Problem 9 A fire-hose is connected to a fire-hydrant at street level. The gauge

Question

Problem 9 A fire-hose is connected to a fire-hydrant at street level. The gauge pressure at the fire-hydrant is 12atm and the diameter of the outlet of the fire-hydrant is 12.0 cm. The nozzle is connected to the fire-hose has a diameter of 4.0cm. Tom, the fire fighter, takes the nozzle up the building to the third floor, so that the nozzle is 30m above the street level. i)What is the gauge pressure at the nozzle, when the nozzle is turned off and the water does not flow.? ii) Next, the hose is turned on, and the gauge pressure at the fire hydrant remains 12atm. Determine the speed of the flow at the fire-hydrant and at the hose.

Explanation / Answer

part i.

let the street level be labeled as point A and third floor be labeled as point B.

gauge pressure at A=12 atm

then absolute pressure at A=Pa=12+1=13 atm

diameter =12 cm

==>radius=rA=0.06 m

fire hose which is at B , has a diameter of 4 cm

==>radius at B=rB=2 cm=0.02 m

height of A=hA=0

height of B=hB=30 m

when water is not flowing speed at both the points are 0.

hence vA=vB=0

using bernouilli’s theorem,

Pa+pho*g*hA+0.5*pho*vA^2=Pb+pho*g*hB+0.5*pho*vB^2

where pho=density of water=1000 kg/m^3

using the given values,

Pb=13 atm+1000*9.8*0-1000*9.8*30

=10.098 atm

so gauge pressure at the hose is 9.098 atm.

part ii.

when the hose is turned on, pB=1 atm

let speed at A and B are vA and vB respectively.

using continuity principle,

rate of flow at A=rate of flow at B

==>vA*area of A=vB*area of B

==>vA*pi*rA^2=vB*pi*rB^2

==>vA*0.06^2=vB*0.02^2

==>vB=9*vA

using this relationship in bernoulli’s eqaution,

13 atm+1000*9.8*0+0.5*1000*vA^2=1 atm+1000*9.8*30+0.5*1000*(9*vA)^2

==>12*101325-1000*9.8*30=0.5*1000*vA^2*(9^2-1)

==>vA=4.8 m/s

then vB=9*vA=43.2 m/s

so speed of the water at the fire hydrant is 4.8 m/s and at the nozzle is 43.2 m/s.

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