Fluids carry 8–12 questions and solids handling another 3–5, together one of the larger blocks of the exam. The fluids are process fluids: piping with fittings, pumps, meters, packed beds, compressible gases and liquids that are not Newtonian.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Chemical
questions on the exam8–12
problems on this page18
of 17 knowledge areas06
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What the specification covers
The outline lists eight fluids subtopics. In our words:
Mechanical energy balance: pipes, valves and fittings, pressure drop through packed beds, pipe networks.
Bernoulli: hydrostatic pressure and velocity head.
Laminar and turbulent flow.
Flow measurement: orifice and Venturi flowmeters.
Pumps, turbines, compressors and vacuum systems.
Compressible flow and non-Newtonian fluids.
18 fluids and solids problems
0 / 18 answered · 0 correct
Question 1 of 18 · Solids Handling · 3–5 items
A conical mass-flow hopper currently features an outlet diameter of 0.50 m and discharges a cohesive powder. At the current consolidation stress, shear cell testing indicates an unconfined yield strength (UYS) of 3.2 kPa. The powder has a bulk density of 850 kg m−3, and the calculated flow factor function for this hopper geometry is H(θ) = 2.0. Assuming standard gravity (g = 9.81 m s−2), what is the minimum required percentage increase in the outlet diameter to safely prevent stable arch formation?
Given
Current outlet diameter = 0.50 m
Unconfined yield strength σc = 3.2 kPa
Bulk density ρb = 850 kg/m3
Flow factor function H(θ) = 2.0
g = 9.81 m/s2
Find
Minimum percentage increase in outlet diameter (%)
increase = (0.768 m − 0.50 m)/0.50 m = 0.535 → 53.5 %
Jenike arching criterion for a conical mass-flow hopper: the outlet must exceed Bmin so the stress in a cohesive arch exceeds the powder's unconfined yield strength.
153.5 % is the new diameter as a percentage of the old one (0.768/0.50); the increase is that ratio minus 100 %.
135.6 % is the growth of outlet area, (0.768/0.50)2 − 1; the arching criterion sets a diameter, so the diameter ratio is what counts.
76.8 % is the critical diameter itself (0.768 m) written as a percentage instead of being compared with the existing 0.50 m outlet.
The critical outlet is B = H(θ)·σc/(ρb·g) = 0.768 m, so the 0.50 m outlet has to grow by (0.768 − 0.50)/0.50 = 53.5 %.
26.8 % is the 0.268 m shortfall (0.768 − 0.50 m) read straight off as a percent; it never divides by the current 0.50 m outlet.
Question 2 of 18 · Solids Handling · 3–5 items
A processing plant handles a highly polydisperse bulk solid with a particle size distribution ratio (d90/d10) of 45. Operators continuously report severe quality control failures due to significant accumulation of fine particles at the end of the discharge cycle. The current silo operates with a stagnant annular region of material near the walls while a vertical central channel moves downward. Which flow regime is occurring, and what is the appropriate geometric redesign required to minimize this segregation?
Solution
A stagnant wall zone with a moving central channel = funnel (core) flow. Wide size distributions sift, and funnel flow discharges the center first, leaving fines in the dead zones. A mass-flow hopper (steep, low-friction walls) gives first-in-first-out discharge and remixes the segregated layers.
Expanded flow is a mass-flow hopper section feeding a funnel-flow bin, which does not match the description, and flow aids treat cohesion, not sifting segregation.
A stagnant zone at the walls with a moving central channel is the opposite of mass flow, and switching to funnel flow would make the segregation worse.
The stagnant annulus with a central channel is funnel flow; a mass-flow hopper moves all material first-in-first-out, so sifted fines are remixed instead of trapped until the end.
Core flow is another name for funnel flow, but a flatter cone is exactly what causes it; mass flow needs steeper, smoother walls.
The regime is named correctly, but a smaller outlet promotes arching and does nothing to make the walls slide; wall angle and friction decide the regime.
Question 3 of 18 · Solids Handling · 3–5 items
An engineer is evaluating a dense slurry pipeline transport system to assess the risk of solids settling out and causing blockages. The horizontal pipeline has an internal diameter of 0.15 m and is designed to transport a total volumetric slurry flow of 0.035 m3/s. Empirical testing provides a single-particle terminal settling velocity of 0.8 m/s, while the dense slurry's bulk minimum transport velocity has been separately evaluated at 2.5 m/s. Which critical velocity criterion governs the plugging assessment, and is this current flow configuration susceptible to plugging?
Given
Pipe diameter D = 0.15 m
Slurry flow Q = 0.035 m3/s
Particle terminal settling velocity = 0.8 m/s
Bulk minimum transport velocity = 2.5 m/s
Find
Governing criterion and plugging risk
Solution
A = πD2/4 = π(0.15 m)2/4 = 0.0177 m2
V = Q/A = 0.035 m3/s ÷ 0.0177 m2 = 1.98 m/s
1.98 m/s < 2.5 m/s (bulk minimum transport velocity), so a settled bed and plugging are likely.
The criterion is right, but 1.98 m/s is below the 2.5 m/s minimum; this answer skips the Q/A calculation or uses the pipe radius as if it were the diameter.
This compares the mean velocity (1.98 m/s) with the 0.8 m/s single-particle value, which ignores hindered settling and gives false comfort.
Cavitation is a local low-pressure phenomenon in pumps and valves; it has nothing to do with solids settling in a horizontal line.
In a dense slurry particles hinder each other, so the single-particle settling velocity is not the design limit, even though the verdict on plugging happens to be right.
For a dense slurry the bulk minimum transport velocity is the criterion, and the mean velocity Q/A = 1.98 m/s is below 2.5 m/s, so solids can settle into a bed.
Question 4 of 18 · Solids Handling · 3–5 items
A ball mill is used to grind material from an initial size of 10 mm to a final size of 0.5 mm. If the power consumption is 100 kW when processing 20 tons/hr, approximately how much power would be required to process the same material at 30 tons/hr with the same size reduction?
Given
Feed size = 10 mm
Product size = 0.5 mm
Power at 20 tons/hr P1 = 100 kW
New throughput = 30 tons/hr
Find
Power required at 30 tons/hr P2 (kW)
Solution
P/ṁ = constant for a fixed size reduction (Bond: W = Wi(10/√P80 − 10/√F80))
50 kW is the increase in power (150 − 100), not the power required at 30 tons/hr.
67 kW scales the power by 20/30, the inverse of the throughput ratio; more tonnage per hour needs more power, not less.
100 kW assumes power depends only on the size reduction; the specific energy (kWh per ton) is fixed, so total power rises with tonnage.
With the same feed and product size the specific energy is unchanged (Bond, Rittinger or Kick all give energy per unit mass), so P = 100 kW × 30/20 = 150 kW.
Question 5 of 18 · Fluid Mechanics/Dynamics · 8–12 items
Which of the following is an example of a non-Newtonian fluid?
Solution
Non-Newtonian: apparent viscosity changes with shear rate (power-law, Bingham). Ketchup is shear-thinning with a yield stress; water, ethanol and air are Newtonian.
Ethanol is a simple low-molecular-weight liquid whose viscosity does not depend on shear rate, so it is Newtonian.
Gases such as air are Newtonian; their viscosity changes with temperature, not with shear rate.
Water is the textbook Newtonian fluid: shear stress is proportional to shear rate.
Ketchup is shear-thinning with a yield stress: it barely moves at rest and thins once sheared, so its viscosity depends on shear rate.
Question 6 of 18 · Solids Handling · 3–5 items
What parameter is most critical when designing a pneumatic conveying system for transporting abrasive solid particles?
Solution
Abrasive wear grows roughly with the cube of impact velocity, and below the saltation velocity solids drop out. The design sets the conveying velocity just above saltation.
Ambient temperature changes gas density slightly but does not govern abrasion or plugging.
Wear rises steeply with particle impact velocity, while too little velocity lets solids drop out (saltation); conveying velocity is the controlling design choice.
Pipe size matters, but it only sets the gas velocity for a given air flow; the word only excludes the velocity itself, which drives wear.
Humidity can affect cohesive or hygroscopic powders, but it is not what controls abrasive wear or saltation.
Question 7 of 18 · Fluid Mechanics/Dynamics · 8–12 items
Fitting Type
Loss Coefficient (K)
Globe valve (fully open)
10.0
Standard elbow
0.9
Gate valve (fully open)
0.2
A short process delivery line consists of 5.0 m of 0.05 m internal diameter piping. The Darcy friction factor is fD = 0.02. The flow velocity through the system is 4.0 m/s. The line includes one fully open globe valve and three standard elbows. Calculate the total head loss for this piping section.
FE Reference Handbook, Fluid Mechanics: head loss in pipes (Darcy-Weisbach) and minor losses in fittings.
hL = (fDL/D + ΣK)·V2/2g = (2.0 + 10.0 + 3 × 0.9) × 0.815 m = 12.00 m; the gate valve is not in this line.
10.76 m divides the Darcy factor by 4 as if it were a Fanning value ((0.5 + 12.7) × 0.815 m).
10.52 m uses one elbow instead of three (ΣK = 10.9), plus friction.
10.36 m counts only the fittings (ΣK = 12.7) and forgets the straight-pipe friction fDL/D = 2.0.
1.63 m is the pipe friction term alone (fDL/D = 2.0 × V2/2g); the globe valve and elbows are left out.
Question 8 of 18 · Fluid Mechanics/Dynamics · 8–12 items
A centrifugal pump delivers 0.05 m3/s of water (ρ= 1000 kg m−3) against a total system head of 40.0 m. The pump has a hydraulic efficiency of 75%, and it is driven by an electric motor with an efficiency of 90%. Calculate the required continuous electrical power input.
Question 9 of 18 · Fluid Mechanics/Dynamics · 8–12 items
What is the flow rate (in m³/s) through an orifice with a diameter of 2 cm, discharge coefficient of 0.62, when the pressure drop across the orifice is 50 kPa and the fluid density is 850 kg/m³?
Question 10 of 18 · Fluid Mechanics/Dynamics · 8–12 items
A high-capacity continuous chemical process requires a flow meter to measure volumetric flow in a main delivery line. The engineering team is evaluating two options based on the specifications provided below. Which flow meter should be selected to minimize long-term pump operating costs?
Meter Type
Capital Cost
Pressure Recovery Fraction
Orifice Plate
$1,500
0.35
Venturi Meter
$8,500
0.88
Solution
permanent loss = (1 − recovery) × Δpmeter
orifice: 0.65·Δp · Venturi: 0.12·Δp
Lower permanent loss means less pump energy; on a continuous high-flow line the energy saving outweighs the $7,000 capital difference.
Equal costs would need equal permanent pressure loss, but the meters recover 88 % and 35 % of their differential.
Pumping cost follows the unrecovered pressure drop; the Venturi recovers 88 % of its differential versus 35 % for the orifice, so it saves energy on a continuous high-flow line.
The orifice wins on purchase price only; it leaves about 65 % of its differential as a permanent loss the pump pays for every hour.
One of the two meters clearly loses less head, so rejecting both does not answer the operating-cost question.
The flow rate changes how much is saved, not which meter wastes less: at any flow the Venturi's permanent loss is the smaller one.
Question 11 of 18 · Fluid Mechanics/Dynamics · 8–12 items
A differential pressure flow meter has an overall discharge coefficient (K) of 0.98 and a throat area of 0.01 m2. The measured pressure drop across the meter is 2500 Pa. The fluid density at standard conditions is 1.0 kg m−3, but the local process density under operating compression is 4.0 kg m−3. Using the simplified flow equation ṁ = K A √(2ρΔP), calculate the mass flow rate.
The density is the one at flowing conditions (4.0 kg/m3); the standard-condition density does not enter.
0.35 kg/s divides the standard-density result by √4 instead of multiplying, applying the density correction backwards.
0.69 kg/s uses the standard-condition density (1.0 kg/m3); the meter responds to the density actually flowing through it.
ṁ = KA√(2ρΔP) with the process density 4.0 kg/m3: 0.98 × 0.01 m2 × √(2 × 4.0 × 2,500) = 1.39 kg/s.
2.77 kg/s scales the standard result linearly with density (× 4); mass flow goes with √ρ, so the factor is 2.
5.54 kg/s applies the density ratio four times over (1.39 × 4); there is no reason to correct an already corrected result.
Question 12 of 18 · Fluid Mechanics/Dynamics · 8–12 items
A 50 m long, 0.05 m diameter pipe carries water (ρ = 1,000 kg/m3, μ = 0.001 Pa·s). The pipe has an absolute roughness of 0.0002 m, and the water flows at 3 m/s. Using the Colebrook equation, what is the approximate Darcy friction factor?
FE Reference Handbook, Fluid Mechanics: Moody (Stanton) diagram and Colebrook equation.
0.007 is the Fanning friction factor (Darcy ÷ 4); the Colebrook equation as written in the Handbook returns the Darcy factor.
0.017 is the smooth-pipe value at Re = 1.5 × 105; it drops the roughness term ε/D = 0.004.
Re = 1.5 × 105 and ε/D = 0.004; iterating Colebrook, 1/√f = −2log(ε/3.7D + 2.51/(Re√f)), gives f ≈ 0.029.
0.065 takes the relative roughness ten times too large (0.04), the result of putting the roughness in millimeters over the diameter in meters.
Question 13 of 18 · Fluid Mechanics/Dynamics · 8–12 items
In a pipe network, two parallel pipes carry water (ρ = 1,000 kg/m3, μ = 0.001 Pa·s) under laminar flow conditions. The first pipe has a diameter of 0.1 m and a length of 50 m; the second has a diameter of 0.05 m and a length of 20 m. For the same pressure drop, what is the ratio of volumetric flow rates Q1/Q2 given by the Hagen-Poiseuille relation?
16.0 is the diameter effect alone, (D1/D2)4; it ignores that pipe 1 is 2.5 times longer.
40.0 multiplies by the length ratio instead of dividing; a longer pipe carries less flow at the same pressure drop.
Question 14 of 18 · Fluid Mechanics/Dynamics · 8–12 items
Air (k = 1.4) flows isentropically through a converging nozzle. If the stagnation pressure is 600 kPa and the nozzle exit pressure is 320 kPa, what is the approximate exit Mach number?
FE Reference Handbook, Fluid Mechanics: compressible flow, isentropic relations. P/P0 = 0.533 > 0.528, so the exit is not choked.
2.09 treats P0/P = 1 + 0.2M2 with no exponent at all; it also ignores that a converging nozzle tops out at Mach 1.
1.68 uses the exponent 1/k instead of (k − 1)/k; a converging nozzle also cannot exceed Mach 1 at its exit.
(P0/P)(k−1)/k = 1.8750.286 = 1.197, so M2 = (1.197 − 1)/0.2 = 0.98 and M ≈ 0.99; the ratio 0.533 is just above the critical 0.528, so the exit is barely subsonic.
0.70 drops the ½ in (k − 1)/2, dividing by 0.4 instead of 0.2 when solving for M2.
Question 15 of 18 · Fluid Mechanics/Dynamics · 8–12 items
A liquid flows downward through a vertical pipe of constant diameter from point B to point A. The pressure at A is 200 kPa, and the pressure at B, 2 m above A, is 180 kPa. If the liquid density is 1,200 kg/m3, what is the head loss between B and A?
Given
Pressure at A PA = 200 kPa
Pressure at B PB = 180 kPa
Elevation of B above A Δz = 2 m
Liquid density ρ = 1,200 kg/m3
Flow direction: downward, B → A; constant diameter
Find
Head loss between B and A hL (m)
Solution
PB/ρg + zB = PA/ρg + zA + hL (VA = VB)
hL = (180,000 − 200,000) Pa/(1,200 kg/m3 × 9.81 m/s2) + 2.0 m = −1.70 m + 2.0 m = 0.30 m
FE Reference Handbook, Fluid Mechanics: energy equation.
With flow from B down to A: hL = (PB − PA)/ρg + (zB − zA) = −1.70 m + 2.0 m = 0.30 m.
1.7 m is the pressure-head difference alone; it ignores the 2 m elevation change in the vertical pipe.
2.0 m is the elevation difference alone, as if the pressures at A and B were equal.
3.7 m adds the pressure head and the elevation head; the 20 kPa pressure rise from B to A works against the elevation drop, so they subtract.
Question 16 of 18 · Fluid Mechanics/Dynamics · 8–12 items
A viscous oil (ρ = 900 kg/m3, μ = 0.5 Pa·s) flows in laminar flow through the annulus between two concentric tubes (inner radius 0.05 m, outer radius 0.1 m, length 10 m) under a pressure drop of 5,000 Pa. Using Q = π·ΔP/(8μL)·[(Ro4 − Ri4) − (Ro2 − Ri2)2/ln(Ro/Ri)] and vavg = Q/[π(Ro2 − Ri2)], what is the average velocity?
vavg = Q/A = 0.210 m/s · Re = ρv(2(Ro − Ri))/μ ≈ 38 (laminar)
0.157 m/s divides the flow by the full circle πRo2; the liquid only occupies the annulus π(Ro2 − Ri2).
Q = 4.95 × 10−3 m3/s through an annular area of 0.0236 m2 gives vavg = 0.210 m/s; Re ≈ 38, so laminar flow holds.
1.56 m/s drops the logarithmic term of the annulus formula, treating the gap as if no inner wall dragged on the fluid.
3.36 m/s puts diameters where the formula needs radii; with fourth powers this inflates the flow sixteen-fold.
Question 17 of 18 · Fluid Mechanics/Dynamics · 8–12 items
Water (ρ = 1,000 kg/m3, μ = 0.001 Pa·s) flows downward through a 2.0 m deep bed of 5 mm spherical catalyst pellets with a void fraction of 0.40. The superficial velocity is 0.10 m/s. Using the Ergun equation, what is the frictional pressure drop across the bed?
FE Reference Handbook, Fluid Mechanics: flow through a packed bed (Ergun equation).
ΔP/L = 3,375 Pa/m (viscous) + 32,810 Pa/m (inertial) = 36,190 Pa/m, so the 2.0 m bed loses 72.4 kPa.
65.6 kPa keeps only the inertial (Burke-Plummer) term and drops the viscous contribution of 6.75 kPa.
36.2 kPa is the pressure gradient per meter of bed; it still has to be multiplied by the 2.0 m depth.
6.75 kPa keeps only the viscous (Blake-Kozeny) term; at a particle Reynolds number near 830 the inertial term dominates.
Question 18 of 18 · Fluid Mechanics/Dynamics · 8–12 items
A vacuum column is held at 5 kPa absolute by a two-stage steam-jet ejector system with a water-cooled intercondenser between the first and second stages. What is the main purpose of the intercondenser?
Solution
Each ejector stage discharges its own motive steam plus the suction load. An intercondenser between stages condenses that steam (and condensable process vapor), so the next stage only compresses the noncondensables and saturated vapor — a smaller, cheaper stage with less steam.
The first-stage discharge is mostly motive steam; condensing it between stages leaves only noncondensables and a little vapor for the second stage, which cuts its steam use sharply.
Condensing column overhead for reflux is the job of the overhead (pre-)condenser upstream of the ejectors, not the intercondenser.
Noncondensables cannot be condensed out; they pass through every stage and leave at the last stage's discharge to atmosphere.
Motive steam pressure is set by the supply header; a condenser lowers the vapor load and cannot raise steam pressure.
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Common traps
Darcy against Fanning. The reference’s friction chart gives the Darcy factor; many chemical engineering texts use Fanning, which is one quarter of it: fD = 4·fF. Plugging a Fanning factor into the reference’s Darcy–Weisbach equation gives a head loss four times too small.
Interstitial velocity in the Ergun equation. The packed-bed equation is written with the superficial velocity and the bed void fraction; feeding it the velocity inside the voids overstates the pressure drop.
One threshold for laminar flow. Sources put the end of laminar pipe flow at Re ≈ 2,100 or 2,300; use the value the problem or the reference gives.
Viscosity units. 1 cP = 1 mPa·s; centipoise left in an SI Reynolds number is off by a thousand.
Sources: the Fluid Mechanics chapter of the FE Reference Handbook; Bird, Stewart and Lightfoot, Transport Phenomena.
Solids handling
Solids Handling is its own area with 3–5 questions, new in the 2020 outline; on this site it shares the page with fluids because slurries, pneumatic lines and packed beds are fluid problems with particles in them. The outline covers particle properties (surface and bulk forces, size distribution), processing (crushing, grinding, crystallization), and transport and storage (belts, pneumatic conveying, slurries, tanks, hoppers).
Which mean size. A size distribution has several means; surface-volume (Sauter) diameter is the one for pressure drop and surface processes. Use the one the method asks for.
Bulk against particle density. A hopper or bin holds bulk density, voids included; settling and slurry calculations use the density of the solid itself.
Crystallization yield with water of hydration. When the crystal is a hydrate, part of the water leaves with the solid; the solute balance must count it.
Sources: McCabe, Smith and Harriott, Unit Operations of Chemical Engineering, sections on particles and size reduction.