Power factor is a term that accounts for the AC waveforms of current and voltage not being synchronized. If you blindly multiply the nominal values (i.e. RMS values, which is a special kind of averaging) of AC voltage and current without thinking about how they are synchronized or not, you get the term that we call apparent power. It has units of VA or kVA, and is commonly represented with the variable S. Apparent power is important, because your distribution equipment needs to be sized for the full voltage and amps, regardless of how well or poorly the waveforms are synchronized That's why transformers are rated in kVA instead of kW.
If you're calculating the load of a heater (resistive load) you can ignore using PF in your calculation.
Nice write up. I think in the above sentence you mean reactive power, rather than apparent power.
I don't think you need calculus for that - it's simple algebra.
Read the first post in that linked comment thread. It was a challenge to show why the power triangle works, from first principles. It's simple algebra to use the power triangle. The calculus comes from how we get the average value of a continuous function, and the theory behind why the RMS nominal values are related to the amplitude with a ratio of sqrt(2).
| Cct No1 - Supply Short Circuit Impedance | 0,64% | ||||||||
| Supply Fault Level | (MVA) | 40 | |||||||
| Supply resistance | (50Hz Ohms) | 0,06534 | 6,53E-02 | 6,53E-02 | 6,53E-02 | 6,53E-02 | 6,53E-02 | 6,53E-02 | |
| Supply reactance | (50Hz ohms) | 0,2640825 | 1,32E+00 | 1,85E+00 | 2,90E+00 | 3,43E+00 | 4,49E+00 | 5,02E+00 | |
| Cct No2 - Drive Transformer | |||||||||
| Rating | (kVA) | 3000 | |||||||
| Impedance | (%) | 5,75 | |||||||
| Resistance | (50Hz Ohms) | 5,01E-02 | 5,01E-02 | 5,01E-02 | 5,01E-02 | 5,01E-02 | 5,01E-02 | 5,01E-02 | |
| Reactance | (50Hz Ohms) | 2,02E-01 | 1,01E+00 | 1,42E+00 | 2,23E+00 | 2,63E+00 | 3,44E+00 | 3,85E+00 | |
| Cct No1 + Cct No2 | |||||||||
| Resistance | (50Hz Ohms) | 1,15E-01 | 1,15E-01 | 1,15E-01 | 1,15E-01 | 1,15E-01 | 1,15E-01 | 1,15E-01 | |
| Reactance | (50Hz Ohms) | 4,67E-01 | 2,33E+00 | 3,27E+00 | 5,13E+00 | 6,07E+00 | 7,93E+00 | 8,86E+00 | |
| Cct No3 - Other Lagging Loads - No details | |||||||||
| Resistance | (50Hz Ohms) | 500 | 5,00E+02 | 5,00E+02 | 5,00E+02 | 5,00E+02 | 5,00E+02 | 5,00E+02 | |
| Reactance | (50Hz Ohms) | 500 | 2,50E+03 | 3,50E+03 | 5,50E+03 | 6,50E+03 | 8,50E+03 | 9,50E+03 |
You are using equations that came from the Laplace transform, which is defined through Calculus. That's what the s stands for, which is the parameter used in the Laplace transform. Working in the s-domain enables Calculus operations of the time domain to be calculated through algebra in the s-domain. Understanding why those equations would work from first principles, is where the Calculus would be involved.
All the time. When you start with the amp ratings that are already published on your equipment datasheets/nameplates, it has already been accounted for.
Clearly I do know that. My point is that you simply don't need calculus for trig calculations.
The power factor is always involved, whether it's included in the calculation or not.
I disagree but what do I know?
That is fine for linear, displacement, power factors, but calculus is unavoidable for distortion power factor, since trig is not sufficient for the calculation for a specified non-sinusoidal waveform.
I do use non-linear harmonics like 5th, 7th, 11th, 13th etc. I don't need calculus for that.
