Normal Shock Relationships

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

14 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 Normal Shock Relationships 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 normal shock relationships describes physically and when it applies.
State every one of the 14 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. Normal Shock Relationships 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: normal shock relationships.
Capstone Studio instructional photograph

Fluid Mechanics — Normal Shock Relationships: 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 14 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

^ k - 1h Ma12 + 2	Quantity produced by "^ k - 1h Ma12 + 2" — read its definition and unit from the handbook line directly above the equation.
2k Ma12 - ^ k - 1h	Quantity produced by "2k Ma12 - ^ k - 1h" — read its definition and unit from the handbook line directly above the equation.
P1	Quantity produced by "P1 = k + 1 1" — read its definition and unit from the handbook line directly above the equation.
^ k - 1h Ma 2 + 2	Quantity produced by "^ k - 1h Ma 2 + 2" — read its definition and unit from the handbook line directly above the equation.
T01	Quantity produced by "T01 = T02" — 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.

A normal shock wave is a physical mechanism that slows a flow from supersonic to subsonic. It occurs over an infinitesimal
distance. The flow upstream of a normal shock wave is always supersonic and the flow downstream is always subsonic as
depicted in the figure.
1 2
Ma > 1 Ma < 1
NORMAL SHOCK
The following equations relate downstream flow conditions to upstream flow conditions for a normal shock wave.

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

Property jumps across a normal shock wave — Normal Shock Relationships

A normal shock stands in an air duct where the upstream Mach number is 2.3 and the static pressure is 50 kPa. Compute the downstream Mach number, static pressure, and the density and temperature ratios (k = 1.4).

Given

M₁ = 2.3
p₁ = 50 kPa
k = 1.4

Find

M₂, p₂, ρ₂/ρ₁ and T₂/T₁

Start with the thinking

A normal shock is always compressive: pressure, density and temperature rise while Mach number falls below 1.
Total pressure drops across the shock even though total temperature is unchanged.

Step-by-step solution

Formula
$M_{2}^{2} = \frac{1 + \frac{k - 1}{2} M _{1}^{2}}{k M _{1}^{2} - \frac{k - 1}{2}}$
Substituting
$M_{2} = (1 + 0.2 (2.3)^{2}) / (1.4 (2.3)^{2} - 0.2) = 0.534$
Formula
$\frac{p _{2}}{p _{1}} = \frac{2 k M _{1}^{2} - ( k - 1 )}{k + 1}$
Substituting
$p_{2} / p_{1} = [2.8 (2.3)^{2} - 0.4] /2.4 = 6.005 \to p_{2} = 300.2 k P a$
Formula
$\frac{ρ _{2}}{ρ _{1}} = \frac{( k + 1 ) M _{1}^{2}}{2 + ( k - 1 ) M _{1}^{2}}$
Substituting
$ρ_{2} / ρ_{1} = 2.4 (2.3)^{2} / [2 + 0.4 (2.3)^{2}] = 3.085$
Temperature
$T_{2} / T_{1} = (p_{2} / p_{1}) / (ρ_{2} / ρ_{1}) = 6.005/3.085 = 1.947$

Answer: M₂ = 0.534, p₂ = 300.2 kPa, ρ₂/ρ₁ = 3.085, T₂/T₁ = 1.947

Why the other options are there

M₂ = 0.435 (reciprocal guess)
p₂ = 8.3 kPa (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Normal Shock Relationships

Example 2

Property jumps across a normal shock wave — Normal Shock Relationships (2)

A normal shock stands in an air duct where the upstream Mach number is 2.2 and the static pressure is 118 kPa. Compute the downstream Mach number, static pressure, and the density and temperature ratios (k = 1.4).

Given

M₁ = 2.2
p₁ = 118 kPa
k = 1.4

Find

M₂, p₂, ρ₂/ρ₁ and T₂/T₁

Start with the thinking

A normal shock is always compressive: pressure, density and temperature rise while Mach number falls below 1.
Total pressure drops across the shock even though total temperature is unchanged.

Step-by-step solution

Formula
$M_{2}^{2} = \frac{1 + \frac{k - 1}{2} M _{1}^{2}}{k M _{1}^{2} - \frac{k - 1}{2}}$
Substituting
$M_{2} = (1 + 0.2 (2.2)^{2}) / (1.4 (2.2)^{2} - 0.2) = 0.547$
Formula
$\frac{p _{2}}{p _{1}} = \frac{2 k M _{1}^{2} - ( k - 1 )}{k + 1}$
Substituting
$p_{2} / p_{1} = [2.8 (2.2)^{2} - 0.4] /2.4 = 5.480 \to p_{2} = 646.6 k P a$
Formula
$\frac{ρ _{2}}{ρ _{1}} = \frac{( k + 1 ) M _{1}^{2}}{2 + ( k - 1 ) M _{1}^{2}}$
Substituting
$ρ_{2} / ρ_{1} = 2.4 (2.2)^{2} / [2 + 0.4 (2.2)^{2}] = 2.951$
Temperature
$T_{2} / T_{1} = (p_{2} / p_{1}) / (ρ_{2} / ρ_{1}) = 5.480/2.951 = 1.857$

Answer: M₂ = 0.547, p₂ = 646.6 kPa, ρ₂/ρ₁ = 2.951, T₂/T₁ = 1.857

Why the other options are there

M₂ = 0.455 (reciprocal guess)
p₂ = 21.5 kPa (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Normal Shock Relationships

Example 3

Property jumps across a normal shock wave — Normal Shock Relationships (3)

A normal shock stands in an air duct where the upstream Mach number is 1.5 and the static pressure is 41 kPa. Compute the downstream Mach number, static pressure, and the density and temperature ratios (k = 1.4).

Given

M₁ = 1.5
p₁ = 41 kPa
k = 1.4

Find

M₂, p₂, ρ₂/ρ₁ and T₂/T₁

Start with the thinking

A normal shock is always compressive: pressure, density and temperature rise while Mach number falls below 1.
Total pressure drops across the shock even though total temperature is unchanged.

Step-by-step solution

Formula
$M_{2}^{2} = \frac{1 + \frac{k - 1}{2} M _{1}^{2}}{k M _{1}^{2} - \frac{k - 1}{2}}$
Substituting
$M_{2} = (1 + 0.2 (1.5)^{2}) / (1.4 (1.5)^{2} - 0.2) = 0.701$
Formula
$\frac{p _{2}}{p _{1}} = \frac{2 k M _{1}^{2} - ( k - 1 )}{k + 1}$
Substituting
$p_{2} / p_{1} = [2.8 (1.5)^{2} - 0.4] /2.4 = 2.458 \to p_{2} = 100.8 k P a$
Formula
$\frac{ρ _{2}}{ρ _{1}} = \frac{( k + 1 ) M _{1}^{2}}{2 + ( k - 1 ) M _{1}^{2}}$
Substituting
$ρ_{2} / ρ_{1} = 2.4 (1.5)^{2} / [2 + 0.4 (1.5)^{2}] = 1.862$
Temperature
$T_{2} / T_{1} = (p_{2} / p_{1}) / (ρ_{2} / ρ_{1}) = 2.458/1.862 = 1.320$

Answer: M₂ = 0.701, p₂ = 100.8 kPa, ρ₂/ρ₁ = 1.862, T₂/T₁ = 1.320

Why the other options are there

M₂ = 0.667 (reciprocal guess)
p₂ = 16.7 kPa (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Normal Shock Relationships

Example 4

Property jumps across a normal shock wave — Normal Shock Relationships (4)

A normal shock stands in an air duct where the upstream Mach number is 1.5 and the static pressure is 37 kPa. Compute the downstream Mach number, static pressure, and the density and temperature ratios (k = 1.4).

Given

M₁ = 1.5
p₁ = 37 kPa
k = 1.4

Find

M₂, p₂, ρ₂/ρ₁ and T₂/T₁

Start with the thinking

A normal shock is always compressive: pressure, density and temperature rise while Mach number falls below 1.
Total pressure drops across the shock even though total temperature is unchanged.

Step-by-step solution

Formula
$M_{2}^{2} = \frac{1 + \frac{k - 1}{2} M _{1}^{2}}{k M _{1}^{2} - \frac{k - 1}{2}}$
Substituting
$M_{2} = (1 + 0.2 (1.5)^{2}) / (1.4 (1.5)^{2} - 0.2) = 0.701$
Formula
$\frac{p _{2}}{p _{1}} = \frac{2 k M _{1}^{2} - ( k - 1 )}{k + 1}$
Substituting
$p_{2} / p_{1} = [2.8 (1.5)^{2} - 0.4] /2.4 = 2.458 \to p_{2} = 91.0 k P a$
Formula
$\frac{ρ _{2}}{ρ _{1}} = \frac{( k + 1 ) M _{1}^{2}}{2 + ( k - 1 ) M _{1}^{2}}$
Substituting
$ρ_{2} / ρ_{1} = 2.4 (1.5)^{2} / [2 + 0.4 (1.5)^{2}] = 1.862$
Temperature
$T_{2} / T_{1} = (p_{2} / p_{1}) / (ρ_{2} / ρ_{1}) = 2.458/1.862 = 1.320$

Answer: M₂ = 0.701, p₂ = 91.0 kPa, ρ₂/ρ₁ = 1.862, T₂/T₁ = 1.320

Why the other options are there

M₂ = 0.667 (reciprocal guess)
p₂ = 15.1 kPa (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Normal Shock Relationships

Example 5

Property jumps across a normal shock wave — Normal Shock Relationships (5)

A normal shock stands in an air duct where the upstream Mach number is 2.6 and the static pressure is 99 kPa. Compute the downstream Mach number, static pressure, and the density and temperature ratios (k = 1.4).

Given

M₁ = 2.6
p₁ = 99 kPa
k = 1.4

Find

M₂, p₂, ρ₂/ρ₁ and T₂/T₁

Start with the thinking

A normal shock is always compressive: pressure, density and temperature rise while Mach number falls below 1.
Total pressure drops across the shock even though total temperature is unchanged.

Step-by-step solution

Formula
$M_{2}^{2} = \frac{1 + \frac{k - 1}{2} M _{1}^{2}}{k M _{1}^{2} - \frac{k - 1}{2}}$
Substituting
$M_{2} = (1 + 0.2 (2.6)^{2}) / (1.4 (2.6)^{2} - 0.2) = 0.504$
Formula
$\frac{p _{2}}{p _{1}} = \frac{2 k M _{1}^{2} - ( k - 1 )}{k + 1}$
Substituting
$p_{2} / p_{1} = [2.8 (2.6)^{2} - 0.4] /2.4 = 7.720 \to p_{2} = 764.3 k P a$
Formula
$\frac{ρ _{2}}{ρ _{1}} = \frac{( k + 1 ) M _{1}^{2}}{2 + ( k - 1 ) M _{1}^{2}}$
Substituting
$ρ_{2} / ρ_{1} = 2.4 (2.6)^{2} / [2 + 0.4 (2.6)^{2}] = 3.449$
Temperature
$T_{2} / T_{1} = (p_{2} / p_{1}) / (ρ_{2} / ρ_{1}) = 7.720/3.449 = 2.238$

Answer: M₂ = 0.504, p₂ = 764.3 kPa, ρ₂/ρ₁ = 3.449, T₂/T₁ = 2.238

Why the other options are there

M₂ = 0.385 (reciprocal guess)
p₂ = 12.8 kPa (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Normal Shock Relationships

Example 6

Property jumps across a normal shock wave — Normal Shock Relationships (6)

A normal shock stands in an air duct where the upstream Mach number is 2.5 and the static pressure is 118 kPa. Compute the downstream Mach number, static pressure, and the density and temperature ratios (k = 1.4).

Given

M₁ = 2.5
p₁ = 118 kPa
k = 1.4

Find

M₂, p₂, ρ₂/ρ₁ and T₂/T₁

Start with the thinking

A normal shock is always compressive: pressure, density and temperature rise while Mach number falls below 1.
Total pressure drops across the shock even though total temperature is unchanged.

Step-by-step solution

Formula
$M_{2}^{2} = \frac{1 + \frac{k - 1}{2} M _{1}^{2}}{k M _{1}^{2} - \frac{k - 1}{2}}$
Substituting
$M_{2} = (1 + 0.2 (2.5)^{2}) / (1.4 (2.5)^{2} - 0.2) = 0.513$
Formula
$\frac{p _{2}}{p _{1}} = \frac{2 k M _{1}^{2} - ( k - 1 )}{k + 1}$
Substituting
$p_{2} / p_{1} = [2.8 (2.5)^{2} - 0.4] /2.4 = 7.125 \to p_{2} = 840.8 k P a$
Formula
$\frac{ρ _{2}}{ρ _{1}} = \frac{( k + 1 ) M _{1}^{2}}{2 + ( k - 1 ) M _{1}^{2}}$
Substituting
$ρ_{2} / ρ_{1} = 2.4 (2.5)^{2} / [2 + 0.4 (2.5)^{2}] = 3.333$
Temperature
$T_{2} / T_{1} = (p_{2} / p_{1}) / (ρ_{2} / ρ_{1}) = 7.125/3.333 = 2.138$

Answer: M₂ = 0.513, p₂ = 840.8 kPa, ρ₂/ρ₁ = 3.333, T₂/T₁ = 2.138

Why the other options are there

M₂ = 0.400 (reciprocal guess)
p₂ = 16.6 kPa (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Normal Shock Relationships

Example 7

Property jumps across a normal shock wave — Normal Shock Relationships (7)

A normal shock stands in an air duct where the upstream Mach number is 2.0 and the static pressure is 113 kPa. Compute the downstream Mach number, static pressure, and the density and temperature ratios (k = 1.4).

Given

M₁ = 2.0
p₁ = 113 kPa
k = 1.4

Find

M₂, p₂, ρ₂/ρ₁ and T₂/T₁

Start with the thinking

A normal shock is always compressive: pressure, density and temperature rise while Mach number falls below 1.
Total pressure drops across the shock even though total temperature is unchanged.

Step-by-step solution

Formula
$M_{2}^{2} = \frac{1 + \frac{k - 1}{2} M _{1}^{2}}{k M _{1}^{2} - \frac{k - 1}{2}}$
Substituting
$M_{2} = (1 + 0.2 (2.0)^{2}) / (1.4 (2.0)^{2} - 0.2) = 0.577$
Formula
$\frac{p _{2}}{p _{1}} = \frac{2 k M _{1}^{2} - ( k - 1 )}{k + 1}$
Substituting
$p_{2} / p_{1} = [2.8 (2.0)^{2} - 0.4] /2.4 = 4.500 \to p_{2} = 508.5 k P a$
Formula
$\frac{ρ _{2}}{ρ _{1}} = \frac{( k + 1 ) M _{1}^{2}}{2 + ( k - 1 ) M _{1}^{2}}$
Substituting
$ρ_{2} / ρ_{1} = 2.4 (2.0)^{2} / [2 + 0.4 (2.0)^{2}] = 2.667$
Temperature
$T_{2} / T_{1} = (p_{2} / p_{1}) / (ρ_{2} / ρ_{1}) = 4.500/2.667 = 1.687$

Answer: M₂ = 0.577, p₂ = 508.5 kPa, ρ₂/ρ₁ = 2.667, T₂/T₁ = 1.687

Why the other options are there

M₂ = 0.500 (reciprocal guess)
p₂ = 25.1 kPa (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Normal Shock Relationships

Example 8

Property jumps across a normal shock wave — Normal Shock Relationships (8)

A normal shock stands in an air duct where the upstream Mach number is 1.4 and the static pressure is 111 kPa. Compute the downstream Mach number, static pressure, and the density and temperature ratios (k = 1.4).

Given

M₁ = 1.4
p₁ = 111 kPa
k = 1.4

Find

M₂, p₂, ρ₂/ρ₁ and T₂/T₁

Start with the thinking

A normal shock is always compressive: pressure, density and temperature rise while Mach number falls below 1.
Total pressure drops across the shock even though total temperature is unchanged.

Step-by-step solution

Formula
$M_{2}^{2} = \frac{1 + \frac{k - 1}{2} M _{1}^{2}}{k M _{1}^{2} - \frac{k - 1}{2}}$
Substituting
$M_{2} = (1 + 0.2 (1.4)^{2}) / (1.4 (1.4)^{2} - 0.2) = 0.740$
Formula
$\frac{p _{2}}{p _{1}} = \frac{2 k M _{1}^{2} - ( k - 1 )}{k + 1}$
Substituting
$p_{2} / p_{1} = [2.8 (1.4)^{2} - 0.4] /2.4 = 2.120 \to p_{2} = 235.3 k P a$
Formula
$\frac{ρ _{2}}{ρ _{1}} = \frac{( k + 1 ) M _{1}^{2}}{2 + ( k - 1 ) M _{1}^{2}}$
Substituting
$ρ_{2} / ρ_{1} = 2.4 (1.4)^{2} / [2 + 0.4 (1.4)^{2}] = 1.690$
Temperature
$T_{2} / T_{1} = (p_{2} / p_{1}) / (ρ_{2} / ρ_{1}) = 2.120/1.690 = 1.255$

Answer: M₂ = 0.740, p₂ = 235.3 kPa, ρ₂/ρ₁ = 1.690, T₂/T₁ = 1.255

Why the other options are there

M₂ = 0.714 (reciprocal guess)
p₂ = 52.4 kPa (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Normal Shock Relationships

Example 9

Property jumps across a normal shock wave — Normal Shock Relationships (9)

A normal shock stands in an air duct where the upstream Mach number is 1.6 and the static pressure is 109 kPa. Compute the downstream Mach number, static pressure, and the density and temperature ratios (k = 1.4).

Given

M₁ = 1.6
p₁ = 109 kPa
k = 1.4

Find

M₂, p₂, ρ₂/ρ₁ and T₂/T₁

Start with the thinking

A normal shock is always compressive: pressure, density and temperature rise while Mach number falls below 1.
Total pressure drops across the shock even though total temperature is unchanged.

Step-by-step solution

Formula
$M_{2}^{2} = \frac{1 + \frac{k - 1}{2} M _{1}^{2}}{k M _{1}^{2} - \frac{k - 1}{2}}$
Substituting
$M_{2} = (1 + 0.2 (1.6)^{2}) / (1.4 (1.6)^{2} - 0.2) = 0.668$
Formula
$\frac{p _{2}}{p _{1}} = \frac{2 k M _{1}^{2} - ( k - 1 )}{k + 1}$
Substituting
$p_{2} / p_{1} = [2.8 (1.6)^{2} - 0.4] /2.4 = 2.820 \to p_{2} = 307.4 k P a$
Formula
$\frac{ρ _{2}}{ρ _{1}} = \frac{( k + 1 ) M _{1}^{2}}{2 + ( k - 1 ) M _{1}^{2}}$
Substituting
$ρ_{2} / ρ_{1} = 2.4 (1.6)^{2} / [2 + 0.4 (1.6)^{2}] = 2.032$
Temperature
$T_{2} / T_{1} = (p_{2} / p_{1}) / (ρ_{2} / ρ_{1}) = 2.820/2.032 = 1.388$

Answer: M₂ = 0.668, p₂ = 307.4 kPa, ρ₂/ρ₁ = 2.032, T₂/T₁ = 1.388

Why the other options are there

M₂ = 0.625 (reciprocal guess)
p₂ = 38.7 kPa (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Normal Shock Relationships

Example 10

Property jumps across a normal shock wave — Normal Shock Relationships (10)

A normal shock stands in an air duct where the upstream Mach number is 2.2 and the static pressure is 87 kPa. Compute the downstream Mach number, static pressure, and the density and temperature ratios (k = 1.4).

Given

M₁ = 2.2
p₁ = 87 kPa
k = 1.4

Find

M₂, p₂, ρ₂/ρ₁ and T₂/T₁

Start with the thinking

A normal shock is always compressive: pressure, density and temperature rise while Mach number falls below 1.
Total pressure drops across the shock even though total temperature is unchanged.

Step-by-step solution

Formula
$M_{2}^{2} = \frac{1 + \frac{k - 1}{2} M _{1}^{2}}{k M _{1}^{2} - \frac{k - 1}{2}}$
Substituting
$M_{2} = (1 + 0.2 (2.2)^{2}) / (1.4 (2.2)^{2} - 0.2) = 0.547$
Formula
$\frac{p _{2}}{p _{1}} = \frac{2 k M _{1}^{2} - ( k - 1 )}{k + 1}$
Substituting
$p_{2} / p_{1} = [2.8 (2.2)^{2} - 0.4] /2.4 = 5.480 \to p_{2} = 476.8 k P a$
Formula
$\frac{ρ _{2}}{ρ _{1}} = \frac{( k + 1 ) M _{1}^{2}}{2 + ( k - 1 ) M _{1}^{2}}$
Substituting
$ρ_{2} / ρ_{1} = 2.4 (2.2)^{2} / [2 + 0.4 (2.2)^{2}] = 2.951$
Temperature
$T_{2} / T_{1} = (p_{2} / p_{1}) / (ρ_{2} / ρ_{1}) = 5.480/2.951 = 1.857$

Answer: M₂ = 0.547, p₂ = 476.8 kPa, ρ₂/ρ₁ = 2.951, T₂/T₁ = 1.857

Why the other options are there

M₂ = 0.455 (reciprocal guess)
p₂ = 15.9 kPa (ratio inverted)

Reference: FE Reference Handbook — Fluid Mechanics → Normal Shock Relationships

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

Normal Shock Relationships contains 14 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.