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A FORUM FOR FIRE PROTECTION QUESTIONS & PE EXAM PROBLEMS | SUBSCRIBE NOW

PE Problem #32

8/18/2020

12 Comments

 
A project site is connected to a water tank through an equivalent 650 feet of 6-inch schedule 10 pipe (c-factor of 140). If the water tank is 165 feet above the project site, the fire flow available at the project site is most nearly:
a. 1,300 gpm
b. 2,100 gpm
c. 2,400 gpm
d. 8,700 gpm

Solution | Posted 08/19/20
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12 Comments
Brenton
8/18/2020 10:19:35 am

b. 2100 gpm

Reply
Matt
8/18/2020 10:52:32 am

The answer is (a). It is the only flow that provides the minimum 20 psi at the fire hydrant.

Static pressure = 0.433 x 165' = 71.5 psi
Required residual / fire flow pressure after friction loss = 20 psi
Maximum allowable friction loss = 51.5 psi

a. [((4.52)(1300)^1.85)/((140^1.85)(6^4.87))] x 650 ft = 30 psi
b. [((4.52)(2100)^1.85)/((140^1.85)(6^4.87))] x 650 ft = 71.5 psi
c. [((4.52)(2400)^1.85)/((140^1.85)(6^4.87))] x 650 ft = 91.5 psi
d. [((4.52)(8700)^1.85)/((140^1.85)(6^4.87))] x 650 ft = 991 psi psi

Reply
Brenton
8/18/2020 11:45:01 am

The actual ID of 6 inch schedule 10 pipe should be used. Recommend using algebra to calculate a Q from the dP; will save time on the exam over solving H-W for each possible answer. Best of luck!

NinaB
8/18/2020 11:40:55 am

C. Q=0.28 C H^0.54 D^2.63/L^0.54=0.28 x 140 x165^0.54 x 6.357^2.63/650^0.54 =2423 gpm

Reply
J
8/18/2020 03:12:43 pm

2444.328431

Gallon per minute

Reply
KK
8/18/2020 06:34:52 pm

c. 2,400 gpm

Reply
Brenton
8/19/2020 07:52:26 am

The solution is calculating the maximum possible flow rate, but “fire flow” is the maximum flow rate with a residual pressure of 20psi. If using the SFPE equation 41.68, the H should be available head, 165 - (20/.433) = 118.8 ft. Right?

Reply
NearEngineer
8/19/2020 04:12:10 pm

I realize in the solution Joe solves for Q using Hazen-Williams formula, but where is the .28 value from?

Reply
B
8/24/2020 06:20:16 pm

It's not exactly obvious from the equation, but with some unit conversions, the Hazen-Williams expression for evaluating Q looks like this: Q=C(deltaP*d^4.87/4,52L)^0.54; however, delta P is pressure loss (in psi) per foot of pipe. In this case, we know the length, so we have to convert from psi to feet -> 0.433*length. The constants in the equation therefore are (0.433/4,52)^0.54, which equals 0.28.

Hope this helps

Reply
Naveen Kumar
8/19/2020 08:45:11 pm

Yes, NCEES 4.5.5 is Piping loops and doesn't have Flow calculation.

Reply
SK
8/26/2020 03:27:00 pm

NCEES, 4.5.2, could be more suitable to use. Head can be 165x.433 due elevation. Then it's coming to atm, so it's 0, then DeltaP = 71.43 psi. Rest plug the values in H-W equation to calculate Q.

Reply
SK
8/26/2020 03:18:53 pm

C.

Reply



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  • Blog
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