Fluid Mechanics · Study · FE Mechanical · FE → PE Prep
Fluid Mechanics
10% of exam
Fluid properties and statics, energy and momentum, internal and external flow, compressible flow and normal shock, and pump, fan, and compressor performance and scaling laws.
6 concepts
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: m˙=ρAv is constant. For an incompressible fluid ρ 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.
m˙=ρAv=const,Q=A1v1=A2v2
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 γ, 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=2gh), or a pitot tube.
γP1+2gv12+z1=γP2+2gv22+z2
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 hf (always added to the downstream side, since losses are positive) and a pump head hp (added on the upstream side, since a pump adds energy). This is the equation you size pumps with: solve for hp, 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 P1−P2=γhf=ρghf.
γP1+2gv12+z1+hp=γP2+2gv22+z2+hf
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 ΣF=m˙(vout−vin)=ρQ(v2−v1), 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.
ΣF=m˙(vout−vin)=ρQ(v2−v1)
Forces on bends, vanes, and nozzles
For a pipe bend the handbook writes the two scalar balances P1A1−P2A2cosα−Fx=ρQ(v2cosα−v1) and Fy−W−P2A2sinα=ρQ(v2sinα), where α is the bend angle and Fx,Fy are the anchoring forces. For a free jet (P=0 gauge) striking a fixed vane that turns it by α, the components reduce to Fx=ρQv(1−cosα) and Fy=ρQvsinα. 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.
The energy grade line (EGL) plots total head P/γ+v2/2g+z above the datum; the hydraulic grade line (HGL) plots only P/γ+z. They are separated everywhere by the velocity head v2/2g, 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).
EGL−HGL=2gv2
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 hf on the downstream side and hp 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 ΣF=ρQ(v2−v1) 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=A1v1=A2v2,m˙=ρAv
Volumetric flow constant when ρ is constant; mass flow always constant in steady flow.
Energy equation with pump and lossγP1+2gv12+z1+hp=γP2+2gv22+z2+hf
Torricelli effluxv=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 pipeP1−P2=γhf=ρghf
Impulse-momentum principleΣF=ρQ(v2−v1)
Vector; apply per component. ΣF
Force on a bend (x, y)P1A1−P2A2cosα−FxFy−W−P2A2sinα=ρQ(v2cosα−v1)=ρQv2sinα
Force on a fixed deflecting vaneFx=ρQv(1−cosα),Fy=ρQvsinα
EGL minus HGLEGL−HGL=2gv2
The two grade lines differ by the velocity head; both drop by hf
Worked examples
Pump head and brake power
Problem. A pump moves 0.05m3/s of water and must add 30m 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: W˙fluid=ρgQhp=1000(9.81)(0.05)(30)=1.47×104W=14.7kW.
Brake (shaft) power: W˙brake=W˙fluid/η=14,715/0.75=1.96×104W=19.6kW.
Sanity: efficiency below 1 means shaft power must exceed fluid power, and 19.6>14.7kW confirms it. Units: (kg/m3)(m/s2)(m3/s)(m)=W. Answers: 14.7kW fluid, 19.6kW brake.
W˙brake=ηρgQhp=19.6kW
Force of a jet on a deflecting vane
Problem. A horizontal water jet of diameter 40mm moving at 20m/s strikes a stationary vane that turns it through 60∘. Find the magnitude of the force on the vane.
Solution. Jet area: A=π(0.04)2/4=1.257×10−3m2
Reaction force on a horizontal nozzle
Problem. Water flows at 0.02m3/s through a horizontal nozzle that contracts from 100mm to 40mm diameter, discharging to atmosphere. Find the anchoring force on the nozzle.
Solution. Areas: A1=π(0.1)2/4=7.854×10−3m2
Common pitfalls
•Using Bernoulli where friction matters. Bernoulli has no hf; for any pipe with appreciable length, switch to the full energy equation.
•Sign of the head-loss and pump terms. hf adds to the downstream side; hp adds to the upstream side. Reversing them inverts the answer.
•Treating momentum as a scalar. ΣF=ρQ(v2−v1) 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 v2/2g; 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.
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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Steady, incompressible, frictionless, no machine; each term has units of length (head).
hp = pump head added (upstream side), hf = total head loss (downstream side).
When z1=z2 and v1=v2, the full energy equation collapses to friction pressure drop.
includes pressure forces, weight, and the boundary reaction.
α = bend angle from inlet axis; Fx,Fy = anchoring force on the fluid (reverse sign for force on the bend).
Free jet (gauge P=0) of speed v turned by angle α; speed unchanged on a frictionless vane.
downstream.
. Flow:
Q=Av=1.257×10−3(20)=0.02513m3/s
. Mass flow:
m˙=ρQ=25.13kg/s
.
On a frictionless vane the speed stays
20m/s
, only the direction changes:
Fx=ρQv(1−cos60∘)=25.13(20)(1−0.5)=251N
.
Fy=ρQvsin60∘=25.13(20)(0.866)=435N
.
Resultant:
F=2512+4352=503N
at
arctan(435/251)=60.0∘
from the original jet direction.
Sanity: for a
60∘
turn the resultant should exceed either component;
503>435N
and the angle equals the turn-bisector geometry. Answer