Calculate Size of Capacitor Bank / Annual Saving & Payback Period
April 1, 2014 43 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(0.69-0.20)=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 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 / 1/3xCn 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= 5439398-5438706 Rs
-
Annual Saving= 692 Rs
- Payback Period= Capital Cost of Capacitor / Annual Saving
- Payback Period= 4912 / 692
-
Payback Period = 7.1 Years
JIGNESH PARMAR
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 )
hi Sir
I am Hussain and working in Australia. i am given a new task of power factor correction of two MCCs (control Panels).
Please guide me how to move forward.
Please email at sudheerraza@gmail.com
regards
Hussain
My MCC total connected load is 3000kw.
Running load is 1800kw.
How much kvar capacitor bank should be installed.
really this is very usefull chapter in eletrical system and it is save our money also ,thanks a lot again sir
Very informative
I wanna know how you derived the formula for finding the size of discharge resistor
If capacitance connected in the circuit is more than required what will be the impact on billing. It will come down or shoot up. Pl write in detail
sir I have a doubt… isn’t the size of cable should be more than the size of C.B….?
which in given case is not
Size of the Circuit Breaker=67Amp
Cables size for Capacitor Connection=64 Amp
please correct me if I’m wrong!
IS STAR CONNECTION MORE EFFICIENT THAN DELTA CONNECTION IN ABOVE EX.?
Could you please inform me whether this is per any standard ?If yes then which one?
Thank you
USEFUL MSG SIR THANK U SIR
450 kvar is need to design .if I connect 150 kvar in star connection per phase .mu question is that is these capacitors will become equal to 450 kvAR connectec in star
its too helpful in explained way to practical techincians
Jignesh,
why voltage across capacitor increases when connected across detuned reactor in capacitor bank.
Why the capacitors in capacitor bank are rated for higher than nominal voltage
KVAR= KW(tan (cos^(-1) (old pf))-tan (cos^(-1) new pf)) is Corrct or not sir
Thanks a lot for this helpful article. However, I have 2 comments:
1- The KVA cost is 60, not 100 Rs.
2- The Capacitor bank cost shouldn’t be added to the annual cost.
Please rectify me if I’m wrong 🙂
Dear Sir,
Thanks for the explanation !
Calculations with ROI for p.f. improvement aspect has been explained in a quite simpler way. Very Easy to understand to an individual concerned with the issue.
Max. Discharge resistance Value (Star Connection) = Ct / Cn x Log (Un x√2/ Dv). is there something missing with this formula? as the same answer can not be obtained for
this example . Maximum Discharge resistance =60 / ((5.96/1000000)x log ( 415x√2 /50)
Maximum Discharge resistance=4087 KΩ.
please inform me when an important artical and comment posted.
The article is good.
Please I need to Know the formulae and how to connect different sizes of capacitors to get the required size/rating of capacitor bank (in MVAr or KVAr) for higher voltage compensation
applications
i neead your help whit the project
sherif1boris@gmail.com
Sir,there are small calculation errors , but overall it is very important and useful information.
41.1x([(√1-(0.82)2) / 0.82]- [(√1-(0.98)2) / 0.98]) this formula please send once again very clearly
because ans missmatch
excellent
Xc = 2×3.14xfxc but the formula used was Xc=2×3.14xfxv, why is that?
Xc = 2×3.14xfxc but the formula used was Xc=2×3.14xfxv, why is that?
Hi sir
Good Evening can help with Pay Calculation in Maharastra MSEB ( KVAH) billing after installation of APFC PANEL.
KVA Demand Charge=82x60Rs?
Electricity Charge is 100Rs/KVA?
How to select capacitor bank , what are the calculation for if we only knows grid incomer breaker current rating . For eg. grid incomer breaker of 500A with load of 250kw for 415vac, 50hz system
your APPLIED KNOWLEDGE REGARDING UTILIZATION OF CAPACITOR BANK IN LT POWER SYSTEM WITH PAYBACK PERIOD IS VERY GOOD FOR THE FIELD ENGINEERS
nice way of problem resolving but the initial calculation for KVARs is incorrect since its mixed between loads KVA1,2,3
Can 6.6 kV, 3 phase Capacitor bank can be charged by 3.3 kV supply ?