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Water hammer when the valve slams

Piping · surge pressure and pipe pressure rating

A 150 mm steel pump main carries 30 L/s to a filling station whose automatic valve, on a power failure, slams shut in well under a second. The pressure wave in water-filled steel of this size travels at about 1250 m/s. The pipe is Schedule 40 — 168.3 mm outside diameter, 7.11 mm wall — and the design review wants the surge set against a hoop-stress rating built on a 120 MPa allowable.

Given
  • Q = 30 L/s — Steady flow
  • D = 150 mm — Inside diameter
  • ρ = 1000 kg/m³ — Water density
  • a = 1250 m/s — Pressure-wave celerity
  • D_o = 168.3 mm — Pipe outside diameter
  • t = 7.11 mm — Wall thickness
  • S = 120 MPa — Allowable hoop stress
Determine
  1. (a)the flow velocity in the main
  2. (b)the Joukowsky surge when the valve slams
  3. (c)that surge expressed as a column of water
  4. (d)the internal pressure the pipe wall is good for
Step 1 of 4(a) · solve for Flow velocity

The surge is bought entirely by the moving water, so its velocity comes first: 30 L/s through the 150 mm bore is 1.70 m/s — a normal, conservative transmission velocity, which makes the number that follows all the more instructive.

Rearranged for v
v=4QπD2v = \tfrac{4Q}{\pi D^{2}}
Your values, in your units
v=4⋅(30 L/s)π⋅(150 mm)2v = \tfrac{4 \cdot \left( 30\ \text{L/s} \right)}{\pi \cdot \left( 150\ \text{mm} \right)^{2}}
Converted to base units
v=4⋅(1,800 L/min)π⋅(150 mm)2v = \tfrac{4 \cdot \left( 1{,}800\ \text{L/min} \right)}{\pi \cdot \left( 150\ \text{mm} \right)^{2}}
Answer
v=1.6977 m/sv = 1.6977\ \text{m/s}

Carried onward at full precision, not this rounded figure.

Open the Pipe Velocity from Flow and Diameter solver →

Step 2 of 4(b) · solve for Surge pressure

Joukowsky: ΔP = ρaΔv, with Δv the FULL 1.70 m/s because a slam takes all of it. The wave speed does the damage — 1250 m/s turns a walking-pace flow into a 2.12 MPa (308 psi) spike ON TOP of line pressure. Note what is absent: the pipe length and the line pressure. A slam is a slam.

Rearranged for ΔP
ΔP=ρ a Δv\Delta P = \rho \, a \, \Delta v
1.6977 m/scarried from step 1
Your values, in your units
ΔP=(1,000 kg/m3)⋅(1,250 m/s)⋅(1.69765 m/s)\Delta P = \left( 1{,}000\ \text{kg/m}^{3} \right) \cdot \left( 1{,}250\ \text{m/s} \right) \cdot \left( 1.69765\ \text{m/s} \right)
Answer
ΔP=2.1221 MPa\Delta P = 2.1221\ \text{MPa}

Carried onward at full precision, not this rounded figure.

Open the Water Hammer Surge (Joukowsky Equation) solver →

Step 3 of 4(c) · solve for Pressure head

Re-read the surge as head: 216 m of water column, appearing and reflecting in under a second. This is the form that explains the symptoms — a 216 m spike against a 10 m check-valve rating is why the clapper hammers, and why gauges installed for 0–60 m die at the filling station first.

Rearranged for h
h=Pρgh = \tfrac{P}{\rho g}
2.1221 MPacarried from step 2
Your values, in your units
h=(2,122,070 Pa)(1,000 kg/m3) (9.80665 m/s2)h = \tfrac{\left( 2{,}122{,}070\ \text{Pa} \right)}{\left( 1{,}000\ \text{kg/m}^{3} \right) \, \left( 9.80665\ \text{m/s}^{2} \right)}
Converted to base units
h=(2,122.07 kPa)(1,000 kg/m3) (9.80665 m/s2)h = \tfrac{\left( 2{,}122.07\ \text{kPa} \right)}{\left( 1{,}000\ \text{kg/m}^{3} \right) \, \left( 9.80665\ \text{m/s}^{2} \right)}
Answer
h=216.39 mh = 216.39\ \text{m}

Carried onward at full precision, not this rounded figure.

Open the Pressure Head (h = P/ρg) solver →

Step 4 of 4(d) · solve for Internal pressure

Barlow closes the review: P = 2St/D ≈ 10.1 MPa of allowable internal pressure. The steel shrugs at a 2.12 MPa surge — the finding is that the PIPE is fine and the fittings, gauges and thin-walled accessories are what the surge will find. Thin-wall plastic would read very differently here.

Rearranged for P
P=2StDP = \tfrac{2 S t}{D}
Your values, in your units
P=2⋅(120 MPa)⋅(7.11 mm)(168.3 mm)P = \tfrac{2 \cdot \left( 120\ \text{MPa} \right) \cdot \left( 7.11\ \text{mm} \right)}{\left( 168.3\ \text{mm} \right)}
Converted to base units
P=2⋅(120,000 kPa)⋅(7.11 mm)(168.3 mm)P = \tfrac{2 \cdot \left( 120{,}000\ \text{kPa} \right) \cdot \left( 7.11\ \text{mm} \right)}{\left( 168.3\ \text{mm} \right)}
Answer
P=10.139 MPaP = 10.139\ \text{MPa}

Carried onward at full precision, not this rounded figure.

Open the Barlow's Formula (Pipe Pressure Rating) solver →

The main runs at 1.70 m/s, so a slammed valve raises a Joukowsky surge of about 2.12 MPa (308 psi, or 216 m of water column) above line pressure; the Schedule 40 steel itself is good for about 10.1 MPa by Barlow, so the surge threatens the accessories, not the pipe wall.

Why this order

The chain is a complete surge screening in four lines, and the order is the argument. Velocity first, because Joukowsky's equation is indifferent to everything except how fast the water was moving and how fast the news of the stoppage travels: ΔP = ρaΔv, derived by Joukowsky on the Moscow water mains in 1898 and unchanged since. The head conversion is the same number in the unit that matches the nameplates on checks, gauges and instrument diaphragms, which is where surge damage actually presents. Barlow last, because the question a review must answer is not 'is there a surge' — there always is — but 'what does it exceed'. Here the wall has a factor of nearly five over the surge-plus-line pressure, and that is the correct, slightly deflating conclusion: in heavy-walled steel the pipe barrel is rarely the victim. Cross-check the celerity intuition: the surge in psi is roughly 0.6 × Δv(ft/s) × a/4660 in field terms, and 5.57 ft/s in steel gives ~308 psi ✓.

What the simple form hides is the closure time, and it forgives nothing twice. Joukowsky's full ΔP applies when the valve closes faster than the wave's round trip 2L/a — on a 500 m main that is 0.8 s, so 'slow' motorised valves an operator would swear are gentle still deliver the full slam; genuinely slow closure (several round trips) sheds the peak in proportion. The second trap is crediting the line's SOFT components everywhere: an air chamber or a length of HDPE drops the effective celerity dramatically — but only downstream of itself; the steel between the valve and the cushion still sees steel numbers. And the fix list runs in the order of this chain: slow the closure, lower the velocity (the surge is linear in v, so half the flow is half the spike), then armour or cushion what remains. Oversizing pipe to slow the water is the one measure that pays twice, in friction every day and in surge on the bad day.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.