**Fire sprinkler systems hydraulic calculations**

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-1-

The 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 that

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 labeling the sprinklers from the most remote one to the nearest , we calculate the pressure

needed at the lst 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 point4 between the 2 parallel branches. Similarly we proceed with the same steps written above for the mirrored branch5-6-7 and we add all the flow rates coming out from all the sprinklers in both parallel branches.

the system above is hydraulically balanced because of the similarity in the branches. Additionally, 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 branches11-10-9 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:

** **

The problem may appear when 1 of the mirrored branches or parallel branches are not alike as the

following figure below:

-It is our job to balance the system, let us take a look at the result when we apply the same steps done previously.

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 K6-5 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 thats 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:

Q6-5=K6-5/ P4 where this result is final and Q total for the 2 branches is equal to:

Q6-5 + Q1,2,3.

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.

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