riaology
Unit 03 · Lesson 04 of 09

How does mechanical energy move with a fluid?

Pressure, speed, and height share a mechanical-energy account.

Explore the model. Derive the result. Check its limits.

What you’ll be able to do
  • State the mechanical-energy relation.
  • Combine continuity with energy.
  • Turn head into measurable pressure.
Before you touch the controls

Does a faster throat necessarily have a higher static pressure?

An experiment you can change

A Venturi trades pressure head for speed

Predict the direction of change, move one control, then explain the response using the equation.

Contraction: inlet speed 2, outlet speed 4.0816326530612255 meters per second2 m/s4.08 m/sNarrower throat: higher speed, lower pressureA Venturi trades pressure head for speed. Diameter ratio versus Pressure drop (kPa). Selected result 6.3299 kPa. The narrow section speeds up because mass flow is conserved; Bernoulli then connects that speed increase to lower static pressure.00.438.0620.776.1251Pressure drop (kPa)Diameter ratio
p1−p2=10002×1000[(2/0.72)2−22]=6.3299 kPap_1-p_2=\frac{1000}{2\times1000}\left[(2/0.7^2)^2-2^2\right]=6.3299\,\mathrm{kPa}

The narrow section speeds up because mass flow is conserved; Bernoulli then connects that speed increase to lower static pressure.

Model assumptions and units

Steady incompressible water, same elevation, negligible losses, no machines, along a streamline; uniform-speed approximation. Pressure-drop result does not itself check cavitation.

pρg+V22g+z=C,V2=V1(D1D2)2,p1−p2=ρ2(V22−V12)\frac p{\rho g}+\frac{V^2}{2g}+z=C,\quad V_2=V_1\left(\frac{D_1}{D_2}\right)^2,\quad p_1-p_2=\frac\rho2(V_2^2-V_1^2)
01

State the mechanical-energy relation.

For steady incompressible flow with negligible viscous effects and no machine work, along a streamline the sum of pressure, kinetic, and elevation heads is constant. Each head is energy per unit weight, measured in meters.

pρg+V22g+z=C\frac p{\rho g}+\frac{V^2}{2g}+z=C
02

Combine continuity with energy.

A Venturi throat has less area, so its speed rises. At equal elevation, the kinetic-head increase is balanced by a pressure-head decrease in the ideal model. Diameter ratio is squared in the area relation.

V2=V1(D1/D2)2V_2=V_1(D_1/D_2)^2
03

Turn head into measurable pressure.

Subtract Bernoulli at the two sections with equal elevations. Multiply by density times g to get the pressure difference. Use absolute pressure to check whether the throat could approach vapor pressure.

p1−p2=ρ2(V22−V12)p_1-p_2=\frac\rho2(V_2^2-V_1^2)
04

Know the limits.

Real devices have losses, and strongly nonuniform profiles may require kinetic-energy correction factors in a section-averaged balance. The ideal relation is a useful estimate when its assumptions fit, not a general statement that fast flow always means low pressure.

A worked example

A water Venturi.

  1. Use V1=2 m/sV_1=2\,\mathrm{m/s} and D2/D1=0.7D_2/D_1=0.7.
  2. Continuity gives V2=2/0.72=4.0816 m/sV_2=2/0.7^2=4.0816\,\mathrm{m/s}.
  3. Then p1−p2=1000(4.08162−22)/2=6.3299 kPap_1-p_2=1000(4.0816^2-2^2)/2=6.3299\,\mathrm{kPa}.
Make the idea your own

Calculate. Compare. Test the assumptions.

Work out your answer, check it, then explore the visual solution. The answer checker uses a 1.5% rounding tolerance; for model-limit questions it checks the displayed model result.

1 · Calculate

Read a Venturi pressure drop

Water enters at 2 m/s; throat/inlet diameter ratio is 0.70. Find pressure drop for the ideal same-height Venturi.

Hint

Square the diameter ratio to get area ratio, then use the speed-squared energy difference.

Explore the visual solution
Contraction: inlet speed 2, outlet speed 4.0816326530612255 meters per second2 m/s4.08 m/sNarrower throat: higher speed, lower pressureA Venturi trades pressure head for speed. Diameter ratio versus Pressure drop (kPa). Selected result 6.3299 kPa. The narrow section speeds up because mass flow is conserved; Bernoulli then connects that speed increase to lower static pressure.00.438.0620.776.1251Pressure drop (kPa)Diameter ratio
p1−p2=10002×1000[(2/0.72)2−22]=6.3299 kPap_1-p_2=\frac{1000}{2\times1000}\left[(2/0.7^2)^2-2^2\right]=6.3299\,\mathrm{kPa}

The narrow section speeds up because mass flow is conserved; Bernoulli then connects that speed increase to lower static pressure.

Model assumptions and units

Steady incompressible water, same elevation, negligible losses, no machines, along a streamline; uniform-speed approximation. Pressure-drop result does not itself check cavitation.

pρg+V22g+z=C,V2=V1(D1D2)2,p1−p2=ρ2(V22−V12)\frac p{\rho g}+\frac{V^2}{2g}+z=C,\quad V_2=V_1\left(\frac{D_1}{D_2}\right)^2,\quad p_1-p_2=\frac\rho2(V_2^2-V_1^2)

Controls explore variations. The steps below retain the original question’s values.

Steady incompressible water, same elevation, negligible losses, no machines, along a streamline; uniform-speed approximation. Pressure-drop result does not itself check cavitation.

pρg+V22g+z=C,V2=V1(D1D2)2,p1−p2=ρ2(V22−V12)\frac p{\rho g}+\frac{V^2}{2g}+z=C,\quad V_2=V_1\left(\frac{D_1}{D_2}\right)^2,\quad p_1-p_2=\frac\rho2(V_2^2-V_1^2)
2 · Compare

A milder contraction

At inlet speed 1 m/s and diameter ratio 0.80, find ideal pressure drop.

Hint

V2 = 1/0.8² m/s.

Explore the visual solution
Contraction: inlet speed 1, outlet speed 1.5624999999999998 meters per second1 m/s1.56 m/sNarrower throat: higher speed, lower pressureA Venturi trades pressure head for speed. Diameter ratio versus Pressure drop (kPa). Selected result 0.7207 kPa. The narrow section speeds up because mass flow is conserved; Bernoulli then connects that speed increase to lower static pressure.00.49.5160.719.0311Pressure drop (kPa)Diameter ratio
p1−p2=10002×1000[(1/0.82)2−12]=0.7207 kPap_1-p_2=\frac{1000}{2\times1000}\left[(1/0.8^2)^2-1^2\right]=0.7207\,\mathrm{kPa}

The narrow section speeds up because mass flow is conserved; Bernoulli then connects that speed increase to lower static pressure.

Model assumptions and units

Steady incompressible water, same elevation, negligible losses, no machines, along a streamline; uniform-speed approximation. Pressure-drop result does not itself check cavitation.

pρg+V22g+z=C,V2=V1(D1D2)2,p1−p2=ρ2(V22−V12)\frac p{\rho g}+\frac{V^2}{2g}+z=C,\quad V_2=V_1\left(\frac{D_1}{D_2}\right)^2,\quad p_1-p_2=\frac\rho2(V_2^2-V_1^2)

Controls explore variations. The steps below retain the original question’s values.

Steady incompressible water, same elevation, negligible losses, no machines, along a streamline; uniform-speed approximation. Pressure-drop result does not itself check cavitation.

pρg+V22g+z=C,V2=V1(D1D2)2,p1−p2=ρ2(V22−V12)\frac p{\rho g}+\frac{V^2}{2g}+z=C,\quad V_2=V_1\left(\frac{D_1}{D_2}\right)^2,\quad p_1-p_2=\frac\rho2(V_2^2-V_1^2)
3 · Challenge the model

Double speed, compare pressure

Use diameter ratio 0.70 and inlet speed 4 m/s. Find pressure drop and compare with the 2 m/s case.

Hint

At fixed geometry all speeds double, so their squared differences quadruple.

Explore the visual solution
Contraction: inlet speed 4, outlet speed 8.163265306122451 meters per second4 m/s8.16 m/sNarrower throat: higher speed, lower pressureA Venturi trades pressure head for speed. Diameter ratio versus Pressure drop (kPa). Selected result 25.3195 kPa. The narrow section speeds up because mass flow is conserved; Bernoulli then connects that speed increase to lower static pressure.00.4152.250.7304.51Pressure drop (kPa)Diameter ratio
p1−p2=10002×1000[(4/0.72)2−42]=25.319 kPap_1-p_2=\frac{1000}{2\times1000}\left[(4/0.7^2)^2-4^2\right]=25.319\,\mathrm{kPa}

The narrow section speeds up because mass flow is conserved; Bernoulli then connects that speed increase to lower static pressure.

Model assumptions and units

Steady incompressible water, same elevation, negligible losses, no machines, along a streamline; uniform-speed approximation. Pressure-drop result does not itself check cavitation.

pρg+V22g+z=C,V2=V1(D1D2)2,p1−p2=ρ2(V22−V12)\frac p{\rho g}+\frac{V^2}{2g}+z=C,\quad V_2=V_1\left(\frac{D_1}{D_2}\right)^2,\quad p_1-p_2=\frac\rho2(V_2^2-V_1^2)

Controls explore variations. The steps below retain the original question’s values.

Steady incompressible water, same elevation, negligible losses, no machines, along a streamline; uniform-speed approximation. Pressure-drop result does not itself check cavitation.

pρg+V22g+z=C,V2=V1(D1D2)2,p1−p2=ρ2(V22−V12)\frac p{\rho g}+\frac{V^2}{2g}+z=C,\quad V_2=V_1\left(\frac{D_1}{D_2}\right)^2,\quad p_1-p_2=\frac\rho2(V_2^2-V_1^2)
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