Question

find the
thrust
developed in N) when
water is pumped through a 300mm diameter
pipe in the bow of a boat v=2m/s and
emitted through a 200mm diameter pipe
in stern.​

Answer 1

Thrust force developed in water is 353.43 N

Explanation:

We are given that:

• Diameter of the pipe= 300 m
• Speed of the boat = 2 m/s
• Diameter through which the water is emitting =  200 m

To Find: Thrust Force "F" = ?

Solution:

Volume of water going in per second  = A1V1

Volume of water coming out per second = A2V2

Now

v1A1 = v2A2

2 x π x (0.3)^2 / 4 = v2 x π x ((0.2)^2 / 4

v2 = 2 x 0.09 / 0.04

v2 = 4.5 m/s

Volume of water going in = Volume of water going out

Mass of water flowing per second

Mass = Volume × density

m = π x (0.3)^2 / 4 x 2 x 1000 Kg/s

m = 141.37 Kg/s

Thrust = change in momentum

= m (v2 - v2)

= 141.37 (4.5 - 2)

= 353.43 N