Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2016 — 04-Agric-A4, Fluid Flow. Three-hour, open-book exam; any non-communicating calculator is permitted. Format: four questions constitute a complete paper (the first four questions appearing in the answer book are marked), each of equal value; all questions require calculation.
Reference texts: White, Fluid Mechanics (7th/8th ed., McGraw-Hill) — pipe-friction networks (Darcy–Weisbach, Colebrook), cavitation number, rotating control volumes (sprinkler reaction), and pump/system energy balances.
Problem 2: Cavitation of a Submersible (equal value)
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 described 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), first 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 the resulting $Ca$ at $V=30$ m/s against $Ca_{\text{crit}}$.
(a) Onset velocity in 20°C water. Setting $Ca=Ca_{\text{crit}}$ and solving for $V$,
$$Ca_{\text{crit}} = \frac{p-p_v}{\tfrac12\rho V^2} \;\Rightarrow\; V = \sqrt{\frac{2(p-p_v)}{Ca_{\text{crit}}\,\rho}} = \sqrt{\frac{2(131{,}000-2337)}{0.25(998)}} = \boxed{V \approx 32.1\ \text{m/s}}.$$
Cavitation bubbles begin to form once the submersible exceeds about 32.1 m/s in this 20°C, 2-m-deep water.
Recover the ambient (surface) pressure. The 131 kPa ambient at 2 m is hydrostatic: $p = p_{\text{atm}}+\rho g h$, so
$$p_{\text{atm}} = 131{,}000 - (998)(9.81)(2) = \boxed{p_{\text{atm}} \approx 111.4\ \text{kPa}}.$$
(b) Ambient pressure at 2 m in cold water. Re-applying hydrostatics with the colder, slightly denser water ($\rho=1000\ \text{kg/m}^3$) at the same depth,
$$p_{2\text{m}} = p_{\text{atm}}+\rho g h = 111{,}400+(1000)(9.81)(2) = \boxed{p_{2\text{m}} \approx 131.0\ \text{kPa}}$$
(essentially unchanged from the 20°C value, since the density difference is only 0.2%).
Actual cavitation number at $V=30$ m/s.
$$Ca_{\text{actual}} = \frac{p_{2\text{m}}-p_v}{\tfrac12\rho V^2} = \frac{131{,}000-863}{\tfrac12(1000)(30^2)} = \boxed{Ca_{\text{actual}} \approx 0.289}.$$
Since $Ca_{\text{actual}}=0.289 > Ca_{\text{crit}}=0.25$, the local pressure margin has not yet been driven down to the vapor pressure, so the body does not cavitate at 30 m/s in the cold water — it would need to reach roughly the same ~32 m/s onset speed found in part (a) (the two ambient pressures are nearly identical).
