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

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

Step-by-step 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

`(d_(1) )/(d_(2) )` =
`(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)` =
`(0.000186)/(1.2* 7.5)`×
`(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