Net Positive Suction Head Available (NPSHA)

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

15 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 Net Positive Suction Head Available (NPSHA) 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 net positive suction head available (npsha) describes physically and when it applies.
State every one of the 15 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. Net Positive Suction Head Available (NPSHA) 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: net positive suction head available (npsha).
Capstone Studio instructional photograph

Fluid Mechanics — Net Positive Suction Head Available (NPSHA): 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 15 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

NPSHA	Quantity produced by "NPSHA = Hpa + Hs - / hL - Hvp = tinlet" — read its definition and unit from the handbook line directly above the equation.
Hpa	Quantity produced by "Hpa = atmospheric pressure head on the surface of the liquid in the sump (ft or m)" — read its definition and unit from the handbook line directly above the equation.
Hs	Quantity produced by "Hs = static suction head of liquid. This is the height of the surface of the liquid above the centerline of the pump" — read its definition and unit from the handbook line directly above the equation.
ΣhL	Quantity produced by "ΣhL = total friction losses in the suction line (ft or m)" — read its definition and unit from the handbook line directly above the equation.
Hvp	Quantity produced by "Hvp = vapor pressure head of the liquid at the operating temperature (ft or m)" — read its definition and unit from the handbook line directly above the equation.
v	Quantity produced by "v = fluid velocity at pump inlet" — read its definition and unit from the handbook line directly above the equation.
Pvapor	Quantity produced by "Pvapor = fluid vapor pressure at pump inlet" — 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.
g	Quantity produced by "g = acceleration due to gravity" — read its definition and unit from the handbook line directly above the equation.
ηpump	Quantity produced by "ηpump = pump efficiency ^0 to 1 h" — read its definition and unit from the handbook line directly above the equation.
ηmotor	Quantity produced by "ηmotor = motor efficiency ^0 to 1 h" — read its definition and unit from the handbook line directly above the equation.
H	Quantity produced by "H = head increase provided by pump" — 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.

P 2
v inlet Pvapor
g + 2g - tg
where
impeller (ft or m).
tgHQ
pump
where

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

Net positive suction head available at a pump inlet — Net Positive Suction Head Available (NPSHA)

A pump draws water with a static lift of -2.5 m (positive = pump above the free surface) through suction piping that loses 2.3 m. Atmospheric pressure is 97.5 kPa and the vapour pressure is 5.2 kPa. Compute NPSHA and check it against the manufacturer's NPSHR of 3.5 m.

Given

p_atm = 97.5 kPa, p_v = 5.2 kPa
z_s = -2.5 m
h_L = 2.3 m
NPSHR = 3.5 m

Find

NPSHA and the cavitation check

Start with the thinking

NPSHA is an absolute-pressure calculation: convert both pressures to head using γ = 9.81 kN/m³.
Cavitation is avoided only when NPSHA exceeds NPSHR, usually with a 0.5–1 m margin.

Step-by-step solution

Formula
$N P S H A = \frac{p _{a t m} - p _{v}}{γ} - z_{s} - h_{L}$
Pressure head
$(97.5 - 5.2) /9.81 = 9.41 m$
Substituting
$N P S H A = 9.41 - - 2.5 - 2.3 = 9.61 m$
Check — NPSHA > NPSHR (3.5 m) → no cavitation expected

Answer: NPSHA = 9.61 m → acceptable

Why the other options are there

10.67 m (vapour pressure added)
4.61 m (lift sign reversed)

Reference: FE Reference Handbook — Fluid Mechanics → Net Positive Suction Head Available (NPSHA)

Example 3

Net positive suction head available at a pump inlet — Net Positive Suction Head Available (NPSHA) (2)

A pump draws water with a static lift of 1.5 m (positive = pump above the free surface) through suction piping that loses 2.5 m. Atmospheric pressure is 97.0 kPa and the vapour pressure is 5.5 kPa. Compute NPSHA and check it against the manufacturer's NPSHR of 4.5 m.

Given

p_atm = 97.0 kPa, p_v = 5.5 kPa
z_s = 1.5 m
h_L = 2.5 m
NPSHR = 4.5 m

Find

NPSHA and the cavitation check

Start with the thinking

NPSHA is an absolute-pressure calculation: convert both pressures to head using γ = 9.81 kN/m³.
Cavitation is avoided only when NPSHA exceeds NPSHR, usually with a 0.5–1 m margin.

Step-by-step solution

Formula
$N P S H A = \frac{p _{a t m} - p _{v}}{γ} - z_{s} - h_{L}$
Pressure head
$(97.0 - 5.5) /9.81 = 9.33 m$
Substituting
$N P S H A = 9.33 - 1.5 - 2.5 = 5.33 m$
Check — NPSHA > NPSHR (4.5 m) → no cavitation expected

Answer: NPSHA = 5.33 m → acceptable

Why the other options are there

6.45 m (vapour pressure added)
8.33 m (lift sign reversed)

Reference: FE Reference Handbook — Fluid Mechanics → Net Positive Suction Head Available (NPSHA)

Example 4

Net positive suction head available at a pump inlet — Net Positive Suction Head Available (NPSHA) (3)

A pump draws water with a static lift of 1.5 m (positive = pump above the free surface) through suction piping that loses 0.4 m. Atmospheric pressure is 100.5 kPa and the vapour pressure is 6.0 kPa. Compute NPSHA and check it against the manufacturer's NPSHR of 2.5 m.

Given

p_atm = 100.5 kPa, p_v = 6.0 kPa
z_s = 1.5 m
h_L = 0.4 m
NPSHR = 2.5 m

Find

NPSHA and the cavitation check

Start with the thinking

NPSHA is an absolute-pressure calculation: convert both pressures to head using γ = 9.81 kN/m³.
Cavitation is avoided only when NPSHA exceeds NPSHR, usually with a 0.5–1 m margin.

Step-by-step solution

Formula
$N P S H A = \frac{p _{a t m} - p _{v}}{γ} - z_{s} - h_{L}$
Pressure head
$(100.5 - 6.0) /9.81 = 9.63 m$
Substituting
$N P S H A = 9.63 - 1.5 - 0.4 = 7.73 m$
Check — NPSHA > NPSHR (2.5 m) → no cavitation expected

Answer: NPSHA = 7.73 m → acceptable

Why the other options are there

8.96 m (vapour pressure added)
10.73 m (lift sign reversed)

Reference: FE Reference Handbook — Fluid Mechanics → Net Positive Suction Head Available (NPSHA)

Example 5

Net positive suction head available at a pump inlet — Net Positive Suction Head Available (NPSHA) (4)

A pump draws water with a static lift of 1.0 m (positive = pump above the free surface) through suction piping that loses 1.3 m. Atmospheric pressure is 96.5 kPa and the vapour pressure is 7.1 kPa. Compute NPSHA and check it against the manufacturer's NPSHR of 2.5 m.

Given

p_atm = 96.5 kPa, p_v = 7.1 kPa
z_s = 1.0 m
h_L = 1.3 m
NPSHR = 2.5 m

Find

NPSHA and the cavitation check

Start with the thinking

NPSHA is an absolute-pressure calculation: convert both pressures to head using γ = 9.81 kN/m³.
Cavitation is avoided only when NPSHA exceeds NPSHR, usually with a 0.5–1 m margin.

Step-by-step solution

Formula
$N P S H A = \frac{p _{a t m} - p _{v}}{γ} - z_{s} - h_{L}$
Pressure head
$(96.5 - 7.1) /9.81 = 9.11 m$
Substituting
$N P S H A = 9.11 - 1.0 - 1.3 = 6.81 m$
Check — NPSHA > NPSHR (2.5 m) → no cavitation expected

Answer: NPSHA = 6.81 m → acceptable

Why the other options are there

8.26 m (vapour pressure added)
8.81 m (lift sign reversed)

Reference: FE Reference Handbook — Fluid Mechanics → Net Positive Suction Head Available (NPSHA)

Example 6

Net positive suction head available at a pump inlet — Net Positive Suction Head Available (NPSHA) (5)

A pump draws water with a static lift of -2.5 m (positive = pump above the free surface) through suction piping that loses 1.0 m. Atmospheric pressure is 96.5 kPa and the vapour pressure is 7.4 kPa. Compute NPSHA and check it against the manufacturer's NPSHR of 3.5 m.

Given

p_atm = 96.5 kPa, p_v = 7.4 kPa
z_s = -2.5 m
h_L = 1.0 m
NPSHR = 3.5 m

Find

NPSHA and the cavitation check

Start with the thinking

NPSHA is an absolute-pressure calculation: convert both pressures to head using γ = 9.81 kN/m³.
Cavitation is avoided only when NPSHA exceeds NPSHR, usually with a 0.5–1 m margin.

Step-by-step solution

Formula
$N P S H A = \frac{p _{a t m} - p _{v}}{γ} - z_{s} - h_{L}$
Pressure head
$(96.5 - 7.4) /9.81 = 9.08 m$
Substituting
$N P S H A = 9.08 - - 2.5 - 1.0 = 10.58 m$
Check — NPSHA > NPSHR (3.5 m) → no cavitation expected

Answer: NPSHA = 10.58 m → acceptable

Why the other options are there

12.09 m (vapour pressure added)
5.58 m (lift sign reversed)

Reference: FE Reference Handbook — Fluid Mechanics → Net Positive Suction Head Available (NPSHA)

Example 7

Net positive suction head available at a pump inlet — Net Positive Suction Head Available (NPSHA) (6)

A pump draws water with a static lift of -1.5 m (positive = pump above the free surface) through suction piping that loses 2.2 m. Atmospheric pressure is 96.5 kPa and the vapour pressure is 3.0 kPa. Compute NPSHA and check it against the manufacturer's NPSHR of 3.0 m.

Given

p_atm = 96.5 kPa, p_v = 3.0 kPa
z_s = -1.5 m
h_L = 2.2 m
NPSHR = 3.0 m

Find

NPSHA and the cavitation check

Start with the thinking

NPSHA is an absolute-pressure calculation: convert both pressures to head using γ = 9.81 kN/m³.
Cavitation is avoided only when NPSHA exceeds NPSHR, usually with a 0.5–1 m margin.

Step-by-step solution

Formula
$N P S H A = \frac{p _{a t m} - p _{v}}{γ} - z_{s} - h_{L}$
Pressure head
$(96.5 - 3.0) /9.81 = 9.53 m$
Substituting
$N P S H A = 9.53 - - 1.5 - 2.2 = 8.83 m$
Check — NPSHA > NPSHR (3.0 m) → no cavitation expected

Answer: NPSHA = 8.83 m → acceptable

Why the other options are there

9.44 m (vapour pressure added)
5.83 m (lift sign reversed)

Reference: FE Reference Handbook — Fluid Mechanics → Net Positive Suction Head Available (NPSHA)

Example 8

Net positive suction head available at a pump inlet — Net Positive Suction Head Available (NPSHA) (7)

A pump draws water with a static lift of 1.0 m (positive = pump above the free surface) through suction piping that loses 2.4 m. Atmospheric pressure is 95.5 kPa and the vapour pressure is 7.3 kPa. Compute NPSHA and check it against the manufacturer's NPSHR of 2.5 m.

Given

p_atm = 95.5 kPa, p_v = 7.3 kPa
z_s = 1.0 m
h_L = 2.4 m
NPSHR = 2.5 m

Find

NPSHA and the cavitation check

Start with the thinking

NPSHA is an absolute-pressure calculation: convert both pressures to head using γ = 9.81 kN/m³.
Cavitation is avoided only when NPSHA exceeds NPSHR, usually with a 0.5–1 m margin.

Step-by-step solution

Formula
$N P S H A = \frac{p _{a t m} - p _{v}}{γ} - z_{s} - h_{L}$
Pressure head
$(95.5 - 7.3) /9.81 = 8.99 m$
Substituting
$N P S H A = 8.99 - 1.0 - 2.4 = 5.59 m$
Check — NPSHA > NPSHR (2.5 m) → no cavitation expected

Answer: NPSHA = 5.59 m → acceptable

Why the other options are there

7.08 m (vapour pressure added)
7.59 m (lift sign reversed)

Reference: FE Reference Handbook — Fluid Mechanics → Net Positive Suction Head Available (NPSHA)

Example 9

Net positive suction head available at a pump inlet — Net Positive Suction Head Available (NPSHA) (8)

A pump draws water with a static lift of 2.0 m (positive = pump above the free surface) through suction piping that loses 0.8 m. Atmospheric pressure is 100.0 kPa and the vapour pressure is 3.2 kPa. Compute NPSHA and check it against the manufacturer's NPSHR of 3.0 m.

Given

p_atm = 100.0 kPa, p_v = 3.2 kPa
z_s = 2.0 m
h_L = 0.8 m
NPSHR = 3.0 m

Find

NPSHA and the cavitation check

Start with the thinking

NPSHA is an absolute-pressure calculation: convert both pressures to head using γ = 9.81 kN/m³.
Cavitation is avoided only when NPSHA exceeds NPSHR, usually with a 0.5–1 m margin.

Step-by-step solution

Formula
$N P S H A = \frac{p _{a t m} - p _{v}}{γ} - z_{s} - h_{L}$
Pressure head
$(100.0 - 3.2) /9.81 = 9.87 m$
Substituting
$N P S H A = 9.87 - 2.0 - 0.8 = 7.07 m$
Check — NPSHA > NPSHR (3.0 m) → no cavitation expected

Answer: NPSHA = 7.07 m → acceptable

Why the other options are there

7.72 m (vapour pressure added)
11.07 m (lift sign reversed)

Reference: FE Reference Handbook — Fluid Mechanics → Net Positive Suction Head Available (NPSHA)

Example 10

Net positive suction head available at a pump inlet — Net Positive Suction Head Available (NPSHA) (9)

A pump draws water with a static lift of -1.0 m (positive = pump above the free surface) through suction piping that loses 1.5 m. Atmospheric pressure is 100.0 kPa and the vapour pressure is 5.5 kPa. Compute NPSHA and check it against the manufacturer's NPSHR of 4.0 m.

Given

p_atm = 100.0 kPa, p_v = 5.5 kPa
z_s = -1.0 m
h_L = 1.5 m
NPSHR = 4.0 m

Find

NPSHA and the cavitation check

Start with the thinking

NPSHA is an absolute-pressure calculation: convert both pressures to head using γ = 9.81 kN/m³.
Cavitation is avoided only when NPSHA exceeds NPSHR, usually with a 0.5–1 m margin.

Step-by-step solution

Formula
$N P S H A = \frac{p _{a t m} - p _{v}}{γ} - z_{s} - h_{L}$
Pressure head
$(100.0 - 5.5) /9.81 = 9.63 m$
Substituting
$N P S H A = 9.63 - - 1.0 - 1.5 = 9.13 m$
Check — NPSHA > NPSHR (4.0 m) → no cavitation expected

Answer: NPSHA = 9.13 m → acceptable

Why the other options are there

10.25 m (vapour pressure added)
7.13 m (lift sign reversed)

Reference: FE Reference Handbook — Fluid Mechanics → Net Positive Suction Head Available (NPSHA)

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

Net Positive Suction Head Available (NPSHA) contains 15 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.