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
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.
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
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.
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
Leave a Reply. |
Free SignupSubscribe and learn something new each day:
CommunityThank You to Our Top
February '21 Contributors!
Your PostThe ToolkitSprinkler Designer or Engineer?
Get all of our tools, including the Sprinkler Database, Friction Loss Calculator, Fire Pump Analyzer and more: Filters
All
Archives
May 2021
Daily
Daily discussions are open-ended fire protection, fire alarm, and life safety questions submitted anonymously for the benefit of sharing expertise and learning from other perspectives. Anyone can submit a question here:
Exam Prep2020 PE Prep Guide
(Available Now!) PE Prep Series
(Available Now!) 2020 PE Prep Series
Current Leaderboard (Click to enlarge) PE Problems
Visit July-October for daily Fire Protection PE Exam sample questions.
Solutions are posted the day after posting. Comment with your solutions, questions or clarifications. Please note that questions posted are unofficial and in accordance with NCEES rules are intended to be similar to actual exam questions, not actual exam questions themselves. |