The volume flow is pi r^2 v both at Y1 and Y2
The water fell 0.39 meters from Y1 to Y2
loss of potential energy = m g h = m g ( .39)
equals gain of kinetic energy = (1/2) m (V2^2 - V1*2)
so
.39 *9.81 = (1/2) (V2^2 - 2.5^2)
solve for V2
R1^2 * V1 = R2^2 * V2 (conservation of mass) solve for R2
A slow running faucet is shown where the water is thinner at the bottom than the top. Location 1 and 2 are outside the faucet and water is in contact with the air. The top starting location is at height y1 = 0.39m and the end location is y2=0. At the top location the water has velocity v1 = 2.5m/s and the stream of water has a radius of r1=.010m. using knowledge of fluids determine a) the volume rage of flow leaving the faucet: m^3/s
B) velocity of the fluid at location 2: v2 =
C) radius of the fluid stream at location 2: r2=
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