Fully Calculated Method for Fire Sprinkler System

Appendix F (Fully Calculated Method) The same fire sprinkler layout is evaluated using the pipe size already determined. In “fully calculated system”, the hydraulically most unfavorable and most favorable locations are attempted. Since the layout plan has false ceiling, there will be upper sprinklers in the ceiling void and sprinklers under false ceiling. We only consider the false ceiling sprinklers (the lower ones) in fully calculated method since when a fire occurs, the closer sprinklers which operate first to put off the fire would be the lower ones. Area served by 1 sprinkler = 3.6x3 = 10.8 m2 Area of Operation (for OH1) = 72m2 Number of sprinklers in operation = 72/10.8 = 6.7 (round off to 7 sprinklers) In the layout, the area of operation where pressure requirement is minimal is hydraulically most favorable whereas the area of operation where pressure requirement is the highest is hydraulically most unfavorable. 4 sprinklers in a square are considered in study. In the most unfavorable region, the last 4 sprinklers in a square are considered. Similarly, in the most favorable region, the most remote 4 sprinklers are considered as shown on layout. Most Unfavorable Area of Operation The last range for the hydraulically most unfavorable location is considered.

Figure 1. The most unfavorable area of operation is shown

Figure 2. Elevation of the left side of the last range

Figure 3. Elevation of the right side of the last range (I)

Calculation of last range

At Sprinkler A Design Density = 5mm/min (as a starting point) for OH1 Area served = 10.8m2 Discharge rate = 10.8 x 5 = 54 L/min Using equation for sprinkler head (k =80 for 15mm orifice) 54 = 80 √𝑃 P(A’)=0.46 bar (54L/min) at Sprinkler A Using Hazen Williams formula (C=120), Equivalent length of a 90 degree screwed elbow, for d=25mm pipe =0.77m from Table 23 of LPC Rules,

Vertical length = 0.5 m 6.05𝑥105 𝑃(𝐴 − 𝐴) = 𝑥 (0.77 + 0.5)𝑥 541.85 1201.85 𝑥254.87 ′

p=0.03bar Pressure loss from P(A’) to P(A) = 0.03 bar P(A) = 0.46+ 0.03 =0.49 bar at 54 L/min At Sprinkler C P(C) = P(A) + P(C-A); range pipe length from C-A=3.6m; equivalent length of a tee in 25mm pipe=1.5m (from Table 23 of LPC Rules) 6.05𝑥105 𝑃(𝐶 − 𝐴) = 𝑥 (5.1)𝑥 541.85 1.85 4.87 120 𝑥25 P(C-A) = 0.11bar P(C) = 0.49 +0.11 = 0.6bar P(C’) = P(C) – P(C-C’) ;vertical distance =0.5m, equivalent length of a tee for 25 mm pipe = 1.5m (from Table 23 of LPC Rules, we use the smaller diameter of the pipe to determine selection of equivalent length of a proper fitting when there is a change in diameter along the pipework) 6.05𝑥105 𝑃(𝐶 − 𝐶′) = 𝑥 (2)𝑥 541.85 1.85 4.87 120 𝑥25 P(C-C’) = 0.04bar P(C’) = 0.6 – 0.04 = 0.56bar Using equation for sprinkler head (k=80 for 15mm orifice) 𝑄(𝐶) = 𝑘 √𝑃 = 80√0.56 =60 L/min which the flow rate of sprinkler C At Distribution Pipe at P (E) P(E ) = P(C) + P(E-C) @ 54+60 =114L/min; range pipe length for 32mm pipe =0.6m, range pipe length for 40mm pipe= 0.9m , equivalent length of a tee for 32mm pipe = 2.1m

𝑃(𝐸 − 𝐶) =

6.05𝑥105 6.05𝑥105 1.85 (2.7)𝑥 𝑥 114 + 𝑥 (0.9)𝑥 1141.85 1201.85 𝑥324.87 1201.85 𝑥404.87

P(E-C) =0.07+0.008 = 0.08bar P(E ) = 0.6 + 0.08 = 0.68bar Consider the other side range pipe Since we only considered false ceiling sprinklers and only 7 of the sprinklers are in operation, upper sprinkler in the ceiling void (sprinkler F) shall not be included in the case. Using the same approach taking a first attempt, and upper sprinkler F is not considered, Sprinkler G is discharging at 5mm/min Sprinkler G discharge = 54 L/min Using equation for sprinkler head (k =80 for 15mm orifice) 54 = 80 √𝑃 P(G’)=0.46 bar (54L/min) at Sprinkler A P(G) = P(G’) + P(G-G’) Using Hazen Williams formula, 6.05𝑥105 𝑝(𝐺′ − 𝐺) = 𝑥 (2)𝑥 541.85 1.85 4.87 120 𝑥25 where 1.5 is the equivalent length of a tee for 25mm pipe (from Table 23 of LPC Rules) & vertical distance =0.5m P(G’-G)=0.04bar P(G) =0.46+ 0.04=0.5 bar at 54 L/min At Sprinkler J P(J) = P(G) + P(J-G) ; range pipe length (dia 25mm) =3m, range pipe length (dia 32mm) = 0.6m, equivalent length of a tee for a 25mm pipe = 2.1m 6.05𝑥105 6.05𝑥105 1.85 𝑃(𝐽 − 𝐺) = 𝑥 (3 + 2.1)𝑥 54 + 𝑥 (0.6)𝑥 541.85 1201.85 𝑥254.87 1201.85 𝑥324.87 P(J-G) = 0.11+0.004bar =0.11bar P(J) = 0.5 +0.11 = 0.61bar P(J’) = P(J) – P(J-J’) ;Equivalent length of Tee for 32mm pipe = 2.1m; vertical distance: 0.5m 𝑃(𝐽 − 𝐽′) =

6.05𝑥105 𝑥 (2.6)𝑥 541.85 1201.85 𝑥324.87

P(J-J’) = 0.02bar P(J’) = 0.61 – 0.02 = 0.59bar Using equation for sprinkler head (k=80 for 15mm orifice) 𝑄(𝐺) = 𝑘 √𝑃 = 80√0.59 =61 L/min which the flow rate of sprinkler J At Distribution Pipe at P(E) P(E ) = P(J) + P(E-J) @ 54+61 =115L/min ; range pipe length = 2.1m 6.05𝑥105 𝑃(𝐸 − 𝐽) = 𝑥 (2.1)𝑥 1151.85 1.85 4.87 120 𝑥40 P(E-D) =0.02bar P(E ) = 0.61 + 0.02 = 0.63bar Balancing of the two ranges Two ranges give different head requirements as 0.68bar and 0.63bar at the same point E. Thus a reattempt of the calculation is required such that both ranges will give the same P(E) of 0.68 bar as higher should be selected. For simplicity, it is possible to assume the relationship ∝ 𝑄 2 , thus 0.63bar @ 115Lmin (original) can be projected to 0.68bar @ Qrev L/min 0.63 115 2 =( ) 0.68 𝑄𝑟𝑒𝑣 𝑄𝑟𝑒𝑣 = 119

𝐿 @ 0.68𝑏𝑎𝑟 𝑚𝑖𝑛

(for pipe range containing sprinkler G, J) Thus, P(E) = 0.68bar Q(E ) = 114+115 = 229 L/min (II)

Calculation for the 2nd last range

Pressure at distribution pipe =P(E) +head loss from E to the 2nd last range(Dia50mm,3m run, 229L/min)

6.05𝑥105 𝑃(𝑓𝑟𝑜𝑚 𝐸 𝑡𝑜 𝑡ℎ𝑒 2𝑛𝑑 𝑙𝑎𝑠𝑡 𝑟𝑎𝑛𝑔𝑒) = 𝑥 (3)𝑥 2291.85 1201.85 𝑥504.87 P=0.03bar Pressure at distribution pipe=0.68bar + 0.03bar=0.71bar In the 2nd last range, two false ceiling sprinklers are on the left, while one false ceiling sprinkler is on the right in operation. Left range: 0.68bar @ 114 L/min (from previous calculation) Right range: 0.48bar @ 54 L/min(calculation elaborated below) The right sprinkler discharges = 54L/min, Using equation for sprinkler head (k =80 for 15mm orifice) 54 = 80 √𝑃 P(sprinkler) = 0.46 bar (54L/min) at that Sprinkler Vertical distance : 0.5m , equivalent length of a tee for a 32mm pipe = 2.4m , range pipe length for a 40mm pipe = 2.1m P(right range) = 0.46 + P(from sprinkler to distribution pipe); range pipe length (dia 25mm) =3m, range pipe length (dia 32mm) = 0.6m, equivalent length of a tee for a 25mm pipe = 2.1m 𝑃(𝑓𝑟𝑜𝑚 𝑆𝑝𝑟. 𝑡𝑜 𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑝𝑖𝑝𝑒) 6.05𝑥105 6.05𝑥105 1.85 = 𝑥 (2.1)𝑥 54 + 𝑥 (2.4 + 0.5)𝑥 541.85 1201.85 𝑥404.87 1201.85 𝑥324.87 P= 0.005+0.019bar =0.024bar P(right range)= 0.46 + 0.024 = 0.48bar Balancing of the two ranges Using 𝑃 ∝ 𝑄 2 0.48 54 2 =( ) 0.68 𝑄𝑟𝑒𝑣 𝑄𝑟𝑒𝑣 = 64.3

𝐿 @ 0.68 𝑏𝑎𝑟 𝑚𝑖𝑛

Thus , flow at distribution pipe : 114 +64.3 = 178.3 L/min @ 0.68bar Using 𝑃 ∝ 𝑄 2 to balance out the pressures from the distribution pipe and last ranges

0.68 178.3 2 =( ) 0.71 𝑄𝑟𝑒𝑣 𝑄𝑟𝑒𝑣 = 182

𝐿 @ 0.71 𝑏𝑎𝑟 𝑚𝑖𝑛

Thus , flow at distribution pipe : 229+182 = 411 L/min @ 0.71bar Pressure at Subsidiary Stop Valve The pressure at the subsidiary stop valve is calculated based on simple hydraulic calculation (@411L/min) Head loss of various pipe sizes at 411 L/min pipe size(mm) Straight length(m)

65 100 total 3 38.2 3.8 3+6x6.1=39.6 Fitting hydraulic length(m) 1 cross 1 elbow+6 tee/cross total hydraulic length (m) 6.8 77.8 hydraulic loss (bar) by Hazen Williams' formula 0.06 0.08 0.14 Hence, the pressure at the outlet of the subsidiary stop valve Pvalve = 0.71 +0.14 = 0.85bar @ 411L/min Using the same method as in pre-calculation installations, the pressure loss for the riser ( from the subsidiary zone valve to the installation alarm valve) is calculated at the flow of 411L/min. Since typical office layout is applied for installation No. 2-4 under OH1, we shall use the above full calculated results to be applied for installation no. 2-4 only. (steps similar to Pre-calculation) I)

Installation No. 2,3,4 (OH1) At 411 L/min, pressure at installation valve = 0.85 bar + ps Where ps =0mH (0bar) assume height of subsidiary stop valve at same level as highest sprinkler head  Installation No.2 (from 5/F-16/F) (a) Static head from highest subsidiary stop valve 16/F to G/F installation valve = 4+3.5x4 + 3.6 x (16-5+1) = 61.2m = 6.1bar With the use of 100mm pipe, (b) Pipe friction loss for vertical rise with use of Hazen Williams Formula With the use of 100mm pipe,

411L/min, 100mm in diameter, 61.2m length = 6.05x105 p= x 61.2 x 4111.85 = 0.065bar 1201.85 x1004.87 At 411 L/min, pressure at installation valve = (0.85 +6.1+0.065) bar = 7.02bar  Installation No.3 (from 17/F-27/F & Refuge/F) (a) Static head from highest subsidiary stop valve 27/F to G/F installation valve = 4+3.5x4 + 3.6 x (27-5+1)+3.6 = 104.4m = 10.4bar With the use of 100mm pipe, (b) Pipe friction loss for vertical rise with use of Hazen Williams Formula With the use of 100mm pipe, 411L/min, 100mm in diameter, 104.4m length = p=

6.05x105 x 104.4 x 4111.85 = 0.11bar 1201.85 x1004.87

At 411 L/min, pressure at installation valve = (0.85 +10.4+0.11) bar = 11.36bar  Installation No.4 (from 28/F-38/F & R/F) (a) Static head from highest subsidiary stop valve R/F to G/F installation valve = 4+3.5x4 + 3.6 x (27-5+1)+3.6 + 3.6x(38-28+1) +4 = 148m = 14.8bar With the use of 100mm pipe , (b) Pipe friction loss for vertical rise with use of Hazen Williams Formula With the use of 100mm pipe, 411L/min, 100mm in diameter, 148m length = 𝑝=

6.05𝑥105 𝑥 148 𝑥 4111.85 = 0.16𝑏𝑎𝑟 1201.85 𝑥1004.87

At 411 L/min, pressure at installation valve = (0.85 +14.8+0.16) bar = 15.81bar

Summary of System Pressure and Flow at Installation Valve by using Fully Calculated Method System pressure and flow at installation valve ("C gauge") (bar)

OH1 ,411 L/min

Installation no.2 (5/F-16/F) Installation no.3 (17/F-27/F& Refuge/F) Installation no.4 (28/F-38/F& R/F)

7.0 11.4 15.8

The fully calculated method verifies that the system pressure and flow at installation valve calculated with pre-calculation is feasible for this sprinkler system. 411L/min is higher than 375 L/min low flow due to the attempt we made that the furthest sprinkler discharging 5 mm/min while other 3 sprinklers in the 4 sprinkler group are having much higher discharge flow rate. The discharge of furthest sprinkler can be reduced as long as the average discharge density of the 4 sprinkler group can be maintained to 5mm/min.

Most Favorable Area of Operation 1st range

2nd range

3rd range

Figure 4. The most favorable area of operation is shown

(I)

Calculation for the first range

Figure 5. Elevation of 1st range in the most favorable area of operation At Sprinkler A Design Density = 5mm/min (as a starting point) for OH1 Area served = 3.6x3= 10.8m2 Discharge rate = 10.8 x 5 = 54 L/min Using equation for sprinkler head (k =80 for 15mm orifice) 54 = 80 √𝑃 P(A’)=0.46 bar (54L/min) at Sprinkler A Using Hazen Williams formula (C=120), for d=25mm pipe , equivalent length of a tee =1.5m from Table 23 of LPC Rules, Vertical length = 0.5 m; 6.05𝑥105 𝑃(𝐴 − 𝐴) = 𝑥 (1.5 + 0.5)𝑥 541.85 1.85 4.87 120 𝑥25 ′

p=0.04bar P(A) = 0.46+ 0.04 =0.5 bar at 54 L/min Distribution pipe P(B) Pressure at distribution pipe P(B) = P(A) + p(B-A) horizontal range pipe = 0.3m 𝑃(𝐵 − 𝐴) =

6.05𝑥105 𝑥 (0.3)𝑥 541.85 1201.85 𝑥254.87

𝑃(𝐵 − 𝐴) = 0.006𝑏𝑎𝑟 P(B) = 0.5+ 0.006 =0.51 bar at 54 L/min Pressure at distribution pipe P(C) =P(B) + P(C-B)(Dia65mm,3m run, 54L/min) 6.05𝑥105 𝑃(𝐶 − 𝐵) = 𝑥 (3)𝑥 541.85 1.85 4.87 120 𝑥65 P=0.0006bar Pressure at distribution pipe P(C) =0.51bar + 0.0006bar≈0.51bar (II)

Calculation for the second range

Figure 6. Elevation of left side of the 2st range in the most favorable region

Figure 7. Elevation of right side of the 2st range in the most favorable region

At Sprinkler D Design Density = 5mm/min Area served = 10.8m2 Discharge rate = 10.8 x 5 = 54 L/min Using equation for sprinkler head (k =80 for 15mm orifice) 54 = 80 √𝑃 P(D’)=0.46 bar (54L/min) at Sprinkler D Using Hazen Williams formula (C=120), Equivalent length of a 90 degree screwed elbow, for d=25mm pipe =0.77m from Table 23 of LPC Rules; Vertical length = 0.5 m 𝑃(𝐷′ − 𝐷) =

6.05𝑥105 𝑥 (0.77 + 0.5)𝑥 541.85 1201.85 𝑥254.87

p=0.03bar Pressure loss from P(D’) to P(D) = 0.03 bar P(D) = 0.46+ 0.03 =0.49 bar at 54 L/min At Sprinkler F P(F) = P(D) + P(F-D); range pipe length from F-D=3.6m; equivalent length of a tee in 25mm pipe=1.5m (from Table 23 of LPC Rules) 6.05𝑥105 𝑃(𝐹 − 𝐷) = 𝑥 (5.1)𝑥 541.85 1201.85 𝑥254.87 P(F-D) = 0.11bar P(F) = 0.49 +0.11 = 0.6bar P(F’) = P(F) – P(F-F’) ;vertical distance =0.5m, equivalent length of a tee for 25 mm pipe = 1.5m 𝑃(𝐹 − 𝐹′) =

6.05𝑥105 𝑥 (2)𝑥 541.85 1201.85 𝑥254.87

P(F-F’) = 0.04bar P(F’) = 0.6 – 0.04 = 0.56bar Using equation for sprinkler head (k=80 for 15mm orifice) 𝑄(𝐹) = 𝑘 √𝑃

= 80√0.56 =60 L/min which the flow rate of sprinkler F At Distribution Pipe at P (C) P(C ) = P(F) + P(C-F) @ 54+60 =114L/min; range pipe length for 25mm pipe =0.3m, 𝑃(𝐶 − 𝐹) =

6.05𝑥105 𝑥 (0.3)𝑥 1141.85 1201.85 𝑥254.87

P(C-F) =0.03bar P(C ) = 0.6 + 0.03 = 0.63bar Consider the other side range pipe Since we only considered false ceiling sprinklers and only 7 of the sprinklers are in operation, upper sprinkler in the ceiling void (sprinkler F) shall not be included in the case. Using the same approach above, upper sprinkler H is not considered, Sprinkler J is discharging at 5mm/min Sprinkler J discharge = 54 L/min Using equation for sprinkler head (k =80 for 15mm orifice) 54 = 80 √𝑃 P (J’) =0.46 bar (54L/min) at Sprinkler A P (J) = P(J’) + P(J-J’) ; equivalent length of 2 elbows and 1 tee = 2x0.77+1.5 = 3.04m (from table 23 of LPC rules) ; horizontal length = 0.5m ; vertical length =0.5m 𝑝(𝐽 − 𝐽′) =

6.05𝑥105 𝑥 (4.04)𝑥 541.85 1201.85 𝑥254.87 P(J-J’)=0.09bar

P (J) =0.46+ 0.09=0.55 bar at 54 L/min At Distribution Pipe at P(C) P(C) = P(J) + P(C-J) @ 54 L/min ; range pipe length = 2.8m; equivalent length of a tee=1.5m (from table 23 of LPC rules) 6.05𝑥105 𝑃(𝐶 − 𝐽) = 𝑥 (4.3)𝑥 541.85 1201.85 𝑥254.87 P(C-J) =0.09bar

P(C) = 0.55 + 0.09 = 0.64bar Balancing of the two ranges Two ranges give different head requirements as 0.64bar and 0.63bar at the same point C. Thus a reattempt of the calculation is required such that both ranges will give the same P(C) of 0.64 bar as higher should be selected. For simplicity, it is possible to assume the relationship ∝ 𝑄 2 , thus 0.63bar @ 114Lmin (original) can be projected to 0.64bar @ Qrev L/min 0.63 114 2 =( ) 0.64 𝑄𝑟𝑒𝑣 𝑄𝑟𝑒𝑣 = 115

𝐿 @ 0.64𝑏𝑎𝑟 𝑚𝑖𝑛

(for pipe range containing sprinkler D&F) Thus, P(C) = 0.64bar Q(C ) = 115+54 = 169 L/min Using 𝑃 ∝ 𝑄 2 From previous calculation, distribution pipe P(C) = 0.51bar @ 54L/min 0.51 54 2 =( ) 0.64 𝑄𝑟𝑒𝑣 𝑄𝑟𝑒𝑣 = 60.5

𝐿 @ 0.64 𝑏𝑎𝑟 𝑚𝑖𝑛

Thus , flow at distribution pipe : 169+60.5 = 230 L/min @ 0.64bar Pressure at distribution pipe =P(C) +head loss from C to the third range in the area of operation (Dia100mm,3m run, 230L/min) 𝑃(𝑓𝑟𝑜𝑚 𝐶 𝑡𝑜 𝑡ℎ𝑒 3𝑛𝑑 𝑟𝑎𝑛𝑔𝑒) =

6.05𝑥105 𝑥 (3)𝑥 2301.85 1201.85 𝑥1004.87

P=0.001bar Pressure at distribution pipe=0.64bar + 0.001bar≈0.64bar

(III)

Calculation for the third range

Same results as the second range, For left range pipe, 0.63bar @114L/min For right range pipe, 0.64bar @ 54L/min Pressure and flow in between both range pipes (distribution pipe) as the following: P(distribution pipe) = 0.64bar Q(distribution pipe) = 114+54+ = 169 L/min Flow at distribution pipe = 169 + 230 = 399 L/min at 0.64bar Most favorable area of operation for 7 sprinklers requires 399 L/min, while most unfavorable area of operation for 7 sprinklers requires 411 L/min. As explained previously, existing uncertainties due to assumption of the further sprinkler discharging at 5mm/min and rounding up digit issue results in the most unfavorable area which should have the smallest amount of water flow having a higher discharge water flow rate than the most favorable area. Pressure at Subsidiary Stop Valve The pressure at the subsidiary stop valve is calculated based on simple hydraulic calculation (@399L/min) Head loss of various pipe sizes at 399 L/min pipe size(mm) Straight length(m) Fitting hydraulic length(m)

100 17.8 3+3x6.1=21.3 1 elbow+3 tee/cross

total hydraulic length (m)

39.1

hydraulic loss (bar) by Hazen Williams' formula

0.04

Hence, the pressure at the outlet of the subsidiary stop valve Pvalve = 0.64 +0.04 = 0.68bar @ 399L/min Using the same method as in pre-calculation installations The pressure loss for the riser (from the subsidiary zone valve to the installation alarm valve) is calculated at the flow of 399L/min. Since typical office layout is applied for installation No. 2-4 under OH1, we shall use the above fully calculated results to be applied for installation no. 2-4 only. (steps similar to Pre-calculation)

II)

Installation No. 2,3,4 (OH1) At 399 L/min, pressure at installation valve = 0.68 bar + ps Where ps =0mH (0bar) assume height of subsidiary stop valve at same level as highest sprinkler head  Installation No.2 (from 5/F-16/F) (a) Static head from lowest subsidiary stop valve 5/F to G/F installation valve = 4+3.5x4 + 3.6 = 21.6m = 2.2bar With the use of 100mm pipe, (b) Pipe friction loss for vertical rise with use of Hazen Williams Formula With the use of 100mm pipe, 399L/min, 100mm in diameter, 21.6m length = 6.05x105 p= x 21.6 x 3991.85 = 0.02bar 1201.85 x1004.87 At 399 L/min, pressure at installation valve = (0.68 +2.2+0.02) bar = 2.9bar  Installation No.3 (from 17/F-27/F & Refuge/F) (a) Static head from lowest subsidiary stop valve 17/F to G/F installation valve = 4+3.5x4 + 3.6 x (17-5+1) = 64.8m = 6.5bar With the use of 100mm pipe, (b) Pipe friction loss for vertical rise with use of Hazen Williams Formula With the use of 100mm pipe, 399L/min, 100mm in diameter, 64.8m length = p=

6.05x105 x 64.8 x 3991.85 = 0.07bar 1201.85 x1004.87

At 399 L/min, pressure at installation valve = (0.68 +6.5+0.07) = 7.3bar

 Installation No.4 (from 28/F-38/F & R/F) (a) Static head from lowest subsidiary stop valve 28/F to G/F installation valve = 4+3.5x4 + 3.6 x (27-5+1)+3.6 + 3.6 = 108m = 10.8bar

With the use of 100mm pipe , (b) Pipe friction loss for vertical rise with use of Hazen Williams Formula With the use of 100mm pipe, 399L/min, 100mm in diameter, 108m length = 𝑝=

6.05𝑥105 𝑥 108 𝑥 3991.85 = 0.11𝑏𝑎𝑟 1201.85 𝑥1004.87

At 399 L/min, pressure at installation valve = (0.68 +10.8+0.11)bar = 11.6bar Summary of the system pressure and flow at installation valve by using fully calculated method System pressure and flow at installation valve ("C gauge") (bar)

OH1 ,399 L/min

Installation no.2 (5/F-16/F) Installation no.3 (17/F-27/F& Refuge/F) Installation no.4 (28/F-38/F& R/F)

2.9 7.3 11.6

Head loss from pump outlet to installation valve ‘C’ gauge For OH1 installation, At 411 L/min, L= 33.5m from pump outlet to installation valve 6.05𝑥105 𝑝= 𝑥 33.5 𝑥 4111.85 1201.85 𝑥1004.87 Head loss from pump outlet to installation valve ‘C’ gauge = 0.04bar

At 399L/min, L= 33.5m from pump outlet to installation valve 6.05𝑥105 𝑝= 𝑥 33.5 𝑥 3991.85 1201.85 𝑥1004.87 Head loss from pump outlet to installation valve ‘C’ gauge = 0.03bar

Summary of pump pressure and flow of Sprinkler pump

Pump Pressure and Flow of Sprinkler Pump (bar) Installation no.2 (5/F-16/F)

OH1 Flow rate Nominal flow 411L/min Obtained after pump 7.0+0.04= selection 7.0

399 L/min 2.9+0.03= 2.9 7.3+0.03= 7.3

Installation no.3 (17/F-27/F& Refuge/F)

Obtained after pump selection

11.4+0.04= 11.4

Installation no.4 (28/F-38/F& R/F)

Obtained after pump selection

15.8+0.04= 11.6+0.03= 15.8 11.6

In practice, if we have these two points (most unfavorable and most favorable at different pressures and different flows), we can select a pump that fits that system and find a corresponding nominal flow data of that pump. If Qmax = 399 L/min, for OH installation, duration of operation is 60 min, then the tank size can be obtained as Qmax x 60 min = tank size (L) 399x 60 = 23940L = 24m3 (fully calculated method) In pre-calculated method, sprinkler water tank = 50 m3 ; hence, Using fully calculated method, we can customize tank size and select a proper pump in a more economical way.