Computation.docx

Hydraulic Calculations for CAFA Given: Discharge density = 0.1 GPM/ft2 Protection area = 1,500 ft2 Protection area per sprinkler = 225 ft2 Sprinkler spacing = 15 ft by 15 ft K‐Factor = 5.6 GPM/psi1/2 Considering the sprinkler system of CAFA arrange using conventional tree system configuration Calculate the total number of sprinklers and number of sprinkler per branch line at the hydraulically most demanding area of operation. 𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑝𝑟𝑖𝑛𝑘𝑙𝑒𝑟𝑠 =

1,500 𝑓𝑡 2 = 7 𝑠𝑝𝑟𝑖𝑛𝑘𝑙𝑒𝑟𝑠 225 𝑓𝑡 2

1.2√1,500 𝑓𝑡 2 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑝𝑟𝑖𝑛𝑘𝑙𝑒𝑟 𝑝𝑒𝑟 𝑏𝑟𝑎𝑛𝑐ℎ 𝑙𝑖𝑛𝑒 = = 3.1 or 4 sprinklers 15 𝑓𝑡 Note: Consider only 3 sprinklers per branch line (referring to the attached Appendix drawing) Step 1. At sprinkler head no. 1 Flow of the most remote sprinkler: 𝑄1 = 0.10

𝐺𝑃𝑀 𝑓𝑡 2

(225 𝑓𝑡 2 )

𝑄1 = 22.5 GPM Pressure to obtain a flow of 22.5 GPM: 𝑄 = 𝐾√𝑃 𝑃1 = 𝑃1 =

𝑄2 𝐾2 (22.5 )2 (5.6) 2

= 16.14 𝑝𝑠𝑖

Step 2: Moving towards sprinkler head no. 2 Friction loss from sprinkler head no. 1 to sprinkler head no. 2: Using Hazen/Williams equation, 𝑓=

4.52𝑄1.85 𝑝𝑠𝑖 ( ) 𝐶 1.85 𝑑 4.87 𝑓𝑡

Where: Q = 22.5 GPM d = 1’’ C = 120 (Galvanized Iron)

𝑓=

4.52(22.5)1.85 𝑝𝑠𝑖 = 0.20426 1.85 4.87 (120) (1) 𝑓𝑡

Pressure loss due to friction @ actual length (L) = 15 ft from sprinkler head no. 1 to 2: 𝑃𝑓=𝐿 𝑥 𝑓 𝑃𝑓=15 𝑥 0.20426

𝑃𝑓= 3.064 psi The pressure loss in this segment of pipe due to friction is added to the pressure required to deliver minimum flow to the end sprinkler head. Pressure required at sprinkler head no. 2: P2 = 16.14 + 3.064 P2 = 19.204 psi Expected flow at sprinkler head no. 2: 𝑄 = 𝐾√𝑃 𝑄 = 5.6√19.204 Q2 = 24.541 GPM Note the increase in flow at sprinkler head no. 2. The pipe feeding sprinkler no. 2 is carrying the combined flow of 22.5 GPM to sprinkler head no. 1 and 24.541 GPM to sprinkler head no. 2. Total flow from sprinkler head no. 2 towards sprinkler no. 1: Qt = 22.5 + 24.541 Qt = 47.041 GPM Step 3: Moving towards sprinkler head no. 3 Friction loss from sprinkler head no. 2 to sprinkler head no. 3: Using Hazen/Williams equation, 𝑓=

4.52𝑄1.85 𝑝𝑠𝑖 ( ) 𝐶 1.85 𝑑 4.87 𝑓𝑡

Where: Q = 47.041 GPM d = 1’’ C = 120 (Galvanized Iron)

𝑓=

4.52(47.041)1.85 𝑝𝑠𝑖 = 0.7993 1.85 4.87 (120) (1) 𝑓𝑡

Pressure loss due to friction @ actual length (L) = 15 ft from sprinkler head no. 2 to 3: 𝑃𝑓=𝐿 𝑥 𝑓 𝑃𝑓=15 𝑥 0.7993 𝑃𝑓 = 11.99 psi The pressure loss in this segment of pipe due to friction is added to the pressure required to deliver minimum flow to the end sprinkler head. Pressure required at sprinkler head no. 3: P3 = 19.204 + 11.99 P3 = 31.194 psi Expected flow at sprinkler head no. 3: 𝑄 = 𝐾√𝑃 𝑄 = 5.6√31.194 Q3 = 31.277 GPM Note the increase in flow at sprinkler head no. 3. The pipe feeding sprinkler no. 3 is carrying the combined flow of 47.041 GPM to sprinkler head no. 1 and 2 and 31.277 GPM to sprinkler head no. 3. Total flow from sprinkler head no. 3 towards sprinkler no. 1 and 2: Qt = 47.041 + 31.227

Qt = 78.268 GPM Step 4: Moving towards A1 Friction loss from sprinkler head no. 2 point A1: Using Hazen/Williams equation, 𝑓=

4.52𝑄1.85 𝑝𝑠𝑖 ( ) 𝐶 1.85 𝑑 4.87 𝑓𝑡

Where: Q = 78.268 GPM d = 1 1/4’’ C = 120 (Galvanized Iron) 4.52(78.268)1.85 𝑝𝑠𝑖 𝑓= = 0.6916 1.85 4.87 (120) (1.25) 𝑓𝑡 Pressure loss due to friction @ actual length (L) = 5.63 ft from sprinkler head no. 3 to point A1: 𝑃𝑓=𝐿 𝑥 𝑓 𝑃𝑓= 5.63 𝑥 0.6916 𝑃𝑓 = 3.894 psi Pressure required at point “A1”: PA1 = 31.194 + 3.894 PA1 = 35.088 psi Since there is no flow at point “A1”: QA1 = 0 GPM Qt = 78.268 + 0 = 78.268 GPM Step 5: At Elbow @ point “A1” Friction loss at elbow is the same with friction loss at sprinkler head no. 3 to point “A1”:

𝑓 = 0.6916

𝑝𝑠𝑖 𝑓𝑡

1

Equivalent length of1 4” 90o long turn elbow: Le = 2 ft C value multiplier for C = 120; multiplying factor = 1.011 (interpolated) LE = 1.011(2) = 2.022 ft Pressure loss due to friction @ equivalent length (LE) = 2.022 ft:

Pƒ = 0.6916 (2.022) = 1. 398 psi Pressure required at elbow: Pat elbow = 1. 398 + 35.088 = 36.486 psi Step 6: Moving towards point “A2” Friction loss at point “A2” is the same with friction loss at sprinkler head no. 3 to point “A1”: 𝑓 = 0.6916

𝑝𝑠𝑖 𝑓𝑡

Pressure loss due to friction @ actual length (L) = 1.64ft from elbow at point “A1” to point “A2”: Pƒ = ƒ x L = 0. 6916 (1.64) Pƒ = 1.134 psi Pressure loss due to change in elevation @ H = 1.64ft: Since pressure loss due to change in elevation is 0.433 psi/ft, Pe = 0.433(1.64) = 0.71 psi Pressure required at point “A2”: PA2 = 36.486 + 1.134 + 0.71 PA2 = 38. 33 psi Determining the K‐factor for this demand will be helpful since this is a significant point in the system and the K‐factor will use in later calculation process. 𝐾𝐴2 =

𝑄𝐴2 √𝑃𝐴2

=

Step 7: Moving towards point “B” Friction loss from point “A2” to point “B”:

78.268 √38.33

= 12.64

𝐺𝑃𝑀 1

𝑝𝑠𝑖 2

Using Hazen/Williams equation,

𝑓=

4.52𝑄1.85 𝑝𝑠𝑖 ( ) 𝐶 1.85 𝑑 4.87 𝑓𝑡

Where: Q = 78.268 GPM d = 2’’ C = 120 (Galvanized Iron)

𝑓=

4.52(78.268)1.85 𝑝𝑠𝑖 = 0.07 1.85 4.87 (120) (2) 𝑓𝑡

Pressure loss due to friction @ actual length (L) = 15 ft: Pƒ = 0.07(15) = 1.05 psi Pressure required at Point “B”: PB = 1.05 + 38. 33 PB = 39.38 psi Step 8: The next step is to determine the flow and pressure at sprinkler head no.4. Again, minimum flow and pressure are established at this point. Examining the system, note that sprinkler head no. 4 through 6 are laid out exactly the same as heads no. 1 through 3. The procedure for calculating heads 4 through 6 is identical and will results in exactly the same as results in heads 1 through 3. The demand at reference point “B” is again 78.268 GPM and 38. 33 psi. This is the second time reference point “B” is reached in the calculation process. There are two calculated pressures at point “B”. However, only one can exist. The most demanding pressure is the

greater of the two or 39.38 psi. Since the flow and pressure demands are the same, it stands to reason that: KB = KA2 Qnew = KA2 √𝑃 Qnew = 12.64√39.38 = 79.32 GPM (adjusted flow at point "B") The total flow leaving at point “B”: QB = 78.268 + 79. 32 = 157.588 GPM Step 9: Moving towards point “C” Friction loss from point “B” to point “C”: Using Hazen/Williams equation,

𝑓=

4.52𝑄1.85 𝑝𝑠𝑖 ( ) 𝐶 1.85 𝑑 4.87 𝑓𝑡

Where: Q = 157.588 GPM d = 2’’ C = 120 (Galvanized Iron)

𝑓=

4.52(157.588 )1.85 𝑝𝑠𝑖 = 0.256 1.85 4.87 (120) (2) 𝑓𝑡

Pressure loss due to friction @ actual length (L) = 15 ft: Pƒ = 0.256(15) = 3.84 psi Pressure required at Point “C”:

PC = 3.84 + 39.38 PC = 43.22 psi Step 10: Again, the next step is to determine the flow and pressure at sprinkler head no. 7. The minimum flow and pressure are established at this point. Examining the system, note that sprinkler head no. 7 through 9 are laid out exactly the same as heads no. 1 through 3. The procedure for calculating heads 7 through 9 is identical and will results in exactly the same as results in heads 1 through 3. The demand at reference point “C” is again 78.268 GPM and 38.33 psi. This is the second time reference point “C” is reached in the calculation process. There are two calculated pressures at point “C”. However, only one can exist. The most demanding pressure is the greater of the two or 43.22 psi. Since the flow and pressure demands are the same, it stands to reason that: KC = KA2 Qnew = KA2 √𝑃 Qnew = 12.64√43.22 = 83.1 GPM (adjusted flow at point "C") The total flow leaving at point “C”: QB = 157.588 + 83.1 QB = 240.69 GPM

Step 10: At sprinkler head no. 7 𝑄1 = 0.10

𝐺𝑃𝑀 (225 𝑓𝑡 2 ) 𝑓𝑡 2

𝑄1 = 22.5 GPM Pressure to obtain a flow of 22.5 GPM: 𝑄 = 𝐾√𝑃 𝑃1 = 𝑃1 =

𝑄2 𝐾2 (22.5 )2 (5.6) 2

= 16.14 𝑝𝑠𝑖

Moving towards point “C” Actual length (include fittings) = 5.63 + 1.64 +2(1.011) = 9.29 ft Lele = 1.64 ft Friction loss from sprinkler head no. 7 to point “C”: Using Hazen/Williams equation, 4.52𝑄1.85 𝑝𝑠𝑖 𝑓 = 1.85 4.87 ( ) 𝐶 𝑑 𝑓𝑡 Where: Q = 22.5 GPM d = 1 1/4’’ C = 120 (Galvanized Iron)

𝑓=

4.52(22.5 )1.85 𝑝𝑠𝑖 = 0.069 1.85 4.87 (120) (1.25) 𝑓𝑡

Pressure loss due to friction @ actual length (L) = 9.29 ft from sprinkler head no. 7 to point C: 𝑃𝑓=𝐿 𝑥 𝑓 𝑃𝑓= 9.29 𝑥 0.069

𝑃𝑓 = 0.641 psi Pressure loss due to elevation @ H = 1.64ft: Pe = 0.433(1.64) = 0.71 psi Pressure required at point “C”: PC = 16.14 + 0.641 + 0.71 PC = 17.491 psi K‐Factor at point “C”: 𝐾𝐶 =

𝑄𝐶 √𝐶

=

22.5 √17.491

= 5.37

𝐺𝑃𝑀 1

𝑝𝑠𝑖 2

This is the second time reference point “C” is reached in the calculation process. There are two calculated pressures at point “C”. However, only one can exist. The most demanding pressure is the greater of the two or 43.22 psi. Since the flow and pressure demands are the same, it stands to reason that: Qnew = KC √𝑃 Qnew = 5.37 √43.22 = 35. 3 GPM (adjusted flow at point "C") The total flow leaving at point “C”: QC = 157.588 + 35. 3 QC = 191.89 GPM Step 11: Moving towards point “I” (base of the riser) Friction loss from point “C” to point “I”: Using Hazen/Williams equation,

𝑓=

Where:

4.52𝑄1.85 𝑝𝑠𝑖 ( ) 𝐶 1.85 𝑑 4.87 𝑓𝑡

Q = 191.89 GPM d = 6’’ C = 120 (Galvanized Iron)

𝑓=

4.52(191.89 )1.85 𝑝𝑠𝑖 = 0.00175 1.85 4.87 (120) (6) 𝑓𝑡

Actual length (include fittings) = 216.5 + 22.5 + 1.011(2(32) + 3 + 14) = 320 ft Lele = 22.5 ft Pressure loss due to friction @ LT = 168 ft: Pƒ = 0.00175 (320) = 0.56 psi Pressure loss due to change in elevation @ H = 22.71ft Pe = 0.433(22.5) = 9.7425 psi Pressure required at point “I” (base of the riser): PI = 0.56 + 9.7425 + 43.22 PI = 53.52 psi

Additional residual pressure of 15 psi, thus: PI = 53.52 + 15 P1 = 68.32 psi The demand at the base of the riser is 200 GPM and 70 psi