Hazen-Williams Equation

Hydraulics · FE Reference Handbook section

Hydraulics

15 formulas

10 exam-style examples

~60 min

All Hydraulics lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Circular Pipe Head Loss Equation (Head Loss Expressed in Feet)
Circular Pipe Head Loss Equation (Head Loss Expressed as Pressure)

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

Hazen-Williams head loss in a water main — Hazen-Williams Equation

A 8 in ductile iron main (C = 100) carries 1258 gpm over 1163 ft. Find the friction head loss and the hydraulic gradient.

Given

$Q = 1258 gpm$
$D = 8 in$
$C = 100$
$L = 1163 ft$

Find

h_f and the slope of the hydraulic grade line

Start with the thinking

The US customary Hazen-Williams form takes Q in gpm and D in inches and returns feet of head.
Head loss scales with Q^1.85 — doubling flow nearly quadruples the loss.

Step-by-step solution

Formula
$h_f = 10.44 L Q^1.85 / (C^1.85 D^4.87)$
Flow term
$Q^1.85 = 1258^1.85 = 542,512$
Coefficient term
$C^1.85 = 100^1.85 = 5,012$
Diameter term
$D^4.87 = 8^4.87 = 25,006$
Substituting
$h_f = 10.44 \times 1163 \times 542,512 / (5,012 \times 25,006)$
Evaluate
$h_f = 52.56 ft$
Gradient
$S = h_f/L = 0.04519 ft/ft$

Answer:

h_f ≈ 52.6 ft over 1163 ft (S ≈ 4.519%)

Why the other options are there

52.6 ft (roughness coefficient mis-scaled)
6.11 ft (exponents dropped)

Reference: FE Reference Handbook — Hydraulics → Hazen-Williams Equation

Example 2

Hazen-Williams head loss in a water main — Hazen-Williams Equation (2)

A 10 in ductile iron main (C = 140) carries 2835 gpm over 2201 ft. Find the friction head loss and the hydraulic gradient.

Given

$Q = 2835 gpm$
$D = 10 in$
$C = 140$
$L = 2201 ft$

Find

h_f and the slope of the hydraulic grade line

Start with the thinking

The US customary Hazen-Williams form takes Q in gpm and D in inches and returns feet of head.
Head loss scales with Q^1.85 — doubling flow nearly quadruples the loss.

Step-by-step solution

Formula
$h_f = 10.44 L Q^1.85 / (C^1.85 D^4.87)$
Flow term
$Q^1.85 = 2835^1.85 = 2,439,064$
Coefficient term
$C^1.85 = 140^1.85 = 9,340$
Diameter term
$D^4.87 = 10^4.87 = 74,131$
Substituting
$h_f = 10.44 \times 2201 \times 2,439,064 / (9,340 \times 74,131)$
Evaluate
$h_f = 80.95 ft$
Gradient
$S = h_f/L = 0.03678 ft/ft$

Answer:

h_f ≈ 80.9 ft over 2201 ft (S ≈ 3.678%)

Why the other options are there

150.8 ft (roughness coefficient mis-scaled)
6.28 ft (exponents dropped)

Reference: FE Reference Handbook — Hydraulics → Hazen-Williams Equation

Example 3

Hazen-Williams head loss in a water main — Hazen-Williams Equation (3)

A 8 in ductile iron main (C = 140) carries 561 gpm over 4390 ft. Find the friction head loss and the hydraulic gradient.

Given

$Q = 561 gpm$
$D = 8 in$
$C = 140$
$L = 4390 ft$

Find

h_f and the slope of the hydraulic grade line

Start with the thinking

The US customary Hazen-Williams form takes Q in gpm and D in inches and returns feet of head.
Head loss scales with Q^1.85 — doubling flow nearly quadruples the loss.

Step-by-step solution

Formula
$h_f = 10.44 L Q^1.85 / (C^1.85 D^4.87)$
Flow term
$Q^1.85 = 561^1.85 = 121,781$
Coefficient term
$C^1.85 = 140^1.85 = 9,340$
Diameter term
$D^4.87 = 8^4.87 = 25,006$
Substituting
$h_f = 10.44 \times 4390 \times 121,781 / (9,340 \times 25,006)$
Evaluate
$h_f = 23.90 ft$
Gradient
$S = h_f/L = 0.00544 ft/ft$

Answer:

h_f ≈ 23.9 ft over 4390 ft (S ≈ 0.544%)

Why the other options are there

44.5 ft (roughness coefficient mis-scaled)
7.34 ft (exponents dropped)

Reference: FE Reference Handbook — Hydraulics → Hazen-Williams Equation

Example 4

Hazen-Williams head loss in a water main — Hazen-Williams Equation (4)

A 10 in ductile iron main (C = 100) carries 331 gpm over 1517 ft. Find the friction head loss and the hydraulic gradient.

Given

$Q = 331 gpm$
$D = 10 in$
$C = 100$
$L = 1517 ft$

Find

h_f and the slope of the hydraulic grade line

Start with the thinking

The US customary Hazen-Williams form takes Q in gpm and D in inches and returns feet of head.
Head loss scales with Q^1.85 — doubling flow nearly quadruples the loss.

Step-by-step solution

Formula
$h_f = 10.44 L Q^1.85 / (C^1.85 D^4.87)$
Flow term
$Q^1.85 = 331^1.85 = 45,886$
Coefficient term
$C^1.85 = 100^1.85 = 5,012$
Diameter term
$D^4.87 = 10^4.87 = 74,131$
Substituting
$h_f = 10.44 \times 1517 \times 45,886 / (5,012 \times 74,131)$
Evaluate
$h_f = 1.96 ft$
Gradient
$S = h_f/L = 0.00129 ft/ft$

Answer:

h_f ≈ 2.0 ft over 1517 ft (S ≈ 0.129%)

Why the other options are there

2.0 ft (roughness coefficient mis-scaled)
0.71 ft (exponents dropped)

Reference: FE Reference Handbook — Hydraulics → Hazen-Williams Equation

Example 5

Hazen-Williams head loss in a water main — Hazen-Williams Equation (5)

A 6 in ductile iron main (C = 140) carries 522 gpm over 2285 ft. Find the friction head loss and the hydraulic gradient.

Given

$Q = 522 gpm$
$D = 6 in$
$C = 140$
$L = 2285 ft$

Find

h_f and the slope of the hydraulic grade line

Start with the thinking

The US customary Hazen-Williams form takes Q in gpm and D in inches and returns feet of head.
Head loss scales with Q^1.85 — doubling flow nearly quadruples the loss.

Step-by-step solution

Formula
$h_f = 10.44 L Q^1.85 / (C^1.85 D^4.87)$
Flow term
$Q^1.85 = 522^1.85 = 106,584$
Coefficient term
$C^1.85 = 140^1.85 = 9,340$
Diameter term
$D^4.87 = 6^4.87 = 6,160$
Substituting
$h_f = 10.44 \times 2285 \times 106,584 / (9,340 \times 6,160)$
Evaluate
$h_f = 44.19 ft$
Gradient
$S = h_f/L = 0.01934 ft/ft$

Answer:

h_f ≈ 44.2 ft over 2285 ft (S ≈ 1.934%)

Why the other options are there

82.4 ft (roughness coefficient mis-scaled)
14.44 ft (exponents dropped)

Reference: FE Reference Handbook — Hydraulics → Hazen-Williams Equation

Example 6

Hazen-Williams head loss in a water main — Hazen-Williams Equation (6)

A 8 in ductile iron main (C = 100) carries 1194 gpm over 4741 ft. Find the friction head loss and the hydraulic gradient.

Given

$Q = 1194 gpm$
$D = 8 in$
$C = 100$
$L = 4741 ft$

Find

h_f and the slope of the hydraulic grade line

Start with the thinking

The US customary Hazen-Williams form takes Q in gpm and D in inches and returns feet of head.
Head loss scales with Q^1.85 — doubling flow nearly quadruples the loss.

Step-by-step solution

Formula
$h_f = 10.44 L Q^1.85 / (C^1.85 D^4.87)$
Flow term
$Q^1.85 = 1194^1.85 = 492,559$
Coefficient term
$C^1.85 = 100^1.85 = 5,012$
Diameter term
$D^4.87 = 8^4.87 = 25,006$
Substituting
$h_f = 10.44 \times 4741 \times 492,559 / (5,012 \times 25,006)$
Evaluate
$h_f = 194.5 ft$
Gradient
$S = h_f/L = 0.04103 ft/ft$

Answer:

h_f ≈ 194.5 ft over 4741 ft (S ≈ 4.103%)

Why the other options are there

194.5 ft (roughness coefficient mis-scaled)
23.63 ft (exponents dropped)

Reference: FE Reference Handbook — Hydraulics → Hazen-Williams Equation

Example 7

Hazen-Williams head loss in a water main — Hazen-Williams Equation (7)

A 10 in ductile iron main (C = 140) carries 1251 gpm over 4376 ft. Find the friction head loss and the hydraulic gradient.

Given

$Q = 1251 gpm$
$D = 10 in$
$C = 140$
$L = 4376 ft$

Find

h_f and the slope of the hydraulic grade line

Start with the thinking

The US customary Hazen-Williams form takes Q in gpm and D in inches and returns feet of head.
Head loss scales with Q^1.85 — doubling flow nearly quadruples the loss.

Step-by-step solution

Formula
$h_f = 10.44 L Q^1.85 / (C^1.85 D^4.87)$
Flow term
$Q^1.85 = 1251^1.85 = 536,940$
Coefficient term
$C^1.85 = 140^1.85 = 9,340$
Diameter term
$D^4.87 = 10^4.87 = 74,131$
Substituting
$h_f = 10.44 \times 4376 \times 536,940 / (9,340 \times 74,131)$
Evaluate
$h_f = 35.43 ft$
Gradient
$S = h_f/L = 0.00810 ft/ft$

Answer:

h_f ≈ 35.4 ft over 4376 ft (S ≈ 0.810%)

Why the other options are there

66.0 ft (roughness coefficient mis-scaled)
5.51 ft (exponents dropped)

Reference: FE Reference Handbook — Hydraulics → Hazen-Williams Equation

Example 8

Hazen-Williams head loss in a water main — Hazen-Williams Equation (8)

A 8 in ductile iron main (C = 100) carries 751 gpm over 4109 ft. Find the friction head loss and the hydraulic gradient.

Given

$Q = 751 gpm$
$D = 8 in$
$C = 100$
$L = 4109 ft$

Find

h_f and the slope of the hydraulic grade line

Start with the thinking

The US customary Hazen-Williams form takes Q in gpm and D in inches and returns feet of head.
Head loss scales with Q^1.85 — doubling flow nearly quadruples the loss.

Step-by-step solution

Formula
$h_f = 10.44 L Q^1.85 / (C^1.85 D^4.87)$
Flow term
$Q^1.85 = 751^1.85 = 208,898$
Coefficient term
$C^1.85 = 100^1.85 = 5,012$
Diameter term
$D^4.87 = 8^4.87 = 25,006$
Substituting
$h_f = 10.44 \times 4109 \times 208,898 / (5,012 \times 25,006)$
Evaluate
$h_f = 71.50 ft$
Gradient
$S = h_f/L = 0.01740 ft/ft$

Answer:

h_f ≈ 71.5 ft over 4109 ft (S ≈ 1.740%)

Why the other options are there

71.5 ft (roughness coefficient mis-scaled)
12.88 ft (exponents dropped)

Reference: FE Reference Handbook — Hydraulics → Hazen-Williams Equation

Example 9

Hazen-Williams head loss in a water main — Hazen-Williams Equation (9)

A 12 in ductile iron main (C = 100) carries 596 gpm over 5388 ft. Find the friction head loss and the hydraulic gradient.

Given

$Q = 596 gpm$
$D = 12 in$
$C = 100$
$L = 5388 ft$

Find

h_f and the slope of the hydraulic grade line

Start with the thinking

The US customary Hazen-Williams form takes Q in gpm and D in inches and returns feet of head.
Head loss scales with Q^1.85 — doubling flow nearly quadruples the loss.

Step-by-step solution

Formula
$h_f = 10.44 L Q^1.85 / (C^1.85 D^4.87)$
Flow term
$Q^1.85 = 596^1.85 = 136,209$
Coefficient term
$C^1.85 = 100^1.85 = 5,012$
Diameter term
$D^4.87 = 12^4.87 = 180,141$
Substituting
$h_f = 10.44 \times 5388 \times 136,209 / (5,012 \times 180,141)$
Evaluate
$h_f = 8.49 ft$
Gradient
$S = h_f/L = 0.00158 ft/ft$

Answer:

h_f ≈ 8.5 ft over 5388 ft (S ≈ 0.158%)

Why the other options are there

8.5 ft (roughness coefficient mis-scaled)
1.86 ft (exponents dropped)

Reference: FE Reference Handbook — Hydraulics → Hazen-Williams Equation

Example 10

Hazen-Williams head loss in a water main — Hazen-Williams Equation (10)

A 8 in ductile iron main (C = 140) carries 1421 gpm over 3728 ft. Find the friction head loss and the hydraulic gradient.

Given

$Q = 1421 gpm$
$D = 8 in$
$C = 140$
$L = 3728 ft$

Find

h_f and the slope of the hydraulic grade line

Start with the thinking

The US customary Hazen-Williams form takes Q in gpm and D in inches and returns feet of head.
Head loss scales with Q^1.85 — doubling flow nearly quadruples the loss.

Step-by-step solution

Formula
$h_f = 10.44 L Q^1.85 / (C^1.85 D^4.87)$
Flow term
$Q^1.85 = 1421^1.85 = 679,671$
Coefficient term
$C^1.85 = 140^1.85 = 9,340$
Diameter term
$D^4.87 = 8^4.87 = 25,006$
Substituting
$h_f = 10.44 \times 3728 \times 679,671 / (9,340 \times 25,006)$
Evaluate
$h_f = 113.3 ft$
Gradient
$S = h_f/L = 0.03038 ft/ft$

Answer:

h_f ≈ 113.3 ft over 3728 ft (S ≈ 3.038%)

Why the other options are there

211.1 ft (roughness coefficient mis-scaled)
15.80 ft (exponents dropped)

Reference: FE Reference Handbook — Hydraulics → Hazen-Williams Equation