Question

Find the thrust developed (in N) when water is pumped through a 300 mm diameter pipe in the bow of a boat at V= 2 m/s and emitted through a 200 mm diameter pipe in the stern

Answer 1

Answer:

the thrust developed when water is pumped is 353.43N

Answer 2

The thrust developed in N is 353.43 N

Explanation:

Given:

Water is pumped through a 300 mm dia pipe and emitted through a 200 mm dia pipe. The velocity of the water is 2m/s

To find out:

The thrust developed

Solution:

Thrust force = Change in momentum of water

Volume of water going in per second

$=v_1A_1$

where $v_1$ is the velocity of inflow and $A_1$ is the area of cross section of pipe 300 mm dia

Similarly volue of water coming out per second

$=v_2A_2$

where $v_2$ is the velocity of inflow and $A_2$ is the area of cross section of pipe 200 mm dia

By continuity equation

$v_1A_1=v_2A_2$

$\implies 2\times\frac{\pi\times 0.3^2}{4}=v_2\frac{\pi\times 0.2^2}{4}$

$\implies 2\times 0.09=v_2\times 0.04$

$\implies v_2=\frac{2\times 0.09}{0.04}$

$\implies v_2=4.5$ m/s

Volume of water going in = Volume of water going out

Mass of water flowing per second

m = Volume × density

$\implies m=\frac{\pi\times 0.3^2}{4}\times 2\times 1000$ kg/s

$\implies m=141.37$ kg/s

Therefore,

Thrust = change in momentum

$=m(v_2-v_1)$

$=141.37(4.5-2)$

$=141.37\times 2.5$

$=353.43$ N

Hope this answer is helpful.

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