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shadmani (Electrical) (OP)
10 Nov 05 8:37
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
BEPC13 (Electrical)
10 Nov 05 9:01
MVA is the aparant power, MW is the real power and, MVAR is reactive power. If you remember your Circuit II course MVA is the Square Root of MW^2+MVAR^2. Just like a right triangle. It is actually called the power triangle. You adjust the MW of a plant based on the load on the grid. Generation = Load at all times. You adjust the MVAR of the machine to adjust grid voltage in the local area. MVARs are the result of the magnetic coupling needed to produce work with a machine. They are not exportable therefor you either have local generation or capacitor banks provide them. You adjust the MW of the machine with the amount of torque you apply to the input shaft of the machine. This usually means more steam to the turbine. This tries to speed up the machine against the grid and produces more power. You adjust the MVAR output of the machine by adjusting the exciter of the machine. That is to raise MVAR output you excite the machine more by increasing the current in the field (rotor). This increases the magnetic coupling of the rotor to the stator and increases the MVAR output.
Helpful Member!  Barrie66 (Electrical)
10 Nov 05 13:46
BEPC13 explained it correctly however, it may be clearer to simply state that an MVAR meter in a power plant is a center zero device. By raising the machine volts the generator will export VARs to the system and conversely by decreasing the machine volts VARs will be imported to the generator. When I was working as a power plant operator shortly after graduating in the early 60s. Each power station on the grid had to maintain its voltage within given parameters. This generally mean't that the voltage on the outgoing line was had to be kept at its nominal kV. The power plant I operated, was on the end of a 132kV spur line some 120 miles from the main 330kV interconnected system, and the voltage used to vary substantially over 24hrs. During the night on light system loads, the voltage would rise and we would have to import VARs by lowering the generator volts and during peak periods on high system loads, raise the generator volts and export VARs to maintain nominal voltage on the 132kV feeder. I guess these days the system controllers spread the VAR loads around much better than we did in the old days.....
rmw (Mechanical)
10 Nov 05 16:26
This is an interesting thread in light of a couple of recent threads on reactive power.  They did not deal with what actually happened at the power station, however.

Now, I have a purely theoretical question.  Since power factor is a function of load, reactive load, and VARS is a function of the PF, (if that is not right correct me right away) in a hypothetical system, a power plant supplies power to a grid that has lots of installed horse power,  electric motors, creating a significant lagging power factor, and hence a given amount of VARS.

What happens when all those workers turn off their motor driven machines at the end of the work day and go home and turn on their electric space heaters, coffee pots, toasters, ovens, and lights.  You get the picture, purely resistive load, approaching unity power factor.

Is something done at the power plant to correct, effect or change either of those conditions?

And, since the power companies are beginning to bill for the effects of reactive power, isn't it in the Utilities interest to not correct the PF?

rmw

PS: As I stated in one of the other threads, I am trying to comprehend this as an ME.  I can make those turbines produce the torque, but I struggle to understand what you EE's do with it to ship it.
Barrie66 (Electrical)
10 Nov 05 17:44
With regards to
"What happens when all those workers turn off their motor driven machines at the end of the work day and go home and turn on their electric space heaters, coffee pots, toasters, ovens, and lights.  You get the picture, purely resistive load, approaching unity power factor" as aked by the previous thread, the following happens.
First of all as the overall system MW decrease, the system  voltage tends to rise. To control the volts, inductive MVARs have to be generated by either switching in reactors or reducing MVARs by reducing volts on the generators supplying MVARs to the system.
With regards to Utility companies charging for MVARs. as it wattless current (except for I2R losses)utility companies do not have fuel costs generating VARs. So should only charge for generator losses etc. which would be very small.
Skogsgurra (Electrical)
10 Nov 05 18:13
The problem with var (yes; var, not VAR or VAr) is that it occupies transformers and transmission lines. Utilities therefore charge to keep var consumption low. If they were to deliver lots of Mvars, there would not be much rom for the MWs.

Gunnar Englund
www.gke.org

cuky2000 (Electrical)
10 Nov 05 21:07

Quote:

Reactive power also happens to be the most misunderstood concept in the electric utility industry. This is mostly because it is usually described as the imaginary part of the load; therefore, people tend to believe that it does not exist or serves not useful purpose. Noting could be further from the truth.  

Reactive power supply is essential for reliably operating the electric transmission system. “Inadequate reactive power has led to voltage collapses and has been a major cause of several recent major power outages worldwide”

One source for “supply” or “absolve” reactive power is a generator. For instance, when a generator is operating at certain limits, a generator can increase its dynamic reactive power is produced from equipment that can quickly change the Mvar level independent of the voltage level. Thus, the equipment can increase its reactive power production level when voltage drops and prevent a voltage collapse.
By convention, reactive power, like real power, is positive when it is “supplied” and negative when it is “consumed.” Consuming reactive power lowers voltage magnitudes, while supplying reactive power increases voltage magnitudes.

We should keep in mind that a production or consumption of reactive power by a generator only could be done by reducing its production of real power. As a result, producing additional reactive power results in reduced revenues associated with reduced real-power production. This may be seen in the generator capability curve similar to the one enclose below.

http://cuky2000.250free.com/Gen_Reactive_Capability_Limits_2.jpg
tinfoil (Electrical)
11 Nov 05 12:51

Quote:

The problem with var (yes; var, not VAR or VAr)...

Nitpick: The unit of Volt-Amp-Reactive is derived from the base units of Volts and Amps, both of which are named after actual people, so SI states that they should be capitalized when abbreviated, whereas 'reactive' is not based on a proper name and should not be capitalized.  

Therefore, it should be formally abbreviated to VAr, but VAR (and flavours like kVAR and MVAR) seem to be the industry norms.
CJCPE (Electrical)
11 Nov 05 17:33

Quote:

The unit of Volt-Amp-Reactive is derived from the base units of Volts and Amps, both of which are named after actual people, so SI states that they should be capitalized when abbreviated...

Nitpick^2: When VAR is used to stand for Volt-Amp-Reactive is that not an acronym rather than an abbreviation? If so, what does SI state re capitalization?
BJC (Electrical)
11 Nov 05 17:47
Here's a good explanation of KW,KVA, KVAR etc.  Remember as you read it, in power tranmission beer mugs are very expensive.

http://www.powerstudies.com/content/resources/Diane%20Power%20Factor.pdf
alehman (Electrical)
11 Nov 05 21:22
Reactive power must be produced by the utility to compensate for reactive power consumed in its own systems and for that consumed by customers. It utilizes T&D capacity and it wastes real power due to the I^2R losses it creates.

Generators, capacitors, synchronous condensers and static VAR compensators are all used to produce reactive power. Capacitors are usually distributed around the network for voltage control and to get VAR production as close to the VAR consumption as possible to minimize I^2R losses and capacity utilization. Switched capacitors are used to help stabilize voltage as reactive loads fluctuate.
Skogsgurra (Electrical)
12 Nov 05 4:11
Nitpick or not. The var (yes, var) is an SI unit on its own and shall be written var. Just like that.

It is not considered to be named after a human being and therefore it isn't spelled with capital letters. That used to be. But isn't any more.

Gunnar Englund
www.gke.org

Skogsgurra (Electrical)
12 Nov 05 4:19
Quote from http://www.iec.ch/zone/si/si_elecmag.htm :

Reactive power, Q
volt ampere, V · A; 1 V · A = 1 kg · m/s3
Usual multiples: MV · A, kV · A

IEC has adopted the name var, var (volt ampere reactive power), for the coherent SI unit volt ampere for reactive power.

Gunnar Englund
www.gke.org

alehman (Electrical)
12 Nov 05 23:20
IEEE concurs it is var and VA.
abeltio (Mechanical)
13 Nov 05 19:06
rmw... still travelling through south america?
regarding your question... in that part of the world several things happen...

1. in Uruguay there are power plants that have the generator working as a "synchronous condenser" i.e. the prime mover (usually a GT) is shutdown and the generator kept synchronized to the grid to absorb var (negative var)
Uruguay has very little industry... and because of the use of fluorescent light the reactive power is very high... therefore they use the synch cond. to improve the power factor (i.e. increase - absorb - var)

2. in Argentina, Chile, Uruguay the grid is not exactly stable... there are noticeable variations in the frequency during the day. usually the frequency falls below 50Hz
So... at night when the load is the lowest the grid dispatcher requests the operators to raise the load (no isochronous driver) so the frequency increases... to compensate for the low frequency during the day... why? so the total number of cycles per day is constant... why? so the electric clocks are always on time when the alarm goes off in the morning... otherwise the error accumulated in one year could be about 10 to 15 min or more.

3. Power factor is a consequence and not a cause... the two independent variables are active and reactive power. some control systems enable the units to keep a constant power factor... but the reason for it is to maintain the generator at the design power factor (usually 0.85 or 0.80) or better (higher).

When the dispatch requests the operator to raise or lower var to maintain the grid's voltage they are using copper of the generator to carry that load... i.e. increased heat generated.
The real danger of very low var (negative) is that you can hit the UEL (under excitation limit) which is normally a trip of the excitation system... otherwise the generator may skip a pole and fall "out of step" with very bad physical consequences for the generator due to the huge electromagnetic torques associated with these events.

HTH

saludos.
a.

rmw (Mechanical)
13 Nov 05 22:25
abeltio,

Thanks for the explanation.  That is what I am trying to come to understand.  What do they physically do to make the copper in the generator carry the load as you put it.

Or, said differently, what they do when they make a power factor shift at the power station to account for power factor changes on the system.

The one thing I do remember about vars from the EE classes I had to take as a ME was that it involved heating up the conductors, but I can't remember why.

And, I do well understand the dangers of under excitation and the potential for 'slipping a pole' and the devastation that can happen there.  I was once a service engineer for a major turbine/generator OEM, but on the mechanical side.  We had to fix the devastation when it (rarely) occurred.

In my thread above, Barrie66 did not understand that I was describing a hypothetical system where the load suddenly shifted from reactive power to resistive power.  Forget about the real world where load begins to drop off when factories shut down at the end of the shift.  Convert all their reactive load from their factory motor driven machinery to resistive load heating their houses and food, and tell me what the power plant does to accommodate the shift in power factor from lagging to unity.

Another question; is part of South America 50 hz?  Somehow, I thought all of our hemisphere was 60 Hz, but I don't know how I got that.

I described in one of the other threads that was discussing this topic about older systems that I used to work on where they maintained their system frequency by having two clocks on the control board, one connected to their system, (usually a municipality or plant with their own power station) and one connected to the grid that they only tied to when their system tripped or was down.

Thanks,

rmw

And, no, I am no longer in South America, but I will be there again within a couple of months.  I'll be fighting the Spanish keyboards again I'm sure.
alehman (Electrical)
13 Nov 05 23:05
In Brazil, the larger cities are 60Hz but I think some areas are 50Hz.
cuky2000 (Electrical)
14 Nov 05 15:12
Not so fast amigo Abeltio.  We should be careful to generalize for one particular isolate case.

We know that is not economical to use the GT to produce vars. Maybe the reason of this operating mode is the abundant hydropower at a particular season.

Uruguay electrical energy generation is highly dependent on the weather conditions. Therefore, the optimal operating mode for generating or consuming var is based on economic decisions and availability of hydro versus thermal power.

Regarding system frequency, Argentina, Chile, Bolivia, Paraguay, Uruguay operate a 50 Hz, while the rest of the continent primarily operate at 60 Hz.
NOTE: Even though larger numbers of consumer  are served from 60Hz, most countries in SA are under the IEC influence. I remember how difficulties was to convince the engineers in Chile to accept the first ring bus and the struggling to bring 500 kV breakers in Argentina.  Most local engineers are not familiar with dead-tank breakers used typically in the ANSI marketplace


abeltio (Mechanical)
14 Nov 05 17:28
cuky2000,
are we sure we r talking about the same thing? in my post i stated that the GT is not run to produce vars... it is used  to bring the generator to synch speed, synch to the grid and then the GT is SHUT DOWN.
btw the synch condenser installation in uruguay is running in synch condenser mode 50% of the time... i know i commissioned the plant.

rmw,
from the operator's point of view it is very simple...
when the excitation control system of the turbine is set to
"nothing" the excitation is DROOPING to the voltage of the grid: i.e. if the voltage drops the excitation increases generating more vars... conversely if the voltage of the line increases the excitation decreases.

in case the excitation control system of the turbine is set to "power factor control": it controls the power factor to a certain setpoint... this is done some times to protect the generator... but does not help the grid.

the other possibility is to set the excitation control to: "var control" then the vars are kept constant to whatever the setpoint is...
this is sometimes required by the grid dispatcher to manage the line voltage.

to carry the load in the copper of the generator given a certain copper section: the oem calculates how many var and mw the generator can carry at a certain cooling condition which could be H2 pressure or cold air temperature depending on the generator.
the result of this calculations are the CAPABILITY CURVES that give the combination of var / mw the generator can carry.

don't keep asking because i do not know much more... i am mechanical too!



saludos.
a.

rmw (Mechanical)
14 Nov 05 21:23
abeltio,

Thanks, you are miles ahead of me, or is that KM's?  It has been several lifetimes since I learned all that stuff, and while most of my work is in the vicinity of the generator's control, I am normally much too busy trying to get mechanical stuff working so that the generator can produce power.

I am trying to brush up on what I learned a long time ago, and haven pretty much taken for granted since then since I didn't need to pay any attention to that part of the plant.

You have been very helpful and I appreciate it.

rmw
Helpful Member!  rcwilson (Electrical)
15 Nov 05 19:38
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 (Electrical)
16 Nov 05 8:07
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 (Mechanical)
16 Nov 05 18:44
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 (Mechanical)
16 Nov 05 19:26
tending to 1.00 (all real power and no reactive power)

saludos.
a.

Helpful Member!  ijl (Electrical)
17 Nov 05 8:30
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 (Electrical)
17 Nov 05 18:22
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 (Electrical)
18 Nov 05 0:41
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
www.gke.org

cuky2000 (Electrical)
18 Nov 05 14:58
…….. 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)

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.


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 (Electrical)
18 Nov 05 15:24
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 (Electrical)
18 Nov 05 17:06
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 (Electrical)
18 Nov 05 20:58
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 (Electrical)
19 Nov 05 13:10
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 (Electrical)
19 Nov 05 22:06
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.

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