DG selection

Basically you need a higher power than 45 Kw electric power at the output of the genset.

You must consider the Coupling losses as well as the alternator losses, decreasing the engine power output.

Calculate 80% for coupling and another 80% for alternator efficiency.

Let's make those simple calculations

Engine output must be (At least) 45 (Kw)/0.8 = 56.25 Kw to compensate the coupling losses. and

Engine output must be (At least) 56.25 Kwe (electric power at the alternator output) /0.8 = 70.32 Kwm (mechanic Power at the engine shaft) to compensate the Alternator losses. (Heat, Hysteresist, Cooling fan etc.)

This is not the end of your calculation. Generator sets are rated for KVA. Relationship KVA to Kw is up to your load's Power Factor. A Common estimation for Power factor in Genset calculations are taken 0.8. This means you need 1.250 KVA power to feed a 1,000 Kilowatts.

For your case the your diesel engine must be rated for 70.32 Kw mechanic shaft output to give a 45 Kw electric power at the generator output. It makes 45 Kw /0.8(PF) = 56 KVA

This is the minimal requirement. You must have a higher rated generator set depending the behaviour of your load.

If you apply the total 45 Kw to generator at once, the speed of the engine suddenly drops down and the generator starts searching.

For an example, direct starting current for the connected motors are approximately 6 times bigger than their normal rating.

An another example, the filaments of incandesant lamps are cold when they are off and they have a lower resistance. They draw much current (instantly) at startup. İf you swich all of them at once, they will behave as a motor, to draw much current to drop the speed of the engine.

Above are the effective factors to calculate the generator set rating.

My personal opinion you must select an engine of which rated for nett 120 Kw(m) diesel engine shaft output. This will enable you to obtain a 76 Kw(e) electric power at the output of the genset.

At 0.8 Power Factor. The output rating for such generator set is 96-100 KVA .

Kindest Regards

Hi Nezihozfirat and good morning from USA

I am not sure what you mean by 80% for coupling loss and 80% for generator loss.

My experience with 'hard' couplings is 0.25% loss maximum, such as flex plates on a single bearing alternator and maybe 2% for stiff elastomeric and 5% for soft elastomeric, all of which is taken into the generator wattage output rating (engine rating) by the DG supplier. Anything more than 7 or 8% loss for a coupling would indicate an unacceptable 'softness' and require re-thinking a bit I should hope.

Also, the worst 45 kW eff I know of is 88% and a modern one might be 91-93% eff. at 80% load +/-.

Do you have other experiences than mine above to share with me so I can learn?

- George

Hi.

80% is not the loss. It is efficiency.

Loss is 20% both for couplings and the alternator conversion from mechanic input to electric output.

80% (Approx) efficiency is indicated in the leaflets of most engine manufacturers, such as Cummins and Perkins. Those figures are their laboratory results.

In similar way the Alternator efficiency is also indicated 80% in the alternator manufacturers own leaflets, 20% loss of which covering mechanical bearing losses+hysterresist losses+cooling fan losses+heat losses in the copper windings.

Kindest Regards

Thanks a ton dear nezihozfirat

G'Day to you also Athreyas,

DGs are rated for output at the terminals, not by engine size (at least you won't typically specify the engine size) and at this size the generator (alternator) will be rated for Power-Factor of 0.8, which is not important for you to understand at this moment.

You must decide if your 45kW need is constant, or temporary, or includes extra power demand just in case everything is totally switched on all at once. Your DG supplier will have a program to calculate how much larger to have the DG set if you intend a block load of X kW all at once. Here are some basics:

This size engine will de-rate for high altitude and high ambient temperatures. If it is turbocharged, it will not de-rate until about 1500-2000 meters (perhaps) but if not turbocharged it will de-rate 3% for each 300 meters altitude above the first 150 meters (approximately). If it is air cooled or has a radiator cooler with fan the fan will also de-rate for altitude.

You may need a unit about 150% size if you need to start all 45 kW in one block load jump.

Next you must decide if the 45 kW load cycle duty is only for a few hours a week with no more than one hour at 45 kW followed by one hour at 80% of 45 kW for 12 hours. This would be termed 'standby' rating for emergency standby. The generator (alternator) will have a temperature rise rating (how hot it gets above ambient) so for standby it may have a 130*C rise above 40*C ambient (the windings will get hot since there is less copper in a standby rated DG set).

If you want the 45 kW for 24 hours duty "continuous" then the engine will be 20% larger (after all de-rating for altitude and high temperature). This is due to lube oil temperature limits at higher load factors and wear and tear on the engine. The alternator will be sized for 80*c or 90*c or maybe 100*C temperature rise over 40*C ambient (it will have more copper and last longer).

Rating can get a bit complex, so ask your DG supplier to explain it all. You can find rating definitions at Cummins, Caterpillar, MTU, Duetz and other websites.

Good luck.

Dear Sir,

Your information is quite informative. Usually we size gensets as according to types of load connected for example : lightnings, pumps + motors, bow thruster (marine), winches etc . Then decide which type of load to run first and how often they run, then we will decide if the genset is deemed to operate as std-by , continuous or prime rating.

I have ever come across the Engine is a smaller capacity compare to the alternator (continuous) operations due to high engine cost vs alternator cost and entire coupling of genset cost. Therefore they size the engine smaller than alternator. But i am still looking for a general formular or method as to how to determine the right engine size . Alternator sizing is no problem as it has to match the load, however as per mentioned earlier some industry people size the engine KW smaller.

Next i think the PF of the load banks for testing gensets also play a part in determining the true power tested. Most load bank are pf 1 but engine and alternator are pf 0.8 lagging.

Some catepillar, mitsubishi, cummins website do not really touch on how they size the genset only briefly touch on it. I suppose most of the software are doing the exact calculations.