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Rating Definitions

A generator set (genset) consists of an engine and a generator and it is best to consider each of them as a system. Individually, each has unique characteristics, but together these qualities have a significant impact on the performance and sizing of the genset system.

Important generator set (genset) rating considerations
Important generator set (genset) rating considerations (photo credit:

Capabilities of both engine and generator are considered individually and collectively when selecting generator sets. Engines produce brake horsepower (or kilowatts) while controlling speed or frequency.

Generators influence engine behavior, but are primarily responsible for changing engine power into kilovolt-amperes (kVA) and electrical kilowatts (kW). They also must satisfy high “magnetizing current” draws (kVAR), or transient conditions from electrical equipment.

Normally, a generator set is furnished with a generator which matches the engine output capability.

Engines are sized according to the actual power in kW required to meet the needs of the facility. The generator, on the other hand, must be capable of handling the maximum apparent power which is measured in kVA. The actual power can be identified in several ways. It can be calculated by adding the nameplate ratings of the equipment to be powered by the generator. If this is done, the efficiencies of the equipment must also be considered.

Genset nameplate
Genset nameplate (photo credit:

The actual power can be determined by performing a load analysis on the facility. This involves making a survey of the power requirements over a period of time.

ekW = pf x kVA

bkW = ekW / eff + Fan Demand


  • kVA = kVA output of generator
  • pf = power factor of connected load
  • ekW = electrical power (electrical kW)
  • bkW = engine power (brake kW)
  • eff = generator efficiency

When kW is qualified as neither electrical (ekW) nor brake (bkW), it is important to clarify between the two when performing calculations or product comparisons.

Engine/Generator Set Load Factor

Load factor of a generator set is used as one criterion for rating a genset. It is calculated by finding the product of various loads:

  • Load Factor = % of time x % of load
  • % of time = time at specific load / total operating time
  • % of load = specific load / rated load

Extended idling time and the time when the generator set is not operating does not enter into the calculation for load factor.

Example //

For example, assume a facility has a genset rated at 550 kW and runs it two hours a week. During those two hours, it runs at 400 kW for 1.5 hours. Find the load factor. The formulas reveal the following:

  • % of load = 400kW / 550kW = 0.73
  • % of time = 90 min. / 120 min. = 0.75
  • Load Factor = 0.73 x 0.75 = 54.75%

This load factor would indicate that the genset could be used as a standby rated genset because it meets the load factor and other criteria of standby.

Power rating definitions for generator sets are based on typical load factor, hours of use per year, peak demand and application use. For example Caterpillar genset ratings for emergency standby, standby, prime and continuous power are listed.

Emergency Standby Power

Typical usage of 50 hours per year with a maximum of 200 hours per year. Typical variable load factor of 70%.

Standby Power

Maximum usage 500 hours per year, up to 300 hours of which may be continuous running with varying loads. No overload is available. Rating is equivalent to Prime +10%. Load factor maximum is 70% of Standby rating.

Prime Power

Unlimited hours of usage. Load factor (70% maximum /80% if no extra cost) of the published Prime power over each 24 hour period.

10% overload limited to 1 in 12 hours but not to exceed 25 hours per year. The 10% overload is available in accordance with ISO 3046-1 (2002). Life to overhaul of the engine is dependent on operation as outlined in ISO8528 (2005) and time spent during operation above
100% load may affect the life to overhaul.

Continuous Power

Unlimited hours of usage. Load factor 100% of the published Continuous Power.

Generally, as ISO 8528 (2005) Continuous power. Note: Operating above these rating definitions will result in shorter life and higher generator and engine costs per year.

The International Standards Organization (ISO)

ISO 8528-1 (2005) defines three types of duty:

  1. Continuous Operating Power (COP)
  2. Prime Running Power (PRP)
  3. Limited-Time running Power (LTP)

1. Continuous Operating Power (COP)

Continuous operating power is the power a generator set can operate at a continuous load for an unlimited number of hours under stated ambient conditions. Maintenance according to the manufacturer must be followed to reach these standards.

2. Prime Running Power (PRP)

Prime running power is the maximum power a generator set has during a variable power sequence for an unlimited number of hours under stated ambient conditions. Maintenance according to the manufacturer must be followed to reach these standards.

3. Limited-Time Running Power (LTP)

Limited-time running power is the maximum power that a generator set delivers for up to 500 hours per year under stated ambient conditions. Only 300 hours can be continuous running. Maintenance according to the manufacturer must be followed to reach these standards.

Specifications are often stated in ISO terms and standards. Table 1 shows ISO genset ratings and correlating Caterpillar genset ratings.

Generator Set Ratings
ISO Caterpillar
ESP Emergency Standby Power
LTP Emergency Standby
PRP Prime
COP Continuous

500KVA Cummins Diesel Generator set cold start/Backfire

500KVA Cummins Diesel Generator set cold start/Backfire, been stood for a while and the temperature was -20º C. The controller shut it down after it revved up at the end.

Cant see this video? Click here to watch it on Youtube.

Reference: Electric Power Application, Engine and Generator Selection – Caterpillar (Download HERE)

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About Author


Edvard Csanyi

Edvard - Electrical engineer, programmer and founder of EEP. Highly specialized for design of LV/MV switchgears and LV high power busbar trunking (<6300A) in power substations, commercial buildings and industry fascilities. Professional in AutoCAD programming. Present on


  1. abraham
    Nov 13, 2015

    Thanks very much I just wana to know about & will send questions.

  2. Faraz Muhammad Fateh
    Sep 10, 2015

    @Juan David,

    You have quoted two cases with changed PF & raised a few queries against the original formula.
    Please note that the Engine power will be prime mover of shaft, which will generate REAL power.
    It is the REACTIVE power which has increased in case-2 thus increasing the KVA from 100 to 266 but the REAL power is still the same (i.e 80KW).

    Since the REAL power in both cases are same therefore the Engine size can also be same irrespective of the the change in KVA ratings.

    While we are working in industries it is difficult sometimes to go back to our old university books due to time constraints & laziness :p . But you are doing a splendid job in refreshing many of our concepts on daily basis in quick time.

    The only thing which bothers me is that i donot see any article written/published here by any other user. Have you restricted its publication or is it that the users themselves are not writing anything??


  3. Essam Drar
    Feb 01, 2015

    it’s very good, but may i find setting calculation example for the different functions of the generator?

  4. Richard Handley
    Jan 31, 2015

    Great information Edvard. I am also interested in information regarding typical fault levels for gen sets in particular single phase diesel generators.

  5. Thanks for great paper, Is very important to keep in touch with our colleagues.

    Another way to see mechanical power becoming electric power

    Beginning with:

    ekW = pf x kVA
    bkW = ekW / eff + Fan Demand

    lets play an example

    bkW =?
    eff =0.7
    Fan Demand usually can be about … = 0.1* bkW


    bkW = (ekW / eff ) + 0.1* bkW

    (1-0.1) * bkW = ekW / eff

    bkW = ekW / (eff * 0.9) = 80/(0.7 * 0.9) = 126.98

    if pf=0.8 then

    ekW = 80
    kVA=100, can be feeded by 127bkW :)

    but it if pf=0.3 then
    ekW = 80
    kVA=266, NOT can be feended by 127bkW ****** :| ********

    So this is not right that same power makes so big kVA with a same kKW
    Common sense means this is impossible; to produce more kVAR is necessary more fuel (brake power from engine)

    That is why I preffer to think this way…

    bkW feeds the electric generator so, it feeds S=kVA not only ekW, so if you guys permit me to think different I would say…

    bkW = ekVA / eff + Fan Demand :) :) :) :)


    bkW = ekW / eff + Fan Demand :( :( :( :(

    best regards

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