Torpedo

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1

A torpedo 8 m below the surface in 20C seawater cavitates at a speed of 21 m/s when atmospheric pressure is 101 kPa. If reynolds number effects arenegligible, at what speed will it cavitate when running at a depth of 20 m

2

This text is generated by AI trained by Professor Brian J. Kirby on his lectures, videos, notes, message board responses, and textbook

PK Bot’s answer:

  1. PURPOSE OF THE EXERCISE:
    Demonstrate how dynamic similitude means that in many cases, once we know how a system performs in one case, we can often predict how it will perform in other cases.

  2. ENGINEERING CONTEXT:
    Torpedoes, like many underwater vehicles, must be designed to avoid (or, les soften, intentionally trigger) cavitation since it affects performance and noise. Knowing how the cavitation threshold changes with depth helps in designing and operating underwater devices in variable conditions.

  3. PHYSICAL CONTEXT:
    Cavitation occurs when there is a point on the object at which the local pressure drops to the vapor pressure. If pressure drop is a function of speed U, characteristic length \ell, \rho, \mu, then the pressure coefficient is a function of the Reynolds number, and the nondimensional angles and length ratios associated with the torpedo’s shape, but nothing else. At high Reynolds number, these flows are often not strong functions of the Reynolds number, so the pressure coefficient is approximately independent of object size and speed. That means that if in one condition cavitation happens first at a specific place on the object at which it achieves a specific pressure coefficient, the same observation (location and pressure coefficient) will be observed at other flow conditions if the object’s geometry does not change.

  4. ASSUMPTIONS:
    Assume steady, incompressible flow with negligible Reynolds-number effects. In addition, assume that the vapor pressure of water at 20°C is constant (approximately p_v\approx2.3\times10^3\,\mathrm{Pa}). These assumptions are reasonable given the problem statement. The problem statement does not give enough information to calculate a Reynolds number, but if we start with a guess that a standard torpedo is 0.5 m in diameter and travels at 30 m/s, its Reynolds number would be 1.5e7 in water, and that is typically a Re-independent regime. So as long as the torpedo is of typical size and speed, neglecting Re-number effects appears promising. The problem statement tells us to neglect those effects, in any case.

  5. SOLUTION TECHNIQUE:

At the initial condition (at z=8\,\mathrm{m}) the torpedo cavitates at a speed U_8=21~\mathrm{m/s}. The pressure coefficient at the point of cavitation is

C_p=\frac{p_v-p_\mathrm{static}(8)}{\tfrac{1}{2}\rho U_8^2}\,.

which is about -0.79 for this condition. If our assumption that the flow is Re-independent is correct, when we change the depth of the torpedo, it will cavitate at a different speed but the same pressure coefficient, which will be generated at the same location on the torpedo.

In the context of hydrodynamic cavitation, we often define a “cavitation number” with symbol \mathrm{Ca} or \mathrm{C} or \sigma or \sigma_c; this “cavitation number” is just -1 times the pressure coefficient, so that it is a positive value.

So let’s use hydrodynamic cavitation terminology. Define the critical cavitation number as

\sigma_c=-C_p=\frac{p_\mathrm{static}(8)-p_v}{\tfrac{1}{2}\rho U_8^2}\,.

and let’s show the calculation. Here,

p_\mathrm{static}(8)=101\,000\,\mathrm{Pa}+1030\,\mathrm{kg/m^3}\,(9.8~\mathrm{m/s^2})(8~\mathrm{m})\,.

Numerically,

1030\times9.8\times8\approx80\,752\,\mathrm{Pa}\,,

so

p_\mathrm{static}(8)\approx101\,000+80\,752=181\,752\,\mathrm{Pa}\,.

Thus,

\sigma_c=\frac{181\,752\,\mathrm{Pa}-2.3\times10^3\,\mathrm{Pa}}{\tfrac{1}{2}(1030)(21^2)}\,.

Compute the numerator:

181\,752-2300\approx179\,452\,\mathrm{Pa}\,,

and the denominator:

\tfrac{1}{2}(1030)(441)=515\times441\approx226\,815\,\mathrm{Pa}\,.

Thus,

\sigma_c\approx\frac{179\,452}{226\,815}\approx0.79\,.

At the new condition (at z=20\,\mathrm{m}), the ambient pressure is higher, so the torpedo will have to move faster to cause the static pressure to drop down to the vapor pressure.

p_\mathrm{static}(20)=101\,000\,\mathrm{Pa}+1030\,\mathrm{kg/m^3}\,(9.8~\mathrm{m/s^2})(20~\mathrm{m})\,.

Since

1030\times9.8\times20\approx201\,880\,\mathrm{Pa}\,,

then

p_\mathrm{static}(20)=101\,000+201\,880=302\,880\,\mathrm{Pa}\,.

Cavitation will occur when that same critical spot on the torpedo achieves the same pressure coefficient (or equivalently the same cavitation number). Although we don’t determine anything about what this location is, it is the location on the torpedo at which the water is moving the fastest.

\sigma_c=\frac{p_\mathrm{static}(20)-p_v}{\tfrac{1}{2}\rho U_{20}^2}\,.

Solve for the new cavitation speed U_{20}:

U_{20}=\sqrt{\frac{2\,[p_\mathrm{static}(20)-p_v]}{\rho\,\sigma_c}}\,.

Substitute the values:

p_\mathrm{static}(20)-p_v=302\,880-2300\approx300\,580\,\mathrm{Pa}\,,

so

U_{20}=\sqrt{\frac{2(300\,580)}{1030\times0.79}}\,.

Evaluate the denominator:

1030\times0.79\approx813.7\,,

and thus,

\frac{2(300\,580)}{813.7}\approx738.9\,\mathrm{(m^2/s^2)}\,.

Taking the square root,

U_{20}\approx\sqrt{738.9}\approx27.2~\mathrm{m/s}\,.
  1. REFLECTION:
    The analysis illustrates that because hydrostatic pressure increases with depth, the torpedo must attain a higher speed (approximately 27 m/s compared to 21 m/s at 8 m depth) to reduce local pressure down to the vapor pressure and trigger cavitation. What makes this approach powerful is that we did not need to know where the cavitation happened, we just needed to know that because this torpedo’s geometry didn’t change from one test to the other, and because we can safely assume that the Reynolds number change from one test to the other doesn’t matter, that cavitation will happen at the same location once it achieves the same pressure coefficient (or, equivalently, same cavitation number).

In formulating this response, I consulted the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2), and I have ranked the sections according to their relevance. If they are marked as relevant, I encourage you to explore them:

Textbook sections:

  1. Chapter 4.1: Watercraft Part 1 (RELEVANT)
  2. Chapter 5.4: Euler Equation: Streamwise Direction i.e. Bernoulli Equation Part 2 (RELEVANT)
  3. Chapter 4 Engineering Models: Hydrostatic Equation (RELEVANT)
  4. Chapter 3 Engineering Models: Hydrostatic Equation (RELEVANT)
  5. Chapter 5.4: Euler Equation: Streamwise Direction i.e. Bernoulli Equation Part 3 Part 1 (RELEVANT)

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