Generator Vs Load

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Hameedulla-Ekhlas

Hameedulla-Ekhlas

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Greeting all,
Hope all are fine and good. My question is regarding to generator size.

I have designed a project and its final Load is equal to 127 kVA. This is a general load, I mean it is a residential buildings.

I have chosen generator size 135kVA with 3ph, 380 volt, pf=0.8. and we know that this generator can provide only 108 kw real power and 80.27 reactive power.

It is possible to increase or decrease the real and reactive power by setting the governer and field winding of generator but we dont have here any automatic control.

So, my question is this. Since this generator can only provide 108kw and the load is a general residential load with 127kVA pf=0.95 which equals to 120.65kW.

Here now at peak load which equals to 120.65 and generator can only provide maximium 108kW. So, according to me this generator can not provide the peak load and it should be increase one step in size.

Am I right or not please advise me.
 
Hameedulla-Ekhlas said:
Greeting all,
Hope all are fine and good. My question is regarding to generator size.

I have designed a project and its final Load is equal to 127 kVA. This is a general load, I mean it is a residential buildings.

I have chosen generator size 135kVA with 3ph, 380 volt, pf=0.8. and we know that this generator can provide only 108 kw real power and 80.27 reactive power.

It is possible to increase or decrease the real and reactive power by setting the governer and field winding of generator but we dont have here any automatic control.

So, my question is this. Since this generator can only provide 108kw and the load is a general residential load with 127kVA pf=0.95 which equals to 120.65kW.

Here now at peak load which equals to 120.65 and generator can only provide maximium 108kW. So, according to me this generator can not provide the peak load and it should be increase one step in size.

Am I right or not please advise me.
Click to expand...

The limitation may be the prime mover, the engine that provides the input power. That will set a limit on your available output power.
 
100126-1554 EST

Hameedulla-Ekhlas:

The generator rating of 135 KVA is based on the temperature rise due to the load current. Since your residential load power factor is higher than the specification on the generator, 0.95 vs 0.8, this means the generator itself can be operated closer to 135 KW without exceeding the temperature limitation of the generator. But as Besoeker indicated the engine driving the generator may not be capable of the 135 KW plus losses.

On the other hand you might want a larger unit for some safety margin on what might be actual loads.

You do very well in your English communication. I have to think about what you write sometimes, but you have to be complemented on your ability. I could never do as well into some other language than English. What is your native language?

.
 
thank you gar,

English is not my native language and my native language is Pashtoo.

As u said "I have to think about what you write sometimes"

Sorry, for this that sometimes I make u trouble to understand my words. But I try my best to explain my question as much as easy and understandable.
 
100126-1757 EST

Hameedulla-Ekhlas:

You do very well.

There are persons on this forum whose native language is English and are harder to understand than you because they do not clearly define the conditions, assumptions, and question.

A further comment on generators. In general magnetic coil and resistor type electrical products can tolerate very large overloads relative to their nominal steady state continuous rating. So I might overload a wire wound resistor with 10 times its rated power for a short time with no adverse effect, and under these overload conditions its resistance would not be much different than when no power was applied. Same with many motors and generators.

By comparison an internal combustion engine has a maximum continuous peak power capability very close to its short time capability. So a 125 KW engine if momentarily loaded to 130 KW will probably just stall with no recovery. A shunt wound DC motor would just slow down a little. A generator would just have a somewhat lower output voltage.
 
yes, good information and I checked generator's catalog. I found that this generator can be overloaded 10% of its output kVA for one houre.
 
gar said:
100126-1757 EST

Hameedulla-Ekhlas:

You do very well in your English communication. I have to think about what you write sometimes, but you have to be complemented on your ability.

There are persons on this forum whose native language is English and are harder to understand than you because they do not clearly define the conditions, assumptions, and question.
Click to expand...

I agree. H-E your English is fine. Where in Afghanistan are you working? Kabul or all over?
 
I concur with the stated compliments directed towards Hameedullah.

Others have as usual provided great comments about generator operation. One point I question is you maximum load calculation. 127kVA may be the total connected load but have you considered demand and diversity factors? In addition, it is my experience that actual connected load is often less than the case assumed by the engineer (which is sound because engineering practice does need to be robust), even assuming very liberal demand/diversity numbers of around 60% each. Of course all varies from case to case and you may well have a reason to assume 127kVA as a possible maximum load, but for a general application I think the 135kVA rated generator should do fine.
 
Thanks for your reply. As I have mentioned in the final load 127kVA, it means that I have already applied NEC Demand factor for all residential equipment.

So, if you have any farther comment please advise me.
 
You seem to have a good understanding of design so I think you're doing fine. It is the correct approach to size a generator set based on kW - I was just pointing out that maximum load is usually less than the calculated worst case. One final question, how do you know the PF of connected load to be .95?
 
Thanks for your reply and there are two reasons I have taken load's p.f=o.95

First:
in this project the highest load is split unit. It is as below per its namplate.

voltage (V) = 220 volt;1ph; P=2000W; I=9.6 ampere

Per above information we can find the Apparant Power (S) = V*I and we know that S=2112 VA

cos(tetha) = P/S = 2000/2112 = 0.9469; p.f = cos(tetha) = 0.9469
almost 0.95

Second:
I am using Digsilent Power factory software for shortcircuit and Power flow calculation.

By default for general load I mean residential this program also uses 0.95 p.f
 
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