A certain torpedo, moving in fresh water at 10°C, has a minimum-pressure point given by the formula

$p_{min} = p_0 - 0.35 ρV^2$ (1)

where $p_0$ = 115 kPa, ρ is the water density, and V is the torpedo velocity. Estimate the velocity at which cavitation bubbles will form on the torpedo. The constant 0.35 is dimensionless.

Question

A certain torpedo, moving in fresh water at 10°C, has a minimum-pressure point given by the formula

[latex]p_{min} = p_0 - 0.35 ρV^2[/latex] (1)

where [latex]p_0[/latex] = 115 kPa, ρ is the water density, and V is the torpedo velocity. Estimate the velocity at which cavitation bubbles will form on the torpedo. The constant 0.35 is dimensionless.

Accepted Answer

• Assumption: Cavitation bubbles form when the minimum pressure equals the vapor pressure [latex]p_{ u}[/latex].

• Approach: Solve Eq. (1) above, which is related to the Bernoulli equation from Example 1.3, for the velocity when [latex]p_{min} = p_{ u}[/latex]. Use SI units (m, N, kg, s).

• Property values: At 10°C, read Table A.1 for ρ = 1000 kg/[latex]m^3[/latex] and Table A.5 for [latex]p_{ u}[/latex] = 1.227 kPa.

Table A.1 Viscosity and Density of Water at 1 atm
T, °C	ρ, kg/[latex]m^3[/latex]	µ, N·s/[latex]m^2[/latex]	[latex] u, m^2[/latex]/s	T, °F	ρ, slug/[latex]ft^3[/latex]	µ, Ib·s/[latex]ft^2[/latex]	[latex] u, ft^2[/latex]/s
0	1000	1.788 E-3	1.788 E-6	32	1.940	3.73 E-5	1.925 E-5
10	1000	1.307 E-3	1.307 E-6	50	1.940	2.73 E-5	1.407 E-5
20	998	1.003 E-3	1.005 E-6	68	1.937	2.09 E-5	1.082 E-5
30	996	0.799 E-3	0.802 E-6	86	1.932	1.67 E-5	0.864 E-5
40	992	0.657 E-3	0.662 E-6	104	1.925	1.37 E-5	0.713 E-5
50	988	0.548 E-3	0.555 E-6	122	1.917	1.14 E-5	0.597 E-5
60	983	0.467 E-3	0.475 E-6	140	1.908	0.975 E-5	0.511 E-5
70	978	0.405 E-3	0.414 E-6	158	1.897	0.846 E-5	0.446 E-5
80	972	0.355 E-3	0.365 E-6	176	1.886	0.741 E-5	0.393 E-5
90	965	0.316 E-3	0.327 E-6	194	1.873	0.660 E-5	0.352 E-5
100	958	0.283 E-3	0.295 E-6	212	1.859	0.591 E-5	0.318 E-5

Table A.5 Surface Tension, Vapor Pressure, and Sound Speed of Water
T, °C	Y, N/m	[latex]p_{ u}[/latex], kPa	a, m/s
0	0.0756	0.611	1402
10	0.0742	1.227	1447
20	0.0728	2.337	1482
30	0.0712	4.242	1509
40	0.0696	7.375	1529
50	0.0679	12.34	1542
60	0.0662	19.92	1551
70	0.0644	31.16	1553
80	0.0626	47.35	1554
90	0.0608	70.11	1550
100	0.0589	101.3	1543
120	0.055	198.5	1518
140	0.0509	361.3	1483
160	0.0466	617.8	1440
180	0.0422	1002	1389
200	0.0377	1554	1334
220	0.0331	2318	1268
240	0.0284	3344	1192
260	0.0237	4688	1110
280	0.019	6412	1022
300	0.0144	8581	920
320	0.0099	11,274	800
340	0.0056	14,586	630
360	0.0019	18,651	370
374[latex]^*[/latex]	0.0[latex]^*[/latex]	22,090[latex]^*[/latex]	0[latex]^*[/latex]

[latex]^*[/latex]Critical point.

• Solution steps: Insert the known data into Eq. (1) and solve for the velocity, using SI units:

[latex]p_{min} = p_{ u} = 1227 Pa = 115,000 Pa - 0.35 \left(1000\frac{kg}{m^3} ight)V^2[/latex], with V in m/s

Solve [latex]V^2=\frac{(115,000-1227)}{0.35(1000)}=325\frac{m^2}{s^2} or V=\sqrt{325} \approx 18.0 m/s[/latex] Ans.

• Comments: Note that the use of SI units requires no conversion factors, as discussed in Example 1.3b. Pressures must be entered in pascals, not kilopascals.

Question 1.9: A certain torpedo, moving in fresh water at 10°C, has a mini...

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