Question

A torpedo is to be designed to be 3 m long with a diameter of 0.5 m. Treating the torpedo as a cylinder, it is to be made to have a velocity of 10 m/s in sea water. Sea water has a dynamic viscosity of 0.00097 Ns/m and a density of 1023 kg/m3 . A 1:15 scale model is going to be tested in air. What velocity will be needed for the model and prototype to be similar

Answer 1

Answer:

The velocity that will be needed for the model and prototype to be similar is 108.97m/s  

Explanation:

length of Torpedo = 3m

diameter, $d_{1} = 0.5m$

velocity of sea water, $v_{1}$= 10m/s

dynamic viscosity of sea water, η$_{1}$ = 0.00097 Ns/m²

density of sea water, ρ$_{1}$ =  1023 kg/m³

Scale model = 1:15

$\frac{d_{1} }{d_{2} }$ = $\frac{1}{15}$

Cross multiplying: d$_{2}$ = $15d_{1} }$ = 15 ×0.5 = 7.5m

Let:

velocity of air, $v_{2}$

viscosity of air, η$_{2}$ =  0.000186Ns/m²

density of air, ρ$_{2}$ =  1.2 kg/m³

For the model and the prototype groups to be equal, Non-dimensional groups should be equal.

Reynold's number:   (ρ$_{2}$ ×$v_{2}$ ×d$_{2}$)/η$_{2}$ =  (ρ$_{1}$ ×$v_{1}$ ×d$_{1}$)/η$_{1}$

$v_{2}$ = η$_{2}$/(ρ$_{2}$ ×d$_{2}$)  ×  (ρ$_{1}$ ×$v_{1}$ ×d$_{1}$)/η$_{1}$

$v_{2}$ =  $\frac{0.000186}{1.2* 7.5}$×$\frac{1023 *10*0.5}{0.00097}$ , note: * means multiplication

$v_{2}$ = 108.97m/s  

velocity that will be needed for the model and prototype to be similar = 108.97m/s