A long horizontal hose of diameter 3.2 cm is connected to a faucet. At the other

Question

A long horizontal hose of diameter 3.2 cm is connected to a faucet. At the other end, there is a nozzle of diameter 1 cm. Water squirts from the nozzle at velocity 12 m/sec. Assume that the water has no viscosity or other form of energy dissipation.

1)What is the velocity of the water in the hose ?

_______________m/s

2)What is the pressure differential between the water in the hose and water in the nozzle ?

_________Pa

3)How long will it take to fill a tub of volume 100 liters with the hose ?

__________sec

Explanation / Answer

A1 v1 = A2 v2
d1^2 v1 = d2^2 v2
v2 = [d1/d2]^2 v1 = (1/4)^2 *12 = 0.75m/s
--------------------------------------...
P + 1/2 ? v^2 = K.
[p2 -p1] =1/2 ? [v1^2 - v2^2] = 1/2* 1000 [12^2 - 0.75^2]
[p2 -p1] = 71718.75 Pa.
--------------------------------------...
Power = force x velocity
and velocity = flow rate /area.
Therefore
Power = force x flow rate /area.
Since force / area is pressure
Power = pressure x flow rate
In the problem pressure = h ? g = 100* 1000* 9.81
Power = 100* 1000* 9.81* 710 = 696510 K.W.= 696.5 M W.
--------------------------------------...
Flow rate = A v= ? [d^2 / 4 ] * v
= ? [0.010 ^2 / 4 ] * 12 = 0.00157 m^3 /s
100 liters = 0.1 m^3
t = 0.1/ 0.00157 = 63.6 s

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