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Closed Conduit Practice Problems — Hazen–Williams, Darcy–Weisbach, Pumps & Networks

Nine original pipe-flow problems with every step shown and every wrong answer traced to the mistake that produces it. Closed-conduit hydraulics is 7–11 of the 80 questions on the PE Civil: WRE exam and a guaranteed presence on the FE Civil exam — mostly tested through the traps: exponents, C factors, curve confusion, and sign errors.

Last reviewed: 2026-10-03

hf = 4.727 L Q1.852/(C1.852D4.87)  ·  hf = f(L/D)(V²/2g)  ·  hm = K V²/2g

Hazen–WilliamsUS units, D in feet; the 1.852/4.87 exponents are the exam's favorite trap
Darcy–Weisbachf from Moody/Haaland; expected when the stem gives roughness ε or a non-water fluid
Energy equationz1 + p1/γ + V1²/2g = z2 + p2/γ + V2²/2g + hf + Σhm — every listed K gets a term

Hazen–Williams

Problem 1 — Head loss in a distribution main

FE Civil Hydraulics & Hydrologic Systems Target pace: ~3 min

A 12-inch ductile-iron water main (C = 130), 2,500 ft long, carries 3.0 cfs. Compute the friction head loss by the Hazen–Williams equation.

  1. 11.0 ft
  2. 12.9 ft
  3. 17.9 ft
  4. 24.8 ft
Full solution
  1. US form with D in feet: D = 12/12 = 1.0 ft.
  2. hf = 4.727 L Q1.852/(C1.852D4.87) = 4.727(2500)(3.0)1.852/(1301.852 × 1.04.87).
  3. 3.01.852 = 7.652; 1301.852 = 8,215. hf = 4.727 × 2500 × 7.652/8,215 = 11.0 ft.

Answer: A — 11.0 ft.

Why the wrong answers are wrong: B (12.9) uses Q² instead of Q1.852 — the exponent slip the exam writes first. C (17.9) picks C = 100 (old pipe) when the stem gives C = 130. D (24.8) uses the SI constant 10.67 with foot units.

Problem 2 — Sizing a pipe to a head-loss limit

PE WRE Closed Conduit Hydraulics (7–11 of 80) Target pace: ~6 min

A 3,000-ft water main (C = 120) must carry 5.0 cfs with friction loss not exceeding 8 ft. What is the minimum commercial diameter (available: 12, 16, 18, 24 in)?

  1. 18 in
  2. 16 in
  3. 42 in
  4. 24 in
Full solution
  1. Solve Hazen–Williams for D: D = [4.727 L Q1.852/(C1.852hf)]1/4.87.
  2. Numerator: 4.727 × 3000 × 5.01.852 = 14,181 × 19.698 = 279,338. Denominator: 1201.852 × 8 = 7,113 × 8 = 56,907.
  3. D = (279,338/56,907)1/4.87 = (4.909)0.20534 = 1.387 ft = 16.6 in. Minimum commercial size at or above this: 18 in. (Check: 18-in gives hf = 5.5 ft < 8 ✓; 16-in gives 9.7 ft > 8 ✗.)

Answer: A — 18 in.

Why the wrong answers are wrong: B (16 in) drops the 1.852 exponents (uses Q²/C²), which undersizes the pipe at 15.1 in — and a 16-in pipe genuinely fails the limit at 9.7 ft. C (42 in) multiplies by hf instead of dividing (the algebra slip). D (24 in) satisfies the limit but isn't the minimum — the question asks for minimum, not comfortable.

Darcy–Weisbach & Moody

Problem 3 — Friction factor by Haaland

FE Civil Hydraulics & Hydrologic Systems Target pace: ~3 min

Water at 60°F (ν = 1.22 × 10−5 ft²/s) flows at 5 ft/s through a 6-inch commercial steel pipe (ε = 0.00015 ft). Estimate the Darcy friction factor.

  1. 0.0175
  2. 0.0154
  3. 0.00031
  4. 0.0268
Full solution
  1. Re = VD/ν = 5 × 0.5/1.22×10−5 = 204,900 (turbulent). ε/D = 0.00015/0.5 = 0.00030.
  2. Haaland: 1/√f = −1.8 log10[(ε/D/3.7)1.11 + 6.9/Re] = −1.8 log10[2.87×10−5 + 3.37×10−5] = −1.8(−4.205) = 7.569.
  3. f = 1/7.569² = 0.0175.

Answer: A — 0.0175.

Why the wrong answers are wrong: B (0.0154) ignores roughness (smooth-pipe). C (0.00031) applies the laminar f = 64/Re to fully turbulent flow — always check the regime first. D (0.0268) uses ε for aged/rusty pipe instead of the given new-steel value.

Problem 4 — Darcy–Weisbach head loss

PE WRE Closed Conduit Hydraulics (7–11 of 80) Target pace: ~6 min

An 18-inch concrete pipe (ε = 0.001 ft), 4,000 ft long, carries 8 cfs of water at 60°F. Compute the friction head loss.

  1. 15.7 ft
  2. 10.9 ft
  3. 1.31 ft
  4. 31.4 ft
Full solution
  1. A = π(1.5)²/4 = 1.767 ft²; V = 8/1.767 = 4.53 ft/s. Re = 4.53 × 1.5/1.22×10−5 = 556,600.
  2. ε/D = 0.001/1.5 = 0.000667. Haaland → f = 0.0185.
  3. hf = f(L/D)(V²/2g) = 0.0185 × (4000/1.5) × (4.53²/64.4) = 0.0185 × 2,666.7 × 0.3183 = 15.7 ft.

Answer: A — 15.7 ft.

Why the wrong answers are wrong: B (10.9) uses the smooth-pipe f. C (1.31) puts D = 18 (inches) in the L/D term. D (31.4) uses V²/g instead of V²/2g — exactly double, exactly wrong.

Minor Losses + Energy Equation

Problem 5 — Reservoir-to-reservoir discharge

PE WRE Closed Conduit Hydraulics (7–11 of 80) Target pace: ~6 min

Two reservoirs differ in surface elevation by 25 ft, connected by 2,000 ft of 12-inch pipe (f = 0.02). Minor losses: entrance K = 0.5, two 90° elbows at K = 0.3 each, exit K = 1.0. Find the discharge.

  1. 4.86 cfs
  2. 6.19 cfs
  3. 4.98 cfs
  4. 2.68 cfs
Full solution
  1. Energy equation between the two surfaces (both V ≈ 0, both atmospheric): 25 ft = [fL/D + ΣK] · V²/2g.
  2. fL/D = 0.02 × 2000/1.0 = 40.0; ΣK = 0.5 + 0.6 + 1.0 = 2.1. Total coefficient = 42.1.
  3. V²/2g = 25/42.1 = 0.5938 → V = √(0.5938 × 64.4) = 6.18 ft/s.
  4. Q = AV = 6.18 × π/4 = 4.86 cfs.

Answer: A — 4.86 cfs.

Why the wrong answers are wrong: B (6.19) stops at velocity — the question asks for discharge. C (4.98) drops every minor loss. D (2.68) runs the whole thing with g = 9.81 m/s².

Problem 6 — Velocity from available head

FE Civil Hydraulics & Hydrologic Systems Target pace: ~3 min

A 4-inch pipe discharges freely to the atmosphere. Entrance K = 0.5, one fully-open gate valve K = 0.15, exit K = 1.0, and the friction term fL/D = 18. The available head is 12 ft. Find the exit velocity.

  1. 6.27 ft/s
  2. 6.44 ft/s
  3. 6.19 ft/s
  4. 4.43 ft/s
Full solution
  1. Total loss coefficient: 18 + 0.5 + 0.15 + 1.0 = 19.65.
  2. V²/2g = 12/19.65 = 0.6107 → V = √(0.6107 × 64.4) = 6.27 ft/s.

Answer: A — 6.27 ft/s.

Why the wrong answers are wrong: B (6.44) forgets the exit loss — the discharge-to-atmosphere exit always costs one velocity head. C (6.19) counts the entrance twice. D (4.43) uses g instead of 2g in the velocity head.

Pump Operating Point

Problem 7 — Operating point from both curves ★ shown in full

PE WRE Closed Conduit Hydraulics (7–11 of 80) Target pace: ~6 min

One problem on each page is shown worked in full, so you can judge the quality before trusting the rest. This one has the highest trap density on the page — both curve equations and the intersection solve, shown end to end.

A pump with characteristic curve Hp = 120 − 25Q² (Q in cfs, H in ft) lifts water through 4,000 ft of 10-inch pipe (C = 120). The static lift is 40 ft and total minor-loss K = 25. Find the operating point.

  1. 706 gpm @ 58.2 ft
  2. 803 gpm @ 40.0 ft
  3. 867 gpm @ 26.8 ft
  4. 706 gpm @ 40.0 ft
Full solution
  1. System curve, friction term (Hazen–Williams, D = 10/12 = 0.8333 ft): hf = 4.727 L Q1.852/(C1.852D4.87) = 4.727(4000)Q1.852/(1201.852 × 0.83334.87) = 6.4806 Q1.852.
  2. System curve, minor-loss term: A = π(0.8333)²/4 = 0.5454 ft²; hm = 25 × Q²/(A² · 2g) = 25Q²/19.158 = 1.3050 Q².
  3. System curve: Hsys = 40 + 6.4806Q1.852 + 1.3050Q². Pump curve: Hp = 120 − 25Q².
  4. Intersect: 120 − 25Q² = 40 + 6.4806Q1.852 + 1.3050Q² → 80 = 26.305Q² + 6.4806Q1.852. Try Q = 1.5: 59.19 + 13.73 = 72.9 (low). Q = 1.6: 67.34 + 15.51 = 82.9 (high). Interpolate → Q = 1.572 cfs; check: 26.305(2.4718) + 6.4806(2.3119) = 65.02 + 14.98 = 80.00 ✓.
  5. Convert and read head: Q = 1.572 × 448.831 = 706 gpm; H = 120 − 25(1.572)² = 120 − 61.8 = 58.2 ft. (System curve at Q = 1.572: 40 + 14.98 + 3.23 = 58.2 ✓.)

Answer: A — 706 gpm @ 58.2 ft.

Why the wrong answers are wrong: B (803 gpm @ 40.0 ft) intersects the pump curve with the static lift alone — no friction, no minors — then reports the static head as the operating head. C (867 gpm @ 26.8 ft) forgets the 40 ft of static lift, so the “system” is pure friction. D (706 gpm @ 40.0 ft) gets the flow right, then reports the static lift instead of evaluating either curve at that flow.

Problem 8 — Affinity laws

FE Civil Hydraulics & Hydrologic Systems Target pace: ~3 min

A pump running at 1,750 rpm delivers 500 gpm at 60 ft of head. The speed is reduced to 1,450 rpm. Estimate the new flow rate (same system).

  1. 414 gpm
  2. 603 gpm
  3. 343 gpm
  4. 500 gpm
Full solution
  1. Affinity laws: Q ∝ N, H ∝ N², P ∝ N³.
  2. Q2 = Q1(N2/N1) = 500 × (1450/1750) = 414 gpm.

Answer: A — 414 gpm.

Why the wrong answers are wrong: B (603) inverts the speed ratio — flow would rise as speed falls, which should fail the smell test. C (343) squares the ratio (that's the head law, misapplied to flow). D (500) assumes speed doesn't matter.

Pipe Networks — Hardy Cross

Problem 9 — One full Hardy Cross iteration

PE WRE Closed Conduit Hydraulics (7–11 of 80) Target pace: ~6 min

A single loop A→B→C→A has three pipes with h = KQ|Q| (n = 2). K values (s²/ft5): AB = 2.0, BC = 5.0, CA = 3.0. Assumed clockwise flows: QAB = 4.00 cfs, QBC = 2.50 cfs, QCA = 1.50 cfs. After one Hardy Cross correction, what is the flow in pipe BC?

  1. 1.80 cfs
  2. 3.20 cfs
  3. 1.10 cfs
  4. 2.50 cfs
Full solution
PipeKQ0 (cfs)h = KQ|Q| (ft)|h/Q|Q1 (cfs)
AB2.0+4.00+32.0016.003.30
BC5.0+2.50+31.2525.001.80
CA3.0+1.50+6.759.000.80
  1. Σh = 32.00 + 31.25 + 6.75 = 70.00 ft (positive — the assumed clockwise flow is too large).
  2. Σ|h/Q| = 16.00 + 25.00 + 9.00 = 50.00.
  3. ΔQ = −Σh/(n Σ|h/Q|) = −70.00/(2 × 50.00) = −0.70 cfs.
  4. QBC,1 = 2.50 − 0.70 = 1.80 cfs.

Answer: A — 1.80 cfs.

Why the wrong answers are wrong: B (3.20) adds the correction instead of subtracting — the minus sign is part of the formula, and Σh > 0 already tells you flows must shrink. C (1.10) uses n = 1 in the denominator. D (2.50) reports the assumed flow, i.e., no correction applied.

Which friction equation does the stem want?

Hazen–WilliamsStem gives a C factor; fluid is water near 60°F; distribution-system context. D in feet, watch the 1.852/4.87 exponents.
Darcy–WeisbachStem gives roughness ε, asks for f explicitly, or the fluid isn't water. f from Moody/Haaland; any consistent units.
Energy equationReservoir-to-reservoir, tanks, or pumps: write the full equation and account for every listed K — entrance, exit, valves, bends.
System vs pump curveOperating point = intersection. System curve starts at static lift and rises with Q; pump curve falls with Q. Confusing either with the other is the whole trap.

Frequently asked questions

When does the exam expect Hazen–Williams vs. Darcy–Weisbach?

Hazen–Williams is the water-distribution workhorse: the stem gives a C factor and the fluid is water near room temperature. Darcy–Weisbach with the Moody chart (or Haaland) is expected when the stem gives roughness ε, a non-water fluid, or asks for the friction factor explicitly.

What are the most common Hazen–Williams mistakes?

Three: the 1.852/4.87 exponents (using 2 and 5 instead), diameter in inches instead of feet, and picking C for new pipe when the stem says the pipe is old — or the reverse. The exam writes one distractor for each.

Is the operating point the same as the best-efficiency point?

No. The operating point is wherever the pump curve crosses the system curve — it is set by the system, not the pump. Best efficiency is a property of the pump alone. A well-designed system puts the operating point near BEP, but the exam will not assume it.

In Hardy Cross, what sign does the flow correction get?

ΔQ = −Σh/(n·Σ|h/Q|). The negative sign is part of the formula — it drives the loop's head-loss imbalance toward zero. If you add a positive ΔQ when the formula says subtract, every corrected flow in the loop is wrong.

Do minor losses matter on the exam, or can I ignore them?

They matter whenever the stem lists them. Entrance, exit, valves, and bends each get their K·V²/2g term in the energy equation, and the exam includes a 'forgot the exit loss' distractor almost every time. Never drop a listed K.

Keep practicing

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