Multipath Pipeline Problems

Fluid Mechanics · FE Reference Handbook section

Fluid Mechanics

6 formulas

10 exam-style examples

~57 min

All Fluid Mechanics lectures

Learning objectives

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

This chapter section covers Multipath Pipeline Problems 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 multipath pipeline problems describes physically and when it applies.
State every one of the 6 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. Multipath Pipeline Problems 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: multipath pipeline problems.
Capstone Studio instructional photograph

Fluid Mechanics — Multipath Pipeline Problems: 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 6 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

hL	Quantity produced by "hL = fA = fB B B" — read its definition and unit from the handbook line directly above the equation.
_rD 2/4i v	Quantity produced by "_rD 2/4i v = `rD A2 /4j v A + `rDB2 /4j vB" — read its definition and unit from the handbook line directly above the equation.
ΣF	Quantity produced by "ΣF = ΣQ2ρ2v2 – ΣQ1ρ1v1" — read its definition and unit from the handbook line directly above the equation.
ΣQ1ρ1v1	Quantity produced by "ΣQ1ρ1v1 = rate of momentum of the fluid flow entering the control volume in the same direction of the force" — read its definition and unit from the handbook line directly above the equation.
ΣQ2ρ2v2	Quantity produced by "ΣQ2ρ2v2 = rate of momentum of the fluid flow leaving the control volume in the same direction of the force" — 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.

L P
v v
Vennard, J.K., Elementary Fluid Mechanics, 6th ed., J.K. Vennard, 1954.
For pipes in parallel, the head loss is the same in each pipe.
L A v A2 L v2
DA 2g DB 2g
The total flowrate Q is the sum of the flowrates in the parallel pipes.
The Impulse-Momentum Principle
The resultant force in a given direction acting on the fluid equals the rate of change of momentum of the fluid.
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

Flow split between two parallel pipeline branches — Multipath Pipeline Problems

Two parallel branches connect the same two junctions. Branch 1 is 608 m of 250.0 mm pipe; branch 2 is 609 m of 225.0 mm pipe; both have f = 0.02. The head loss across the loop is 3.5 m. Compute each branch flow and the total delivered discharge.

Given

h_f = 3.5 m (identical for both branches)
Branch 1: L = 608 m, D = 250.0 mm
Branch 2: L = 609 m, D = 225.0 mm
f = 0.02

Find

Q₁, Q₂ and Q_total

Start with the thinking

Parallel branches share the same head loss; the flows differ. In series it is the reverse.
Solve each branch's Darcy-Weisbach equation for V, then convert to discharge.

Step-by-step solution

Formula
$h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g} \Rightarrow V = \frac{2 g D h _{f}}{f L}$
Branch 1
$V_{1} = 2 (9.81) (0.250) (3.5) / (0.02 \times 608) = 1.19 m / s$
Branch 1
$Q_{1} = 1.19 (0.04909) = 0.0583 m^{3} / s$
Branch 2
$V_{2} = 2 (9.81) (0.225) (3.5) / (0.02 \times 609) = 1.13 m / s$
Branch 2
$Q_{2} = 1.13 (0.03976) = 0.0448 m^{3} / s$
Total
$Q = 0.0583 + 0.0448 = 0.1031 m^{3} / s$

Answer: Q₁ = 0.0583 m³/s, Q₂ = 0.0448 m³/s, total 0.1031 m³/s

Why the other options are there

Head losses added in parallel (7.0 m)
Q split evenly at 0.0516 m³/s each

Reference: FE Reference Handbook — Fluid Mechanics → Multipath Pipeline Problems

Example 2

Flow split between two parallel pipeline branches — Multipath Pipeline Problems (2)

Two parallel branches connect the same two junctions. Branch 1 is 484 m of 300.0 mm pipe; branch 2 is 1053 m of 150.0 mm pipe; both have f = 0.02. The head loss across the loop is 14.0 m. Compute each branch flow and the total delivered discharge.

Given

h_f = 14.0 m (identical for both branches)
Branch 1: L = 484 m, D = 300.0 mm
Branch 2: L = 1053 m, D = 150.0 mm
f = 0.02

Find

Q₁, Q₂ and Q_total

Start with the thinking

Parallel branches share the same head loss; the flows differ. In series it is the reverse.
Solve each branch's Darcy-Weisbach equation for V, then convert to discharge.

Step-by-step solution

Formula
$h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g} \Rightarrow V = \frac{2 g D h _{f}}{f L}$
Branch 1
$V_{1} = 2 (9.81) (0.300) (14.0) / (0.02 \times 484) = 2.92 m / s$
Branch 1
$Q_{1} = 2.92 (0.07069) = 0.2062 m^{3} / s$
Branch 2
$V_{2} = 2 (9.81) (0.150) (14.0) / (0.02 \times 1053) = 1.40 m / s$
Branch 2
$Q_{2} = 1.40 (0.01767) = 0.0247 m^{3} / s$
Total
$Q = 0.2062 + 0.0247 = 0.2310 m^{3} / s$

Answer: Q₁ = 0.2062 m³/s, Q₂ = 0.0247 m³/s, total 0.2310 m³/s

Why the other options are there

Head losses added in parallel (28.0 m)
Q split evenly at 0.1155 m³/s each

Reference: FE Reference Handbook — Fluid Mechanics → Multipath Pipeline Problems

Example 3

Flow split between two parallel pipeline branches — Multipath Pipeline Problems (3)

Two parallel branches connect the same two junctions. Branch 1 is 317 m of 225.0 mm pipe; branch 2 is 802 m of 300.0 mm pipe; both have f = 0.02. The head loss across the loop is 7.0 m. Compute each branch flow and the total delivered discharge.

Given

h_f = 7.0 m (identical for both branches)
Branch 1: L = 317 m, D = 225.0 mm
Branch 2: L = 802 m, D = 300.0 mm
f = 0.02

Find

Q₁, Q₂ and Q_total

Start with the thinking

Parallel branches share the same head loss; the flows differ. In series it is the reverse.
Solve each branch's Darcy-Weisbach equation for V, then convert to discharge.

Step-by-step solution

Formula
$h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g} \Rightarrow V = \frac{2 g D h _{f}}{f L}$
Branch 1
$V_{1} = 2 (9.81) (0.225) (7.0) / (0.02 \times 317) = 2.21 m / s$
Branch 1
$Q_{1} = 2.21 (0.03976) = 0.0878 m^{3} / s$
Branch 2
$V_{2} = 2 (9.81) (0.300) (7.0) / (0.02 \times 802) = 1.60 m / s$
Branch 2
$Q_{2} = 1.60 (0.07069) = 0.1133 m^{3} / s$
Total
$Q = 0.0878 + 0.1133 = 0.2011 m^{3} / s$

Answer: Q₁ = 0.0878 m³/s, Q₂ = 0.1133 m³/s, total 0.2011 m³/s

Why the other options are there

Head losses added in parallel (14.0 m)
Q split evenly at 0.1005 m³/s each

Reference: FE Reference Handbook — Fluid Mechanics → Multipath Pipeline Problems

Example 4

Flow split between two parallel pipeline branches — Multipath Pipeline Problems (4)

Two parallel branches connect the same two junctions. Branch 1 is 344 m of 275.0 mm pipe; branch 2 is 342 m of 275.0 mm pipe; both have f = 0.02. The head loss across the loop is 3.5 m. Compute each branch flow and the total delivered discharge.

Given

h_f = 3.5 m (identical for both branches)
Branch 1: L = 344 m, D = 275.0 mm
Branch 2: L = 342 m, D = 275.0 mm
f = 0.02

Find

Q₁, Q₂ and Q_total

Start with the thinking

Parallel branches share the same head loss; the flows differ. In series it is the reverse.
Solve each branch's Darcy-Weisbach equation for V, then convert to discharge.

Step-by-step solution

Formula
$h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g} \Rightarrow V = \frac{2 g D h _{f}}{f L}$
Branch 1
$V_{1} = 2 (9.81) (0.275) (3.5) / (0.02 \times 344) = 1.66 m / s$
Branch 1
$Q_{1} = 1.66 (0.05940) = 0.0984 m^{3} / s$
Branch 2
$V_{2} = 2 (9.81) (0.275) (3.5) / (0.02 \times 342) = 1.66 m / s$
Branch 2
$Q_{2} = 1.66 (0.05940) = 0.0987 m^{3} / s$
Total
$Q = 0.0984 + 0.0987 = 0.1971 m^{3} / s$

Answer: Q₁ = 0.0984 m³/s, Q₂ = 0.0987 m³/s, total 0.1971 m³/s

Why the other options are there

Head losses added in parallel (7.0 m)
Q split evenly at 0.0985 m³/s each

Reference: FE Reference Handbook — Fluid Mechanics → Multipath Pipeline Problems

Example 5

Flow split between two parallel pipeline branches — Multipath Pipeline Problems (5)

Two parallel branches connect the same two junctions. Branch 1 is 392 m of 250.0 mm pipe; branch 2 is 306 m of 275.0 mm pipe; both have f = 0.02. The head loss across the loop is 8.0 m. Compute each branch flow and the total delivered discharge.

Given

h_f = 8.0 m (identical for both branches)
Branch 1: L = 392 m, D = 250.0 mm
Branch 2: L = 306 m, D = 275.0 mm
f = 0.02

Find

Q₁, Q₂ and Q_total

Start with the thinking

Parallel branches share the same head loss; the flows differ. In series it is the reverse.
Solve each branch's Darcy-Weisbach equation for V, then convert to discharge.

Step-by-step solution

Formula
$h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g} \Rightarrow V = \frac{2 g D h _{f}}{f L}$
Branch 1
$V_{1} = 2 (9.81) (0.250) (8.0) / (0.02 \times 392) = 2.24 m / s$
Branch 1
$Q_{1} = 2.24 (0.04909) = 0.1098 m^{3} / s$
Branch 2
$V_{2} = 2 (9.81) (0.275) (8.0) / (0.02 \times 306) = 2.66 m / s$
Branch 2
$Q_{2} = 2.66 (0.05940) = 0.1577 m^{3} / s$
Total
$Q = 0.1098 + 0.1577 = 0.2676 m^{3} / s$

Answer: Q₁ = 0.1098 m³/s, Q₂ = 0.1577 m³/s, total 0.2676 m³/s

Why the other options are there

Head losses added in parallel (16.0 m)
Q split evenly at 0.1338 m³/s each

Reference: FE Reference Handbook — Fluid Mechanics → Multipath Pipeline Problems

Example 6

Flow split between two parallel pipeline branches — Multipath Pipeline Problems (6)

Two parallel branches connect the same two junctions. Branch 1 is 667 m of 250.0 mm pipe; branch 2 is 926 m of 275.0 mm pipe; both have f = 0.02. The head loss across the loop is 13.0 m. Compute each branch flow and the total delivered discharge.

Given

h_f = 13.0 m (identical for both branches)
Branch 1: L = 667 m, D = 250.0 mm
Branch 2: L = 926 m, D = 275.0 mm
f = 0.02

Find

Q₁, Q₂ and Q_total

Start with the thinking

Parallel branches share the same head loss; the flows differ. In series it is the reverse.
Solve each branch's Darcy-Weisbach equation for V, then convert to discharge.

Step-by-step solution

Formula
$h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g} \Rightarrow V = \frac{2 g D h _{f}}{f L}$
Branch 1
$V_{1} = 2 (9.81) (0.250) (13.0) / (0.02 \times 667) = 2.19 m / s$
Branch 1
$Q_{1} = 2.19 (0.04909) = 0.1073 m^{3} / s$
Branch 2
$V_{2} = 2 (9.81) (0.275) (13.0) / (0.02 \times 926) = 1.95 m / s$
Branch 2
$Q_{2} = 1.95 (0.05940) = 0.1156 m^{3} / s$
Total
$Q = 0.1073 + 0.1156 = 0.2229 m^{3} / s$

Answer: Q₁ = 0.1073 m³/s, Q₂ = 0.1156 m³/s, total 0.2229 m³/s

Why the other options are there

Head losses added in parallel (26.0 m)
Q split evenly at 0.1115 m³/s each

Reference: FE Reference Handbook — Fluid Mechanics → Multipath Pipeline Problems

Example 7

Flow split between two parallel pipeline branches — Multipath Pipeline Problems (7)

Two parallel branches connect the same two junctions. Branch 1 is 298 m of 250.0 mm pipe; branch 2 is 1181 m of 200.0 mm pipe; both have f = 0.02. The head loss across the loop is 3.0 m. Compute each branch flow and the total delivered discharge.

Given

h_f = 3.0 m (identical for both branches)
Branch 1: L = 298 m, D = 250.0 mm
Branch 2: L = 1181 m, D = 200.0 mm
f = 0.02

Find

Q₁, Q₂ and Q_total

Start with the thinking

Parallel branches share the same head loss; the flows differ. In series it is the reverse.
Solve each branch's Darcy-Weisbach equation for V, then convert to discharge.

Step-by-step solution

Formula
$h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g} \Rightarrow V = \frac{2 g D h _{f}}{f L}$
Branch 1
$V_{1} = 2 (9.81) (0.250) (3.0) / (0.02 \times 298) = 1.57 m / s$
Branch 1
$Q_{1} = 1.57 (0.04909) = 0.0771 m^{3} / s$
Branch 2
$V_{2} = 2 (9.81) (0.200) (3.0) / (0.02 \times 1181) = 0.71 m / s$
Branch 2
$Q_{2} = 0.71 (0.03142) = 0.0222 m^{3} / s$
Total
$Q = 0.0771 + 0.0222 = 0.0993 m^{3} / s$

Answer: Q₁ = 0.0771 m³/s, Q₂ = 0.0222 m³/s, total 0.0993 m³/s

Why the other options are there

Head losses added in parallel (6.0 m)
Q split evenly at 0.0497 m³/s each

Reference: FE Reference Handbook — Fluid Mechanics → Multipath Pipeline Problems

Example 8

Flow split between two parallel pipeline branches — Multipath Pipeline Problems (8)

Two parallel branches connect the same two junctions. Branch 1 is 697 m of 225.0 mm pipe; branch 2 is 338 m of 300.0 mm pipe; both have f = 0.02. The head loss across the loop is 7.5 m. Compute each branch flow and the total delivered discharge.

Given

h_f = 7.5 m (identical for both branches)
Branch 1: L = 697 m, D = 225.0 mm
Branch 2: L = 338 m, D = 300.0 mm
f = 0.02

Find

Q₁, Q₂ and Q_total

Start with the thinking

Parallel branches share the same head loss; the flows differ. In series it is the reverse.
Solve each branch's Darcy-Weisbach equation for V, then convert to discharge.

Step-by-step solution

Formula
$h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g} \Rightarrow V = \frac{2 g D h _{f}}{f L}$
Branch 1
$V_{1} = 2 (9.81) (0.225) (7.5) / (0.02 \times 697) = 1.54 m / s$
Branch 1
$Q_{1} = 1.54 (0.03976) = 0.0613 m^{3} / s$
Branch 2
$V_{2} = 2 (9.81) (0.300) (7.5) / (0.02 \times 338) = 2.56 m / s$
Branch 2
$Q_{2} = 2.56 (0.07069) = 0.1806 m^{3} / s$
Total
$Q = 0.0613 + 0.1806 = 0.2419 m^{3} / s$

Answer: Q₁ = 0.0613 m³/s, Q₂ = 0.1806 m³/s, total 0.2419 m³/s

Why the other options are there

Head losses added in parallel (15.0 m)
Q split evenly at 0.1210 m³/s each

Reference: FE Reference Handbook — Fluid Mechanics → Multipath Pipeline Problems

Example 9

Flow split between two parallel pipeline branches — Multipath Pipeline Problems (9)

Two parallel branches connect the same two junctions. Branch 1 is 434 m of 225.0 mm pipe; branch 2 is 779 m of 200.0 mm pipe; both have f = 0.02. The head loss across the loop is 9.0 m. Compute each branch flow and the total delivered discharge.

Given

h_f = 9.0 m (identical for both branches)
Branch 1: L = 434 m, D = 225.0 mm
Branch 2: L = 779 m, D = 200.0 mm
f = 0.02

Find

Q₁, Q₂ and Q_total

Start with the thinking

Parallel branches share the same head loss; the flows differ. In series it is the reverse.
Solve each branch's Darcy-Weisbach equation for V, then convert to discharge.

Step-by-step solution

Formula
$h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g} \Rightarrow V = \frac{2 g D h _{f}}{f L}$
Branch 1
$V_{1} = 2 (9.81) (0.225) (9.0) / (0.02 \times 434) = 2.14 m / s$
Branch 1
$Q_{1} = 2.14 (0.03976) = 0.0851 m^{3} / s$
Branch 2
$V_{2} = 2 (9.81) (0.200) (9.0) / (0.02 \times 779) = 1.51 m / s$
Branch 2
$Q_{2} = 1.51 (0.03142) = 0.0473 m^{3} / s$
Total
$Q = 0.0851 + 0.0473 = 0.1324 m^{3} / s$

Answer: Q₁ = 0.0851 m³/s, Q₂ = 0.0473 m³/s, total 0.1324 m³/s

Why the other options are there

Head losses added in parallel (18.0 m)
Q split evenly at 0.0662 m³/s each

Reference: FE Reference Handbook — Fluid Mechanics → Multipath Pipeline Problems

Example 10

Flow split between two parallel pipeline branches — Multipath Pipeline Problems (10)

Two parallel branches connect the same two junctions. Branch 1 is 571 m of 275.0 mm pipe; branch 2 is 783 m of 225.0 mm pipe; both have f = 0.02. The head loss across the loop is 17.0 m. Compute each branch flow and the total delivered discharge.

Given

h_f = 17.0 m (identical for both branches)
Branch 1: L = 571 m, D = 275.0 mm
Branch 2: L = 783 m, D = 225.0 mm
f = 0.02

Find

Q₁, Q₂ and Q_total

Start with the thinking

Parallel branches share the same head loss; the flows differ. In series it is the reverse.
Solve each branch's Darcy-Weisbach equation for V, then convert to discharge.

Step-by-step solution

Formula
$h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g} \Rightarrow V = \frac{2 g D h _{f}}{f L}$
Branch 1
$V_{1} = 2 (9.81) (0.275) (17.0) / (0.02 \times 571) = 2.83 m / s$
Branch 1
$Q_{1} = 2.83 (0.05940) = 0.1683 m^{3} / s$
Branch 2
$V_{2} = 2 (9.81) (0.225) (17.0) / (0.02 \times 783) = 2.19 m / s$
Branch 2
$Q_{2} = 2.19 (0.03976) = 0.0870 m^{3} / s$
Total
$Q = 0.1683 + 0.0870 = 0.2554 m^{3} / s$

Answer: Q₁ = 0.1683 m³/s, Q₂ = 0.0870 m³/s, total 0.2554 m³/s

Why the other options are there

Head losses added in parallel (34.0 m)
Q split evenly at 0.1277 m³/s each

Reference: FE Reference Handbook — Fluid Mechanics → Multipath Pipeline Problems

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

Multipath Pipeline Problems contains 6 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.