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5803 NEWTON DR; ; AS120168; Permit
12/13/24, 4:18 PM AS120168 Permit Data City of Carlsbad Sprinkler Permit Permit No: AS120168 Status: APPROVED Applied 12/20/2012 Approved: 12/26/2012 Issued: 12/26/2012 Job Address: 5803 NEWTON DR Permit Type: SPRINK Parcel No: 2121304800 Lot #: 0 Reference No.: PC #: Inspector: Project Title: SIGNET ARMORLITE STE AA/S Ti. NEW.STORAGE AND MEZZANINE Applicant: AMERICAN AUTOMATIC FIRE SUPPRESSION SUITE B 9510 PATHWAY STREET SANTEE CA 619-562-3010 Owner: RI F IV NEWTON L L C 11620 WILSHIRE BLVD #300 LOS ANGELES CA Fees ($) Add'I Fees ($) Total ($) Balance ($) 334 0 334 0 about:blank 1/1 Parsley Consulting 350 West 9th Avenue, Suite 206 Escondido, California 92025-5053 Telephone (760) 745-6181 Fax(760) 745-0537 December 18, 2012 Fire Code Official City of Carlsbad Fire Department Office of Fire Protection Services 1635 Faraday Avenue Carlsbad, California 92008 Subject: Plan Review for Signet Armorlite TI, 5803 Newton Drive, Carlsbad, California Dear Sir: We have reviewed the submittal as prepared by American Automatic Fire Suppression, Inc.. Based on our review it is our opinion that the submittal is in substantial conformance with the applicable codes and standards in force at this time, including local adopted ordinances. Further submittal should not be required. Sincerely, Xenneth EYCONSULTJNG id W. agone7T Office Manager End (1) Working Drawings - 2 sets Hydraulic Calculations - 2 sets Review Comments - 1 set Note: A copy of this letter has been transmitted via e-mail to American Automatic Fire Suppression, Inc. on December 18, 2012 file:\C:\Documents and Settings\Ken\Rev2 12405.O1.wpd 5/LO /61 Signet Armorlite TI Review Comments. Page 2 Review Comments Signet Armorlite TI, 5803 Newton Drive, Carlsbad, California 1. In the event that the t-bar ceilings are not rigidly braced per ASME E580, the sprinklers to be installed in the suspended panels will require either a 2" clearance around the penetration through the panel, or the use of a flexible drop fitting, per ASCE7-05, section 13.5.6.2.2(e), which is attached. However, gypsum board ceilings are assumed to be rigidly braced. All the above conditions can be field verified. 2. Field verify that the grooved coupling at the bottom of the drop to the mezzanine is of the "flexible" type, as is required by section 9.3.2.3(1). 3. Verify that either: A barrier exists between the high pile storage area capable of preventing heat from that area from activating sprinklers in the adjacent design areas, per 12.3 .2(2), or The storage protection design and installation is extended at least 15'-0" into the adjacent area, per 12.3(1), or The ceiling elevation difference between the to areas is at least 24" in depth, per 12.3(3) 4. As the modifications to the existing system cannot be isolated for hydrostatic testing to 200 psi, the required testing should only be done at system pressure, per 24.2.1.6. Advisory text from the NFPA-13 Handbook notes: In many cases, segregating new work from existing work is difficult. To require the entire system to undergo another hydrostatic test when only relatively minor changes are made is unreasonable. When new portions of a system cannot be isolated, such as relocated drops, NFPA 13 provides some flexibility and permits hydrostatic tests to be conducted at the system's normal static pressure. In general, existing portions of the system do not need to be subjected to a new hydrostatic test. EXCEPTION: Rigid braces are permitted to be used instead of di- agonal splay wires. Braces and attachments to the structural system above shall be adequate to limit relative lateral deflections at point of attachment of ceiling grid to less than 0.25 in. (6 mm) for the loads prescribed in Section 13.3.1. For ceiling areas exceeding 2,500 ft2 (232 m2), a seismic sepa- ration joint or full height partition that breaks the ceiling up into areas not exceeding 2,500 ft2 shall be provided unless struc- tural analyses are performed of the ceiling bracing system for the prescribed seismic forces that demonstrate ceiling system penetrations and closure angles provide sufficient clearance to accommodate the anticipated lateral displacement. Each area shall be provided with closure angles in accordance with item 2 and horizontal restraints or bracing in accordance with item 3. Except where rigid brae-es are used to limit lateral deflec- tions, sprinkler heads and other penetrations shall have a 2 in. (50 mm) oversize ring, sleeve, or adapter through the ceiling tile to allow for free movement of at least 1 in. (25 mm) in all horizontal directions. Alternatively, a swing joint that can accommodate 1 in. (25 mm) of ceiling movement in all hori- zontal directions is permitted to be provided at the top of the The tributary areas imately equal. Changes in ceiling plan elevation shall be provided live bracing. Cable trays and electrical conduits shall be supported indepen- dently of the ceiling. Suspended ceilings shall be subject to the special inspection requirements of Section 11 A. 1.3.9 of this standard. Integral construction. As an aiternate to providing large clearances around sprinkler system penetrations through ceiling systems, the sprinkler system and ceiling grid are per- mitted to be designed and tied together as an integral unit. Such a design shall consider the mass and flexibility of all elements involved, including the ceiling system, sprinkler system, light fixtures, and mechanical (HVAC) appurtenances. Such design shall be performed by a registered design professional. 13.5.7 Access Floors. 13.5.7.1 General. The weight of the access floor, Wi.,, shall in- clude the weight of the floor system, 100 percent of the weight of all equipment fastened to the floor, and 25 percent of the weight of all equipment supported by, but not fastened to the floor. The seismic force, F,,, shall be transmitted from the top surface of the access floor to the supporting structure. Overturning effects of equipment fastened to the access floor panels also shall be considered. The ability of "slip on" heads for pedestals shall be evaluated for suitability to transfer overturning effects of equipment. Where checking individual pedestals for overturning effects, the maximum concurrent axial load shall not exceed the portion of W, assigned to the pedestal under consideration. 13.5.7.2 Special Access Floors. Access floors shall be consid- ered to be "special access floors" if they are designed to comply with the following considerations: I. Connections transmitting seismic loads consist of me- chanical fasteners, anchors satisfying the requirements of Appendix D of ACI 318, welding, or bearing. Design load capacities comply with recognized design codes and/or certified test results. 147 In the event that the t bar ceilings are not rigidly drift braced per ASME E580, the sprinklers to be din installed in the suspended panels will require either a 2" clearance around the penetration I for through the panel, or the use of a flexible drop size teel, fitting, per ASCE7-05, section 1 3.5.6.2.2(e), cflle which is attached However, gypsum board ceilings are assumed to be rigidly braced All It e or the above conditions can be field verified. erts, ibe values of R,, and a, taken i\ eh 1 11 -1 applied at the center of mass of the panel. e. Where anchorage is achieveg flat straps embe concrete or masonry, such str' be attached to or around reinforcing steel or otse'jminated so as to effec- tively transfer forces to the rcineel or to assure that pullout of anchorage is not tial faile mechanism. 13.5.4 Glass. Glass in glazed c walls aitorefronts shall be designed and installed in accce with Sedn 13.5.9. 13.5.5 Out-of-Plane Bending.verse or out-ojane bend- ing or deformation of a compoor system that is'%jibjected to forces as determined in Sec13.5.2 shall not exed the deflection capability of the comt or system. 13.5.6 Suspended Ceilings. 13.5.6.1 Seismic Forces. The weight of the ceiling, W,,, shall' include the ceiling grid and panels; light fixtures if attached clipped to, or laterally supported by the ceiling grid; and other components that are laterally supported by the ceiling. W,, shall be taken as not less than 4 psf (19 N/rn2). The seismic force, F,,, shall be transmitted through the ceiling attachments to the building structural elements or the ceiling- structure boundary. 13.5.6.2 Industry Standard Construction. Unless designed in accordance with Section 13.5.6.3, suspended ceilings shall be designed and constructed in accordance with this section. 13.5.6.2.1 Seismic Design Category C. Suspended ceilings in structures assigned to Seismic Design Category C shall be designed and installed in accordance with ASTM C635, ASTM C636, and the CISCA for Seismic Zones 0-2, except that seismic forces shall be determined in accordance with Sections 13.3.1 and 13.5.6.1. pended ceilings in Seismic Design Categories D, E, and F shall be designed and installed in accordance with ASTM C635, ASTM C636, and the CISCA for Seismic Zones 3-4 as modified by the following: A heavy duty T-bar grid system shall be used. The width of the perimeter supporting closure angle shall be not less than 2.0 in. (50 mm). In each orthogonal horizontal direction, one end of the ceiling grid shall be attached to the closure angle. The other end in each horizontal direction shall have a 0.75 in. (19 mm) clearance from the wall and shall rest upon and be free to slide on a closure angle. For ceiling areas exceeding 1,000 ft2 (92.9 m2), horizontal re- straint of the ceiling to the structural system shall be provided. Minimum Design Loads for Buildings and Other Structures 5803 Newton Drive, Carlsbad, CA Conterminous 48 States 2003 NEHRP Seismic Design Provisions Latitude = 33.138143 Longitude = -117.289301 Spectral Response Accelerations Ss and Si Ss and Si = Mapped Spectral Acceleration Values Site Class B - Fa = 1.0 ,Fv = 1.0 Data are based on a 0.01 deg grid spacing Period Sa (sec) (g) 0.2 1.170 (Ss, Site Class B) 1.0 0.443 (51, Site Class B) These plaits have been evtexved lot anpaicnt coil ipliarice %% ith the current applicable codes and stttllilaflhs ill totce ill the uiidiciioit. The owner or contractor lila iitihice res etc in support of an application br a hnibdinn pert nit oh the cx press understand up 111111 1 h I rex cxx ii ii tvt\ relietes the otriier tsr coisracior tie their olessiniab eticitirer or aichitect or sole rctiinsibilttv iii hull compliance with said codes and sltidards. illclLt(II JIL the build in code or or atty respotisthiltv or deficiencies. errors. intissions or negligence ill these plaits xxhieihcr or not noted in this review. j1jJTReviewer 4i) RI Ct )M helEN) A h'lR( )VA I. 11 REC(thlMENl) AI'PROVAI.. AS NOTED CEFY OF CARLSBAD I FIRE DEPARTMENT I APPROVED I Subject to field inspection and required I tests notations hereon conditions in correspondence and conformance with applicable regulations. The stamping of these p)Qis sf1I not be held to permit I or appr d lation of any law. By Date12hJ' Hydraulic Calculations for Project: Signet Armorlight, Inc 5803 Newton Dr. Carlsbad, CA Drawing no.: 12059 Date: 11/28/2012 Design Remote area number: Remote area location: Occupancy classification: Density: Area of application: Coverage per sprinkler: Type of sprinklers calculated: No. of sprinklers calculated: In rack demand: Hose streams: Total water required (including hose streams): Total water required at base of system riser: Type of system: Volume of dry or preaction system: 1 Open Lab high piled storage 0.30 gpm 2500 sf 100 s Standard Coverage SR pendent 11.2k high temp 35 na 500 gpm outside + 0 inside 1730.99 gpm at 76.49 psi [27.53 psi safety margin] 1230.99 gpm at 54.73 psi wet pipe na Water Supply Information Date: 11-1-2012 Location: 5803 Newton Dr Source: city water main Contractor: Name of designer: Authority having jurisdiction: American Automatic Fire Suppresion, Inc. 9510 Pathway Street, Suite B Santee, Ca 92071 C-16,731167 (619) 562-3010 Debbie Seeman City of Carlsbad Fire Department Notes Static and residual pressures reduced by 10% 11/28/2012 Page 2 Hydraulic Demand Graph 140 Water Source: A) 113.4 psi Static 120 B) 1400 gpm at 107.06 psi Source at BOR: C) 104.45 psi Static 100 D) 1231 gpm at 82.26 psi E) 1400 gpm at 76.3 psi 80 P S 60 (D) KU Demand at BOR: F) 1231 gpm at 54.73 psi (F) 1 Ew 20 [suiuI. so 0 M win 01, 600 800 1000 1200 1400 GPM report created by the 'Simple Hydraulic Calculator" version 2.3.2 from www.igneusinc.com 11/28/2012 Page 3 SuDDlv Analysis Static Residual Available Total Required Node at Pressure Pressure Flow Pressure Demand Pressure [psi] [psi] [gpm] [psi] [gpm] [psi] SRC 113.4 106.2 1500.0 104.02 1730,99 76.49 Node Analysis Elev Pressure Req Disch Discharge Node Tag [ft] Type [psi] [gpm] [gpm] SRC -3.000 source 76.485 -1730.99 FH -3.000 ref 68.844 500.000 500.000 Ui -3.000 r of 68.245 0.000 U2 -3.000 ref 58.725 0.000 BF1 3.000 iet 05. 351 0.000 BF2 3.000 re f 56.42 0.000 A -3,000 ref 56.976 0.000 BOB 1.500 ref 54.729 0.000 TOP 23 . 333 rot 43,998 0.900 Ni 23 . 331, ref 40.765 0.000 Nt 23 .333 ref 38.536 0.000 N3 2 .333 r f 38.211 0.000 N4 23. 33 ref 35.067 0,000 N5 2 3 . 333 ref '7,541 0.000 N6 23 . ief 37.976 0.000 737 23. 333 ref 37.849 0.000 738 23,3 ret 37.84t 0.000 RN 26.167 ref 33 .459 0.000 9732 )6.1o7 r e 3.14 0.000 73733 7 6.1i,i ref 32. 372 0.000 Rh 1 6.r67 ref '1.627 0.000 RN 5 26.i / ref 31.863 9.000 73736 26.1(7 ref 31.48-5 0.000 73117 21.1 7 ref 37.270 0.000 PNO 26.167 re f 33 .t 4 2.000 Fr .13 . 333 ef 15.' 0 0 900 Fl 23.33 ref 5,09 U73 90' 73, 333 rrt ,5.206 0.000 F4 23 . ro f 15.201 0.000 F 23 .333 ref 15.204 0.00 F6 23,773 ro' 15 .214 0.000 737 " 3 .333 ret 11,247 0.030 88 r . e f, 15.t49 0.000 FF1 26.167 ref 4.689 0.000 FF2 26. 1 17 ref 13 .98 0.000 FF3 26.f67 ref 13 .991 0.000 FF4 26.167 itt 1.5i9 0.000 FF5 26.167 ref 12.629 0.000 RF6 2r.167 ref 12.561 0.000 9737 . 26.167 ref 15 . 8 91 0.000 FF8 26.167 ref 15.898 0.000 Li 26.167 re f 16.296 0.000 L2 26,167 ref 16.016 0.000 L'3 26.1r7 ref, 15.323 0.000 L4 26.167 ref 15.760 0.000 735 76.167 r e I _4.728 0.000 736 26.167 ref 18.238 0.000 737 26.167 ref 13.772 0.0'00 Elev Pressure Req Disch Discharge Node Tag [ft] Type [psi] [gpm] [gpm] 738 26.167 ref 15.751 0.000 739 2e.167 ref 1' .744 0.000 7310 26.167 ref 14.257 0.000 L11 26.167 ref 1 .780 0.000 7312 26.1 7 ref 1 .0r7 0.000 L13 26.167 ref 12.967 0.000 7314 26.167 r of 12.296 0.000 7315 26.167 r ef 12.282 0.000 LiE 2 6.167 ref 12.270 0.000 L17 26.167 rot 12.230 0.000 L18 26.167 re t 11.805 0.000 L19 26.167 ref 11.860 0.000 7320 26.167 ref 14.558 0.000 L21 26.167 ref 13.491 0.000 L22 26.167 ref 12.950 0.000 L23 28.167 ref 9 . 94 0.000 L24 26.167 rot 8. 0.000 L2' 2 .167 ref 8.280 0.000 L26 26.167 ref 8.26' 0.000 L2 7 21.16' rot 8.2r 0.000 L28 26.167 ref i 3.076 0.009 L24 '6. 67 ref 13 .016 0.000 L30 26.167 ref 1. 19 0.000 7331 26.67 rtr 12.3 3 5 OOO '.67 re f 12 .b2 9.099 L33 26.67 ref '2. '02 0.00 7374 26.11' re f 16.'O' L'S 26.167 ret 11.'02 0.000 1 9.000 K=1 1.20 1.211( 30.009 39.1 4 2 2.009 Ke1i.20 I . ii 3').000 37.668 3 9.000 K11.20 10.63 '0.090 37.084 4 9.000 K1i.2(1 11.949 30.000 38.7 5 5 2.000 73=11.20 11,107 90000 37.828 6 9.000 K=1.20 11.146 30.000 3 92 7 9.000 73=1.20 10,8t7 30.000 36.973 8 9.000 K=11.20 11.49 30.000 '7.979 9 9.000 73=71,20 10.°80 30.000 77.113 10 9.000 Koi1.20 10. 126 39.000 36.681 11 9.000 K1 I .20 10.688 30.000 36.266 12 9,000 K=1.20 10.098 30.000 35.591 13 9.000 73=11.20 9.696 30.000 34.871 14 9.000 73=] L.20 °.723 30.000 4.923,' 15 9.000 73=11.20 9.355 30.000 3.257 16 , 9.000 73=11.20 9.789 30.000 34,83' 17 9.000 K=1i.20 9. 0 3(1.000 34.210 18 9.000 Koii.20 9,697 30.000 34.'16 14 9,000 K=11.2O 9.142 30.000 33.864 20 9.600 K=11. 2 0 11.317 10.000 37.677 report created by the "Simple Hydraulic Calculator" version 2.3.2 from www.igneusinc.com 11/28/2012 Page 4 Node Analysis. cont. Elev Pressure Req Disch Discharge Node Tag IN Type [psi] [gpm] [gpm] 21 9.000 K=11.20 10.78 30.000 36.718 -22 9.000 K=11.20 10.45 30.000 36. 21 21 9.000 K=11.20 9.678 30.000 22.993 24 3.000 K=11,20 7.704 10 000 31.086 29 0.000 K=11.20 i.74 20.000 10.178 069 .000 K= 11, .20 7.311 30.000 30.23 27 9.000 K=11.20 7.279 0.001 10.00fl 21 °.000 14=11.20 9.748 30.000 34. 9 6 8 29 9.0Q K=11.20 9.70 30.0)0 3 4.°1° 0 0.°0 9 K=11.20 9.476 30.000 313 9.000 K=11.0 995 30.000 34.14' 32 9.010 K=11.20 9 . 44 30.000 34.417 33 9.000 K=11.20 °.26 30.000 14.099 4 9.000 K=11.2,0 9.2 6 30.000 34.075 35 9.000 K=11.20 9.083 30.000 33.755 report created by the "Simple Hydraulic Calculator version 2.3.2 from www.igneusinc.com 11/28/2012 Page 5 PiDe Information negative pipe flow (Q) indicates flow is from node 2 towards node I Node I Discharge Nom Fittings I [ft] total (Pt) Pipe Elev & Flow l.d. num & length F [ft] C factor elev (Pe) Node 2 [ft] K-factor [gpm] [in] [ftj T [ft] psi/ft frict (Pf) Notes SRC -3.000 q=-1730.99 8 2T110.709 300.000 Pt= 76.488 Mat='CD152" 1, Q=17 3 0.995 0.27 10=6.376 173.971, C=140 Pe= 0.000 FH -7.000 2E=56.976 4 73.9i1 0.011 Pf= 7.641 PH - 000 q= 500.000 8 1G=6. 326 35.000 P5= 68.814 OOGPM HON 2 Q=1230.99-9 8.27 1E=28.468 34 794 C=11 0 Pe= 0.000 Matfl7111" Hi - 00 0 °. 4 0.11 p5 Q32 u - .000 q= 0.000 8 1E=28.468 6.000 Pt= 68.24 5 Mat"C0152 3 0=1230.99 8.27 28.468 C=140 Pe= 2.528 Bi .000 . 4.468 0.007 Pf= 0.296 BFI 3 .000 q= 0 . fl 00 8 1E=2 8 4(7 ) Pt= 65. DCDA 4 Q=1230 995 8 11 28.468 C-140 Pe= 0.000 Mat="00111" BF2 7.000 37.468 0.009 P5= 0. 2 8? Pdev.64 p1 BF2 3 .000 q= 0.000 8 1E=28.468 5.000 Pt= 55.411 Mat="flDI52" 5 Q=17 3 0.995 8.27 25.468 C=140 Pe= -21198 P2 .000 74.408 0.009 p5= 0.295 U2 -.000 q= 0.000 8 3E=85.404 118.500 Pt= -58 . 725 Mat=00152" 6 . 0=1230 995 8.27 85.404 C=140 Pe= 0 000 A -3.000 . . 203.904 0 . 009 P5= 1.750 A -3.000 q= 0.000 8 1E=2 1 .111 4.6(7 P5= 56. 76 Mat=" S 10" 7 Q=1230 . 99 8.249 11.141 C=120 Pe- 1.948 BOR 1.500 25.808 0.012 P1= 0.29$ BOR 1.500 q= 0.000 8 iG=4.628 21.873 P5= 54.729 FLOW SW B Q-1230.995 8.24Q 4 598 C=120 Fe= 9.454 Mat='SlO" TOP 23.373 26.531 0.0 1 21 Pf= 0.707 Pdev=1.0 psi TOR 2 . "7 q= 0.300 6 2 E=3 5.z65 4/711 P5= 43 .9:2 Mat=" S 10" 9 Q=1230.9°5 6.357 3" .205 C=120 Pe= 0.000 Ni 2. - 77.255 0.041 Pf= 3.204 Ni 23.733 q= 0.000 6 2E=35.205 77.i67 Pt= 40.765 Mat'SiO" 10 Q=1053 .15 6. 357 35.205 C=20 Pe= 0.000 N2 23.73: 2 . 3-2 0.9 3 1 Pf= 2.228 N2 23.33. q= 0.000 6 17.000 Pt= 38.536 Mat=" S 10" ii. Q= 89:1165 6. 357 0.000 C=120 Pe= ;0.000. N3 23.733 12.000 0.023 Pf= 0.275 - N3 2 .311 q= 0.070 8 L .000 Pt= 38.261 Mat&'610" 12 0= 4 31 .8 0. 57 0.000 1112 7; Pe= 0.320 N4 2 3 . i2.000 0.016 P5= 0.194 N4 23.113 q= 0.000 6 12.000 Pt= 38.067 WatS10" 13 Q= 580. 51 6. 757 . 0.000 C=120 Pe= 0. 000 - N5 2 . 333 12.000 0 .01 P5= 0.127 N5 11.7 33 q= 0.000 6 1' .600 Pt= 37.944 Mat51fl" 14 Q= 422.669 6.357 0.000 C=120 Pe= 0.000 N6 23.333 - . 12.000 0.006 Pf= 0.068 report created by the "Simple Hydraulic Calculator" version 2.3.2 from www.igneusinc.com 11/28/2012 Page 6 Pipe Information. cont. Node I Discharge Nom Fittings I [ft] total (Pt) Pipe Elev & Flow id. num & length F [ft] C factor elev (Pe) Node 2 [ft] K-factor [gpm] [in] [ft] T [ft] psi/ft frict (Pf) Notes N6 23.333 q= 0.000 6 12.000 Pt= 37.876 Mat="SlO" 15 Q 259.833 6.357 0000 C=120 Fe=, 0.000 N7 23.333 12.000 0 . 002 Pf= 0.028 N7 2 .333 q= 0.000 1.000 Pt= 37.848 Mat=" S 10" 16 Q= 129.823 6 °57 0.000 0=120 Pe= 0.000 NB .0°0 0.001 Pf= 0 00 Fl 23. 333 q= 0.702 o 12.000 06= 15.O Ma 6=". LO" 17 Q= 63.884 6.57 0. iOO C=.20 Pe= 0.000 F2 213.7 7 1 000 0,000 Pf= u.00 21 F2 23.333 q= 0.000 6 12.000 Pt= 15.208 Mat-S10" 18 ,. Q= 68.177 6.357 0.000 0=120 Pe= '0.000 F323.333 12.000 0.000 Pf= 0.002 F3 23.337 q= 0.000 7 12.000 Pt= 15.206 Mat' '10" 19 Q= 74.223 6.357 0.000 C=120 P= 0.000 F4 23 333 12.000 0.000 Pf= 0.003 F'S 23 . 3 33 q= 0.000 U 00 Pt= 15.20 Mat="10" 20 Q- 39 .143 6.37 0.000 C=120 Pe= 0.000 F4 23 . 3 12.000 0 Pf= 0.001 F6 22. 333 q= 0.002 1 i.000 06= 17.214 MaL="10" 21 Q= 147.817 6. 357 0.000 0= 2 Ci Pe= 0.000 F5 23.7 3 12.000 0.001 Pf= 0.010 F7 23. 333 q= 0.000 6 12.000 Pt= 15.242 Mat'S10" 22 , Q= 9.833 6.377 0.000 0=120 Pe F6 23.373 12.000 0.002 Pf= 0.028 F'S 23. 33 q= 0.000 6 12.000 Pt= 15.249 Nat' S 10" 23 Q= 129.80 6.357 0.000 0=120 Pe= 0.000 57 .r.3.3 3 12 -000 0.001 , Pf= 0.008 Ni 2 . 33 q=- 0.009 2T=4. 1 2.8 3 Pt= 40.775 . Mat=" 10" 24 Q= 1 7.87° 2 .1 7 2 613 C=I 2 0 Pe- RN1 /7.77 27.44° 0.21 P5= 6.079 N2 23.733 q= 0.000 2 2T=24.613 2 . 833 Pt= 38.36 Mat="SlO" 25 Q= 15 7.270 2. 157 24 . 613 C=20 Pe= 1.227 RN2 26. 167 , . 27 . 4 0. 172 Pf= 4.725 N3 23.333 . q= 0.000 2 25=24.613 .833 Pt= 38.261 Nat'S10" 26 ' ' ,. ', Q= 1 4.089 2.157 'G 13 C=120 Pe= 1.227' RN3 26.167 , 27.447 0.17 Pf= 4.663 N4 23.333 q= 0.000 2 2T=24.613 1.8 3 Pt= 38.067 Mat'Si0" 27 Q= lb . 78 ' 2.57 74.613 0=120 Pe 1,227 RN4 26.167 27.47 0.1° Pf= 5.217 N5 27.3 77 q= 0.000 2 2T4.213 /.7 3 Pt 37.944 Mat="010" 28 Q= I 7.52 2, .127 24•7 C=12 0 Pe= 1. 73' RN5 2 6.i67 17.447 0.177 05= 41. B55 report created by the "Simple Hydraulic Calculator" version 2.3.2 from www.igneusinc.com Page 7 1 1/28/2012 Pipe Information. cont. Node I Discharge Nom Fittings L [ft] total (Pt) Pipe Elev & Flow Id. num & length F [ft] C factor elev (Pe) Node 2 [ft] K-factor [gpm] [in] [ft] I [ft] psi/ft frict (Pf) Notes N6 23.333 0.000 2 2T=24.613 2 833 Pt= 37.876 Mat&'SlO" 29 Q= 16.836 2.187 24.613 C=120 Pe= ,1 .227 RN 6, 26.167 27.447 0.188 Pf= 5.165 N7 23.3 3 q= 0.000 2 2T=24.613 2 . 833 Pt= 37.848 Mat=S10" Q= 129.940 2.157 ,24.613 C=lO Pe= l.227 RN7 26.167 . - 27.447 0.124 Pf= 3.402 NB 23.333 q= 0.000 2 2T=24.617 2 . B33 Pt= 37.841 Mat="S10" Q= 129.893,2.13? 24.613 C120 RN8 26.167 27.447 0.174 Pf= 3.400 RF1 26.167 q= 0.000 2 1T=12.3 07 2.833 Pt= 14.488 MatS10" 32 - Q= 63.884 2.15 7 12.307 C=1 20 Pe'--1z-22_-7z,'-- Fl 23.333 ., , , 15.240 0.033 Pf= 0.505 RF2 26.167 q= 0.000 2 1T=12.707 2.837 Pt= 13.985 Mat&'SlO" Q= 4.292 2.107 1 07 C=120 Pe -1.227 F2 23.333 15.140 0.000 Pf=,0.003 RF3 26.167 q= 0.000 2 1T=12.307 2.833 ' Pt= 13.986 MatS10" 6.047 7 157 2: 07 Ci2b- Pe= 1. 227 F3 23.333 15.140 0.000 Pf= 0.006 F4 23.333 q= 0. 000 2 1T12.107 2.833 Pt= 15.203 Mat'S10" 35 Q 113.366 2.?57 1 2 .307 , C 120 Re= 1.227 . RF4 26.167 15.10 0.096 Pf= 1.458 F5 23.333 q= 0.000 2 1T12.307 2.8 3 Pt= 15.204 Nat'S10" 36' -. ., ' -d=-.:l D 8 . 674 2.157 12.307 C120Pe= 1.227 RF526.167 15.10 0.089 Pf= 1.348 F6 23.33 q= 0.000 2 1T=12.307 2 . 833 t= 15.214 Mat" S 10' Q 12.017 2.157 ' . 12.307 C=12'0 'Pe 1.227 26.167 18.140 0.094 Pf= 1.426 RF7 26.167 - q= 0.000 2 1T=12.307 2.83 - Pt= 15.891 Mat'S10" . . 'Q= 129.940 2.157- - , 12. 07 C=120 - Pe -1.227 F7 23.333 15.10 0.124 Pf= 1.877 RFB 26.167 q= 0.000 2 1T=12.307 2.87 Pt= 15.898 Mat="SlO" 39 - Q= 129.893 2.157 1.307 C=10 - Pe -1.227 F8 23.333 15.140 0.124 Pf= 1.875 RNl 26.167 q= - 0.000 2 77.500 Pt= 33.459 Mat="SlO" T"Q= 177.830 2.157 ' ,. 0o606 c=12"d = 0.000 - L3 26.167 .................................................- 77.500 0.221 Pf= 17.164 Li 26.167 q= 0.000 2 ' - 2.000 - - Pt 16.296 Mat="SlO" 41. - . - . - Q= 138.636 . 2.1-5'7 0.000 C 2.0. P'e=- 0.000. -, - L2 26.167 2.000 0.14 Pf= 0.279 - L2 26.167 q= 0.000 2 8.917 - Pt= 16.016 MatS10" Q= 100.969 2.157 - 0.000 .C=120 - PLO = 0.000 L3 26.167 - - 0.078 Pf= 0.693 - report created by the "Simple Hydraulic Calculator" version 2.3.2 from www.igneusinc.com 11/28I2012 Page 8 Pipe Information. cont. Node I Discharge Nom Fittings L[ft] total (Pt) Pipe Elev & Flow i.d. num & length F [ft] C factor elev (Pe) Node 2 [ft] K-factor [gpm] [in] [ft] T [ft] psi/ft frict (Pf) Notes L3 28.167 q= 0.000 2 1T12.307 12.750 Pt= 15.323 Mat="S1O" 43 Q= 63.884 2.157 12.307 0=120 Pe-L .000 RF1 26. 167 25. 57 0 .033 P6= 0.835 RN2 26 .167 q= 0.000 2 97 .833 P0= 3 2 . 5 9_4 1at="S10" 44 Q= 155.200 2.157 0.000 0=120 Pe= 0 000 L4 26.127 97.83 0.172 Pf= 16.844 L4 2 .167 q= 0.000 2 10.000 Pt= l.740 Mat- S0" 45 Q= 116.470 2.157 0.000 C-=120 Pe= 0.000 L5 26.167 10.000 0.101 06= 1 017 L5 '6.167 q= 0.000 2 0.000 Pt= 74.728 Mat="$0" 46 Q= 78.707 2.107 7.000 C=10 Pe- 0.000 56 27.167 . ,.- 10.000 0,042 Pf= 0.490 L6 26.167 q= 0.000 2 1T=12.307 4.7r0 Pt= 14.238 Mat'S10" 47 Q= 41.265 2. 57 12.307 0=120 Pe= 0.000 0F2 2 .16/ 17.07 p.01 Pf= O.53 052 26.167 q= 0.000 2 1T=12.307 5.250 P0= 13.985 Mat=" S 10" 48 Q= 36.973 2.i57 12.307 0=120 Pe- 0.000 L7 26.167 17. 07 0.012 06= 0.213 0N3 26.117 q= 0.000 2 97 . 833 P0= 32. 372 Mat="S1O" 49 Q= 154.086 2.157 0.000 0=120 Pe= 0.000 L8 26.167 . 97.837 0.17 P6=16.621 LB 2' .167 q= 0.000 2 in.000 P0= 15.751 Mat="S1O" 50 Q= 16.07 2 15 7 0.000 0=120 Pe= 0.000 LB 26. 167 , 10.000 0.101 Pf= 1.006 LB 2(117 q= 0.000 2 10.000 Pt= 14.744 Mat="SlO" 51 , Q= 78.094 7.157 0.090 0=120 Pe= 0.000 LiD 21.i67 , i0.006 0.049 P6= Q•494 L10 21.67 q= 0.000 2 1T=12.707 4.750 Pt= 14.251 Mat="S1O" 52 Q=- 42. 3 12 4 157 12.307 0=120 Pe= 0.000 RF3 26. 17 17.007 0.010 P6= 0.265 RF3 20. 167 q= 0.106 2 1T=12. 3 07 5.250 Pt= 1 .986 Mat="S1O" 53 Q= 36.261 2.1 7 12. 307 0=120 Pe- 0.000 Lii 26.107 , 17.557 0.012 Pf= 0.205 RN4 26. 67 q= , 0.00 2 97 .833 P0= 1.12 Mat.,10" 54 Q= 163.728 2.157 0.000 C=I. 20 Pe= 0.000 L12 20.167 97.833 0.19 Pf= 18.596 L12 26167 q= 0.000 2 0.500 Pt= 10.077 3atS10" 55 Q= 128.138 2.1p7 0.000 0-120 Pe= 0.000 L13 26 .1 67 0.096 0.21 Pf= 0.060 L13 26 167 q= 0.000 2 10.000 Pt= 12.967 Mat="SlO" 56 Q= 93. 266 2.157 ' 0.000 0=120 Pe= 0,000 L14 26.167 10.000 0 067 Pf= 0.671 report created by the "Simple Hydraulic Calculator" version 2.3.2 from www.igneusinc.com 111128/2012 Page 9 Pipe Information. cont. Node I Discharge Nom Fittings I [ft] total (Pt) Pipe Elev & Flow i.d. num & length F [ft] C factor elev (Pe) Node 2 [ft] K-factor [gpm] [in] (ft] T [ft] psi/ft frict (Pf) Notes L14 26.167 q= 0.000 2 0.500 Pt= 12.296 Mat="S10" 57 Q= OR. 343 2.251 0.000 C=120 9e= 0.000 L15 26.167 0.500 0.028 Pf= 0.014 L15 26.167 q= 0.000 2 9. 500 Pt= 12.292 Mat510" 58 Q== 24.08o 7 0.000 0=129 Pe- 0.000 L16 2' .167 0 0.05 P6= 0.052 L17 26.1 7 q= 0.000 2 0.500 Pt= 12.270 Mat'S10" 59 Q= 10.76 2.157 0.000 0=120 Pe= 0.000 L16 26.167 0.500 0.001 P6= 0.00 RF4 26. 167 q= 0 . 000 2 IT-12 . 20 4 .200 Pt= I . ° Mat="010" 60 Q= 44.98o 2.107 12. 307 C=10 Pe= 0.000 L17 2 6.67 i 1.007 0 . 017 Pf= 0.288 P. FA 26.167 q= 0.000 2 1T=12.307 2.250 Pt= 17.519 Mat="S]O" 61 Q= 68.80 2.157 12.307 C=120 Pe= 0.000 L18 7 17.557 0.039 P6= 0.56q L18 2. 167 q= 0.000 2 0.000 P0= 11. 855 Mat=" S 10" Q= 33.8 4 2. 157 0,000 C=20 Pe= 0.000 L19 26.167 . . 0.500 0.01 P6= 0.005 RN5 26.167 q= 0.000 2 97.83 0 Pt= 31.867 Mat0 0" 63 Q= 157.482 2. 7 0.000 C=12 0 Pe= 0.000 020 21 6.167 27,9'7 0 .i77 P6= 17. 305 L20 26.157 q= 0.000 2 10.000 Pt= 1.558 Mat="S10" 64 Q= 11°,800 2,157 0.000 0=170 Pe= .0.000 021 26.167 10 000 0.107 Pf= 1.067 621 26.167 q= 0.2/0 2 12 000 Pt= 12.421 Mat="010" 65 Q= 83.081 2.157 0.000 C=120 Pe= 0.000 622 26.167 10.000 0.04 Pf= 0.542 L22 26.167 q= 0.000 2 1T=12. 07 4.750 Pt= 1.900 Mat 'S10" 66 Q= 9.866 2.57 12.07 0=120 Pe= 0.000 RFS 26 167 1.057 0.0 0 P6= 0.7z0 RF5 26 167 q= 0.000 2 T-1 2 07 5.o00 Pt= 12.52 9 Ma t-S 10 67 Q= 155. 540 2.157 12. 07 0=120 Pe= 0.000 L23 26.167 17.557 0.173 P6= 3.0 5 L23 2 0 •(7 q= 0.000 2 6.000 Pt= 9,94 Mat=". 50" 68 Q= 122.5 47 2 157 0 00 C=I 20 Pe ' 0.00 0 024 2.1o7 6.000 0.111 Pf= 0.6k 7 624 26.167 q= 0.000 2 10.000 Pt= 8.J27 Mat=" S 10" 69 Q= 91 461 2.17 0.000 0=120 Pe= 3.000 L25 26.67 10.000 0.065 Pf= 0.647 L25 26 . '7 q= 6.000 2 0. /0 Pt= 8.280 Mat="S 10" 70 Q= 60.283 2.157 0.000 C=20 Pe= 0.000 L26 26.167 , , 0.500 0.0' Pf= 0.01 report created by the "Simple Hydraulic Calculator" version 2.3.2 from www.igneusinc.com 11/28/2012 Page 10 Pipe Information. cont. Node I Discharge Nom Fittings I [ft] total (Pt) Pipe Elev & Flow i.d. num & length F [ft] C factor elev (Pe) Node 2 [ft] K-factor [gpm] [in] (ft] T [ft] psi/ft frict (Pf) Notes L26 26.167 q= 0.000 2 6.167 Pt= 8.265 Mat"S1O" 71 Q= 30.0d0 2.157 0.000 C=120 Pe= 0.000 L27 26.767 6.177 0.008 Pf= 0.051 RN6 27. 167 q= 0.000 2 J7.8 5= 3, 1 .485 Mat='SlO" 72 Q= 162.836 1.157 0.000 C=120 Pe= 0.000 L28 26.1(7 97.323 0.188 P6= 18.409 L28 26.167 q= 0.000 2 0.500 P5= 12.076 Mat='SlO" 73 Q= 677.855 2.657 0.000 C=120 P e = 0.000 L29 26.167 0.500 0.12 P6= 0.00 L29 26.s7 q= 0 0 2 10.000 Pt= i 3 .Ow6 Mat=" S 10" 74 Q= 92. 993 2.15 0.000 C=120 Pe= 0.000 L30 25.167 i 0.000 0 .067 P6= 0.667 L30 26.167 q= 0.000 2 0.500 Pt= 12.349 Mat="S10" 75 Q= 5 8.475 2.157 0.000 C=120 Pe= 0.000 L31 26.167 0.500 0.023 Pf= 0.014 L31 2 6 .7(7 q= 0.000 9 . 50 Pt= 12.7 5 Mat='710" 76 Q= 24. 2 .157 0.00s C=1 20 Pe= 0.000 L32 25. 1 67 5.00 0.000 P6= 0.05' L33 26.167 q= 0.000 2 0.500 Pt= 12.282 Mat- S10" 77 Q= 10.087 2.157 0.000 0=120 Pe= 0.000 L32 25.167 - 0.500 0 . 001 P6= 0.001 RF6 2 6.107 q= 0. 000 2 ST=i 2 . 70 / . 250 Pt= 1 . i Mat=" S 10 78 Q= 44.186 2. 7 _2. 07 C=120 Pe= 0.000 L33 26.1 7 o6.57 0.017 P6= 0.279 RFO 2 6.167 q= 0.000 2 1T=12.307 5.20 P5= 12.561 Mat=" S 10" 79 Q= 67.8 0 2.157 17.307 0=120 Pe= 0.000 L34 26 167 17.557 0.037 P6= 0.654 L34 /0.17 q= 0.000 2 0.500 P6= 11.17 Ma t=,, S tO" 80 Q= 33.75 .157 0 .000 C=10 Pe= 0. 000 L35 /6.167 0. 00 0.01 P6= 0.005 RN7 26.167 q= 0.000 2 17=12.307 1r7.500 Pt= 33.20 Mat="S10" 81 Q- 129.940 2.157 . 12 307 C=120 Pe- 0.000 RF7 6.167 129 507 0.124 P6= 17. 326 RN8 2 6 .167 q= 0.000 2 iT=72.07 127.D00 Pt= 33 .214 Mat in 82 Q= 129.897 ,'.157 2.307 C-120 Pe= 0.000 RFB 28.167 = 13 2.307 0.124 Pf= 17.317 Li 26.167 q= 0.000 1 175.000 18.417 Pt= 16.296 Nat=" S 10' 83 Q-= 39.194 1.u59 1E=2.000 7.000 C=120 Pe= -7.433 1 9.000 25.4/7 0.4 2 P6= 11.483 L2 2 i(2 q= 0. 00 1 iT5.000 / . 7 P0= 16.016 Mat=2Z40" 84 Q= -, 669 1.049 1E=2.000 7.000 C=20 Pe= -7.433 2 9.000 23.'17 0.2 P6= 12.138 report created by the "Simple Hydraulic Calculator" version 2.3.2 from www.igneusinc.com 111/28/2612 Page 11 Pipe Information. cont. Node I Discharge Nom Fittings I [ft] total (Pt) Pipe Elev & Flow Id. num & length F [ft] C factor elev (Pe) Node 2 [tt] K-factor [gpm] [in] (ft] T [ft] psi/ft frict (Pt) Notes L3 26.167 q= 0.000 1 1T5.000 21.917 Pt= 15.323 Mat="S40" 85 3 7.084 1.,b49 16=2.000 7.000 C=120 Pe= -7.433 3 9.000 2 B.17 0.408 Pf= 1.79 L4 2 6.]67 q= 0.000 1 1T=5.000 18.417 Pt= 18.740 Mat='340" 86 Q= 3 8.715 1.040 1E-2.000 7.000 6=120 Pe= -7.433 4 9.000 25.417 0.442 P1= 11.25 L5 26 .i7 q= 0.000 1 1T= .000 18.487 Pt= 14.720 Mat="O" 87 . Q= 37.828 1.049 1E=.000 7.000 C=120 Pe- -7.433 5 9.000 25.417 0.23 P1= 10. 754 L6 26. 167 q= 0. 000 3 1T= . 000 8 .4 7 Pt= 1 . 23 8 Mat=" $ 0" 88 Q= 37.3 1.09 16=2.000 7.0(0 C=120 Pe- -7.'q 3 6 9.000 25.417 0.±4 Pf= 10 525 L7 2(167 q= 0.000 1 1T=5.000 18.417 Pt= 1 .772 Ma5="540" 89 Q= 36 . 973 049 1E=2.000 7.000 C=120 Pe= -7 433 7 2 1.417 0 . 406 P6= 553 LB 28.167 q= 0.000 1 1T=5.000 20.417 Pt= 1.751 t4a8="S40" 90 Q= 7.979 1.049 1 E =2 000 7.000 C=120 Pe= -1.433 8 9.000 27.417 0.426 P6=11.695 L9 26.167 q= 0.000 1 1T=9 .000 20.417 Pt= 1.744 Mat="540" 37.1 13 1.049 1E=2.000 7.000 C-120 Pe= 4 33 9 9.000 2 7.417 0.408 P 11 f=11.197 L10 2 6.167 q= 0.000 1 11=5.000 20.4 17 Pt= 14.25 1 Mat="S40" 92 Q= 36.681 1 .049 1E-.900 7.00 C=120 Pe- -7.43 10 9.000 27 . 417 0.4 P6= 10.9 8 Lii 2 6.167 q= 0.000 1 1T=5.000 20.417 P8= 13. 780 Mat="040" 93 Q= 36.266 1 .09 16=2.000 /000 C=120 Pe= -7.4 3 11 9.000 27 .4 17 0.3°1 P6= 10. /29 L12 7o.167 q= 0.000 1 1T=5.000 20.417 P5= 1 3 .027 Mat="540" 94 Q= 35 91 1.049 1E=2.000 7.000 C=120 Pe= -7.433 12 9.00 21.417 9. 378 P1= 10.362 L13 26.167 q= 0.000 1. 11=5.000 22 .417 Pt= 12.967 Mat="S40" 95 Q= 3 4 871 1.049 1E=2 000 7.000 C=120 Pe= -7.4 3 13 9.000 . 29.411 0. 364 Pf= 10.706 L14 6 .167 q= 8.001 1 1T=8.000 20.417 P8= 12.29 6 Mat="S40" 96 Q= 14.93 1 .049 16=2.000 2.000 C=12 0 Fe= -7.4 3 - 14 9.000 . 77.417 0.365 P1= 10.006 L15 26.187 q= 0.000 1 1T=5.000 2 2.417 Pt= l.82 Mat='540" 97 Q= A.257 1.049 1E=2.090 7.000 C=120 Pe= -7.43 15 9.000 2D.17 0.022 Pf= 10.360 L16 26. 67 q= 0.000 1 1T=5.000 20.417 Pt= 12.230 Nat'S40" 98 Q= 34 863 1.049 16=2.000 7.000 0=120 Pe= -7.433 16 9.000 27.417 9.364 Pf= 9. 974 report created by the "Simple Hydraulic Calculator" version 2.3.2 from www.igneusinc.com 11/28/2b12 Page 12 Pipe Information. cont. Node I Discharge Nom Fittings I [ft] total (Pt) Pipe Elev & Flow id. num & length F [if] C factor elev (Pe) Node 2 [ft] K-factor (gpm] [in] (ft] T [ft] psi/ft frict (Pf) Notes L17 26.167 q= 0.000 1 1T=5.000 22.417 Pt= 12.230 Mat="S40' 99 Q= 34.210 1.249 15=2.000 7 000 C-120 e- -7.433 17 9.000 29.417 0.351 Pf= 10.334 518 76.167 q= 0.000 1 1T=0.000 20.4i7 P6= 11 855 Mat="548" 100 Q= 4.16 1 .049 1E=7.000 7.000 0=±20 Fe= .433 18 9.020 ±7.417 0.J5 P5= °.791 L19 26.127 q= 0.000 1 1T=5.000 22.417 Pt= 12.850 Mat="540" 101 Q= 3.864 1 .049 10=2.000 7.000 C=120 Pe= 7.433 19 9.000 ±9.41 0 .3,45 P5= 10.141 L20 26.167 q= 0.000 1 1T=2.000 18.417 Pt= 12.558 Mat="S40" 102 Q= 57.677 1.042 15=2.000 7.000 C=120 Pe= -7.433 20 9 .000 . - 5.47 0 42 P5= 10.674 L21 26.167 q= 0.000 1 11=5.000 18.417 P5= 1 .491 Mat'S40" 103 Q= 76.18 1 .049 1E=2.000 7.000 0=120 Pe- -7.43 21 9 .005 25.4 17 0.4 P5= 10.177 L22 26.167 q= 0.000 1 1T= .000 18.417 P2= 12.950 Mat=" 40" 104 . Q= 36 221 1.049 10=2.000 7.000 0=170 Pe= -7.433 22 9.00 ±.4l 0 .39 Pf= 0.974 L23 26.167 q= 0.000 1 1T=5.000 18.417 Pt= 9592 Mat=' 40" 105 . Q= 32 . 963 1.049 10=2.000 7.000 0=120 Pe- _7433 23 0.000 2 .417 0. 329 Pf= 8.350 L24 2 . 67 q= 0.000 1 1T.000 22.2 17 Pt= 8.927 Mat="SO0" 10.6 Q= 71.082 1.049 10=2.000 7.000 C=1. 20 Pe= 4 33 24 9.000 2 9 .47 3.204 Pf= 6.156 L25 26.167 q= 0.000 1 1T=5.000 19.017 Pt= 8.280 MatS40" 107 Q= 31.278 1.049 1E=2.000 7.000 0=170 Pe= -7.433 25 7.0o0 26. 917 0.296 p5= 7. 96 2 L26 26.167 q= 0.000 1 1T=".000 22.217 Pt= 8.285 Mat='S40" ioe Q= 0.±$3 ±949 1E=2.000 7.000 C-140 Pe= -7.43 26 0 050 2 9.0 1 7 0.22 Pf= 6. 387 L27 26.162 q= 0.000 1 11= .000 23.750 Pt= 8.2 2 Mat' 2 0" 109 Q= 00.000 1.02 9 10-2.000 7.000 C=120 Pe= _7•433- 27 0 .000 3 0.780 979 p5= 8 .4 72 L28 11 6.167 q= 0.000 1 11=5.000 22.417 Pt= 13. 0 7 6 14at='740" 110 Q 4.9 8 1.049 1E2 000 7.000 0=170 Pe= -7.43 28 0 .000 29 •417 0.3 0 P5= 10.781 L29 ±6. 16 7 q= 0.000 1 1T= .000 22.417 P5= 1 .016 Ma t=2'S 40 111 Q= 34.975 1.049 10=7.000 7.000 0=120 Pe= -7.433 29 0.000 - 29.4l 0. 36 5 P5= 10 .73 L30 26.167 q= 0.000 1 1T=5.000 21.917 Pt= lo.342 Mat="520" 112 - Q= 34.478 1 049 1E=2.000 7.000 C=120 Pe= 433 30 9.000 ±9.91 0.358 P5=10.300 report created by the "Simple Hydraulic Calculator" version 2.3.2 from www.igneusinc.com 11/28/2012 Page 13 Pipe Information. cont. Node I Discharge Nom Fittings L [ft] total (Pt) Pipe Elev & Flow id. num & length F [ft] C factor elev (Pe) Node 2 [ft] K-factor [gpm] [in] [ft] T [ft] psi/ft frict (Pt) Notes L31 26.167 q= 0 00 1 1T=5.000 22.917 Pt= 12.75 ?4at'S40" 113 Q= 34 1,46 1.049 1E=2.000 7.000 C=1O Pe- 7.433 31 9.000 29.917 0.35 Pf= 10.473 L32 7b. 16 q= 0 . 000 I 11=.000 21.9L7 Pt= ].202 Mat=" S 4 0" 114 Q= 34.4571.049 1E=.000 7.000 C=120 Pe- 433, 32 9.000 28.917 0.355 Pf= 10.27 L33 - 26.167 q= 0.000 1 1T=5.000 22.91 Pt= 12.02 Mat= 40 115 Q= 34.09 1.047 1E-2 000 7.000 C-120 Pe= 433 33 0.000 - 29.917 0. 349 Pf7 10.46 L34 26.167 q= 0.000 1 1T=5.000 21.17 Pt= 11.977 Ma t"i40" 116 Q 7.D7 5 1.049 1E-2.000 7.000 C120 Pe -74 3 3 34 9.000 28.917 0.349 Pf= 10.004 L35 26.167 q= 0.000 1 1T=5.000 L2.9 7 Pt= 11.002 Mat40° 117 Q= 33.75j 1 .09 1E2.000 7.005 C 20 Pe= -7.433 35 9.000 23.917 0.34 33 Pf 10.5 Material Codes Pipe Material Fittings Sb - Schedule 10 Steel - Standard 90 degree elbow S40 - Schedule 40 Steel - Gate Valve CD152 - Cement Lined Ductile Iron Thickness Class 52 - Tee - Flow turn 90 degrees report created by the "Simple Hydraulic Calculator" version 2.3.2 from www.igneusinc.com Hydraulic Calculations for Project: Signet Armorlight, Inc 5803 Newton Dr. Carlsbad, CA ing no.: 12059 Date: 11/2812012 Design Remote area number: 2 Remote area location: Utility Room Occupancy classification: ordinary hazard group II Density: 0.20 gpm Area of application: 1500 sf Coverage per sprinkler: 130 sf Type of sprinklers calculated: Standard Spray SR pendent 5.6k ordinary temp No. of sprinklers calculated: 13 In rack demand: na Hose streams: 250 gpm outside + 0 inside Total water required (including hose streams): 615.55 gpm at 96.76 psi [15.25 psi safety margin Total water required at base of system riser: 365.55 gpm at 85.31 psi Type of system: wet pipe Volume of dry or preaction system: na Water SuDDlv Information Date: 11-1-2012 Location: 5803 Newton Dr Source: city water main Contractor: American Automatic Fire Suppresion, Inc. 9510 Pathway Street, Suite B Santee, Ca 92071 C-16, 731167 (619) 562-3010 Name of designer: Debbie Seeman Authority having jurisdiction: City of Carlsbad Fire Department Notes Static and residual pressures reduced by 10% KRI 11/28/2012 Hydraulic Demand Graph (D) a a a aas4.is_---- 80 P S 60 Page 2 Water Source: A)113.4 psi Static B)700 gpm at 111.64 psi Source at BOR: 104.45 psi Static 365.6 gpm at 100.56 psi 700 gpm at 91.52 psi Demand at BOR: 365.6 gpm at 85-31 psi 140 120 100 Ew 20 i1tIII!4sII] 300 400 500 700 GPM report created by the Simple Hydraulic Calculator" version 2.3.2 from www.igneusinc.com 111/28/2012 Page 3 SuDDly Analysis Static Residual Available Total Required Node at Pressure Pressure Flow Pressure Demand Pressure [psi] [psi] [gpm] [psi] [gpm] [psi] SRC 113.4 106.2 1500.0 112.01 615.55 96.76 Node Analysis Elev, Pressure Req Disch Discharge Elev, Pressure Req Disch Discharge Node Tag [ft] Type [psi] [gpmI (gpm] Node Tag (ftj Type [psi] (gpmj [gpm] SRC -3.000 source 96.760 -615.554 M7 11.333 ref 25.230 0.000 - .000 ref 95.3 050.000 2.000 58 11.333 ref 25.155 0.030 01 -3.000 ref 95.058 0.000 01 11.333 ref 45,441 0.000 32 ref, 87.475 0.000 B2 11 .333 re f 33.17 0.000 B '1 000 ref 92.939 0. u0fl 33 i 1 .379 re 27 9 9 0.000 B 2 3 .000 ref. 84.908 0.000 34 11.337 ref 27.920 0.000 A -3.000 ref. 87.0 0.000 BS 11.337 ref 24.239 0.000 BOR 1 . 00 ref 85. 310 0.000 BO 11 333 r e 24. 06 0.000 TOP 2 797 r=f 74.824 8.000 B 7 11.333 ref 22.7 3 7 0.000 Ni 27.339 ref 74.485 0.000 B8 11.333 re f 22.802 0.000 52 23 . 333 r ef 74.246 0.000 09 11 . 333 ref 22.180 0.000 53 27.3 3 ref 74 211 0.000 510 11. .333 rf 22.214 0.000 N4 73,3 7 re f 74.190 0 .000 311 11.33 ref 22.10 0.000 N5 23.333 ref 74. 81 0.000 101 10.000 5=5.60 42.517 26.000 36.515 NO 23.333 ref 74.373 0.000 102 10.000 5=5.60 31.2 2 26,000 71. 21 57 73. 33 r e 74. 69 0.000 103 1.0.000 5=5.60 25,3 6 26.000 28.188 N8 23.33 ref 74.168 0.000 104 10.000 5=5.60 75.00° 25.000 28.005 3111 /6,167 ref 72.659 0.000 105 10.000 K=5.60 25.40/ 26.000 78.224 3112 2 6.i67 ref 72.539 0.000 106 10.002 5°.60 20.0° 3 26.000 28.041 RN7 26.187 ref 72.513 0.000 107 10.000 5=0.60 23.118 26.000 /5.926 3114 76.16/ ref 72 . 493 0.000 109 10.000 5= .60 23.132 26.000 26.962 RN5 /6.167 ref 72.479 0.000 10 10.000 K=5.60 27.210 26.000 2.3°1 RN6 72.i6 7 ref 7r.469 0.000 1/0 10.000 R.60 2°. 72 7.000 76.428 RN7 26.i67 ref 72.462, 0.000 in 10.000 5=5.60 21.986 26.000 26.258 ROB 7(167 ref 72.462 0.000 112 10000 5=7.60 22.048 26.000 26.29 31 23. 333 f e 71.°4 0.000 113 10.000 5=5.60 2 . 56 26.000 26.000 32 23.333 ef 71.053 0.000 F3 Z3 333 ref 71.049 0.000 r4 2 .333 ref 71.040 0.000 5° /3. 333 =1 7i.05 0.000 - 26 23.333 ref 71.003 0.000 57 23.333 r° 70.983 0.000 59 77. . re f 70.78z 0. 00° T 113 .333 r ef 70.980 0.000 R F 1 2r.167 re f 70.158 0.000 3112 /6.167 r1 70.091 0.000 R F3 26 157 r 70.10° 0.020 324 26.167 ref 70 075 0.000 R 7/ .367 r 6 70.000 0.000 356 n .167 ref 70.079 0.000 337 26.167 ref 70.019 0.000 3118 26.167 ref 70.020 0.000 Ml 23.331 rLf 50,306 o.000 52 _l.3°3 r ef 51.50 9 .020 i1.73 ref 37.0 6 0.000 M4 1 1.333 ref 30.201 0.000 35 1 1. 1133 ref 21.34 0.000 56 1.333 ref 35.8 7 0.000 report created by the "Simple Hydraulic Calculator" version 2.3.2 from www.igneusinc.com 11/28/2012 Page 4 Pipe Information negative pipe flow (Q) indicates flow is from node 2 towards node I Node I Discharge Nom Fittings L [ft] total (Pt) Pipe Elev. & Flow i.d. num & length F [ft] C factor elev (Pe) Node 2 [ft] K-factor [gpm] [in] [ft] T [ft] psi/ft frict (P0 Notes SRC -3.000 q=-615.554 B 2T=110.709 300.000 Pt= 96.760Mat='CD152" 1 Q= 611 . 534 8.27 1G=6. 26 17 .97W 0=140 Pe= 0.000 FH -3.000 2E=56.931 47 .971 0.002 pf= 1.128 FH - .000 q= 2c0.000 0 1G6. 26 35.000 Pt= 95. 631 230 PM HOSE 2 Q= 365.54 2,7 1E=28.468 34.794 C=±40 Pe= 0.000 Mat='flDI52" Ui -3.000 - 69.7°4 0.001 06= 0.063 Ui -3.900 q= 0.000 5 1E=28.468 6.000 Pt= 95.568MatC13157' 3 Q= 365.54 8.27 28.468 0=140 Pe= 2.598 BFI 3.000 - 4.462 0.001 06= 0.031 NFl 3 .000 q= 0.000 8 1E=28,468 5.000 Pt= 92.9 9 8" DCDA 4 Q= 365.554 . -7 28.968 C=40 Pe= 0.000 MatCDI52" NF2 3.000 -. /3.468 0.001 Pf= 0.030Pdev,0si BF2 3.000 q= 0.000 8 1E=28.468 6.000 Pt 2.908 Mat="CDIS2" 5 Q= 365. 5,54 8.27 28- 4 68 0=140 Pe = - 12 .598 02 -3.000 34.468 0.001 06= 0.031 02 -3. 000 q= 0.000 8 3E5.40 118. 500 Pt= 87. 475 MaL=00152' 6 Q= 3 s5,5 4 8.2i 8.404 0=140 Re= 0.000 A -3.000 , 203.Q04 0.001 P6= 2.185 A -3.000 q= 0.000 8 1E=21.11 4.667 Pt= 57.290 Mat=" S 10 7 Q= 35,554 5,29 21.741 C=L0 Pe= I .4B BOO 1.500 - . 25.808 0.001 Pf= 0.03= BOO 1.500 q= 0.000 8 1G=4.698 21.833 Pt= 8.310 FLOW SW 8 Q= 36.554 8,29 4.698 C=120 Pe= 9,454 Mat="10" - TOO 2 . 33 , 26.5 1 0.001 06= 0.0 20dev=i.0 pi TOO 23. 333 q= 0.000 6 2E=3 .205 42.750 Pt= 74.824 Mat="S10" 9 Q= 365 . 554 6. 3J57 35.205 C=120 Re= 0.000 Ni 23.33 , 77.955 0.00 P6= 0. 339 Ni 2 . q= 0.000 6 2E=35.205 37.167 Pt= 74.485 M&t'Sl0' 10 Q= 314.731 6. 57 35 .205 0=120 Pe= 0.000 N2 23,033 , 72. 3 7 2 0.003 P6= 0.239 N2 2 . 3 3 q= 0.000 6 ' 12.000 Pt= 7.246 Nat=" S 10" 11 Q= 269.u22 6. 3 57 0.000 C=0 Re= 0.200 N3 23 .333 12,000 0.90/ P6= 0.0 0 - N3 2. 333 q= 0.000 6 12.000 Pt= i.21 Mat=" S 10" 12 Q= 224,7fl7 s.357 0.000 C=20 Pe- 0.000 N4 2 3 . 12.000 0.00/ 06= 0.021 N4 23 . 333 q= 0.000 6 12.000 Pt= 74.195 Mat="S1O' 13 Q= 179.857 6.357 0.000 0=120 Re= 0.000 N5 2 . 33 1/000 0.001 Pf= 0, .014 N5 23 . 333 q= 0.000 6 12.000 Pt= 74.181 MatS1 0" 14 Q= 135 .059 6.357 0.000 0=120 Pe= 0.000 N6 2' .3 3 - 12.000 0.001 Pf= 0.008 report created by the "Simple Hydraulic Calculator" version 2.3.2 from www.igneusinc.com Page 5 11/28/2012 Pipe Information. cont. Node I 1 Discharge Nom Fittings I [ft] total (Pt) Pipe Elev. & Flow i.d. num & length F [ft] C factor elev (Pe) Node 2 [ft] K-factor [gpm] [in] [ft] T [ft] psi/ft frict (P0 Notes N6 23.333 q= 0.000 6 12.000 Pt= 74.173 Mat'SOO" 15 Q= 90.125 6.357 0.000 C=290 Pe= 0.300 N7 27.333 12.000 0.000 Pf= 0.004 N7 2 3 . 3 3_3 q= 0.000 1 12.000 Pt= 74.169 Nat=' 10 16 Q= 45.094 6 . 17 0.000 C=20 Pe= 0.000 N8 2 . 333 12.000 9 Pf= 0.001 51 73633 q= 0.000 6 12.000 Pt= 71.054 Mat="S10" 17 Q= 50 . 823 6.357 0.000 0120 Pe- 0.000 52 2 . 73 12 .000 0.000 Pf= 7.001 52 2 . 333 q= 0.000 6 12.000 Pt= 71.053 MatS10" 18 Q= 99.932 1.757 0.000 C=I 2 0 Pe= 0.000 F3 2 .333 12.000 0.030 Pf= 0.004 53 2 . 333 q= 0.000 6 12 300 Pt= 71. •349 Mat="10" 19 Q= 140 847 6.357 0.000 C=120 Pe= 0.000 F4 23.333 12.000 0.001 Pf= 0.009 54 2 . q= 0.000 1 1 .000 Pt= 7 .040 Mat="610" 20 Q= 185.668 6.357 0.000 C=120 Pe= 0.000 F5 23.3 3 . 12.000 0.001 Pf= 0.015 55 2?. 333 q= 0.000 1 1.O00 Pt= 71.025 Mat="910" 21 Q= 2 30.495 6.7 0.000 C=1 20 Pe= 0.000 FE 23. 3 37 12.000 0.002 Pf= 0.022 FE 23 . 333 q= 0.000 6 8.83 Pt= 71.003 Mat="S10" 21A Q= 275.429 0.357 0.000 C=120 Pe- 0.000 T 2?.333 8 833 0.003 Pf= 0.02 57 23 333 - q= 0.000 ' 3.167 Pt= 70.B1 Mat10" 22 Q= 90.125 6.3o7 0.000 0=120 Pe- 0.000 T 23 333 3 . 67 0.000 Pf= 0.001 58 23.3 q= 0.000 6 12.000 Pt= 70 . 961, Mat="010" 23 Q 45.054 6 3 57 0.000 C=120 Pe= 0.000 57 23 . 333 12.000 0 Pf= 0.001 Ni 2 •33 q= 0.000 2 2T=24.61 .8 3 Pt= 7.483 Mat="6 0" 24 Q= 30.823 24.6k? C=120 Pe= 1 .227 RN1 26.167 a.447 0.072 Pf= 0.599 N2 27. 333 q= 0.000 2 2T=24.613 2.833 Pt= 7.241 Mat=" 10" 25 Q= q.209 2.197 24.(1? C-1,20 Pe= 1.227 RN2 26.167 0.05 Pf= 0.480 N3 23. 333 q= 0.000 2 2T=24.613 2.8 3 Pt= 74.216 Mat= 10" 26 Q= 44.915 2.157 24.613 0=120 Pe= 1.227 RN3 26.167 27.447 0.717 Pf= 0.47 N4 23.733 q= 0.000 2 2T24.613 2 . 833 Pt= 74.195 Mat=" S 10 27 Q 44.821 2.i57 24 613 C120 2e= 1.227 RN4 z6 167 27.447 0.017 Pf= 0.475 report created by the "Simple Hydraulic Calculator" version 2.3.2 from www.igneusinc.com 11/28/2012 Pipe Information, cont. Node I Discharge Nom Fittings L (ft] total (Pt) Pipe Elev & Flow i.d. num & length F [ft] C factor elev (Pe) Node 2 [ftj K-factor [gpm] [in] [ftj T [ft] psi/ft frict (Pf) Notes N5 23.133 q= 0.000 2 2T=24.613 2.833 Pt= 74.181 Mat="clO" 28 Q= 44.87 2.1 24.612 C=±20 Pe= 1.227 RN5 27.47 0.017 P6= 0.676 N6 21. 33,3 q= 0.000 2 2T=24.613 2 . 833 Pt= 74 . 173 Mat=" S10" 29 Q= 44 234 2.157 24.613 C=I 20 Pe= 1.227 RN6 26,16/ 7.447 0.017 Pf= 0.477 N7 . 13 q= 0.000 2 2T=24.013 2 . 833 Pt= 76.169 Mat='blO 30 . Q= 45.071 2 157 2 4.613 C=20 Pe= 1 .227 RN7 26.167 27.447 0.017 P6= 0.480 N8 z7.13 q= 0.700 2 2T=24 . 6132.833 Pt= i. 16 8 Mat="S10" 31 Q= 45 04 2.157 24.61 3 C=120 Pe- 1.227 RN8 26.167 . 27,467 0,017 Pf= 0.479 RF1 20.167 q= 0.000 2 1T=12.307 2.8 7 P6= 70.158 Mat="S10" 32 Q= 50 . 823 2.157 12.307 C=/20 Pe= - .227 Fl 2 . 3 15.40 0. 022 P6= 0.R 0 RF2 26. 67 q= 0.000 2 1T=12. 307 2 . 833 Pt= 70.091 MatS10" 33 Q= 45.10 2.157 12.307 0=126 Pe= -1.227 F2 23.331, 16.140 0.018 P6= 0.265 RF3 26.i67 q= 0.000 2 1T-1, 2 .3 07 2.823 P6= 70 . 085 Mat- 210" 34 Q= 44.9 5 2.57 12.307 C=120 Pe= -1.227 F3 23 . 33 15.140 0.017 Pf= 0.263 RF4 26. 161 q= 0.000 2 1T=12.707 2 . 833 Pt= 70,075 Mat="S10" 35 Q= 44.821 ±.167 12307 0= 0 Pe= -1.227 P4 2 333 15.140 0.017 P6= 0.262 RFS 26.167 q= 0.000 2 1T=1. 307 2.833 Pt= 70.060 Mat="SlO" 36 Q= 44.877 2.1 7 12.307 0=120 Pe= -1.227 P5 2 .3 3 1.140 0 . 01.7 P6= 1,6/ RF6 26.167 q= 0.000 2 1 T=I .307 2 633 Pt= 70.039 Mat= 10 37 . Q= 44,94 2.157 12.107 C=120 Pe= -1.227 FE 23.313 15.140 0.017 Pf= 0.263 RF7 2C.67 q= 0.0"O 2 1T=1. 307 2 . 833 P6= 7 0.019 Ma t=" S 10" 38 Q= 45.071 2.177 12. 307 0=120 Pe= -1.227 F7 23 . 3133 15.140 0.017 P6= 0.26 RF8 2 6.'I7 q= 0.000 2 1T=52. 307 2.8 3 3 P6= 70.020 Ma6="S10" 39 Q= 45.04 - 15 2.307 P=120 Pe= -_.227 F8 21.333 15.140 0 017 Pf= 0±64 RN1 26,167 q= 0.000 2 1T12.307 10±.250 P6= i2.59 Mat=" S 10" 40 Q= 50. 82 3 2.157 12.307 C=20 Pe= 0.000 RF1 ±.167 1±4.5 57 0.02± Pf= 2.501 RN2 ±6.167 q= 0.000 2 1T=12.307 127.500 Pt= 72.539 Mat='510" 41 Q_ 45,i09,12 .157 12.107 0=120 P = 0.000 RF2 26.167 139.807 0.018 Pf= 2.447 report created by the "Simple Hydraulic Calculator" version 2.3.2 from www.igneusinc.com Page 6 Page 7 11/28/2012 Pipe Information. cont. Node I Discharge Nom Fittings L [ft] total (Pt) Pipe Elev & Flow i.d. num & length F [ft] C factor elev (Pe) Node 2 [ft] K-factor [gpm] [in] (ft] T [ft] psi/ft frict (Pf) Notes RN3 26.167 q= 0.000 2 1T=12.307 127.500 Pt= 72.513 Mat='SlO' 42 Q= 44.915 2.17 12.307 C=120 Pe= 0.000 RF3 26.167 139.807 0.017 Pf= 2.428 RN4 26.167 q= 0.000 2 1T=12.307 17.500 Pt= 72.493 Mat- 010" 43 Q= 44.821 2.157 12.307 C=123 Pe= 7.000 RF4 26.167 13 9.797 0,017 Pf= 2.41 RNS 26.167 q= 0.000 2 1T=12.107 127.500 Pt= 7.479 Mat="S1O" 44 - Q= 44.827 2.157 12.307 ,C=l 20 Pe= 0.000 RFS 26.167 139.807 0.017 05= 2.419 RN6 6.117 q= 0,000 2 1T=22. 07 127.500 Pt= 72. 4 69 Mat'S10" ,45 Q= 44 934 2 157 12. 107 C=I 20 Pe- 0.000 RF6 26. 67 139. 807 07017 Pf= 2.4110 RN7 26,167 q= 0.000 2 1T=12.707 127.500 Pt= 72462 Mat'S10" 1 Q= 45.07 2.157 , 12.307 0= 20 -,.0,. 000 R07 26.i57 1 9.707 0.917 P5= 7.444 RN8 207167 q= 0.000 2 1T=12.307 127,070 Pt= 7.' .072 Mat="S0" 47. Q= 4 054 2.077 12.3 07 C=120 Pe=. 0.000 008 26.167 . 139.807 0.017 Pf= 2.442 T . 333 q= 0.000 2 5 1T=16.474 24. 750 Pt= 70.980 Mat="070" 48 Q= 650754 2.635 1E=8.2 7 207711 0=120 Pe= 0.000 Ml 23. 3 3 47 461 0,3i7 P5=15.674 Ml 23. 333 q= 0.000 2.5 12.000 Pt= 07.306 Mat='SlO" 49 Q= 165.554 2.65 0.000 0=070 Pe- -5.196 M2 11 . 333 12.000 0. 317 Pf= 3.203 M2 11.333 q= 0.000 2 1E=1.15 .250 Pt= 51.697 Mat0710" 50 Q= 3 65.5 5 4 2.077 6.153 0=120 Pe= 07000 Bi I 1.337 13407 0.84 P5=11.258 BI 11. 333 q= 0.000 2 1E=6.153 5.717 Pt= 45.441 Mat"SlO" 51 Q= 3207039 2.157 .073 0070 Pe= 0.000 M3 15, 33 12.070 0.091 P5= 8. 40 M3 11 .333 q= 0.000 2 1000 Pt= 37.0°6 Mat-, - S 10" Q= 297.718 2.157 0.000 C=120 Pe- 0.000 M4 11.3 3 , - 12.000 0.575 Pf= 6.895 M4 11.3 1 q= 0.000 2 52.090 Pt=30 201 Mat="610 53 Q= 075.760 2.157 0.000 0=120 Pe= 0.000 MS 11. 33 07.000 072 .° P5= 2.867 MS 11.373 q= 0.000 2 12.000 Pt= 27. 33 4 Mat'S10" 54 Q= 1 1.372 2.257 0.000 0=120 Pe= 0.000 M6 37 2.000 0.076 P5= 1. A M6 I 1 .333 q= 0.010 2 107000 Pt= ?5 . 8 17 .007 Mat=" S 10" 55 Q= 70755 2.077 0.000 0=120 Pe- 07000 M7 11:333 . 12.000 0.044 P1= 0.586 report created by the "Simple Hydraulic Calculator" version 2.3.2 from www.igneusinc.com 11/28/2012 Pipe Information. cont. Node I Discharge Nom Fittings L [ft] total (Pt) Pipe Elev. & Flow i.d. num & length F [ft] C factor elev (Pc) Node 2 [ft] K-factor [gpm] [in] (ft] T [ft] psi/ft frict (Pt) Notes M7 11.333 q= 0.000 2 12.000 Pt= 25.230 Mat"S10" 56 Q 26.000 2 . 57 0.000 0=123 Pe= 0.000 M8 11.33 12.000 0.006 Pf= 0.076 Bi 11.733 q= 0.000 1 1T=5.000 1.833 Pt= 45.441 Mat'540' 57 Q= 36.65 1.042 1E=2.000 7.000 C=120 Pe- -0.577 101 10.000 8 . 833 0. 396 Pf= 3.001 M3 11. 03' q= 0.000 1 2T=10 .000 0 333 Pt= 77.026 Mat'.40" 58 Q= 31.321 1.042 10.000 0=120 Pe= 0.000 B2 11.3 3 3 13 .33 0.298 P5= 3.97° B2 1 1.3 q= 0.000 1 2E=4.000 4.08 1 Pt= 33.117 Ma t=" S 40' 59 Q= 3 1 .721 1.040 4.000 C=120 Pe= -0. 7 102 10.000 . 8 083 0.98 Pf= 2.412 M4 ii. 33 q= 0.000 1.25 1T=6.000 4.167 Pt= 30.201 Mat="540" 60 Q= i 56.i0 3 1. 8 6.000 0=120 Pe= 0.000 B3 11. 333 0. 67 0.2 1 P5= /. 3 5 P3 11.3 3 q= 0.000 1 1T=5.000 6.583 Pt= 27.849 Mat="S00" 61 . Q= 28.188 1.049 1E=2.000 7.000 C=120 Pe-- -0.577 103 10.900 12 583 0. 46 Pf= 3.089 B3 11. 3 3 q= 0.000 1 15= .000 7.083 Pt= 2 7.249 Mat="540' 62 Q= 28.0051 049 1E=2.000 7.000 C=120 Pe= -0.577 104 10.000 . 14 . 083 0.243 Pf7 3 416 M4 11. 333 q= 0.000 i .2i 15=(.000 3 .8 33 3 Pt= 30.201 Mat=" S4 0" 63 Q= 05 5 1. 38 6.000 0=120 Pe= 0.000 B4 10. 333 9.833 0 232 P5= 2.281 B4 11.3 3 q= 0.000 1 1T=5.000 5.583 Pt= 27.90 MatS40" 64 Q= 28.22 i.09 1E=2.000 7.009 0=120 Pe= -0.577 105 1 U.000 12.083 0.240 P5= 7.097 04 11.3 3 q= 0.000 1 1T=5.000 7 083 P6= 27.920 Mat 40' 65 Q= 28.041 1.049 1E=7.000 7.000 0=120 Pe= -0.577 106 i0.000 14.083 0.24 3, P5= 3 405 MS 1 31 q= 0.000 1 2T=10.000 4.167 Pt= 27. 34 Mat="S40" 66 Q= 09 6 1.049 0.000 C=120 Pe= 0.000 05 11.3 0 . 14.167 0.276 P5= 3.196 BS 11 . 33 q= (1.000 1 1E=2.000 5 083 P6= 24 13 ° Ma2="540" 67 Q= 26 .32 1.049 2 000 C=20 Pe= -0.577 107 10.000 7.083 0.226Pf= 1.598 MS 11.033 q= 0.000 1 2T=10.000 '.803 P6= z7.334 Mat="540" 68 Q= 26 '02 1.049 /0.000 C=10 Pe= 0.000 B6 I 1 333 13.93 0.226 Pf= 3.128 B6 11.333 q= 0.000 1 1E=2.000 5.08 3, Pt= 24.206 Mat="S40" 69 Q 26.92 1.042 2.000 C120 Pe= -0.577 108 13.000 7.083 0.226 25= 1.602 report created by the Simple Hydraulic Calculator" version 2.3.2 from www.igneusinc.com Page 8 11/28/2012 Page 9 Pipe Information. cont. Node I I Discharge Nom Fittings L [ft] total (Pt) Pipe Elev , & Flow l.d. num & length F [ft] C factor elev (Pe) Node 2 [ft] K-factor [gpm] [in] [ft] T [ft] psi/ft frict (Pf) Notes M6 11.333 q= 0.000 1 2T10.000 4.167 Pt= 25.817 Mat'S40" Q= 26.391 1.0-49. 10.000 c=126.. Pe= 0O00 B7 11.33 - 1.167 0.217 Pf= 3.079 B7 11.333 q= 0.000 1 16=2.600 3.083 Pt= 22.737 Mat 40" 71 Q= 26.391 1.049 .2.000 C=12QPe=.O,577 - 109 10.000 5..083 0.217 Pf= 1.105 M6 11.333 q= 0.000 1 2T=10.000 3. 833 Pt= 25.817 Mat='340" :72 -. Q 26.428 .049 10.000 C=120 Pe=.0,O0 B8 11.333 13.833 0.218 Pf= 3.015 B8 11 333 q= 0.000 1 - 1E=2.000 3 .083 Pt= 22.802 MatS40" 73 z Q= 26.428 1.049 2.000 C=170 Pe= -0.577 110 10.000 . 5.083 0.218 Pf= 1.108 M7 11.333 q= 0.000 1 2T=10.000 4.167 Pt= 25.230MatS40" 74 Q= 26.258 1.049 10.000 C=120 Pe-- 0.000 B9 11.333 14.167 0 215 Pf= 3.051 B9 11. 333 q= 0.000 1 1E2.000 1.583 Pt= 22.180 Mat="S40" 75 . . Q= 26.258 1.049 2 2.000 C=120 Pe= -0.577 . iii 10 000 . 3: 583 0.215 ?,. 0.772 M7 11.333 q= 0.000 1 2T=10.000 3.8 3 Pt= 25.230 Mat=" 40" Q= 26.295 1.049 10.000 0=120 P= 0.000 B 1 11.3 . 13.833 0.216 20= 2.986 B10 11.333 q= 0.000 1 1E=2.000 1.583 Pt= 22.244 Mat="540" 77 . Q= 26.25 1 049 200O .Cl20-, Pe= -0.577 112 10.000 -, .3.583 0.216 Pf= 0774 8 11. 333 q= 0.000 3 - 2T=10.000 4.167 Pt= 25.155 Mat'S40" 78- - Q= 26.000 1.049 . . 10.000 C120.. Pe= 0', 000 Bli 11.333 . ...- 14.16i 0.211 Pf= 2.995 - Bli 11.333 q= 0 000 1 1E=2 000 3 583 Pt= 22.159 Mat S40 79.1.049 . - 2.000 C=120-Pe= -0;577 . 113 10.000 5.583 0.211 Pf= 1.181 Material Codes Pipe Material S10 - Schedule 10 Steel S40 - Schedule 40 Steel Fittings E - Standard 90 degree elbow G - Gate Valve CD152 - Cement Lined Ductile Iron Thickness Class 52 T - Tee - Flow turn 90 degrees report created by the "Simple Hydraulic Calculator" version 2.3.2 from www.igneusinc.com