Fluid Mechanics

B. Fluid statics

Fluid Properties and Statics: Pressure, Manometry and Buoyancy

Density, viscosity and surface tension, the hydrostatic pressure field, manometry, hydrostatic force on submerged surfaces with center of pressure, and Archimedes' buoyancy.

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C. Energy, impulse, and momentum

Energy, Continuity and Momentum (Bernoulli)

Conservation of mass, the energy (Bernoulli) equation with pump and head-loss terms, the impulse-momentum principle for forces on bends and vanes, and the EGL/HGL.

Three conservation laws — mass, energy, and momentum — carry the bulk of the fluid mechanics exam, and Bernoulli's equation is the energy law in its most quotable form. The trap is treating Bernoulli as universal: it is the energy equation with friction, pumps, and turbines deleted, valid only along a streamline for steady, incompressible, inviscid flow. Add back a head-loss term and a pump head and you have the engineering energy equation that actually governs pipe systems. Momentum is the separate, vector law that tells you the force a moving fluid exerts on a bend, a vane, or a nozzle. The NCEES FE Reference Handbook places all of this under Fluid Mechanics ('Principles of One-Dimensional Fluid Flow' and 'The Impulse-Momentum Principle'); the equations below match its notation.

Continuity: conservation of mass

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

\dot m=\rho A v

is constant. For an incompressible fluid

\rho

cancels and the volumetric flow

Q=Av

is constant — when a pipe narrows, the fluid speeds up in inverse proportion to area. This single relation links the velocities you plug into Bernoulli and momentum, and it is usually the first equation you write.

\dot m = \rho A v = \text{const}, \qquad Q = A_1 v_1 = A_2 v_2

The Bernoulli equation

For steady, incompressible, frictionless flow with no pump or turbine, the sum of pressure head, velocity head, and elevation head is constant along a streamline. Each term has units of length (a 'head') when divided by

\gamma

, which is why the equation reads naturally as an energy-per-weight balance. Use it for accelerating, low-loss flows: a nozzle, a venturi, flow from a tank (Torricelli,

v=\sqrt{2gh}

), or a pitot tube.

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

The energy equation with pump and head loss

Real pipe flow loses energy to friction and may gain energy from a pump. The full energy equation inserts a head-loss term

h_f

(always added to the downstream side, since losses are positive) and a pump head

h_p

(added on the upstream side, since a pump adds energy). This is the equation you size pumps with: solve for

h_p

, then the required power follows from the pump-power relation. When the pipe diameter and elevation are equal at both sections, the equation collapses to

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

.

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

The impulse-momentum principle

Newton's second law for a control volume: the resultant external force equals the net rate at which momentum leaves. In one dimension

\Sigma F=\dot m(v_\text{out}-v_\text{in})=\rho Q(v_2-v_1)

, applied component-by-component because momentum is a vector. The external forces include pressure forces

PA

on the inlet and outlet faces, gravity, and the reaction force the solid boundary exerts on the fluid. Solve for that reaction, then flip its sign to get the force the fluid exerts on the bend, vane, or nozzle.

\Sigma F = \dot m\,(v_\text{out} - v_\text{in}) = \rho Q\,(v_2 - v_1)

Forces on bends, vanes, and nozzles

For a pipe bend the handbook writes the two scalar balances

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

and

F_y - W - P_2A_2\sin\alpha = \rho Q(v_2\sin\alpha)

, where

\alpha

is the bend angle and

F_x,F_y

are the anchoring forces. For a free jet (

P=0

gauge) striking a fixed vane that turns it by

\alpha

, the components reduce to

F_x=\rho Q v(1-\cos\alpha)

and

F_y=\rho Q v\sin\alpha

. A nozzle that accelerates the flow develops a backward reaction, which is why a fire hose kicks. Always sketch the control volume and pick a consistent positive direction before substituting.

\begin{aligned} P_1A_1 - P_2A_2\cos\alpha - F_x &= \rho Q(v_2\cos\alpha - v_1)\\ F_y - W - P_2A_2\sin\alpha &= \rho Q\,v_2\sin\alpha \end{aligned}

Energy and hydraulic grade lines

The energy grade line (EGL) plots total head

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

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

P/\gamma+z

. They are separated everywhere by the velocity head

v^2/2g

, so the HGL sits below the EGL by that gap, and both slope downward in the flow direction by the head loss. A pump steps both lines up; a turbine steps them down; a sudden velocity increase widens the EGL-HGL gap. Reading these lines tells you at a glance where pressure is low (HGL near or below the pipe means cavitation or sub-atmospheric pressure).

\text{EGL} - \text{HGL} = \dfrac{v^2}{2g}

Exam strategy

Start with continuity to relate velocities, then decide between Bernoulli (no loss, no machine) and the full energy equation (loss and/or pump present). Put

h_f

on the downstream side and

h_p

on the upstream side — getting the signs right is half the battle. For force problems, momentum is vector: draw the control volume, choose positive directions, account for pressure forces

PA

on the cut faces, and solve

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

per component before combining. Remember Bernoulli is gauge-pressure friendly because atmosphere cancels in most setups; switch to absolute only when a gas law enters.

Key equations

Continuity (incompressible)

Q = A_1 v_1 = A_2 v_2, \qquad \dot m = \rho A v

Volumetric flow constant when $\rho$ is constant; mass flow always constant in steady flow.

Bernoulli equation

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

Steady, incompressible, frictionless, no machine; each term has units of length (head).

Energy equation with pump and loss

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

$h_p$ = pump head added (upstream side), $h_f$ = total head loss (downstream side).

Torricelli efflux

v = \sqrt{2gh}

Free-jet speed from a tank with surface height $h$ above the opening (Bernoulli, surface and jet at atmospheric).

Pressure drop in equal-area pipe

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

When $z_1=z_2$ and $v_1=v_2$ , the full energy equation collapses to friction pressure drop.

Impulse-momentum principle

\Sigma F = \rho Q\,(v_2 - v_1)

Vector; apply per component. $\Sigma F$ includes pressure forces, weight, and the boundary reaction.

Force on a bend (x, y)

\begin{aligned} P_1A_1 - P_2A_2\cos\alpha - F_x &= \rho Q(v_2\cos\alpha - v_1)\\ F_y - W - P_2A_2\sin\alpha &= \rho Q\,v_2\sin\alpha \end{aligned}

$\alpha$ = bend angle from inlet axis; $F_x,F_y$ = anchoring force on the fluid (reverse sign for force on the bend).

Force on a fixed deflecting vane

F_x = \rho Q\,v(1-\cos\alpha), \qquad F_y = \rho Q\,v\sin\alpha

Free jet (gauge $P=0$ ) of speed $v$ turned by angle $\alpha$ ; speed unchanged on a frictionless vane.

EGL minus HGL

\text{EGL} - \text{HGL} = \dfrac{v^2}{2g}

The two grade lines differ by the velocity head; both drop by $h_f$ downstream.

Worked examples

Pump head and brake power

Problem. A pump moves $0.05\,\text{m}^3/\text{s}$ of water and must add $30\,\text{m}$ of head to the flow. If the pump efficiency is $75\%$ , find the fluid (hydraulic) power and the brake power.

Solution. Hydraulic power delivered to the fluid:

\dot W_\text{fluid} = \rho g Q h_p = 1000(9.81)(0.05)(30) = 1.47\times 10^4\,\text{W} = 14.7\,\text{kW}

. Brake (shaft) power:

\dot W_\text{brake} = \dot W_\text{fluid}/\eta = 14{,}715/0.75 = 1.96\times 10^4\,\text{W} = 19.6\,\text{kW}

. Sanity: efficiency below 1 means shaft power must exceed fluid power, and

19.6>14.7\,\text{kW}

confirms it. Units:

(\text{kg/m}^3)(\text{m/s}^2)(\text{m}^3/\text{s})(\text{m})=\text{W}

. Answers:

14.7\,\text{kW}

fluid,

19.6\,\text{kW}

brake.

\dot W_\text{brake} = \dfrac{\rho g Q h_p}{\eta} = 19.6\ \text{kW}

Force of a jet on a deflecting vane

Problem. A horizontal water jet of diameter $40\,\text{mm}$ moving at $20\,\text{m/s}$ strikes a stationary vane that turns it through $60^\circ$ . Find the magnitude of the force on the vane.

Solution. Jet area:

A=\pi(0.04)^2/4 = 1.257\times 10^{-3}\,\text{m}^2

. Flow:

Q=Av=1.257\times10^{-3}(20)=0.02513\,\text{m}^3/\text{s}

. Mass flow:

\dot m=\rho Q = 25.13\,\text{kg/s}

. On a frictionless vane the speed stays

20\,\text{m/s}

, only the direction changes:

F_x=\rho Q v(1-\cos 60^\circ)=25.13(20)(1-0.5)=251\,\text{N}

.

F_y=\rho Q v\sin 60^\circ=25.13(20)(0.866)=435\,\text{N}

. Resultant:

F=\sqrt{251^2+435^2}=503\,\text{N}

at

\arctan(435/251)=60.0^\circ

from the original jet direction. Sanity: for a

60^\circ

turn the resultant should exceed either component;

503>435\,\text{N}

and the angle equals the turn-bisector geometry. Answer

503\,\text{N}

.

F = \rho Q v\sqrt{(1-\cos\alpha)^2+\sin^2\alpha} = 503\ \text{N}

Reaction force on a horizontal nozzle

Problem. Water flows at $0.02\,\text{m}^3/\text{s}$ through a horizontal nozzle that contracts from $100\,\text{mm}$ to $40\,\text{mm}$ diameter, discharging to atmosphere. Find the anchoring force on the nozzle.

Solution. Areas:

A_1=\pi(0.1)^2/4=7.854\times10^{-3}\,\text{m}^2

,

A_2=\pi(0.04)^2/4=1.257\times10^{-3}\,\text{m}^2

. Velocities:

v_1=Q/A_1=2.546\,\text{m/s}

,

v_2=Q/A_2=15.92\,\text{m/s}

. Inlet gauge pressure from Bernoulli (

P_2=0

gauge, horizontal):

P_1=\tfrac12\rho(v_2^2-v_1^2)=\tfrac12(1000)(15.92^2-2.546^2)=1.234\times10^5\,\text{Pa}=123\,\text{kPa}

. Momentum in flow direction (

P_2=0

):

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

, so

F_x=P_1A_1-\rho Q(v_2-v_1)=123{,}409(7.854\times10^{-3})-1000(0.02)(15.92-2.546)

.

F_x = 969.2 - 267.5 = 702\,\text{N}

(the anchor holds the nozzle back against the upstream pressure). Sanity: the pressure force (

969\,\text{N}

) dominates the momentum flux (

268\,\text{N}

), so the net force is positive and of order several hundred newtons. Answer

702\,\text{N}

.

F_x = P_1 A_1 - \rho Q(v_2 - v_1) = 702\ \text{N}

Common pitfalls

•Using Bernoulli where friction matters. Bernoulli has no $h_f$ ; for any pipe with appreciable length, switch to the full energy equation.
•Sign of the head-loss and pump terms. $h_f$ adds to the downstream side; $h_p$ adds to the upstream side. Reversing them inverts the answer.
•Treating momentum as a scalar. $\Sigma F=\rho Q(v_2-v_1)$ is a vector equation — resolve into x and y before combining.
•Forgetting the pressure forces $PA$ on the control-volume faces in a bend or contraction. They are usually larger than the momentum flux term.
•Confusing the force on the fluid with the force on the structure. The boundary reaction in the momentum balance is the force on the fluid; the force on the bend is equal and opposite.
•Mixing the EGL and HGL. They differ by the velocity head $v^2/2g$ ; the HGL is what a piezometer reads, the EGL includes kinetic energy.
•Using absolute pressure in a gauge-based jet problem (or vice versa). Free jets discharge at atmospheric, i.e. zero gauge — keep one convention throughout.

References

NCEES FE Reference Handbook — Fluid Mechanics
Vennard & Street, Elementary Fluid Mechanics — momentum and grade-line treatment

D. Internal flow

Internal Flow: Reynolds Number, Friction and Head Loss

Laminar versus turbulent regimes, the Reynolds number, Darcy-Weisbach head loss with the Moody chart, minor (fitting) losses, and the hydraulic diameter for noncircular ducts.

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E. External flow

External Flow: Drag, Lift and Boundary Layers

The drag and lift equations and their coefficients, the boundary layer, flow over a flat plate, cylinder and sphere, and terminal velocity from a force balance.

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F. Compressible flow

Compressible Flow: Mach Number, Isentropic Relations and Nozzles

The speed of sound and Mach number, stagnation properties, isentropic nozzle area-velocity behavior, choked flow at the throat, and the normal shock.

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H. Performance curves

Pumps, Fans and Compressors: Performance Curves and Affinity Laws

Pump and system curves and the operating point, NPSH and cavitation, hydraulic power and efficiency, and the affinity (scaling) laws for fans, pumps, and compressors.

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