Hydraulic Gradient (Grade Line)

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 Hydraulic Gradient (Grade Line) 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 hydraulic gradient (grade line) 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. Hydraulic Gradient (Grade Line) 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.

Row of centrifugal pumps and valved steel piping inside a water pumping station. — Photo 1. Where this shows up in practice: hydraulic gradient (grade line).
Capstone Studio instructional photograph

Fluid Mechanics — Hydraulic Gradient (Grade Line): 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.

Hydraulic grade line is the line connecting the sum of pressure and elevation heads at different points in conveyance systems.
If a row of piezometers were placed at intervals along the pipe, the grade line would join the water levels in the piezometer
water columns.

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

Locating the hydraulic grade line and energy grade line on a pipeline — Hydraulic Gradient (Grade Line)

A 175.0 mm pipeline carries 0.130 m³/s over 550 m with a friction factor of 0.023. At the upstream gauge the pipe centreline is at elevation 20.0 m and the pressure is 385 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 = 0.130 m³/s
L = 550 m, f = 0.023
z = 20.0 m, p = 385 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
$V = Q / A = 0.130/0.02405 = 5.40 m / s$
Formula
$h_{v} = \frac{V ^{2}}{2 g}$
Substituting
$h_{v} = 5.4 0^{2} / (2 \times 9.81) = 1.489 m$
Formula
$H G L = z + p / γ$
Substituting
$H G L = 20.0 + 385/9.81 = 59.25 m$
EGL
$E G L = H G L + h_{v} = 59.25 + 1.489 = 60.73 m$
Formula
$h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g}$
Substituting
$h_{f} = 0.023 (550/0.175) (1.489) = 107.6 m$
Gradient
$S = h_{f} / L = 107.6/550 = 0.19568 m / m$

Answer: HGL = 59.25 m, EGL = 60.73 m, h_f = 107.6 m, S = 0.19568

Why the other options are there

EGL = 57.76 m (velocity head subtracted)
S = 107.6 (loss not divided by length)

Reference: FE Reference Handbook — Fluid Mechanics → Hydraulic Gradient (Grade Line)

Example 2

Locating the hydraulic grade line and energy grade line on a pipeline — Hydraulic Gradient (Grade Line) (2)

A 175.0 mm pipeline carries 0.055 m³/s over 883 m with a friction factor of 0.023. At the upstream gauge the pipe centreline is at elevation 19.5 m and the pressure is 275 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 = 0.055 m³/s
L = 883 m, f = 0.023
z = 19.5 m, p = 275 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
$V = Q / A = 0.055/0.02405 = 2.29 m / s$
Formula
$h_{v} = \frac{V ^{2}}{2 g}$
Substituting
$h_{v} = 2.2 9^{2} / (2 \times 9.81) = 0.266 m$
Formula
$H G L = z + p / γ$
Substituting
$H G L = 19.5 + 275/9.81 = 47.53 m$
EGL
$E G L = H G L + h_{v} = 47.53 + 0.266 = 47.80 m$
Formula
$h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g}$
Substituting
$h_{f} = 0.023 (883/0.175) (0.266) = 30.93 m$
Gradient
$S = h_{f} / L = 30.93/883 = 0.03503 m / m$

Answer: HGL = 47.53 m, EGL = 47.80 m, h_f = 30.93 m, S = 0.03503

Why the other options are there

EGL = 47.27 m (velocity head subtracted)
S = 30.9275 (loss not divided by length)

Reference: FE Reference Handbook — Fluid Mechanics → Hydraulic Gradient (Grade Line)

Example 3

Locating the hydraulic grade line and energy grade line on a pipeline — Hydraulic Gradient (Grade Line) (3)

A 225.0 mm pipeline carries 0.095 m³/s over 551 m with a friction factor of 0.024. At the upstream gauge the pipe centreline is at elevation 27.5 m and the pressure is 237 kPa. Compute the velocity head, the HGL and EGL elevations at that gauge, the friction loss and the hydraulic gradient.

Given

D = 225.0 mm
Q = 0.095 m³/s
L = 551 m, f = 0.024
z = 27.5 m, p = 237 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
$V = Q / A = 0.095/0.03976 = 2.39 m / s$
Formula
$h_{v} = \frac{V ^{2}}{2 g}$
Substituting
$h_{v} = 2.3 9^{2} / (2 \times 9.81) = 0.291 m$
Formula
$H G L = z + p / γ$
Substituting
$H G L = 27.5 + 237/9.81 = 51.66 m$
EGL
$E G L = H G L + h_{v} = 51.66 + 0.291 = 51.95 m$
Formula
$h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g}$
Substituting
$h_{f} = 0.024 (551/0.225) (0.291) = 17.10 m$
Gradient
$S = h_{f} / L = 17.10/551 = 0.03104 m / m$

Answer: HGL = 51.66 m, EGL = 51.95 m, h_f = 17.10 m, S = 0.03104

Why the other options are there

EGL = 51.37 m (velocity head subtracted)
S = 17.1009 (loss not divided by length)

Reference: FE Reference Handbook — Fluid Mechanics → Hydraulic Gradient (Grade Line)

Example 4

Locating the hydraulic grade line and energy grade line on a pipeline — Hydraulic Gradient (Grade Line) (4)

A 275.0 mm pipeline carries 0.040 m³/s over 186 m with a friction factor of 0.027. At the upstream gauge the pipe centreline is at elevation 23.0 m and the pressure is 260 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 = 0.040 m³/s
L = 186 m, f = 0.027
z = 23.0 m, p = 260 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
$V = Q / A = 0.040/0.05940 = 0.67 m / s$
Formula
$h_{v} = \frac{V ^{2}}{2 g}$
Substituting
$h_{v} = 0.6 7^{2} / (2 \times 9.81) = 0.023 m$
Formula
$H G L = z + p / γ$
Substituting
$H G L = 23.0 + 260/9.81 = 49.50 m$
EGL
$E G L = H G L + h_{v} = 49.50 + 0.023 = 49.53 m$
Formula
$h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g}$
Substituting
$h_{f} = 0.027 (186/0.275) (0.023) = 0.42 m$
Gradient
$S = h_{f} / L = 0.42/186 = 0.00227 m / m$

Answer: HGL = 49.50 m, EGL = 49.53 m, h_f = 0.42 m, S = 0.00227

Why the other options are there

EGL = 49.48 m (velocity head subtracted)
S = 0.4221 (loss not divided by length)

Reference: FE Reference Handbook — Fluid Mechanics → Hydraulic Gradient (Grade Line)

Example 5

Locating the hydraulic grade line and energy grade line on a pipeline — Hydraulic Gradient (Grade Line) (5)

A 300.0 mm pipeline carries 0.085 m³/s over 686 m with a friction factor of 0.022. At the upstream gauge the pipe centreline is at elevation 17.0 m and the pressure is 280 kPa. Compute the velocity head, the HGL and EGL elevations at that gauge, the friction loss and the hydraulic gradient.

Given

D = 300.0 mm
Q = 0.085 m³/s
L = 686 m, f = 0.022
z = 17.0 m, p = 280 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
$V = Q / A = 0.085/0.07069 = 1.20 m / s$
Formula
$h_{v} = \frac{V ^{2}}{2 g}$
Substituting
$h_{v} = 1.2 0^{2} / (2 \times 9.81) = 0.074 m$
Formula
$H G L = z + p / γ$
Substituting
$H G L = 17.0 + 280/9.81 = 45.54 m$
EGL
$E G L = H G L + h_{v} = 45.54 + 0.074 = 45.62 m$
Formula
$h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g}$
Substituting
$h_{f} = 0.022 (686/0.300) (0.074) = 3.71 m$
Gradient
$S = h_{f} / L = 3.71/686 = 0.00540 m / m$

Answer: HGL = 45.54 m, EGL = 45.62 m, h_f = 3.71 m, S = 0.00540

Why the other options are there

EGL = 45.47 m (velocity head subtracted)
S = 3.7077 (loss not divided by length)

Reference: FE Reference Handbook — Fluid Mechanics → Hydraulic Gradient (Grade Line)

Example 6

Locating the hydraulic grade line and energy grade line on a pipeline — Hydraulic Gradient (Grade Line) (6)

A 200.0 mm pipeline carries 0.065 m³/s over 665 m with a friction factor of 0.017. At the upstream gauge the pipe centreline is at elevation 14.5 m and the pressure is 296 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 = 0.065 m³/s
L = 665 m, f = 0.017
z = 14.5 m, p = 296 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
$V = Q / A = 0.065/0.03142 = 2.07 m / s$
Formula
$h_{v} = \frac{V ^{2}}{2 g}$
Substituting
$h_{v} = 2.0 7^{2} / (2 \times 9.81) = 0.218 m$
Formula
$H G L = z + p / γ$
Substituting
$H G L = 14.5 + 296/9.81 = 44.67 m$
EGL
$E G L = H G L + h_{v} = 44.67 + 0.218 = 44.89 m$
Formula
$h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g}$
Substituting
$h_{f} = 0.017 (665/0.200) (0.218) = 12.33 m$
Gradient
$S = h_{f} / L = 12.33/665 = 0.01855 m / m$

Answer: HGL = 44.67 m, EGL = 44.89 m, h_f = 12.33 m, S = 0.01855

Why the other options are there

EGL = 44.46 m (velocity head subtracted)
S = 12.3330 (loss not divided by length)

Reference: FE Reference Handbook — Fluid Mechanics → Hydraulic Gradient (Grade Line)

Example 7

Locating the hydraulic grade line and energy grade line on a pipeline — Hydraulic Gradient (Grade Line) (7)

A 325.0 mm pipeline carries 0.140 m³/s over 882 m with a friction factor of 0.023. At the upstream gauge the pipe centreline is at elevation 22.5 m and the pressure is 391 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 = 0.140 m³/s
L = 882 m, f = 0.023
z = 22.5 m, p = 391 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
$V = Q / A = 0.140/0.08296 = 1.69 m / s$
Formula
$h_{v} = \frac{V ^{2}}{2 g}$
Substituting
$h_{v} = 1.6 9^{2} / (2 \times 9.81) = 0.145 m$
Formula
$H G L = z + p / γ$
Substituting
$H G L = 22.5 + 391/9.81 = 62.36 m$
EGL
$E G L = H G L + h_{v} = 62.36 + 0.145 = 62.50 m$
Formula
$h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g}$
Substituting
$h_{f} = 0.023 (882/0.325) (0.145) = 9.06 m$
Gradient
$S = h_{f} / L = 9.06/882 = 0.01027 m / m$

Answer: HGL = 62.36 m, EGL = 62.50 m, h_f = 9.06 m, S = 0.01027

Why the other options are there

EGL = 62.21 m (velocity head subtracted)
S = 9.0606 (loss not divided by length)

Reference: FE Reference Handbook — Fluid Mechanics → Hydraulic Gradient (Grade Line)

Example 8

Locating the hydraulic grade line and energy grade line on a pipeline — Hydraulic Gradient (Grade Line) (8)

A 300.0 mm pipeline carries 0.085 m³/s over 827 m with a friction factor of 0.018. At the upstream gauge the pipe centreline is at elevation 17.0 m and the pressure is 243 kPa. Compute the velocity head, the HGL and EGL elevations at that gauge, the friction loss and the hydraulic gradient.

Given

D = 300.0 mm
Q = 0.085 m³/s
L = 827 m, f = 0.018
z = 17.0 m, p = 243 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
$V = Q / A = 0.085/0.07069 = 1.20 m / s$
Formula
$h_{v} = \frac{V ^{2}}{2 g}$
Substituting
$h_{v} = 1.2 0^{2} / (2 \times 9.81) = 0.074 m$
Formula
$H G L = z + p / γ$
Substituting
$H G L = 17.0 + 243/9.81 = 41.77 m$
EGL
$E G L = H G L + h_{v} = 41.77 + 0.074 = 41.84 m$
Formula
$h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g}$
Substituting
$h_{f} = 0.018 (827/0.300) (0.074) = 3.66 m$
Gradient
$S = h_{f} / L = 3.66/827 = 0.00442 m / m$

Answer: HGL = 41.77 m, EGL = 41.84 m, h_f = 3.66 m, S = 0.00442

Why the other options are there

EGL = 41.70 m (velocity head subtracted)
S = 3.6570 (loss not divided by length)

Reference: FE Reference Handbook — Fluid Mechanics → Hydraulic Gradient (Grade Line)

Example 9

Locating the hydraulic grade line and energy grade line on a pipeline — Hydraulic Gradient (Grade Line) (9)

A 325.0 mm pipeline carries 0.080 m³/s over 573 m with a friction factor of 0.018. At the upstream gauge the pipe centreline is at elevation 25.0 m and the pressure is 336 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 = 0.080 m³/s
L = 573 m, f = 0.018
z = 25.0 m, p = 336 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
$V = Q / A = 0.080/0.08296 = 0.96 m / s$
Formula
$h_{v} = \frac{V ^{2}}{2 g}$
Substituting
$h_{v} = 0.9 6^{2} / (2 \times 9.81) = 0.047 m$
Formula
$H G L = z + p / γ$
Substituting
$H G L = 25.0 + 336/9.81 = 59.25 m$
EGL
$E G L = H G L + h_{v} = 59.25 + 0.047 = 59.30 m$
Formula
$h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g}$
Substituting
$h_{f} = 0.018 (573/0.325) (0.047) = 1.50 m$
Gradient
$S = h_{f} / L = 1.50/573 = 0.00263 m / m$

Answer: HGL = 59.25 m, EGL = 59.30 m, h_f = 1.50 m, S = 0.00263

Why the other options are there

EGL = 59.20 m (velocity head subtracted)
S = 1.5042 (loss not divided by length)

Reference: FE Reference Handbook — Fluid Mechanics → Hydraulic Gradient (Grade Line)

Example 10

Locating the hydraulic grade line and energy grade line on a pipeline — Hydraulic Gradient (Grade Line) (10)

A 225.0 mm pipeline carries 0.145 m³/s over 899 m with a friction factor of 0.020. At the upstream gauge the pipe centreline is at elevation 14.5 m and the pressure is 206 kPa. Compute the velocity head, the HGL and EGL elevations at that gauge, the friction loss and the hydraulic gradient.

Given

D = 225.0 mm
Q = 0.145 m³/s
L = 899 m, f = 0.020
z = 14.5 m, p = 206 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
$V = Q / A = 0.145/0.03976 = 3.65 m / s$
Formula
$h_{v} = \frac{V ^{2}}{2 g}$
Substituting
$h_{v} = 3.6 5^{2} / (2 \times 9.81) = 0.678 m$
Formula
$H G L = z + p / γ$
Substituting
$H G L = 14.5 + 206/9.81 = 35.50 m$
EGL
$E G L = H G L + h_{v} = 35.50 + 0.678 = 36.18 m$
Formula
$h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g}$
Substituting
$h_{f} = 0.020 (899/0.225) (0.678) = 54.17 m$
Gradient
$S = h_{f} / L = 54.17/899 = 0.06025 m / m$

Answer: HGL = 35.50 m, EGL = 36.18 m, h_f = 54.17 m, S = 0.06025

Why the other options are there

EGL = 34.82 m (velocity head subtracted)
S = 54.1669 (loss not divided by length)

Reference: FE Reference Handbook — Fluid Mechanics → Hydraulic Gradient (Grade Line)

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

Hydraulic Gradient (Grade Line) 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.