Bernoulli & the Venturi Effect

Narrow a pipe and watch the speed rise and the pressure fall, until the water reaches the point where it boils instead.

Worked example — Bernoulli & the Venturi Effect

Speed ratio	4.00×
Throat velocity	8.00 m/s
Pressure change	-29.95 kPa
Throat absolute pressure	71.38 kPa
Throat Mach number	0.005
Flow rate	3.93 L/s
Cavitates below this throat	18.75 mm
Cavitates above this velocity	3.64 m/s

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Bernoulli & the Venturi Effect

Two statements about a pipe, and everything on this page is their consequence. Continuity: what goes in comes out, so A₁v₁ = A₂v₂ and a narrower throat means faster water. Bernoulli: along a streamline the total of pressure and kinetic energy per unit volume is fixed, so p + ½ρv² = constant and the faster water has to be at lower pressure. Not because anything pushes it there — because the energy is already spent on speed.

What each symbol means

A: the cross-section of the pipe
v: the speed of the fluid through it
ρ: the density of the fluid

Assumes: Incompressible, inviscid, steady flow along a single streamline. Viscosity is the one that matters most here: a real venturi does not recover all of its pressure downstream, and the difference is the pressure permanently lost to friction — which is what the meter costs to run.

The speed squares the ratio; the pressure squares the speed 🖖

Set the inlet to 50 mm and the throat to 25 mm, with water at 2 m/s. The throat velocity row reads 8.00 m/s and the pressure change reads −29.95 kPa. Continuity is responsible for the first of those: the same volume through a quarter of the area has to move four times as fast, and the speed ratio row says 4.00×, the square of 50/25. Bernoulli does the rest, and its kinetic term carries v², so four times the speed is sixteen times the kinetic pressure.

Now type 10 into the throat box. Speed ratio 25.00×, throat speed 50.00 m/s, pressure change −1245.75 kPa. The inlet is now five times the throat rather than twice, and the drop is 41.6 times the one you had just read. The second panel is where this is visible rather than arithmetic: it holds the inlet and the fluid fixed and sweeps the throat alone, narrow to the right, and the curve stays almost level across the wide throats before it plunges.

Where the arithmetic keeps going and the water does not 🖖

That −1245.75 kPa comes off an ambient 101.325 kPa, which puts the throat at −1144.43 kPa absolute. No fluid of any kind can be at a negative absolute pressure, and water gives up a long way above zero: at 20 °C it boils at 2339 Pa, and the shaded band along the bottom of both panels is drawn at exactly that line. Watch the pressure curve cross into it and turn red.

The row prints it anyway, marked as not physical, because the number is the lesson. And the tool solves the boundary from the other side. With a 50 mm inlet, water at 2 m/s and sea-level ambient, the Cavitates below this throat row reads 18.75 mm: squeeze past that and the liquid tears rather than thins. Type 19 into the throat box, then 18, and the verdict flips between the two.

Ten metres of water costs almost a whole atmosphere 🖖

Ten metres of height costs almost the whole atmosphere, and it costs it whether or not anything is narrowing. Set both diameters to 50 mm, then drag the height slider, which sets how far the throat sits above the inlet, to +10.0 m. The pressure change row reads −97.92 kPa, and every part of that is ρgh: water at 998.2 kg/m³, g = 9.81 m/s², ten metres. Set against the 101.325 kPa the atmosphere supplies, the height alone has spent 96.6% of it, through a pipe that is still perfectly straight. The throat absolute pressure row reads 3.40 kPa, about one kPa above the vapour line.

The slider stops at 10 m and the water stops at 10.11 m, where ρgh finishes off what is left. A suction pump lifts nothing; it lowers the pressure above the water and lets the atmosphere push the column up. The lowest pressure it can leave behind is the 2339 Pa from two blocks ago, where the water boils rather than rising further, so ten metres is the whole budget and the quality of the pump is not what is being tested.

Problem solved in full

Reading the flow off a Venturi meter, and the flow it can never read 8 steps
A 50 mm water main carries a Venturi insert with a 25 mm throat. Press Classic Venturi to set it up. The differential gauge across the two tappings reads 29.95 kPa. What is the flow, and what is the largest flow this meter could ever report?

Open this problem in the tool
1. Whatever volume passes the inlet each second passes the throat as well, so where the area falls the speed rises to keep the product constant. Area goes as the square of the diameter, which is why halving the bore does not double the speed but quadruples it. Speed ratio reads 4.00× and Throat velocity 8.00 m/s.
  
  $\frac{A _{1}}{A _{2}} = (\frac{d _{1}}{d _{2}})^{2} = (\frac{50}{25})^{2} = 4.00 v_{2} = v_{1} (\frac{d _{1}}{d _{2}})^{2} = 2 \times 4.00 = 8.00 m s^{- 1}$
2. Bernoulli's equation pays for that speed out of the pressure. The pipe is level, so the two ρgh terms cancel and only the kinetic terms are left. The kinetic term carries v², so four times the speed costs sixteen times the inlet's dynamic pressure, one of which was already being spent: the drop is fifteen times ½ρv₁², and ½ρv₁² here is 1996.4 Pa. Pressure change reads −29.95 kPa.
  
  $Δ p = \frac{1}{2} ρ (v_{1}^{2} - v_{2}^{2}) = \frac{1}{2} (998.2) (2^{2} - 8^{2}) = - 29, 946 Pa$
3. Take that off the atmosphere at the inlet tapping and the throat sits at 71.38 kPa absolute. There is room. Water at 20 °C boils at 2339 Pa, so 69 kPa of margin is still unspent. Throat Mach number reads 0.005, which is the other assumption inside Bernoulli's equation reporting that it has not been strained either.
  
  $p_{2} = p_{amb} + Δ p \Rightarrow 101.325 - 29.946 = 71.379 kPa M = \frac{v _{2}}{c} = \frac{8.00}{1481} = 0.005$
4. None of that is what a Venturi meter does. A meter has no velocity slider. It has two tappings and a differential gauge, and flow is the unknown it was built to find. So run the same equation the other way, with Δp known and v₁ wanted. Feed the gauge's 29.95 kPa back in and the inlet velocity comes out at 2.000 m/s, which is where the inlet was set.
  
  $Δ p = - \frac{1}{2} ρ v_{1}^{2} [(\frac{d _{1}}{d _{2}})^{4} - 1] \Rightarrow v_{1} = \frac{2 ∣ Δ p ∣}{ρ [ ( d _{1} / d _{2} ) ^{4} - 1 ]} = \frac{2 ( 29 , 950 )}{( 998.2 ) ( 15 )} = 4.0005 = 2.000 m s^{- 1}$
5. Multiply by the inlet area, π(0.050)²/4 = 1.9635×10⁻³ m², and the Flow rate row's 3.93 L/s has been obtained from a pressure reading instead of from a velocity. A meter datasheet writes the two steps as one equation with β for the diameter ratio d₂/d₁, which is 0.5 here. That equation is the reason the device exists, and it is the one thing on this page the tool never evaluates.
  
  $Q = v_{1} A_{1} = 2.000 \times 1.9635 \times 1 0^{- 3} = 3.927 \times 1 0^{- 3} m^{3} s^{- 1} Q = A_{2} \frac{2 ∣ Δ p ∣}{ρ ( 1 - β ^{4} )} β = \frac{d _{2}}{d _{1}} = 0.5$
6. Every meter has a ceiling, and this one belongs to the water rather than to the gauge. Squeeze the throat and the pressure there falls as the fourth power of the diameter ratio until it reaches 2339 Pa, at which point the liquid boils instead of accelerating. Solve for the throat where that happens at today's flow: the bracket comes to 50.582, its fourth root is 2.6669, and Cavitates below this throat reads 18.75 mm. The insert's 25 mm throat is a third wider than that.
  
  $d_{2}^{*} = d_{1} [1 + \frac{2 ( p _{amb} - p _{v} )}{ρ v _{1}^{2}}]^{- 1/4} = d_{1} (50.582)^{- 1/4} \frac{50}{2.6669} = 18.75 mm$
7. The same boundary from the other side, with the throat held at 25 mm. What the fluid has to spend is ambient less the vapour pressure, 98,986 Pa, and what spends it is ρ[(d₁/d₂)⁴ − 1]. Cavitates above this velocity reads 3.64 m/s.
  
  $v_{1}^{m a x} = \frac{2 ( p _{amb} - p _{v} )}{ρ [ ( d _{1} / d _{2} ) ^{4} - 1 ]} = \frac{2 ( 98 , 986 )}{( 998.2 ) ( 15 )} = 13.222 = 3.636 m s^{- 1}$
8. No row turns that velocity into a flow, so turn it. 3.636 m/s through a 50 mm bore is 7.14×10⁻³ m³/s, which is 7.14 L/s. The meter's full scale is 1.82 times the flow it is reading now, and the largest differential it will ever generate is 98.99 kPa: ambient less the vapour pressure, with nothing available past it. Push harder and the throat fills with vapour, the differential stops tracking the flow, and the gauge is reading a boiling pocket.
  
  $Q_{m a x} = v_{1}^{m a x} A_{1} = 3.636 \times 1.9635 \times 1 0^{- 3} = 7.14 \times 1 0^{- 3} m^{3} s^{- 1} \frac{7.14}{3.927} = 1.818$
Answer

3.93 L/s now, and 7.14 L/s ever. The tool prints neither, and the second is the one that decides the design.

Press Gentle narrowing to open the throat to 40 mm. Cavitates above this velocity rises from 3.64 to 11.73 m/s, which is 23.0 L/s, while Pressure change collapses from −29.95 kPa to −2.88 kPa. Both figures come out of the same bracket, (d₁/d₂)⁴ − 1, sitting in the numerator of one and under a square root in the denominator of the other, so range can only ever grow as the inverse square root of the signal surrendered: ten times less differential buys 3.2 times more flow. Choosing β is that trade, and a throat wide enough never to cavitate is a throat asking the gauge to resolve a few hundred pascals.

$Q_{m a x} = A_{1} \frac{2 ( p _{amb} - p _{v} )}{ρ [ ( d _{1} / d _{2} ) ^{4} - 1 ]}$

By the 42Math team · Reviewed against published sources. Corrections welcome.