Fluid Mechanics and Hydraulics

A. Fluid statics

Fluid Statics and Manometry

Hydrostatic pressure variation, manometer equations, force and center of pressure on submerged plane and curved surfaces, and buoyancy by Archimedes' principle.

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B. Closed conduits

Pipe Friction: Darcy–Weisbach, Hazen–Williams, and Minor Losses

Friction head loss in closed conduits via Darcy-Weisbach with the Moody factor, the Hazen-Williams C-factor form, and minor losses by K-factor and equivalent length.

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C. Open channel

Open-Channel Flow and Pump Application

Manning uniform flow and normal depth, specific energy and critical depth via the Froude number, the pump-vs-system operating point, NPSH and cavitation, and pumps in series and parallel.

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E. Flow measurement

Flow Measurement: Weirs, Orifices, and Venturi Meters

Rectangular and V-notch weirs, the orifice equation and vena contracta, venturi and differential meters, discharge coefficients, and the head-discharge relationship.

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F. Blowers

Blowers, Aeration, and Unsteady Flow

Blower power and adiabatic compression, diffused-aeration airflow and oxygen transfer (SOTR/AOTR), the fan/blower affinity laws, and basic unsteady-flow ideas like tank draining.

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G. Fluid dynamics

Continuity, Bernoulli, and the Energy Equation

Conservation of mass and energy along a streamline, the head form with pump, turbine, and friction terms, EGL/HGL, and the impulse-momentum principle.

Almost every hydraulics problem on the FE Environmental exam rests on two conservation laws applied to a flowing fluid: mass cannot accumulate in a steady pipe, and mechanical energy is conserved except for what friction degrades and what pumps add or turbines remove. Master the head form of the energy equation and you can size a pump, find a pressure at a junction, or predict a velocity at a nozzle without ever leaving the NCEES FE Reference Handbook ("Fluid Mechanics — Principles of One-Dimensional Fluid Flow"). Points are lost here by mixing energy units, dropping a velocity-head term, or putting the pump on the wrong side of the equation.

Continuity: conservation of mass

For steady one-dimensional flow the mass flow rate

\dot m = \rho A v

is the same at every cross section. When the fluid is incompressible — true for water and essentially true for the low-pressure air streams in environmental work — density cancels and the volumetric flow

Q = Av

is constant. The immediate consequence you use constantly: where a pipe narrows, the velocity must rise in inverse proportion to area, so halving the diameter quadruples the velocity.

\dot m_1 = \dot m_2 \;\Rightarrow\; \rho_1 A_1 v_1 = \rho_2 A_2 v_2, \qquad Q = A_1 v_1 = A_2 v_2

The energy equation in head form

Dividing the steady-flow energy balance by weight expresses every term as a length — a "head" in meters or feet. Pressure head

p/\gamma

, velocity head

v^2/2g

, and elevation head

z

at section 1 equal the same three at section 2, plus any friction loss

h_f

, with a pump adding head

h_p

and a turbine extracting

h_t

. This single equation is the workhorse: choose two sections, write all six potential terms, and cancel the ones that are zero.

\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 as the frictionless special case

When no pump or turbine sits between the two sections and friction is negligible, the energy equation collapses to the Bernoulli equation: the sum of pressure, velocity, and elevation head is constant along the streamline. Use Bernoulli for short accelerating reaches — a nozzle, a venturi throat, flow from a tank — where the friction loss is genuinely small. The moment a long pipe, a fitting, or a pump enters the picture, return to the full energy equation.

\frac{p_1}{\gamma} + \frac{v_1^2}{2g} + z_1 = \frac{p_2}{\gamma} + \frac{v_2^2}{2g} + z_2 = \text{constant}

EGL and HGL

The energy grade line (EGL) plots the total head

p/\gamma + v^2/2g + z

above a datum; the hydraulic grade line (HGL) plots only

p/\gamma + z

. The two are separated everywhere by exactly the velocity head

v^2/2g

, so they run parallel in a constant-diameter pipe and spread apart where the flow speeds up. The EGL always slopes downward in the direction of flow (friction degrades energy) and steps up across a pump; where the HGL drops below the pipe, the gauge pressure is negative — a cavitation warning.

The impulse-momentum principle

Energy methods give heads and velocities but never the force a moving fluid exerts on a bend, nozzle, or plate — for that you need the momentum equation. The resultant external force in any direction equals the rate of change of momentum flux,

\Sigma F = \rho Q (v_{\text{out}} - v_{\text{in}})

taken component by component. A free jet striking a fixed plate reverses or stops, so the plate feels

F = \rho Q v

; a reducing bend feels both pressure forces and the momentum change. Always draw the control volume and pick consistent positive directions.

\Sigma F_x = \rho Q\,(v_{2x} - v_{1x}), \qquad \Sigma F_y = \rho Q\,(v_{2y} - v_{1y})

Putting it together

A reliable routine: (1) sketch the system and number two sections where you know the most; (2) write the energy equation and strike the zero terms — an open reservoir surface has

p=0

gauge and

v\approx 0

; (3) solve for the unknown head, velocity, or pressure; (4) if a force is asked, switch to momentum with the velocities you just found. Keep

\gamma_{\text{water}} = 9{,}810\ \text{N/m}^3

(62.4 lbf/ft

^3

) and

g = 9.81\ \text{m/s}^2

(32.2 ft/s

^2

) at hand and watch that every term lands in the same unit of length.

Exam strategy

Decide first whether friction or a machine is present: if yes, use the full energy equation; if not, Bernoulli is faster. Pump head is added on the upstream side (it raises the EGL), turbine head is subtracted. When two sections are both open to atmosphere, the pressure-head terms drop and the problem reduces to elevation and velocity heads plus losses. For any force question, abandon energy and use

\Sigma F = \rho Q\,\Delta v

— and remember the pressure-times-area terms on a bend. Carry units explicitly; a velocity head left in

\text{m}^2/\text{s}^2

instead of meters is the classic silent error.

Key equations

Continuity (volumetric)

Q = A_1 v_1 = A_2 v_2

Incompressible steady flow; $Q$ in m $^3$ /s, $A$ in m $^2$ , $v$ in m/s. Velocity rises where area shrinks.

Continuity (mass)

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

General steady form; use when density changes (gas/air streams), $\dot m$ in kg/s.

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

Each term a length; $h_p$ pump head added, $h_t$ turbine head removed, $h_f$ friction + minor losses.

Bernoulli equation

\frac{p}{\gamma} + \frac{v^2}{2g} + z = \text{constant}

Frictionless, no machine, along a streamline. Use for nozzles, venturis, tank efflux.

Velocity head

h_v = \frac{v^2}{2g}

Kinetic energy per unit weight; the vertical gap between EGL and HGL (m or ft).

Pressure-head / pressure relation

p_1 - p_2 = \gamma\,h_f = \rho g\,h_f

Pressure drop in a horizontal constant-area pipe equals the friction head times specific weight.

Pump head and fluid power

\dot W_{\text{fluid}} = \gamma Q h_p, \qquad \dot W_{\text{brake}} = \frac{\gamma Q h_p}{\eta}

Fluid (hydraulic) power vs. shaft/brake power; $\eta$ is pump efficiency (0-1).

Impulse-momentum

\Sigma F = \rho Q\,(v_{\text{out}} - v_{\text{in}})

Per direction; gives force on bends, nozzles, plates. Include $pA$ pressure forces on bends.

Force of a jet on a fixed plate

F = \rho Q v

Normal flat plate stops the jet ( $v_{\text{out}}=0$ in flow direction); $F$ in N.

Worked examples

Velocity and pressure drop in a contraction

Problem. Water at $1000\ \text{kg/m}^3$ flows at $Q = 0.10\ \text{m}^3/\text{s}$ through a horizontal contraction from $300\ \text{mm}$ to $150\ \text{mm}$ diameter. Find the two velocities and the pressure drop, assuming negligible friction.

Solution. Areas:

A_1 = \tfrac{\pi}{4}(0.30)^2 = 0.0707\ \text{m}^2

,

A_2 = \tfrac{\pi}{4}(0.15)^2 = 0.01767\ \text{m}^2

. Continuity:

v_1 = Q/A_1 = 0.10/0.0707 = 1.41\ \text{m/s}

;

v_2 = 0.10/0.01767 = 5.66\ \text{m/s}

(quadruples, as area drops fourfold). Horizontal Bernoulli (

z_1=z_2

):

p_1 - p_2 = \tfrac{\rho}{2}(v_2^2 - v_1^2) = \tfrac{1000}{2}(5.66^2 - 1.41^2) = 500(32.0 - 2.0)

.

\boxed{p_1 - p_2 = 1.50\times 10^4\ \text{Pa} = 15.0\ \text{kPa}}

. Sanity: faster flow at section 2 means lower pressure there, so the drop is positive — consistent.

p_1 - p_2 = \frac{\rho}{2}\left(v_2^2 - v_1^2\right)

Pump brake power for a lift

Problem. A pump delivers $Q = 0.050\ \text{m}^3/\text{s}$ of water against a total dynamic head of $25\ \text{m}$ (elevation plus all losses). The pump efficiency is $75\%$ . Find the brake power required.

Solution. Fluid power:

\dot W_{\text{fluid}} = \gamma Q h_p = 9810 \times 0.050 \times 25 = 12{,}260\ \text{W}

. Brake power:

\dot W_{\text{brake}} = \dfrac{\dot W_{\text{fluid}}}{\eta} = \dfrac{12{,}260}{0.75}

.

\boxed{\dot W_{\text{brake}} = 1.64\times 10^4\ \text{W} = 16.4\ \text{kW}}

(about

21.9\ \text{hp}

). Sanity:

\gamma Q h

has units

\tfrac{\text{N}}{\text{m}^3}\cdot\tfrac{\text{m}^3}{\text{s}}\cdot\text{m} = \text{N·m/s} = \text{W}

— correct.

\dot W_{\text{brake}} = \frac{\gamma Q h_p}{\eta}

Force of a water jet on a wall

Problem. A $50\ \text{mm}$ nozzle discharges $Q = 0.020\ \text{m}^3/\text{s}$ of water horizontally onto a large fixed vertical plate. Find the force the jet exerts on the plate.

Solution. Jet velocity:

v = Q/A = 0.020 / (\tfrac{\pi}{4}(0.050)^2) = 0.020/0.001963 = 10.2\ \text{m/s}

. The plate stops the jet in the flow direction (

v_{\text{out},x}=0

), so by momentum

F = \rho Q v = 1000 \times 0.020 \times 10.2

.

\boxed{F = 204\ \text{N}}

. Sanity:

\rho Q v

gives

\tfrac{\text{kg}}{\text{m}^3}\cdot\tfrac{\text{m}^3}{\text{s}}\cdot\tfrac{\text{m}}{\text{s}} = \tfrac{\text{kg·m}}{\text{s}^2} = \text{N}

— correct.

F = \rho Q v

Common pitfalls

•Mixing energy units: every term in the head form must be a length (m or ft). Leaving $v^2/2g$ as $\text{m}^2/\text{s}^2$ (forgetting to divide by $g$ ) is the most common silent error.
•Putting the pump head on the wrong side: a pump ADDS head ( $+h_p$ on the upstream side) and raises the EGL; a turbine SUBTRACTS it. Reversing them flips the answer's sign.
•Using Bernoulli where friction matters. Over a long pipe or through a fitting you must use the full energy equation with $h_f$ ; Bernoulli only holds for short accelerating reaches.
•Confusing EGL and HGL: they differ by exactly the velocity head $v^2/2g$ . The EGL is the total head; the HGL omits the kinetic term and is what a piezometer reads.
•Forgetting the pressure-area ( $pA$ ) terms in a momentum balance on a bend. Energy gives velocities, but the force balance also needs the static pressure forces on the inlet and outlet faces.
•Assuming velocity is negligible at both sections. At a reservoir surface $v\approx 0$ , but at a pipe outlet or nozzle the velocity head can dominate — keep it.

References

NCEES FE Reference Handbook — Fluid Mechanics
NCEES FE Reference Handbook — Civil Engineering
Munson, Young & Okiishi, Fundamentals of Fluid Mechanics — energy line / control-volume momentum background