Fluid mechanics and hydraulics is tied for the largest area at 12–18 questions, and it is the water-flow base of the whole exam: the same head-loss and channel equations come back in treatment plants, culverts and stream channels.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Environmental
questions on the exam12–18
problems on this page18
areas in the specification15
Checked
What the area covers
Eight headings in the specification, from statics to unsteady flow:
Fluid statics: pressure with depth, manometers, forces on submerged surfaces.
Closed conduits: Darcy-Weisbach with the Moody diagram, Hazen-Williams, minor losses.
Open channels: Manning's equation, subcritical and supercritical flow, culverts, hydraulic elements.
Pumps: power, operating point, pumps in series and in parallel.
Blowers: power, inlet and outlet pressure, efficiency, operating point, series and parallel.
Flow measurement: weirs, orifices, flumes.
Fluid dynamics: continuity, Bernoulli, laminar and turbulent flow.
Steady and unsteady flow.
18 pipe, channel and pump problems
0 / 18 answered · 0 correct
Question 1 of 18 · Fluid Mechanics and Hydraulics · 12–18 items
Water flows through a horizontal pipe that reduces from a 20 cm diameter to a 10 cm diameter. If the velocity in the larger pipe is 2 m/s, what is the velocity in the smaller pipe?
Given
Larger pipe diameter D1 = 20 cm
Smaller pipe diameter D2 = 10 cm
Velocity in the larger pipe V1 = 2 m/s
Find
Velocity in the smaller pipe V2 (m/s)
Solution
Continuity for an incompressible fluid: the flow rate is the same in both sections (Fluid mechanics: continuity).
V2 = V1(D1/D2)2 = 2 m/s × (20 cm/10 cm)2 = 8 m/s
0.5 m/s squares the diameter ratio the wrong way up; the smaller pipe must carry the same flow faster.
1 m/s uses the inverted diameter ratio without squaring it; velocity goes up, not down, in the smaller pipe.
4 m/s scales velocity by the diameter ratio (2×); flow area, and so velocity, scales with the diameter squared.
Question 2 of 18 · Fluid Mechanics and Hydraulics · 12–18 items
What is the Reynolds number for water (kinematic viscosity = 10−6 m²/s) flowing at 2 m/s through a 10 cm diameter pipe?
Given
Kinematic viscosity ν = 10−6 m2/s
Velocity V = 2 m/s
Pipe diameter D = 10 cm
Find
Reynolds number Re
Solution
Reynolds number with the kinematic viscosity (Fluid mechanics: dimensionless numbers).
Re = VD/ν = (2 m/s × 0.10 m)/(1 × 10−6 m2/s) = 2.0 × 105
2,000 enters the diameter as 0.001 m, a slip of two decimal places.
20,000 enters the 10 cm diameter as 0.01 m instead of 0.10 m.
Re = VD/ν = (2 m/s × 0.10 m)/(10−6 m2/s) = 200,000.
2,000,000 enters the 10 cm diameter as 1 m instead of 0.10 m.
Question 3 of 18 · Fluid Mechanics and Hydraulics · 12–18 items
Which of the following flow measuring devices creates a pressure drop by constricting the flow?
Solution
Differential-pressure meters (venturi, orifice, flow nozzle) create a pressure drop by narrowing the flow area (Fluid mechanics: flow measurement).
A venturi narrows the pipe through a throat; velocity rises and pressure falls there, and that differential gives the flow rate.
A pitot tube measures stagnation minus static pressure at a point; it does not constrict the flow.
A magnetic flowmeter senses the voltage induced in a conductive liquid and has a full, unobstructed bore.
A current meter infers velocity from the spin rate of a propeller or cups in an open stream; no constriction is involved.
Question 4 of 18 · Fluid Mechanics and Hydraulics · 12–18 items
In an open channel flow, the hydraulic depth is defined as:
Solution
Hydraulic depth D = A/T (flow area over top width), used in Fr = V/√(gD) (Hydraulics: open-channel flow).
P/A is the reciprocal of the hydraulic radius and has no standard use in channel flow.
Hydraulic depth D = A/T, the depth of a rectangle with the same area and surface width; it enters the Froude number.
A/P is the hydraulic radius, used in Manning's equation, not the hydraulic depth.
The maximum depth equals A/T only for a rectangular section; for any other shape it is larger.
Question 5 of 18 · Fluid Mechanics and Hydraulics · 12–18 items
A horizontal pipe undergoes a smooth contraction from a diameter of 0.30 m to 0.15 m. Water flows in the larger section at 3 m/s. Using the continuity equation and Bernoulli's principle for inviscid flow, determine the pressure drop between the two sections.
Given
Upstream diameter D1 = 0.30 m
Downstream diameter D2 = 0.15 m
Upstream velocity v1 = 3 m/s
Find
Pressure drop Δp (kPa)
Solution
Continuity gives the throat velocity; Bernoulli for a horizontal pipe gives the pressure drop (Fluid mechanics: Bernoulli equation).
Question 6 of 18 · Fluid Mechanics and Hydraulics · 12–18 items
A 200 mm diameter pipe system includes five fittings: two standard elbows (K=0.9 each), two gate valves (K = 0.2 each), and one check valve (K = 2.5). If the friction factor f is 0.025, calculate the total equivalent length of these fittings.
Given
Pipe diameter D = 200 mm
Standard elbows: 2, K = 0.9 each
Gate valves: 2, K = 0.2 each
Check valve: 1, K = 2.5
Friction factor f = 0.025
Find
Total equivalent length of fittings (m)
Solution
Equivalent length makes a fitting's loss look like straight pipe: K·V2/2g = f(Le/D)V2/2g (Fluid mechanics: minor losses).
ΣK = 2 × 0.9 + 2 × 0.2 + 2.5 = 4.7
Le = ΣK·D/f = 4.7 × 0.200 m / 0.025 = 37.6 m
0.940 m is ΣK × D; it still has to be divided by f = 0.025.
4.70 is ΣK, a dimensionless sum, not a length.
28.8 m counts each fitting type once (ΣK = 3.6); there are two elbows and two gate valves.
ΣK = 2(0.9) + 2(0.2) + 2.5 = 4.7, and Le = ΣK·D/f = 4.7 × 0.200 m/0.025 = 37.6 m.
188 m is ΣK/f without the diameter; Le = K·D/f.
Question 7 of 18 · Fluid Mechanics and Hydraulics · 12–18 items
A student estimates the head loss in a rough concrete pipe (ε/D = 0.01) at Re = 200,000 using the Blasius approximation f = 0.316·Re−0.25. If the correct friction factor from the Colebrook equation is 0.038, what is the percent error in the calculated head loss?
Given
Relative roughness ε/D = 0.01
Reynolds number Re = 200,000
Blasius f = 0.316 Re−0.25
Colebrook friction factor = 0.038
Find
Percent error in head loss
Solution
Head loss is proportional to f at fixed V, L and D, so its error equals the friction-factor error (Fluid mechanics: Darcy–Weisbach).
−154% divides the difference by the Blasius value; percent error is referenced to the correct value, 0.038.
0% assumes Blasius applies here; it is a smooth-pipe correlation and ignores ε/D = 0.01.
+154% takes the correct value relative to the Blasius value, with the wrong sign and the wrong reference.
+61% flips the sign: Blasius underpredicts f for a rough pipe, so the head loss comes out too low.
Blasius gives f = 0.316 × (200,000)−0.25 = 0.0149; since hf ∝ f, the error is (0.0149 − 0.038)/0.038 = −61%.
Question 8 of 18 · Fluid Mechanics and Hydraulics · 12–18 items
A pump lifts water 76 ft vertically through 1,000 ft of 6-inch diameter pipe (D = 0.5 ft). The flow rate is 449 gpm (1.0 cfs), resulting in a velocity head of 0.4 ft. If the friction factor f is 0.03 and minor losses are negligible, calculate the Total Dynamic Head (TDH).
Given
Static lift = 76 ft
Pipe length L = 1,000 ft
Pipe diameter D = 0.5 ft
Flow rate Q = 1.0 cfs
Velocity head V2/2g = 0.4 ft
Friction factor f = 0.03
Find
Total dynamic head TDH (ft)
Solution
TDH = static lift + friction loss, with minor losses neglected (Fluid mechanics: energy equation).
hf = f(L/D)(V2/2g) = 0.03 × (1,000 ft/0.5 ft) × 0.4 ft = 24 ft
TDH = 76 ft + 24 ft = 100 ft
24 ft is the friction loss alone; the pump must also supply the 76 ft lift.
76 ft is the static lift alone, with no friction loss.
78 ft enters D as 6 (inches) in L/D, making the friction loss 2 ft instead of 24 ft.
88 ft leaves D out of L/D, so the friction loss is 0.03 × 1,000 × 0.4 = 12 ft.
Question 9 of 18 · Fluid Mechanics and Hydraulics · 12–18 items
A gravity pipeline originally operates with a friction factor of 0.020. Due to aging and roughness buildup, the friction factor increases to 0.03125. Assuming the driving head remains constant and the flow is fully turbulent, what is the percentage reduction in flow rate?
Given
Original friction factor f1 = 0.020
Aged friction factor f2 = 0.03125
Driving head = constant
Find
Percentage reduction in flow rate (%)
Solution
With the driving head fixed, Darcy–Weisbach gives h ∝ fQ2 (Fluid mechanics: Darcy–Weisbach).
Q2/Q1 = √(f1/f2) = √(0.020/0.03125) = 0.80
reduction = 1 − 0.80 = 20%
At constant head fQ2 stays constant, so Qaged/Qnew = √(0.020/0.03125) = 0.80, a 20% reduction.
25% is √(0.03125/0.020) − 1, which measures the change relative to the reduced flow, not the original.
36% is 1 − 0.020/0.03125; it forgets that Q varies with the square root of 1/f.
56% is the increase in friction factor (0.03125/0.020 − 1), not the change in flow.
80% is the fraction of the original flow that remains, not the reduction.
Question 10 of 18 · Fluid Mechanics and Hydraulics · 12–18 items
A differential manometer with mercury (SG=13.6) connects water pipe A (SG=1.0) to oil pipe B (SG=0.88) at the same elevation. The mercury level is 0.40 m below A and 0.80 m below B. Calculate the pressure difference PB - PA.
Given
Mercury specific gravity SG = 13.6
Pipe A fluid: water SG = 1.0
Pipe B fluid: oil SG = 0.88
Mercury level below A = 0.40 m
Mercury level below B = 0.80 m
Find
Pressure difference PB − PA (kPa)
Solution
Walk the manometer from A to B, adding γh going down and subtracting going up (Fluid mechanics: manometers).
PA + γw(0.40) + 13.6γw(0.40) − 0.88γw(0.80) = PB
PB − PA = 9.81 kN/m3 × (0.40 + 5.44 − 0.704) m = 50.4 kPa
49.4 kPa treats the oil column as water (SG 1.0 instead of 0.88).
From A: PA + 9.81(0.40) + 9.81(13.6)(0.40) − 9.81(0.88)(0.80) = PB, so PB − PA = 9.81 × 5.136 = 50.4 kPa.
53.4 kPa keeps only the 0.40 m mercury column and drops both the water and oil legs.
57.3 kPa leaves out the 0.80 m oil column on the B side.
64.2 kPa adds the oil column; moving up through the oil toward B, pressure decreases.
Question 11 of 18 · Fluid Mechanics and Hydraulics · 12–18 items
A 40 kN concrete anchor (SG=2.4) is attached to a foam buoy (SG=0.15). Calculate the foam volume for neutral buoyancy in water.
Given
Anchor weight = 40 kN
Concrete SG = 2.4
Foam SG = 0.15
Find
Foam volume (m3)
Solution
Neutral buoyancy: total buoyant force = total weight (Fluid mechanics: buoyancy).
1.70 m3 is the volume of the concrete anchor, not of the foam.
2.38 m3 ignores the foam's own weight; each cubic meter of foam lifts only 9.81 × (1 − 0.15) = 8.34 kN net.
The anchor's submerged weight is 40 − 16.7 = 23.3 kN; foam lifts 9.81 × 0.85 = 8.34 kN/m3 net, so V = 23.3/8.34 = 2.80 m3.
4.08 m3 is 40/9.81: it ignores both the buoyancy on the concrete and the foam's weight.
4.80 m3 accounts for the foam's weight but not the 16.7 kN of buoyancy acting on the submerged concrete.
Question 12 of 18 · Fluid Mechanics and Hydraulics · 12–18 items
An open tank contains a 1.8 m layer of oil (SG = 0.85) floating on a 2.5 m layer of water. Calculate the gauge pressure at the bottom of the tank in psi.
Given
Oil layer thickness = 1.8 m
Oil SG = 0.85
Water layer thickness = 2.5 m
Find
Gauge pressure at tank bottom (psi)
Solution
Gauge pressure at the bottom sums the weight of each layer (Fluid mechanics: hydrostatic pressure).
p = γw(SGoilhoil + hw) = 9.81 kN/m3 × (0.85 × 1.8 + 2.5) m = 39.5 kPa
p = 39.5 kPa ÷ 6.895 kPa/psi = 5.73 psi
2.18 psi counts the oil layer only.
3.56 psi counts the water layer only; the oil above it adds its weight too.
6.12 psi treats the oil as water (SG 1.0) for the full 4.3 m.
39.5 is the pressure in kPa; it was never converted to psi.
Question 13 of 18 · Fluid Mechanics and Hydraulics · 12–18 items
A sharp-crested rectangular weir with a crest elevation of 100.0 m spans a 3.0 m wide channel. The upstream water surface elevation is 100.6 m, and the downstream tailwater elevation is 99.8 m. Using a weir discharge coefficient Cw = 1.84 (SI units), what is the discharge over the weir?
7.6 m3/s multiplies by (2/3)√(2g) = 2.95 on top of Cw = 1.84, which already contains that factor.
4.6 m3/s uses the US customary coefficient 3.33 with SI lengths; in SI the coefficient is 1.84.
4.0 m3/s takes the head as upstream minus tailwater (0.8 m); the head is measured above the crest.
3.3 m3/s uses H to the first power; weir discharge goes as H3/2.
Tailwater (99.8 m) is below the crest, so flow is free; H = 100.6 − 100.0 = 0.6 m and Q = 1.84 × 3.0 × 0.63/2 = 2.6 m3/s.
Question 14 of 18 · Fluid Mechanics and Hydraulics · 12–18 items
Water discharges from a large reservoir through a sluice gate into a horizontal rectangular channel. The depth of water in the reservoir is 5.0 m, and the depth at the vena contracta immediately downstream of the gate is 1.0 m. Assuming energy is conserved and the velocity in the reservoir is negligible, what is the flow velocity at the vena contracta?
Given
Reservoir depth = 5.0 m
Depth at the vena contracta = 1.0 m
Reservoir velocity ≈ 0
Find
Velocity at the vena contracta (m/s)
Solution
Specific energy is conserved from the reservoir to the vena contracta (Hydraulics: energy equation).
y1 = y2 + V22/2g → 5.0 = 1.0 + V22/(2 × 9.81)
V2 = √(2 × 9.81 m/s2 × 4.0 m) = 8.86 m/s
4.43 m/s uses the 1.0 m downstream depth as the head; the driving head is the drop from 5.0 m to 1.0 m.
6.26 m/s uses √(gΔh) and drops the factor 2 from V2/2g.
Energy from the reservoir to the vena contracta: 5.0 = 1.0 + V2/2g, so V = √(2 × 9.81 × 4.0) = 8.86 m/s.
9.90 m/s uses the full 5.0 m reservoir depth and ignores the 1.0 m depth at the vena contracta.
17.7 m/s takes the square root of 2g only and multiplies by Δh = 4.0 m.
Question 15 of 18 · Fluid Mechanics and Hydraulics · 12–18 items
A valve at the downstream end of a pipe is closed instantaneously, reducing the water velocity from 2.5 m/s to zero. The wave celerity in the pipe is 1,200 m/s, and the density of water is 1,000 kg/m3. What is the maximum pressure rise expected at the valve?
Given
Velocity change ΔV = 2.5 m/s
Wave celerity a = 1,200 m/s
Density ρ = 1,000 kg/m3
Find
Maximum pressure rise (kPa)
Solution
Instantaneous closure: Joukowsky equation (Fluid mechanics: water hammer).
Joukowsky: Δp = ρaΔV = 1,000 kg/m3 × 1,200 m/s × 2.5 m/s = 3.0 × 106 Pa = 3,000 kPa.
306 kPa divides ρaΔV by g, as if the formula were in head units.
29,400 kPa uses the unit weight 9,810 N/m3 in place of the density 1,000 kg/m3.
1,500 kPa halves the Joukowsky rise; the full ρaΔV applies to an instantaneous closure.
3,000,000 is the result in Pa; labeling it kPa overstates it 1,000 times.
Question 16 of 18 · Fluid Mechanics and Hydraulics · 12–18 items
In the design of a water hammer protection system for a steel pipeline, an engineer needs to reduce the pressure wave celerity (a) to minimize surge pressures. Which modification results in the greatest decrease in wave celerity?
Solution
a = √[(K/ρ)/(1 + (K/E)(D/e))]: the pipe wall's elasticity E dominates; a flexible plastic wall lowers a sharply (Fluid mechanics: water hammer).
A thicker wall makes the pipe stiffer (smaller D/e), which raises the celerity.
A smaller diameter at the same wall thickness lowers D/e and stiffens the pipe, so celerity rises.
Water's bulk modulus changes only a few percent between 4 °C and 20 °C, a minor effect next to the wall material.
Anchoring changes the restraint factor only slightly and, if anything, raises the celerity.
Celerity falls as the pipe wall gets more elastic; HDPE's modulus is roughly 200 times lower than steel's, which can cut a from over 1,000 m/s to a few hundred.
Question 17 of 18 · Fluid Mechanics and Hydraulics · 12–18 items
A positive pressure transient wave travels upstream through a pipeline and encounters a constant-head reservoir. Based on the boundary conditions required to maintain fixed head at the reservoir, how is the incident pressure wave reflected back into the pipe?
Solution
At a constant-head boundary the reflection coefficient is −1: a pressure rise returns as an equal pressure drop (Fluid mechanics: water hammer).
A reservoir holds the head fixed, so the incident pressure rise must be canceled: the wave returns as a negative (relief) wave.
Same-sign reflection happens at a closed end such as a shut valve, where velocity, not head, is fixed.
Full absorption would need a boundary matched to the pipe's impedance; a reservoir reflects the wave completely.
A constant-head boundary reflects the whole wave (coefficient −1), not half of it.
Pressure doubles at a closed dead end; a constant-head reservoir allows no pressure change at all.
Question 18 of 18 · Fluid Mechanics and Hydraulics · 12–18 items
A 3,000 m penstock connects a reservoir to a valve. Wave celerity is 1,000 m/s. The valve closes linearly in 8 s. How does the Joukowsky equation (ΔP = -ρa ΔV) prediction compare to the actual maximum pressure rise?
Given
Penstock length L = 3,000 m
Wave celerity a = 1,000 m/s
Valve closure time = 8 s
Find
How Joukowsky compares with the actual pressure rise
Solution
Compare the closure time with the wave round trip (Fluid mechanics: water hammer).
2L/a = 2 × 3,000 m ÷ 1,000 m/s = 6 s < tc = 8 s → slow closure, peak < ρaΔV
Friction can add line-packing in very long lines, but here the controlling effect is the slow closure, which lowers the peak.
Closure is "instantaneous" only when it is shorter than 2L/a = 6 s; 8 s is slower.
The direction is right, but Joukowsky does not neglect compressibility — the celerity a exists only because water is compressible.
2L/a = 2 × 3,000/1,000 = 6 s < 8 s, so relief waves return from the reservoir before closure ends and the actual rise is below ρaΔV.
The reservoir reflection returns as a negative wave that relieves pressure; it does not add to it.
Answer every question to see your score
Common traps
Gauge or absolute pressure. Hydraulics problems run in gauge pressure; switch to absolute only where the physics needs it, such as gas laws and vapor pressure. Pump suction checks (NPSH) are worked on FE Civil water resources.
Darcy or Fanning: the handbook's Moody chart gives the Darcy factor; the mix-up between the two is explained on FE Chemical fluid mechanics. With Darcy f:
hf = f L V2 / (2g D)
Manning's constant follows the units. In V = (k/n) R2/3 S1/2, k = 1.0 in SI and 1.49 in USCS, with the hydraulic radius R in meters or feet to match.
Hazen-Williams is for water only. Its C is a smoothness rating, not a friction factor: a higher C means less loss.
Read the regime first. In a rectangular channel Fr = V/√(gy); below 1 the flow is subcritical, above 1 supercritical, and which root of the specific-energy equation you keep depends on it.
Pumps in series add head; in parallel they add flow. The operating point is where the pump curve meets the system curve, which is not necessarily the best-efficiency point.
A blower is not a pump. Air compresses across a blower, so its power comes from the adiabatic compression relation, not from γQH.
Other pages that use these equations
This page is the site's home for the flow of water in pipes, channels and pumps. Other pages pick up the rest: