Water Distribution Systems for the FE Civil Exam: Hazen-Williams, Pressure and Demand

NCEES FE Civil spec: Water Resources and Environmental Engineering, D. Water distribution systems. FE Reference Handbook: Civil Engineering chapter, Hydrology/Water Resources section (Hazen-Williams, fire hydrant discharge).

What the FE Civil Spec Covers

Section D covers water distribution systems: the pipes, pumps, storage tanks and hydrants that move treated water to customers. FE questions usually ask for the friction loss in a main, the pressure at a node, the flow split at a junction or in parallel pipes, or the effect of demand (average day, maximum day, peak hour and fire flow) on pipe sizing.

Key Concepts

Hydraulic grade line (HGL): elevation plus pressure head. Pressure head at a node equals HGL minus ground elevation.
Continuity at a junction: flow in equals flow out plus the demand taken at that node.
Loops: around any closed loop the head losses sum to zero (Hardy Cross principle). In parallel pipes the head loss is the same in each branch.
Demand: maximum day is often about 1.5 to 2 times average day, and peak hour about 2 to 4 times. Mains are checked for maximum day plus fire flow.
Pressure: normal service pressure is commonly about 40 to 80 psi, and a minimum residual of 20 psi is a typical requirement during fire flow.

Core Formulas

Hazen-Williams, velocity form (k = 0.849 for SI, 1.318 for US units):

V = k C R_{H}^{0.63} S^{0.54} S = \frac{h _{f}}{L}

For a full circular pipe, $R_{H} = D /4$ . A common US head-loss form (h and L in ft, Q in cfs, D in ft):

h_{f} = \frac{4.73 L Q ^{1.852}}{C ^{1.852} D ^{4.87}}

Pressure from head:

p = γ h 1 psi = 2.31 ft of water

Where It Is in the FE Reference Handbook

Look in the Civil Engineering chapter, Hydrology/Water Resources section. It has the Hazen-Williams equation, a table of C values, and fire hydrant and sprinkler discharge equations. The Fluid Mechanics chapter also lists Hazen-Williams next to Darcy-Weisbach and minor loss coefficients.

Worked Example

A 300 mm main, 1,500 m long, with C = 120, carries 0.09 m³/s from a reservoir whose water surface is at elevation 150 m to a node at ground elevation 95 m. Ignoring minor losses and velocity head, find the friction loss and the pressure at the node.

Step 1, velocity and hydraulic radius:

A = \frac{π ( 0.30 ) ^{2}}{4} = 0.0707 m^{2} V = \frac{0.09}{0.0707} = 1.273 m/s

R_{H} = \frac{0.30}{4} = 0.075 m

Step 2, solve Hazen-Williams for S:

S = (\frac{V}{0.849 C R _{H}^{0.63}})^{1/0.54} = (\frac{1.273}{0.849 ( 120 ) ( 0.075 ) ^{0.63}})^{1.852} = 0.00614

h_{f} = S L = 0.00614 (1500) = 9.2 m

Step 3, pressure at the node:

\frac{p}{γ} = 150 - 9.2 - 95 = 45.8 m

p = 9.81 (45.8) \approx 449 kPa (\approx 65 psi)

Answer: friction loss about 9.2 m; node pressure about 449 kPa.

How FE Questions Are Framed

Distribution questions often chain two steps: compute the head loss in a main, then use the energy equation to get the pressure at a hydrant or a high point. Keep a running HGL: start at the reservoir or tank water surface, subtract each pipe loss, and subtract the node elevation only at the end to get pressure head. Pump stations add head to the HGL; pressure reducing valves remove it.

For parallel pipes, assume a head loss, compute the flow in each branch from Hazen-Williams, and scale until the branch flows add up to the total; because flow varies with head loss to the 0.54 power, the split is the same for any head loss. For a network loop, check that the clockwise and counterclockwise losses balance. Storage tanks are sized to cover peak hour demand above the maximum day supply, plus fire and emergency reserve, so demand factors matter in sizing questions.

Common Mistakes

Using D instead of D/4 for the hydraulic radius.
Mixing the SI constant 0.849 with US units, or 1.318 with SI units.
Measuring pressure head from the reservoir bottom or pipe invert instead of from the node elevation.
Forgetting node demands when applying continuity at a junction.
Assuming flow splits equally in parallel pipes. Equal head loss, not equal flow, is the rule.

Practice Problems

Try these from the free PolyCalc FE Civil problem bank. Each one has a full worked solution.

A 300 mm diameter ductile iron water main 1,200 m long has a Hazen-Williams coefficient C = 130 and carries 0.10 m³/s. The friction headloss over this length is most nearly: See the worked solution or practice the Fluids & Hydraulics deck
At a pipe junction, 0.30 m³/s and 0.20 m³/s flow in from two pipes and 0.15 m³/s flows out in a third. The flow in the fourth pipe, which leaves the junction, is: See the worked solution or practice the Fluids & Hydraulics deck
For a pipe with Darcy friction factor f = 0.02, length L = 100 m, diameter D = 0.20 m, and average velocity V = 2 m/s, what is the Darcy-Weisbach head loss? Use g = 9.81 m/s². See the worked solution or practice the Fluids & Hydraulics deck
Water flows through a 0.20-m-diameter pipe at 0.060 m³/s. The average velocity is most nearly: See the worked solution or practice the Fluids & Hydraulics deck

Related Calculators and Examples

Frequently Asked Questions

Why do water utilities use Hazen-Williams instead of Darcy-Weisbach?

Hazen-Williams uses a single roughness coefficient C that does not depend on velocity or Reynolds number, which makes hand calculations and network models quick. It is only valid for water at ordinary temperatures in turbulent flow, while Darcy-Weisbach works for any fluid.

Does a higher Hazen-Williams C mean more or less head loss?

Less. C measures smoothness, so new plastic or lined pipe has C around 130 to 150 while old unlined cast iron can drop below 100. Head loss varies with C to the power negative 1.85.

How do I convert pressure head to psi?

For water, divide feet of head by 2.31 to get psi, or multiply psi by 2.31 to get feet. In SI units, multiply meters of head by 9.81 to get kilopascals.

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