(a) V = (Volume flow rate)/(faucet area)
=25*10^3 cm^3/(30 s* pi cm^2)= 265 cm/s = 2.7 m/s Correct
(b) The pressure is ambient (Po) at the faucet exit, where the water is flowing. Use this form of the Bernolli equation:
P + (1/2) rho V^2 + rho*g y = constant
Let Po be ambient pressure, P1 be the pressure in the entrance pipe, and V1 be the velocity there, which will be 1/9 of the velocity in the faucet, due to the larger diameter. (since V*Area = constant)
Therefore
Po + (1/2) rho V2^2 =
P1 + (1/2)rho V1^2 + rho g h
The gauge pressure is P1-Po
= (1/2)rho (V2^2 - V1^2) - rho g h
Substitute V1 = 2.7 m/s, V2 = 0.3 m/s and solve. h = 2m
The water supply of a building is fed through a main entrance pipe 6cm in diameter. A 2cm diameter faucet tap positioned 2m above the main pipe fills a 25 liter container in 30s. (a) What is the speed at which the water leaves the faucet? (b) What is the gauge pressure in the main pipe? (Assume that the faucet is the only outlet in the system.)
(a) (25x10^3cm^3)/(pix30s)=2.7m/s
(b) P-Pa=pgh
I am unsure what to plug in for h.
(1000kg/m^3)(9.8m/s^2)(h)
(answer (b): 2.3x10^5 Pa)
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