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How to calculate a ៵�re sprinkler system
Category: Hydraulic calculation for ៵�re protection engineers (/Table/Support/Hydraulic-calculation-for-៵�re-protection-engineers/)
The long way - by hand In this article, will demonstrate some of the basics for carrying out 玘�re sprinkler calculations by the long hand method with just the aid of a scienti玘�c calculator or our own hydraulic calculator - Hcal2 (/products/hcalc.html) which you can freely download from our website. We will for this example use simple three sprinklers and three pipes which would of course be part of a much larger 玘�re sprinkler system. These basic procedures can also be used for calculating many other types of systems such as 玘�re hydrant, hose reel or the discharge from a water cannon or monitor we can also use the same principal for almost all other water-based 玘�re protection systems if we have a k-factor for the output device (玘�re sprinkler, water mist nozzle and so on). In this example, will we use a very simple system with just three sprinklers and three pipes this is often called a range pipe or branch pipe, which is part of a larger tree system. A tree system is end feed, that is water is only fed from one direction as opposed to a grid or loop system when water may arrive at the sprinkler head from more than one direction. Below is a diagram of the three sprinklers and three pipes which we will calculate. We have dimensioned the pipe lengths and given each junction point a unique node reference number which we use throughout the calculations.
For each pipe, we need to know the pipe length, internal diameter (ID) of the pipe and the pipe material so we can determine the pipes c-factor, the table below summarises the pipe data which we will need for the calculation for this example:
Node Ref
Pipe Size ID (mm)
Length (m)
C-factor
130-120
27.30 3.20
120
120-110
27.30 3.20
120
110-100
36.00 3.20
120
We will also we will need some additional information such as the type sprinkler head, the area each head is covering, the design density for each sprinkler head in the system. For this example, we will use the following design parameters: design density: 7.50 mm/min sprinkler head: K-factor of 70 with a minimum pressure 0.5 bar head area: 10.20 m2 In this example, we have kept it very simple and used the same sprinkler head for all three sprinklers but this may not always be the case so again it way be useful to summarises the information in a table such as this:
Node Ref
Design Density
Sprinkler k-factor
(mm/min)
Sprinkler minimum
Head area (m2)
pressure (Bar)
130
7.50
70
0.5
10.20
120
7.50
70
0.5
10.20
110
7.5
70
0.5
10.20
The 玘�rst step is to calculate the minimum �ow which will be required at the most remote sprinkler which in this case is at node [130], this is a two-step process as will need to calculate the minimum �ow required to satisfy the 7.50 mm/min design density and then 玘�nd the �ow rate from the sprinkler given the sprinklers minimum pressure requirement, whichever is the greater �ow will become our initial �ow from the 玘�rst sprinkler at node [130]. We will 玘�rst calculate the �ow given the design density of 7.50 mm/min and the area the head is covering, we do this by multiplying the design density (/support-topmenu/support-basichydraulics/54-designdensity.html) by the head area: Equation 1: q1 = (design density) x (area per sprinkler) In this example, this gives: q1 = 7.50 mm/min x 10.20 m2 = 76.50 L/min The second step is to calculate the minimum �ow from the sprinkler given the K-Factor and the minimum head pressure by using the standard K-Factor formula (/support-topmenu/supportbasichydraulics/27-k-factor-formula.html):
Equation 2: q = kp0.5 Where p = the required pressure q = the required �ow from the 玘�rst sprinkler k = the discharge coe㇝�cient of the sprinkler (k-factor) In this example, this gives: q = 70 x 0.50.5 = 49.50 L/min By comparing the two calculations above we can see that the minimum �ow required from the sprinkler head will be 76.50 L/min as this is the highest �ow rate from the two calculations and is required to meet the 7.50 mm/min design density. We can also see that the minimum sprinkler pressure of 0.5 bar is not su㇝�cient to produce the require �ow rate so the next step will be to determine what pressure will be required to produce the require �ow of 76.50 L/min at the 玘�rst sprinkler head at node [130] we can do this by using equation 3. Equation 3 p = (q/k)2 In the example, this gives: p = (76.50 / 70)0.5 = 1.194 bar We have now determined the minimum pressure and �ow for the 玘�rst sprinkler at node [130] which will be 76.50 L/min @ 1.19 bar the next step is to calculate the pressure drop in the pipe between node [130] and [120] and for this we will use the Hazen-Williams pressure loss formula (/support-topmenu/support-basichydraulics/26-the-hazen-williams-formula-for-use-in-៵�re-sprinkler-systems.html). Equation 4
Where p = pressure loss in bar per meter Q = �ow through the pipe in L/min C = friction loss coe㇝�cient d = internal diameter of the pipe in mm We know that the �ow rate from the sprinkler at node [130] is 76.50 L/min and this will be the �ow rate in the 玘�rst pipe between nodes [130]-[120]. As the pipe has an internal diameter of 27.30 mm and has a C value of 120 this will give us:
The pressure loss in the 玘�rst pipe is 0.027 Bar/m and the total pressure loss in the pipe is 0.086 bar. We now need to add the pressure loss in the pipe to the start pressure at the sprinkler head at node [130] which was 1.19 bar to 玘�nd to pressure at node [120] and at the seconded sprinkler head at node [120] this gives us 1.194 + 0.086 = 1.28 bar. The next step is to 玘�nd the �ow from the seconded sprinkler head at node [120] to do this we will use the K-Factor formula Equation 5
This gives 70 x 1.2800.5 = 79.20 L/min from the sprinkler head at node [120] which we now add to the �ow in the 玘�rst pipe node [130]-[120] to 玘�nd the total �ow in the second pipe [120]-[110] to 玘�nd the total �ow in the seconded pipe which is 155.70 L/min.
Having found the total �ow in the seconded pipe [120]-[110] we can now 玘�nd the pressure loss in, to do this we will use the Hazen-Williams pressure loss, formula 4 which we used above this gives us:
We now add the pressure loss 0.317 bar to the pressure at node [120] to 玘�nd the pressure at node [110] this give us: 0.317 + 1.280 = 1.597 bar
We now need to 玘�nd the �ow from the sprinkler at node [110] we do this by using the k-factor given in equation 5 as we now know the pressure at node [110] is 1.597 bar, this gives 70 x 1.5970.5 = 88.50 L/min from the sprinkler head at node [110]. We now add this �ow to the �ow in the seconded pipe [120]-[110] to 玘�nd the total �ow in the third pipe [110]-[100] which will give us the �ow of 244.20 L/min.
The last step is to 玘�nd the pressure loss in the third pipe [110]-[100] and again we will use Hazen-Williams pressure loss formula given is formula 4 above. However, the last pipe has an internal diameter of 36.0 mm so this gives us:
We now add the pressure loss in this pipe to the pressure at node [110] to 玘�nd the pressure at node [100] this will be 0.189 + 1.597 = 1.786 bar. We have now complete the calculation for all three sprinkler heads and have found the source pressure and �ow required for this system is: 244.20 L/min @ 1.786 Bar This pressure and �ow is offer referred to as the source requirement for the system and is the minimum pressure and �ow required for the system for it to be able to provide the required design density (in this example 7.50 mm/min) at the most remote head [MRH] at node [130]. You should also be able to see that only the Most Remote Head has the minimum requirement of 7.50 mm/min design density and all the other sprinklers will have a higher pressure as they are hydraulically closer to the water source so they will have a higher pressure and will discharge more water through the sprinkler this can be seen in the table below:
Node Ref
min Design Density
Pressure
Flow from sprinkler
Head Area
Actual
(mm/min)
(Bar)
(L/min)
(m2)
Design Density
130 [MRH]
7.50
1.194
76.50
10.20
7.50
120
7.50
1.280
79.20
10.20
7.76
130
7.50
1.597
88.50
10.20
8.68
Sprinkler calculation step by step 1. Calculate minimum �ow from the MRH with the sprinkler minimum pressure and k-factor 2. Calculate the minimum �ow given the system design density and sprinkler head area. 3. If the calculation in step 2 is the highest �ow demand, then calculate the required head pressure otherwise we can use the minimum sprinkler pressure in step 1. 4. Calculate the pressure loss in the pipe. 5. Add the head pressure to the pressure loss in step 4 to determine the pressure at the next sprinkler. 6. Use the k-factor formula to determine the �ow from the sprinkler head. 7. Repeat step 4 to 6 until you do not have any more sprinklers or pipes
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