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

The velocity in the water tunnel is 128.717m/s

Step-by-step explanation:

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

`(V_m I_m)/(\upsilon_m)= (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,

`(V_m I_m)/(\upsilon_m)= (VI)/(\upsilon)`

Solving to V_m

`V_m = (I)/(I_m)(\upsilon_m)/(\upsilon)V`

`V_m= (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