Question

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

Answer 1

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