Energy Line (Bernoulli Equation)

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

0 formulas

10 exam-style examples

~45 min

All Fluid Mechanics lectures

Learning objectives

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

This chapter section covers Energy Line (Bernoulli 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 energy line (bernoulli equation) describes physically and when it applies.
State every one of the 0 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. Energy Line (Bernoulli 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: energy line (bernoulli equation).
Capstone Studio instructional photograph

Fluid Mechanics — Energy Line (Bernoulli 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 0 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.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

The Bernoulli equation states that the sum of the pressure, velocity, and elevation heads is constant. The energy line is this sum
or the "total head line" above a horizontal datum. The difference between the hydraulic grade line and the energy line is the
v2/2g term.

Core formulas for this FE topic

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

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

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

Example 1

Bernoulli between two pipe sections

Water flows at 0.25 m³/s from a 300 mm pipe (p₁ = 250 kPa) into a 200 mm pipe at the same elevation. Neglecting losses, what is p₂?

Given

Q = 0.25 m³/s
D₁ = 300 mm, D₂ = 200 mm
p₁ = 250 kPa
ρ = 1,000 kg/m³

Find

p₂

Start with the thinking

Continuity gives both velocities; Bernoulli converts the velocity change to pressure.
Same elevation means the z terms cancel.

Figure for Bernoulli between two pipe sections

Step-by-step solution

Areas
$A_{1} = π (0.30)^{2} /4 = 0.0707 m^{2} an d A_{2} = π (0.20)^{2} /4 = 0.0314 m^{2}$
Velocities
$V_{1} = 0.25/0.0707 = 3.54 m / s an d V_{2} = 0.25/0.0314 = 7.96 m / s$
Bernoulli
$p_{2} = p_{1} + ρ (V_{1}^{2} - V_{2}^{2}) /2$
Velocity terms
$(3.5 4^{2} - 7.9 6^{2}) /2 = (12.5 - 63.4) /2 = - 25.4 m^{2} / s^{2}$
Result
$p_{2} = 250, 000 + 1, 000 (- 25.4) = 224, 600 P a = 225 k P a$

Answer: p₂ ≈ 225 kPa

Why the other options are there

275 kPa (sign of the velocity term reversed)
250 kPa (velocity change ignored)

Reference: FE Reference Handbook — Fluid Mechanics — Energy equation

Example 2

Reynolds number and flow regime

Water at 20 °C (ν = 1.00 × 10⁻⁶ m²/s) flows at 1.8 m/s in a 150 mm pipe. Classify the flow.

Given

V = 1.8 m/s
D = 0.150 m
ν = 1.00 × 10⁻⁶ m²/s

Find

Re and the regime

Start with the thinking

Re > 4,000 is turbulent for pipe flow.
Use the pipe diameter, not the radius.

Step-by-step solution

Reynolds number
$R e = V D / ν$
Substitute
$R e = 1.8 (0.150) /1.00 \times 1 0^{- 6}$
Result
$R e = 2.70 \times 1 0^{5}$
Classification — Re ≫ 4,000, so the flow is fully turbulent

Answer: Re = 2.7 × 10⁵ — turbulent

Why the other options are there

1.35 × 10⁵ (radius used)
2.70 × 10² (viscosity exponent slipped)

Reference: FE Reference Handbook — Fluid Mechanics — Reynolds number

Example 3

Bernoulli between two points in a pipeline — Energy Line (Bernoulli Equation)

Water at point 1 has p₁ = 50 psi and v₁ = 7.0 ft/s. Point 2 sits 26 ft higher with v₂ = 21.5 ft/s. Neglecting losses, find p₂.

Given

p₁ = 50 psi
v₁ = 7.0 ft/s
Δz = 26 ft
v₂ = 21.5 ft/s
γ = 62.4 lb/ft³

Find

p₂

Start with the thinking

Every term must be in feet of head before combining.
Rising elevation and rising velocity both cost pressure.

Step-by-step solution

Energy equation
$p_{1} / γ + v_{1}^{2} /2 g + z_{1} = p_{2} / γ + v_{2}^{2} /2 g + z_{2}$
Pressure head 1
$50 (144) /62.4 = 115.4 f t$
Velocity heads
$v_{1}^{2} /2 g = 0.761 f t, v_{2}^{2} /2 g = 7.178 f t$
Solve
$p_{2} / γ = 115.4 + 0.761 - 7.178 - 26 = 82.97 f t$
Convert
$p_{2} = 82.97 (62.4) /144 = 35.95 p s i$

Answer: p₂ ≈ 36.0 psi

Why the other options are there

24.0 psi (elevation subtracted directly in psi)
50 psi (velocity and elevation ignored)

Reference: FE Reference Handbook — Fluid Mechanics → Energy Line (Bernoulli Equation)

Example 4

Locating the hydraulic grade line and energy grade line on a pipeline — Energy Line (Bernoulli Equation)

A 175.0 mm pipeline carries 0.150 m³/s over 886 m with a friction factor of 0.019. At the upstream gauge the pipe centreline is at elevation 13.0 m and the pressure is 198 kPa. Compute the velocity head, the HGL and EGL elevations at that gauge, the friction loss and the hydraulic gradient.

Given

D = 175.0 mm
Q = 0.150 m³/s
L = 886 m, f = 0.019
z = 13.0 m, p = 198 kPa

Find

Velocity head, HGL, EGL, h_f and S

Start with the thinking

The HGL plots pressure head above the pipe elevation; the EGL sits one velocity head above the HGL.
The hydraulic gradient is the slope of the HGL, which for uniform pipe equals h_f/L.

Step-by-step solution

Velocity
$V = Q / A = 0.150/0.02405 = 6.24 m / s$
Formula
$h_{v} = \frac{V ^{2}}{2 g}$
Substituting
$h_{v} = 6.2 4^{2} / (2 \times 9.81) = 1.982 m$
Formula
$H G L = z + p / γ$
Substituting
$H G L = 13.0 + 198/9.81 = 33.18 m$
EGL
$E G L = H G L + h_{v} = 33.18 + 1.982 = 35.17 m$
Formula
$h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g}$
Substituting
$h_{f} = 0.019 (886/0.175) (1.982) = 190.7 m$
Gradient
$S = h_{f} / L = 190.7/886 = 0.21521 m / m$

Answer: HGL = 33.18 m, EGL = 35.17 m, h_f = 190.7 m, S = 0.21521

Why the other options are there

EGL = 31.20 m (velocity head subtracted)
S = 190.7 (loss not divided by length)

Reference: FE Reference Handbook — Fluid Mechanics → Energy Line (Bernoulli Equation)

Example 5

Bernoulli between two points in a pipeline — Energy Line (Bernoulli Equation) (2)

Water at point 1 has p₁ = 81 psi and v₁ = 7.5 ft/s. Point 2 sits 7 ft higher with v₂ = 15.0 ft/s. Neglecting losses, find p₂.

Given

p₁ = 81 psi
v₁ = 7.5 ft/s
Δz = 7 ft
v₂ = 15.0 ft/s
γ = 62.4 lb/ft³

Find

p₂

Start with the thinking

Every term must be in feet of head before combining.
Rising elevation and rising velocity both cost pressure.

Step-by-step solution

Energy equation
$p_{1} / γ + v_{1}^{2} /2 g + z_{1} = p_{2} / γ + v_{2}^{2} /2 g + z_{2}$
Pressure head 1
$81 (144) /62.4 = 186.9 f t$
Velocity heads
$v_{1}^{2} /2 g = 0.873 f t, v_{2}^{2} /2 g = 3.494 f t$
Solve
$p_{2} / γ = 186.9 + 0.873 - 3.494 - 7 = 177.3 f t$
Convert
$p_{2} = 177.3 (62.4) /144 = 76.83 p s i$

Answer: p₂ ≈ 76.8 psi

Why the other options are there

74.0 psi (elevation subtracted directly in psi)
81 psi (velocity and elevation ignored)

Reference: FE Reference Handbook — Fluid Mechanics → Energy Line (Bernoulli Equation)

Example 6

Locating the hydraulic grade line and energy grade line on a pipeline — Energy Line (Bernoulli Equation) (2)

A 200.0 mm pipeline carries 0.110 m³/s over 708 m with a friction factor of 0.026. At the upstream gauge the pipe centreline is at elevation 21.5 m and the pressure is 181 kPa. Compute the velocity head, the HGL and EGL elevations at that gauge, the friction loss and the hydraulic gradient.

Given

D = 200.0 mm
Q = 0.110 m³/s
L = 708 m, f = 0.026
z = 21.5 m, p = 181 kPa

Find

Velocity head, HGL, EGL, h_f and S

Start with the thinking

The HGL plots pressure head above the pipe elevation; the EGL sits one velocity head above the HGL.
The hydraulic gradient is the slope of the HGL, which for uniform pipe equals h_f/L.

Step-by-step solution

Velocity
$V = Q / A = 0.110/0.03142 = 3.50 m / s$
Formula
$h_{v} = \frac{V ^{2}}{2 g}$
Substituting
$h_{v} = 3.5 0^{2} / (2 \times 9.81) = 0.625 m$
Formula
$H G L = z + p / γ$
Substituting
$H G L = 21.5 + 181/9.81 = 39.95 m$
EGL
$E G L = H G L + h_{v} = 39.95 + 0.625 = 40.58 m$
Formula
$h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g}$
Substituting
$h_{f} = 0.026 (708/0.200) (0.625) = 57.51 m$
Gradient
$S = h_{f} / L = 57.51/708 = 0.08123 m / m$

Answer: HGL = 39.95 m, EGL = 40.58 m, h_f = 57.51 m, S = 0.08123

Why the other options are there

EGL = 39.33 m (velocity head subtracted)
S = 57.5126 (loss not divided by length)

Reference: FE Reference Handbook — Fluid Mechanics → Energy Line (Bernoulli Equation)

Example 7

Bernoulli between two points in a pipeline — Energy Line (Bernoulli Equation) (3)

Water at point 1 has p₁ = 35 psi and v₁ = 6.0 ft/s. Point 2 sits 10 ft higher with v₂ = 8.5 ft/s. Neglecting losses, find p₂.

Given

p₁ = 35 psi
v₁ = 6.0 ft/s
Δz = 10 ft
v₂ = 8.5 ft/s
γ = 62.4 lb/ft³

Find

p₂

Start with the thinking

Every term must be in feet of head before combining.
Rising elevation and rising velocity both cost pressure.

Step-by-step solution

Energy equation
$p_{1} / γ + v_{1}^{2} /2 g + z_{1} = p_{2} / γ + v_{2}^{2} /2 g + z_{2}$
Pressure head 1
$35 (144) /62.4 = 80.77 f t$
Velocity heads
$v_{1}^{2} /2 g = 0.559 f t, v_{2}^{2} /2 g = 1.122 f t$
Solve
$p_{2} / γ = 80.77 + 0.559 - 1.122 - 10 = 70.21 f t$
Convert
$p_{2} = 70.21 (62.4) /144 = 30.42 p s i$

Answer: p₂ ≈ 30.4 psi

Why the other options are there

25.0 psi (elevation subtracted directly in psi)
35 psi (velocity and elevation ignored)

Reference: FE Reference Handbook — Fluid Mechanics → Energy Line (Bernoulli Equation)

Example 8

Locating the hydraulic grade line and energy grade line on a pipeline — Energy Line (Bernoulli Equation) (3)

A 275.0 mm pipeline carries 0.080 m³/s over 720 m with a friction factor of 0.025. At the upstream gauge the pipe centreline is at elevation 19.5 m and the pressure is 141 kPa. Compute the velocity head, the HGL and EGL elevations at that gauge, the friction loss and the hydraulic gradient.

Given

D = 275.0 mm
Q = 0.080 m³/s
L = 720 m, f = 0.025
z = 19.5 m, p = 141 kPa

Find

Velocity head, HGL, EGL, h_f and S

Start with the thinking

The HGL plots pressure head above the pipe elevation; the EGL sits one velocity head above the HGL.
The hydraulic gradient is the slope of the HGL, which for uniform pipe equals h_f/L.

Step-by-step solution

Velocity
$V = Q / A = 0.080/0.05940 = 1.35 m / s$
Formula
$h_{v} = \frac{V ^{2}}{2 g}$
Substituting
$h_{v} = 1.3 5^{2} / (2 \times 9.81) = 0.092 m$
Formula
$H G L = z + p / γ$
Substituting
$H G L = 19.5 + 141/9.81 = 33.87 m$
EGL
$E G L = H G L + h_{v} = 33.87 + 0.092 = 33.97 m$
Formula
$h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g}$
Substituting
$h_{f} = 0.025 (720/0.275) (0.092) = 6.05 m$
Gradient
$S = h_{f} / L = 6.05/720 = 0.00841 m / m$

Answer: HGL = 33.87 m, EGL = 33.97 m, h_f = 6.05 m, S = 0.00841

Why the other options are there

EGL = 33.78 m (velocity head subtracted)
S = 6.0522 (loss not divided by length)

Reference: FE Reference Handbook — Fluid Mechanics → Energy Line (Bernoulli Equation)

Example 9

Bernoulli between two points in a pipeline — Energy Line (Bernoulli Equation) (4)

Water at point 1 has p₁ = 36 psi and v₁ = 5.5 ft/s. Point 2 sits 13 ft higher with v₂ = 8.5 ft/s. Neglecting losses, find p₂.

Given

p₁ = 36 psi
v₁ = 5.5 ft/s
Δz = 13 ft
v₂ = 8.5 ft/s
γ = 62.4 lb/ft³

Find

p₂

Start with the thinking

Every term must be in feet of head before combining.
Rising elevation and rising velocity both cost pressure.

Step-by-step solution

Energy equation
$p_{1} / γ + v_{1}^{2} /2 g + z_{1} = p_{2} / γ + v_{2}^{2} /2 g + z_{2}$
Pressure head 1
$36 (144) /62.4 = 83.08 f t$
Velocity heads
$v_{1}^{2} /2 g = 0.470 f t, v_{2}^{2} /2 g = 1.122 f t$
Solve
$p_{2} / γ = 83.08 + 0.470 - 1.122 - 13 = 69.42 f t$
Convert
$p_{2} = 69.42 (62.4) /144 = 30.08 p s i$

Answer: p₂ ≈ 30.1 psi

Why the other options are there

23.0 psi (elevation subtracted directly in psi)
36 psi (velocity and elevation ignored)

Reference: FE Reference Handbook — Fluid Mechanics → Energy Line (Bernoulli Equation)

Example 10

Locating the hydraulic grade line and energy grade line on a pipeline — Energy Line (Bernoulli Equation) (4)

A 325.0 mm pipeline carries 0.045 m³/s over 250 m with a friction factor of 0.016. At the upstream gauge the pipe centreline is at elevation 28.0 m and the pressure is 317 kPa. Compute the velocity head, the HGL and EGL elevations at that gauge, the friction loss and the hydraulic gradient.

Given

D = 325.0 mm
Q = 0.045 m³/s
L = 250 m, f = 0.016
z = 28.0 m, p = 317 kPa

Find

Velocity head, HGL, EGL, h_f and S

Start with the thinking

The HGL plots pressure head above the pipe elevation; the EGL sits one velocity head above the HGL.
The hydraulic gradient is the slope of the HGL, which for uniform pipe equals h_f/L.

Step-by-step solution

Velocity
$V = Q / A = 0.045/0.08296 = 0.54 m / s$
Formula
$h_{v} = \frac{V ^{2}}{2 g}$
Substituting
$h_{v} = 0.5 4^{2} / (2 \times 9.81) = 0.015 m$
Formula
$H G L = z + p / γ$
Substituting
$H G L = 28.0 + 317/9.81 = 60.31 m$
EGL
$E G L = H G L + h_{v} = 60.31 + 0.015 = 60.33 m$
Formula
$h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g}$
Substituting
$h_{f} = 0.016 (250/0.325) (0.015) = 0.18 m$
Gradient
$S = h_{f} / L = 0.18/250 = 0.00074 m / m$

Answer: HGL = 60.31 m, EGL = 60.33 m, h_f = 0.18 m, S = 0.00074

Why the other options are there

EGL = 60.30 m (velocity head subtracted)
S = 0.1846 (loss not divided by length)

Reference: FE Reference Handbook — Fluid Mechanics → Energy Line (Bernoulli 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

Energy Line (Bernoulli Equation) contains 0 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.