Calculate Size of Capacitor Bank / Annual Saving & Payback Period
April 1, 2014 10 Comments
 Calculate Size of Capacitor Bank Annual Saving in Bills and Payback Period for Capacitor Bank.
 Electrical Load of (1) 2 No’s of 18.5KW,415V motor ,90% efficiency,0.82 Power Factor ,(2) 2 No’s of 7.5KW,415V motor ,90% efficiency,0.82 Power Factor,(3) 10KW ,415V Lighting Load. The Targeted Power Factor for System is 0.98.
 Electrical Load is connected 24 Hours, Electricity Charge is 100Rs/KVA and 10Rs/KW.
 Calculate size of Discharge Resistor for discharging of capacitor Bank. Discharge rate of Capacitor is 50v in less than 1 minute.
 Also Calculate reduction in KVAR rating of Capacitor if Capacitor Bank is operated at frequency of 40Hz instead of 50Hz and If Operating Voltage 400V instead of 415V.
 Capacitor is connected in star Connection, Capacitor voltage 415V, Capacitor Cost is 60Rs/Kvar. Annual Deprecation Cost of Capacitor is 12%.
Calculation:
 For Connection (1):
 Total Load KW for Connection(1) =Kw / Efficiency=(18.5×2) / 90%=41.1KW
 Total Load KVA (old) for Connection(1)= KW /Old Power Factor= 41.1 /0.82=50.1 KVA
 Total Load KVA (new) for Connection(1)= KW /New Power Factor= 41.1 /0.98= 41.9KVA
 Total Load KVAR= KWX([(√1(old p.f)2) / old p.f] [(√1(New p.f)2) / New p.f])
 Total Load KVAR1=41.1x([(√1(0.82)2) / 0.82] [(√1(0.98)2) / 0.98])
 Total Load KVAR1=20.35 KVAR
 OR
 tanǾ1=Arcos(0.82)=0.69
 tanǾ2=Arcos(0.98)=0.20
 Total Load KVAR1= KWX (tanǾ1 tanǾ2) =41.1(0.690.20)=20.35KVAR
 For Connection (2):
 Total Load KW for Connection(2) =Kw / Efficiency=(7.5×2) / 90%=16.66KW
 Total Load KVA (old) for Connection(1)= KW /Old Power Factor= 16.66 /0.83=20.08 KVA
 Total Load KVA (new) for Connection(1)= KW /New Power Factor= 16.66 /0.98= 17.01KVA
 Total Load KVAR2= KWX([(√1(old p.f)2) / old p.f] [(√1(New p.f)2) / New p.f])
 Total Load KVAR2=20.35x([(√1(0.83)2) / 0.83] [(√1(0.98)2) / 0.98])
 Total Load KVAR2=7.82 KVAR
 For Connection (3):
 Total Load KW for Connection(3) =Kw =10KW
 Total Load KVA (old) for Connection(1)= KW /Old Power Factor= 10/0.85=11.76 KVA
 Total Load KVA (new) for Connection(1)= KW /New Power Factor= 10 /0.98= 10.20KVA
 Total Load KVAR3= KWX([(√1(old p.f)2) / old p.f] [(√1(New p.f)2) / New p.f])
 Total Load KVAR3=20.35x([(√1(0.85)2) / 0.85] [(√1(0.98)2) / 0.98])
 Total Load KVAR1=4.17 KVAR
 Total KVAR=KVAR1+ KVAR2+KVAR3
 Total KVAR=20.35+7.82+4.17
 Total KVAR=32 Kvar
Size of Capacitor Bank:
 Site of Capacitor Bank=32 Kvar.
 Leading KVAR supplied by each Phase= Kvar/No of Phase
 Leading KVAR supplied by each Phase =32/3=10.8Kvar/Phase
 Capacitor Charging Current (Ic)= (Kvar/Phase x1000)/Volt
 Capacitor Charging Current (Ic)= (10.8×1000)/(415/√3)
 Capacitor Charging Current (Ic)=44.9Amp
 Capacitance of Capacitor = Capacitor Charging Current (Ic)/ Xc
 Xc=2 x 3.14 x f x v=2×3.14x50x(415/√3)=75362
 Capacitance of Capacitor=44.9/75362= 5.96µF
 Required 3 No’s of 10.8 Kvar Capacitors and

Total Size of Capacitor Bank is 32Kvar
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=2×44.9Amp
 Size of the fuse=90Amp
Size of Circuit Breaker for Capacitor Protection:
 Size of the Circuit Breaker =135% to 150% of Capacitor Charging current.
 Size of the Circuit Breaker=1.5×44.9Amp
 Size of the Circuit Breaker=67Amp
 Thermal relay setting between 1.3 and 1.5of Capacitor Charging current.
 Thermal relay setting of C.B=1.5×44.9 Amp
 Thermal relay setting of C.B=67 Amp
 Magnetic relay setting between 5 and 10 of Capacitor Charging current.
 Magnetic relay setting of C.B=10×44.9Amp
 Magnetic relay setting of C.B=449Amp
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 x1.1 x nominal capacitor Current
 Cables size for Capacitor Connection = 1.43 x nominal capacitor Current
 Cables size for Capacitor Connection=1.43×44.9Amp
 Cables size for Capacitor Connection=64 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 IECstandard 60831).
 Discharge resistors have to be connected directly to the capacitors. There shall be no switch, fuse cutout 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 / 1/3xCn x Log (Un x√2/ Dv)
 Where Ct =Capacitor Discharge Time (sec)
 Cn=Capacitance Farad.
 Un = Line Voltage
 Dv=Capacitor Discharge voltage.
 Maximum Discharge resistance =60 / ((5.96/1000000)x log ( 415x√2 /50)
 Maximum Discharge resistance=4087 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
 Reduced in Kvar size of Capacitor when operating 50 Hz unit at 40 Hz
 Actual KVAR = Rated KVAR x(Operating Frequency / Rated Frequency)
 Actual KVAR = Rated KVAR x(40/50)
 Actual KVAR = 80% of Rated KVAR
 Hence 32 Kvar Capacitor works as 80%x32Kvar= 26.6Kvar
 Reduced in Kvar size of Capacitor when operating 415V unit at 400V
 Actual KVAR = Rated KVAR x(Operating voltage / Rated voltage)^2
 Actual KVAR = Rated KVAR x(400/415)^2
 Actual KVAR=93% of Rated KVAR
 Hence 32 Kvar Capacitor works as 93%x32Kvar= 23.0Kvar
Annual Saving and Pay Back Period
Before Power Factor Correction:
 Total electrical load KVA (old)= KVA1+KVA2+KVA3
 Total electrical load= 50.1+20.08+11.76
 Total electrical load=82 KVA
 Total electrical Load KW=kW1+KW2+KW3
 Total electrical Load KW=37+15+10
 Total electrical Load KW =62kw
 Load Current=KVA/V=80×1000/(415/1.732)
 Load Current=114.1 Amp
 KVA Demand Charge=KVA X Charge
 KVA Demand Charge=82x60Rs
 KVA Demand Charge=8198 Rs
 Annual Unit Consumption=KWx Daily usesx365
 Annual Unit Consumption=62x24x365 =543120 Kwh
 Annual charges =543120×10=5431200 Rs
 Total Annual Cost= 8198+5431200

Total Annual Cost before Power Factor Correction= 5439398 Rs
After Power Factor Correction:
 Total electrical load KVA (new)= KVA1+KVA2+KVA3
 Total electrical load= 41.95+17.01+10.20
 Total electrical load=69 KVA
 Total electrical Load KW=kW1+KW2+KW3
 Total electrical Load KW=37+15+10
 Total electrical Load KW =62kw
 Load Current=KVA/V=69×1000/(415/1.732)
 Load Current=96.2 Amp
 KVA Demand Charge=KVA X Charge
 KVA Demand Charge=69x60Rs =6916 Rs————(1)
 Annual Unit Consumption=KWx Daily usesx365
 Annual Unit Consumption=62x24x365 =543120 Kwh
 Annual charges =543120×10=5431200 Rs—————–(2)
 Capital Cost of capacitor= Kvar x Capacitor cost/Kvar = 82 x 60= 4919 Rs—(3)
 Annual Interest and Deprecation Cost =4919 x 12%=590 Rs—–(4)
 Total Annual Cost= 6916+5431200+4919+590

Total Annual Cost After Power Factor Correction =5438706 Rs
Pay Back Period:
 Total Annual Cost before Power Factor Correction= 5439398 Rs
 Total Annual Cost After Power Factor Correction =5438706 Rs
 Annual Saving= 54393985438706 Rs

Annual Saving= 692 Rs
 Payback Period= Capital Cost of Capacitor / Annual Saving
 Payback Period= 4912 / 692

Payback Period = 7.1 Years
very informative
thank you/
the conversion from farad to micro farad is incorrect…………..capacitance =44.9/75362 is equal to 595.75 micro farad
Dear Sir,
Could you please tell me what value does a D.C. measuring instrument indicate r.m.s.,average,maximum or average value and why
very much informative and useful calculation..
You have also uploaded the XLS sheet of this calculation also…
Thanks for ur this valuable service..
Prashat K. Soni
excelente…gracias…
Dear Sir,
Can you share types of electrical loads and colour code indicated on panels respective loads.
Hi Pls we need calculate the no of secondary turns for current transformers .and the size of tornado cor
Thank u so much sir…
respected sir
i want to know about how to select capacitor bank in control panel design and which basis its
apply i..e load? can u explain with examples ? i m tired to find out its process. mail id : vaibhav12aug@gmail.com
vaibhav
Dear sir your explanation is very good & i need the same calculation for HT line (including Transformer inrush,Line fault etc )