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Water is flowing through a horizontal pipe whose one end is closed with a valve and the pressure reads 3x10^5 N/m.This pressure reading is changes 1x10^5 when the valve is opened.calculate the speed of water flowing through the pipe.

siddhant , 9 Years ago
Grade 11
anser 1 Answers
Rituraj Tiwari
Using Bernoulli's principle we know that,

P₁/ρ + V₁²/2 = P₂/ρ + V₂²/2

where P₁ = initial pressure = 3.5 x 10⁵

V₁ = initial velocity = 0

P₂ = final pressure = 3 x 10⁵

V₂ = final velocity

ρ = density of water = 1000

Hence putting these value in eq we get,

3.5 x 10⁵/1000 + 0 = 3x10⁵/1000 + V₂²/2

=> V₂²/2 = 350 - 300 = 50

=> V₂² = 100

=> V₂ = 10 m/s.

Hence the velocity of the water after opening valve will be10 m/s
Last Activity: 5 Years ago
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