Hoja11.-BELT CONVEYOR ARRANGEMENT NRO. 30 CV 00118.5312.5208.33333333332.-DATA39.216584Q =44.20TPH44.2B =72.00inchs0.6985032074137.5L =87.24ft145.56C =121.80lb/ft3V =283.00FPMH =0.00ft7.983520Lb =46.26ft45.92Vo =0FPM(Belt Feeder)3.-WEIGHT OF MATERIAL, LBS PER FT OF BELT LENGTH4800Wm =Q x 2000=5lb /ft60 x V4.-EFFECTIVE BELT TENSIONTe =L x Kt ( Kx + Ky.Wb + 0.015.Wb) + Wm.(L.Ky + H) + Tp + Tam + Tac (lbs)Wb - Weight ofbelt in pound per foot of belt lengthWb =26lb/ft(Tab. 6.1 - CEMA / 5ta Ed.)Kt - Ambient Temperature Correction FactorKt =1.1(Fig. 6.1 - CEMA / 5ta Ed.)Kx - Idler Friction FactorKx =0.00068 ( Wb + Wm) + ( Ai / Si )Ai =1.8for 4 - inches idler spacigSi =4.5fttroughing idler spacigKx =0.421Ky - Factor for Calculating the Forces of Belt and Load Flexure Over The IdlersKy =( Wb + Wm ) x A x 10-4 + B x 10-2A =2.15Table 6.4 Idler Spacing 3 ftB =1.565Average belt tension 1000 LbsKy =0.022Tp - Tension resulting from resistance of belt to flexure aroud pulleys and the resistanceof pulleys to rotation on their bearings, total for all pulleysTp =200lbs(Tab. 6.5 - CEMA / 5ta Ed.)Tam - Tension resulting from the force to accelerate the material continuously as it is fedonto the beltTam =2.8755 x 10-4 x Q x VTam =3.60lbsTac - Total of the Tensions from conveyors accessoriesTac =Tsb + Tpl + Ttr + Tbc lbsSkirtboard friction Tsb :Tsb =(Cs x D2 + 6) x LCs =2 x dm ( 1 - sin f)288 ( 1+ sin f)dm =121.80lb/ft3f =36oangle of repose of material, degrees.Cs =0.220D =0,1 x B7.2inchsTsb =1608lbsTension resulting from belt sag Betwen Idlers To :To =6.25 x Si x (Wb + Wm)To =878lbsTension resulting from belt pull requerid for belt-cleanig devices Tbc :Tbc =3 x B =432lbsTac =2040lbsBelt Tension Calculus :Te =2388lbs6.-POWER REQUIREMENTSBelt hp :Te x V(hp)hp1 =20.5hp33000Drive pulley hp := 200 x Vhp2 =1.7hp330005% for speed reduction shaft= 0,05 ( hp1 + hp2)hp2 =1.1Horspower at motor shaftShp23.3hp11.25HP Motor Select :HPM =28.0hp7.-BELT TENSIONST2a =To + Tb - TyrT2b =Cw x TeChoice the larger value of T2Tension resulting from belt sag Betwen Idlers To :To =6.25 x Si x (Wb + Wm)To =878lbsTension resulting from the force needed to lift or lower the belt Tb :Tb =H x WbTb =0.00lbsTension resulting frrom the resistance of the belt as it rides over the return idlers Tyr :Tyr =L x 0.015 x Wb x KtTyr =37lbsWrap Factor Cw :Cw =0.8Wrap angle 180 no SnubT2a =840lbsT2b =1910lbsT2 =840lbsT1 =Te + T2T1 =3228lbsTt =T2 + Tyr - Tb =878Tt =878lbs8.-SELECTION OF PULLEY SIZE AND BELTBelt Tension =T1=45PIW1206310B43From Table 1 ANSI/CEMA B105.1-19922.7906976744With 180 arc of contact for 16" pulley diameter is acceptable, Max. 195PIW68.4Diameter Drive Pulley =24.75inchesDiameter Tail Pulley =20inchesBelt150PIW9.-DRIVE EQUIPMENTN =12 x V44RPM1097.5772438609p x Dp17501.75027624311.5944208116REDUCER :Mark:Sumitomo114.3Type:SM Shaft Mount114.935ClassClass II (conveyors)Model215GRatio25Speed Out44RPMWeight:74.00Kg.MOTOR:Mark:WEGType:Efficiency EstndarModel:213THP:10N:1760V:440 V 3FWeight:72.00Kg.10.-SHAFTINGSDrive Pulley Shaftb=18R =T1 + T2 + W(Vectorial)W =210lbsR =4137.93lbsD ={ 32.FS/p [ (K.M/Sf)2 + 3/4 (T/Sy)2 ]1/2 }1/3FS =1.5N =6inchsA =7.25inchsB =33inchsC =8inchsM =R. (N + 1,25)=15000lb - inch2T =(T1 - T2 ) x Dp29548lb - inch2P =321.2=321.2lbJ =P.C(B-A)=0.07RBAK =1.11Table 5 ANSI CEMA B 105,1-1992Sf =ka.kb.kc.kd.ke.kf.kg.Sf*ka =0.8kb=0.7909kf =0.63kc=0.897kd =1ke=1Sf* =41000PSIAISI 1045Sy =45000PSIAISI 1045Calculating :Sf =14660PSID =2.72inchsDiam. at hub2 3/4"Diam. at Bearing2 1/4"Tail Pulley ShaftW=150lbsR =2191.73lbsMb =R. A=7945lb - inch2Calculating :Sf =14660PSID=32*Mbp*SbD =1.77inchDiam. at hub2 1/4"Diam. at Bearing1 3/4"

&LFYAGA ING. S.A.C.&R&14PRELIMINARPgina &P de &N3

Hoja1 (2)1.-BELT CONVEYOR ARRANGEMENT NRO. 218.52.-DATA39.216584Q =312.00TPHB =24inchs0.6985032074137.5L =62.39ft60.6862.3973727422C =178lb/ft3V =300FPMH =12.1ft3706H12.15568Lb =11.48ft11.4817983LhVo =0FPM(Belt Feeder)3.-WEIGHT OF MATERIAL, LBS PER FT OF BELT LENGTHWm =Q x 2000=34.7lb /ft60 x V4.-EFFECTIVE BELT TENSIONTe =L x Kt ( Kx + Ky.Wb + 0.015.Wb) + Wm.(L.Ky + H) + Tp + Tam + Tac (lbs)Wb - Weight ofbelt in pound per foot of belt lengthWb =6lb/ft(Tab. 6.1 - CEMA / 5ta Ed.)Kt - Ambient Temperature Correction FactorKt =1(Fig. 6.1 - CEMA / 5ta Ed.)Kx - Idler Friction FactorKx =0.00068 ( Wb + Wm) + ( Ai / Si )Ai =2.3for 4 - inches idler spacigSi =4fttroughing idler spacigKx =0.603Ky - Factor for Calculating the Forces of Belt and Load Flexure Over The IdlersKy =( Wb + Wm ) x A x 10-4 + B x 10-2A =2.2Table 6.4 Idler Spacing 3 ftB =2.25Average belt tension 1000 LbsKy =0.031Tp - Tension resulting from resistance of belt to flexure aroud pulleys and the resistanceof pulleys to rotation on their bearings, total for all pulleysTp =350lbs(Tab. 6.5 - CEMA / 5ta Ed.)Tam - Tension resulting from the force to accelerate the material continuously as it is fedonto the beltTam =2.8755 x 10-4 x Q x VTam =26.91lbsTac - Total of the Tensions from conveyors accessoriesTac =Tsb + Tpl + Ttr + Tbc lbsSkirtboard friction Tsb :Tsb =(Cs x D2 + 6) x LCs =2 x dm ( 1 - sin f)288 ( 1+ sin f)dm =178lb/ft3f =38oangle of repose of material, degrees.Cs =0.294D =0,1 x B2.4inchsTsb =177lbsTension resulting from belt sag Betwen Idlers To :To =6.25 x Si x (Wb + Wm)To =1017lbsTension resulting from belt pull requerid for belt-cleanig devices Tbc :Tbc =3 x B =72lbsTac =249lbsBelt Tension Calculus :Te =1168lbs6.-POWER REQUIREMENTSBelt hp :Te x V(hp)hp1 =10.6hp33000Drive pulley hp := 200 x Vhp2 =1.8hp330005% for speed reduction shaft= 0,05 ( hp1 + hp2)hp2 =0.6Horspower at motor shaftShp13.1hpHP Motor at 4500 m.s.n.m= ShpHPMIN =16.3hp0.8HP Motor Select :HPM =17.5hp7.-BELT TENSIONST2a =To + Tb - TyrT2b =Cw x TeChoice the larger value of T2Tension resulting from belt sag Betwen Idlers To :To =6.25 x Si x (Wb + Wm)To =1017lbsTension resulting from the force needed to lift or lower the belt Tb :Tb =H x WbTb =72.6lbsTension resulting frrom the resistance of the belt as it rides over the return idlers Tyr :Tyr =L x 0.015 x Wb x KtTyr =6lbsWrap Factor Cw :Cw =0.8Wrap angle 180 no SnubT2a =1084lbsT2b =934lbsT2 =934lbsT1 =Te + T2T1 =2102lbsTt =T2 + Tyr - Tb =867Tt =867lbs8.-SELECTION OF PULLEY SIZE AND BELTBelt Tension =T1=88PIWBFrom Table 1 ANSI/CEMA B105.1-1992With 180 arc of contact for 20" pulley diameter is acceptable, Max. 345PIW68.4Diameter Drive Pulley =20inchesDiameter Tail Pulley =16inchesBelt150PIW9.-DRIVE EQUIPMENTN =12 x V57RPMp x DpREDUCER :Mark:SumitomoType:SM Shaft MountClassClass II (conveyors)Model215GRatio25Speed Out57RPMWeight:74.00Kg.MOTOR:Mark:WEGType:Efficiency EstndarModel:213THP:10N:1760V:440 V 3FWeight:72.00Kg.10.-SHAFTINGSDrive Pulley Shaftb=18R =T1 + T2 + W(Vectorial)W =210lbsR =3108.20lbsD ={ 32.FS/p [ (K.M/Sf)2 + 3/4 (T/Sy)2 ]1/2 }1/3FS =1.5N =6inchsA =7.25inchsB =33inchsC =8inchsM =R. (N + 1,25)=11267lb - inch2T =(T1 - T2 ) x Dp11680lb - inch2P =321.2=321.2lbJ =P.C(B-A)=0.09RBAK =1.11Table 5 ANSI CEMA B 105,1-1992Sf =ka.kb.kc.kd.ke.kf.kg.Sf*ka =0.8kb=0.7909kf =0.63kc=0.897kd =1ke=1Sf* =41000PSIAISI 1045Sy =45000PSIAISI 1045Calculating :Sf =14660PSID =2.41inchsDiam. at hub2 3/4"Diam. at Bearing2 1/4"Tail Pulley ShaftW=150lbsR =2166.18lbsMb =R. A=7852lb - inch2Calculating :Sf =14660PSID=32*Mbp*SbD =1.76inchDiam. at hub2 1/4"Diam. at Bearing1 3/4"

&LFYAGA ING. S.A.C.&R&14PRELIMINARPgina &P de &N3

Hoja1 (3)1.-BELT CONVEYOR ARRANGEMENT NRO. 218.52.-DATA39.216584Q =18.00TPHB =24inchs0.6985032074137.5L =62.39ft60.6862.3973727422C =178lb/ft3V =300FPMH =12.1ft3706H12.15568Lb =11.48ft11.4817983LhVo =0FPM(Belt Feeder)3.-WEIGHT OF MATERIAL, LBS PER FT OF BELT LENGTHWm =Q x 2000=2.0lb /ft60 x V4.-EFFECTIVE BELT TENSIONTe =L x Kt ( Kx + Ky.Wb + 0.015.Wb) + Wm.(L.Ky + H) + Tp + Tam + Tac (lbs)Wb - Weight ofbelt in pound per foot of belt lengthWb =6lb/ft(Tab. 6.1 - CEMA / 5ta Ed.)Kt - Ambient Temperature Correction FactorKt =1(Fig. 6.1 - CEMA / 5ta Ed.)Kx - Idler Friction FactorKx =0.00068 ( Wb + Wm) + ( Ai / Si )Ai =2.3for 4 - inches idler spacigSi =4fttroughing idler spacigKx =0.580Ky - Factor for Calculating the Forces of Belt and Load Flexure Over The IdlersKy =( Wb + Wm ) x A x 10-4 + B x 10-2A =2.2Table 6.4 Idler Spacing 3 ftB =2.25Average belt tension 1000 LbsKy =0.024Tp - Tension resulting from resistance of belt to flexure aroud pulleys and the resistanceof pulleys to rotation on their bearings, total for all pulleysTp =350lbs(Tab. 6.5 - CEMA / 5ta Ed.)Tam - Tension resulting from the force to accelerate the material continuously as it is fedonto the beltTam =2.8755 x 10-4 x Q x VTam =1.55lbsTac - Total of the Tensions from conveyors accessoriesTac =Tsb + Tpl + Ttr + Tbc lbsSkirtboard friction Tsb :Tsb =(Cs x D2 + 6) x LCs =2 x dm ( 1 - sin f)288 ( 1+ sin f)dm =178lb/ft3f =38oangle of repose of material, degrees.Cs =0.294D =0,1 x B2.4inchsTsb =177lbsTension resulting from belt sag Betwen Idlers To :To =6.25 x Si x (Wb + Wm)To =200lbsTension resulting from belt pull requerid for belt-cleanig devices Tbc :Tbc =3 x B =72lbsTac =249lbsBelt Tension Calculus :Te =678lbs6.-POWER REQUIREMENTSBelt hp :Te x V(hp)hp1 =6.2hp33000Drive pulley hp := 200 x Vhp2 =1.8hp330005% for speed reduction shaft= 0,05 ( hp1 + hp2)hp2 =0.4Horspower at motor shaftShp8.4hpHP Motor at 4500 m.s.n.m= ShpHPMIN =10.5hp0.8HP Motor Select :HPM =5hp7.-BELT TENSIONST2a =To + Tb - TyrT2b =Cw x TeChoice the larger value of T2Tension resulting from belt sag Betwen Idlers To :To =6.25 x Si x (Wb + Wm)To =200lbsTension resulting from the force needed to lift or lower the belt Tb :Tb =H x WbTb =72.6lbsTension resulting frrom the resistance of the belt as it rides over the return idlers Tyr :Tyr =L x 0.015 x Wb x KtTyr =6lbsWrap Factor Cw :Cw =0.8Wrap angle 180 no SnubT2a =267lbsT2b =543lbsT2 =543lbsT1 =Te + T2T1 =1221lbsTt =T2 + Tyr - Tb =476Tt =476lbs8.-SELECTION OF PULLEY SIZE AND BELTBelt Tension =T1=51PIWBFrom Table 1 ANSI/CEMA B105.1-1992With 180 arc of contact for 20" pulley diameter is acceptable, Max. 345PIW68.4Diameter Drive Pulley =20inchesDiameter Tail Pulley =16inchesBelt150PIW9.-DRIVE EQUIPMENTN =12 x V57RPMp x DpREDUCER :Mark:SumitomoType:SM Shaft MountClassClass II (conveyors)Model215GRatio25Speed Out57RPMWeight:74.00Kg.MOTOR:Mark:WEGType:Efficiency EstndarModel:213THP:10N:1760V:440 V 3FWeight:72.00Kg.10.-SHAFTINGSDrive Pulley Shaftb=18R =T1 + T2 + W(Vectorial)W =210lbsR =1839.45lbsD ={ 32.FS/p [ (K.M/Sf)2 + 3/4 (T/Sy)2 ]1/2 }1/3FS =1.5N =6inchsA =7.25inchsB =33inchsC =8inchsM =R. (N + 1,25)=6668lb - inch2T =(T1 - T2 ) x Dp6783lb - inch2P =321.2=321.2lbJ =P.C(B-A)=0.15RBAK =1.11Table 5 ANSI CEMA B 105,1-1992Sf =ka.kb.kc.kd.ke.kf.kg.Sf*ka =0.8kb=0.7909kf =0.63kc=0.897kd =1ke=1Sf* =41000PSIAISI 1045Sy =45000PSIAISI 1045Calculating :Sf =14660PSID =2.02inchsDiam. at hub2 3/4"Diam. at Bearing2 1/4"Tail Pulley ShaftW=150lbsR =1187.89lbsMb =R. A=4306lb - inch2Calculating :Sf =14660PSID=32*Mbp*SbD =1.44inchDiam. at hub2 1/4"Diam. at Bearing1 3/4"

&LFYAGA ING. S.A.C.&R&14PRELIMINARPgina &P de &N3

Hoja1 (4)1.-BELT CONVEYOR ARRANGEMENT NRO. 218.52.-DATA39.216584Q =350.00TPHB =24inchs0.6985032074137.5L =62.39ft60.6862.3973727422C =178lb/ft3V =250FPMH =12.1ft3706H12.15568Lb =11.48ft11.4817983LhVo =0FPM(Belt Feeder)3.-WEIGHT OF MATERIAL, LBS PER FT OF BELT LENGTHWm =Q x 2000=46.7lb /ft60 x V4.-EFFECTIVE BELT TENSIONTe =L x Kt ( Kx + Ky.Wb + 0.015.Wb) + Wm.(L.Ky + H) + Tp + Tam + Tac (lbs)Wb - Weight ofbelt in pound per foot of belt lengthWb =6lb/ft(Tab. 6.1 - CEMA / 5ta Ed.)Kt - Ambient Temperature Correction FactorKt =1(Fig. 6.1 - CEMA / 5ta Ed.)Kx - Idler Friction FactorKx =0.00068 ( Wb + Wm) + ( Ai / Si )Ai =2.3for 4 - inches idler spacigSi =4fttroughing idler spacigKx =0.611Ky - Factor for Calculating the Forces of Belt and Load Flexure Over The IdlersKy =( Wb + Wm ) x A x 10-4 + B x 10-2A =2.2Table 6.4 Idler Spacing 3 ftB =2.25Average belt tension 1000 LbsKy =0.034Tp - Tension resulting from resistance of belt to flexure aroud pulleys and the resistanceof pulleys to rotation on their bearings, total for all pulleysTp =350lbs(Tab. 6.5 - CEMA / 5ta Ed.)Tam - Tension resulting from the force to accelerate the material continuously as it is fedonto the beltTam =2.8755 x 10-4 x Q x VTam =25.16lbsTac - Total of the Tensions from conveyors accessoriesTac =Tsb + Tpl + Ttr + Tbc lbsSkirtboard friction Tsb :Tsb =(Cs x D2 + 6) x LCs =2 x dm ( 1 - sin f)288 ( 1+ sin f)dm =178lb/ft3f =38oangle of repose of material, degrees.Cs =0.294D =0,1 x B2.4inchsTsb =177lbsTension resulting from belt sag Betwen Idlers To :To =6.25 x Si x (Wb + Wm)To =1317lbsTension resulting from belt pull requerid for belt-cleanig devices Tbc :Tbc =3 x B =72lbsTac =249lbsBelt Tension Calculus :Te =1344lbs6.-POWER REQUIREMENTSBelt hp :Te x V(hp)hp1 =10.2hp33000Drive pulley hp := 200 x Vhp2 =1.5hp330005% for speed reduction shaft= 0,05 ( hp1 + hp2)hp2 =0.6Horspower at motor shaftShp12.3hpHP Motor at 4500 m.s.n.m= ShpHPMIN =15.4hp0.8HP Motor Select :HPM =5hp7.-BELT TENSIONST2a =To + Tb - TyrT2b =Cw x TeChoice the larger value of T2Tension resulting from belt sag Betwen Idlers To :To =6.25 x Si x (Wb + Wm)To =1317lbsTension resulting from the force needed to lift or lower the belt Tb :Tb =H x WbTb =72.6lbsTension resulting frrom the resistance of the belt as it rides over the return idlers Tyr :Tyr =L x 0.015 x Wb x KtTyr =6lbsWrap Factor Cw :Cw =0.8Wrap angle 180 no SnubT2a =1384lbsT2b =1075lbsT2 =1075lbsT1 =Te + T2T1 =2420lbsTt =T2 + Tyr - Tb =1008Tt =1008lbs8.-SELECTION OF PULLEY SIZE AND BELTBelt Tension =T1=101PIWBFrom Table 1 ANSI/CEMA B105.1-1992With 180 arc of contact for 20" pulley diameter is acceptable, Max. 345PIW68.4Diameter Drive Pulley =20inchesDiameter Tail Pulley =16inchesBelt150PIW9.-DRIVE EQUIPMENTVel Faja HuarEstimado MoroN =12 x V48RPM63.661828367742.441218911853.0785562633p x Dp199.8924REDUCER :Mark:Sumitomo83.3 10600 1200 1.46 II 3370 15000 15 6165 Y B 21 AV[H] A-128 A-180 A-208Type:SM Shaft Mount15 HP, 11 kW, 60 Hz, 1750 RPMClassClass II (conveyors)Model215GRatio25Speed Out48RPMWeight:74.00Kg.MOTOR:Mark:WEGType:Efficiency EstndarModel:213THP:10N:1760V:440 V 3FWeight:72.00Kg.10.-SHAFTINGSDrive Pulley Shaftb=18R =T1 + T2 + W(Vectorial)W =210lbsR =3565.42lbsD ={ 32.FS/p [ (K.M/Sf)2 + 3/4 (T/Sy)2 ]1/2 }1/3FS =1.5N =6inchsA =7.25inchsB =33inchsC =8inchsM =R. (N + 1,25)=12925lb - inch2T =(T1 - T2 ) x Dp13442lb - inch2P =321.2=321.2lbJ =P.C(B-A)=0.08RBAK =1.11Table 5 ANSI CEMA B 105,1-1992Sf =ka.kb.kc.kd.ke.kf.kg.Sf*ka =0.8kb=0.7909kf =0.63kc=0.897kd =1ke=1Sf* =41000PSIAISI 1045Sy =45000PSIAISI 1045Calculating :Sf =14660PSID =2.52inchsDiam. at hub2 3/4"Diam. at Bearing2 1/4"Tail Pulley ShaftW=150lbsR =2518.13lbsMb =R. A=9128lb - inch2Calculating :Sf =14660PSID=32*Mbp*SbD =1.85inchDiam. at hub2 1/4"Diam. at Bearing1 3/4"

&LFYAGA ING. S.A.C.&R&14PRELIMINARPgina &P de &N3

Hoja2

Hoja3

MBD0006DC70.unknown

MBD0006DC72.unknown

MBD001E3CBC.unknown

MBD001E3CBD.unknown

MBD001E3CBA.unknown

MBD001E3CBB.unknown

MBD0006DC73.unknown

MBD0006DC71.unknown

MBD0005EBDA.unknown

MBD0005EBDC.unknown

MBD0005EBDD.unknown

MBD0005EBDB.unknown

MBD00018FAF.unknown

MBD0004EFB3.unknown

MBD00047868.unknown