MVA, MVAR & MW ?? 3

[shadmani]

Controls guy here, trying to understand the hardcore power stuff. I recently visited a power plant and was kind a surprised the way the power is dispatached. Not only they were trying to maintain a certain MW load output from the station, from time to time, they were also somehow adjusting the MVAR on the machine (operators did not knew how exactly how MVAR is manipulated, they did tell me it is required based on the load on the grid and is requested to be adjusted by the dispatcher).
Could someone please, in simple terms, explain the relationship between MW, MVAR and MVA, why we talk all terms in power dispatch and what/why do we adjust the MVAR.
Thanks a bunch

 

[rcw retired EE]

To adjust MW - adjust throttle input to the turbine (increase power input to generator). To adjust Vars or reactive power - adjust excitation to the generator - raise & lower voltage.

To visualize this, take the reactive capability curve that has MW on the x axis and vars on the y axis. The semi-circle arc of radius = Rated MVA is the machine limit.

Imagine pasting this chart on an Etch-A-Sketch (kid's toy with two knobs that control a stylus to draw pictures). Label the left knob (the horizontal position) "Throttle" or "Speed". Label the right knob "Excitation" or "Voltage"

Now to increase KW output, turn the left knob. To increase excitation or Var output, turn the right knob. The best use of this analogy is when a turbine is running in power factor control. A constant power factor is a straight line from the origin to the capability limit drawn at an angle = cos-1(pf) Try to draw a straight line. It is hard, every time MW output changes, you have to change excitation to keep on the line.

 

[fsmyth]

rc, I really do like your Etch-a-Sketch analogy.
I had pictured something similar in my head (albeit
with the other axis orientation), but never considered
putting KNOBS on it. Kudos.
<als>

 

[rmw]

So if you crank on your thottle knob increasing the power input to the generator without a corresponding adjustment to the excitation knob (I assume this would be to raise the exicitation voltage), what would the resulting generator PF be, tending toward lagging?

rmw

 

[abeltio]

tending to 1.00 (all real power and no reactive power)

saludos.
a.

 

[ijl]

The reactive power was discussed in thread238-138258. But how to control the reactive power at the generating end? The answer is given in

http://www.ece.lsu.edu/mendrela/EE3410Electricgenerators.pdf

, for example, with figures and several equations.

The generator is a voltage source, when it is considered as an electrical component in a network. The parameters of a voltage source, or, in this case, a generator in a large ("infinite") network are the magnitude and the phase of the internal voltage. This internal voltage is induced in the stator windings by the rotating magnetic field of the rotor. The voltage source, i.e. the generator has also an internal impedance. The induced voltage can be thought to be behind this fixed internal impedance.

A generator has several variables that one wants to control: The magnitude and phase angle of the internal voltage, the power, and the reactive power, at least. There are two control variables: the throttle position of the prime mover and the magnetisation current. How do you control those four variables with these two control variables? Well, you don't, at least not at the same time.

The effects of the throttle and the magnetisation current are not quite simple. When one opens the throttle a little from the present position, the generator attempts to increase the rotating speed, i.e. the frequency. But if the generator is a part of a large network containing several generators, it cannot increase the speed, because it has not enough power to force all the other generators to accept the same speed. The result is that the relative position of the rotor only advances a little. This means that the phase angle (the "power angle") of the induced voltage changes in the positive direction, because it is determined by the relative rotor position.

When the magnetisation current is increased, the magnet of the rotor becomes stronger and the magnitude of the induced voltage in the stator windings increases. But a larger voltage means a larger current and a larger power and a larger loading of the prime mover. This loading attempts to slow down the prime mover and the rotor. Again, it is not possible to reduce the rotating speed because of the large network. If the throttle position and thus power are kept constant, the relative position of the rotor retards a little, so that the "power angle" becomes smaller.

In a summary: The magnetisation current controls both the magnitude and phase of the internal voltage. The throttle position controls the phase of the internal voltage. These controls must be used together in practice, as rcwilson writes.

The change in the power and the reactive power depends on how these two control variables are operated. The analysis is fortunately simplified by the assumption that the generator sits in a large network containing several generators. This means that the voltage at the terminals of the generator can be assumed to be constant. The reason is the same as that for the constant frequency: One generator cannot do much about the voltage at the terminals, it can only adjust the internal voltage.

The internal impedance Zs of a generator is typically almost purely inductive, the resistance is very small, Zs = jXs. If Ef is the induced internal voltage, and V is the voltage at the terminals, then the generator current is simply Is = (Ef – V)/jXs = -j(Ef – V)/Xs. This current is 90 degrees behind the voltage difference Ef – V, but it may lead or lag the voltage V at the terminals.

It is known that 1) the voltage V is given, 2) the reactance Xs is given and fixed, 3) the magnitude and phase of the voltage Ef are controlled by the magnetisation current, and 4) the phase of Ef is controlled by the throttle position. It should now be possible (but not necessarily easy) to see what happens to the real and reactive power in different control operations. (Remember, the complex power P + jQ = voltage times the complex conjugate of current.)

So, what happens, when the power is increased, but the magnetisation current is kept constant, as rmw asks? See figure 2.6b in the above reference. The induced voltage vector Ef turns counter-clockwise, with a constant magnitudes, i.e. the phase moves in the positive direction. As a consequence, the phase of the current moves also in the positive direction and its magnitude increases. The reactive power Q becomes less negative first, then zero, and eventually positive. See figure 2.8 in the link above for the opposite case of a constant power and varying magnetisation current.

That was a lengthy explanation, and I hope that I got it right.

 

[fsmyth]

Sounds good to me!
Like an old welder friend once told me:
"Find a good story, and stick with it."

Although I would have like to hear about the effects on
the generator when a load was suddenly removed, and
returned while the operator(s) was making adjustments.
And who or what controls the frequency on a large network
(Is there a "master" generator? Do all utilities refer
back to a master clock somewhere, similar to data networks?)

<als>

 

[Skogsgurra]

There is something called "Central Operation Desk" or similar in your language (Centrala Driftledningen in Sweden). They tell every plant on the grid how they shall run their machines to have an optimum performance all over the grid and to keep frequency.

And, yes. There is a central clock. In the older days, there were actually two clocks. One master clock telling true time and one electric clock run off the grid. They both hung on the wall in the Centrala Driftledningen - and may be there even today.

The goal of the frequency adjustment was to keep the "electric" clock within +/- 1 second of master time. That is why the frequency in large grids are quite stable. The variation is seldom more than +/- 0.1 Hz.

Unprioritized loads are normally separated from the grid when load is high and frequency lags more than about 0.15 Hz - in return they pay a lower energy price.

The frequency normally goes down a bit during the day and is then caught up in low load periods like night and early morning.

Gunnar Englund

http://www.gke.org

 

[cuky2000]

…….. effects on the generator when a load was suddenly removed and returned while the operator(s) was making adjustments. And who or what controls the frequency on a large network.

Here is an attempt to explain in a simplified manner what may happen after the sudden loss of load (ex. Loss of a transmission line or disconnection of large load)

[blue]Background:
Immediately following that, a transient redistribution of the power flow will be initiated with significant reduction of voltage adjacent to the load center due to extra reactive power demand. [/blue]

The generator conditions will not act instantaneous do to the inertia of the rotating mass and the EMF energy stored in the winding causing a time-delay for the sensor to notice any change for sudden loss of load. After that, the generators automatic voltage regulators (AVRs) will command the control to increase excitation to elevate the voltage terminals and the governors would respond to regulate the frequency by reducing the MW output.

Two scenarios may happen depending upon the size of the load, system parameters and location of the generator(s) with respect to the sudden loss of load.

SCENARIO 1: The above conditions happens within the generator limits of the reactive capability of the unit the system parameter will be swinging until the oscillation will damp down stabilizing after a few second in a new steady-state conditions.

SCENARIO 2: Synchronism will be loss pulling the generator out of step because one or more of the following conditions happens:
- Sudden changes above the limits of the generator
- voltage below system limits (trip undervoltage protections)
- Frequency out of limits (trip synchro-check protection).

 

[fsmyth]

I'm guessing that "P" is voltage magnitude, and "Q" is
service factor?

I just had pictured in my head all the assorted service
providers (generators) scrambling to correct a major,
but not critical, fault somewhere on the network.
I feel pretty sure that someone has worked out a set of
procedures for this; surely there is some type of
co-ordination between plants. Or does the biggest guy
set the order of march, and instruct the smaller plants
to "fine tune" the balance?

I certainly do not know enough about power distribution
to understand it, and probably not enough to ask decent
questions. But that has never stopped me before.
How much time lapses from the time a fault occurs (say
a breaker trips on a 500A load 10 miles away from the
substation, which is in turn 50 miles away from the
generating plant). Cycles? Milliseconds? Minutes?

<als>

 

[davidbeach]

P is real power, Q is reactive power.

500A load at what voltage? 500A at any utilization voltage 4160V or below is a rounding error as far as grid capacity is concerned. Electric power is the one true just-in-time production and delivery system. When the load goes away, power ceases to be generated and the power angle of the generators increases. Where this is loss of load is small, the difference in power angle is insignificant. 500A at 480V is 415kW, less than 1/2 of 1% of the capacity of a 100MW generator.

When a sudden loss of load is significant (maybe a 500A load at 230kV or higher), the excess power going into the generators causes the generators to increase in speed. One of three things can happen at that point: the generator operates at an increased, but stable, power angle until the input power is decreased to match the remaining load; the generator power angle increases at a rate that would lead to instability if allowed to continue, but the input power is reduced sufficiently fast enough, or load returns soon enough that stability is not lost; or the generator power angle increases to the point of instability and the generator starts slipping poles. It would take two 100MW generators running at full capacity to produce 500A at 230kV, so a plant running a few 100MW generators would have serious stability problems if it suddenly lost 500A at 230kW.

The generator would know of its loss of load at a time equal to distance divided by the propagation velocity of the lines; for overhead lines this is very close to the speed of light.

 

[fsmyth]

First off, I want to thank all of you for having the
patience to explain a very large subject to a very
small mind.

I just picked an arbitrary number (500A) because I
figured it would be a noticable change at any location.
I had figured the HV distribution to be some large
multiple of 10kV, but did not know it was 230 kV.
I had already picked up on most of what davidbeach
posted from various other posts, and was/am more
interested in what actually occurs in the real world.
What actually got me wondering, was how in the heck
all those wind generators are kept in concert (don't
start with the inverter explanations, I pretty much
got a grip on that). Which sorta led to wondering
how all the generators tied to one grid were kept in
sync. Surely there is added delay in substation
transformers and other effects that come into play
with miles of wire.

I may be asking for more than I really want to know.
<als>

 

[davidbeach]

Figure the effects are essentially instantaneous. Even if the propagation velocity through a transformer were reduced to 3 x 10^6 m/s (1/100 the speed in an overhead line), the transformer is so small that it doesn't matter.

Also, fsmyth, please don't hit return at the end of each line when typing your posts. The forum software will handle the word wrap at the end of the lines and your posts will be easier to read without all the short lines. Thanks.

 

[alehman]

Every generator connected to the 'grid' is physically forced to be in sychronism. If it starts to get off, it will be pulled back by the grid or will be forced off-line by its protection. There is nothing other than the transmission system itself to keep the generators synchornized. Over a very large distance, there may be a slight phase difference due to propagation velocity (essentially speed of light), but at 60Hz, the wavelength is 5000km. Phase shifts due to system impedances are much larger.

 

[rmw]

http://powermarketers.netcontentinc.net/newsreader.asp?ppa=8kowu][fghfoyuVUgc"EN&bfek\v

rmw