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Presentation made by Raef Ali kobeissifor more information pleasecontact me :[email protected] Using the hydraulic calculation method for a 1 side branched tree system was previously explained .-We used to add the pressure losses in sprinklers1-2-3-4-We added them to the losses in the pipes1-2,2-3,3-4,4-5-We found the New K Factor at node 6 that declared the 2ndbranch as a 1 big sprinkler.-we calculated the flow rate in the 2ndbranch 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-williamequation at the convenient C value , usually 120.22QPK=-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 1.854.8711.1101.10 .( ) .qpc 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-PipePipe lengthFor pipe lossesFlow on node/in pipePressure on node/loss in pipeNotes1* Sprinkler calculation Q=0.15x130=19.5GPM 12.1psi1-2 Pipe calculationLpipe=13ft19.5gpm 1.6P2=1.6+12.1=13.7 psihazenwillamDiameter of pipe is given2( )QPK=Node-PipePipe lengthFor pipe lossesFlow on node/in pipePressure on node/loss in pipeNotes2* Sprinkler calculation Q=K2.20.7gpm 13.7psi P1+Ploss(1-2)2-3 Pipe calculationLpipe=13ft20.7+19.5=40.2gpmQ1+Q21.6psiP3= 13.7+1.6=15.3psihazenwillamDiameter of pipe is given2PNode-PipePipe lengthFor pipe lossesFlow on node/in pipePressure on node/loss in pipeNotes3* Sprinkler calculation Q=K3.=21.9GPM 15.3psi P2+Ploss(2-3)3-4 Pipe calculationLpipe=13ft40.2+21.9=62.1gpmQ1+Q2+Q31.7psiP4= 15.3+1.7=17psihazenwillamDiameter of pipe is given3PNode-PipePipe lengthFor pipe lossesFlow on node/in pipePressure on node/loss in pipeNotes4* Sprinkler calculation Q=K4.=23.1GPM 17psi P3+Ploss(3-4)4-5 Pipe calculationLpipe=13ft62.1+23.1=85.2gpmQ1+Q2+Q3+Q49 psiP5= 17+9=26psihazenwillamDiameter of pipe is given4P-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 6-7As 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 gpmNode-PipePipe lengthFor pipe lossesFlow on node/in pipePressure on node/loss in pipeNotes7* Sprinkler calculation Q=0.15x130=19.5GPM 12.1psi6-7 Pipe calculationLpipe=13ft19.5gpm 1.6P6=1.6+12.1=13.7 psihazenwillamDiameter of pipe is given2( )QPK=Node-PipePipe lengthFor pipe lossesFlow on node/in pipePressure on node/loss in pipeNotes6* Sprinkler calculation Q=K6.20.7gpm 13.7psi P7+Ploss(6-7)6-5 Pipe calculationLpipe=13ft20.7+19.5=40.2gpmQ7+Q61.6psiP5= 13.7+1.6=15.3psihazenwillamDiameter of pipe is given6P-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 6-7Is not the real pressure at nipple 5 , the real pressure is already calculated = 26 psi from branch 1-2-3-4-5Ktot is equal to 10.25QKtotP=5 P -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 branch1-2-3-4-55 P-Q6-7 = 10.2. = 52 Gpm-Total Q at pipe 8-5 = 52 +85.2=137Gpm.-Total pressure at 5 =26 psi.26We continue backward by calculating the flow rate atevery branch , because of the smiliratiy in branch12-11-10-9-8And branch 8,13,14,15,16 , the balance is already achievedA 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 nipple5 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 unknowas well ???? -The solution for this problem is to create 2 equations with 2 unknowns-We know that P6 = P5- .-We know that Q6 = . Equation 15 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 4.875 611.1101.10 .( ) .qpc D= Equation210 1.8565 6 4.875 611.1101.10 .( ) .qpc D= Equation26. 6totK PEquation 1 Q6=-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 26 6. Q K P = 6 5 5 6P p p = 2 2 10 1.8566 6 5 5 64.871. 1.1101.10 .( ) . . PipeQQ K P LC D ( (| |= ( | (\ . -We replace Equation 2 in equation 1 :2 26 6. Q K P = 6 5 5 6P p p = ( )2 1.856 61 2. Q Cst Cst Q ( =