(1) 2 No’s of 120KW, 415V motor, 90% Efficiency, 0.8 P.F.
(2) 2 No’s of 150KW, 415V motor, 90% Efficiency, 0.8 P.F.
(3) 100KW, 415V Lighting Load. The targeted P.F. Of system is 0.98
Total Load KVAR1=145.98 KVAR
OR
Total KVAR=KVAR1+ KVAR2+KVAR3
Total KVAR=145.98+182.33+54.7
Total KVAR=383.01 KVAR
Leading KVAR supplied by each Phase= KVAR/No of Phase
Leading KVAR supplied by each Phase =383.01/3
Capacitance of Capacitor = (Ic)/ 2πfVc
Size of the fuse = 165% to 200% of Capacitor Charging current.
Size of the fuse= 2×532.84 Amp
Size of the fuse= 1065.68 Amp
Size of the Circuit Breaker=1.5×532.84 Amp
Size of the Circuit Breaker=799.26 Amp
Thermal relay setting between 1.3 and 1.5 of Capacitor Charging current.
Thermal relay setting of C.B. = 1.5×532.84 Amp
Thermal relay setting of C.B= 799.26 Amp
Magnetic relay setting between 5 and 10 of Capacitor Charging current.
Magnetic relay setting of C.B. = 10×532.84 Amp
Magnetic relay setting of C.B.= 5328.4 Amp
Cables size for Capacitor Connection= (1.3 + 1.1) x nominal capacitor Current
Cables size for Capacitor Connection = 2.4 × nominal capacitor Current
Cables size for Capacitor Connection= 2.4 × 532.84 Amp
Cables size for Capacitor Connection= 1278.816 Amp
Maximum size of discharge Resistor for Capacitor:
Capacitors will be discharge by discharging resistors.
After the capacitor is disconnected from the source of supply, discharge resistors are required for discharging each unit within 3 min to 75 V or less from initial nominal peak voltage (according to IEC-standard 60831). Discharge resistors have to be connected directly to the capacitors. There shall be no switch, fuse cut-out or any other isolating device between the capacitor unit and the discharge resistors.
Max. Discharge resistance Value (Star Connection) = Ct / Cn x Log (Un x√2/ Dv).
Max. Discharge resistance Value (Delta Connection) = Ct / 1/3×Cn×log(Un×√2/ Dv)
Reduction in KVAR size of Capacitor when operating 415V unit at 390V
Actual KVAR = Rated KVAR x (Operating voltage / Rated voltage)^2
Actual KVAR = Rated KVAR x (390/415)^2
Actual KVAR=88.3% of Rated KVAR
Hence 383.01 KVAR Capacitor works as 88.3% × 383.01KVAR = 338.19KVAR
(2) 2 No’s of 150KW, 415V motor, 90% Efficiency, 0.8 P.F.
(3) 100KW, 415V Lighting Load. The targeted P.F. Of system is 0.98
Calculate Size of Discharge Resister for discharging of Capacitor Bank Discharge rate of capacitor is 55v in less than 1 minute. Also calculate reduction in KVAR Rating of capacitor if Capacitor bank is operated at frequency of 30Hz instead of 50Hz and if operating voltage is 390V instead of 415V
Total Load KVAR1= KW×([(√1-(Old P.F.)^2) / Old P.F.] - [(√1-(New P.F.)^2 / New P.F])
Calculation:
For Connection (1):
Total Load KW for Connection (1) =KW / Efficiency=(120×2) / 0.9=266.67KWTotal Load KVAR1= KW×([(√1-(Old P.F.)^2) / Old P.F.] - [(√1-(New P.F.)^2 / New P.F])
Total Load KVAR1=266.67×([(√1-(0.8)^2) / 0.8] - [(√1-(0.98)^2) / 0.98])
Total Load KVAR1=145.98 KVAR OR
Total Load KVAR1= KW×(tanϕ1- tanϕ2) =266.67×(tan 36.87 - tan 11.48)
= 266.67×(0.75 - 0.203)
= 145.98KVAR
For Connection (2):
Total Load KW for Connection (2) =KW / Efficiency=(150×2) / 0.9=333.33KW
Total Load KVAR2= KW×([(√1-(Old P.F.)∧2) / Old P.F.] - [(√1-(New P.F.)∧2 / New P.F])
Total Load KVAR2= KW×([(√1-(Old P.F.)∧2) / Old P.F.] - [(√1-(New P.F.)∧2 / New P.F])
Total Load KVAR2=333.33×([(√1-(0.8)∧2) / 0.8] - [(√1-(0.98)∧2) / 0.98])
Total Load KVAR2=182.33 KVAR
OR
Total Load KVAR2= KW×(tanϕ1- tanϕ2) =333.33×(tan 36.87 - tan 11.48)
= 333.33×(0.75 - 0.203)
= 182.33KVAR
For Connection (3):
Total Load KW for Connection (3) = KW = 100KW
Total Load KVAR3= KW×([(√1-(Old P.F.)∧2) / Old P.F.] - [(√1-(New P.F.)∧2 / New P.F])
Total Load KVAR3= KW×([(√1-(Old P.F.)∧2) / Old P.F.] - [(√1-(New P.F.)∧2 / New P.F])
Total Load KVAR3=100×([(√1-(0.8)∧2) / 0.8] - [(√1-(0.98)∧2) / 0.98])
Total Load KVAR3=54.7 KVAR
OR
Total Load KVAR3= KW×(tanϕ1- tanϕ2) =100×(tan 36.87 - tan 11.48)
= 100×(0.75 - 0.203)
=54.7KVAR
Total KVAR=KVAR1+ KVAR2+KVAR3
Total KVAR=145.98+182.33+54.7
Total KVAR=383.01 KVAR
Size of Capacitor Bank
Size of Capacitor Bank=383.01 KVARLeading KVAR supplied by each Phase= KVAR/No of Phase
Leading KVAR supplied by each Phase =383.01/3
=127.67KVAR per Phase
Capacitor Charging Current (Ic) = (KVAR per Phase×1000)/Volt
Capacitor Charging Current (Ic )= (127.67×1000)/(415/√3)
Capacitor Charging Current (Ic) = 532.84 Amp
Capacitor Charging Current (Ic) = 532.84 Amp Capacitance of Capacitor = (Ic)/ 2πfVc
C = 532.842 ÷ (2×3.14×50×(415/√3))
= 7070.4 µF
Protection of Capacitor Bank
Size of HRC Fuse for Capacitor Bank Protection:Size of the fuse = 165% to 200% of Capacitor Charging current.
Size of the fuse= 2×532.84 Amp
Size of the fuse= 1065.68 Amp
Size of Circuit Breaker for Capacitor Protection:
Size of the Circuit Breaker =135% to 150% of Capacitor Charging current.Size of the Circuit Breaker=1.5×532.84 Amp
Size of the Circuit Breaker=799.26 Amp
Thermal relay setting between 1.3 and 1.5 of Capacitor Charging current.
Thermal relay setting of C.B. = 1.5×532.84 Amp
Thermal relay setting of C.B= 799.26 Amp
Magnetic relay setting between 5 and 10 of Capacitor Charging current.
Magnetic relay setting of C.B. = 10×532.84 Amp
Magnetic relay setting of C.B.= 5328.4 Amp
Sizing of cables for capacitor Connection:
Capacitors can withstand a permanent over current of 30% + tolerance of 10% on capacitor Current.Cables size for Capacitor Connection= (1.3 + 1.1) x nominal capacitor Current
Cables size for Capacitor Connection = 2.4 × nominal capacitor Current
Cables size for Capacitor Connection= 2.4 × 532.84 Amp
Cables size for Capacitor Connection= 1278.816 Amp
Maximum size of discharge Resistor for Capacitor:
Capacitors will be discharge by discharging resistors.
After the capacitor is disconnected from the source of supply, discharge resistors are required for discharging each unit within 3 min to 75 V or less from initial nominal peak voltage (according to IEC-standard 60831). Discharge resistors have to be connected directly to the capacitors. There shall be no switch, fuse cut-out or any other isolating device between the capacitor unit and the discharge resistors.
Max. Discharge resistance Value (Star Connection) = Ct / Cn x Log (Un x√2/ Dv).
Max. Discharge resistance Value (Delta Connection) = Ct / 1/3×Cn×log(Un×√2/ Dv)
Where
Ct =Capacitor Discharge Time (sec)
Cn=Capacitance Farad
Un = Line Voltage
Dv=Capacitor Discharge voltage
Maximum Discharge resistance = 90/((7070.4/1000000) × log( 415×√2 /55)
Cn=Capacitance Farad
Un = Line Voltage
Dv=Capacitor Discharge voltage
Maximum Discharge resistance = 90/((7070.4/1000000) × log( 415×√2 /55)
Maximum Discharge resistance = 12.38 KΩ
Effect of Decreasing Voltage & Frequency on Rating of Capacitor:
The KVAR of capacitor will not be same if voltage applied to the capacitor and frequency changes
Reduction in KVAR size of Capacitor when operating 50 Hz unit at 40 Hz
Actual KVAR = Rated KVAR ×(Operating Frequency / Rated Frequency)
Actual KVAR = Rated KVAR ×(30/50)
Effect of Decreasing Voltage & Frequency on Rating of Capacitor:
The KVAR of capacitor will not be same if voltage applied to the capacitor and frequency changes
Reduction in KVAR size of Capacitor when operating 50 Hz unit at 40 Hz
Actual KVAR = Rated KVAR ×(Operating Frequency / Rated Frequency)
Actual KVAR = Rated KVAR ×(30/50)
Actual KVAR = 60% of Rated KVAR
Hence 383.01 KVAR Capacitor works as 60% × 383.01KVAR = 229.8KVAR
Hence 383.01 KVAR Capacitor works as 60% × 383.01KVAR = 229.8KVAR
Reduction in KVAR size of Capacitor when operating 415V unit at 390V
Actual KVAR = Rated KVAR x (Operating voltage / Rated voltage)^2
Actual KVAR = Rated KVAR x (390/415)^2
Actual KVAR=88.3% of Rated KVAR
Hence 383.01 KVAR Capacitor works as 88.3% × 383.01KVAR = 338.19KVAR
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