Q = flow rate = pi r^2 V
loses rho g h = 1000 kg/m^3 * 9.81 m/s^2 * 5.2 m in N/m^2 or Pascals
it will go faster through the smaller pipe so loses pressure by Bernoulli
Q2 = Q = pi r^2 V2
so
V2/1.8 = 6.5^2/5.5^2
calculate V2
then pressure loss due to v increase is
(1/2)(1000) (v2^2 -v^2)
add up the two pressure losses and subtract from 2.2*10^5
The pressure of water flowing through a 6.5×10−2−m -radius pipe at a speed of 1.8m/s is 2.2 × 105 N/m2
What is the flow rate of the water?
What is the pressure in the water after it goes up a 5.2−m -high hill and flows in a 5.0×10−2−m -radius pipe?
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