Orifices

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

3 formulas

10 exam-style examples

~51 min

All Fluid Mechanics lectures

Learning objectives

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

This chapter section covers Orifices 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 orifices describes physically and when it applies.
State every one of the 3 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. Orifices 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: orifices.
Capstone Studio instructional photograph

Fluid Mechanics — Orifices: 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 3 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

Q	Quantity produced by "Q = CA0" — 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.

The cross-sectional area at the vena contracta A2 is characterized by a coefficient of contraction Cc and given by Cc A0.
A1 A0 A2
Vennard, J.K., Elementary Fluid Mechanics, 6th ed., J.K. Vennard, 1954.
2g d c1 + z1 − c2 − z2 n
P P
where C, the coefficient of the meter (orifice coefficient), is given by
CvCc
1 - Cc2 _ A0 A1i
Bober, W., and R.A. Kenyon, Fluid Mechanics, Wiley, 1980. Diagrams reprinted by permission of William Bober and Richard A. Kenyon.
For incompressible flow through a horizontal orifice meter installation
t _ P1 − P 2 i

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

Discharge from a sharp-edged orifice venting to atmosphere — Orifices

A 75 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 2.7 m. With a discharge coefficient of 0.60, compute the theoretical velocity, the actual discharge and the jet's volume delivered in one minute.

Given

d = 75 mm
H = 2.7 m
C_d = 0.60

Find

V_theoretical, Q and the one-minute volume

Start with the thinking

Torricelli's result comes straight from Bernoulli with atmospheric pressure on both ends.
The discharge coefficient bundles the vena-contracta area reduction with velocity losses.

Step-by-step solution

Formula
$V = 2 g H$
Substituting
$V = (2 \times 9.81 \times 2.7) = 7.28 m / s$
Orifice area
$A = π (0.075)^{2} /4 = 0.00442 m^{2}$
Formula
$Q = C_{d} A 2 g H$
Substituting
$Q = 0.60 (0.00442) (7.28) = 0.0193 m^{3} / s$
Volume in 60 s
$\forall = 0.0193 (60) = 1.16 m^{3}$

Answer: V = 7.28 m/s, Q = 0.0193 m³/s (19.3 L/s)

Why the other options are there

0.0322 m³/s (C_d ignored)
0.1404 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifices

Example 2

Discharge from a sharp-edged orifice venting to atmosphere — Orifices (2)

A 75 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 4.5 m. With a discharge coefficient of 0.59, compute the theoretical velocity, the actual discharge and the jet's volume delivered in one minute.

Given

d = 75 mm
H = 4.5 m
C_d = 0.59

Find

V_theoretical, Q and the one-minute volume

Start with the thinking

Torricelli's result comes straight from Bernoulli with atmospheric pressure on both ends.
The discharge coefficient bundles the vena-contracta area reduction with velocity losses.

Step-by-step solution

Formula
$V = 2 g H$
Substituting
$V = (2 \times 9.81 \times 4.5) = 9.40 m / s$
Orifice area
$A = π (0.075)^{2} /4 = 0.00442 m^{2}$
Formula
$Q = C_{d} A 2 g H$
Substituting
$Q = 0.59 (0.00442) (9.40) = 0.0245 m^{3} / s$
Volume in 60 s
$\forall = 0.0245 (60) = 1.47 m^{3}$

Answer: V = 9.40 m/s, Q = 0.0245 m³/s (24.5 L/s)

Why the other options are there

0.0415 m³/s (C_d ignored)
0.2301 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifices

Example 3

Discharge from a sharp-edged orifice venting to atmosphere — Orifices (3)

A 35 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 1.6 m. With a discharge coefficient of 0.61, compute the theoretical velocity, the actual discharge and the jet's volume delivered in one minute.

Given

d = 35 mm
H = 1.6 m
C_d = 0.61

Find

V_theoretical, Q and the one-minute volume

Start with the thinking

Torricelli's result comes straight from Bernoulli with atmospheric pressure on both ends.
The discharge coefficient bundles the vena-contracta area reduction with velocity losses.

Step-by-step solution

Formula
$V = 2 g H$
Substituting
$V = (2 \times 9.81 \times 1.6) = 5.60 m / s$
Orifice area
$A = π (0.035)^{2} /4 = 0.00096 m^{2}$
Formula
$Q = C_{d} A 2 g H$
Substituting
$Q = 0.61 (0.00096) (5.60) = 0.0033 m^{3} / s$
Volume in 60 s
$\forall = 0.0033 (60) = 0.20 m^{3}$

Answer: V = 5.60 m/s, Q = 0.0033 m³/s (3.3 L/s)

Why the other options are there

0.0054 m³/s (C_d ignored)
0.0184 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifices

Example 4

Discharge from a sharp-edged orifice venting to atmosphere — Orifices (4)

A 35 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 4.9 m. With a discharge coefficient of 0.62, compute the theoretical velocity, the actual discharge and the jet's volume delivered in one minute.

Given

d = 35 mm
H = 4.9 m
C_d = 0.62

Find

V_theoretical, Q and the one-minute volume

Start with the thinking

Torricelli's result comes straight from Bernoulli with atmospheric pressure on both ends.
The discharge coefficient bundles the vena-contracta area reduction with velocity losses.

Step-by-step solution

Formula
$V = 2 g H$
Substituting
$V = (2 \times 9.81 \times 4.9) = 9.80 m / s$
Orifice area
$A = π (0.035)^{2} /4 = 0.00096 m^{2}$
Formula
$Q = C_{d} A 2 g H$
Substituting
$Q = 0.62 (0.00096) (9.80) = 0.0058 m^{3} / s$
Volume in 60 s
$\forall = 0.0058 (60) = 0.35 m^{3}$

Answer: V = 9.80 m/s, Q = 0.0058 m³/s (5.8 L/s)

Why the other options are there

0.0094 m³/s (C_d ignored)
0.0573 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifices

Example 5

Discharge from a sharp-edged orifice venting to atmosphere — Orifices (5)

A 70 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 6.2 m. With a discharge coefficient of 0.59, compute the theoretical velocity, the actual discharge and the jet's volume delivered in one minute.

Given

d = 70 mm
H = 6.2 m
C_d = 0.59

Find

V_theoretical, Q and the one-minute volume

Start with the thinking

Torricelli's result comes straight from Bernoulli with atmospheric pressure on both ends.
The discharge coefficient bundles the vena-contracta area reduction with velocity losses.

Step-by-step solution

Formula
$V = 2 g H$
Substituting
$V = (2 \times 9.81 \times 6.2) = 11.03 m / s$
Orifice area
$A = π (0.070)^{2} /4 = 0.00385 m^{2}$
Formula
$Q = C_{d} A 2 g H$
Substituting
$Q = 0.59 (0.00385) (11.03) = 0.0250 m^{3} / s$
Volume in 60 s
$\forall = 0.0250 (60) = 1.50 m^{3}$

Answer: V = 11.03 m/s, Q = 0.0250 m³/s (25.0 L/s)

Why the other options are there

0.0424 m³/s (C_d ignored)
0.2762 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifices

Example 6

Discharge from a sharp-edged orifice venting to atmosphere — Orifices (6)

A 30 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 4.0 m. With a discharge coefficient of 0.65, compute the theoretical velocity, the actual discharge and the jet's volume delivered in one minute.

Given

d = 30 mm
H = 4.0 m
C_d = 0.65

Find

V_theoretical, Q and the one-minute volume

Start with the thinking

Torricelli's result comes straight from Bernoulli with atmospheric pressure on both ends.
The discharge coefficient bundles the vena-contracta area reduction with velocity losses.

Step-by-step solution

Formula
$V = 2 g H$
Substituting
$V = (2 \times 9.81 \times 4.0) = 8.86 m / s$
Orifice area
$A = π (0.030)^{2} /4 = 0.00071 m^{2}$
Formula
$Q = C_{d} A 2 g H$
Substituting
$Q = 0.65 (0.00071) (8.86) = 0.0041 m^{3} / s$
Volume in 60 s
$\forall = 0.0041 (60) = 0.24 m^{3}$

Answer: V = 8.86 m/s, Q = 0.0041 m³/s (4.1 L/s)

Why the other options are there

0.0063 m³/s (C_d ignored)
0.0361 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifices

Example 7

Discharge from a sharp-edged orifice venting to atmosphere — Orifices (7)

A 25 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 3.8 m. With a discharge coefficient of 0.65, compute the theoretical velocity, the actual discharge and the jet's volume delivered in one minute.

Given

d = 25 mm
H = 3.8 m
C_d = 0.65

Find

V_theoretical, Q and the one-minute volume

Start with the thinking

Torricelli's result comes straight from Bernoulli with atmospheric pressure on both ends.
The discharge coefficient bundles the vena-contracta area reduction with velocity losses.

Step-by-step solution

Formula
$V = 2 g H$
Substituting
$V = (2 \times 9.81 \times 3.8) = 8.63 m / s$
Orifice area
$A = π (0.025)^{2} /4 = 0.00049 m^{2}$
Formula
$Q = C_{d} A 2 g H$
Substituting
$Q = 0.65 (0.00049) (8.63) = 0.0028 m^{3} / s$
Volume in 60 s
$\forall = 0.0028 (60) = 0.17 m^{3}$

Answer: V = 8.63 m/s, Q = 0.0028 m³/s (2.8 L/s)

Why the other options are there

0.0042 m³/s (C_d ignored)
0.0238 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifices

Example 8

Discharge from a sharp-edged orifice venting to atmosphere — Orifices (8)

A 95 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 8.3 m. With a discharge coefficient of 0.64, compute the theoretical velocity, the actual discharge and the jet's volume delivered in one minute.

Given

d = 95 mm
H = 8.3 m
C_d = 0.64

Find

V_theoretical, Q and the one-minute volume

Start with the thinking

Torricelli's result comes straight from Bernoulli with atmospheric pressure on both ends.
The discharge coefficient bundles the vena-contracta area reduction with velocity losses.

Step-by-step solution

Formula
$V = 2 g H$
Substituting
$V = (2 \times 9.81 \times 8.3) = 12.76 m / s$
Orifice area
$A = π (0.095)^{2} /4 = 0.00709 m^{2}$
Formula
$Q = C_{d} A 2 g H$
Substituting
$Q = 0.64 (0.00709) (12.76) = 0.0579 m^{3} / s$
Volume in 60 s
$\forall = 0.0579 (60) = 3.47 m^{3}$

Answer: V = 12.76 m/s, Q = 0.0579 m³/s (57.9 L/s)

Why the other options are there

0.0905 m³/s (C_d ignored)
0.7387 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifices

Example 9

Discharge from a sharp-edged orifice venting to atmosphere — Orifices (9)

A 60 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 4.1 m. With a discharge coefficient of 0.64, compute the theoretical velocity, the actual discharge and the jet's volume delivered in one minute.

Given

d = 60 mm
H = 4.1 m
C_d = 0.64

Find

V_theoretical, Q and the one-minute volume

Start with the thinking

Torricelli's result comes straight from Bernoulli with atmospheric pressure on both ends.
The discharge coefficient bundles the vena-contracta area reduction with velocity losses.

Step-by-step solution

Formula
$V = 2 g H$
Substituting
$V = (2 \times 9.81 \times 4.1) = 8.97 m / s$
Orifice area
$A = π (0.060)^{2} /4 = 0.00283 m^{2}$
Formula
$Q = C_{d} A 2 g H$
Substituting
$Q = 0.64 (0.00283) (8.97) = 0.0162 m^{3} / s$
Volume in 60 s
$\forall = 0.0162 (60) = 0.97 m^{3}$

Answer: V = 8.97 m/s, Q = 0.0162 m³/s (16.2 L/s)

Why the other options are there

0.0254 m³/s (C_d ignored)
0.1456 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifices

Example 10

Discharge from a sharp-edged orifice venting to atmosphere — Orifices (10)

A 95 mm sharp-edged orifice in the side of a tank discharges freely to the atmosphere under a head of 4.0 m. With a discharge coefficient of 0.65, compute the theoretical velocity, the actual discharge and the jet's volume delivered in one minute.

Given

d = 95 mm
H = 4.0 m
C_d = 0.65

Find

V_theoretical, Q and the one-minute volume

Start with the thinking

Torricelli's result comes straight from Bernoulli with atmospheric pressure on both ends.
The discharge coefficient bundles the vena-contracta area reduction with velocity losses.

Step-by-step solution

Formula
$V = 2 g H$
Substituting
$V = (2 \times 9.81 \times 4.0) = 8.86 m / s$
Orifice area
$A = π (0.095)^{2} /4 = 0.00709 m^{2}$
Formula
$Q = C_{d} A 2 g H$
Substituting
$Q = 0.65 (0.00709) (8.86) = 0.0408 m^{3} / s$
Volume in 60 s
$\forall = 0.0408 (60) = 2.45 m^{3}$

Answer: V = 8.86 m/s, Q = 0.0408 m³/s (40.8 L/s)

Why the other options are there

0.0628 m³/s (C_d ignored)
0.3616 m³/s (square root omitted)

Reference: FE Reference Handbook — Fluid Mechanics → Orifices

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

Orifices contains 3 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.

Orifices

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

Q=CA0​

C=Needs technical review

Q=CA0​

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section