08-26-2011 02:32 PM
The State is asking for the Hp ratings of the generators existing on-site. Our records have the kW rating, and I see some of the Manufacturer literature gives a range of Hp depending upon use level. How do I answer the Hp question without over stating the Hp, and the emissions, from the generators? Many thanks for your comments!
08-26-2011 07:02 PM
Since most air districts will only accept what is on a unit nameplate, to get a BHP number if you only have a kW rating you'll need to to do the following. kW X 1.34 / gen eff (.94 is generally accepted if you don't have exact data). If the unithas an engine driven radiator fan you will also need to add that as well, although most air quality folks won't know to ask and for many years I've used the above formual and been ok.
Hope that helps, Mike L.
11-22-2011 12:08 PM - edited 11-22-2011 12:14 PM
A good rule of thumb to calculate engine horsepower on a package generator set with an engine-driven fan is multiply the generator nameplate kW rating by 1.5 and the result is the engine power in brake-horsepower - BHP (i.e. 400 kWe generator set x 1.5 equals 600 brake-horsepower - the engine's power rating needed to produce 400 kWe is approximately 600 BHP).
Conversely if you know an engine's hosepower rating then divide that by 1.5 and the result is the power you can expect out of the generator expressed as kiloWatts-electric (kWe).
11-28-2011 10:16 AM
If you want a more exact answer, you can contact your Cat dealer who can research your generator efficiency using an ESO (Engine Shipping Order) number and approximate ship date of the gensets. You can use the dealer locator on cat.com to find your Cat dealer. The link is http://www.cat.com/dealer-locator and select Electric Power on the left-hand side of the page. The Cat dealer can give you the generator efficiency for your generator(s) to use in the calculation kW x 1.34 / gen efficiency. The Cat dealer may (should) also have information on your radiator.
01-20-2012 01:25 PM
The 1.341 is a direct conversion for kW - HP at the flywheel (work done)
As stated by Mike L you need to factor in the efficiency of the alternator and any other parasitic loads on the unit such as the engine driven fan.
This calculation will get you real close to the HP requirements of the prime mover at a specific electrical load on the alternator.
The 1.5 ratio is a rule of thumb for units with engine driven fans and some wiggle room for parasitic losses. This is not as precise.
As stated, this is a calculation of the HP requirement of the prime mover to produce a specific kW output from the alternator.
The actual engine rating may differ as the packager may have over or under sized the prime mover for an unknown reason (altitude, high ambient temp,transient response etc.)
Hope this helps