Question

Water flows at 0.61 m/s through a 3.0 cm diameter hose that terminates in a 0.35 cm diameter nozzle. Assume laminar non-viscous steady-state flow. (a) At what speed does the water pass through the nozzle? 44.816 Correct: Your answer is correct. m/s (b) If the pump at one end of the hose and the nozzle at the other end are at the same height, and if the pressure at the nozzle is 1 atm, what is the pressure at the pump? 10.70097467 Incorrect: Your answer is incorrect. atm( I have question for question 2)( 10.7 is wrong)

Answer 1

The hose area is
A = Pi*r^2 = 3.14*(0.03/2m)^2 = 7.07E-4m^2

The flow rate is
Q = A.v = 7.07E-4m^2 * 0.61m/s = 4.31E-4 m^3/s

The nozzle area is
An = Pi*r^2 = 3.14*(0.0035/2m)^2 = 9.62E-6m^2

a) The speed of water at the nozzle is
Vn = Q/An = 4.38E-4m^3/s/9.61E-6m^2 = 45.4m/s

b) The pressure right before the nozzle is given by
ΔP = P1-P2 = 1/2ρV2² - 1/2ρV1²
where
ΔP, P1 and P2 are in Pa
ρ is water density = 1000 kg/m^3
V1 and V2 are in m/s

In your case
ΔP = 1/2*1000*(45.5m/s)² - 1/2*1000*(0.61m/s)²
ΔP = 1 034 933 Pa = 10.2 atm

So right before the nozzle, the pressure is 11.2 atm.

And because the speed is so slow, it can be assumed that the pipe friction losses are minimum. For example at this speed, a plastic pipe 100meter long will cause losses of 0.014 atm.

So the pressure at the pump is 11.2 atm.