10 Pipe Flow Problems Solved Step by Step (Fluid Mechanics)

Pipe Flow Problems for Engineering Students

These 10 problems cover the core calculations in pipe flow hydraulics. Each includes full step-by-step solutions. Perfect for studying fluid mechanics, hydraulic engineering, or preparing for the FE/PE exam.

Problem 1: Reynolds Number

Water at 20°C flows through a 150 mm steel pipe at 3 L/s. Determine the flow regime.

Solution:

A = pi × 0.15² / 4 = 0.01767 m²
V = 0.003 / 0.01767 = 0.170 m/s
Re = V × D / v = 0.170 × 0.15 / (1.004×10⁻⁶)
Re = 25,398 → Turbulent (Re > 4000)

Problem 2: Friction Factor (Swamee-Jain)

For Problem 1, calculate the Darcy friction factor. Steel roughness ε = 0.045 mm.

Solution:

ε/D = 0.000045 / 0.15 = 0.0003
f = 0.25 / [log10(0.0003/3.7 + 5.74/25398^0.9)]²
f = 0.25 / [log10(0.0000811 + 0.000476)]²
f = 0.25 / [-3.253]²
f = 0.0236

Problem 3: Head Loss (Darcy-Weisbach)

Calculate the friction head loss for 200 m of the pipe from Problems 1-2.

Solution:

hf = f × (L/D) × V²/(2g)
hf = 0.0236 × (200/0.15) × 0.170²/(2×9.81)
hf = 0.0236 × 1333.3 × 0.001474
hf = 0.046 m

Very low head loss because the velocity is low (0.17 m/s). This pipe is oversized.

Problem 4: Head Loss (Hazen-Williams)

Repeat Problem 3 using Hazen-Williams. C = 120 for steel.

Solution:

hf = 10.67 × 200 × 0.003^1.852 / (120^1.852 × 0.15^4.87)
hf = 2134 × 1.664×10⁻⁵ / (6932.5 × 6.97×10⁻⁵)
hf = 0.0355 / 0.000483
hf = 0.073 m

Hazen-Williams gives a slightly different result (0.073 m vs 0.046 m). Both are acceptable for this low-velocity case.

Problem 5: Fitting Losses

A pipe section has 3 standard 90° elbows, 1 gate valve (open), and 1 check valve. V = 2.0 m/s. Calculate total fitting losses.

Solution:

K = 3×0.30 + 0.20 + 2.50 = 3.60
hm = 3.60 × 2.0²/(2×9.81)
hm = 3.60 × 0.2039
hm = 0.73 m

Problem 6: Total Dynamic Head (TDH)

A pump must deliver 5 L/s from a ground-level cistern to a tank 20 m above. Piping: 50 m of 3" PVC (C=150). Fittings: K_total = 6.5. Calculate TDH.

Solution:

Static head: Hg = 20 m

Friction (H-W):
D = 0.0762 m, Q = 0.005 m³/s
hf = 10.67 × 50 × 0.005^1.852 / (150^1.852 × 0.0762^4.87)
hf = 1.24 m

Velocity: V = 0.005/0.00456 = 1.10 m/s
Fittings: hm = 6.5 × 1.10²/(2×9.81) = 0.40 m

TDH = 20 + 1.24 + 0.40 = 21.64 m
With 15% safety: TDH = 24.9 m

Problem 7: Pump Power (WHP and BHP)

For Problem 6, calculate the required pump power. Pump efficiency = 65%.

Solution:

WHP = Q × TDH / 76 = 5 × 24.9 / 76 = 1.64 HP
BHP = WHP / η = 1.64 / 0.65 = 2.52 HP
Select motor: 3 HP (next standard size)

Problem 8: NPSH Available

Pump is 3 m above the water surface. Suction pipe: 4 m of 3" PVC with foot valve (K=2.5) and 1 elbow (K=0.3). Water at 25°C, sea level. Calculate NPSHa.

Solution:

Pa/(ρg) = 10.34 m
Pv/(ρg) = 0.324 m (at 25°C)
Hs = -3 m (pump above water)

Suction velocity = 1.10 m/s
hf_suction = 0.10 m (4 m PVC friction)
hm_suction = (2.5 + 0.3) × 1.10²/(2×9.81) = 0.17 m

NPSHa = 10.34 - 0.324 - 3 - 0.10 - 0.17 = 6.75 m

If pump NPSHr = 3.0 m → margin = 3.75 m ✅ Safe.

Problem 9: Optimal Pipe Diameter

Find the optimal discharge pipe diameter for 10 L/s at recommended velocity 1.5 m/s.

Solution:

D = sqrt(4Q / (π×V))
D = sqrt(4×0.010 / (π×1.5))
D = sqrt(0.00849)
D = 0.092 m = 92 mm

Select next commercial size: 100 mm (4"). Check velocity: V = 0.010/0.00785 = 1.27 m/s ✅ Within range.

Problem 10: Water Hammer Pressure Surge

Steel pipe 300 m long, velocity 2.0 m/s. Valve closes instantly. Wave speed c = 1100 m/s. Calculate pressure surge and critical closing time.

Solution:

Pressure surge:
ΔH = c × ΔV / g = 1100 × 2.0 / 9.81 = 224.3 m

Critical closing time:
tc = 2L / c = 2 × 300 / 1100 = 0.545 s

224 m of water hammer! Close the valve in at least 5 × tc = 2.7 seconds to reduce the surge.

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