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WARRIOR [948]
3 years ago
7

Given: The drag characteristics of a torpedo are to be studied in a water tunnel using a 1:5 scale model. The tunnel operates wi

th fresh water at 20°C, whereas the prototype torpedo is to be used in seawater at 15.6°C. To correctly simulate the behavior of the prototype moving with the velocity of 30 m/s, what velocity (in m/s) is required in the water tunnel?

Physics
1 answer:
Gnoma [55]3 years ago
4 0

Answer:

The velocity in the water tunnel is 128.717m/s

Explanation:

To find the result we need to Apply dynamic symiliraty, that is,

\frac{V_m I_m}{\upsilon_m}= \frac{VI}{\upsilon}

Where \upsilon_i is the kinematic viscosity, V the velocity and I the lenght.

To find the kinematic viscosity it is necessary search in the table with this properties (I attached it)

To this kind of fluids, the properties to 15.6°c (Seawater) and 20°c (Water)are:

\upsilon_m= 1.17*10^{-6}m^2/s (Seawater)

\upsilon_m = 1.004*10^{-6}m^2/s (Water)

Replacing,

\frac{V_m I_m}{\upsilon_m}= \frac{VI}{\upsilon}

Solving to V_m

V_m = \frac{I}{I_m}\frac{\upsilon_m}{\upsilon}V

V_m= \frac{5}{1} * \frac {1.004*10^{-6}}{1.17*10^{-6}} 30

V_m = 128.717m/s

The velocity in the water tunnel is 128.717m/s

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