor in the other direction.
  • 1.4888 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fans, Blowers, and Compressors

Example 7
Blower / compressor fluid power — solve for shaft power (case 3) — Compressor Isentropic Efficiency (7)

a positive-displacement blower feeding an aeration basin Given volumetric flow (Q) = 8.4000 m^3/s; pressure rise (\Delta p) = 26,900 Pa; efficiency (\eta) = 0.6100, determine the shaft power (W) in W.

Given

  • volumetricflow(Q)=8.4000m3/svolumetric flow (Q) = 8.4000 m^3/s
  • pressurerise(Δp)=26,900Papressure rise (\Delta p) = 26,900 Pa
  • efficiency(η)=0.6100efficiency (\eta) = 0.6100

Find

shaft power (W), in W

Start with the thinking

  • The governing relation printed in this handbook section is Blower / compressor fluid power.
  • Everything except W is given, so isolate W symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • A blower raises the pressure of an air stream across an aeration basin diffuser system.

Step-by-step solution

  1. Step 1 — State the governing relation:

    W=QΔpηW = \dfrac{Q \Delta p}{\eta}
  2. Step 2 — Rearrange symbolically for W:

    W=QΔpηW = \dfrac{Q \Delta p}{\eta}
  3. Step 3 — List the givens: volumetric flow (Q) = 8.4000 m^3/s, pressure rise (\Delta p) = 26,900 Pa, efficiency (\eta) = 0.6100.

  4. Step 4 — Substitute the given values:

    W=8.4000269000.6100W = \dfrac{8.4000 26900}{0.6100}
  5. Step 5 — Evaluate:

    W=370426 WW = 370426\ \text{W}
  6. Step 6 — Check: returning W = 370,426 W to

    W=QΔpηW = \dfrac{Q \Delta p}{\eta}

    reproduces the given quantities, and both sides carry the same units.

Answer:
W=370426 WW = 370426\ \text{W}

Why the other options are there

  • 740,852 — kept a factor of two that cancels in the correct rearrangement.
  • 185,213 — dropped that same factor in the other direction.
  • 407,469 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fans, Blowers, and Compressors

Example 8
Blower / compressor fluid power — solve for volumetric flow (case 3) — Compressor Isentropic Efficiency (8)

a centrifugal blower on a dust-collection duct Given shaft power (W) = 170,637 W; pressure rise (\Delta p) = 11,500 Pa; efficiency (\eta) = 0.5700, determine the volumetric flow (Q) in m^3/s.

Given

  • shaftpower(W)=170,637Wshaft power (W) = 170,637 W
  • pressurerise(Δp)=11,500Papressure rise (\Delta p) = 11,500 Pa
  • efficiency(η)=0.5700efficiency (\eta) = 0.5700

Find

volumetric flow (Q), in m^3/s

Start with the thinking

  • The governing relation printed in this handbook section is Blower / compressor fluid power.
  • Everything except Q is given, so isolate Q symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • A blower raises the pressure of an air stream across an aeration basin diffuser system.

Step-by-step solution

  1. Step 1 — State the governing relation:

    W=QΔpηW = \dfrac{Q \Delta p}{\eta}
  2. Step 2 — Rearrange symbolically for Q:

    Q=WηΔpQ = \dfrac{W \eta}{\Delta p}
  3. Step 3 — List the givens: shaft power (W) = 170,637 W, pressure rise (\Delta p) = 11,500 Pa, efficiency (\eta) = 0.5700.

  4. Step 4 — Substitute the given values:

    Q=1706370.570011500Q = \dfrac{170637 0.5700}{11500}
  5. Step 5 — Evaluate:

    Q = 8.4577\ \text{m^3/s}
  6. Step 6 — Check: returning Q = 8.4577 m^3/s to

    W=QΔpηW = \dfrac{Q \Delta p}{\eta}

    reproduces the given quantities, and both sides carry the same units.

Answer:
Q = 8.4577\ \text{m^3/s}

Why the other options are there

  • 16.9153 — kept a factor of two that cancels in the correct rearrangement.
  • 4.2288 — dropped that same factor in the other direction.
  • 9.3034 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fans, Blowers, and Compressors

Example 9
Blower / compressor fluid power — solve for efficiency (case 3) — Compressor Isentropic Efficiency (9)

a compressor supplying pneumatic construction tools Given shaft power (W) = 104,396 W; volumetric flow (Q) = 6.6000 m^3/s; pressure rise (\Delta p) = 50,600 Pa, determine the efficiency (\eta).

Given

  • shaftpower(W)=104,396Wshaft power (W) = 104,396 W
  • volumetricflow(Q)=6.6000m3/svolumetric flow (Q) = 6.6000 m^3/s
  • pressurerise(Δp)=50,600Papressure rise (\Delta p) = 50,600 Pa

Find

efficiency (\eta)

Start with the thinking

  • The governing relation printed in this handbook section is Blower / compressor fluid power.
  • Everything except \eta is given, so isolate \eta symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • A blower raises the pressure of an air stream across an aeration basin diffuser system.

Step-by-step solution

  1. Step 1 — State the governing relation:

    W=QΔpηW = \dfrac{Q \Delta p}{\eta}
  2. Step 2 — Rearrange symbolically for \eta:

    η=QΔpW\eta = \dfrac{Q \Delta p}{W}
  3. Step 3 — List the givens: shaft power (W) = 104,396 W, volumetric flow (Q) = 6.6000 m^3/s, pressure rise (\Delta p) = 50,600 Pa.

  4. Step 4 — Substitute the given values:

    η=6.600050600104396\eta = \dfrac{6.6000 50600}{104396}
  5. Step 5 — Evaluate:

    η=3.1990\eta = 3.1990
  6. Step 6 — Check: returning \eta = 3.1990 to

    W=QΔpηW = \dfrac{Q \Delta p}{\eta}

    reproduces the given quantities, and both sides carry the same units.

Answer:
η=3.1990\eta = 3.1990

Why the other options are there

  • 6.3979 — kept a factor of two that cancels in the correct rearrangement.
  • 1.5995 — dropped that same factor in the other direction.
  • 3.5189 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fans, Blowers, and Compressors

Example 10
Blower / compressor fluid power — solve for shaft power (case 4) — Compressor Isentropic Efficiency (10)

a positive-displacement blower feeding an aeration basin Given volumetric flow (Q) = 7.4000 m^3/s; pressure rise (\Delta p) = 19,800 Pa; efficiency (\eta) = 0.7700, determine the shaft power (W) in W.

Given

  • volumetricflow(Q)=7.4000m3/svolumetric flow (Q) = 7.4000 m^3/s
  • pressurerise(Δp)=19,800Papressure rise (\Delta p) = 19,800 Pa
  • efficiency(η)=0.7700efficiency (\eta) = 0.7700

Find

shaft power (W), in W

Start with the thinking

  • The governing relation printed in this handbook section is Blower / compressor fluid power.
  • Everything except W is given, so isolate W symbolically first — never rearrange after the numbers are in.
  • Tabulate each given with its unit and confirm the units are consistent with the relation before substituting.
  • A blower raises the pressure of an air stream across an aeration basin diffuser system.

Step-by-step solution

  1. Step 1 — State the governing relation:

    W=QΔpηW = \dfrac{Q \Delta p}{\eta}
  2. Step 2 — Rearrange symbolically for W:

    W=QΔpηW = \dfrac{Q \Delta p}{\eta}
  3. Step 3 — List the givens: volumetric flow (Q) = 7.4000 m^3/s, pressure rise (\Delta p) = 19,800 Pa, efficiency (\eta) = 0.7700.

  4. Step 4 — Substitute the given values:

    W=7.4000198000.7700W = \dfrac{7.4000 19800}{0.7700}
  5. Step 5 — Evaluate:

    W=190286 WW = 190286\ \text{W}
  6. Step 6 — Check: returning W = 190,286 W to

    W=QΔpηW = \dfrac{Q \Delta p}{\eta}

    reproduces the given quantities, and both sides carry the same units.

Answer:
W=190286 WW = 190286\ \text{W}

Why the other options are there

  • 380,571 — kept a factor of two that cancels in the correct rearrangement.
  • 95,143 — dropped that same factor in the other direction.
  • 209,314 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fans, Blowers, and Compressors

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