A 250 ft high building has a 6065indiameter streel standpipe

A 250 ft high building has a 6.065-in-diameter streel standpipe and a 100-ft-long 2 9/16 in diameter fire hose on each floor. The nearest fireplug is 100 ft from the standpipe\'s ground-level connection. Assume that firefighters connect a 6-in. diameter, 50-ft-long fire hose from the fireplug to the fire truck and a 4-in-diameter, 50-ft-long fire hose from the fire truck to the standpipe\'s ground-level connection. The National Fire Protection Association (NFPA) requires that a minimum pressure of 65 psig be maintained at the connection of the 2 9/16 in-diameter hose and the standpipe while 16 maintaining a flow rate of 500 gal/min through the fire hose. What pressure rise must the pump on the fire engine supply to satisfy the NFPA requirement for this building? The fire hydrant water pressure is 80 psig and the water temperature is 60 degree F. The connections are threaded.

Solution

solution:

1)here pressure equation for flow is given as follows

(start pressure)-(end pressure)-(friction loss in hose between plug to pump)-(fitting loss)-(friction loss in hose from pump to standpipe)-(elevation head)+pump head=0

2) start pressure=80 psi

3)end pressure=65 psi

4)pressure loss in firehose is given by following formula

Hf=C*(Q/100)^2*L/100

Q=500 gpm

for pipe between plug to pump=Hfpp=C*(Q/100)^2*L/100

for 4 in pipe value of c=.2

L=50ft

so we get

Hfpp=.2*(500/100)^2*50/100=2.5 psi

5)for pipe between pump to standpipe is=Hfps=C*(Q/100)^2*L/100

d=6 in,C=.05,L=50 ft

Hfps=.625 psi

6)fitting losses are minimum and can be neglected as all fitting are made by fire hose and connection are threaded so losses of fitting is neglected.

7)elevation=end elevation-start elevation=250-0=250 ft=108 psi

8)friction losses in steel stand pipe is given by hazen william equation as follows

Hfs=.002083*L*(100/C)^1.85*Q^1.85/d^4.8655

for steel pipe,c=120

L=250 ft

d=6.065 in

Q=500 gpm

Hfs=5.6803 ft or 2.46 psi

9)now we have computed all lossses between start to end so pump head which is required at fire truck is given by following relation

pump head+(start pressure)-(end pressure)-(friction loss in hose between plug to pump)-(fitting loss)-(friction loss in hose from pump to standpipe)-(elevation head)=0

pump head+80-65-.625-2.5-2.46-108=0

pump head=98.585 psi

so pump head required is given by 98.585 psi, so we will set pump pressure to 99 psi to take care of additional fitting losses

so pump head=99 psi