Fluid Mechanics

A. Fluid properties

Fluid Properties and Flow Regimes

Density, specific weight, specific gravity, viscosity, surface tension, compressibility, and the Reynolds-number split between laminar and turbulent flow.

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B. Dimensionless numbers

Dimensionless Numbers and Similitude

Reynolds, Froude, Mach, and Weber numbers as force ratios, the Buckingham Pi theorem, and how to scale a model to a prototype with dynamic similarity.

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D. Fluid statics

Fluid Statics

Hydrostatic pressure, manometry, resultant force and center of pressure on submerged surfaces, and Archimedes' buoyancy and floating stability.

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E. Energy, impulse, and momentum equations

Continuity, Bernoulli, and Momentum

The three control-volume workhorses — conservation of mass, the Bernoulli energy equation in head form, and the impulse-momentum principle for forces on bends, nozzles, and jets.

Three conservation laws — mass, energy, and momentum — carry most of the Fluid Mechanics points on the FE-OTH, and the handbook's "Principles of One-Dimensional Fluid Flow" section is where they live. Continuity tells you how velocity changes when a pipe narrows; Bernoulli trades pressure, velocity, and elevation head along a streamline; and the impulse-momentum principle gives the force a moving fluid exerts on a bend, nozzle, or vane. The recurring exam errors are mechanical, not conceptual: dropping a head term, using gauge where absolute is needed, or — most often — forgetting that momentum is a ·vector· so pressure forces and momentum flux must be resolved by component. Build the habit of writing the full equation, then crossing out the terms the problem zeroes.

Conservation of mass — continuity

Mass cannot accumulate in a steady flow, so the mass flow rate is the same at every cross section:

\dot m = \rho A v

is constant. When the fluid is incompressible (

\rho

constant) this reduces to constant volumetric flow

Q = Av

, which means velocity rises where the pipe narrows. This single relation links the two velocities in almost every Bernoulli problem.

\dot m = \rho_1 A_1 v_1 = \rho_2 A_2 v_2, \qquad Q = A_1 v_1 = A_2 v_2

The Bernoulli equation

For steady, incompressible, frictionless flow along a streamline with no pump or turbine, the sum of pressure energy, kinetic energy, and potential energy per unit weight is constant between two points. Each term is a ·head· — a height of fluid — so the whole equation has units of length. It is the energy budget of a streamline: speed up and the pressure must fall.

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

The energy equation with losses and machines

Real systems lose head to friction and may add or remove it with pumps and turbines. The extended energy equation inserts a pump head

h_p

(energy in), a turbine head

h_t

(energy out), and a friction head loss

h_f

. This is the form you actually use for pump-and-pipe problems; Bernoulli is just the special case with

h_p = h_t = h_f = 0

.

\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

Energy grade line versus hydraulic grade line

The energy grade line (EGL) plots the total head — pressure + velocity + elevation head — along the system; the hydraulic grade line (HGL) plots only pressure + elevation head. The two differ by exactly the velocity head

v^2/2g

, so the HGL sits below the EGL by that amount and the gap widens where the flow speeds up. In frictionless flow the EGL is horizontal; friction makes it slope downward in the direction of flow.

EGL - HGL = \frac{v^2}{2g}

Torricelli's result and stagnation

Two famous special cases fall straight out of Bernoulli. A tank draining through a small hole a depth

h

below the surface gives the efflux speed

v = \sqrt{2gh}

(atmospheric on both ends, negligible surface velocity). And bringing a stream to rest at a stagnation point converts all velocity head to pressure: the stagnation pressure exceeds the static pressure by the dynamic pressure

\tfrac12\rho v^2

— the basis of the pitot tube.

v = \sqrt{2gh}, \qquad P_0 = P_s + \tfrac{1}{2}\rho v^2

The impulse-momentum principle

Newton's second law for a control volume says the net external force equals the rate of change of momentum flux through it. For steady one-inlet, one-outlet flow this is

\sum \vec F = \dot m(\vec v_2 - \vec v_1) = \rho Q(\vec v_2 - \vec v_1)

. Because it is a vector equation, you must write a separate scalar equation for each direction, including the pressure forces

PA

on the inlet and outlet faces.

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

Forces on bends, nozzles, and jets

To find the anchoring force on a bend or nozzle, draw the control volume, mark inlet/outlet pressures and velocities, and apply momentum component-by-component. In the

x

-direction,

P_1 A_1 - P_2 A_2\cos\alpha - F_x = \rho Q(v_2\cos\alpha - v_1)

; the reaction

F_x

is the force the ·bend· exerts on the fluid, and the fluid pushes back on the bend equal and opposite. A jet striking a fixed plate normal to it (

v_2

has no component back along the jet) exerts

F = \rho Q v

.

P_1 A_1 - P_2 A_2\cos\alpha - F_x = \rho Q\,(v_2\cos\alpha - v_1)

Exam strategy

Pick your two points where you know the most (free surfaces and atmospheric exits zero out pressure and velocity heads). Write the full energy equation, then strike the zero terms. Use continuity to relate the two velocities ·before· substituting. For force problems, switch tools: continuity + Bernoulli give the velocities and pressures, then momentum (resolved into

x

and

y

, with

PA

terms included) gives the force — and remember the force on the structure is the negative of the force on the fluid. Keep everything in absolute or gauge consistently, and carry units of head (m or ft) throughout.

Key equations

Continuity (mass)

\dot m = \rho A v = \text{const}

Steady-flow mass conservation; $\dot m$ in kg/s. Same at every cross section.

Continuity (volume)

Q = A_1 v_1 = A_2 v_2

Incompressible flow; velocity scales as $1/A$ , i.e. as $(D_1/D_2)^2$ for round pipes.

Bernoulli equation (head form)

\frac{P}{\gamma} + \frac{v^2}{2g} + z = \text{const}

Steady, incompressible, frictionless, no machine. Each term is a head (length).

Energy equation with machines and losses

\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

$h_p$ pump head added, $h_t$ turbine head removed, $h_f$ friction loss; all in head units.

EGL minus HGL

EGL - HGL = \frac{v^2}{2g}

Hydraulic grade line lies below the energy grade line by the velocity head.

Torricelli efflux velocity

v = \sqrt{2gh}

Speed of fluid leaving a small orifice a depth $h$ below a large free surface.

Stagnation (dynamic) pressure

P_0 = P_s + \tfrac{1}{2}\rho v^2

Pitot-tube relation; $\tfrac12\rho v^2$ is the dynamic pressure, valid for $Ma<0.3$ .

Impulse-momentum principle

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

Vector equation; resolve by component. Include $PA$ pressure forces on inlet/outlet faces.

Force on a pipe bend (x-component)

P_1 A_1 - P_2 A_2\cos\alpha - F_x = \rho Q\,(v_2\cos\alpha - v_1)

$F_x$ = force of bend on fluid; the fluid's force on the bend is equal and opposite.

Jet force on a fixed flat plate (normal)

F = \rho Q v = \rho A v^2

Force on a plate held perpendicular to a free jet of velocity $v$ , area $A$ .

Worked examples

Continuity through a contraction

Problem. Water flows in a horizontal pipe that contracts from $D_1 = 100\,\text{mm}$ to $D_2 = 50\,\text{mm}$ . The inlet velocity is $2\,\text{m/s}$ . Find the volumetric flow rate and the outlet velocity.

Solution.

A_1 = \tfrac{\pi}{4}(0.100)^2 = 7.854\times10^{-3}\,\text{m}^2

.

Q = A_1 v_1 = (7.854\times10^{-3})(2) = 1.57\times10^{-2}\,\text{m}^3/\text{s}

.

v_2 = v_1 (D_1/D_2)^2 = 2(100/50)^2 = 2(4) = 8.0\,\text{m/s}

. Sanity check: area drops by

4\times

, so velocity rises by

4\times

— consistent. Answer:

Q = 1.57\times10^{-2}\,\text{m}^3/\text{s}

,

v_2 = 8.00\,\text{m/s}

.

v_2 = v_1\left(\frac{D_1}{D_2}\right)^2 = 2(2)^2 = 8.00\ \text{m/s}

Bernoulli pressure drop in a contraction

Problem. For the pipe above (horizontal, frictionless), the inlet gauge pressure is $200\,\text{kPa}$ . Find the outlet pressure.

Solution. Horizontal, so

z_1 = z_2

; no friction or machine. Bernoulli:

P_2 = P_1 + \tfrac12\rho(v_1^2 - v_2^2)

.

P_2 = 200{,}000 + \tfrac12(1000)(2^2 - 8^2) = 200{,}000 + 500(4 - 64) = 200{,}000 - 30{,}000

.

P_2 = 1.70\times10^{5}\,\text{Pa} = 170\,\text{kPa}

(gauge). Sanity check: flow speeds up, so pressure must fall — and it dropped by

30\,\text{kPa}

. Answer:

170\,\text{kPa}

.

P_2 = 200 + \tfrac{1}{2}(1)(2^2-8^2)\ \text{kPa} = 170\ \text{kPa}

Anchoring force on a nozzle

Problem. A horizontal nozzle contracts from $D_1 = 80\,\text{mm}$ to $D_2 = 30\,\text{mm}$ and discharges water to the atmosphere at $Q = 0.012\,\text{m}^3/\text{s}$ . Find the force required to hold the nozzle.

Solution. Areas:

A_1 = \tfrac{\pi}{4}(0.080)^2 = 5.027\times10^{-3}\,\text{m}^2

,

A_2 = \tfrac{\pi}{4}(0.030)^2 = 7.069\times10^{-4}\,\text{m}^2

. Velocities:

v_1 = Q/A_1 = 0.012/5.027\times10^{-3} = 2.39\,\text{m/s}

;

v_2 = Q/A_2 = 0.012/7.069\times10^{-4} = 16.98\,\text{m/s}

. Inlet gauge pressure (Bernoulli, outlet to atmosphere

P_2=0

):

P_1 = \tfrac12\rho(v_2^2 - v_1^2) = \tfrac12(1000)(16.98^2 - 2.39^2) = 1.412\times10^{5}\,\text{Pa}

. Momentum (x): the anchoring force on the nozzle is

F = P_1 A_1 - \rho Q(v_2 - v_1)

.

F = (1.412\times10^{5})(5.027\times10^{-3}) - (1000)(0.012)(16.98 - 2.39) = 709.8 - 175.1 = 535\,\text{N}

. Sanity check: the pressure term dominates and the net force is positive (the nozzle must be held back against the flow). Answer:

535\,\text{N}

.

F = P_1 A_1 - \rho Q(v_2 - v_1) = 709.8 - 175.1 = 535\ \text{N}

Tank drain velocity (Torricelli)

Problem. A large open tank has a small hole $5.0\,\text{m}$ below the water surface. Find the velocity of the escaping jet.

Solution. Bernoulli from surface (point 1,

v\approx0

,

P=P_{atm}

) to jet (point 2,

P=P_{atm}

):

v = \sqrt{2gh}

.

v = \sqrt{2(9.81)(5.0)} = \sqrt{98.1} = 9.90\,\text{m/s}

. Sanity check: this is the same as free-fall speed from

5\,\text{m}

— the energy-conservation hallmark of Torricelli. Answer:

9.90\,\text{m/s}

.

v = \sqrt{2(9.81)(5.0)} = 9.90\ \text{m/s}

Common pitfalls

•Applying Bernoulli where it does not hold: it requires steady, incompressible, frictionless flow with no pump/turbine. Across a pump, fitting, or shock, use the energy or momentum equation instead.
•Dropping a head term — forgetting the elevation head on a non-horizontal run, or the velocity head when areas differ — unbalances the equation.
•Mixing gauge and absolute pressure between the two points. Pick one convention and keep it.
•Treating momentum as a scalar. It is a vector: resolve into $x$ and $y$ and include the $PA$ pressure forces on each face.
•Sign error on the reaction force: the force on the structure is equal and opposite to the force on the fluid.
•Using continuity backwards — velocity is inversely proportional to area, so a smaller pipe means a ·faster·, not slower, flow.
•Reading the EGL/HGL gap wrong: the HGL is below the EGL by $v^2/2g$ , and that gap grows where the fluid accelerates.

References

NCEES FE Reference Handbook — Fluid Mechanics (One-Dimensional Fluid Flow)
NCEES FE Reference Handbook — Fluid Mechanics (Impulse-Momentum Principle)

F. Pipe and duct flow and friction losses

Pipe Flow and Friction Losses

Darcy-Weisbach head loss, the friction factor from the Moody chart and the laminar f=64/Re law, minor losses with K-coefficients, and the hydraulic diameter for noncircular ducts.

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G. Open-channel flow

Open-Channel Flow

Manning's equation and hydraulic radius for normal depth, specific energy and critical depth, and the Froude-number split between subcritical and supercritical flow.

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H. Fluid transport systems

Fluid Transport Systems

How series and parallel pipes combine, equivalent-pipe reduction, and finding the pump-system operating point where the pump curve meets the system curve.

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I. Flow measurement

Flow Measurement

Pitot and pitot-static tubes for velocity, venturi, orifice, and nozzle meters for pipe flow, weirs for open channels, and the role of the discharge coefficient.

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J. Turbomachinery

Pumps, Turbines, and Affinity Laws

Pump and turbine head and power, hydraulic efficiency, the affinity laws for speed and size scaling, NPSH and cavitation, and specific speed with characteristic curves.

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K. Ideal gas law

Ideal and Real Gas Behavior

The ideal gas law in mass and molar form, gas mixtures via Dalton partial pressures and Amagat volumes, molar mass, and the real-gas compressibility z-factor.

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