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 theFormula : Q=Ktot . 6P

-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

QP

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

11.1101.10 .( ) .

qp

c D

The Problem may arise when the system is double branched and the 2 branches do not have the same

number of sprinklers.

-At the furthest run the 2 branches must together be balanced

-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.-Sprinkler K factor = 5.6.

Node-Pipe Pipe length

For pipe

losses

Flow on

node/in pipe

Pressure on

node/loss in

pipe

Notes

1* Sprinkler

calculationQ=0.15x130

=19.5GPM12.1psi

1-2 Pipe

calculation

Lpipe=13ft

19.5gpm 1.6

P2=1.6+

12.1=

13.7 psi

hazen

willam

“Diameter

of pipe is

given

2( )Q

PK

Node-Pipe Pipe length

For pipe

losses

Flow on node/in

pipe

Pressure on

node/loss in

pipe

Notes

2* Sprinkler

calculationQ=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

2P

Node-Pipe Pipe length

For pipe

losses

Flow on node/in

pipe

Pressure on

node/loss in

pipe

Notes

3* Sprinkler

calculationQ=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

3P

Node-Pipe Pipe length

For pipe

losses

Flow on node/in

pipe

Pressure on

node/loss in

pipe

Notes

4* Sprinkler

calculationQ=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

4P

-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

calculationQ=0.15x130

=19.5GPM12.1psi

6-7 Pipe

calculation

Lpipe=13ft

19.5gpm 1.6

P6=1.6+

12.1=

13.7 psi

hazen

willam

“Diameter

of pipe is

given

2( )Q

PK

Node-Pipe Pipe length

For pipe

losses

Flow on

node/in pipe

Pressure on

node/loss in

pipe

Notes

6* Sprinkler

calculationQ=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

6P

-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

QKtot

P

5P

-Now we have to balance the system by finding the realFlow rate that is going to branch 6-7Q6-7 = Ktot.

Caution: the pressure P5 in the equation above is equal to 26 psi which belong to the real pressureAt nipple 5 calculated from the longest branch“1-2-3-4-5”

5P

-Q6-7 = 10.2. = 52 Gpm-Total Q at pipe 8-5 = 52 +85.2=137Gpm.-Total pressure at 5 =26 psi.

26

We continue backward by calculating the flow rate atevery 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 ,henceThe 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 6p

6. 6totK P

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

10 1.8565 6 4.87

5 6

11.1101.10 .( ) .

qp

c D

Equation2

10 1.8565 6 4.87

5 6

11.1101.10 .( ) .

qp

c D

Equation2

6. 6totK P Equation 1Q6=

-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 6.Q K P 6 5 5 6P p p

2 2 10 1.8566 6 5 5 64.87

1. 1.1101.10 .( ) . . Pipe

QQ K P L

C D

-We replace Equation 2 in equation 1 :

2 2

6 6.Q K P 6 5 5 6P p p

2 1.856 61 2.Q Cst Cst Q