Adiabatic Compression

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

8 formulas

10 exam-style examples

~60 min

All Fluid Mechanics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Adiabatic Compression within Fluid Mechanics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what adiabatic compression describes physically and when it applies.
State every one of the 8 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early.

Lecture

Why this section exists. Adiabatic Compression is the part of Fluid Mechanics that lets you connect a pipeline, jet or submerged surface to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as continuity plus energy, with one head-loss or force term. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Crane lowering a steel plate girder onto bridge bearings while ironworkers guide it. — Photo 1. Where this shows up in practice: adiabatic compression.
Capstone Studio instructional photograph

Fluid Mechanics — Adiabatic Compression: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a pipeline, jet or submerged surface. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 8 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

Wo comp	Quantity produced by "Wo comp = mo Pi k >d Pe n - 1H" — read its definition and unit from the handbook line directly above the equation.
^ k - 1h ti hc Pi	Quantity produced by "^ k - 1h ti hc Pi" — read its definition and unit from the handbook line directly above the equation.
Pi	Quantity produced by "Pi = inlet or suction pressure (N/m2)" — read its definition and unit from the handbook line directly above the equation.
Pe	Quantity produced by "Pe = exit or discharge pressure (N/m2)" — read its definition and unit from the handbook line directly above the equation.
k	Quantity produced by "k = ratio of specific heats = cp/cv" — read its definition and unit from the handbook line directly above the equation.
ρi	Quantity produced by "ρi = inlet gas density (kg/m3)" — read its definition and unit from the handbook line directly above the equation.
ηc	Quantity produced by "ηc = isentropic compressor efficiency" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

1-1 k
where

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

Isothermal versus adiabatic compressor power for an air blower — Adiabatic Compression

A blower takes 2.8 m³/s of air at 101.3 kPa and raises it to 515 kPa. Compute the ideal isothermal power, the ideal adiabatic power (k = 1.4) and the shaft power at 66% efficiency.

Given

Q = 2.8 m³/s
p₁ = 101.3 kPa, p₂ = 515 kPa
k = 1.4
η = 0.66

Find

W_isothermal, W_adiabatic and the shaft power

Start with the thinking

Isothermal compression is the minimum ideal work; adiabatic compression always requires more.
Efficiency divides the ideal work — never multiplies it.

Step-by-step solution

Formula
$W_{i so} = p_{1} Q ln \frac{p _{2}}{p _{1}}$
Substituting
$W = 101.3 (2.8) ln (515/101.3) = 461.2 k W$
Formula
$W_{a d i} = \frac{k}{k - 1} p_{1} Q [(\frac{p _{2}}{p _{1}})^{(k - 1) / k} - 1]$
Pressure ratio term
$(515/101.3)^{0.2857} - 1 = 0.5913$
Substituting
$W = 3.5 (101.3) (2.8) (0.5913) = 587.1 k W$
Shaft power
$P = 587.1/0.66 = 889.5 k W$

Answer: W_iso = 461.2 kW, W_adi = 587.1 kW, shaft = 889.5 kW

Why the other options are there

387.5 kW (efficiency multiplied)
1,158 kW (linear Δp × Q used)

Reference: FE Reference Handbook — Fluid Mechanics → Adiabatic Compression

Example 2

Isothermal versus adiabatic compressor power for an air blower — Adiabatic Compression (2)

A blower takes 1.1 m³/s of air at 101.3 kPa and raises it to 610 kPa. Compute the ideal isothermal power, the ideal adiabatic power (k = 1.4) and the shaft power at 74% efficiency.

Given

Q = 1.1 m³/s
p₁ = 101.3 kPa, p₂ = 610 kPa
k = 1.4
η = 0.74

Find

W_isothermal, W_adiabatic and the shaft power

Start with the thinking

Isothermal compression is the minimum ideal work; adiabatic compression always requires more.
Efficiency divides the ideal work — never multiplies it.

Step-by-step solution

Formula
$W_{i so} = p_{1} Q ln \frac{p _{2}}{p _{1}}$
Substituting
$W = 101.3 (1.1) ln (610/101.3) = 200.1 k W$
Formula
$W_{a d i} = \frac{k}{k - 1} p_{1} Q [(\frac{p _{2}}{p _{1}})^{(k - 1) / k} - 1]$
Pressure ratio term
$(610/101.3)^{0.2857} - 1 = 0.6702$
Substituting
$W = 3.5 (101.3) (1.1) (0.6702) = 261.4 k W$
Shaft power
$P = 261.4/0.74 = 353.2 k W$

Answer: W_iso = 200.1 kW, W_adi = 261.4 kW, shaft = 353.2 kW

Why the other options are there

193.4 kW (efficiency multiplied)
559.6 kW (linear Δp × Q used)

Reference: FE Reference Handbook — Fluid Mechanics → Adiabatic Compression

Example 3

Isothermal versus adiabatic compressor power for an air blower — Adiabatic Compression (3)

A blower takes 2.9 m³/s of air at 101.3 kPa and raises it to 414 kPa. Compute the ideal isothermal power, the ideal adiabatic power (k = 1.4) and the shaft power at 84% efficiency.

Given

Q = 2.9 m³/s
p₁ = 101.3 kPa, p₂ = 414 kPa
k = 1.4
η = 0.84

Find

W_isothermal, W_adiabatic and the shaft power

Start with the thinking

Isothermal compression is the minimum ideal work; adiabatic compression always requires more.
Efficiency divides the ideal work — never multiplies it.

Step-by-step solution

Formula
$W_{i so} = p_{1} Q ln \frac{p _{2}}{p _{1}}$
Substituting
$W = 101.3 (2.9) ln (414/101.3) = 413.6 k W$
Formula
$W_{a d i} = \frac{k}{k - 1} p_{1} Q [(\frac{p _{2}}{p _{1}})^{(k - 1) / k} - 1]$
Pressure ratio term
$(414/101.3)^{0.2857} - 1 = 0.4951$
Substituting
$W = 3.5 (101.3) (2.9) (0.4951) = 509.1 k W$
Shaft power
$P = 509.1/0.84 = 606.1 k W$

Answer: W_iso = 413.6 kW, W_adi = 509.1 kW, shaft = 606.1 kW

Why the other options are there

427.6 kW (efficiency multiplied)
906.8 kW (linear Δp × Q used)

Reference: FE Reference Handbook — Fluid Mechanics → Adiabatic Compression

Example 4

Isothermal versus adiabatic compressor power for an air blower — Adiabatic Compression (4)

A blower takes 0.2 m³/s of air at 101.3 kPa and raises it to 477 kPa. Compute the ideal isothermal power, the ideal adiabatic power (k = 1.4) and the shaft power at 82% efficiency.

Given

Q = 0.2 m³/s
p₁ = 101.3 kPa, p₂ = 477 kPa
k = 1.4
η = 0.82

Find

W_isothermal, W_adiabatic and the shaft power

Start with the thinking

Isothermal compression is the minimum ideal work; adiabatic compression always requires more.
Efficiency divides the ideal work — never multiplies it.

Step-by-step solution

Formula
$W_{i so} = p_{1} Q ln \frac{p _{2}}{p _{1}}$
Substituting
$W = 101.3 (0.2) ln (477/101.3) = 31.4 k W$
Formula
$W_{a d i} = \frac{k}{k - 1} p_{1} Q [(\frac{p _{2}}{p _{1}})^{(k - 1) / k} - 1]$
Pressure ratio term
$(477/101.3)^{0.2857} - 1 = 0.5569$
Substituting
$W = 3.5 (101.3) (0.2) (0.5569) = 39.5 k W$
Shaft power
$P = 39.5/0.82 = 48.2 k W$

Answer: W_iso = 31.4 kW, W_adi = 39.5 kW, shaft = 48.2 kW

Why the other options are there

32.4 kW (efficiency multiplied)
75.1 kW (linear Δp × Q used)

Reference: FE Reference Handbook — Fluid Mechanics → Adiabatic Compression

Example 5

Isothermal versus adiabatic compressor power for an air blower — Adiabatic Compression (5)

A blower takes 1.7 m³/s of air at 101.3 kPa and raises it to 397 kPa. Compute the ideal isothermal power, the ideal adiabatic power (k = 1.4) and the shaft power at 79% efficiency.

Given

Q = 1.7 m³/s
p₁ = 101.3 kPa, p₂ = 397 kPa
k = 1.4
η = 0.79

Find

W_isothermal, W_adiabatic and the shaft power

Start with the thinking

Isothermal compression is the minimum ideal work; adiabatic compression always requires more.
Efficiency divides the ideal work — never multiplies it.

Step-by-step solution

Formula
$W_{i so} = p_{1} Q ln \frac{p _{2}}{p _{1}}$
Substituting
$W = 101.3 (1.7) ln (397/101.3) = 235.2 k W$
Formula
$W_{a d i} = \frac{k}{k - 1} p_{1} Q [(\frac{p _{2}}{p _{1}})^{(k - 1) / k} - 1]$
Pressure ratio term
$(397/101.3)^{0.2857} - 1 = 0.4773$
Substituting
$W = 3.5 (101.3) (1.7) (0.4773) = 287.7 k W$
Shaft power
$P = 287.7/0.79 = 364.2 k W$

Answer: W_iso = 235.2 kW, W_adi = 287.7 kW, shaft = 364.2 kW

Why the other options are there

227.3 kW (efficiency multiplied)
502.7 kW (linear Δp × Q used)

Reference: FE Reference Handbook — Fluid Mechanics → Adiabatic Compression

Example 6

Isothermal versus adiabatic compressor power for an air blower — Adiabatic Compression (6)

A blower takes 2.6 m³/s of air at 101.3 kPa and raises it to 361 kPa. Compute the ideal isothermal power, the ideal adiabatic power (k = 1.4) and the shaft power at 71% efficiency.

Given

Q = 2.6 m³/s
p₁ = 101.3 kPa, p₂ = 361 kPa
k = 1.4
η = 0.71

Find

W_isothermal, W_adiabatic and the shaft power

Start with the thinking

Isothermal compression is the minimum ideal work; adiabatic compression always requires more.
Efficiency divides the ideal work — never multiplies it.

Step-by-step solution

Formula
$W_{i so} = p_{1} Q ln \frac{p _{2}}{p _{1}}$
Substituting
$W = 101.3 (2.6) ln (361/101.3) = 334.7 k W$
Formula
$W_{a d i} = \frac{k}{k - 1} p_{1} Q [(\frac{p _{2}}{p _{1}})^{(k - 1) / k} - 1]$
Pressure ratio term
$(361/101.3)^{0.2857} - 1 = 0.4377$
Substituting
$W = 3.5 (101.3) (2.6) (0.4377) = 403.5 k W$
Shaft power
$P = 403.5/0.71 = 568.4 k W$

Answer: W_iso = 334.7 kW, W_adi = 403.5 kW, shaft = 568.4 kW

Why the other options are there

286.5 kW (efficiency multiplied)
675.2 kW (linear Δp × Q used)

Reference: FE Reference Handbook — Fluid Mechanics → Adiabatic Compression

Example 7

Isothermal versus adiabatic compressor power for an air blower — Adiabatic Compression (7)

A blower takes 2.6 m³/s of air at 101.3 kPa and raises it to 243 kPa. Compute the ideal isothermal power, the ideal adiabatic power (k = 1.4) and the shaft power at 77% efficiency.

Given

Q = 2.6 m³/s
p₁ = 101.3 kPa, p₂ = 243 kPa
k = 1.4
η = 0.77

Find

W_isothermal, W_adiabatic and the shaft power

Start with the thinking

Isothermal compression is the minimum ideal work; adiabatic compression always requires more.
Efficiency divides the ideal work — never multiplies it.

Step-by-step solution

Formula
$W_{i so} = p_{1} Q ln \frac{p _{2}}{p _{1}}$
Substituting
$W = 101.3 (2.6) ln (243/101.3) = 230.5 k W$
Formula
$W_{a d i} = \frac{k}{k - 1} p_{1} Q [(\frac{p _{2}}{p _{1}})^{(k - 1) / k} - 1]$
Pressure ratio term
$(243/101.3)^{0.2857} - 1 = 0.2840$
Substituting
$W = 3.5 (101.3) (2.6) (0.2840) = 261.8 k W$
Shaft power
$P = 261.8/0.77 = 340.0 k W$

Answer: W_iso = 230.5 kW, W_adi = 261.8 kW, shaft = 340.0 kW

Why the other options are there

201.6 kW (efficiency multiplied)
368.4 kW (linear Δp × Q used)

Reference: FE Reference Handbook — Fluid Mechanics → Adiabatic Compression

Example 8

Isothermal versus adiabatic compressor power for an air blower — Adiabatic Compression (8)

A blower takes 2.6 m³/s of air at 101.3 kPa and raises it to 157 kPa. Compute the ideal isothermal power, the ideal adiabatic power (k = 1.4) and the shaft power at 79% efficiency.

Given

Q = 2.6 m³/s
p₁ = 101.3 kPa, p₂ = 157 kPa
k = 1.4
η = 0.79

Find

W_isothermal, W_adiabatic and the shaft power

Start with the thinking

Isothermal compression is the minimum ideal work; adiabatic compression always requires more.
Efficiency divides the ideal work — never multiplies it.

Step-by-step solution

Formula
$W_{i so} = p_{1} Q ln \frac{p _{2}}{p _{1}}$
Substituting
$W = 101.3 (2.6) ln (157/101.3) = 115.4 k W$
Formula
$W_{a d i} = \frac{k}{k - 1} p_{1} Q [(\frac{p _{2}}{p _{1}})^{(k - 1) / k} - 1]$
Pressure ratio term
$(157/101.3)^{0.2857} - 1 = 0.1334$
Substituting
$W = 3.5 (101.3) (2.6) (0.1334) = 122.9 k W$
Shaft power
$P = 122.9/0.79 = 155.6 k W$

Answer: W_iso = 115.4 kW, W_adi = 122.9 kW, shaft = 155.6 kW

Why the other options are there

97.1 kW (efficiency multiplied)
144.8 kW (linear Δp × Q used)

Reference: FE Reference Handbook — Fluid Mechanics → Adiabatic Compression

Example 9

Isothermal versus adiabatic compressor power for an air blower — Adiabatic Compression (9)

A blower takes 1.5 m³/s of air at 101.3 kPa and raises it to 700 kPa. Compute the ideal isothermal power, the ideal adiabatic power (k = 1.4) and the shaft power at 64% efficiency.

Given

Q = 1.5 m³/s
p₁ = 101.3 kPa, p₂ = 700 kPa
k = 1.4
η = 0.64

Find

W_isothermal, W_adiabatic and the shaft power

Start with the thinking

Isothermal compression is the minimum ideal work; adiabatic compression always requires more.
Efficiency divides the ideal work — never multiplies it.

Step-by-step solution

Formula
$W_{i so} = p_{1} Q ln \frac{p _{2}}{p _{1}}$
Substituting
$W = 101.3 (1.5) ln (700/101.3) = 293.7 k W$
Formula
$W_{a d i} = \frac{k}{k - 1} p_{1} Q [(\frac{p _{2}}{p _{1}})^{(k - 1) / k} - 1]$
Pressure ratio term
$(700/101.3)^{0.2857} - 1 = 0.7372$
Substituting
$W = 3.5 (101.3) (1.5) (0.7372) = 392.1 k W$
Shaft power
$P = 392.1/0.64 = 612.6 k W$

Answer: W_iso = 293.7 kW, W_adi = 392.1 kW, shaft = 612.6 kW

Why the other options are there

250.9 kW (efficiency multiplied)
898.1 kW (linear Δp × Q used)

Reference: FE Reference Handbook — Fluid Mechanics → Adiabatic Compression

Example 10

Isothermal versus adiabatic compressor power for an air blower — Adiabatic Compression (10)

A blower takes 2.4 m³/s of air at 101.3 kPa and raises it to 232 kPa. Compute the ideal isothermal power, the ideal adiabatic power (k = 1.4) and the shaft power at 63% efficiency.

Given

Q = 2.4 m³/s
p₁ = 101.3 kPa, p₂ = 232 kPa
k = 1.4
η = 0.63

Find

W_isothermal, W_adiabatic and the shaft power

Start with the thinking

Isothermal compression is the minimum ideal work; adiabatic compression always requires more.
Efficiency divides the ideal work — never multiplies it.

Step-by-step solution

Formula
$W_{i so} = p_{1} Q ln \frac{p _{2}}{p _{1}}$
Substituting
$W = 101.3 (2.4) ln (232/101.3) = 201.5 k W$
Formula
$W_{a d i} = \frac{k}{k - 1} p_{1} Q [(\frac{p _{2}}{p _{1}})^{(k - 1) / k} - 1]$
Pressure ratio term
$(232/101.3)^{0.2857} - 1 = 0.2671$
Substituting
$W = 3.5 (101.3) (2.4) (0.2671) = 227.3 k W$
Shaft power
$P = 227.3/0.63 = 360.8 k W$

Answer: W_iso = 201.5 kW, W_adi = 227.3 kW, shaft = 360.8 kW

Why the other options are there

143.2 kW (efficiency multiplied)
313.7 kW (linear Δp × Q used)

Reference: FE Reference Handbook — Fluid Mechanics → Adiabatic Compression

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a pipeline, jet or submerged surface, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Adiabatic Compression contains 8 relations; you must be able to find this page in under 15 seconds.
Exam style: continuity plus energy, with one head-loss or force term.
Unit rule: γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

γ = 62.4 lb/ft³ or 9.81 kN/m³; convert psi to feet of head early
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.