Euler's Equation

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

7 formulas

10 exam-style examples

~59 min

All Fluid Mechanics lectures

Learning objectives

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

This chapter section covers Euler's Equation 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 euler's equation describes physically and when it applies.
State every one of the 7 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. Euler's Equation 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: euler's equation.
Capstone Studio instructional photograph

Fluid Mechanics — Euler's Equation: 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 7 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

P1, P2	Quantity produced by "P1, P2 = pressure at Locations 1 and 2" — read its definition and unit from the handbook line directly above the equation.
γ	Quantity produced by "γ = specific weight of the fluid (ρg)" — read its definition and unit from the handbook line directly above the equation.
z1, z2	Quantity produced by "z1, z2 = elevation at Locations 1 and 2" — read its definition and unit from the handbook line directly above the equation.
ρ	Quantity produced by "ρ = fluid density" — read its definition and unit from the handbook line directly above the equation.
ax	Quantity produced by "ax = local (temporal) acceleration of fluid in the x-direction" — read its definition and unit from the handbook line directly above the equation.
∆x	Quantity produced by "∆x = distance between Locations 1 and 2 in the x-direction" — 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.

z 1 ∆x ax
For unsteady flow due to local acceleration (i.e., temporal acceleration) in the x-direction, the change in pressure between two
points in a fluid can be determined by Euler's equation:
where
Crowe, Clayton T., Engineering Fluid Mechanics, 2nd ed., New York: John Wiley and Sons, 1980, p. 144.

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

One-dimensional flow: continuity plus Euler's equation along a streamline — Euler's Equation

Water flows steadily at 0.045 m³/s through a reducer from 290.0 mm to 130.0 mm diameter; the outlet is 3.0 m above the inlet. The inlet gauge pressure is 467 kPa. Using continuity and Euler's (frictionless) equation along the centre streamline, compute both velocities and the outlet pressure.

Given

Q = 0.045 m³/s
D₁ = 290.0 mm, D₂ = 130.0 mm
p₁ = 467 kPa
Δz = 3.0 m

Find

V₁, V₂ and p₂

Start with the thinking

Continuity fixes the velocities from areas alone — pressure never enters that step.
Euler integrated along a streamline for incompressible flow is Bernoulli: pressure falls where velocity or elevation rises.

Step-by-step solution

Formula
$Q = A_{1} V_{1} = A_{2} V_{2}$
Areas
$A_{1} = π (0.290)^{2} /4 = 0.06605 m^{2}, A_{2} = 0.01327 m^{2}$
Substituting
$V_{1} = 0.045/0.06605 = 0.68 m / s, V_{2} = 0.045/0.01327 = 3.39 m / s$
Formula
$\frac{p _{1}}{ρ} + \frac{V _{1}^{2}}{2} + g z_{1} = \frac{p _{2}}{ρ} + \frac{V _{2}^{2}}{2} + g z_{2}$
Velocity term
$ρ (V_{2}^{2} - V_{1}^{2}) /2 = 1000 (3.3 9^{2} - 0.6 8^{2}) /2 = 5.5 k P a$
Elevation term — ρgΔz = 1000(9.81)(3.0) = 29.4 kPa
Substituting
$p_{2} = 467 - 5.5 - 29.4 = 432.1 k P a$

Answer: V₁ = 0.68 m/s, V₂ = 3.39 m/s, p₂ = 432.1 kPa

Why the other options are there

p₂ = 472.5 kPa (sign of the velocity term reversed)
V₂ = 0.31 m/s (diameter ratio not squared)

Reference: FE Reference Handbook — Fluid Mechanics → Euler's Equation

Example 2

One-dimensional flow: continuity plus Euler's equation along a streamline — Euler's Equation (2)

Water flows steadily at 0.095 m³/s through a reducer from 330.0 mm to 110.0 mm diameter; the outlet is 0.5 m above the inlet. The inlet gauge pressure is 512 kPa. Using continuity and Euler's (frictionless) equation along the centre streamline, compute both velocities and the outlet pressure.

Given

Q = 0.095 m³/s
D₁ = 330.0 mm, D₂ = 110.0 mm
p₁ = 512 kPa
Δz = 0.5 m

Find

V₁, V₂ and p₂

Start with the thinking

Continuity fixes the velocities from areas alone — pressure never enters that step.
Euler integrated along a streamline for incompressible flow is Bernoulli: pressure falls where velocity or elevation rises.

Step-by-step solution

Formula
$Q = A_{1} V_{1} = A_{2} V_{2}$
Areas
$A_{1} = π (0.330)^{2} /4 = 0.08553 m^{2}, A_{2} = 0.00950 m^{2}$
Substituting
$V_{1} = 0.095/0.08553 = 1.11 m / s, V_{2} = 0.095/0.00950 = 10.00 m / s$
Formula
$\frac{p _{1}}{ρ} + \frac{V _{1}^{2}}{2} + g z_{1} = \frac{p _{2}}{ρ} + \frac{V _{2}^{2}}{2} + g z_{2}$
Velocity term
$ρ (V_{2}^{2} - V_{1}^{2}) /2 = 1000 (10.0 0^{2} - 1.1 1^{2}) /2 = 49.3 k P a$
Elevation term — ρgΔz = 1000(9.81)(0.5) = 4.9 kPa
Substituting
$p_{2} = 512 - 49.3 - 4.9 = 457.7 k P a$

Answer: V₁ = 1.11 m/s, V₂ = 10.00 m/s, p₂ = 457.7 kPa

Why the other options are there

p₂ = 561.3 kPa (sign of the velocity term reversed)
V₂ = 0.37 m/s (diameter ratio not squared)

Reference: FE Reference Handbook — Fluid Mechanics → Euler's Equation

Example 3

One-dimensional flow: continuity plus Euler's equation along a streamline — Euler's Equation (3)

Water flows steadily at 0.145 m³/s through a reducer from 350.0 mm to 110.0 mm diameter; the outlet is 0.5 m above the inlet. The inlet gauge pressure is 330 kPa. Using continuity and Euler's (frictionless) equation along the centre streamline, compute both velocities and the outlet pressure.

Given

Q = 0.145 m³/s
D₁ = 350.0 mm, D₂ = 110.0 mm
p₁ = 330 kPa
Δz = 0.5 m

Find

V₁, V₂ and p₂

Start with the thinking

Continuity fixes the velocities from areas alone — pressure never enters that step.
Euler integrated along a streamline for incompressible flow is Bernoulli: pressure falls where velocity or elevation rises.

Step-by-step solution

Formula
$Q = A_{1} V_{1} = A_{2} V_{2}$
Areas
$A_{1} = π (0.350)^{2} /4 = 0.09621 m^{2}, A_{2} = 0.00950 m^{2}$
Substituting
$V_{1} = 0.145/0.09621 = 1.51 m / s, V_{2} = 0.145/0.00950 = 15.26 m / s$
Formula
$\frac{p _{1}}{ρ} + \frac{V _{1}^{2}}{2} + g z_{1} = \frac{p _{2}}{ρ} + \frac{V _{2}^{2}}{2} + g z_{2}$
Velocity term
$ρ (V_{2}^{2} - V_{1}^{2}) /2 = 1000 (15.2 6^{2} - 1.5 1^{2}) /2 = 115.3 k P a$
Elevation term — ρgΔz = 1000(9.81)(0.5) = 4.9 kPa
Substituting
$p_{2} = 330 - 115.3 - 4.9 = 209.8 k P a$

Answer: V₁ = 1.51 m/s, V₂ = 15.26 m/s, p₂ = 209.8 kPa

Why the other options are there

p₂ = 445.3 kPa (sign of the velocity term reversed)
V₂ = 0.47 m/s (diameter ratio not squared)

Reference: FE Reference Handbook — Fluid Mechanics → Euler's Equation

Example 4

One-dimensional flow: continuity plus Euler's equation along a streamline — Euler's Equation (4)

Water flows steadily at 0.100 m³/s through a reducer from 180.0 mm to 100.0 mm diameter; the outlet is 0.0 m above the inlet. The inlet gauge pressure is 189 kPa. Using continuity and Euler's (frictionless) equation along the centre streamline, compute both velocities and the outlet pressure.

Given

Q = 0.100 m³/s
D₁ = 180.0 mm, D₂ = 100.0 mm
p₁ = 189 kPa
Δz = 0.0 m

Find

V₁, V₂ and p₂

Start with the thinking

Continuity fixes the velocities from areas alone — pressure never enters that step.
Euler integrated along a streamline for incompressible flow is Bernoulli: pressure falls where velocity or elevation rises.

Step-by-step solution

Formula
$Q = A_{1} V_{1} = A_{2} V_{2}$
Areas
$A_{1} = π (0.180)^{2} /4 = 0.02545 m^{2}, A_{2} = 0.00785 m^{2}$
Substituting
$V_{1} = 0.100/0.02545 = 3.93 m / s, V_{2} = 0.100/0.00785 = 12.73 m / s$
Formula
$\frac{p _{1}}{ρ} + \frac{V _{1}^{2}}{2} + g z_{1} = \frac{p _{2}}{ρ} + \frac{V _{2}^{2}}{2} + g z_{2}$
Velocity term
$ρ (V_{2}^{2} - V_{1}^{2}) /2 = 1000 (12.7 3^{2} - 3.9 3^{2}) /2 = 73.3 k P a$
Elevation term — ρgΔz = 1000(9.81)(0.0) = 0.0 kPa
Substituting
$p_{2} = 189 - 73.3 - 0.0 = 115.7 k P a$

Answer: V₁ = 3.93 m/s, V₂ = 12.73 m/s, p₂ = 115.7 kPa

Why the other options are there

p₂ = 262.3 kPa (sign of the velocity term reversed)
V₂ = 2.18 m/s (diameter ratio not squared)

Reference: FE Reference Handbook — Fluid Mechanics → Euler's Equation

Example 5

One-dimensional flow: continuity plus Euler's equation along a streamline — Euler's Equation (5)

Water flows steadily at 0.055 m³/s through a reducer from 220.0 mm to 140.0 mm diameter; the outlet is 2.0 m above the inlet. The inlet gauge pressure is 180 kPa. Using continuity and Euler's (frictionless) equation along the centre streamline, compute both velocities and the outlet pressure.

Given

Q = 0.055 m³/s
D₁ = 220.0 mm, D₂ = 140.0 mm
p₁ = 180 kPa
Δz = 2.0 m

Find

V₁, V₂ and p₂

Start with the thinking

Continuity fixes the velocities from areas alone — pressure never enters that step.
Euler integrated along a streamline for incompressible flow is Bernoulli: pressure falls where velocity or elevation rises.

Step-by-step solution

Formula
$Q = A_{1} V_{1} = A_{2} V_{2}$
Areas
$A_{1} = π (0.220)^{2} /4 = 0.03801 m^{2}, A_{2} = 0.01539 m^{2}$
Substituting
$V_{1} = 0.055/0.03801 = 1.45 m / s, V_{2} = 0.055/0.01539 = 3.57 m / s$
Formula
$\frac{p _{1}}{ρ} + \frac{V _{1}^{2}}{2} + g z_{1} = \frac{p _{2}}{ρ} + \frac{V _{2}^{2}}{2} + g z_{2}$
Velocity term
$ρ (V_{2}^{2} - V_{1}^{2}) /2 = 1000 (3.5 7^{2} - 1.4 5^{2}) /2 = 5.3 k P a$
Elevation term — ρgΔz = 1000(9.81)(2.0) = 19.6 kPa
Substituting
$p_{2} = 180 - 5.3 - 19.6 = 155.0 k P a$

Answer: V₁ = 1.45 m/s, V₂ = 3.57 m/s, p₂ = 155.0 kPa

Why the other options are there

p₂ = 185.3 kPa (sign of the velocity term reversed)
V₂ = 0.92 m/s (diameter ratio not squared)

Reference: FE Reference Handbook — Fluid Mechanics → Euler's Equation

Example 6

One-dimensional flow: continuity plus Euler's equation along a streamline — Euler's Equation (6)

Water flows steadily at 0.100 m³/s through a reducer from 300.0 mm to 90 mm diameter; the outlet is 0.5 m above the inlet. The inlet gauge pressure is 266 kPa. Using continuity and Euler's (frictionless) equation along the centre streamline, compute both velocities and the outlet pressure.

Given

Q = 0.100 m³/s
D₁ = 300.0 mm, D₂ = 90 mm
p₁ = 266 kPa
Δz = 0.5 m

Find

V₁, V₂ and p₂

Start with the thinking

Continuity fixes the velocities from areas alone — pressure never enters that step.
Euler integrated along a streamline for incompressible flow is Bernoulli: pressure falls where velocity or elevation rises.

Step-by-step solution

Formula
$Q = A_{1} V_{1} = A_{2} V_{2}$
Areas
$A_{1} = π (0.300)^{2} /4 = 0.07069 m^{2}, A_{2} = 0.00636 m^{2}$
Substituting
$V_{1} = 0.100/0.07069 = 1.41 m / s, V_{2} = 0.100/0.00636 = 15.72 m / s$
Formula
$\frac{p _{1}}{ρ} + \frac{V _{1}^{2}}{2} + g z_{1} = \frac{p _{2}}{ρ} + \frac{V _{2}^{2}}{2} + g z_{2}$
Velocity term
$ρ (V_{2}^{2} - V_{1}^{2}) /2 = 1000 (15.7 2^{2} - 1.4 1^{2}) /2 = 122.5 k P a$
Elevation term — ρgΔz = 1000(9.81)(0.5) = 4.9 kPa
Substituting
$p_{2} = 266 - 122.5 - 4.9 = 138.6 k P a$

Answer: V₁ = 1.41 m/s, V₂ = 15.72 m/s, p₂ = 138.6 kPa

Why the other options are there

p₂ = 388.5 kPa (sign of the velocity term reversed)
V₂ = 0.42 m/s (diameter ratio not squared)

Reference: FE Reference Handbook — Fluid Mechanics → Euler's Equation

Example 7

One-dimensional flow: continuity plus Euler's equation along a streamline — Euler's Equation (7)

Water flows steadily at 0.110 m³/s through a reducer from 300.0 mm to 90 mm diameter; the outlet is 2.0 m above the inlet. The inlet gauge pressure is 204 kPa. Using continuity and Euler's (frictionless) equation along the centre streamline, compute both velocities and the outlet pressure.

Given

Q = 0.110 m³/s
D₁ = 300.0 mm, D₂ = 90 mm
p₁ = 204 kPa
Δz = 2.0 m

Find

V₁, V₂ and p₂

Start with the thinking

Continuity fixes the velocities from areas alone — pressure never enters that step.
Euler integrated along a streamline for incompressible flow is Bernoulli: pressure falls where velocity or elevation rises.

Step-by-step solution

Formula
$Q = A_{1} V_{1} = A_{2} V_{2}$
Areas
$A_{1} = π (0.300)^{2} /4 = 0.07069 m^{2}, A_{2} = 0.00636 m^{2}$
Substituting
$V_{1} = 0.110/0.07069 = 1.56 m / s, V_{2} = 0.110/0.00636 = 17.29 m / s$
Formula
$\frac{p _{1}}{ρ} + \frac{V _{1}^{2}}{2} + g z_{1} = \frac{p _{2}}{ρ} + \frac{V _{2}^{2}}{2} + g z_{2}$
Velocity term
$ρ (V_{2}^{2} - V_{1}^{2}) /2 = 1000 (17.2 9^{2} - 1.5 6^{2}) /2 = 148.3 k P a$
Elevation term — ρgΔz = 1000(9.81)(2.0) = 19.6 kPa
Substituting
$p_{2} = 204 - 148.3 - 19.6 = 36.1 k P a$

Answer: V₁ = 1.56 m/s, V₂ = 17.29 m/s, p₂ = 36.1 kPa

Why the other options are there

p₂ = 352.3 kPa (sign of the velocity term reversed)
V₂ = 0.47 m/s (diameter ratio not squared)

Reference: FE Reference Handbook — Fluid Mechanics → Euler's Equation

Example 8

One-dimensional flow: continuity plus Euler's equation along a streamline — Euler's Equation (8)

Water flows steadily at 0.035 m³/s through a reducer from 190.0 mm to 90 mm diameter; the outlet is 3.5 m above the inlet. The inlet gauge pressure is 385 kPa. Using continuity and Euler's (frictionless) equation along the centre streamline, compute both velocities and the outlet pressure.

Given

Q = 0.035 m³/s
D₁ = 190.0 mm, D₂ = 90 mm
p₁ = 385 kPa
Δz = 3.5 m

Find

V₁, V₂ and p₂

Start with the thinking

Continuity fixes the velocities from areas alone — pressure never enters that step.
Euler integrated along a streamline for incompressible flow is Bernoulli: pressure falls where velocity or elevation rises.

Step-by-step solution

Formula
$Q = A_{1} V_{1} = A_{2} V_{2}$
Areas
$A_{1} = π (0.190)^{2} /4 = 0.02835 m^{2}, A_{2} = 0.00636 m^{2}$
Substituting
$V_{1} = 0.035/0.02835 = 1.23 m / s, V_{2} = 0.035/0.00636 = 5.50 m / s$
Formula
$\frac{p _{1}}{ρ} + \frac{V _{1}^{2}}{2} + g z_{1} = \frac{p _{2}}{ρ} + \frac{V _{2}^{2}}{2} + g z_{2}$
Velocity term
$ρ (V_{2}^{2} - V_{1}^{2}) /2 = 1000 (5.5 0^{2} - 1.2 3^{2}) /2 = 14.4 k P a$
Elevation term — ρgΔz = 1000(9.81)(3.5) = 34.3 kPa
Substituting
$p_{2} = 385 - 14.4 - 34.3 = 336.3 k P a$

Answer: V₁ = 1.23 m/s, V₂ = 5.50 m/s, p₂ = 336.3 kPa

Why the other options are there

p₂ = 399.4 kPa (sign of the velocity term reversed)
V₂ = 0.58 m/s (diameter ratio not squared)

Reference: FE Reference Handbook — Fluid Mechanics → Euler's Equation

Example 9

One-dimensional flow: continuity plus Euler's equation along a streamline — Euler's Equation (9)

Water flows steadily at 0.085 m³/s through a reducer from 370.0 mm to 90 mm diameter; the outlet is 0.0 m above the inlet. The inlet gauge pressure is 339 kPa. Using continuity and Euler's (frictionless) equation along the centre streamline, compute both velocities and the outlet pressure.

Given

Q = 0.085 m³/s
D₁ = 370.0 mm, D₂ = 90 mm
p₁ = 339 kPa
Δz = 0.0 m

Find

V₁, V₂ and p₂

Start with the thinking

Continuity fixes the velocities from areas alone — pressure never enters that step.
Euler integrated along a streamline for incompressible flow is Bernoulli: pressure falls where velocity or elevation rises.

Step-by-step solution

Formula
$Q = A_{1} V_{1} = A_{2} V_{2}$
Areas
$A_{1} = π (0.370)^{2} /4 = 0.10752 m^{2}, A_{2} = 0.00636 m^{2}$
Substituting
$V_{1} = 0.085/0.10752 = 0.79 m / s, V_{2} = 0.085/0.00636 = 13.36 m / s$
Formula
$\frac{p _{1}}{ρ} + \frac{V _{1}^{2}}{2} + g z_{1} = \frac{p _{2}}{ρ} + \frac{V _{2}^{2}}{2} + g z_{2}$
Velocity term
$ρ (V_{2}^{2} - V_{1}^{2}) /2 = 1000 (13.3 6^{2} - 0.7 9^{2}) /2 = 88.9 k P a$
Elevation term — ρgΔz = 1000(9.81)(0.0) = 0.0 kPa
Substituting
$p_{2} = 339 - 88.9 - 0.0 = 250.1 k P a$

Answer: V₁ = 0.79 m/s, V₂ = 13.36 m/s, p₂ = 250.1 kPa

Why the other options are there

p₂ = 427.9 kPa (sign of the velocity term reversed)
V₂ = 0.19 m/s (diameter ratio not squared)

Reference: FE Reference Handbook — Fluid Mechanics → Euler's Equation

Example 10

One-dimensional flow: continuity plus Euler's equation along a streamline — Euler's Equation (10)

Water flows steadily at 0.160 m³/s through a reducer from 290.0 mm to 100.0 mm diameter; the outlet is 3.0 m above the inlet. The inlet gauge pressure is 383 kPa. Using continuity and Euler's (frictionless) equation along the centre streamline, compute both velocities and the outlet pressure.

Given

Q = 0.160 m³/s
D₁ = 290.0 mm, D₂ = 100.0 mm
p₁ = 383 kPa
Δz = 3.0 m

Find

V₁, V₂ and p₂

Start with the thinking

Continuity fixes the velocities from areas alone — pressure never enters that step.
Euler integrated along a streamline for incompressible flow is Bernoulli: pressure falls where velocity or elevation rises.

Step-by-step solution

Formula
$Q = A_{1} V_{1} = A_{2} V_{2}$
Areas
$A_{1} = π (0.290)^{2} /4 = 0.06605 m^{2}, A_{2} = 0.00785 m^{2}$
Substituting
$V_{1} = 0.160/0.06605 = 2.42 m / s, V_{2} = 0.160/0.00785 = 20.37 m / s$
Formula
$\frac{p _{1}}{ρ} + \frac{V _{1}^{2}}{2} + g z_{1} = \frac{p _{2}}{ρ} + \frac{V _{2}^{2}}{2} + g z_{2}$
Velocity term
$ρ (V_{2}^{2} - V_{1}^{2}) /2 = 1000 (20.3 7^{2} - 2.4 2^{2}) /2 = 204.6 k P a$
Elevation term — ρgΔz = 1000(9.81)(3.0) = 29.4 kPa
Substituting
$p_{2} = 383 - 204.6 - 29.4 = 149.0 k P a$

Answer: V₁ = 2.42 m/s, V₂ = 20.37 m/s, p₂ = 149.0 kPa

Why the other options are there

p₂ = 587.6 kPa (sign of the velocity term reversed)
V₂ = 0.84 m/s (diameter ratio not squared)

Reference: FE Reference Handbook — Fluid Mechanics → Euler's Equation

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

Euler's Equation contains 7 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.

Euler's Equation

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

‘P2​+c:z2​j−‘P1​+c:z1​j=−Dx:t:ax

P1​,P2​=pressureatLocations1and2

γ = specific weight of the fluid (ρg)Needs technical review

z1​,z2​=elevationatLocations1and2

ρ=fluiddensity

ax=local(temporal)accelerationoffluidinthex−direction

∆x=distancebetweenLocations1and2inthex−direction

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section