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If for the pipe carrying water in a building, h= 158 meters, v1= 3.78 m/s and the cross sectional area at 1 is 3.6 times that at 2, what must P1 be, to the nearest KPa, in order that P2= 101000 Pa?

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Bernoulli's Equation :

P1 + 0.5 \rho v12 = P2 + \rho gh + 0.5 \rho v22

P1 + 500 v12 = 101000 + 1548400 + 500 v22

Equation of continuity :

A1v1 = A2v2 ; where (A1 = 3.6 A2)

3.6 v1 = v2

Substituting in the above equation :

P1 = 101000 + 1548400 + 500 (3.6v1)2 -Â Â 500 v12 = 1734844.632 Pa

P1 = 1734.84 kPa
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