Question

Water is flowing in a horizontal pipe of non-uniform area of cross section At one place on it, the
radius of the pipe is 0.3 m and the velocity of water is 1.0 m/s. Calculate the velocity of water
second place where the radius of the pipe is 0.15 m.​

Answer 1

The speed of the water is  31 .62 cm/s

Explanation:

By using Bernoulli equation.

P + (1/2)dV^2 + hdg= Constant

Where,

P = pressure

d = density of the liquid

V = velocity

h = height

g = acceleration due to gravity

but this question states that there is a horizontal pipe. so height should be zero.  Thus

P + (1/2)dV^{2} = Constant

but we know that, pressure (P) = hdg

hdg  +  (1/2)dV^{2} = Constant

hg  +  (1/2)V^{2} = Constant

substituting values for above equation,

h1g + (1/2)v1² = h2g + (1/2)v2²

10 x 10 + (1/2) x (30)² = 5 x 10 + (1/2) x (V2)²

100 + 450 = 50 + (v2²/2)

500 = v2²/2

1000 =v2²

31.62 cm/s =v2

Thus the speed of the water is  31 .62 cm/s

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