How to understand negative real and reactive power?

[narke]

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.

 

[daestrom]

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.

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

 

[narke]

What is it you don't understand?

Thanks for the good explaination, but I still have some difficulties in
understanding, please see my comments below.

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)

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?

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.

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.

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').

I don't understant what you mean 'infinite bus'.

 

[narke]

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.

Is there an answer? Thanks in advance.

 

[Guest]

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? ----------------------------
Yes
------------------

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.

---------------
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?

 

[narke]

---------------
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?

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!

 

[Andrew Gabriel]

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.

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.

http://blogs.sun.com/agabriel/entry/ac_power_power_factor_explained

 

[Guest]

ply

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!

----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.

 

[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.

http://blogs.sun.com/agabriel/entry/ac_power_power_factor_explained

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.

 

[narke]

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

Cleared for this, many thanks!

Reactive (KVARH) same as without L and C for meter at 1 or 3 (zero)

Also cleared! Thanks.

Reactive negative with meter at 3

I assume you meant to say 'Reactive negative with meter at 2', right?

Meter reads KWH and KVARH "downstream" of its location.

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.

 

[narke]

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.

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.

 

[Andrew Gabriel]

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?

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.

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.

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.

 

[narke]

Hi, Andrew, I just came back for a travel. Sorry for so late following
the thread.

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.

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.

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.

As stated above, yes I don't see a sign that I can prove it. So please
help

-
narke