[quote]SkyzykS wrote:
Getstrength, is this a matter of being hung up on the definitions and the function they serve in a system, or just solving the formula?
Sometimes with stuff like this you just have to solve the problem in abstract and apply the numbers to the formula, then figure out the theory later.
[/quote]
Hi SkyzykS,
Solving the equations is fine, just trying to understand what it is/means.
I was hoping this thread would be that later time when I figure out the theory.
[quote]waldo21212 wrote:
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).
[/quote]
So say we make a power supply and the load needs 3KW, the reactive components will store 2KVAr.
How much power do we need to produce at the supply (ignoring losses in the lines and transformers etc) ?
[quote]
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.[/quote]
[quote]Edgy wrote:
< -------------------------- not anywheres near smart enuf to contribute to this thread, but hate be be left out.[/quote]
Same here dude haha
Honestly, I believe the best scenario for you to figure out your problem would be face to face with one of your teachers. Thats what you’re paying them for, and thats their job. Internet discussions could go on for ages without you finding an answer. A conversation with someone who is teaching this stuff to you should take less than half an hour.
I hope that was contributing enough to class as useful! Dont be scared you’re there to understand. If you dont understand, keep pushing and get everything you can out of it!
jw154505’s explanation is excellent. I would give it a few days to sink in. If you are struggling with your classes this early on you may want to find a new major because this will only get harder, but if this is all you are having trouble understanding don’t worry so much about understanding everything and just learn it as you are told.
I didn’t truly understand many of the concepts I learned as an undergrad until my last semester when I started studying for the FE exam.
[quote]quaffloid wrote:
jw154505’s explanation is excellent. I would give it a few days to sink in. If you are struggling with your classes this early on you may want to find a new major because this will only get harder, but if this is all you are having trouble understanding don’t worry so much about understanding everything and just learn it as you are told.
I didn’t truly understand many of the concepts I learned as an undergrad until my last semester when I started studying for the FE exam.[/quote]
Hey quaffloid,
jw154505’s explanation didn’t really explain much (no offense jw), I already know how to calculate all the different powers for different systems such as star and delta. It was essentially just the same stuff about vectors and its the vector sum or the hypotenuse. None of this actually explains what apparent power is. This is the only thing I haven’t understood so far and I don’t think anyone else would either but since they generally take the attitude of only learning enough to pass the exams I can’t say confidently if they would understand it or not. My grades are an A- average so far (in my first yr of EE but had a yr of physics and calculus/differential equations and c programming etc last year). I’m just really interested and so always think
more deeply about what is taught and believe this is part of the reason I get good grades.
I am confused as to why in general people are saying to essentially forget about it until later. It seems like no one in here actually knows and don’t want anyone asking.
No one has actually said they understand the question, can see where the confusion is and offer
an explanation. I don’t mean to sound rude or anything but does anyone in here actually understand it or are you guys just regurgitating what you have been told? It seems to me that if anyone knew they could explain it as I’m sure anyone who got taught this stuff and thought about it would ask the same thing I am asking.
Again I don’t mean to sound rude but I’m not looking for advice on passing a course or who to ask or anything else, I’m asking if anyone can understand what I am saying, see the confusion and explain. Bit of a ramble but I’m sure you guys get the idea. This isn’t just directed at you quaffloid.
[quote]GETSTRENGTH wrote:
So say we make a power supply and the load needs 3KW, the reactive components will store 2KVAr.
How much power do we need to produce at the supply (ignoring losses in the lines and transformers etc) ?
[/quote]
The supply would need to be rated at 3 kW and 2 kVAR, or 3.6 kVA @ 0.8333 leading PF (assuming the power supply is supplying VARs and the load is absorbing VARs).
Giving a rating in kW and kVAR or kVA w/PF is really 2 ways of saying the same information.
i feel like im getting trolled…
[quote]waldo21212 wrote:
[quote]GETSTRENGTH wrote:
So say we make a power supply and the load needs 3KW, the reactive components will store 2KVAr.
How much power do we need to produce at the supply (ignoring losses in the lines and transformers etc) ?
[/quote]
The supply would need to be rated at 3 kW and 2 kVAR, or 3.6 kVA @ 0.8333 leading PF (assuming the power supply is supplying VARs and the load is absorbing VARs).
Giving a rating in kW and kVAR or kVA w/PF is really 2 ways of saying the same information.[/quote]
Ok cool, I get what you’re saying. Thanks Waldo thats half my confusion cleared up.
How would you describe why the supply has to provide less than the system needs in our example?
On first inspection you would thinkit would need 5KW supplied all up…
Do you know what I mean? where is the difference betwen the 3.6KVA being supplied and the 3+2=5 being taken by the circuit?
I was thinking that it could be something along the lines of since KVAR isn’t actually used and it just gets stored in the field say in an inductive load that we would only have to actually create this power on start up and it would then not need to be continuously createdand supplied tothe circuit because it nevers get used by the resistive load. I also thought that maybe if that was the case that the 0.6 difference got burnt up in the lines or something?
You pop a kilowatt into the system, at the end when you pop your switch on you get 900watts. you lost 100watt due to stuff like inductors (ooils) in the system. so the apparant power is a kilo watt but you only get to use 900 watts of it.
the losses get eaten up or stored by inductors for abit then go back into the system later
[quote]kickingking wrote:
You pop a kilowatt into the system, at the end when you pop your switch on you get 900watts. you lost 100watt due to stuff like inductors (ooils) in the system. so the apparant power is a kilo watt but you only get to use 900 watts of it.
the losses get eaten up or stored by inductors for abit then go back into the system later[/quote]
To follow up with that, as a former commercial electrician one of the things I did to counter the inductive reactance referenced above was to install a capacitor bank after the main switch in parallel with the building load. It brought the power factor back down to a much more palatable figure. Today’s buildings are VERY heavily inductive due to all of the coils associated with modern lighting (ballasts), electronics - even a greater 120/208v demand which needs additional transformers. All of these devices/equipment contain coils thus creating inductance/inductive reactance which will cause some serious lag on a three phase system. A capacitor bank introduces a bit of leading (capacitive reactance) and brings the power factor back to where it should be.
I’m not an engineer, but that’s how I’ve applied some of what you guys are talking about IRL. I’m sure if I got the leading/lagging part wrong, someone will correct me! (It’s been a minute since I studied this shit)
[quote]angry chicken wrote:
[quote]kickingking wrote:
You pop a kilowatt into the system, at the end when you pop your switch on you get 900watts. you lost 100watt due to stuff like inductors (ooils) in the system. so the apparant power is a kilo watt but you only get to use 900 watts of it.
the losses get eaten up or stored by inductors for abit then go back into the system later[/quote]
To follow up with that, as a former commercial electrician one of the things I did to counter the inductive reactance referenced above was to install a capacitor bank after the main switch in parallel with the building load. It brought the power factor back down to a much more palatable figure. Today’s buildings are VERY heavily inductive due to all of the coils associated with modern lighting (ballasts), electronics - even a greater 120/208v demand which needs additional transformers. All of these devices/equipment contain coils thus creating inductance/inductive reactance which will cause some serious lag on a three phase system. A capacitor bank introduces a bit of leading (capacitive reactance) and brings the power factor back to where it should be.
I’m not an engineer, but that’s how I’ve applied some of what you guys are talking about IRL. I’m sure if I got the leading/lagging part wrong, someone will correct me! (It’s been a minute since I studied this shit)
[/quote]
Former electrician here aswell 
I think you have the lead/lag right as thats how I remember it.
[quote]GETSTRENGTH wrote:
[quote]waldo21212 wrote:
[quote]GETSTRENGTH wrote:
So say we make a power supply and the load needs 3KW, the reactive components will store 2KVAr.
How much power do we need to produce at the supply (ignoring losses in the lines and transformers etc) ?
[/quote]
The supply would need to be rated at 3 kW and 2 kVAR, or 3.6 kVA @ 0.8333 leading PF (assuming the power supply is supplying VARs and the load is absorbing VARs).
Giving a rating in kW and kVAR or kVA w/PF is really 2 ways of saying the same information.[/quote]
Ok cool, I get what you’re saying. Thanks Waldo thats half my confusion cleared up.
How would you describe why the supply has to provide less than the system needs in our example?
On first inspection you would thinkit would need 5KW supplied all up…
Do you know what I mean? where is the difference betwen the 3.6KVA being supplied and the 3+2=5 being taken by the circuit?
I was thinking that it could be something along the lines of since KVAR isn’t actually used and it just gets stored in the field say in an inductive load that we would only have to actually create this power on start up and it would then not need to be continuously createdand supplied tothe circuit because it nevers get used by the resistive load. I also thought that maybe if that was the case that the 0.6 difference got burnt up in the lines or something?[/quote]
Again, you cannot simply add P + Q = S
Since they are vectors, it is: |S| = Sqrt(P^2+Q^2)
or S = P + jQ (where j is the imaginary operator i, but in electrical engineering i is current so we use j)
so S can be expressed in the example as: S = 3 + j2 in rectangular form, which again does not equal 5, but equals 3.6/_33.55 degrees espressed in polar form (magnatude and angle. And, cos(33.55) = 0.833, which is the Power factor in the example.
[quote]angry chicken wrote:
To follow up with that, as a former commercial electrician one of the things I did to counter the inductive reactance referenced above was to install a capacitor bank after the main switch in parallel with the building load. It brought the power factor back down to a much more palatable figure. Today’s buildings are VERY heavily inductive due to all of the coils associated with modern lighting (ballasts), electronics - even a greater 120/208v demand which needs additional transformers. All of these devices/equipment contain coils thus creating inductance/inductive reactance which will cause some serious lag on a three phase system. A capacitor bank introduces a bit of leading (capacitive reactance) and brings the power factor back to where it should be.
I’m not an engineer, but that’s how I’ve applied some of what you guys are talking about IRL. I’m sure if I got the leading/lagging part wrong, someone will correct me! (It’s been a minute since I studied this shit)
[/quote]
Yeah, there’s a difference between what the engineers need (to make the thing work) and what the power co. needs (to calculate your bill); what electrical power “is” can be different things depending on who wants to know and why…
OP you may find this site interesting (or maddening): What is Electricity?