Fluid Dynamics and Bernoullis Equation Questions and Answers

Answer: A

The correct option is \(P_{o} \gt P_{i}\).

Fluid going round a bend needs a net force towards the centre of curvature to supply its centripetal acceleration. That force comes from a pressure gradient across the section, with pressure rising outward: \(\partial p/\partial r = \rho V^{2}/r\). So the outer wall sees the higher pressure.

This makes \(P_{o} \lt P_{i}\) wrong. The difference grows roughly as \(\rho V^{2}/R\) times the tube width, so it is inversely, not directly, proportional to the bend radius, and it increases with velocity rather than being inversely proportional to it.

Answer: A

Bernoulli's equation cannot be applied between any two arbitrary points in the flow field for steady incompressible rotational flow.

For steady, incompressible, inviscid flow, the Bernoulli constant is the same along a streamline. Only if the flow is also irrotational is it the same throughout the field, so that any two points can be related.

The other three are valid uses: arbitrary points in irrotational flow, and two points on the same streamline, whether the flow is rotational or not.

Answer: C

The outlet velocity is about 5.3 m/s.

Mass conservation: inflow equals outflow plus the rate of storage in the tank.

\[ \begin{aligned} Q_{out} &= 0.02 - (1)(0.005) = 0.015\ \text{m}^{3}\text{/s} \\ A_{pipe} &= \tfrac{\pi}{4}(0.06)^{2} = 2.827 \times 10^{-3}\ \text{m}^{2} \\ V &= \frac{0.015}{2.827 \times 10^{-3}} \approx 5.3\ \text{m/s} \end{aligned} \]

Ignoring the rising level and using the full 0.02 m³/s would give about 7.1 m/s, which is not offered. 0.005 is the rate of rise of the level itself, and 0.06 is the pipe diameter in metres, not a velocity.

Answer: D

The maximum velocity is 12 m/s.

Apply Bernoulli's equation from the free surface in the large tank (velocity negligible, pressure 130 kPa, 4.2 m above the pipe centerline) to a point in the horizontal pipe where the pressure is just atmospheric:

\[ \begin{aligned} p_{gas} + \rho g h &= p_{atm} + \tfrac{1}{2}\rho V^{2} \\ 130000 + 1000(10)(4.2) &= 100000 + 500V^{2} \\ 500V^{2} &= 72000 \Rightarrow V = 12\ \text{m/s} \end{aligned} \]

Any faster and the pressure would fall below atmospheric. 10 m/s ignores the 30 kPa of gas overpressure (\(\sqrt{2gh} \approx 9.2\)), and 4.2 is the head in metres, not a velocity.

Answer: A

The extensive property is the angular momentum \(\vec{r} \times m\vec{V}\).

The Reynolds Transport Theorem converts the rate of change of any extensive property B of a system into control-volume terms. Choosing \(B = \vec{r} \times m\vec{V}\), the moment of momentum, with its intensive form \(b = \vec{r} \times \vec{V}\), gives the angular momentum equation, whose system-level statement is that the net torque equals the rate of change of angular momentum.

\(\vec{r} \times \vec{V}\) is the intensive (per unit mass) form, not the extensive property. \(m\vec{V}\) gives the linear momentum equation and \(m\) gives continuity.

Answer: A

The blanks are system, control volume, exiting.

The Reynolds transport theorem links a system, a fixed collection of matter, to a control volume, a fixed region of space:

\[ \begin{aligned} \frac{dB_{sys}}{dt} &= \frac{d}{dt}\int_{CV} \rho b\,dV \\ &+ \int_{CS} \rho b\,(\vec{V} \cdot \vec{n})\,dA \end{aligned} \]

The left side is the rate of change of the system's extensive property; the first term on the right is the rate of change inside the control volume; the last term is the net rate at which the property leaves (exits) through the control surface, since \(\vec{n}\) points outward.

Swapping system and control volume reverses the theorem's meaning, and 'entering' would give the wrong sign for the flux term.

Answer: A

The connectors are converging, constant area, diverging.

The vertical gap between the TEL and the HGL is the velocity head \(V^{2}/2g\). The TEL falls only slightly (friction), so changes in that gap show how the velocity, and hence the area, changes.

• A to B: the HGL drops away from the TEL, so the velocity head grows and the velocity rises: the pipe converges.
• B to C: the HGL runs nearly parallel to the TEL, so the velocity head is constant: constant area.
• C to D: the HGL climbs back toward the TEL, so the velocity falls: the pipe diverges.

The other sequences reverse at least one of these trends; for example a diverging first connector would make the HGL rise from A to B.

Answer: C

The discharge velocities are equal: \(V_{I} = V_{II} = V_{III}\).

Apply Bernoulli's equation along a streamline from the free surface of the tank (at rest, atmospheric pressure) to the outlet jet (atmospheric pressure). With no friction, the shape and highest point of the tube do not appear:

\[ \begin{aligned} z_{s} &= z_{out}+\frac{V^{2}}{2g} \\ V &= \sqrt{2gh} \end{aligned} \]

In all three siphons the outlet is the same depth h below the free surface, so all three velocities are \(\sqrt{2gh}\). The heights a and b only govern the pressure inside the tube at its crest, which must stay above vapour pressure for the siphon to run, not the outlet speed. The orderings in the other options assume those heights matter.

Answer: C

The power input is \(P = Q \cdot H_m \cdot Y\).

Each second a volume Q of fluid passes, weighing \(YQ\) newtons, and each newton of it gains \(H_m\) metres of head, that is \(H_m\) joules of energy. So the energy added per second is

\[ \begin{aligned} P &= Y Q H_m \\ &\to \text{N/m}^3 \times \text{m}^3/\text{s} \times \text{m} \\ &= \text{N m/s} = \text{W} \end{aligned} \]

The dimensions confirm it: the product gives N·m/s, which is watts. Every option that divides one of the three quantities by another gives the wrong units, so only the plain product can be power.

Answer: B

The Board's key is a-II, b-I, c-III, d-IV.

d matches IV at once: integrating the equation of motion along a streamline for steady ideal flow gives Bernoulli's equation. Every option pairs a with II, since Euler's equation is the momentum equation for an ideal fluid, so the choice is between B and C.

The key takes b with I: the pressure of a gas on a wall is explained by the change of momentum of molecules striking it, the principle of momentum. The rotating lawn sprinkler (c) is the standard application of the momentum equation in its moment-of-momentum form, matching III. Options pairing d with III are wrong because Bernoulli's equation belongs to the streamline.