Pressure Drop for Laminar Flow

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

1 formulas

10 exam-style examples

~47 min

All Fluid Mechanics lectures

Learning objectives

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

This chapter section covers Pressure Drop for Laminar Flow 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 pressure drop for laminar flow describes physically and when it applies.
State every one of the 1 relation 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. Pressure Drop for Laminar Flow 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: pressure drop for laminar flow.
Capstone Studio instructional photograph

Fluid Mechanics — Pressure Drop for Laminar Flow: 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 1 relation 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 equation for Q in terms of the pressure drop ∆Pf is the Hagen-Poiseuille equation. This relation is valid only for flow in the
laminar region.
rR 4 DPf rD 4 DPf
8n L 128nL

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

Hydrostatic pressure at depth — Pressure Drop for Laminar Flow

Find the gauge pressure 22.5 ft below the surface of a fluid with specific gravity 7.80.

Given

h = 22.5 ft
SG = 7.80
γ_water = 62.4 lb/ft³

Find

Gauge pressure in psf and psi

Start with the thinking

Pressure grows linearly with depth, independent of container shape.
Specific weight = SG × 62.4.

Figure for Hydrostatic pressure at depth — Pressure Drop for Laminar Flow

Step-by-step solution

Specific weight
$γ = S G \times γ_{w} = 7.80 (62.4) = 486.7 l b / f t^{3}$
Hydrostatic — p = γh
Substituting
$p = 486.7 (22.5) = 10, 951 p s f$
Convert
$p = 10, 951/144 = 76.05 p s i$

Answer: p ≈ 10,951 psf (76.05 psi)

Why the other options are there

1,404 psf (specific gravity ignored)
1,576,973 psf (conversion applied backwards)

Reference: FE Reference Handbook — Fluid Mechanics → Pressure Drop for Laminar Flow

Example 2

Reynolds number and flow regime — Pressure Drop for Laminar Flow

Water (ν = 1.08 × 10⁻⁵ ft²/s) flows at 5.00 ft/s in a 0.25 ft pipe. Compute Re and classify the flow.

Given

v = 5.00 ft/s
D = 0.25 ft
ν = 1.08e−5 ft²/s

Find

Reynolds number and regime

Start with the thinking

Re < 2,100 is laminar; above ~4,000 is turbulent.
Use the kinematic viscosity form to avoid density bookkeeping.

Step-by-step solution

Reynolds
$R e = v D / ν$
Substituting
$R e = 5.00 (0.25) /1.08 e - 5$
Evaluate
$R e = 115, 741$
Classify — turbulent

Answer: Re ≈ 115,741 → turbulent

Why the other options are there

9,645 (diameter left in inches)
1,851,852 (diameter divided instead of multiplied)

Reference: FE Reference Handbook — Fluid Mechanics → Pressure Drop for Laminar Flow

Example 3

Hydrostatic pressure at depth — Pressure Drop for Laminar Flow (2)

Find the gauge pressure 7.5 ft below the surface of a fluid with specific gravity 12.70.

Given

h = 7.5 ft
SG = 12.70
γ_water = 62.4 lb/ft³

Find

Gauge pressure in psf and psi

Start with the thinking

Pressure grows linearly with depth, independent of container shape.
Specific weight = SG × 62.4.

Figure for Hydrostatic pressure at depth — Pressure Drop for Laminar Flow (2)

Step-by-step solution

Specific weight
$γ = S G \times γ_{w} = 12.70 (62.4) = 792.5 l b / f t^{3}$
Hydrostatic — p = γh
Substituting
$p = 792.5 (7.5) = 5, 944 p s f$
Convert
$p = 5, 944/144 = 41.28 p s i$

Answer: p ≈ 5,944 psf (41.28 psi)

Why the other options are there

468.0 psf (specific gravity ignored)
855,878 psf (conversion applied backwards)

Reference: FE Reference Handbook — Fluid Mechanics → Pressure Drop for Laminar Flow

Example 4

Reynolds number and flow regime — Pressure Drop for Laminar Flow (2)

Water (ν = 1.08 × 10⁻⁵ ft²/s) flows at 5.25 ft/s in a 0.50 ft pipe. Compute Re and classify the flow.

Given

v = 5.25 ft/s
D = 0.50 ft
ν = 1.08e−5 ft²/s

Find

Reynolds number and regime

Start with the thinking

Re < 2,100 is laminar; above ~4,000 is turbulent.
Use the kinematic viscosity form to avoid density bookkeeping.

Step-by-step solution

Reynolds
$R e = v D / ν$
Substituting
$R e = 5.25 (0.50) /1.08 e - 5$
Evaluate
$R e = 243, 056$
Classify — turbulent

Answer: Re ≈ 243,056 → turbulent

Why the other options are there

20,255 (diameter left in inches)
972,222 (diameter divided instead of multiplied)

Reference: FE Reference Handbook — Fluid Mechanics → Pressure Drop for Laminar Flow

Example 5

Hydrostatic pressure at depth — Pressure Drop for Laminar Flow (3)

Find the gauge pressure 4.0 ft below the surface of a fluid with specific gravity 10.40.

Given

h = 4.0 ft
SG = 10.40
γ_water = 62.4 lb/ft³

Find

Gauge pressure in psf and psi

Start with the thinking

Pressure grows linearly with depth, independent of container shape.
Specific weight = SG × 62.4.

Figure for Hydrostatic pressure at depth — Pressure Drop for Laminar Flow (3)

Step-by-step solution

Specific weight
$γ = S G \times γ_{w} = 10.40 (62.4) = 649.0 l b / f t^{3}$
Hydrostatic — p = γh
Substituting
$p = 649.0 (4.0) = 2, 596 p s f$
Convert
$p = 2, 596/144 = 18.03 p s i$

Answer: p ≈ 2,596 psf (18.03 psi)

Why the other options are there

249.6 psf (specific gravity ignored)
373,801 psf (conversion applied backwards)

Reference: FE Reference Handbook — Fluid Mechanics → Pressure Drop for Laminar Flow

Example 6

Reynolds number and flow regime — Pressure Drop for Laminar Flow (3)

Water (ν = 1.08 × 10⁻⁵ ft²/s) flows at 1.50 ft/s in a 1.25 ft pipe. Compute Re and classify the flow.

Given

v = 1.50 ft/s
D = 1.25 ft
ν = 1.08e−5 ft²/s

Find

Reynolds number and regime

Start with the thinking

Re < 2,100 is laminar; above ~4,000 is turbulent.
Use the kinematic viscosity form to avoid density bookkeeping.

Step-by-step solution

Reynolds
$R e = v D / ν$
Substituting
$R e = 1.50 (1.25) /1.08 e - 5$
Evaluate
$R e = 173, 611$
Classify — turbulent

Answer: Re ≈ 173,611 → turbulent

Why the other options are there

14,468 (diameter left in inches)
111,111 (diameter divided instead of multiplied)

Reference: FE Reference Handbook — Fluid Mechanics → Pressure Drop for Laminar Flow

Example 7

Hydrostatic pressure at depth — Pressure Drop for Laminar Flow (4)

Find the gauge pressure 20.5 ft below the surface of a fluid with specific gravity 0.90.

Given

h = 20.5 ft
SG = 0.90
γ_water = 62.4 lb/ft³

Find

Gauge pressure in psf and psi

Start with the thinking

Pressure grows linearly with depth, independent of container shape.
Specific weight = SG × 62.4.

Figure for Hydrostatic pressure at depth — Pressure Drop for Laminar Flow (4)

Step-by-step solution

Specific weight
$γ = S G \times γ_{w} = 0.90 (62.4) = 56.16 l b / f t^{3}$
Hydrostatic — p = γh
Substituting
$p = 56.16 (20.5) = 1, 151 p s f$
Convert
$p = 1, 151/144 = 8.00 p s i$

Answer: p ≈ 1,151 psf (8.00 psi)

Why the other options are there

1,279 psf (specific gravity ignored)
165,784 psf (conversion applied backwards)

Reference: FE Reference Handbook — Fluid Mechanics → Pressure Drop for Laminar Flow

Example 8

Reynolds number and flow regime — Pressure Drop for Laminar Flow (4)

Water (ν = 1.08 × 10⁻⁵ ft²/s) flows at 4.50 ft/s in a 0.50 ft pipe. Compute Re and classify the flow.

Given

v = 4.50 ft/s
D = 0.50 ft
ν = 1.08e−5 ft²/s

Find

Reynolds number and regime

Start with the thinking

Re < 2,100 is laminar; above ~4,000 is turbulent.
Use the kinematic viscosity form to avoid density bookkeeping.

Step-by-step solution

Reynolds
$R e = v D / ν$
Substituting
$R e = 4.50 (0.50) /1.08 e - 5$
Evaluate
$R e = 208, 333$
Classify — turbulent

Answer: Re ≈ 208,333 → turbulent

Why the other options are there

17,361 (diameter left in inches)
833,333 (diameter divided instead of multiplied)

Reference: FE Reference Handbook — Fluid Mechanics → Pressure Drop for Laminar Flow

Example 9

Hydrostatic pressure at depth — Pressure Drop for Laminar Flow (5)

Find the gauge pressure 4.0 ft below the surface of a fluid with specific gravity 0.90.

Given

h = 4.0 ft
SG = 0.90
γ_water = 62.4 lb/ft³

Find

Gauge pressure in psf and psi

Start with the thinking

Pressure grows linearly with depth, independent of container shape.
Specific weight = SG × 62.4.

Figure for Hydrostatic pressure at depth — Pressure Drop for Laminar Flow (5)

Step-by-step solution

Specific weight
$γ = S G \times γ_{w} = 0.90 (62.4) = 56.16 l b / f t^{3}$
Hydrostatic — p = γh
Substituting
$p = 56.16 (4.0) = 224.6 p s f$
Convert
$p = 224.6/144 = 1.56 p s i$

Answer: p ≈ 224.6 psf (1.56 psi)

Why the other options are there

249.6 psf (specific gravity ignored)
32,348 psf (conversion applied backwards)

Reference: FE Reference Handbook — Fluid Mechanics → Pressure Drop for Laminar Flow

Example 10

Reynolds number and flow regime — Pressure Drop for Laminar Flow (5)

Water (ν = 1.08 × 10⁻⁵ ft²/s) flows at 3.75 ft/s in a 0.50 ft pipe. Compute Re and classify the flow.

Given

v = 3.75 ft/s
D = 0.50 ft
ν = 1.08e−5 ft²/s

Find

Reynolds number and regime

Start with the thinking

Re < 2,100 is laminar; above ~4,000 is turbulent.
Use the kinematic viscosity form to avoid density bookkeeping.

Step-by-step solution

Reynolds
$R e = v D / ν$
Substituting
$R e = 3.75 (0.50) /1.08 e - 5$
Evaluate
$R e = 173, 611$
Classify — turbulent

Answer: Re ≈ 173,611 → turbulent

Why the other options are there

14,468 (diameter left in inches)
694,444 (diameter divided instead of multiplied)

Reference: FE Reference Handbook — Fluid Mechanics → Pressure Drop for Laminar Flow

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

Pressure Drop for Laminar Flow contains 1 relation; 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.