Fluid Mechanics
6–9 of 110 itemsNCEES draws between 6 and 9 of the 110 from this area on any legal paper. Its outline lists 4 sub-topics; the bank holds 32 verified templates against them, which expand to 937 questions.
Sub-topics, from the NCEES outline
- AFlow measurement
- BFluid properties
- CFluid statics
- DEnergy, impulse, and momentum of fluids
5 worked problems
Real questions from the bank, at the depth the exam asks them. Pick an answer first: the reasoning opens when you do, along with why each wrong answer is tempting.
Question 1 of 5
A nozzle discharges water to atmosphere from a line at 90 psi. With a velocity coefficient of 0.95, the exit velocity is most nearly:
Nozzle on the end of a line pressurised to 90 psi, discharging a free jet to atmosphere. Show the answer and the reasoning
Why C
Given: line pressure 90 psi, Cv = 0.95, discharge to atmosphere. Relation: the pressure head becomes velocity head, V = Cv sqrt(2gh) with h = p/gamma. Substitute: h = 90 x 144/62.4 = 207.69 ft; V = 0.95 sqrt(64.4 x 207.69). Result: V = 109.9 ft/s.
Why A is tempting
Left out the 144 that converts square inches to square feet.
Why B is tempting
Used the pressure in psi directly as a head in feet. It must be converted with 144/gamma.
Why D is tempting
Left out the velocity coefficient.
Question 2 of 5
A vertical rectangular gate 7 ft wide and 5 ft tall has its top edge 10 ft below the water surface. The depth to the centre of pressure is most nearly:
Vertical rectangular gate 7 ft wide and 5 ft tall with its top edge 10 ft below the water surface; the pressure diagram over its face is triangular, so the resultant acts below mid-height. Show the answer and the reasoning
Why A
Given: a gate 7 ft wide and 5 ft tall, its top edge 10 ft below the surface. Relation: y_cp = y_bar + I/(y_bar A), with I = bh^3/12 about the centroid. Substitute: y_bar = 12.50 ft; I/(y_bar A) = 0.1667 ft. Result: y_cp = 12.67 ft. Check: the centre of pressure always sits below the centroid.
Why B is tempting
Used the moment of inertia about the top edge rather than the centroid.
Why C is tempting
Left the centroidal depth out of the correction term.
Why D is tempting
Placed the resultant at the bottom edge.
Question 3 of 5
Water at 1.94 slug/ft^3 flows at 10 cfs and 13 ft/s through a 180-degree bend. The magnitude of the momentum force on the bend is most nearly:
Pipe bend turning the flow through 180 degrees, carrying 10 cubic feet per second at 13 ft/s so that the inlet and outlet momentum fluxes both push in the same direction. Show the answer and the reasoning
Why B
Given: rho = 1.94 slug/ft^3, Q = 10 cfs, V = 13 ft/s, turned through 180 degrees. Relation: F = rho Q (change in V); reversing the flow makes the change 2V. Substitute: F = 2(1.94)(10)(13). Result: F = 504.4 lb. Check: pressure forces add to this in a real thrust-block calculation.
Why A is tempting
Left the fluid density out.
Why C is tempting
Used the velocity squared in place of the discharge times velocity.
Why D is tempting
Multiplied by gravity. With density in slugs the result is already in pounds.
Question 4 of 5
Water flows at 5.5 ft/s through a 700-ft length of pipe with an inside diameter of 12 in. The Darcy friction factor is 0.03. The head loss due to friction is most nearly:
Straight pipe run 700 ft long with an inside diameter of 12 in, carrying water at 5.5 ft/s. Show the answer and the reasoning
Why D
Given: V = 5.5 ft/s, L = 700 ft, D = 12 in = 1.000 ft, f = 0.03. Relation: hf = f (L/D) V^2/2g. Substitute: V^2/2g = 0.470 ft; hf = 0.03(700/1.000)(0.470). Result: hf = 9.86 ft. Check: the factor given is the Darcy factor, so no factor of four applies.
Why A is tempting
Left the diameter in inches while the length is in feet.
Why B is tempting
Used V instead of V^2. Friction loss scales with the square of velocity.
Why C is tempting
Divided by four as if the Darcy factor were a Fanning factor. The Darcy-Weisbach form takes the Darcy factor directly.
Question 5 of 5
A fluid has a dynamic viscosity of 1.0e-5 lb-s/ft^2 and a density of 1.7 slug/ft^3. Its kinematic viscosity is most nearly:
Show the answer and the reasoning
Why D
Given: mu = 1.0e-5 lb-s/ft^2, rho = 1.7 slug/ft^3. Relation: nu = mu/rho. Substitute: nu = 1.0e-5/1.7. Result: nu = 0.00000588 ft^2/s. Check: the units are area over time, which is why it is called kinematic.
Why A is tempting
Divided by the specific weight rather than the density.
Why B is tempting
Divided by the density twice over.
Why C is tempting
Divided by the density of water, 1.94 slug/ft^3, rather than the density of the fluid given.
Four mistakes that cost the question
Each of these lands on an answer that is offered, so it costs the question outright. The minutes are our estimate of the time each one burns on top, against an average of 2.9 minutes a question (320 minutes for 110).
Costs the question and about 3 min
Mixing gauge and absolute pressure in one equation
The energy equation works in gauge pressure if every term is gauge; vapour pressure, net positive suction head and gas laws need absolute. Absolute is gauge plus atmospheric, 14.7 psi at sea level. Mixing the two in one equation puts the head off by about 34 ft of water.
Costs the question and about 3 min
Putting the centre of pressure at the centroid
On a submerged plane surface the resultant force equals the pressure at the centroid times the area, but it acts below the centroid, by the centroidal moment of inertia divided by the centroid depth times the area. Using the centroid depth for the line of action is the standard distractor in gate and hinge questions.
Costs the question and about 3 min
Treating velocities as speeds in a momentum problem
The force on a bend or a plate is the mass flow rate times the change in velocity, taken as vectors in each direction. A 180-degree bend reverses the flow, so the change is twice the velocity, not zero and not the velocity once. Treating the velocities as speeds gives the answer for a different bend.
Costs the question and about 2 min
Forgetting the specific gravity when converting pressure to head
Pressure head is pressure divided by the specific weight of the fluid in the pipe, not of water: 1 psi is 2.31 ft of water but 2.31 divided by the specific gravity of any other liquid. Questions on oil lines and manometers offer the water value beside the right one.
Questions about Fluid Mechanics
How does Fluid Mechanics differ from the hydraulics in Water Resources?
Fluid Mechanics covers fluid properties, statics, flow measurement and the energy and momentum equations in general. Water Resources applies them to civil systems: open channels, pipe networks, pumps, weirs and stormwater. Bernoulli's equation appears in both.
Which flow meters should I know?
Pitot tubes, venturi meters, orifice meters and flow nozzles. All four rest on the energy equation between two points with a coefficient for losses; what changes between them is the coefficient and the area ratio.
Do I need to read the Moody diagram?
Yes. Darcy-Weisbach head loss needs a friction factor, and for turbulent flow that comes from the Moody diagram, or the Colebrook equation, using the Reynolds number and the relative roughness. Many questions give the friction factor outright, so check the stem before reaching for a chart.
What decides whether flow is laminar or turbulent?
The Reynolds number, velocity times diameter divided by kinematic viscosity. In a pipe, below about 2,000 is laminar and the friction factor is 64 divided by the Reynolds number; well above that it is turbulent and the factor depends on roughness as well.
How many study hours this area is worth · Using the handbook under the clock