How to understand negative real and reactive power?

Hi,
In a AC power quadrant diagram, how to understand the physical meanings
of powers that falls in each quadrant?
I: +P, +Q
II: -P, +Q
II: -P, -Q
IV: +P, -Q
where P denotes real power and Q denotes reactive power.
Thanks in advance.
Reply to
narke
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What is it you don't understand?
Negative real power simply means that power is flowing in the direction opposite from convention. For a generator, it would mean that power is flowing from the grid/bus into the generator to keep it spinning. This happens when the engine/turbine is not generating enough power to overcome friction/windage losses and the electrical bus has to supply power into the generator to keep it spinning. (it does NOT mean the generator shaft has reversed its direction of rotation)
Similarly, negative reactive power means reactive power is flowing in the direction opposite from convention. Normally a generator supplies reactive power to a bus to 'feed' the reactive loads on the bus. Convention is that inductive loads consume 'positive reactive power' and capacitive loads are said to supply 'positive reactive power'. You could also argue that capacitive loads supply 'negative' reactive power which cancels out the 'positive' reactive power of inductive loads.
For a generator, reactive power is usually labeled 'positive' when it is over-excited and supplying reactive power to inductive loads. If under-excited, it actually draws reactive power from an infinite bus (its reactive power is 'negative').
daestrom
Reply to
daestrom
Thanks for the good explaination, but I still have some difficulties in understanding, please see my comments below.
Because my domain is three-phase electricity meters that are installed in utilities and high load users. So I like to ask, when the meters observed negative real power, does it mean that the energy is flowing back to from consumer side to the grid side?
That sounds clear. And, for the follow two AC circuits,
Setup A: ------L------+ | +----R------Electricity Meter-- | ------C------+
Setup B: ----R------Electricity Meter--
where L is ideal inductor and C is ideal capacitor, and impedance L is same as impedance C in maganitute.
So, can I deduce followings?
1. the apparent energy observed in A is same as that in B; 2. real energy in A is same as real energy in B, and that amounts to I^2 * R; 3. reactive energy in A is same as reactive energy in B, and that amounts to zero.
I don't understant what you mean 'infinite bus'.
Reply to
narke
Is there an answer? Thanks in advance.
Reply to
narke
---------------------------- Yes ------------------
--------------- Your circuit A is hard to follow. It doesn't appear to be either a 3 phase or a single phase circuit. Source and return paths are omitted.
Possibly this single phase circuit may help. ----1------+---2-----+----3----+ source | | | V L C R Impedance conditions as above | | | -----------+----------+--------- +
Meter connected at 1 (and return) , 2 or 3 real power and energy is measured At 1,3 there will be no reactive measured. At 2 there will be reactive measured and it will be negative (capacitive load).
Does this help?
---------------- This means an ideal voltage source (no internal impedance) and implies, in practice, that the source is so large that what you connect to it doesn't measurably affect its voltage.
Reply to
<dhky
I fell it can be very helpful ... just now the format of the graph not good, it's hard to get the information. Would you please redraw the circuit to make it clear? (yes, single phase is assumed)
Many thanks!
Reply to
narke
I wrote a java applet which allows you to play with phase shift and see the effect on the reactive power draw. You can see how the system effectively borrows energy from the source and returns it at different points in the cycle. This excess energy you repeatedly borrow and return is the reactive power, whereas the energy you take, use, and don't return is the real power.
formatting link
Reply to
Andrew Gabriel
ply
----1------+---2-----+----3----+ source | | | V L C R | | | -----------+----------+--------- +
L, C and R in parallel: Source at 1 and L between 1 and 2, C between 2 and 3 and R after 3 meter at 1, 2 or 3 with its voltage leads to bottom line
KWH reading the same at all locations and the same as if L and C didn't exist Reactive (KVARH) same as without L and C for meter at 1 or 3 (zero) Reactive negative with meter at 3 Meter reads KWH and KVARH "downstream" of its location.
Reply to
<dhky
That's so much wonderful!!! Actualy, in the way of learning these concepts these days, I also came with an idea that is to do exactyly what you did after I really understand the knowledge. But I plan to implement the demo application in other language. So I am thinking, could you share your source code to me, so I can see the algorithm. If you don't want to share, it is okay. You've already help anyway.
Thanks again.
Reply to
narke
Cleared for this, many thanks!
Also cleared! Thanks.
I assume you meant to say 'Reactive negative with meter at 2', right?
Can you explain what mean "downstream" ?
For the serial connection of L,C,R in the following circuit:
--- AC Source V -------L-------R-------C--------- 1 2 3
Please also give me an answer for meter readings of kWh and kVAh in point 1, 2, 3 respectively.
Thanks.
Reply to
narke
Andrew,
I have another qeustion after read your web page. When you demostrate the concepts, your were using the areas of the curve. The areas (red, grean), i think, are energes. And, I feel it can be decuded from your tratement, that VAH = kWh + kvarh (1) where, * kWh is (red area) - (green area) * kvarh = (green area) * VAH = (red area)
Am I right?
However, we know S^2 = P^2 + Q^2 (2) where, S is apparent power, P is real power or true power, Q is reactive power.
Now I get problem, since (2) can not be deduced from (1). Did you see it? Please help.
Reply to
narke
No. It's simply red area is energy drawn from supply (quadrants I and III in your original post), and green is energy returned to supply (quadrants II and IV in your original post). So red includes both real and reactive energy draw, but green can only be reactive. Since the reactive energy returned to the supply (green) will be same as reactive energy drawn from supply, you can assume that a part of the red area equal in size to the green area is also reactive energy, and the remainder of the red is real power. However, it doesn't make any sense to try and identify that part because you can't sensibly say which particular bit of the energy drawn from the supply (red) is reactive and will be returned, verses another bit of the red which is not reactive and will be burned in the load. i.e. at a phase angle of 60° (PF=0.5), 1/3rd of the red area is reactive and 2/3rds is real, but there's no way to say which of the red pixels is reactive or which is real.
So: Wh is fixed in this example. It's same as red area at phase angle = 0. VAh will be red + green. VArh will be 2*green.
That comes from the vector diagram equalatral triangle. It should be same result, but at 01:05 in the middle of the night here, I can't immediately think how you could prove it.
Reply to
Andrew Gabriel
Hi, Andrew, I just came back for a travel. Sorry for so late following the thread.
Okay, you fixed my error in understanding what is apparent power. So, let me rearrange the result as:
Wh = red - green VAh = (red - green) + (green + green) = Wh + WArh VArh = 2*green
So, it still shows that VAh = Wh + WArh. To my mind, it still conflicts with S^2 = P^2 + Q^2, becaseu I think S is WAh/h, P is Wh/h, Q is WArh/h.
As stated above, yes I don't see a sign that I can prove it. So please help :)
- narke
Reply to
narke

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