FE Mechanical · knowledge area 10 · 10–15 questions
FE fluid mechanics review, pipes to nozzles
Fluid mechanics carries 10–15 questions, and the mechanical outline spells out compressible flow and turbomachinery: Mach number, nozzles, shocks, and the curves and scaling laws of pumps, fans and compressors.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Mechanical
questions on the exam10–15
worked problems here18
of 14 knowledge areas10
Checked
What the area covers
Nine topics in the specification. In plain terms:
Fluid properties: density, specific weight, dynamic and kinematic viscosity, surface tension, vapor pressure.
Fluid statics: pressure with depth, manometers, buoyancy, forces on submerged surfaces.
Energy, impulse and momentum: the energy equation with losses, and momentum forces on jets, vanes and pipe bends.
Internal flow: laminar and turbulent pipe flow, friction factor, minor losses.
External flow: drag and lift on immersed bodies, boundary layers.
Compressible flow: Mach number, isentropic relations, normal shocks.
Machines: power and efficiency, performance curves, and scaling laws for fans, pumps and compressors.
18 fluid mechanics problems, every option explained
0 / 18 answered · 0 correct
Question 1 of 18 · Fluid Mechanics · 10–15 items
A fan's rotational speed is increased by 20%. According to the fan laws, what is the approximate percentage increase in power consumption?
Given
Speed increase = 20%
Find
Increase in fan power (%)
Solution
P2/P1 = (N2/N1)3 = 1.23 = 1.728 → +73%
FE Reference Handbook, Fluid Mechanics: fan and pump laws.
Only flow rises in direct proportion to speed; power rises with its cube.
1.22 = 1.44 is the pressure (head) ratio; power has one more power of speed.
Tripling the 20% speed increase (3 × 20 = 60%) treats the cube law as linear; 1.23 = 1.728.
Power varies with the cube of speed: 1.23 = 1.728, an increase of about 73%.
Question 2 of 18 · Fluid Mechanics · 10–15 items
A 1:5 scale model of a vehicle is tested in a water channel to determine its aerodynamic drag. The prototype travels in air at 30 m/s. Water has a kinematic viscosity of 1.0 × 10−6 m2/s and a density of 1000 kg/m3; air has a kinematic viscosity of 1.5 × 10−5 m2/s and a density of 1.2 kg/m3. Under complete Reynolds number similitude, the measured drag on the model is 250 N. What is the predicted drag force on the prototype?
Given
Scale Lm/Lp = 1:5
Prototype velocity Vp = 30 m/s
Water kinematic viscosity νm = 1.0 × 10−6 m2/s
Water density ρm = 1000 kg/m3
Air kinematic viscosity νp = 1.5 × 10−5 m2/s
Air density ρp = 1.2 kg/m3
Model drag Fm = 250 N
Find
Prototype drag Fp (N)
Solution
Rem = Rep: Vm = Vp(Lp/Lm)(νm/νp) = 30 × 5 × (1.0 × 10−6/1.5 × 10−5) = 10 m/s
FE Reference Handbook, Fluid Mechanics: similitude.
This scales by the density ratio alone, 250 × 1.2/1000 = 0.3 N, and drops the velocity and length ratios.
This scales density and velocity but leaves out the area ratio (Lp/Lm)2 = 25.
This leaves out the velocity ratio squared; the prototype moves three times faster than the model.
The velocity ratio enters squared in ρV2L2; using it to the first power gives a third of the answer.
Re matching gives Vm = 10 m/s; with equal drag coefficients F ∝ ρV2L2, so Fp = 250 × (1.2/1000) × 32 × 52 = 67.5 N.
Question 3 of 18 · Fluid Mechanics · 10–15 items
Flow (L/s)
Single Pump Head (m)
0
60
10
50
20
30
A system requires pumping fluid against a static head of 75 m. A single pump has the performance shown in the table, with a shutoff head of 60 m. Two identical pumps are placed in series. The system friction curve is Hsys = 75 + 0.25 Q2 (Q in L/s). What is the operating flow rate?
FE Reference Handbook, Fluid Mechanics: pumps in series and the system curve.
This is beyond the pump curve; adding pumps in series raises head, not the flow range of each pump.
At 20 L/s the pair gives only 60 m, far below the 175 m the system requires.
At 15 L/s the pumps give 2 × 40 = 80 m against a system demand of about 131 m; the flow cannot reach this point.
Two pumps in series double the head: at 10 L/s, 2 × 50 = 100 m, and the system needs 75 + 0.25(10)2 = 100 m, so the curves cross there.
At 5 L/s the pumps give 2 × 55 = 110 m but the system needs only about 81 m, so the flow would keep rising.
Question 4 of 18 · Fluid Mechanics · 10–15 items
Calculate the annual electricity cost for a pump operating according to the schedule below. The pump efficiency is 80%, the motor efficiency is 90%, and the electricity rate is $0.10/kWh.
Parameter
Value
Flow Rate
1,000 gpm
Total Head
50 ft
Operation
3,000 hrs/year
Specific Gravity
1.0
Given
Flow = 1,000 gpm
Total head = 50 ft
Specific gravity = 1.0
Pump efficiency = 80 %
Motor efficiency = 90 %
Operation = 3,000 h/yr
Electricity rate = $0.10/kWh
Find
Annual electricity cost ($)
Solution
WHP = Q·H·SG/3960 = (1000 gpm)(50 ft)(1.0)/3960 = 12.63 hp
FE Reference Handbook, Fluid Mechanics: pump power equation.
This multiplies by the efficiencies instead of dividing; losses make the input power larger than the water power.
This prices only the water power and ignores both the pump and motor losses.
This divides by the motor efficiency only; the pump efficiency of 80% must also be applied.
This divides by the pump efficiency only; the motor's 90% efficiency adds more input power.
Water power = 1000 × 50/3960 = 12.6 hp; dividing by 0.80 × 0.90 gives 17.5 hp = 13.1 kW, and 13.1 kW × 3000 h × $0.10/kWh ≈ $3,923.
Question 5 of 18 · Fluid Mechanics · 10–15 items
A piping system operates with a static head of 20 m and a dynamic head of 15 m at a flow rate of 100 L/s. The pump speed is increased to achieve a new scaled flow rate of 200 L/s. What is the new required system head to maintain this scaled flow rate?
Given
Static head Hs = 20 m
Dynamic head Hd = 15 m
Flow rate Q1 = 100 L/s
New flow rate Q2 = 200 L/s
Find
New system head H2 (m)
Solution
H2 = Hstatic + Hdyn(Q2/Q1)2 = 20 + 15 × 22 = 80 m
FE Reference Handbook, Fluid Mechanics: pumps (system curve).
This doubles the dynamic head with the flow; friction head grows with the square of flow.
Doubling the whole 35 m wrongly scales the static head too, and linearly.
Only the friction (dynamic) part scales with Q2: 20 + 15 × (200/100)2 = 20 + 60 = 80 m.
Multiplying the whole 35 m by four squares the static head as well; elevation head does not change with flow.
This applies the cube law for power to head, and to the static part too.
Question 6 of 18 · Fluid Mechanics · 10–15 items
A high-pressure natural gas metering station uses a standard orifice meter to calculate steady mass flow. The meter consistently overestimates the mass flow compared with downstream custody-transfer measurements. The automation system uses the standard incompressible orifice equation with a correctly calibrated discharge coefficient. What is the most likely root cause of the error across the high-pressure-drop orifice?
Solution
ṁ = C·Y·A0√(2ρ1Δp/(1 − β4)), Y < 1 for a gas; Y = 1 only for liquids
Gas flow in a high-pressure station is highly turbulent, and the orifice equation does not assume a laminar regime.
The orifice equation is built on the kinetic energy change across the restriction; assuming it constant would give no flow at all.
A static-pressure error would bias density in either direction; it does not explain a consistent overestimate tied to a large pressure drop.
Wall roughness affects friction loss in the pipe, not the orifice equation that relates Δp to flow.
A gas expands as it accelerates through the orifice; the expansion factor Y < 1 corrects the incompressible equation, and leaving it out overstates the flow.
Question 7 of 18 · Fluid Mechanics · 10–15 items
For isentropic flow of a perfect gas in a converging nozzle that is choked at the throat, which expression gives the ratio of the throat static pressure p* to the stagnation pressure p0, with k the ratio of specific heats?
Solution
p*/p0 = (2/(k + 1))k/(k − 1) = (2/2.4)3.5 = 0.528 for k = 1.4
FE Reference Handbook, Fluid Mechanics: compressible flow (isentropic relations).
At the sonic throat T*/T0 = 2/(k + 1), and the isentropic relation raises it to k/(k − 1) for pressure: 0.528 for air.
The exponent 1/(k − 1) gives the density ratio ρ*/ρ0, not the pressure ratio.
With 2k in the numerator the base exceeds 1 and the 'ratio' would exceed 1, impossible for static over stagnation pressure.
This rearranges the terms incorrectly; for air it gives about 0.21, not the familiar 0.528.
Question 8 of 18 · Fluid Mechanics · 10–15 items
A 1:10 scale model of a wind turbine is tested in water (ρ = 1000 kg/m3) and produces 5.0 kW of power at a fluid velocity of 2.0 m/s. The full-scale prototype will operate in air (ρ = 1.2 kg/m3) at a velocity of 8.0 m/s. What is the estimated power output of the prototype?
Scaling by D3 instead of D2 overstates the power tenfold; the swept area grows with D2.
This scales size and velocity but drops the density ratio; air is about 830 times lighter than water.
Question 9 of 18 · Fluid Mechanics · 10–15 items
Flow Rate (L/s)
Pump Head (m)
Efficiency
100
50
75%
150
45
80%
200
30
60%
250
20
40%
A centrifugal pump operates steadily at 150 L/s. A downstream control valve suddenly fails open, driving the flow rate up to 250 L/s. Moments later, the pump motor trips off line due to a severe power overload. The original design calculations assumed constant pump efficiency across the operating range. Using the performance data provided, determine the most direct root cause of the motor overload.
A normal shock does not stop exactly at sonic speed; the downstream flow is always subsonic and depends on M1.
The Mach number cannot stay at 2.5; a normal shock always makes the flow subsonic.
Question 11 of 18 · Fluid Mechanics · 10–15 items
An engineer calculates the pump power required for a water pipeline using US Customary units. The proposed formula for power (P) is given as: P = (ṁ g H)/η where ṁ is the mass flow rate in lbm/s, g is the acceleration due to gravity in ft/s2, H is the pump head in ft, and η is the dimensionless pump efficiency. The engineer claims the resulting power P will be in units of ft · lbf/s, which can then be directly converted to horsepower. What is the fundamental flaw in this dimensional analysis?
Solution
P = ṁ(g/gc)H/η, gc = 32.2 lbm·ft/(lbf·s2)
Without gc: (lbm/s)(ft/s2)(ft) = lbm·ft2/s3, which is not ft·lbf/s.
For a pump, the shaft power exceeds the fluid power, so efficiency divides; and it is dimensionless either way, so it is not the units flaw.
Head in feet is already consistent with ṁgH; pressure units are not required, and absolute pressure is irrelevant to added head.
Head enters power linearly in every unit system; squaring it breaks dimensional consistency.
Mass flow in lbm/s carries no force; g (with gc) is what converts mass to weight.
In USCS, lbm × ft/s2 is not lbf; dividing by gc = 32.2 lbm·ft/(lbf·s2) turns ṁgH into ft·lbf/s.
Question 12 of 18 · Fluid Mechanics · 10–15 items
For a perfect gas passing through a stationary normal shock, how does the stagnation pressure ratio p02/p01 behave?
Solution
Δs = −R ln(p02/p01) > 0 → p02/p01 < 1 (T02 = T01)
FE Reference Handbook, Fluid Mechanics: compressible flow (normal shock).
A normal shock is irreversible: entropy rises and stagnation pressure falls, while stagnation temperature stays constant for an adiabatic shock.
The size of the loss depends on M1, but its direction does not: p02 is always below p01.
Constant stagnation pressure holds only in isentropic flow; the shock itself is not isentropic.
An increase in stagnation pressure would mean an entropy decrease in an adiabatic process, which the second law forbids.
Question 13 of 18 · Fluid Mechanics · 10–15 items
For flow around a cylinder, at what approximate Reynolds number does vortex shedding begin to occur?
Solution
Cylinder wake: Re < 5 attached; 5 < Re < 40 steady twin eddies; Re ≈ 40–50 onset of periodic shedding.
At Re below about 5 the flow stays attached (creeping flow), and up to about 40 a steady pair of eddies sits behind the cylinder.
Above Re ≈ 40–50 the twin recirculating eddies behind a cylinder become unstable and shed alternately (Kármán vortex street).
Shedding is already regular well below this range; Re in the hundreds is laminar shedding moving toward a turbulent wake.
This is the subcritical range, where the wake is turbulent and shedding has long been established.
Question 14 of 18 · Fluid Mechanics · 10–15 items
In a turbulent boundary layer, how does the thickness of the viscous sublayer compare to the overall boundary layer thickness?
The inner layer (including the log region) spans roughly 10–20% of δ; the viscous sublayer is much thinner.
The viscous sublayer reaches only y+ ≈ 5, a very small fraction of δ, typically well under 1% at engineering Reynolds numbers.
Half the layer is the outer, wake region of turbulent eddies, not the viscous sublayer.
Viscous effects dominate only right at the wall; a quarter of δ is well inside the turbulent region.
Question 15 of 18 · Fluid Mechanics · 10–15 items
In a converging-diverging nozzle with subsonic inlet flow, the back pressure is lowered just to the value at which the throat first reaches Mach 1 (the first critical back pressure). What is the flow condition?
Solution
Lowering back pressure in a C-D nozzle: subsonic throughout → first critical (M = 1 at throat, subsonic after) → shock in diverging section → design condition (supersonic exit).
At the first critical back pressure the throat just reaches Mach 1 and the diverging section acts as a subsonic diffuser.
Mach 1 is reached only where the area is a minimum; elsewhere the flow is subsonic or supersonic.
A shock inside the diverging section appears only for back pressures between the first and third critical values.
Fully supersonic diverging flow needs a much lower back pressure, the design (third critical) value.
Question 16 of 18 · Fluid Mechanics · 10–15 items
In a converging-diverging nozzle undergoing compressible flow, if the back pressure is increased above the design condition while choked flow persists at the throat, what happens to the mass flow rate?
Solution
ṁmax = A*p0√(k/(RT0))·(2/(k + 1))(k + 1)/(2(k − 1)) (independent of back pressure)
Flow falls with back pressure only after the throat unchokes; while choked it does not change.
Raising back pressure never increases flow; in an unchoked nozzle it would reduce it.
Once the throat is sonic, pressure changes downstream cannot travel upstream, so the mass flow depends only on stagnation conditions and throat area.
Choked mass flow is steady and set by upstream conditions; back pressure changes only move the shock position.
Question 17 of 18 · Fluid Mechanics · 10–15 items
In high-speed external flow over a curved surface, shock-boundary layer interaction can lead to flow separation. What is the primary effect of this interaction on the aerodynamic drag?
Solution
Shock–boundary-layer interaction: adverse pressure jump → separation → larger wake → pressure (form) drag up, plus wave drag.
Separation changes both the lift and the drag; drag rises sharply.
A shock decelerates the flow and the separation it causes adds drag.
Shock-induced separation is a main source of transonic drag rise.
The shock's adverse pressure gradient separates the boundary layer, thickening the wake and adding pressure drag.
Question 18 of 18 · Fluid Mechanics · 10–15 items
Water flows in a steady-flow mechanical energy system from an upper reservoir at an elevation of 100 m to a lower reservoir at 20 m. The total piping friction head loss is 15 m. If the volumetric flow rate is 0.5 m3/s and the turbine overall efficiency is 85%, what is most nearly the electrical power generated? Assume the fluid density is 1000 kg/m3 and gravitational acceleration is 9.81 m/s2.
FE Reference Handbook, Fluid Mechanics: energy equation (turbines).
The turbine gets the net head 80 − 15 = 65 m: P = ηρgQh = 0.85 × 1000 × 9.81 × 0.5 × 65 ≈ 271.0 kW.
This is the hydraulic power ρgQh at 65 m before the 85% turbine efficiency is applied.
This applies the efficiency but uses the full 80 m elevation difference; friction losses reduce the head reaching the turbine.
Dividing by the efficiency is the pump form; a turbine delivers less than the hydraulic power, so η multiplies.
This is ρgQ times the full 80 m with neither friction nor efficiency taken into account.
Answer every question to see your score
Common traps
Checked against the fluid mechanics material of the FE Reference Handbook and Munson's Fundamentals of Fluid Mechanics. The errors that cost the most points:
Gauge or absolute, NPSH. Both have one home on this site: gauge versus absolute pressure on FE Environmental fluid mechanics, NPSH on the civil water resources page. Compressible-flow relations here need absolute pressure throughout.
Two viscosities. Dynamic viscosity μ (Pa·s) goes with shear stress; kinematic viscosity ν = μ/ρ (m2/s) goes straight into the Reynolds number. Using the wrong one changes Re by the density.
Hydraulic diameter for ducts. A rectangular duct uses Dh = 4A/P, where P is the wetted perimeter, not the side length.
Stagnation versus static. In compressible flow, tables list ratios to stagnation conditions. A thermometer in a fast stream reads close to stagnation temperature, not static temperature.
Temperatures in kelvin or rankine
The speed of sound and every isentropic relation use absolute temperature. A nozzle problem worked in °C or °F gives a Mach number that looks plausible and is wrong.
Which fluid mechanics lives where on this site
The same section name appears on four FE exams, with different problems. This page owns fluid properties, gauge and absolute pressure, momentum, external flow, compressible flow and pump, fan and compressor performance.
Yes. The specification names Mach number, isentropic flow relationships and normal shocks. The isentropic and normal-shock relations are in the on-screen Handbook; the skill tested is using absolute temperatures and pressures and choosing the right ratio.
How many fluid mechanics questions are on the FE Mechanical exam?
10–15. Together with thermodynamics and heat transfer, the thermal-fluid areas are about as large a block as statics, dynamics and mechanics of materials.
More fluid mechanics problems on your phone
FE Mechanical questions, answers and worked solutions in a study app.