Embed Size (px)
Citation preview
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
