Fluid Dynamics and Bernoullis Equation Questions and Answers

Answer: C

The pump imparts 28 kW of kinetic energy.

In the boat's frame, water enters at 5 m/s and is discharged at 9 m/s, so the pump raises its kinetic energy per second by

\[ \begin{aligned} \dot{E} &= \tfrac{1}{2}\rho Q(V_{j}^{2} - V^{2}) \\ &= \tfrac{1}{2}(1000)(1)(81 - 25) = 28000\ \text{W} \end{aligned} \]

that is 28 kW. Of this, 20 kW goes into propelling the boat (the previous question) and the remaining 8 kW is left in the wake, which gives a propulsive efficiency of 20/28, about 71 percent. The option 81 uses only the jet speed, and 25 only the inlet speed.

Answer: D

Bernoulli's equation is derived from the conservation of energy. It is the energy equation for a flowing fluid along a streamline, in which pressure energy, kinetic energy and potential energy add up to a constant. The conservation of mass leads to the continuity equation and the conservation of momentum to the momentum equation, so the answer is energy.

Answer: A

Bernoulli's equation applies to incompressible and non-viscous flows, and in its usual form along a streamline in steady flow. It is derived for an ideal fluid that has constant density and has no friction, so energy is conserved with no loss to viscosity. In viscous or compressible flow additional terms are required (head loss, change of density), so those options are wrong.

The equation states that the sum of pressure head, velocity head and elevation head stays constant along a streamline.

Answer: D

The momentum flux remains constant before and after a hydraulic jump, as the key says. In the analysis of a hydraulic jump the momentum equation is applied to a control volume, and the sum of the pressure force and momentum flux is the same on both sides, while energy is lost in the jump. The depth increases and the velocity decreases across the jump, but this does not alter the momentum flux. The options of a decrease, increase or zero momentum flux are wrong.

Answer: B

The jet strikes the floor 1.0 m from the tank.

The hole is 0.5 m below the water surface, so by Torricelli's theorem the jet leaves horizontally at \(v = \sqrt{2g \times 0.5} = \sqrt{g}\). It then falls the 0.5 m from the hole to the floor:

\[ \begin{aligned} t &= \sqrt{\frac{2 \times 0.5}{g}} = \frac{1}{\sqrt{g}} \\ X &= vt = \sqrt{g} \times \frac{1}{\sqrt{g}} = 1.0\ \text{m} \end{aligned} \]

In general X = \(2\sqrt{h_{1}h_{2}}\), with \(h_{1}\) the depth of the hole below the surface and \(h_{2}\) its height above the floor; here \(2\sqrt{0.5 \times 0.5} = 1\). 0.5 m halves this, and 2.0 or 4.0 m would need a much greater head or drop.

Answer: A

The pressure change is -1.28 kPa.

Continuity gives the velocities from Q = 1.5 m³/s:

\[ \begin{aligned} V_{1} &= \frac{1.5}{\pi(0.2)^{2}} = 11.94\ \text{m/s} \\ V_{2} &= \frac{1.5}{\pi(0.1)^{2}} = 47.75\ \text{m/s} \end{aligned} \]

Bernoulli's equation for a horizontal pipe then gives

\[ \begin{aligned} p_{2} - p_{1} &= \frac{\rho}{2}\left(V_{1}^{2} - V_{2}^{2}\right) \\ &= 0.6\,(142.5 - 2280.0) \\ &= -1282\ \text{Pa} \approx -1.28\ \text{kPa} \end{aligned} \]

The pressure falls as the flow speeds up in the narrower section, so the change is negative; +1.28 has the sign reversed. 2.56 doubles the result by dropping the 1/2, and -2.13 does not follow from these data.

Answer: D

The flow rate is \(\tfrac{\pi}{4}D^{2}\sqrt{V^{2} - 2gh_{2}}\).

The flow is steady and inviscid, so Bernoulli's equation holds along the streamline that enters the tube mouth and leaves at the top. Work in heads, with gauge pressure and the free surface as datum. Far upstream at the depth of the mouth, the pressure head is \(h_{1}\), the elevation is \(-h_{1}\) and the speed is \(V\). At the outlet, the pressure head is zero (atmospheric), the elevation is \(h_{2}\) and the speed is \(v\):

\[ \begin{aligned} h_{1} - h_{1} + \frac{V^{2}}{2g} &= h_{2} + \frac{v^{2}}{2g} \\ v &= \sqrt{V^{2} - 2gh_{2}} \end{aligned} \]

The depth \(h_{1}\) cancels: the extra hydrostatic pressure at the mouth exactly offsets the extra height the liquid must climb. Multiplying by the area \(\pi D^{2}/4\) gives the answer.

\(\tfrac{\pi}{4}D^{2}V\) ignores the rise of \(h_{2}\) above the surface. \(\sqrt{2gh_{2}}\) and \(\sqrt{2g(h_{1} + h_{2})}\) treat the liquid as if it fell through a height, when here it is the stream's kinetic energy that lifts it.

Answer: C

The pressure gradient is \(-6\hat{i} - 6\hat{j}\).

For an inviscid fluid with no body force, Euler's equation gives \(\nabla p = -\rho\,\vec{a}\), where \(\vec{a}\) is the convective acceleration of the steady field. With \(u = y^{2} - x^{2}\) and \(v = 2xy\), at (1, 1) we have \(u = 0\), \(v = 2\), \(\partial u/\partial y = 2y = 2\) and \(\partial v/\partial y = 2x = 2\):

\[ \begin{aligned} a_{x} &= u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} = 0 + 2 \times 2 = 4 \\ a_{y} &= u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y} = 0 + 2 \times 2 = 4 \end{aligned} \]

\[ \begin{aligned} \nabla p &= -1.5\,(4\hat{i} + 4\hat{j}) \\ &= -6\hat{i} - 6\hat{j} \end{aligned} \]

\(-4\hat{i} - 4\hat{j}\) is the acceleration with its sign reversed but without the density factor. \(10\hat{j}\) and \(20\hat{i}\) have no basis in the field: at (1, 1) both components of the acceleration are equal, so the gradient must have equal components too.

Answer: A

The discharge velocity is (A):

\[ \begin{aligned} v &= \sqrt{2gh_3 X} \\ X &= 1+\frac{\rho_1}{\rho_3}\frac{h_1}{h_3} \\ &\quad +\frac{\rho_2}{\rho_3}\frac{h_2}{h_3} \end{aligned} \]

The static pressure at the nozzle axis is the weight of all three layers above it:

\[ p = \rho_{1}gh_{1}+\rho_{2}gh_{2}+\rho_{3}gh_{3} \]

The fluid leaving the nozzle is fluid 3, so applying Bernoulli's equation to it, from rest in the tank to the jet at atmospheric pressure:

\[ \begin{aligned} \tfrac{1}{2}\rho_{3}v^{2} &= p \\ v^{2} &= 2g\left(h_{3}+\frac{\rho_{1}h_{1}}{\rho_{3}}+\frac{\rho_{2}h_{2}}{\rho_{3}}\right) \end{aligned} \]

Taking \(h_{3}\) out of the bracket gives option (A). Option (B), \(\sqrt{2g(h_{1}+h_{2}+h_{3})}\), is right only if all three densities are equal. Options (C) and (D) average the densities in ways that do not come from the pressure balance. They reduce to the right result only in special cases, not for three layers of different density in general.

Answer: B

The exit velocities are equal: \(V_{2} = V_{1}\).

Neglecting losses, Torricelli's result from Bernoulli's equation gives the jet speed as \(V = \sqrt{2gh}\). Density cancels from both sides of the energy balance, so the velocity depends only on the head \(h\), which is the same for both tanks.

Viscosity would make the oil jet slower in reality, but the question says to neglect all losses, so \(V_{2} \lt V_{1}\) is not the answer intended. The data are sufficient, and nothing makes the oil faster.