Energy, Impulse and Momentum of Fluids

NCEES FE Civil spec: Fluid Mechanics, D. Energy, impulse, and momentum of fluids. FE Reference Handbook: Fluid Mechanics chapter, energy equation and impulse-momentum principle.

Moving fluids obey three conservation laws: mass (continuity), energy (Bernoulli and the energy equation) and momentum. Continuity and energy give velocities and pressures in pipes and channels; momentum gives the forces fluids exert on bends, nozzles and plates. Together they are the core of the FE Civil fluids and hydraulics questions.

What the NCEES spec covers

Section D, "Energy, impulse, and momentum of fluids", includes continuity, the Bernoulli equation, the energy equation with head loss, pump and turbine head, hydraulic and energy grade lines, and the linear momentum equation for jets, bends and nozzles.

Key concepts

Continuity. For incompressible flow, Q = AV is constant along a pipe, and flows into a junction equal flows out.
Heads. Every term of the energy equation is energy per unit weight, in meters or feet: pressure head p/γ, velocity head V²/2g and elevation head z.
Bernoulli vs energy equation. Bernoulli assumes no losses and no pumps. The energy equation adds pump head, turbine head and friction and minor losses.
Grade lines. The energy grade line (EGL) is the total head; the hydraulic grade line (HGL) sits one velocity head below it.
Momentum. The net external force on a control volume equals the rate of momentum flowing out minus flowing in. It is a vector equation, so work each direction separately.

Core formulas

Q = A_{1} V_{1} = A_{2} V_{2}

\frac{p _{1}}{γ} + \frac{V _{1}^{2}}{2 g} + z_{1} + h_{p} = \frac{p _{2}}{γ} + \frac{V _{2}^{2}}{2 g} + z_{2} + h_{t} + h_{L}

h_{f} = f \frac{L}{D} \frac{V ^{2}}{2 g} P_{pump} = \frac{γ Q h _{p}}{η}

\sum F = ρQ (V_{2} - V_{1})

Where it is in the FE Reference Handbook

The Fluid Mechanics chapter covers the continuity equation, the energy equation, head loss (Darcy and minor losses with the Moody diagram), pump power and the impulse-momentum principle with examples for pipe bends, jets and nozzles.

Worked example

Water flows at 0.080 m³/s through a 200 mm pipe of constant diameter. At point 1 the pressure is 300 kPa and the elevation is 0. Point 2 is 10 m higher, and the head loss between the points is 4.0 m. (a) Find the pressure at point 2. (b) Separately, a 0.010 m³/s water jet at 20 m/s strikes a fixed flat plate head-on and spreads out sideways. Find the force on the plate.

(a) Velocity is the same at both points (same diameter), so velocity heads cancel:

V = \frac{0.080}{π ( 0.20 ) ^{2} /4} = 2.55 m/s at both points

\frac{300}{9.81} + 0 = \frac{p _{2}}{9.81} + 10 + 4.0 \Rightarrow \frac{p _{2}}{9.81} = 30.58 - 14.0 = 16.58 m

p_{2} = 16.58 (9.81) = 163 kPa

(b) Momentum in the jet direction. The outflow leaves sideways, so its velocity in the jet direction is zero:

F = ρQ (V_{1} - 0) = 1, 000 (0.010) (20) = 200 N

The plate must push back on the water with 200 N, so the jet exerts 200 N on the plate.

Common mistakes

Mixing pressure in kPa with heads in meters. Divide kPa by 9.81 kN/m³ to get meters of water.
Adding head loss on the wrong side of the energy equation. Losses go on the downstream side.
Dropping velocity heads when the diameters are different.
Using Bernoulli across a pump or a long pipe with significant friction.
Treating the momentum equation as scalar on a bend. Resolve x and y separately and include pressure forces.

See the Bernoulli reservoir example.

Practice Problems

Try these from the free PolyCalc FE Civil problem bank. Each one has a full worked solution.

A circular pipe's diameter is halved while its volumetric flow rate remains constant. How does its average velocity change? See the worked solution or practice the Fluids & Hydraulics deck
For steady incompressible flow along a streamline, neglecting losses and pump or turbine work, Bernoulli's equation expresses conservation of: See the worked solution or practice the Fluids & Hydraulics deck
Steady water flow of 0.10 m³/s changes average velocity from 2 to 5 m/s in the same direction. What is the magnitude of the net external force on the water in that direction? Use ρ = 1,000 kg/m³ and uniform velocity profiles. See the worked solution or practice the Fluids & Hydraulics deck
For a pipe with Darcy friction factor f = 0.02, length L = 100 m, diameter D = 0.20 m, and average velocity V = 2 m/s, what is the Darcy-Weisbach head loss? Use g = 9.81 m/s². See the worked solution or practice the Fluids & Hydraulics deck
A pump moves water between two large reservoirs open to the atmosphere. The static elevation rise is 20 m and total pipe losses are 5 m. What pump head is required? See the worked solution or practice the Fluids & Hydraulics deck
A pump moves water at Q = 0.05 m³/s through a total head of 20 m with 80% pump efficiency. What is the required shaft input power? Use ρ = 1,000 kg/m³ and g = 9.81 m/s². See the worked solution or practice the Fluids & Hydraulics deck

Related Calculators and Examples

Frequently Asked Questions

What is the difference between the Bernoulli equation and the energy equation?

Bernoulli assumes steady, incompressible, frictionless flow with no pumps or turbines. The energy equation adds pump head, turbine head and head losses, so it applies to real pipe systems.

How do you calculate the force of a water jet on a plate?

Apply the momentum equation in the jet direction: force equals density times flow rate times the change in velocity. For a jet that strikes a flat plate head-on and leaves sideways, F = rho Q V.

What is the hydraulic grade line?

It is the line of pressure head plus elevation head along a pipe. The energy grade line lies one velocity head above it, and both slope downward in the flow direction because of head loss.

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