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# Water is flowing with a speed of 4 m/s in a horizontal pipe of non-uniform area of cross-section decreasing from 0.04 m² to 0.01 m² at pressure 40 x 104 Pa. What will be the pressure at narrower end?

Asked by Topperlearning User 31st July 2014, 3:30 PM

v1 = 4 m/s, a1 = 0.04 m², P1 = 40 x 104 Pa

a2 = 0.01 m², P2 = ?

According to equation of continuity, we have
a1v1 = a2v2

v= a1v1 / a2 = 0.04 x 4/ 0.01 = 16 m/s

According to Bernoulli's theorem, we have
P1+ ½ρv12 = P2 + ½ρv22
P2 = P1+ ½ρ(v12 - v22) = 40 x 104 + ½ x 10³ (4² - 16²)
= 40 x 104  - ½ x 10³ x 240
= (400 - 120) x 10³
= 280 x 10³ Pa
Answered by Expert 31st July 2014, 5:30 PM
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