Fluid Mechanics practice questions

Answer: A) 0.133 m

Hydraulic diameter is D_h = 4A/P. For a 0.1 m × 0.2 m duct flowing full, D_h = 0.08/0.6 ≈ 0.133 m, which is four times the hydraulic radius.

• ACorrect. D_h = 4A/P = 4(0.1 × 0.2)/[2(0.1 + 0.2)] = 0.08/0.6 ≈ 0.133 m.
• BIncorrect. This is the hydraulic radius, R_h = A/P = 0.02/0.6. The hydraulic diameter is four times that.
• CIncorrect. This is the average of the two sides. Hydraulic diameter depends on the ratio of area to wetted perimeter, not on the average dimension.
• DIncorrect. This is the diameter of a circle with the same area, √(4A/π). That ignores the wetted perimeter, which controls friction behavior.

Answer: B) 2.55 m

Darcy-Weisbach: hf = f(L/D)(v²/2g) = 0.025 × 500 × 0.204 m ≈ 2.55 m.

• AIncorrect. This value comes from dividing v² by g rather than by 2g in the velocity head term.
• BCorrect. hf = f(L/D)(v²/2g) = 0.025 × (50/0.1) × (2²/(2 × 9.81)) ≈ 2.55 m.
• CIncorrect. This is low by a factor of 10, as happens if L/D is taken as 50 instead of 500.
• DIncorrect. This is f(L/D) alone; it omits multiplication by the velocity head v²/2g.

Answer: A) Decreases

• ACorrect. Bernoulli's equation trades pressure for velocity.
• BIncorrect. Higher velocity means lower pressure.
• CIncorrect. Bernoulli's equation links them.
• DIncorrect. It decreases, not to zero.

Answer: B) Flow is the same in each pipe and head losses add

Pipes in series carry the same flow rate, and their head losses add. Pipes in parallel have equal head losses, and their flows add.

• AIncorrect. This describes pipes in parallel: each branch has the same head loss, and the branch flows add up to the total flow.
• BCorrect. In series, the same flow rate passes through every pipe (continuity), and the total head loss is the sum of the losses in each pipe.
• CIncorrect. The flow rate is constant, but velocity V = Q/A changes with each pipe's diameter.
• DIncorrect. In series, the individual head losses add together.

Answer: D) 4.9 kW

Hydraulic power is P = γQH = ρgQH = 9,810 N/m³ × 0.05 m³/s × 10 m ≈ 4.9 kW.

• AIncorrect. This uses ρQH without multiplying by g, which gives a mass-based quantity rather than power.
• BIncorrect. This incorrectly halves the result, as if a velocity-head factor of 1/2 applied; hydraulic power is simply ρgQH.
• CIncorrect. This is ten times too large, typically from a decimal error in the flow rate.
• DCorrect. P = ρgQH = 1000 × 9.81 × 0.05 × 10 ≈ 4,905 W ≈ 4.9 kW.

Answer: B) 13.3 kPa

• AIncorrect. Use h = 0.1 m.
• BCorrect. Δp = ρgh = 13,600(9.81)(0.1) Pa.
• CIncorrect. Include g = 9.81.
• DIncorrect. That's for water, not mercury.

Answer: C) 1,000 N

• AIncorrect. Force scales with area.
• BIncorrect. The larger piston gives more force.
• CCorrect. Equal pressure: F₂ = F₁(A₂/A₁).
• DIncorrect. The area ratio is 10.

Answer: C) Elevation head, pressure head and velocity head

• AIncorrect. That isn't total head.
• BIncorrect. Elevation and velocity also count.
• CCorrect. Losses reduce it along the flow.
• DIncorrect. Those aren't heads.

Answer: D) 0.82 m

Velocity head is the kinetic-energy term of the energy equation written as a height: v²/(2g) = 4²/(2 × 9.81) ≈ 0.82 m.

• AIncorrect. 8.0 is v²/2 = 16/2. This leaves out g, so the result isn't even a length in meters.
• BIncorrect. 0.41 is v/g = 4/9.81. This fails to square the velocity and also drops the factor of 2.
• CIncorrect. 1.63 is v²/g = 16/9.81. This leaves out the factor of 2 in the denominator.
• DCorrect. Velocity head = v²/(2g) = 16/(2 × 9.81) = 16/19.62 ≈ 0.82 m.

Answer: D) Doubles

• AIncorrect. Take the square root.
• BIncorrect. Take the square root.
• CIncorrect. A steeper slope speeds the flow.
• DCorrect. Velocity scales with √S, and √4 = 2.