Hi everyone,
I’m a first yr electrical engineer. We have been covering 3 phase power supplies.
I understand true power and reactuve power but I am struggling to understand the relevance of
apparent power. This also means I don’t understand power factor. I understand that apparent power is the vector sum of true and reactive power. I just don’t understand what it really is.
I also would have thought that we would need a total power more than apparent power. That way we could work out what our total power supplied is and how much of that is true vs reactive and so on. I know I must be wrong about this. Can somebody please explain this shit to me?
Power Factor measures the power efficiency within the system. It’s basically a ratio between real and apparent power. The apparent power is important because ideally you want a pf of 1 (purely resistive) meaning a smaller system can be implemented. If your pf is lower than one, its basically more expensive to run the system because it requires more current and causes more voltage drops.
The “relevance” of apparent power here is to build your system close to a pf of 1 and since pf is real power/apparent power. I am an Electrical Engineer but I don’t deal with 3-phase systems too much so I would look into some power focused books, you can download a few for free from (http://artikel-software.com/blog/ebooks/electrical-engineering/). Hope that helps.
[quote]Edevus wrote:
Why don’t you appoint a meeting with a teacher so he can explain you?
I did two years of electronical engineering and what you are saying is very familiar, but that was years ago so I can’t help you much.[/quote]
Hi Edevus,
Good advice but I have transferred from another uni recently and there have been many complications and so have required a lot of my lecturers time lately and don’t want to
become a pain in the ass. I asked in class but the answer was a bit odd.
Thanks anyway.
You mentioned doing two years, was the the whole course or did you move on for some reason?
[quote]jw154505 wrote:
Power Factor measures the power efficiency within the system. It’s basically a ratio between real and apparent power. The apparent power is important because ideally you want a pf of 1 (purely resistive) meaning a smaller system can be implemented. If your pf is lower than one, its basically more expensive to run the system because it requires more current and causes more voltage drops. The “relevance” of apparent power here is to build your system close to a pf of 1 and since pf is real power/apparent power.[/quote]
Hi Jw
I know understand what you are saying but my question is what is apparent power?
I understand power factor from the point of view of real power/apparent but I feel like
it should be real power/total power. I don’t understand from the point of view that apparent power has no meaning to me and so the ratio of real power/apparent doesn’t make sense to me.
I appreciate your input and added information. I hope I am making sense in my response.
[quote]Edevus wrote:
Why don’t you appoint a meeting with a teacher so he can explain you?
I did two years of electronical engineering and what you are saying is very familiar, but that was years ago so I can’t help you much.[/quote]
Hi Edevus,
Good advice but I have transferred from another uni recently and there have been many complications and so have required a lot of my lecturers time lately and don’t want to
become a pain in the ass. I asked in class but the answer was a bit odd.
Thanks anyway.
You mentioned doing two years, was the the whole course or did you move on for some reason?
[/quote]
Looks like jw154505 gave you some valuable information.
You shouldn’t be afraid of asking teachers again and again. Do it in private, so there are no hurries and full focus on the matter. It’s their job after all.
I didn’t finish because I became extremely frustrated with many things, including teachers. So I wish you good luck, those engineering degrees are quite complicated.
Ok, I guess to understand the actually meaning of apparent power, look at the system as a whole:
True Power (P) is power dissipated by a load
Reactive Power (Q) is power absorbed and returned in load
Apparent Power (S) is the total power in an AC circuit, both dissipated and absorbed/returned
The “power triangle” is all you really need to know, especially for solving them, which you said you already understand.
Hope that helps in terms of “defining” what it actually is.
[quote]jw154505 wrote:
Ok, I guess to understand the actually meaning of apparent power, look at the system as a whole:
True Power (P) is power dissipated by a load
Reactive Power (Q) is power absorbed and returned in load
Apparent Power (S) is the total power in an AC circuit, both dissipated and absorbed/returned
The “power triangle” is all you really need to know, especially for solving them, which you said you already understand.
Hope that helps in terms of “defining” what it actually is.[/quote]
Thanks for the response.
In relation to the power triangle, I have this question:
If apparent power is the total power, how can it be less than the sum of the
true power and the reactive power?
Thats the crux of my confusion right there.
[quote]jw154505 wrote:
They are all different units: W, VAR and VA.
If they were all the same unit, say watts, then yes, the sum could not be less.
The resistances in each give them the different unit.
P=I^2R
Q=I^2X
S=I^2Z[/quote]
But if this were the case how could we add vectors which have different units?
I would argue that they are the same units…they all come from the supply and are joules/second
Aren’t they all watts? we just call them Watts,KVA and KVAr to distunguish between them?
I am a physics student so I do not know the specifics about electrical engineering but it seems to me this is just a simple Pythagoras triangle where the phase represents the Impedance.
If these are all vectors in the same plane they must have the same units and the Apparent Power(S) squared is the sum of the True Power(P) and Reactive Power(Q) squared, so S^2=P^2+Q^2. This is why the Apparent Power is less than the normal sum of the other two.
[quote]jfarts wrote:
I am a physics student so I do not know the specifics about electrical engineering but it seems to me this is just a simple Pythagoras triangle where the phase represents the Impedance.
If these are all vectors in the same plane they must have the same units and the Apparent Power(S) squared is the sum of the True Power(P) and Reactive Power(Q) squared, so S^2=P^2+Q^2. This is why the Apparent Power is less than the normal sum of the other two.[/quote]
That’s what I think as well, which is why I am confused as to why the apparent power matters
or what it actually is.
Look at this wikipedia page: Electric power - Wikipedia. The apparent power is what you might think is the power of a source, but due to a phase difference between the current and voltage there flows an amount of power back into the source. What is left is the true power.
Apparent power (VA) is important measure in power systems, because equipment ratings are usually in MVA (Mega Volt-Amperes), and both Watts and Vars have current associated with them, and thermal limits on equipment are derived from the maximum allowable current flowing through them (current x voltage = VA rating).
Power factor is also an important measure, because the closer you are to unity power factor, the more Watts you can have flowing through a given piece of equipment (unity PF = 0 VARs, so VA = Watts in that case).
I am a power system engineer in the utilty industry, so I deal with this quite frequently.
[quote]jfarts wrote:
Look at this wikipedia page: Electric power - Wikipedia. The apparent power is what you might think is the power of a source, but due to a phase difference between the current and voltage there flows an amount of power back into the source. What is left is the true power.[/quote]
Hi jfarts,
when I open that page there is no information, just some stuff about no info
being there. In regards to what you said: If true power was what is left, shouldn’t it
be the difference between the apparent power and the reactive power? but it isn’t
[quote]waldo21212 wrote:
Apparent power (VA) is important measure in power systems, because equipment ratings are usually in MVA (Mega Volt-Amperes), and both Watts and Vars have current associated with them, and thermal limits on equipment are derived from the maximum allowable current flowing through them (current x voltage = VA rating).
Power factor is also an important measure, because the closer you are to unity power factor, the more Watts you can have flowing through a given piece of equipment (unity PF = 0 VARs, so VA = Watts in that case).
I am a power system engineer in the utilty industry, so I deal with this quite frequently.[/quote]
Hi Waldo,
Thanks for the response. Can you define what apparent power is?
[quote]jfarts wrote:
Look at this wikipedia page: Electric power - Wikipedia. The apparent power is what you might think is the power of a source, but due to a phase difference between the current and voltage there flows an amount of power back into the source. What is left is the true power.[/quote]
Hi jfarts,
when I open that page there is no information, just some stuff about no info
being there. In regards to what you said: If true power was what is left, shouldn’t it
be the difference between the apparent power and the reactive power? but it isn’t
[/quote]
As I said before, I’m not very familiar with electrical engineering. But in a vector plane you can not just subtract two values when they are not in the same direction.
Apparent power could also be called the total power. As stated above, the reason you cannot add real power and reactive power to get apparent power is that the 3 quanities are acutally vectors (magnitude and direction).
The reason direction is needed is because you can have a load which absorbs power (Watts, Vars), a generator that supplys power (Watts, Vars), a capacitor that supplys Vars, or a reactor that absorbs Vars.
Like the guy above said, it may be best just to ignore the theory and follow the formula, and eventually the theory will start to make sense once you start to see practical application.
