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Fluid Mechanics covers fluid properties, hydrostatics, control-volume analysis, Bernoulli and energy equations, internal flow, and dimensional analysis. Exams mix conceptual traps (gauge vs absolute pressure, laminar vs turbulent) with multi-step pipe and momentum problems. Full worked practice problems are the fastest way to learn which equation applies — and which assumptions you're allowed to make.

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Topics typically tested

  • Fluid properties & viscosity
  • Hydrostatic pressure & forces on surfaces
  • Bernoulli & energy equations
  • Control-volume momentum analysis
  • Laminar & turbulent pipe flow (Moody chart)
  • Dimensional analysis & similitude

Sample Fluid Mechanics exam questions

Three original questions in the style of a real Fluid Mechanics exam, each with a worked solution. Your own practice exam can be built from your course's notes and syllabus.

Question 1Multiple choiceHydrostatic pressure & forces on surfaces

A vertical rectangular gate in the wall of a water reservoir is 2.0 m wide and 3.0 m tall. Its top edge is horizontal and lies 1.0 m below the free surface of the water, so its bottom edge is 4.0 m below the surface. The water is at rest and has a density of 1000 kg/m³; the air above the free surface and on the dry side of the gate is at atmospheric pressure. Take g=9.81 m/s2g = 9.81\,\text{m/s}^2. At what depth below the free surface does the resultant hydrostatic force on the gate act?

  1. A.

    2.80 m

  2. B.

    2.50 m

  3. C.

    2.67 m

  4. D.

    3.00 m

Worked solution
Answer:A

For a plane surface the resultant force equals the pressure at the centroid times the area, but it acts below the centroid because the pressure grows with depth.

Centroid depth and area: the gate spans depths 1.0 m to 4.0 m, so hc=2.5h_c = 2.5 m and A=2.0(3.0)=6.0 m2A = 2.0(3.0) = 6.0\,\text{m}^2.

Resultant force (gauge, since atmospheric pressure acts on both sides and cancels):

FR=ρghcA=1000(9.81)(2.5)(6.0)=147.2 kNF_R = \rho g h_c A = 1000(9.81)(2.5)(6.0) = 147.2\,\text{kN}

Centre of pressure. Because the gate is vertical, the distance yy measured along the plate from the free surface equals the depth. The second moment of area of the rectangle about its own horizontal centroidal axis is

Ixc=bh312=2.0(3.0)312=4.5 m4I_{xc} = \frac{bh^3}{12} = \frac{2.0(3.0)^3}{12} = 4.5\,\text{m}^4 ycp=yc+IxcycA=2.5+4.52.5(6.0)=2.5+0.30=2.80 my_{cp} = y_c + \frac{I_{xc}}{y_c A} = 2.5 + \frac{4.5}{2.5(6.0)} = 2.5 + 0.30 = 2.80\,\text{m}

The pressure prism gives the same answer: the gauge pressure varies linearly from 9.819.81 kPa at the top edge to 39.2439.24 kPa at the bottom edge, and the centroid of that trapezoid lies 1.801.80 m below the top edge of the gate, which is 2.80 m below the surface.

The tempting answer 2.50 m is the centroid, where the resultant would act only if the pressure were uniform. 2.67 m is two-thirds of the bottom depth, which is correct only for a plate that starts at the free surface; 3.00 m comes from measuring ycy_c from the top of the gate instead of from the free surface.

The resultant acts 2.80 m below the free surface (0.30 m below the centroid of the gate).

Question 2Short answerBernoulli & energy equations

Water (density 1000 kg/m³) flows steadily through a horizontal pipe that contracts smoothly from a diameter of 100 mm to a diameter of 50 mm. The volume flow rate is 0.010 m³/s and the gauge pressure in the 100 mm section is 200 kPa. Neglect viscous losses. What is the gauge pressure in the 50 mm section?

Worked solution
Answer:

188 kPa (gauge)

Continuity gives the velocity in each section. The areas are

A1=π(0.100)24=7.854×10−3 m2,A2=π(0.050)24=1.963×10−3 m2A_1 = \frac{\pi(0.100)^2}{4} = 7.854 \times 10^{-3}\,\text{m}^2,\qquad A_2 = \frac{\pi(0.050)^2}{4} = 1.963 \times 10^{-3}\,\text{m}^2 V1=QA1=0.0107.854×10−3=1.273 m/s,V2=QA2=0.0101.963×10−3=5.093 m/sV_1 = \frac{Q}{A_1} = \frac{0.010}{7.854 \times 10^{-3}} = 1.273\,\text{m/s},\qquad V_2 = \frac{Q}{A_2} = \frac{0.010}{1.963 \times 10^{-3}} = 5.093\,\text{m/s}

Halving the diameter quarters the area, so the velocity quadruples. Apply Bernoulli along a streamline between the two sections; the pipe is horizontal, so the elevation terms cancel:

P1+12ρV12=P2+12ρV22P_1 + \tfrac{1}{2}\rho V_1^2 = P_2 + \tfrac{1}{2}\rho V_2^2 P2=P1−12ρ(V22−V12)=200 000−12(1000)(5.0932−1.2732)P_2 = P_1 - \tfrac{1}{2}\rho\left(V_2^2 - V_1^2\right) = 200\,000 - \tfrac{1}{2}(1000)(5.093^2 - 1.273^2) P2=200 000−500(25.94−1.62)=200 000−12 160=187 840 PaP_2 = 200\,000 - 500(25.94 - 1.62) = 200\,000 - 12\,160 = 187\,840\,\text{Pa}

The pressure drops by about 12 kPa because the fluid must accelerate into the narrower section, and that acceleration is driven by a pressure gradient. Both pressures are gauge: the atmospheric term would appear on both sides of Bernoulli and cancel, so the answer is gauge as well.

P2≈188 kPaP_2 \approx 188\,\text{kPa} (gauge)

Question 3Worked problemControl-volume momentum analysis

A horizontal jet of water (density 1000 kg/m³) with a diameter of 40 mm and a speed of 25 m/s strikes a fixed, smooth, curved vane. The vane turns the jet through 150∘150^\circ in the horizontal plane, so the water leaves at 25 m/s in a direction that makes 30∘30^\circ with the reversed incoming direction. The jet is open to the atmosphere throughout and friction is negligible, so the speed is unchanged. Take the incoming jet direction as the positive xx axis and let the outgoing jet have a positive yy component. Find the magnitude and direction of the force that the water exerts on the vane.

Worked solution
Answer:

1.52 kN, directed 15.0° below the +x axis (on the opposite side of the x axis from the outgoing jet)

Mass flow rate of the jet:

m˙=ρAV=1000[π(0.040)24](25)=1000(1.257×10−3)(25)=31.42 kg/s\dot m = \rho A V = 1000\left[\frac{\pi(0.040)^2}{4}\right](25) = 1000(1.257 \times 10^{-3})(25) = 31.42\,\text{kg/s}

Velocity components. Inlet: V1x=25V_{1x} = 25 m/s, V1y=0V_{1y} = 0. Outlet, turned 150∘150^\circ from the xx axis:

V2x=25cos⁡150∘=−21.65 m/s,V2y=25sin⁡150∘=12.50 m/sV_{2x} = 25\cos 150^\circ = -21.65\,\text{m/s},\qquad V_{2y} = 25\sin 150^\circ = 12.50\,\text{m/s}

Take a control volume around the water in contact with the vane. The jet is at atmospheric pressure everywhere, so pressure forces cancel and the only force on the water is the vane force. The steady linear momentum equation, ∑F=m˙(V2−V1)\sum \mathbf{F} = \dot m(\mathbf{V}_2 - \mathbf{V}_1), gives the force on the water:

Fx=31.42(−21.65−25)=−1466 N,Fy=31.42(12.50−0)=393 NF_x = 31.42(-21.65 - 25) = -1466\,\text{N},\qquad F_y = 31.42(12.50 - 0) = 393\,\text{N}

By Newton's third law the force of the water on the vane is equal and opposite: Rx=+1466R_x = +1466 N, Ry=−393R_y = -393 N.

∣R∣=14662+3932=1517 N,θ=tan⁡−13931466=15.0∘|\mathbf{R}| = \sqrt{1466^2 + 393^2} = 1517\,\text{N},\qquad \theta = \tan^{-1}\frac{393}{1466} = 15.0^\circ

The force on the vane acts 15.0° below the positive xx axis: it pushes the vane along the original jet direction and away from the side on which the water leaves. For a turn of angle β\beta at constant speed the change in velocity has magnitude 2Vsin⁡(β/2)=2(25)sin⁡75∘=48.32V\sin(\beta/2) = 2(25)\sin 75^\circ = 48.3 m/s, and 31.42(48.3)=151731.42(48.3) = 1517 N, in agreement.

R=1.52 kNR = 1.52\,\text{kN}, directed 15.0∘15.0^\circ below the incoming-jet (+x+x) direction

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