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Adiabatic Compression

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics
8 formulas
10 exam-style examples
~60 min
All Fluid Mechanics lectures

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 — Adiabatic Compression

a positive-displacement blower feeding an aeration basin Given volumetric flow (Q) = 5.4000 m^3/s; pressure rise (\Delta p) = 18,400 Pa; efficiency (\eta) = 0.5800, determine the shaft power (W) in W.

Given

  • volumetricflow(Q)=5.4000m3/svolumetric flow (Q) = 5.4000 m^3/s
  • pressurerise(Δp)=18,400Papressure rise (\Delta p) = 18,400 Pa
  • efficiency(η)=0.5800efficiency (\eta) = 0.5800

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) = 5.4000 m^3/s, pressure rise (\Delta p) = 18,400 Pa, efficiency (\eta) = 0.5800.

  4. Step 4 — Substitute the given values:

    W=5.4000184000.5800W = \dfrac{5.4000 18400}{0.5800}
  5. Step 5 — Evaluate:

    W=171310 WW = 171310\ \text{W}
  6. Step 6 — Check: returning W = 171,310 W to

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

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

Answer:
W=171310 WW = 171310\ \text{W}

Why the other options are there

  • 342,621 — kept a factor of two that cancels in the correct rearrangement.
  • 85,655 — dropped that same factor in the other direction.
  • 188,441 — 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 — Adiabatic Compression (2)

a centrifugal blower on a dust-collection duct Given shaft power (W) = 81,366 W; pressure rise (\Delta p) = 9,400 Pa; efficiency (\eta) = 0.6000, determine the volumetric flow (Q) in m^3/s.

Given

  • shaftpower(W)=81,366Wshaft power (W) = 81,366 W
  • pressurerise(Δp)=9,400Papressure rise (\Delta p) = 9,400 Pa
  • efficiency(η)=0.6000efficiency (\eta) = 0.6000

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) = 81,366 W, pressure rise (\Delta p) = 9,400 Pa, efficiency (\eta) = 0.6000.

  4. Step 4 — Substitute the given values:

    Q=813660.60009400Q = \dfrac{81366 0.6000}{9400}
  5. Step 5 — Evaluate:

    Q = 5.1936\ \text{m^3/s}
  6. Step 6 — Check: returning Q = 5.1936 m^3/s to

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

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

Answer:
Q = 5.1936\ \text{m^3/s}

Why the other options are there

  • 10.3871 — kept a factor of two that cancels in the correct rearrangement.
  • 2.5968 — dropped that same factor in the other direction.
  • 5.7129 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fans, Blowers, and Compressors

Example 3
Blower / compressor fluid power — solve for efficiency — Adiabatic Compression (3)

a compressor supplying pneumatic construction tools Given shaft power (W) = 184,681 W; volumetric flow (Q) = 3.7000 m^3/s; pressure rise (\Delta p) = 28,400 Pa, determine the efficiency (\eta).

Given

  • shaftpower(W)=184,681Wshaft power (W) = 184,681 W
  • volumetricflow(Q)=3.7000m3/svolumetric flow (Q) = 3.7000 m^3/s
  • pressurerise(Δp)=28,400Papressure rise (\Delta p) = 28,400 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) = 184,681 W, volumetric flow (Q) = 3.7000 m^3/s, pressure rise (\Delta p) = 28,400 Pa.

  4. Step 4 — Substitute the given values:

    η=3.700028400184681\eta = \dfrac{3.7000 28400}{184681}
  5. Step 5 — Evaluate:

    η=0.5690\eta = 0.5690
  6. Step 6 — Check: returning \eta = 0.5690 to

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

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

Answer:
η=0.5690\eta = 0.5690

Why the other options are there

  • 1.1380 — kept a factor of two that cancels in the correct rearrangement.
  • 0.2845 — dropped that same factor in the other direction.
  • 0.6259 — 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) — Adiabatic Compression (4)

a positive-displacement blower feeding an aeration basin Given volumetric flow (Q) = 7.1000 m^3/s; pressure rise (\Delta p) = 40,700 Pa; efficiency (\eta) = 0.6400, determine the shaft power (W) in W.

Given

  • volumetricflow(Q)=7.1000m3/svolumetric flow (Q) = 7.1000 m^3/s
  • pressurerise(Δp)=40,700Papressure rise (\Delta p) = 40,700 Pa
  • efficiency(η)=0.6400efficiency (\eta) = 0.6400

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) = 7.1000 m^3/s, pressure rise (\Delta p) = 40,700 Pa, efficiency (\eta) = 0.6400.

  4. Step 4 — Substitute the given values:

    W=7.1000407000.6400W = \dfrac{7.1000 40700}{0.6400}
  5. Step 5 — Evaluate:

    W=451516 WW = 451516\ \text{W}
  6. Step 6 — Check: returning W = 451,516 W to

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

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

Answer:
W=451516 WW = 451516\ \text{W}

Why the other options are there

  • 903,031 — kept a factor of two that cancels in the correct rearrangement.
  • 225,758 — dropped that same factor in the other direction.
  • 496,667 — 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) — Adiabatic Compression (5)

a centrifugal blower on a dust-collection duct Given shaft power (W) = 160,051 W; pressure rise (\Delta p) = 59,600 Pa; efficiency (\eta) = 0.7600, determine the volumetric flow (Q) in m^3/s.

Given

  • shaftpower(W)=160,051Wshaft power (W) = 160,051 W
  • pressurerise(Δp)=59,600Papressure rise (\Delta p) = 59,600 Pa
  • efficiency(η)=0.7600efficiency (\eta) = 0.7600

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) = 160,051 W, pressure rise (\Delta p) = 59,600 Pa, efficiency (\eta) = 0.7600.

  4. Step 4 — Substitute the given values:

    Q=1600510.760059600Q = \dfrac{160051 0.7600}{59600}
  5. Step 5 — Evaluate:

    Q = 2.0409\ \text{m^3/s}
  6. Step 6 — Check: returning Q = 2.0409 m^3/s to

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

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

Answer:
Q = 2.0409\ \text{m^3/s}

Why the other options are there

  • 4.0818 — kept a factor of two that cancels in the correct rearrangement.
  • 1.0205 — dropped that same factor in the other direction.
  • 2.2450 — 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) — Adiabatic Compression (6)

a compressor supplying pneumatic construction tools Given shaft power (W) = 24,108 W; volumetric flow (Q) = 1.4000 m^3/s; pressure rise (\Delta p) = 16,100 Pa, determine the efficiency (\eta).

Given

  • shaftpower(W)=24,108Wshaft power (W) = 24,108 W
  • volumetricflow(Q)=1.4000m3/svolumetric flow (Q) = 1.4000 m^3/s
  • pressurerise(Δp)=16,100Papressure rise (\Delta p) = 16,100 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) = 24,108 W, volumetric flow (Q) = 1.4000 m^3/s, pressure rise (\Delta p) = 16,100 Pa.

  4. Step 4 — Substitute the given values:

    η=1.40001610024108\eta = \dfrac{1.4000 16100}{24108}
  5. Step 5 — Evaluate:

    η=0.9350\eta = 0.9350
  6. Step 6 — Check: returning \eta = 0.9350 to

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

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

Answer:
η=0.9350\eta = 0.9350

Why the other options are there

  • 1.8699 — kept a factor of two that cancels in the correct rearrangement.
  • 0.4675 — dropped that same factor in the other direction.
  • 1.0285 — 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) — Adiabatic Compression (7)

a positive-displacement blower feeding an aeration basin Given volumetric flow (Q) = 3.9000 m^3/s; pressure rise (\Delta p) = 18,400 Pa; efficiency (\eta) = 0.8500, determine the shaft power (W) in W.

Given

  • volumetricflow(Q)=3.9000m3/svolumetric flow (Q) = 3.9000 m^3/s
  • pressurerise(Δp)=18,400Papressure rise (\Delta p) = 18,400 Pa
  • efficiency(η)=0.8500efficiency (\eta) = 0.8500

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) = 3.9000 m^3/s, pressure rise (\Delta p) = 18,400 Pa, efficiency (\eta) = 0.8500.

  4. Step 4 — Substitute the given values:

    W=3.9000184000.8500W = \dfrac{3.9000 18400}{0.8500}
  5. Step 5 — Evaluate:

    W=84424 WW = 84424\ \text{W}
  6. Step 6 — Check: returning W = 84,424 W to

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

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

Answer:
W=84424 WW = 84424\ \text{W}

Why the other options are there

  • 168,847 — kept a factor of two that cancels in the correct rearrangement.
  • 42,212 — dropped that same factor in the other direction.
  • 92,866 — 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) — Adiabatic Compression (8)

a centrifugal blower on a dust-collection duct Given shaft power (W) = 20,876 W; pressure rise (\Delta p) = 37,000 Pa; efficiency (\eta) = 0.8200, determine the volumetric flow (Q) in m^3/s.

Given

  • shaftpower(W)=20,876Wshaft power (W) = 20,876 W
  • pressurerise(Δp)=37,000Papressure rise (\Delta p) = 37,000 Pa
  • efficiency(η)=0.8200efficiency (\eta) = 0.8200

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) = 20,876 W, pressure rise (\Delta p) = 37,000 Pa, efficiency (\eta) = 0.8200.

  4. Step 4 — Substitute the given values:

    Q=208760.820037000Q = \dfrac{20876 0.8200}{37000}
  5. Step 5 — Evaluate:

    Q = 0.4627\ \text{m^3/s}
  6. Step 6 — Check: returning Q = 0.4627 m^3/s to

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

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

Answer:
Q = 0.4627\ \text{m^3/s}

Why the other options are there

  • 0.9253 — kept a factor of two that cancels in the correct rearrangement.
  • 0.2313 — dropped that same factor in the other direction.
  • 0.5089 — 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) — Adiabatic Compression (9)

a compressor supplying pneumatic construction tools Given shaft power (W) = 49,868 W; volumetric flow (Q) = 4.4000 m^3/s; pressure rise (\Delta p) = 27,700 Pa, determine the efficiency (\eta).

Given

  • shaftpower(W)=49,868Wshaft power (W) = 49,868 W
  • volumetricflow(Q)=4.4000m3/svolumetric flow (Q) = 4.4000 m^3/s
  • pressurerise(Δp)=27,700Papressure rise (\Delta p) = 27,700 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) = 49,868 W, volumetric flow (Q) = 4.4000 m^3/s, pressure rise (\Delta p) = 27,700 Pa.

  4. Step 4 — Substitute the given values:

    η=4.40002770049868\eta = \dfrac{4.4000 27700}{49868}
  5. Step 5 — Evaluate:

    η=2.4441\eta = 2.4441
  6. Step 6 — Check: returning \eta = 2.4441 to

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

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

Answer:
η=2.4441\eta = 2.4441

Why the other options are there

  • 4.8881 — kept a factor of two that cancels in the correct rearrangement.
  • 1.2220 — dropped that same factor in the other direction.
  • 2.6885 — 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) — Adiabatic Compression (10)

a positive-displacement blower feeding an aeration basin Given volumetric flow (Q) = 6.7000 m^3/s; pressure rise (\Delta p) = 56,300 Pa; efficiency (\eta) = 0.8200, determine the shaft power (W) in W.

Given

  • volumetricflow(Q)=6.7000m3/svolumetric flow (Q) = 6.7000 m^3/s
  • pressurerise(Δp)=56,300Papressure rise (\Delta p) = 56,300 Pa
  • efficiency(η)=0.8200efficiency (\eta) = 0.8200

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) = 6.7000 m^3/s, pressure rise (\Delta p) = 56,300 Pa, efficiency (\eta) = 0.8200.

  4. Step 4 — Substitute the given values:

    W=6.7000563000.8200W = \dfrac{6.7000 56300}{0.8200}
  5. Step 5 — Evaluate:

    W=460012 WW = 460012\ \text{W}
  6. Step 6 — Check: returning W = 460,012 W to

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

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

Answer:
W=460012 WW = 460012\ \text{W}

Why the other options are there

  • 920,024 — kept a factor of two that cancels in the correct rearrangement.
  • 230,006 — dropped that same factor in the other direction.
  • 506,013 — rounded an intermediate value before the final step.

Reference: FE Handbook — Fans, Blowers, and Compressors

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