Fluid Statics Questions and Answers

Answer: A

Cut the jet along a diameter and consider a length L. The pressure force on the half jet is \(p \times d \times L\), acting over the projected area. It is balanced by the surface tension acting on the two edges of the cut, \(2\sigma L\):

The factors pd/4, pd/6 and pd/8 are not given by this balance. For a spherical droplet the relation is \(p = 4\sigma/d\), so \(\sigma = pd/4\) belongs to a droplet, not a jet, and pd/8 is the relation for a soap bubble with its two surfaces.

Answer: B

The key gives the shape of the dam face towards the upstream side, because the hydrostatic pressure acts on the face in contact with the water, and the total force on a unit length depends on the water depth and on how that face is shaped (its slope or curvature fixes both the magnitude and the direction of the resultant). The density of the dam material, the shape of the downstream face and the overall length of the dam do not alter the pressure of the water on the upstream face. I have followed the official key for this question.

Answer: C

Mechanical gauges, such as the Bourdon tube gauge, the diaphragm gauge and the dead-weight gauge, measure pressure by balancing the pressure of the fluid against a spring or a dead weight. Simple manometers, differential manometers and piezometers are all manometric devices that balance the pressure against a column of liquid, and they have no spring or weight. The piezometer is the simplest of these and measures the pressure in a pipe by the height of the liquid in a tube.

Answer: A

Both limbs hold water above the mercury, so the pressure difference between the two points equals the unbalanced column of mercury less the water displaced by it. The mercury surface in the right limb is 0.4 m below that in the left. In terms of metres of water:

The value 12.6 m is the factor without the 0.4 m, and 5.84 m would come from adding the specific gravity of water instead of subtracting it (0.4 x 14.6). The figure of 14.6 is the sum of the specific gravities, not an answer.

Answer: A

The gage reads about 61.26 kPa.

The actuator acts along its own axis, inclined at \(60^{\circ}\), so it carries the component of the weight along that axis:

\[ \begin{aligned} F &= mg\sin 60^{\circ} = 5000 \times 0.866 = 4330\ \text{N} \\ A &= \pi(0.3)^{2}/4 = 0.07069\ \text{m}^{2} \\ p &= 4330/0.07069 = 61.26\ \text{kPa} \end{aligned} \]

70.71 kPa is the full weight over the area, ignoring the inclination, and 35.35 kPa uses \(\cos 60^{\circ}\) instead of \(\sin 60^{\circ}\). 6.13 is the correct figure divided by 10.

Answer: A

The gauge pressure at A is about \(118.4 \times 10^{3}\) Pa.

Start at the oil-mercury interface at the bottom of the U. The open limb carries mercury a height \(h_{2}\) above it, open to the atmosphere; the oil side rises \(h_{1}\) from the interface to the level of A. Equating pressures at the interface level:

\[ \begin{aligned} p_{A} + \rho_{oil} g h_{1} &= \rho_{Hg} g h_{2} \\ p_{A} &= 13600(10)(0.9) - 800(10)(0.5) \\ &= 122400 - 4000 = 118400\ \text{Pa} \end{aligned} \]

That is \(118.4 \times 10^{3}\) Pa, about 118 kPa gauge. The options 118.4, 11.84 and 1.184 are the same digits with the wrong power of ten.

Answer: B

The vertical force is \(\frac{\pi D^{2}}{8}\rho g\) per unit length, acting upward.

The liquid touches the left half of the cylinder, from the floor contact at the bottom up to the top, where the free surface lies. The vertical force on a curved surface equals the weight of liquid, real or imaginary, between the surface and the free-surface level.

• On the upper-left quarter the liquid lies above the surface and pushes down.
• On the lower-left quarter the liquid pushes up, with the imaginary liquid column reaching up to the free surface.

The two columns overlap except for the half-circle itself, so the net force is the weight of liquid that would fill the left half of the cylinder:

\[ \begin{aligned} F_{V} &= \rho g \cdot \frac{1}{2}\cdot\frac{\pi D^{2}}{4} \\ &= \frac{\pi D^{2}}{8}\rho g \end{aligned} \]

\(\pi D^{2}/4\) is the buoyancy of the whole cylinder, which would need liquid all round it; the other two options do not correspond to any part of the wetted surface.

Answer: B

The pressure on the bottom is about 2.38 bar.

Add the gas pressure and the weight of each liquid column above the bottom. The water below the oil layer is 1 + 1 + 1 = 3 m deep.

\[ \begin{aligned} p &= p_{gas} + \rho_{oil}g h_{oil} + \rho_{w}g h_{w} \\ &= 2 \times 10^{5} + 800(10)(1) + 1000(10)(3) \\ &= 200000 + 8000 + 30000 = 238000\ \text{Pa} \end{aligned} \]

That is 2.38 bar (1 bar = 10⁵ Pa). The options 0.238, 23.8 and 238 are the same digits with the bar-to-pascal conversion slipped by a factor of ten or more.

Answer: C

The force on the gate is about \(2.23 \times 10^{5}\) N.

For a plane vertical gate the resultant force is the pressure at its centroid times its area. The centroid is 1.5 m below the oil-water interface (1 m to the top of the gate plus half its 1 m height).

\[ \begin{aligned} p_{c} &= 2 \times 10^{5} + 800(10)(1) + 1000(10)(1.5) \\ &= 223000\ \text{Pa} \\ F &= p_{c}A = 223000 \times (1 \times 1) = 2.23 \times 10^{5}\ \text{N} \end{aligned} \]

The gas pressure dominates, which is why the answer is so large. The other options differ only by powers of ten, the result of dropping the gas pressure's 10⁵ or mis-converting bar.

Answer: A

The actual pressure is called absolute pressure.

Absolute pressure is measured from a perfect vacuum (zero pressure), so it is the true pressure at the point. Gauge pressure is measured relative to the local atmosphere: absolute = gauge + atmospheric.

Atmospheric pressure is just the pressure of the surrounding air, and vacuum pressure is the amount by which a pressure falls short of atmospheric. Neither is the actual pressure at an arbitrary position.