Fire sprinkler systems hydraulic calculations
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Fire fighting systems have many forms and layouts. We will try in this paper to pick 1
layout(the
tree
layout)
and interpret it thoroughly in details with a study case.
Tree Layout:
Figure-1The figure above shows us the standard tree sprinkler layout where each crossed circle represents a
sprinkler head and the big circle at the bottom represents a pump and the branches are the pipelines tha t
connect the sprinklers together.
The distance between the sprinklers and the distance between each branch are determined through
simple calculations
found in the NFPA standard.
The density and the number of sprinklers used are based on previous studies and researches where charts
were made for this purpose and they are available in the NFPA manual book.
The hydraulic calculations is based on Hazen Williams equation that determines the pressure loss inside
the pipe-lines.
2 equations needed for any design:
Q: Flow rate in “ GPM” .
K: Nominal factor in “ GPM/(PSI1/2 )
Symmetrical layout
In symmetrical layouts the design area is covered by equal number of sprinklers from right and left
branches and parallel branches. ” Check figure 1 “
-We start numerating the sprinklers from the most remote one to the nearest , we calculate the pressure
needed at the l’ st sprinkler using the first equation where K-factor is given based on the sprinkler
type used and the flow rate which is also dependent on the density “ gpm/ ft 2 ” of the sprinkler and its
coverage area .
-The second step is finding the losses in the pipe that sits behind the last sprinkler and here we must use
Hazen William equation that states:
hf = friction head loss in feet of water per 100 feet of pipe(fth20/100 ft pipe)
c = Hazen-Williams roughness constant
q = volume flow (gal/min)
dh = inside hydraulic diameter (inches)
hf must be multiplied by the length of the calculated pipe and then divided by 100.
-We repeat the same procedure at the 2nd sprinkler head ” 2” using equation 1 but this time we have
the pressure at the head of sprinkler from the sum of all the losses in the pipe and the most remote
sprinkler and from this value we can calculate the needed flow rate.
-We continue going backward until we reach the intersection point” 4” between the 2 parallel branches.
-Similarly we proceed with the same steps written above for the mirrored branch ” 5-6-7” and we add
all the flow rates coming out from all the sprinklers in both parallel branches.
Figure -2
-the system above is hydraulically balanced because of the similarity in the branches.
-The final interesting step is calculating the pressure loss in the pipe set between the 2 intersections “ 4
& 8” , the total flow rate needed in both branches” 11-10-9” and “ 14-13-12” is :
Where P8 is equal to P4+ P losses in pipe “ 4-8” Calculated using Hazen William equation
-Ktot8 is obtained by considering the branch “ 11-10-9” as the most remote branch and because we
have a balanced system the procedure is easy where K8= Qtot4/ P4 and surely you will be
wondering why?
Well the answer is simple, if we do consider this branch as the most remote branch then Qtot8 is going
to be similarly equal to Qtot4 and P8 is equal to P4, take notice this consideration is not applicable in
reality it is used only for calculating the nominal factor K8 ,afterward the true total flow rate going
into the intersection “ 8” is found through Ktot8. P4  Plosses(“4,8”) .
-What is necessary for us is to calculate the total loss in the system and the total flow rate needed to
eventually select an appropriate pump.
-The loss path is marked from the beginning of the riser till the most remote hydraulic branch as
shown below:
Figure – 3
The problem may appear when 1 of the mirrored branches or parallel branches are not alike as the
following figure below:
Figure 4-It is our job to balance the system, let us take a look to the result when we apply the same steps we
have done previously, well clearly we will notice a difference in pressure at intersection “ 4” when the
calculation
is
made
both
ways
:
“ 1,2,3”
and
“ 6,5” .
-In order to compensate this difference in pressure we start calculating the losses in the longest path of
the last branch using the same steps as before until we reach point “ 4” that will indicate the needed
pressure
.
-It is very obvious that if we do the same calculation within the mirrored branch we would get a
smaller value of pressure at “ 4” due to the decrease in the number of sprinklers at this branch
relatively to the branch “ 1,2,3” but this procedure is necessary in order to calculate the major nominal
factor
of
branch
“ 6,5”
to
represent
it
as
1
big
sprinkler.
-Major K” 6-5” is equal to Q6-5* / P4* where P4* is the pressure value at point 4 from the
perspective of branch “ 6-5” and not the pressure P4 calculated from the branch “ 1,2,3” that’ s why I
called it P4* and not P4 which is the real pressure at the intersection calculated through the path
“ 1,2,3” ,
remember
P4*
<
P4.
-The
next
step
must
be
decisive
and
final
for
achieving
the
balance:
Q” 6-5” =K” 6-5/ P4 where this result is final and Q total for the 2 branches is equal to:
Q” 6-5” + Q” 1,2,3” .
-The
rest
of
the
operation
becomes
smooth
and
usual.
Good example on symmetrical layouts are found in NFPA 13-annex A explanatory materials A.14.3.2
2002 meeting edition.
Non-symmetrical layout full example
Assuming the following:
-Parking car total area: 13200 ft2.
-Density, 0.15 gpm/ft
2
calculated over the most remote 1500 sq.ft.
Figure-5- Area-density chart.
-In this example take S=13 ft &L= 10ft and As:” area of discharge of sprinkler” = 130 ft
2
.
-K=5.6.
-13 sprinklers in most remote area.
-The minimum flow at the most remote sprinkler =19.5gpm.
-The minimum required pressure at the most remote sprinkler=12.1 Psi.
We have to size the piping systems & determine the pump duty.
Knowing that, the branch lines are schedule 40 black steel. (C=120 ) crossmain lines are schedule 10
black steel.
Fully calculated systems can be pipe-sized through hydraulic calculation. It is also possible to pipe
size using a tabulated information.
Figure-6- Plan figure of the sprinkler system, the area in red is the most remote area, 1500 ft2.
The expected number of sprinklers is determined as follows :
Number of sprinklers =Total area of the car parking / As
Area of the car parking 13200 ft2
( 110 ft x 120 ft) ,divided by the sprinkler coverage area As =120 ft2
we get 110 sprinklers.
Figure-7- Numeration of most remote area sprinklers cluster.
Hydraulic calculation
Figure-8- Flow distribution.
Figure-9- At: top of riser nipple, Ab: bottom of riser nipple.
Pressure drop due to elevation 1.5 ‘ , Pe= Z x 0.433 Psi/ft =1.5 x0.433 =0.65 Psi.
F = Pressure drop due Tee = 10 ft.
Friction loss Psi/100 ft from Hazen William equation with Q= 84.8 gpm & d=2 ” we get Pf=0.064
Psi.
The total equivalent length T = (1.5 +10) = 11.5 ft
The pressure drop Pf= 11.5x 0.064= 0.74 ft
The total pressure drop Pt = 23.12+ 0.65+ 0.74 =24.51 psi.
Branch Line “ K” calculation: Imagine a huge sprinkler installed at nod B
B
having K= 16.68.This “ K”
represents the total flow of sprinklers 5 thru 8 @ node B
B
with an orifice capable of discharging the
total flow for 4 sprinklers at nodes 5 thru 8.
Q=K .√P => K=
84.8
24.51
= 17.1.
For Pt = 24.51+0.07=24.58 Psi and K=17.1 the flow in branch B-C, Q= 17.1. 24.58 =84.77 gpm.
Figure-10- Flow distribution.
-Pressure drop form D to E: L=5+10+10+10+10+10=55 ft
-Fittings = 15 ft , 1 tee 3’ ’ .
-L effective = 70 ft .
-Pressure drop = 70 x 0.062 = 4.32 ft.
-We add all the losses from Point E to
-We then add the fire hose flow rate
-The total loss is the sum of losses in the longest path from the infl
furthest sprinkler: n=1, the total losses and the total flow rate are the
appropriate pump.
l exit pump node.
d to the total flow rate needed for the sprinkler system.
inflow point:” pump node” to the
results needed for selecting an appropriate pump.