Fluid Mechanics

6% of exam

Fluid properties and statics, flow measurement, and energy, impulse, and momentum of fluids.

3 concepts

Exam areas

A. Flow measurement

Flow Measurement

Pitot and pitot-static tubes, venturi and orifice meters, nozzles and the discharge coefficient, rectangular and V-notch weirs, and velocity-area stream gauging.

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B. Fluid properties

Fluid Properties and Statics

Density, specific weight, and viscosity, then the hydrostatic pressure field, manometry, forces on plane and curved surfaces, center of pressure, and buoyancy.

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D. Energy, impulse, and momentum of fluids

Energy Equation, Bernoulli, and Momentum

Continuity, the energy and Bernoulli equations in head form, energy and hydraulic grade lines, head loss, and the momentum equation for forces on bends and nozzles.

If fluid statics is about pressure at rest, the heart of fluid mechanics — and the densest cluster of FE Civil points — is what happens when the fluid moves. Two conservation laws do almost all of the work: conservation of mass gives continuity, and conservation of energy gives the Bernoulli/energy equation. A third, conservation of momentum, hands you the force a moving stream exerts on bends and nozzles. The FE Reference Handbook's Fluid Mechanics chapter writes the energy equation in head form (everything in meters or feet of fluid), and learning to think in heads — pressure head, velocity head, elevation head — is the single most valuable habit for this topic.

Continuity: conservation of mass

For steady flow, mass is neither created nor destroyed, so the mass flow rate

\dot m=\rho A v

is the same at every section. For an incompressible fluid (all liquids on the FE) density cancels and the volumetric flow rate

Q=Av

is constant. The immediate consequence is the one students forget under pressure: where a pipe narrows, the velocity must rise in inverse proportion to area, so halving the diameter quadruples the velocity. Continuity is what couples the two ends of a Bernoulli problem.

\dot m=\rho A v=\text{const}\;\Rightarrow\; Q=A_1v_1=A_2v_2

The energy equation in head form

Between two sections of a streamtube, the steady-flow energy equation balances pressure head

P/\gamma

, velocity head

v^2/2g

, and elevation head

z

, with sources and sinks added in: a pump adds head

h_p

, a turbine extracts

h_t

, and friction dissipates

h_f

. Every term carries units of length, so you can read the equation as an elevation budget for the fluid's mechanical energy. This is the workhorse — most pipe-system problems are one careful application of it.

\frac{P_1}{\gamma}+\frac{v_1^2}{2g}+z_1+h_p=\frac{P_2}{\gamma}+\frac{v_2^2}{2g}+z_2+h_t+h_f

Bernoulli: the frictionless special case

Drop the pump, turbine, and head loss and the energy equation collapses to Bernoulli's equation: the sum of the three heads is constant along a streamline. It is exact only for steady, incompressible, frictionless flow with no machine between the sections, yet it is astonishingly useful for short accelerating runs — nozzles, contractions, jets, and flow-measurement devices. Treat Bernoulli as the idealization and the energy equation as reality with

h_f

restored.

\frac{P_1}{\gamma}+\frac{v_1^2}{2g}+z_1=\frac{P_2}{\gamma}+\frac{v_2^2}{2g}+z_2

Energy and hydraulic grade lines

These two lines make a pipe system visible. The hydraulic grade line (HGL) plots the piezometric head

P/\gamma+z

— the level water would rise to in a piezometer tapped into the pipe. The energy grade line (EGL) lies a velocity head

v^2/2g

above the HGL and represents the total head. In steady flow with friction the EGL always slopes downward in the direction of flow (energy is lost); a pump steps it up, a turbine steps it down. Where the HGL drops below the pipe, pressure is sub-atmospheric — a cavitation warning.

\text{HGL}=\frac{P}{\gamma}+z,\qquad \text{EGL}=\frac{P}{\gamma}+z+\frac{v^2}{2g}

Head loss

Friction in a full pipe is captured by the Darcy-Weisbach equation,

h_f=f\,(L/D)(v^2/2g)

, where the friction factor

f

comes from the Moody diagram as a function of Reynolds number and relative roughness

\varepsilon/D

. Minor losses at fittings, entrances, exits, and area changes are written as a loss coefficient times the velocity head,

h_f=C\,v^2/2g

. Both forms feed the

h_f

slot in the energy equation; the velocity head

v^2/2g

is the common currency, so always compute it once and reuse it.

h_f=f\,\frac{L}{D}\,\frac{v^2}{2g},\qquad h_{f,\,minor}=C\,\frac{v^2}{2g}

The momentum equation: forces on bends and nozzles

Energy tells you pressures and velocities; momentum tells you forces. The impulse-momentum principle says the net external force on a control volume equals the rate of change of momentum flux,

\sum F=\rho Q(v_2-v_1)

, applied component by component. For a pipe bend, the external forces are the pressure forces on the two faces plus the unknown anchoring force from the pipe wall. Solve each direction:

P_1A_1-P_2A_2\cos\alpha-F_x=\rho Q(v_2\cos\alpha-v_1)

, and likewise in

y

. The reaction the fluid exerts on the bend is equal and opposite to

\boldsymbol F

— this sizes the thrust block.

P_1A_1-P_2A_2\cos\alpha-F_x=\rho Q\,(v_2\cos\alpha-v_1)

Exam strategy

Number your two sections, pick a datum, and write the energy equation in head form before touching numbers; cross out terms that are zero (open to atmosphere

\Rightarrow P=0

gauge; large reservoir

\Rightarrow v\approx0

). Use continuity to relate

v_1

and

v_2

— this is what closes the system at a contraction or nozzle. Keep gauge pressures throughout so atmospheric cancels. For force problems, FIRST get velocities from continuity and pressures from Bernoulli, THEN apply momentum per axis; watch the sign of

F_x

and remember the force on the fitting is the negative of the force on the fluid. Power is

\dot W=\gamma Q h

, divided by efficiency for brake/input power.

Key equations

Continuity (incompressible)

Q=A_1v_1=A_2v_2

Volumetric flow rate constant. $Q$ in $\text{m}^3/\text{s}$ or cfs; velocity rises where area falls.

Mass flow rate

\dot m=\rho A v=\rho Q

Conserved in steady flow for any fluid; reduces to constant $Q$ when $\rho$ is constant.

Energy equation (head form)

\frac{P_1}{\gamma}+\frac{v_1^2}{2g}+z_1+h_p=\frac{P_2}{\gamma}+\frac{v_2^2}{2g}+z_2+h_t+h_f

Steady incompressible flow with pump head $h_p$ , turbine head $h_t$ , friction head $h_f$ . Every term in length units.

Bernoulli equation

\frac{P_1}{\gamma}+\frac{v_1^2}{2g}+z_1=\frac{P_2}{\gamma}+\frac{v_2^2}{2g}+z_2

Frictionless, no-machine special case. Valid along a streamline for steady incompressible flow.

Hydraulic and energy grade lines

\text{HGL}=\frac{P}{\gamma}+z,\qquad \text{EGL}=\text{HGL}+\frac{v^2}{2g}

HGL = piezometric head; EGL = total head, one velocity head higher. EGL falls in the flow direction.

Pressure drop from head loss

P_1-P_2=\gamma h_f=\rho g h_f

For equal area and elevation, the pressure drop equals $\gamma$ times the friction head.

Darcy-Weisbach head loss

h_f=f\,\frac{L}{D}\,\frac{v^2}{2g}

Major friction loss; $f$ from the Moody diagram via $Re$ and $\varepsilon/D$ . $L$ , $D$ pipe length and diameter.

Minor (fitting) loss

h_{f}=C\,\frac{v^2}{2g}

Loss at valves, elbows, entrances, exits; $C$ (or $K$ ) given. Sudden exit to reservoir $C\approx1$ .

Reynolds number

Re=\frac{\rho v D}{\mu}=\frac{v D}{\nu}

Laminar $Re<2100$ , transitional $2100<Re<10^4$ , fully turbulent $Re>10^4$ . Sets $f$ and the flow regime.

Impulse-momentum principle

\textstyle\sum F=\rho Q\,(v_2-v_1)

Net external force = momentum flux out minus in. Apply per component.

Force on a pipe bend (x-component)

P_1A_1-P_2A_2\cos\alpha-F_x=\rho Q(v_2\cos\alpha-v_1)

$\alpha$ = bend angle; $F_x$ = anchoring force on fluid. Reaction on bend is $-F_x$ .

Hydraulic power

\dot W=\gamma Q h=\rho g Q h

Fluid power for a pump/turbine head $h$ ; divide by efficiency for brake/input power.

Worked examples

Pressure change through a contraction (Bernoulli + continuity)

Problem. Water flows at $Q=0.060\,\text{m}^3/\text{s}$ through a horizontal pipe that contracts from $200\,\text{mm}$ to $100\,\text{mm}$ diameter. The pressure upstream is $300\,\text{kPa}$ . Neglecting losses, find the downstream pressure.

Solution. Areas:

A_1=\tfrac{\pi}{4}(0.20)^2=0.03142\,\text{m}^2

A_2=\tfrac{\pi}{4}(0.10)^2=0.007854\,\text{m}^2

. Velocities (continuity):

v_1=Q/A_1=0.060/0.03142=1.910\,\text{m/s}

;

v_2=0.060/0.007854=7.639\,\text{m/s}

. Horizontal, so

z_1=z_2

; Bernoulli gives

P_2=P_1+\tfrac{\rho}{2}(v_1^2-v_2^2)=300{,}000+\tfrac{1000}{2}(1.910^2-7.639^2)

P_2=300{,}000+500(3.648-58.36)=300{,}000-27{,}357=272{,}643\,\text{Pa}=273\,\text{kPa}

. Sanity check: the stream speeds up, so pressure must fall — and it does, by about

27\,\text{kPa}

, the gain in velocity head.

P_2=P_1+\tfrac{\rho}{2}\left(v_1^2-v_2^2\right)

Pump head and power between two reservoirs

Problem. A pump lifts water at $Q=0.050\,\text{m}^3/\text{s}$ from a lower reservoir to one $30\,\text{m}$ higher. Total friction head loss in the line is $8.0\,\text{m}$ . Both surfaces are open and essentially still. Find the required pump head and the brake power if the pump efficiency is $75\%$ .

Solution. Energy equation between the two free surfaces:

P_1=P_2=0

gauge and

v_1\approx v_2\approx0

, so

h_p=(z_2-z_1)+h_f=30+8.0=38.0\,\text{m}

. Fluid power:

\dot W_{fluid}=\gamma Q h_p=(9810)(0.050)(38.0)=18{,}639\,\text{W}

. Brake power:

\dot W_{brake}=\dot W_{fluid}/\eta=18{,}639/0.75=24{,}852\,\text{W}=24.9\,\text{kW}

. Sanity check: the pump must overcome both static lift and friction, so

h_p>30\,\text{m}

;

\approx25\,\text{kW}

for

50\,\text{L/s}

38\,\text{m}

is a reasonable mid-size pump.

h_p=(z_2-z_1)+h_f,\qquad \dot W_{brake}=\frac{\gamma Q h_p}{\eta}

Anchoring force on a horizontal nozzle (momentum)

Problem. A horizontal nozzle reduces from $100\,\text{mm}$ to $50\,\text{mm}$ and discharges $Q=0.020\,\text{m}^3/\text{s}$ of water to the atmosphere. Find the magnitude of the axial force the flange bolts must resist.

Solution. Areas/velocities:

A_1=0.007854\,\text{m}^2

A_2=0.001963\,\text{m}^2

;

v_1=0.020/0.007854=2.546\,\text{m/s}

v_2=0.020/0.001963=10.19\,\text{m/s}

. Upstream pressure from Bernoulli (exit

P_2=0

gauge):

P_1=\tfrac{\rho}{2}(v_2^2-v_1^2)=500(103.8-6.48)=48{,}634\,\text{Pa}

. Momentum in flow direction (

\alpha=0

P_1A_1-F_x=\rho Q(v_2-v_1)

P_1A_1=48{,}634\times0.007854=382.0\,\text{N}

;

\rho Q(v_2-v_1)=1000(0.020)(10.19-2.546)=152.8\,\text{N}

F_x=382.0-152.8=229\,\text{N}

. The fluid pushes the nozzle forward with

229\,\text{N}

, so the bolts carry

\approx229\,\text{N}

. Sanity check: pressure thrust (

382\,\text{N}

) exceeds the momentum term, so the net bolt force points downstream and is a couple hundred newtons — sensible for this small jet.

F_x=P_1A_1-\rho Q\,(v_2-v_1)

Common pitfalls

•Applying Bernoulli across a pump, turbine, or long lossy run. The instant a machine or significant friction sits between your sections, you must use the full energy equation with $h_p$ , $h_t$ , or $h_f$ .
•Forgetting continuity at area changes — leaving $v_2$ unknown. The two ends are coupled by $Q=A_1v_1=A_2v_2$ ; without it a contraction problem has too many unknowns.
•Mixing gauge and absolute pressure between sections. Keep both gauge so atmospheric cancels; a jet discharging to air has $P=0$ gauge at the exit.
•Confusing EGL and HGL: the EGL is the total head and lies a velocity head ABOVE the HGL. The HGL dipping below the pipe signals sub-atmospheric pressure, not the EGL.
•Sign errors in the momentum equation: use a consistent positive axis, include pressure forces on BOTH faces, and remember the force on the fitting is the negative of the force on the fluid.
•Dropping the velocity-head change in nozzle force problems — you must get $P_1$ from Bernoulli first; assuming $P_1=0$ understates the bolt force badly.
•Power without efficiency: $\gamma Q h$ is fluid (hydraulic) power. Divide by $\eta$ for a pump's brake/input power; multiply by $\eta$ for a turbine's output.

References

NCEES FE Reference Handbook — Fluid Mechanics (Continuity, Energy, and Bernoulli Equations)
NCEES FE Reference Handbook — Fluid Mechanics (Impulse-Momentum Principle; Pipe Bends)
NCEES FE Reference Handbook — Fluid Mechanics (Head Loss; Hydraulic and Energy Grade Lines)