physics
posted by rock .
The intake in the figure has crosssectional area of 0.75 m2 and water flow at 0.41 m/s. At the outlet, distance D = 180 m below the intake, the crosssectional area is smaller than at the intake and the water flows out at 9.6 m/s. What is the pressure difference between inlet and outlet?

p + rho g h + (1/2)rho v^2 = constant
Vin = .41 m/s
Vout = 9.6 m/s
hinlet = 180
houtlet = 0
rho = 1000 kg/m^2
g = 9.8 m/s^2
so
Pinlet + 1000(9.8)(180)+(1/2)1000(.41)^2
= Poutlet +1000(9.8)(0)+(1/2)1000(9.8)^2
solve for Poutlet  Pinlet
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