# Calculate Size of Capacitor Bank / Annual Saving & Payback Period

• 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])
• 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])
• 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 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

## 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)
• 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 KW =62kw
• 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

### After Power Factor Correction:

•  Total electrical load KVA (new)= KVA1+KVA2+KVA3
• Total electrical Load KW =62kw
• 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

## 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 has completed M.Tech (Power System Control), B.E(Electrical). He is member of Institution of Engineers (MIE) and CEng,India. Membership No:M-1473586.He has more than 16 years experience in Transmission -Distribution-Electrical Energy theft detection-Electrical Maintenance-Electrical Projects (Planning-Designing-Technical Review-coordination -Execution). He is Presently associate with one of the leading business group as a Deputy Manager at Ahmedabad,India. He has published numbers of Technical Articles in “Electrical Mirror”, “Electrical India”, “Lighting India”,”Smart Energy”, “Industrial Electrix”(Australian Power Publications) Magazines. He is Freelancer Programmer of Advance Excel and design useful Excel base Electrical Programs as per IS, NEC, IEC,IEEE codes. He is Technical Blogger and Familiar with English, Hindi, Gujarati, French languages. He wants to Share his experience & Knowledge and help technical enthusiasts to find suitable solutions and updating themselves on various Engineering Topics.

### 27 Responses to Calculate Size of Capacitor Bank / Annual Saving & Payback Period

1. Jonryl Novicio says:

very informative

thank you/

• Ghani Ullah Khan says:

2. Nabojit Bhowal says:

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

3. PRASHANT SONI says:

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

4. gustavo rafael faquira soto says:

excelente…gracias…

5. Hanumanth says:

Dear Sir,

Can you share types of electrical loads and colour code indicated on panels respective loads.

6. dream egy says:

Hi Pls we need calculate the no of secondary turns for current transformers .and the size of tornado cor

7. Aman Aggarwal says:

Thank u so much sir…

8. vaibhav says:

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

9. RUBESH says:

Dear sir your explanation is very good & i need the same calculation for HT line (including Transformer inrush,Line fault etc )

10. sidh187 says:

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.
regards
Hussain

11. chandandhir says:

really this is very usefull chapter in eletrical system and it is save our money also ,thanks a lot again sir

12. Binu Joy says:

Very informative

13. dan says:

I wanna know how you derived the formula for finding the size of discharge resistor

14. krishna naik says:

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

15. 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!

16. M D BHOSALE says:

IS STAR CONNECTION MORE EFFICIENT THAN DELTA CONNECTION IN ABOVE EX.?

17. netfreak says:

Could you please inform me whether this is per any standard ?If yes then which one?

18. sam says:

Thank you

19. GNANAGRI says:

USEFUL MSG SIR THANK U SIR

20. haji says:

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

21. dipesh nidhi says:

its too helpful in explained way to practical techincians

22. Rajesh says:

Jignesh,
why voltage across capacitor increases when connected across detuned reactor in capacitor bank.

23. Rajesh21 says:

Why the capacitors in capacitor bank are rated for higher than nominal voltage

24. hari says:

KVAR= KW(tan⁡ (cos^(-1)⁡ (old pf))-tan⁡ (cos^(-1)⁡ new pf)) is Corrct or not sir

25. Ahmed Mo'men says:

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 🙂

26. Mongia DK says:

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.