Documents

132kV SAG Calculation

Description
132kV SAG Calculation
Categories
Published
of 14
All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.
Related Documents
Share
Transcript
  DOC NO : HZL-BTN-ELE-DS-SY-028 DESIGN INPUTSystem Parameters Bay Location132kV TRANSFORMER FEEDERConductor type & strandsSINGLE BEAR ACSRInitial Tension (Max.)kg(1T per phase)c/c distance of tower (Maximum Span)mmGirder Width mmTower height mmHeight of the equipment below the conductormmNumber of Conductors Nos.Number of Insulator Strings Nos.Basic Wind Speed m/s (As per IS: 875 -1987,  Part : 3) Span (c/c tower - lg) mmMaximum Temperature °C (As per Clause 10.2, IS-802,pageno.9) Minimum Temperature°C  ACSR Conductor  Conductor unit weight kg/mm Conductor Area mm 2 Conductor overall diameter mmExpansion coefficient of conductor /°CElasticity modulus kg/mm 2 Tension Insulator   Number of discs per string Weight of each disc kg (As per Vendor drawing) Weight of hardwarekgMean Diameter of Insulator mmLength of each disc mmLength of hardware mmWidth of the hardwaremm1.1.1= SAG TENSION CALCULATION FOR 26.68 m SPAN SINGLE BEAR ACSR - 132kV TRANSFORMER FEEDER 1.0.01.1.01.1.21.1.31.1.41.1.51.1.61.1.71.1.81.1.111.1.91.1.101.2.11.2.21.1.121.1.131.2.01.2.31.2.41.2.51.3.01.3.41.3.11.3.21.3.31.3.71.3.51.3.6T 1 LL g H 1 H 2 n c n s V b L s T o T min m's A c d c α EW d W h n d 7508000=L d L h d i ====100026680=5955=1=1=47=25930=85=0=0.001213 =325.6=23.45=1.78E-05=8.2E+03=12=7.5=17.02=255=145=750=250d h 1 of 14  DOC NO : HZL-BTN-ELE-DS-SY-028 CALCULATION OF BASIC DESIGN PARAMETERSWeight of Disc insulator string (W wi )Length of the Disc insulator string (L str  )Conductor Chord length (L c )DESIGN CALCULTIONDesign wind speed (V d ) Wind zoneBasic wind speedm/sReliability level of structurefactor(As per IS 802, Clause 8.2, Pg no:3)Meteorological wind speedm/sRisk co efficientTerrain roughness co efficientDesign wind speedV r  x k 1 x k 2 34.19 x 1 x 1m/s Design Wind pressure (P d )Wind pressure on conductor (P c ) Drag Co efficient for conductor=(As per IS 802, Clause 9.2, Pg no:7)Gust response factor for 8000mm level=(As per IS 802, Table 7, Pg no:9)(For reliability level 1, Terrain category 2)Full wind pressure on conductor, Wind pressure on Insulator (P i ) Drag Co efficient for insulator=(As per IS 802, Clause 9.3, Pg no:9)Gust response factor for level=(As per IS 802, Table 6, Pg no:9)(For reliability level 1, Terrain category 2)Full wind pressure on insulator,34.190.0000715 x 1 x 1.660.0000715 x 1.2 x 1.864kg/mm2 kg/mm 2 kg/mm 2 0.00007150.00011871.6644711.375V b / k 0 11=34.19701.38N/m 2 ==kgn d  x L d 12 x 145W d  x n d  x n s =7.5 x 12 x 1=17402.0.12.0.32.0.43.0.03.1.0P d  x C d  x G c =1G c C d =C di 1.23.4.0V d ==0.6 x (34.19^2)==k 0 ==k 1 V r G ci 1.8642.0.0P c =3.2.03.3.0W wi =90mmL str ===L c =V b ==(L - L g ) - 2 x (L str   + L h )mm===20950(26680 - 750) - 2 x (1740 + 750)0.00016=P i =P d  x C di  x G ci =P d =0.6 x V d2 ==k 2 2 of 14  DOC NO : HZL-BTN-ELE-DS-SY-028 Equivalent weight of Conductor in loaded conditionFull wind load on conductor (W c )Equivalent weight of conductor at full wind (W 2 )Equivalent weight of insulator in loaded conditionFull wind load on insulator (W s ) (As per IS 802, Clause 9.3, Pg no:9) Equivalent weight of insulator at full wind (W i2 )Resultant insulator load on each sub conductor (W i )Equivalent load of insulator hardware in loaded conditionFull wind load on insulator hardwareEquivalent weight of hardware at full wind (W hT )Resultant Hardware load on each sub conductor (W hr  ) 3.5.0=W c 3.5.13.5.20.00279=kg/mmkg=30=P c  x d c =0.0001187 x 23.453.6.03.6.1W 2 ==3.7.13.6.23.6.3=/n c =0.003050.5 x P i  x d i  x L str   x n s m'sc 2 + W c2 =W i kg/mm=35.5W wi2 + W s2 0.5 x 0.00016 x 255 x 1740 x 1(0.001213^2) + (0.00279^2)kg(90^2) + (35.5^2)96.75=kgW s =W i2 ==96.75 / 13.7.0=96.75=W i2 3.7.2W wh W hT =W wh2 + W 2T kgW T =17.02 x 1=kg17.02=P i  x d h  x L h =34.492kg=(30^2) + (17.02^2)3.7.3=W hr =W hT /n c =34.492 / 10.00016 x 250 x 75034.492kg3 of 14  DOC NO : HZL-BTN-ELE-DS-SY-028 FULL WIND CONDITIONLoad distributionReaction at each endShear force diagram +163.1908 kg+66.4408 kg+31.9488 kg-31.9488 kg-66.4408 kgNOTE: LENGTH IN mm -163.1908 kgMaximum Sag occurs at the centre of the span Cross force area (Upto maximum sag)Cross Force moments kg 2 mmTS=(S 1  + S 2  + S 3  + S 4 ) x 2=32611885.8588191386.0915kg 2 mm1782017.1633kg 2 mm11584588.1840kg 2 mm2747951.4906kg 2 mm====141975.996 x 163.1908 x 0.582718.796 x 66.4408 x 0.511980.8 x 31.9488 x 0.5167331.84 x 31.9488 / 3S 3 =S 4 =S 1 =S 2 =4.1.4kg.mmkg.mmkg.mmkg.mm141975.99682718.79611980.8167331.840.5 x 31.9488 x 10475====4.1.3=I 4 ==I 2 =I 1 kg.mmSI 1 =404007.432163.1908 x 87066.4408 x 124531.9488 x 375I 3 4.1.2870124537510475104754.0.04.1.14.1.0=1740.0 mm96.750 kg96.750 kg34.492 kg0.003050 kg/mm34.492 kg A=96.75 + 34.492 + ((0.00305 x 20950)/2)kg1740.0 mm750 mm20950.000 mm=R  A R B 870163.1908kg750 mm1245375B4 of 14
We Need Your Support
Thank you for visiting our website and your interest in our free products and services. We are nonprofit website to share and download documents. To the running of this website, we need your help to support us.

Thanks to everyone for your continued support.

No, Thanks