A village maintains a large tank with an open top, containing water for emergencies. The water can drain from the tank through a hose of diameter 6.60cm. the hose ends with a nozzle of diameter 2.20 cm. A rubber stopper is inserted into the nozzle. the water level in the tank is kept 7.50 m above the nozzle. a) Calculate the friction force exerted on the stopper by the nozzle. b) the stopper is removed. What mass of water flows from the nozzle in 2.00hours? c)Calculate the Gauge pressure of the flowing water in the hose just behind the nozzle.
Bernoulli's equation, Torricelli's equation, Pressure= Force/Area, Area = (∏/4)(Diameter)2, P-P0=ρgh, Continuity of flow
The Attempt at a Solution
A) The pressure pushing the plug out is atmospheric pressure and the pressure of the water in the tank, PTank=P0+ρgh
the pressure keeping the plug in is the atmospheric pressure and the force of friction acting over the area of the nozzle, P0+F/A
They must be equal, otherwise the plug would pop out, so
[STRIKE]P0[/STRIKE] + ρgh =[STRIKE] P0[/STRIKE] + F/A
Aρgh = F = 27.94
B) Torricelli's equation, Vhose=sqrt(2gh)
Use it to find the flow rate, VhoseAhose= .004607 m3/s
Multiply the flow rate by 2*60*60 seconds to get 33.17 m3
Multiply by the density of water to get the total mass, m= 33170 kg
C) This is where I'm having trouble.
Vhose = 12.12 m/s, I can take that from the last part.
I can find Vnozzle = 1.3467 m/s using the continuity of flow A1V1=A2V2
I should plug the velocities, density, and pressures into the Bernoulli equation, P1 + (.5)ρV12 + ρgh1 = P2 + (.5)ρV22 + ρgh2, but I'm not sure where to put which pressures and if I should only be considering the hose or if I should include the tank.
I found some other explanations and solutions but none of them have helped me very much, so if someone could set up the equation for me and explain how you did, that would be amazing!