Asked by Luke

A garden hose is used to fill a drum with water. The hose is held 0.88 m above the ground. The cross-sectional area of the 2.0 cm diameter hose is two times the cross-sectional area at the spout end where the water is coming out and the gage pressure inside the hose at a section laying on the ground is 29.7 kPa. Based on this information one can determine that the water velocity at the spout end is [ FILL IN BLANK ]m/s, the water velocity inside the hose is [ FILL IN BLANK ]m/s, the volumetric flowrate of the water is [ FILL IN BLANK ]× 10-3 m3/s, and the time it takes to fill the 0.210 m3 drum to its maximum capacity is [ FILL IN BLANK ]min. Note that the water density is ρw = 1000 kg/m3.
Answered by Damon
loss in pressure from inside hose to exit is 29.7*10^3 N/m^2
pin + (1/2) rho Vin^2 = Pout + (1/2) rho Vout^2
29.7 * 10^3 = (1/2)(10^3)(Vout^2-Vin^2)
but if Aout = .5 Ain
then Vout = 2 Vin
so
2*29.7 = (4-1)Vin^2 = 3 Vin^2
solve for Vin
then Vout

Vi * Area inside = Q, volume flow

that should get you going.

Answered by Anonymous
why you have not considered datum height of 0.88 m?
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