# Calculate Size of Diesel Generator

November 2, 2015 13 Comments

Calculate Size of Diesel Generator having following Electrical Load. Consider Future Expansion ratio is 10%. Average use of Equipment is 0.8 (1 is Full Time Use)

- 4 No’s of 1Ph, 230V, 80Watt CFL Bulbs, Diversity Factor is 0.8; Starting & Running P.F is 0.8.
- 2 No’s of 1Ph, 230V, 3000Watt Air Condition, Diversity Factor is 1, Starting & Running P.F is 0.8.
- 2 No’s of 1Ph, 230V, 500Watt Halogen Lights Diversity Factor is 0.8 Starting & Running P.F is 0.8.
- 1 No’s of 1Ph,230V, 10KW Motor with Y-D Starter, Diversity Factor is 0.8,Starting P.F is 0.7 & Running P.F is 0.8
- 1 No’s of 3Ph,430V, 130KW Motor with Soft Starter , Diversity Factor is 0.8,Starting P.F is 0.7 & Running P.F is 0.8

__ Calculation:__

__Calculation:__

Type of Load |
Equipment |
Starting Current |

Linear Load | General Equipment |
100% of Full Load Current |

Non-Linear Load | UPS, Inverter, Computer, Ballast | 160% of Full Load Current |

__ __

Type of Starter |
Starting Current |

DOL | 6 X Full Load Current |

Star-Delta | 4 X Full Load Current |

Auto Transformer | 3 X Full Load Current |

Soft Starter | 2 X Full Load Current |

VFD | 1.5 X Full Load Current |

__ __

__Load Calculation-1:__

__Load Calculation-1:__

- Full Load KW of CFL Bulb=(No X Watt X Diversity Factor) /1000
- Full Load KW of CFL Bulb=(4x80x0.8)/1000=0.3KW
- Full Load KVA of CFL Bulb=KW / P.F
**Full Load KVA of CFL Bulb=0.3 / 0.8=0.4KVA————(H)**- Full Load current of CFL Bulb=(No X Watt X Diversity Factor) / (Volt x P.F)
**Full Load current of CFL Bulb=(4x80x0.8) / (230×0.8) =2 Amp—-(M)**- Type of Load=Linear
- Starting KVA of CFL Bulb=1 X (KW / Starting P.F)
**Starting KVA of CFL Bulb=0.3 / 0.8=0.4KVA———————-(1)**- Starting Current=100% of Full Load Current.
**Starting Current=1 X 2= 2 Amp.————–(A)**

__Load Calculation-2:__

__Load Calculation-2:__

- Total Full Load KW of A.C=(No X Watt X Diversity Factor) /1000
- Total Full Load KW of A.C =(2x3000x0.8)/1000=4.8KW
- Total Full Load KVA of A.C =KW / P.F
**Total Full Load KVA of A.C =4.8 / 0.8=6KVA————(I)**- Total Full Load current of A.C =(No X Watt X Diversity Factor) / (Volt x P.F)
**Total Full Load current of A.C =(2x3000x0.8) / (230×0.8) =26 Amp——-(N)**- Type of Load=Non Linear
- Starting KVA of A.C=1.6 X (KW / Starting P.F)
**Starting KVA of A.C =1.6 X (4.8 / 0.8)=9.6KVA———————-(2)**- Starting Current=160% of Full Load Current.
**Starting Current=1.6 X 26= 42 Amp. ————–(B)**

__Load Calculation-3:__

__Load Calculation-3:__

- Full Load KW of Halogen Bulb=(No X Watt X Diversity Factor) /1000
- Full Load KW of Halogen Bulb=(2x500x0.8)/1000=0.8KW
- Full Load KVA of Halogen Bulb=KW / P.F
**Full Load KVA of Halogen Bulb=0.8 / 0.8=1KVA————(J)**- Full Load current of Halogen Bulb=(No X Watt X Diversity Factor) / (Volt x P.F)
**Full Load current of Halogen Bulb=(2x500x0.8) / (230×0.8) =4 Amp—-(O)**- Type of Load=Non Linear
- Starting KVA of Halogen Bulb =1.6 X (KW / Starting P.F) / Starting P.F
**Starting KVA of Halogen Bulb =1.6 X (0.8 / 0.8)=1.6KVA———————-(3)**- Starting Current=160% of Full Load Current.
**Starting Current=1.6 X 4= 7 Amp .————–(C)**

__Load Calculation-4:__

__Load Calculation-4:__

- Full Load KW of Motor=(No X Watt X Diversity Factor) /1000
- Full Load KW of Motor =(1x10000x0.8)/1000=8KW
- Full Load KVA of Motor =KW / P.F
**Full Load KVA of Motor =8 / 0.8=10KVA————(K)**- Full Load current of Motor =(No X Watt X Diversity Factor) / (Volt x P.F)
**Full Load current of Motor =(1x10000x0.8) / (230×0.8) =43 Amp—-(P)**- Type of Starter=Star-Delta
- Starting KVA of Motor =4 X (KW / Starting P.F)
**Starting KVA of Motor=4X (8 / 0.7)=45.7KVA————————(4)**- Starting Current=4 X Full Load Current
**Starting Current=4 X 11.4 = 174 Amp.————–(D)**

__Load Calculation-5:__

__Load Calculation-5:__

- Full Load KW of Motor=(No X Watt X Diversity Factor) /1000
- Full Load KW of Motor =(1x120000x0.8)/1000=96KW
- Full Load KVA of Motor =KW / P.F
**Full Load KVA of Motor =96 / 0.8=120KVA————(L)**- Full Load current of Motor =(No X Watt X Diversity Factor) / (Volt x P.F)
**Full Load current of Motor =(1x120000x0.8) / (1.732x430x0.8) =167 Amp—-(Q)**- Type of Starter=Auto Transformer
- Starting KVA of Motor =3 X (KW / Starting P.F)
**Starting KVA of Motor=3 X (96 / 0.7)=411.4KVA—————(5)**- Starting Current=3 X Full Load Current
**Starting Current=3 X 167 = 501 Amp.————–(E)**

__Total Load Calculation:__

__Total Load Calculation:__

**Total Starting KVA =**(1) + (2) +(3) + (4) + (5)**Total Starting KVA**=0.4+9.6+1.6+45.7+411.4 =**468.7 KVA****Total Starting Current =**(A) + (B) +(C) + (D) + E**Total Starting Current**=2+42+7+174+501=**725 Amp****Total Running KVA =**(H)+(I)+(J)+(K)+(L)**Total Running KVA =**4+6+1+10+120=**137KVA****Total Running Current=**(M)+(N)+(O)+(P)+(Q)**Total Running Current=**2+26+4+43+167**= 242 Amp****Size of Diesel Generator= Starting KVA X Future Expansion X Average Use of Equipments**- Size of Diesel Generator=468.7 X 1.1 X 0.8
**Size of Diesel Generator= 412 KVA**

__Summary:__

__Summary:__

**Total Starting KVA**=**468.7 KVA****Total Starting Current**=**725 Amp****Total Running KVA**=**137KVA****Total Running Current= 242 Amp****Size of Diesel Generator= 412 KVA**

Dear sir,

Thanks for your valuable Generator calculation tips and your advice ,but now i need the calculation of lighting power density, please send to me on the below mail ID ,

Sir,

Thanks , this is very useful for step 5 tpm pasing stages .

As all the associates along with officer should know how to calculate .

All your mails are very useful to us .i want to share with you one product called Ultra .

We took trials and our team saying there is no power saving I can share the details .

Regards

SANjeev

Sr manager

Mahindra

91 9922900891

Dear jignesh sir,

i want to know some doubts about electrical system,

1.why powerfactor defined by “cos phi” not tan and sin?

2.why we are goes to higher voltage to step down 433v…11kv to 433?

Following points may be noticed 1) PF should be considered as 0.95 as we normally use APFC but don’t consider it for calculation of capacity of DG and Substation. 2)Loading for Motors is taken as 100% which never designed for. Get some cases studies only then we may know. I have done study for three buildings where Substation capacities were 3 to 4 times than peak due to which not only huge initial investment but very high maintenance cost also.

dear sir pls ans me calculation in transformer load

wery nice and easy calculation method thanks

thank u very much sir

mahesh

Sir may I want to know that single phase supply 230 v,8kw motor with star delta starter available. Or ??

Dear Sir,

Is it possible to have star /delta starter for 1 Ph motor?

Regards

Kuldip Singh

Sir

Sent from Yahoo Mail on Android

Nice presentation.

Could you please tell me why the starting kVAs of all the loads in the example have been added to arrive at the genset capacity? Does this approach not imply that all the loads are being simultaneously started up? If this is not intended, then will we not land up with an oversized generator?

On the other hand if all the loads are not proposed to be started up simultaneously, then I think the right approach would be to add up the running kVAs of all the loads except that of the biggest load (130 kW motor or is it 120 kW as it appears in the calculation ?) which would be 17 kVA and then to this fig add the starting kVA of the biggest load which would be 411.4 kVA. This would give us a genset capacity of 428.4 kVA.

Thanks

800 kva DG is their…….? i want how much current it takes?

800kva DG how much current it takes?