Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-Agric-A4, Fluid Flow — National Exams, December 2019. Open-book, 3-hour exam; four questions of equal value, all requiring calculation.
Reference texts. White, Fluid Mechanics (pipe friction/laminar duct flow, cavitation, open-channel flow, pumps & the energy equation) — the standard undergraduate text for this subject.
Problem 2: Cavitation of a Submersible (25 points)
Check: this exact submersible-cavitation dataset (2-m depth, 131 kPa ambient, $Ca_{\text{crit}}=0.25$, 30 m/s cold-water check) also appears, unchanged. The steps below reproduce that verified solution.
Given. A submersible body's critical cavitation number is $Ca_{\text{crit}}=0.25$ — the value at which the local low-pressure region on the body first drops to the vapor pressure.
Quantity
Value
Depth
2 m
Ambient pressure at 2 m, 20°C water
131 kPa
Vapor pressure, 20°C
2.337 kPa
Critical cavitation number, $Ca_{\text{crit}}$
0.25
Cold-water case: T, $p_v$, $\rho$
5°C, 863 Pa, 1000 kg/m³
Find. (a) The velocity $V$ at which cavitation bubbles first form in the 20°C water above; (b) whether the body cavitates at $V=30$ m/s in cold (5°C) water at the same 2-m depth.
Approach. The cavitation number $Ca=(p-p_v)/(\tfrac12\rho V^2)$ falls as speed rises; cavitation begins exactly when $Ca$ reaches $Ca_{\text{crit}}$. Solve that condition for $V$ in part (a). For part (b), recover the (depth-independent) atmospheric pressure implied by the 20°C data, use it to get the true ambient pressure at 2 m in the colder, slightly denser water, then compare $Ca$ at $V=30$ m/s against $Ca_{\text{crit}}$.
(a) Onset velocity in 20°C water. Setting $Ca=Ca_{\text{crit}}$ and solving for $V$,
$$V = \sqrt{\frac{2(p-p_v)}{Ca_{\text{crit}}\,\rho}} = \sqrt{\frac{2(131{,}000-2337)}{0.25(998)}} = \boxed{V \approx 32.1\ \text{m/s}}.$$
Recover the ambient (surface) pressure. The 131 kPa ambient at 2 m is hydrostatic, $p=p_{\text{atm}}+\rho g h$, so
$$p_{\text{atm}} = 131{,}000-(998)(9.81)(2) = \boxed{p_{\text{atm}}\approx111.4\ \text{kPa}}.$$
(b) Ambient pressure at 2 m in cold water. Re-applying hydrostatics with the colder, slightly denser water,
$$p_{2\text{m}} = 111{,}400+(1000)(9.81)(2) = \boxed{p_{2\text{m}}\approx131.0\ \text{kPa}}$$
(essentially unchanged, since the density difference is only 0.2%).
Actual cavitation number at $V=30$ m/s.
$$Ca_{\text{actual}} = \frac{p_{2\text{m}}-p_v}{\tfrac12\rho V^2} = \frac{131{,}000-863}{\tfrac12(1000)(30^2)} = \boxed{Ca_{\text{actual}}\approx0.289}.$$
Since $0.289>Ca_{\text{crit}}=0.25$, the pressure margin has not yet been driven down to vapor pressure, so the body does not cavitate at 30 m/s in cold water — it would need to reach roughly the same ~32 m/s onset speed found in part (a).
