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Home My WebLink AboutFire Flow ReportHampton Inn Hotel Fire Flow Analysis t "OD S. AC 4MS 63351 �- ,....,,t':;` • ,af t; Municipal Development Group 2551 Texas Avenue South, Suite A College Station, Texas 77840 Ph: 979.693.5359 Fax: 979.693.4243 Email: mdgcstx @yahoo.com The project consists of developing a 3 -acre lot with a 3 story, 79 room hotel with 1200 sf of conference space on Lot 1, Block 1 of the Greensworld Subdivision, Phase III currently zoned C -2. It is undergoing rezoning to weed UDO requirements. The site is now served by an 18" DIP main, with two existing tees on -site. We intend to utilize these tees with an 8" C900 fire line looped, with 2 -8" laterals for fire and domestic to the hotel. The proposed fire hydrant at the northernmost tee is for additional capacity to the hotel and future development on the site. A fire flow analysis was conducted in conjunction with EPANET for the three proposed fire hydrants at Hampton Inn. Appendix B of the 2000 International Fire Code was used to determine the flow rates and pressures for the proposed hydrants. From Appendix B, the proposed hydrants must have a combined design flow rate of 4750 gpm. However, because the site has a fire sprinkler system with a maximum demand of 80 gpm, the flow rate may have up to a 75% reduction, yielding a combined flow rate of 1187 gpm. Despite the reduction, a minimum combined design flow rate of 1500 gpm will be used, as required in the 2000 International Fire Code. Flow data for the existing fire hydrant number G -033 at Douglas Nissan was obtained from the City of College Station Department of Public Works. The information for this hydrant is as follows: Static Pressure: 100 psi Residual Pressure: 80 psi Flow Rate: 1175 gpm For the initial analysis, a pressure of 80 psi was used. However, according to Appendix B, it must also be determined if the proposed fire hydrants can hold a minimum pressure of 20 psi over a four hour time period. For this analysis, a pressure of 20 psi was also used. From the gathered data on fire hydrant number G -033, the Hydraulic Grade Line (HGL) for the hydrant was determined for each of the two time periods using the equation: HGL = z + P/d Where HGL => Hydraulic Grade Line (ft) Z => Elevation (ft) P => Pressure (psi) d => Density (62.4 Ibs /cfl Knowing HGL for the known hydrant at the two time periods, the HGL for every node in the network at both time periods was obtained using the calculation: HGL = HGL, - KQ Where K => Head loss coefficient Q => Flow rate (cfs) HGL => Downstream HGL (ft) HGL _> Upstream HGL (ft) For this calculation, it is important to start at a known HGL, and move down the network computing the HGL in order from upstream to downstream. However, if the known HGL is located downstream and upstream HGLs need to be determined, the calculation can be reversed to accommodate the situation. The equation for this instance is as follows: HGL = HGL + KQ The head loss coefficient (K) for each pipe section is calculated using: K = (f ,) /(2 *DA Where f => Friction factor L => Length of pipe (ft) D => Diameter of pipe (ft) A => Area of pipe (ft) g => Gravity (32.2 ft/S2) The HGL for each of the three proposed fire hydrants was determined using this method. The pressure at each of the hydrants was determined by rearranging the above equation that solved for the HGL. Knowing the pressures, the flow rates of the hydrants can be determined using the following equation: Q = 29.83cD ^2 *sgrt(P) Where Q => Flow Rate (gpm) c => Friction Coefficient (0.7 -0.9) D => Diameter of Outlet (in) P => Pressure (psi) The proposed hydrants were assumed to have outlet diameters of 2.5 in. Further analysis of the network was also performed using the computer simulation program EPANET. This program can perform time simulations of the system by entering the information for each node and pipe. However, for the program to properly work, the known fire hydrant must be entered as a tank rather than a valve or junction. The demand for the existing fire hydrant, as a tank, must be enough to supply the entire network. To satisfy this requirement, a demand of 4500 gpm was used. Because of this high demand, the program computed excessive velocities for pipes L and M that are faulty to this analysis. After completion of the analysis, it was determined that the three proposed fire hydrants will supply adequate flow rates to the site. Initially, the hydrants can provide a total of 3448.83 gpm, which is well over the minimal requirement of 1500 gpm. After four hours, the hydrants will provide a total of 1666.62 gpm. Taking into account the 80 gpm demand for the sprinkler system, the hydrants, after four hours will still be above the 1500 gpm requirement with a total of 1586.62 gpm. 12 J8 re ' FH3 EPANET 2 Pagel J1 J2 g FH1 Fire Flow Analysis Initial Junction Elevation (ft) 1 258.0 2 258.0 3 257.6 4 258.8 5 256.5 6 257.0 7 257.0 8 257.0 9 253.8 10 254.3 11 256.8 12 256.9 FH1 258.0 FH2 259.0 FH3 257.1 EX FH 254.1 HGL =z +P /d Pipe Length (ft) Dia. (ft) A 6 0.667 B 14 0.500 C 62 0.667 D 264 0.667 E 25 0.500 F 115 0.667 G 126 0.667 H 3 0.667 I 5 0.667 J 16 0.500 K 172 1.50 L 133 0.667 M 37 0.667 N 473 1.50 O 8 0.667 P 50 0.667 HGL => Hydraulic Grade Line (ft) Z => Elevation (ft) P => Pressure (psi) d => Density (Ibs /cf) HGL of EX FH Z = 254.1 ft P = 80 psi d = 62.4 Ibs /cf HGL = 438.72 ft HGL, = HGL - KO` Fire H ydrant Numb G-033 (EX FH) Static Pressure: 100 psi Residual Pressure: 80 psi Flowrate: 1175 gpm 2.618 cfs K => Headloss coefficient Q => Flowrate (cfs) HGL => Downstream HGL (ft) HGL, _> Upstream HGL (ft) Junction 10 Junction 9 K = 0.0919 0.3305 Q = 2.618 cfs HGL = 438.72 ft HGL = 439.35 ft P = 80.19 psi Junction 11 Junction 9 K = K = 0.3305 Q = Q = 2.618 cfs HGL = 439.35 ft HGL = 441.61 ft 1 P= 81.38 psi Junction 11 Friction factor K = 0.0235 D => Q = 2.618 cfs HGL = 441.61 ft HGL = 441.45 ft P = 80.01 psi K = (fl) /( "DA f => Friction factor L => Length of pipe (ft) D => Diameter of pipe (ft) A => Area of pipe (ft) g => Gravity (32.2 ft/9) T = U.U13 L= 6 ft D = 0.667 ft A = 0.349 ft` g = 32.2 ft/s K = 0.0149 f= 0.013 L= 14 ft D = 0.500 ft A = 0.196 11 2 g = 32.2 ft/s` K = 0.1466 T = U.0 "13 L = 62 ft D = 0.667 ft A = 0.349 ft 2 g = 32.2 ft/s K = 0.1541 T = U.U13 L = 264 ft D = 0.667 ft A = 0.349 fl? g = 32.2 ft/s K = 0.6561 Hampton Inn Hotel Project Manager: Lee Adams 8/28/2003 Job # 000756 -3518 Calculations: Mark Taylor Fire Flow Analysis Initial Junction 12 Junction 5 K = 0.0199 0.1243 Q = 2.618 cfs HGL = 441.45 ft HGL = 441.31 ft J P= 79.91 psi K = Junction 5 K = 0.1243 cfs Q = 2.618 cfs HGL = 441.31 ft HGL = 440.46 ft P = 79.72 psi K = Junction 6 K = 0.3131 cfs Q = 2.618 cfs HGL = 440.46 ft HGL = 438.32 ft P = 78.57 psi K = Junction 7 K = 0.0075 cfs Q = 2.618 cfs HGL = 438.32 ft HGL = 438.26 ft P = 78.55 psi K = Junction 8 K = 0.0124 cfs Q = 2.618 cfs HGL = 438.26 ft HGL = 438.18 ft P = 78.51 psi K = 0.1676 Q = 2.618 cfs HGL = 438.26 ft HGL = 437.12 ft P = 78.01 psi Q = 1152.65 gpm t = U.U1J L = 25 It D = 0.500 ft A = 0.196 ft Z g = 32.2 ft/s' K = 0.2618 ft/S T = K F f = 0.013 ft L = 115 ft D = 0.667 ft A = 0.349 ft 2 g = 32.2 ft/S K = 0.2858 T = U.U13 L = 126 ft D = 0.667 ft A = 0.349 ft' g = 32.2 ft/s K = 0.3131 T= U.UIs L= 3 ft D = 0.667 ft A = 0.349 ft Z g = 32.2 ft/s K = 0.0075 f = 0.013 L= 5 ft D = 0.667 ft A = 0.349 ft Z g = 32.2 ft/s K = 0.0124 1 = U.U1J L= 16 ft D = 0.500 ft A = 0.196 ft Z g = 32.2 ft/S K = 0.1676 Hampton Inn Hotel Project Manager: Lee Adams 8/28/2003 Job # 000756 -3518 Calculations: Mark Taylor Fire Flow Analysis Initial �FH1rau Juncti 1 0.1466 K = 0.0086 2.618 Q = 2.618 cfs HGL = 441.45 ft HGL = 441.39 ft P = 79.47 psi �FH1rau Junction 2 0.1466 K = 0.0149 2.618 Q = 2.618 cfs HGL = 441.39 ft HGL = 441.29 ft J P= 79.43 psi �FH1rau K = 0.1466 K = Q = 2.618 cfs HGL = 441.29 ft HGL = 440.28 ft P = 78.99 psi I Q= 1159.89 gpm T= Junction 3 K K = 0.1541 2.618 cfs Q = 2.618 cfs HGL = 441.29 ft HGL = 440.23 ft P = 79.14 psi T= Junction 4 K K = 0.6561 2.618 cfs Q = 2.618 cfs HGL = 440.23 ft HGL2 = 435.74 ft 1 P= 76.67 psi T= U.0 10 K 0.2618 Q = 2.618 cfs HGL = 435.74 ft HGL2 = 433.94 ft P = 75.81 psi I Q= 1136.29 gpm T= U.0 10 L = 172 ft D = 1.50 ft A = 1.767 ft g = 32.2 ft/s K = 0.0086 T= U.UIa L = 133 ft D = 0.667 ft A = 0.349 ft g = 32.2 ft/s K = 0.3305 t = U.U13 L = 37 ft D = 0.667 ft A = 0.349 ft g = 32.2 ft/s K = 0.0919 T = U.u15 L = 473 ft D= 1.50 ft A = 1.767 ft g = 32.2 We K = 0.0235 1= u.uIJ L= 8 ft D = 0.667 ft A = 0.349 ft g = 32.2 ft/s K = 0.0199 T = U.U13 L = 50 ft D = 0.667 ft A = 0.349 ft g = 32.2 ft/s K = 0.1243 Hampton Inn Hotel Project Manager: Lee Adams 8/28/2003 Job # 000756 -3518 Calculations: Mark Taylor Fire Flow Analysis After 4 Hours Junction Elevation (ft) 1 258.0 2 258.0 3 257.6 4 258.8 5 256.5 6 257.0 7 257.0 8 257.0 9 253.8 10 254.3 11 256.8 12 256.9 FH1 258.0 FH2 259.0 FH3 257.1 EX FH 254.1 HGL =z +P /d Pipe Length (ft) Dia. (ft) A 6 0.667 B 14 0.500 C 62 0.667 D 264 0.667 E 25 0.500 F 115 0.667 G 126 0.667 H 3 0.667 1 5 0.667 J 16 0.500 K 172 1.500 L 133 0.667 M 37 0.667 N 473 1.500 O 8 0.667 P 50 0.667 HGL => Hydraulic Grade Line (ft) Z => Elevation (ft) P => Pressure (psi) d => Density (Ibs /cf) HGL = HGL, - KV Fire Hydrant Number G -033 (EX FH) Static Pressure: 100 psi Residual Pressure: 20 psi Fiowrate: 583.64 gpm 1.300 cfs K=> HGL of EX FH Q => Z = 254.1 ft P = 20 psi d = 62.4 Ibs /cf HGL = 300.25 ft HGL = HGL, - KV Fire Hydrant Number G -033 (EX FH) Static Pressure: 100 psi Residual Pressure: 20 psi Fiowrate: 583.64 gpm 1.300 cfs K=> Headloss coefficient Q => Flowrate (cfs) HGL => Downstream HGL (ft) HGL, => Upstream HGL (ft) Junction 10 Junction 9 K = 0.0919 0.3305 Q = 1.300 cfs HGL = 300.25 ft HGL = 300.41 ft P = 19.98 psi Junction 11 Junction 9 K = K = 0.3305 Q = Q = 1.300 cfs HGL _ .300.41 ft HGL = 300.97 ft 1 P= 20.44 psi Junction 11 Friction factor K = 0.0235 D => Q = 1.300 cfs HGL = 300.97 ft HGL = 300.93 ft P = 19.12 psi K = (fL) /(2 "DA f => Friction factor L => Length of pipe (ft) D => Diameter of pipe (ft) A => Area of pipe (ft) g => Gravity (32.2 ftd) t= u.u13 L= 6 ft D = 0.667 ft A = 0.349 W g = 32.2 ft/s K = 0.0149 T = u.ui s L= 14 ft D = 0.500 ft A = 0.196 ft 2 g = 32.2 ft/s` K = 0.1466 T = U.Ul6 L = 62 ft D = 0.667 ft A = 0.349 ft 2 g = 32.2 ft /s' K = 0.1541 f= 0.013 L = 264 ft D = 0.667 ft A = 0.349 ft 2 g = 32.2 ft/s K = 0.6561 Hampton Inn Hotel Project Manager: Lee Adams 8/28/2003 Job # 000756 -3518 Calculations: Mark Taylor Fire Flow Analysis After 4 Hours Junction 12 Junction 5 K = 0.0199 0.1243 Q = 1.300 cfs HGL = 300.93 ft HGL = 300.89 ft P = 19.06 psi K = Junction 5 K = 0.1243 cfs Q = 1.300 cfs HGL = 300.89 ft HGL = 300.68 ft P = 19.15 psi K = Junction 6 K = 0.3131 cfs Q = 1.300 cfs HGL = 300.68 ft HGL = 300.16 ft P = 18.70 psi K = Junction 7 K = 0.0075 cfs Q = 1.300 cfs HGL = 300.16 ft HGL = 300.14 ft P = 18.70 psi K = Junction 8 K = 0.0124 cfs Q = 1.300 cfs HGL = 300.14 ft HGL = 300.12 ft J P= 18.69 psi K = U.lb /D Q = 1.300 cfs HGL = 300.14 ft HGL = 299.86 ft P = 18.53 psi Q = 561.77 apm T= U.0 13 L = 25 ft D = 0.500 ft A = 0.196 ft 2 g = 32.2 ft/s K = 0.2618 f = 0.013 L = 115 ft D = 0.667 ft A = 0.349 ft 2 g = 32.2 ft/s K = 0.2858 t = U.M.5 L = 126 ft D = 0.667 ft A = 0.349 ft 2 g = 32.2 ft/s K = 0.3131 f = 0.013 L= 3 ft D = 0.667 ft A = 0.349 ft 2 g = 32.2 ft/s2 K = 0.0075 t= U.UIJ L= 5 ft D = 0.667 ft A = 0.349 ft 2 g = 32.2 We K = 0.0124 t= U.U1s L= 16 ft D = 0.500 ft A = 0.196 ft g = 32.2 ft/s K = 0.1676 Hampton Inn Hotel Project Manager: Lee Adams 8128/2003 Job # 000756 -3518 Calculations: Mark Taylor Fire Flow Analysis After 4 Hours K = U.UUBo Q = 1.300 cfs HGL = 300.93 ft HGL = 300.91 ft P = 18.60 psi K = Junction 2 1.300 K = 0.0149 cfs Q = 1.300 cfs HGL = 300.91 ft HGL = 300.89 ft P = 18.59 psi K = 0.1466 1.300 Q = 1.300 cfs HGL = 300.89 ft HGL = 300.64 ft P = 18.48 psi I Q= 560.99 gpm K = 0.1541 Q = 1.300 cfs HGL = 300.89 ft HGL = 300.63 ft P = 18.65 psi K= Junction 4 K = 0.6561 cfs Q = 1.300 cfs HGL = 300.63 ft HGL = 299.52 ft P = 17.64 psi K= U.zo10 Q = 1.300 cfs HGL = 299.52 ft HGLZ = 299.08 ft P = 17.37 psi Q = 543.86 gpm T = U.U10 L = 172 ft D = 1.50 ft A = 1.767 ft 2 g = 32.2 ft/s K = 0.0086 T= u.u1s L = 133 ft D = 0.667 ft A = 0.349 ft 2 g = 32.2 ft/s K = 0.3305 t= U.Ul3 L = 37 ft D = 0.667 ft A = 0.349 ft 2 g = 32.2 ft/S K = 0.0919 t= U.uiS L = 473 ft D = 1.50 ft A = 1.767 ft 2 g = 32.2 ft/s K = 0.0235 T= U.0 13 L= 8 ft D = 0.667 ft A = 0.349 ft 2 g = 32.2 ft/s2 K = 0.0199 t= U.Uis L = 50 ft D = 0.667 ft A = 0.349 ft 2 g = 32.2 ft/s K = 0.1243 Hampton Inn Hotel Project Manager: Lee Adams 8/28/2003 Job # 000756 -3518 Calculations: Mark Taylor Network Table - Nodes at 0:00 Hrs Node ID Demand GPM Head R Pressure psi Quality Junc J 1 0.00 386.70 55.76 0.00 Junc J2 0.00 386.19 55.54 0.00 Junc FH1 1500.00 385.67 55.32 0.00 Junc J3 0.00 385.42 55.38 0.00 Junc J4 0.00 375.57 50.60 0.00 Junc FH2 1500.00 374.64 50.11 0.00 Junc J 12 0.00 386.36 56.14 0.00 Junc J5 0.00 384.50 55.46 0.00 Junc J6 0.00 379.80 53.21 0.00 Junc J7 0.00 379.69 53.16 0.00 Junc J8 0.00 379.69 53.16 0.00 Junc FH3 1500.00 379.09 52.86 0.00 Junc J 1 1 0.00 386.95 56.39 0.00 Junc J9 0.00 389.31 58.72 0.00 Junc J 10 0.00 427.87 75.21 0.00 Tank EXFH - 4500.00 438.60 79.94 0.00 EPANET 2 Pagel Network Table - Nodes at 4:00 Hrs Node ID 1 Junc J 1 Quality Junc J2 R June FH1 Junc J3 313.17 Junc J4 0.00 Junc FH2 312.66 Junc J12 0.00 Junc J5 312.14 Junc J6 0.00 Junc J7 311.89 Junc J8 0.00 Junc FH3 302.04 Junc J 1 I 0.00 Junc J9 301.11 Junc J 10 0.00 Tank EXFH 312.83 )emand Head Pressure Quality GPM R psi 0.00 313.17 23.90 0.00 0.00 312.66 23.68 0.00 1500.00 312.14 23.46 0.00 0.00 311.89 23.52 0.00 0.00 302.04 18.74 0.00 1500.00 301.11 18.25 0.00 0.00 312.83 24.28 0.00 0.00 310.97 23.60 0.00 0.00 306.27 21.35 0.00 0.00 306.16 21.30 0.00 0.00 306.16 21.30 0.00 1500.00 305.56 21.00 0.00 0.00 313.42 24.53 0.00 0.00 315.78 26.86 0.00 0.00 354.34 43.35 0.00 - 4500.00 365.07 48.08 0.00 Page 1 EPANET 2 Pipe Analysis Pipe Analysis - Initial Pipe Length (ft) Diameter (in) Flow (gpm) Velocity (ft/s) A 6 8 1790.02 11.43 B 14 6 1159.89 13.16 C 62 8 630.13 4.02 D 264 8 1136.29 7.25 E 25 6 1136.29 12.89 F 115 8 506.16 3.23 G 126 8 1152.65 7.36 H 3 8 1152.65 7.36 1 5 8 0.00 0.00 J 16 6 1152.65 13.08 K 172 18 1790.02 2.26 L 133 8 3448.83 22.01 M 37 8 3448.83 22.01 N 473 18 3448.83 4.35 0 8 8 1658.81 10.59 P 50 8 1152.65 7.36 Pipe Analysis - After 4 Hours Pipe Length (ft) Diameter (in) Flow (gpm) Velocity (ft/s) A 6 8 862.67 5.51 B 14 6 560.99 6.37 C 62 8 301.68 1.93 D 264 8 543.86 3.47 E 25 6 543.86 6.17 F 115 8 242.18 1.55 G 126 8 560.99 3.58 H 3 8 560.99 3.58 1 5 8 0.00 0.00 J 16 6 560.99 6.37 K 172 18 862.67 1.09 L 133 8 1665.84 10.63 M 37 8 1665.84 10.63 N 473 18 1665.84 2.10 0 8 8 803.17 5.13 P 50 8 560.99 3.58 Hampton Inn Hotel Project Manager: Lee Adams 8/28/2003 Job # 000756 -3518 Calculations: Mark Taylor