so does this mean it should carry a 69 amp single phase load running at 133% ?
Based on the assumptions I made above, is the quote correct at a .8 power or a 1.0 PF . From what I understand about power factor the limitations are the engines HP... I think?
Having the Tm state the the genset should run with the %rated current meter at 80% and the power factory being .8 are these two related? IE. If the %rated current meter is at 100% is the PF 1.0 ?
My main quest is to understand what KW i will get at 133% and at a 300% surge
Thanks All for your input
One of the experts needs to clarify this but I'll try hoping I don't mis speak.
My 3kw 016B has the similar specs. I think what they are saying is that if you have an inductive load, such as a motor, and if that motor has a power factor of .8 then you will be at full load on the generator when the meter is at 80%.
Likewise, if you have a pure resistive load like an electric heater then your power factor is 1.0 and you are at full load when the meter is at 100%
If you have a mixture of loads, in reality, the actual PF is somewhere in the middle.
The reason for why the full load is made at 80% on the meter when using motors is because of what the motor does to the generator. AC current is like a wave as it is transmitted in the wire. It has peaks and valleys. As a resistive load (like a heater) travels the voltage peak and the current peak are coincident or ride together which gives a PF of 1.0 Inductive loads such as motors or transformers create a situation where the peak current does not occur at the peak voltage. (If I remember right the current lags the voltage) The Power, (Kw) = the voltage X the current. Therefore, the 80% max meter reading at full power and 0.8 PF is pushing the generator windings as much as 100% meter at 1.0 PF. If the gen was run at 100% meter with a motor load with .8 PF then the gen would be overloaded.
Stated another way the load meter and the PF are not directly related.
The PF is determined by the type of and combination of equipment connected to the gen. The more inductive the load is the more it is going to shift the voltage peaks from the current peaks and the lower the reading on the power meter will be to make 100% load. When hooking up heaters, light bulbs, small electronics the PF will be 1.0 and the gen will be at 100% rated load at 100% meter indication. Larger motors will usually have a PF number on their nameplate. If one was to run only a large motor as the gen load then the PF of the motor would determine where 100% load is met by the meter. The TM uses .8 PF at 80% as the guideline, but I have seen plenty of motors with .75 PF. Generally, inductive loads that have less than 1.0 PF are any devices with coils or windings in them and operate by induction such as motors or transformers.
Of course these gens can be overloaded beyond 80% at .8 or 100% at 1.0 which is where you are going with your question. I just wanted to clarify what the PF is and how it relates to the load meter and why they give us different meter limits for 1.0 and 0.8 PF.
In order to get to the maximum Amp limits you are looking for or to relate your Amps to the % load or overload you will first have to know if you have any significant motors or transformers on your load and what the PF will be. For backfeeding a house I would say you are close to 1.0 PF unless you are running an A/C unit or heat pump. For running a bunch of motors in a workshop or a 240vac pressure washer or a big air compressor the PF will need to be considered.
If the gen was to only power inductive loads, like a big motor, and these loads were always the same every time, then large capacitors can be added between the generator and the load to make the PF closer to 1.0. Electric Utility companys do this on their poles where their transformers feed customers with a lot of motors or the like.
Back to your main question: (for single phase, three phase is different)
If your loads create a PF of 1.0 and 52A is considered 100% load, then Kw = 52 x 120vac = 6240w or 6.24Kw
133% = 52 x 1.33 = 69A 69A x 120vac = 8280w or 8.3Kw
If your loads create a PF of 0.8 then 100% load is 52A x 0.8 = 41.6A and Kw = 41.6 x 120vac = 4992w or 5Kw (the official rating)
133% = 41.6A x 1.33 = 55.3A 55.3A x 120vac = 6636w or 6.6Kw.
So for 133%, depending on the PF you would be between 6.6Kw and 8.3Kw
Similar calculation for 300%. Edit: Something else I just thought of. In 120vac single phase mode you will probably exceed the breaker capacity before you reach 300%. Check your TM for the breaker rating and multiply it by 3.00. I bet it is less than the amps required to make the 300% which would be 52 x 3 = 156A at 1.0 PF