Impulse Turbine

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 Impulse Turbine 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 impulse turbine 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. Impulse Turbine 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: impulse turbine.
Capstone Studio instructional photograph

Fluid Mechanics — Impulse Turbine: 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

⋅ v	Quantity produced by "⋅ v" — read its definition and unit from the handbook line directly above the equation.
⋅	Quantity produced by "⋅" — read its definition and unit from the handbook line directly above the equation.
v α	Quantity produced by "v α" — read its definition and unit from the handbook line directly above the equation.
Wo	Quantity produced by "Wo = Qρ _v1 − v i_1 − cos α i v" — read its definition and unit from the handbook line directly above the equation.
where Womax	Quantity produced by "where Womax = Qρ ` v12 /4 j_1 − cos α i" — read its definition and unit from the handbook line directly above the equation.
When α	Quantity produced by "When α = 180c," — read its definition and unit from the handbook line directly above the equation.
Womax	Quantity produced by "Womax = `Qρv12 j /2 = `Qγv12 j /2g" — 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.

v v v v
v v v
Vennard, J.K., Elementary Fluid Mechanics, 6th ed., J.K. Vennard, 1954.

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

Pump brake power

A pump delivers 0.06 m³/s against a total head of 32 m at 74% efficiency. What brake power is required?

Given

Q = 0.06 m³/s
H = 32 m
η = 0.74
γ = 9.81 kN/m³

Find

Brake power (kW)

Start with the thinking

Water power first, then divide by efficiency.
Dividing by efficiency increases the power — a common sign check.

Step-by-step solution

Water power — P_w = γQH = 9.81(0.06)(32)
Evaluate
$P_{w} = 18.8 k W$
Brake power
$P = P_{w} / η = 18.8/0.74$
Result
$P = 25.5 k W$

Answer: P ≈ 25.5 kW

Why the other options are there

13.9 kW (multiplied by efficiency)
18.8 kW (efficiency ignored)

Reference: FE Reference Handbook — Fluid Mechanics — Pump power

Example 2

Pump brake horsepower — Impulse Turbine

A pump delivers 12.0 cfs against 35 ft of head at 74% efficiency. Find the water and brake horsepower.

Given

Q = 12.0 cfs
H = 35 ft
η = 0.74

Find

WHP and BHP

Start with the thinking

Water horsepower is γQH/550; efficiency only affects input power.
Efficiency divides — the motor always draws more.

Step-by-step solution

Water power — WHP = γQH/550
Substituting
$W H P = 62.4 (12.0) (35) /550 = 47.65 h p$
Brake power
$B H P = W H P / η$
Substituting
$B H P = 47.65/0.74 = 64.39 h p$

Answer: WHP ≈ 47.7 hp; BHP ≈ 64.4 hp

Why the other options are there

35.3 hp (efficiency multiplied)
0.76 hp (specific weight omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Impulse Turbine

Example 3

Water horsepower, shaft power and pump operating cost — Impulse Turbine

A pump delivers 0.300 m³/s against a total dynamic head of 17 m at 84% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 13 h/day and $0.16/kWh.

Given

Q = 0.300 m³/s
H = 17 m
η = 0.84
13 h/day at $0.16/kWh

Find

Water power, shaft power and annual cost

Start with the thinking

Water power is the useful hydraulic output; shaft power is larger because efficiency is below one.
Never multiply by efficiency when moving from water power to shaft power — divide.

Step-by-step solution

Formula
$P_{w} a t er = ρ g Q H$
Substituting
$P = 1000 (9.81) (0.300) (17) = 50.03 k W$
Formula
$P_{s} ha f t = P_{w} a t er / η$
Substituting
$P_{s} ha f t = 50.03/0.84 = 59.56 k W$
Annual energy
$59.56 k W \times 13 h / d a y \times 365 = 282, 616 k W h$
Cost — 282,616 kWh × $0.16 = $45,218 per year

Answer: P_water = 50.0 kW, P_shaft = 59.6 kW, cost ≈ $45,218/yr

Why the other options are there

42.0 kW (efficiency multiplied)
67.1 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Impulse Turbine

Example 4

Pump brake horsepower — Impulse Turbine (2)

A pump delivers 10.0 cfs against 92 ft of head at 74% efficiency. Find the water and brake horsepower.

Given

Q = 10.0 cfs
H = 92 ft
η = 0.74

Find

WHP and BHP

Start with the thinking

Water horsepower is γQH/550; efficiency only affects input power.
Efficiency divides — the motor always draws more.

Step-by-step solution

Water power — WHP = γQH/550
Substituting
$W H P = 62.4 (10.0) (92) /550 = 104.4 h p$
Brake power
$B H P = W H P / η$
Substituting
$B H P = 104.4/0.74 = 141.1 h p$

Answer: WHP ≈ 104.4 hp; BHP ≈ 141.1 hp

Why the other options are there

77.2 hp (efficiency multiplied)
1.67 hp (specific weight omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Impulse Turbine

Example 5

Water horsepower, shaft power and pump operating cost — Impulse Turbine (2)

A pump delivers 0.040 m³/s against a total dynamic head of 17 m at 68% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 16 h/day and $0.14/kWh.

Given

Q = 0.040 m³/s
H = 17 m
η = 0.68
16 h/day at $0.14/kWh

Find

Water power, shaft power and annual cost

Start with the thinking

Water power is the useful hydraulic output; shaft power is larger because efficiency is below one.
Never multiply by efficiency when moving from water power to shaft power — divide.

Step-by-step solution

Formula
$P_{w} a t er = ρ g Q H$
Substituting
$P = 1000 (9.81) (0.040) (17) = 6.67 k W$
Formula
$P_{s} ha f t = P_{w} a t er / η$
Substituting
$P_{s} ha f t = 6.67/0.68 = 9.81 k W$
Annual energy
$9.81 k W \times 16 h / d a y \times 365 = 57, 290 k W h$
Cost — 57,290 kWh × $0.14 = $8,021 per year

Answer: P_water = 6.7 kW, P_shaft = 9.8 kW, cost ≈ $8,021/yr

Why the other options are there

4.5 kW (efficiency multiplied)
8.9 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Impulse Turbine

Example 6

Pump brake horsepower — Impulse Turbine (3)

A pump delivers 4.0 cfs against 195 ft of head at 62% efficiency. Find the water and brake horsepower.

Given

Q = 4.0 cfs
H = 195 ft
η = 0.62

Find

WHP and BHP

Start with the thinking

Water horsepower is γQH/550; efficiency only affects input power.
Efficiency divides — the motor always draws more.

Step-by-step solution

Water power — WHP = γQH/550
Substituting
$W H P = 62.4 (4.0) (195) /550 = 88.49 h p$
Brake power
$B H P = W H P / η$
Substituting
$B H P = 88.49/0.62 = 142.7 h p$

Answer: WHP ≈ 88.5 hp; BHP ≈ 142.7 hp

Why the other options are there

54.9 hp (efficiency multiplied)
1.42 hp (specific weight omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Impulse Turbine

Example 7

Water horsepower, shaft power and pump operating cost — Impulse Turbine (3)

A pump delivers 0.240 m³/s against a total dynamic head of 34 m at 74% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 7 h/day and $0.10/kWh.

Given

Q = 0.240 m³/s
H = 34 m
η = 0.74
7 h/day at $0.10/kWh

Find

Water power, shaft power and annual cost

Start with the thinking

Water power is the useful hydraulic output; shaft power is larger because efficiency is below one.
Never multiply by efficiency when moving from water power to shaft power — divide.

Step-by-step solution

Formula
$P_{w} a t er = ρ g Q H$
Substituting
$P = 1000 (9.81) (0.240) (34) = 80.05 k W$
Formula
$P_{s} ha f t = P_{w} a t er / η$
Substituting
$P_{s} ha f t = 80.05/0.74 = 108.2 k W$
Annual energy
$108.2 k W \times 7 h / d a y \times 365 = 276, 387 k W h$
Cost — 276,387 kWh × $0.10 = $27,639 per year

Answer: P_water = 80.0 kW, P_shaft = 108.2 kW, cost ≈ $27,639/yr

Why the other options are there

59.2 kW (efficiency multiplied)
107.3 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Impulse Turbine

Example 8

Pump brake horsepower — Impulse Turbine (4)

A pump delivers 15.0 cfs against 148 ft of head at 60% efficiency. Find the water and brake horsepower.

Given

Q = 15.0 cfs
H = 148 ft
η = 0.60

Find

WHP and BHP

Start with the thinking

Water horsepower is γQH/550; efficiency only affects input power.
Efficiency divides — the motor always draws more.

Step-by-step solution

Water power — WHP = γQH/550
Substituting
$W H P = 62.4 (15.0) (148) /550 = 251.9 h p$
Brake power
$B H P = W H P / η$
Substituting
$B H P = 251.9/0.60 = 419.8 h p$

Answer: WHP ≈ 251.9 hp; BHP ≈ 419.8 hp

Why the other options are there

151.1 hp (efficiency multiplied)
4.04 hp (specific weight omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Impulse Turbine

Example 9

Water horsepower, shaft power and pump operating cost — Impulse Turbine (4)

A pump delivers 0.360 m³/s against a total dynamic head of 28 m at 84% efficiency. Compute the water power and the shaft power required, then estimate the annual energy cost at 10 h/day and $0.09/kWh.

Given

Q = 0.360 m³/s
H = 28 m
η = 0.84
10 h/day at $0.09/kWh

Find

Water power, shaft power and annual cost

Start with the thinking

Water power is the useful hydraulic output; shaft power is larger because efficiency is below one.
Never multiply by efficiency when moving from water power to shaft power — divide.

Step-by-step solution

Formula
$P_{w} a t er = ρ g Q H$
Substituting
$P = 1000 (9.81) (0.360) (28) = 98.88 k W$
Formula
$P_{s} ha f t = P_{w} a t er / η$
Substituting
$P_{s} ha f t = 98.88/0.84 = 117.7 k W$
Annual energy
$117.7 k W \times 10 h / d a y \times 365 = 429, 678 k W h$
Cost — 429,678 kWh × $0.09 = $38,671 per year

Answer: P_water = 98.9 kW, P_shaft = 117.7 kW, cost ≈ $38,671/yr

Why the other options are there

83.1 kW (efficiency multiplied)
132.6 hp (unit mix-up)

Reference: FE Reference Handbook — Fluid Mechanics → Impulse Turbine

Example 10

Pump brake horsepower — Impulse Turbine (5)

A pump delivers 7.5 cfs against 98 ft of head at 86% efficiency. Find the water and brake horsepower.

Given

Q = 7.5 cfs
H = 98 ft
η = 0.86

Find

WHP and BHP

Start with the thinking

Water horsepower is γQH/550; efficiency only affects input power.
Efficiency divides — the motor always draws more.

Step-by-step solution

Water power — WHP = γQH/550
Substituting
$W H P = 62.4 (7.5) (98) /550 = 83.39 h p$
Brake power
$B H P = W H P / η$
Substituting
$B H P = 83.39/0.86 = 96.96 h p$

Answer: WHP ≈ 83.4 hp; BHP ≈ 97.0 hp

Why the other options are there

71.7 hp (efficiency multiplied)
1.34 hp (specific weight omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Impulse Turbine

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

Impulse Turbine 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.

Impulse Turbine

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

⋅v

⋅

vα

Wo = Qρ _v1 − v i_1 − cos α i vNeeds technical review

Wo=poweroftheturbine.

whereWomax=Qρ‘v12/4j1​−cosαi

Whenα=180c,

Womax = `Qρv12 j /2 = `Qγv12 j /2gNeeds technical review

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section