Firefighting Hydraulic Calculation

Using the hydraulic calculation method for a 1 side branched tree system was previously explained .

-We used to add the pressure losses in sprinklers”1-2-3-4” -We added them to the losses in the pipes”1-2,2-3,3-4,4-5” -We found the New “K Factor at node “6” that declared the 2nd _{branch as a 1 big sprinkler.}

-we calculated the flow rate in the 2nd _{branch using the}

-in 1 side branched system we used to start with the furthest sprinkler “1” by calculating the flow rate which is equal to the density x area of coverage .

-and then we calculated the pressure on the mentioned sprinkler using:

-Then we have calculated the pressure loss in the pipe using Hazen-william equation at the convenient C value , usually 120. 2 2 Q P K 

-The pressure losses which must be covered by the pump are the losses that occur on the longest run of the whole system and the flow rate covered is the sum of flow rate of all sprinklers .

-The Hazen William Equation is equal to : 10 1.85 4.87

1

1.1101.10 .( )

q

.

p

c

D



The Problem may arise when the system is double

branched and the 2 branches do not have the same

number of sprinklers.

-Considering this system being: ordinary hazard 1 with Area of operation equal to 1500 ft2.

-Density of Hydraulic most demand sprinkler is 0.15gpm/ft2 -Area of coverage of sprinkler = 130 ft2.

Node-Pipe Pipe length For pipe losses Flow on node/in pipe Pressure on node/loss in pipe Notes 1* Sprinkler calculation Q=0.15x130 =19.5GPM

12.1psi

Node-Pipe Pipe length For pipe losses Flow on node/in pipe Pressure on node/loss in pipe Notes 2* Sprinkler calculation Q=K2. 20.7gpm 13.7psi P1+ Ploss(1-2) 2-3 Pipe calculation Lpipe=13ft 20.7+19.5=40.2gp m Q1+Q2 1.6psi P3= 13.7+1.6 =15.3psi hazen willam “Diameter of pipe is given 2 P

Node-Pipe Pipe length For pipe losses Flow on node/in pipe Pressure on node/loss in pipe Notes 3* Sprinkler calculation Q=K3. =21.9GPM 15.3psi P2+ Ploss(2-3) 3-4 Pipe calculation Lpipe=13ft 40.2+21.9= 62.1gpm Q1+Q2+Q3 1.7psi P4= 15.3+1.7 =17psi hazen willam “Diameter of pipe is given 3 P

Node-Pipe Pipe length For pipe losses Flow on node/in pipe Pressure on node/loss in pipe Notes 4* Sprinkler calculation Q=K4. =23.1GPM 17psi P3+ Ploss(3-4) 4-5 Pipe calculation Lpipe=13ft 62.1+23.1= 85.2gpm Q1+Q2+Q3+Q4 9 psi P5= 17+9 =26psi hazen willam “Diameter of pipe is given 4 P

-The final pressure at nipple 5 =26 psi and the flow rate Going to branch “4-3-2-1” is equal to 85.2 gpm

-Now we have to identify how much flow rate shall go to branch “6-7”.

-We start hydraulic calculation with “6-7” As if this branch is the longest branch.

-We start with sprinkler 7 as if it is the hydraulic most demand sprinkler, so k=5.6 , q=19.5 gpm

Node-Pipe Pipe length For pipe losses Flow on node/in pipe Pressure on node/loss in pipe Notes 7* Sprinkler calculation Q=0.15x130 =19.5GPM

12.1psi

Node-Pipe Pipe length For pipe losses Flow on node/in pipe Pressure on node/loss in pipe Notes 6* Sprinkler calculation Q=K6. 20.7gpm 13.7psi P7+ Ploss(6-7) 6-5 Pipe calculation Lpipe=13ft 20.7+19.5=40.2g pm Q7+Q6 1.6psi P5= 13.7+1.6 =15.3psi hazen willam “Diameter of pipe is given 6 P

-According to Branch “6-7” Pressure at nipple “5” Is equal to =15.3 psi and flow rate to “6-7” = 40.2gpm -We find now the new K factor for the whole branch “6-7”

Is not the real pressure at nipple 5 , the real pressure is already calculated = 26 psi from branch “1-2-3-4-5”

Ktot is equal to 10.2 5 Q Ktot P  

5

P



-Now we have to balance the system by finding the real Flow rate that is going to branch 6-7

Q6-7 = Ktot.

Caution: the pressure P5 in the equation above is

equal to 26 psi which belong to the real pressure

At nipple 5 calculated from the longest branch

“1-2-3-4-5”

5

-Q6-7 = 10.2. = 52 Gpm

-Total Q at pipe 8-5 = 52 +85.2=137Gpm.

-Total pressure at 5 =26 psi.

We continue backward by calculating the flow rate at

every branch , because of the smiliratiy in branch”12-11-10-9-8” And branch “8,13,14,15,16” , the balance is already achieved

A new Problem may arise when the most remote

Sprinkler is not at the last branch of the tree system

-How then we can calculate the demanded flow rate and The pressure ?

-As we see above sprinkler “1” will cost the pump the maximum pressure

-we can use the hydraulic calulation to calculate the Pressure at nipple”5” and the flow rate sent to branch: ‘1-2-3-4-5” But the pressure at node 6 is unknown ,hence The flow rate at node 6 is unknow as well ????

-The solution for this problem is to create 2 equations with 2 unknowns

-We know that P6 = P5- .

-We know that Q6 = . Equation 1

5 6

p

6. 6

tot

-Ktot 6 can be always calculated by assuming the 2 Branched “7-8” and “9-10” as the furthest branches.

10 ₆ 1.85 5 6 _4.87 5 6

1

1.1101.10 .(

q

)

.

p

c

D



10 ₆ 1.85 5 6 _4.87 5 6

1

1.1101.10 .(

q

)

.

p

c

D



-Ktot is known , D5-6 is known , C is known &P5 is known -P6 and Q6 are unknon

-We replace Equation 2 in equation 1 : 2 2 6

.

Q



K P

P



p



p

-We replace Equation 2 in equation 1 : 2 2 6

.

Q



K P

P



p



p





1

2.

Q



Cst





_

Cst

Q



_