New power plant engineer here with a few questions. Any help is greatly
appreciated as I have never had any type of mentor to help explain
aspects of plant operation.
1) Firstly - the plant I work at has (2) 35 MW brush generators with an
aeroderivative turbine on each generator. The operators typically run
these generators at 0.9 to 0.95 lagging power factor. I have been
looking for the theory behind how over and under exciting the generator
yields a change in that specific units power factor but have not come
up with anything. For example, by reducing the excitation voltage on a
generator - why does the power factor want to transition from lagging
over to leading?
2) Why does the plant typically run in lagging pf mode? It would seem
logical to me that a lagging pf is induced to overcome the capacitive
line losses associated with transmission. Is this correct? The
transmission handler always calls on us for VAR support (which we do by
adjusting the substation auto transformer).
Thanks in advance.
You don't mention whether the plant is connected to a grid. First consider a
single machine connected to a large (i.e. "infinite") grid which presents a
constant voltage and negligable impedance. You try to raise the voltage of
the generator but the grid says "no way". the result is that as the internal
generator voltage is increased and the power is held constant, the reactive
output changes. The machine generates more reactive and for a generator,
this implies going more lagging. The grid obliges by reducing reactive
generation elsewhere as the load reactive is unchanged. Draw phasor
diagrams- that is the easiest, assuming Vt and P are constant. and the
internal generator voltage varies (are you familiar with V curves for a
synchronous motor?- similar to this).
Overexcited generator goes lagging and underexcited goes leading. I can give
the equations for this. Let me know if you need them or any examples. I do
have a DOS based Turbo Basic program which allows jockeying between two
machines feeding a common load.
Now if your two machines are the only machines on the system, they must
supply the power and reactive of the load. If the load is 10MW at 0.8 pf
lag, the load requires 6MVAR reactive (note that loads generally absorb
reactive- hence the need to generate it). This can be split between the
machines and in your situation it is good to split the power (8MW) and the
reactive equally between the machines. If you try to reduce the voltage on
one machine, the terminal voltage drops a bit (because the "grid" is finite)
but the current distribution changes. The power is determined by the
governor settings so the main thing that changes is the reactive. The result
is that one machine can take 1MVAR (at 4MW and o.97 pf lag) while the other
picks up the slack and carries 5MVAR , 4MW at 0.62 pf lag. Go further and
one could get one machine at -1MVAR (leading) while the other is at
7MVAR -supplying the load but also uselessly sending reactive to the other
machine. Go too far this way and stability problems can occur due to
relative phase shifts between machines.
Transmission line reactive can be lagging or leading depending on line
length and load. Your boosting of voltage at your system connection is
sending more reactive into the system to compensate for reactive losses. You
have to supply this. Adjusting the substation autotransformer can try to
boost voltage and so increase reactive flow. Your generators will have to
supply this reactive.
What is your background as knowing this can help considerably in attempts to
Don Kelly email@example.com
remove the X to answer
Thanks for the great response. To answer a few unclear portions of my
original post - yes the machines are in parallel and connected to the
grid. I am still very wet behind the ears regarding power plant
engineering (I'm an EE with less than 5 years experience). I've not
seen any V curves but that is most likely something I can google for
the theory behind it. But yes - I would love the equations for the
over/under excited generators yielding lagging/leading pf. An example
probably wouldn't hurt either. Thanks!
I follow your logic regarding the loading and reactive power sharing.
Regarding the transmission lines - the overall length is about 7
miles. Fundamentally, it would make sense that the transmission lines
act as big capacitors (with the air between the lines being the
dielectric) hence the need to transmit at a lagging power factor to
compensate. It is still not clear to me why an inductive power factor
is being injected into a grid which likely (cant say for sure) has
massive inductive loads on it.
I guess I am also getting stuck with the concept of voltage increasing
yielding more reactive power. Is this voltage increase basically
elongating the hypotenuse of the power triangle thus increasing the pf?
Your response was great and I appreciate you taking the time to help
me. Thank you.
I will pump this book:
(Amazon.com product link shortened)
If you're going to work in the generation field, I would recommend you learn
synchronous machines real well. I'm sure you also have lots of motors in
the plant, so understanding those well also would be good.
Don did a good explanation, but one other thing you should understand is the
load capability curve. This is a curve of the generators VAR and W output
capability. Obviously the turbine sets a limit on the power output. But
when the load is reduced, the VAR output can go up because the limit near
full load is usually the stator winding current. With lower power out, the
VAR output can increase without overloading the stator. But two more limits
occur. The leading VAR is limited by how much current can be pumped through
the rotor winding - either the winding limit, or the limit of the exciter
supply. The lagging VAR limit is from two things - one is stability, the
other is rotor iron heating. With lower rotor field, the generator coupling
to the grid is reduced. It can fall out of sync easier. The rotor iron
also heats at low field current because the 60Hz current now starts going
through the edge of the rotor, heating it up.
You may be confusing 'lagging pf' for a generator and 'lagging pf' for a
load. As Don mentioned, when a load (such as an induction motor) has a
lagging pf, the current is lagging behind the voltage waveform, literally.
But in a generator, the 'real' component of current is the opposite of
voltage waveform (power is flowing *out* of the generator, where in a motor
power is flowing *in*). So current that is lagging the voltage waveform in
a generator has reactive power flowing *out* of it, while in a motor with
current lagging the voltage waveform, the reactive power is flowing *into*
it. (where, by long convention, lagging reactive power is called 'positive'
and leading reactive power is called 'negative').
In an isolated system, the total load pf determines the amount of reactive
power being consumed, and the generator(s) must supply it. If it does not,
then voltage on the terminals will change, changing the generator current
until the reactive power flowing out of the generator does match the load.
In an 'infinite bus' situation, the voltage on the bus is fixed. So the
infinite bus itself can supply the loads with whatever reactive power they
need. The infinite bus can also absorb any amount of reactive power the
generator wants to generate. So the reactive power flow out of the
generator is determined solely by the excitation applied, the machine's
characturistics (synchronous impedance for steady-state stuff) and the real
power being carried by the machine.
If you draw a single phase circuit with an 'ideal' voltage source, an
internal inductance and then the generator terminals, finally connected to
the 'infinite bus' (another ideal voltage source that is fixed voltage and
frequency), you'll probably get the idea. The difference between 'ideal'
voltage source inside the generator and the infinite bus voltage is simply
the voltage drop across the internal inductance. The current through the
machine is a function of this voltage difference and the phase difference
between the 'ideal' source and the infinite bus. Changing excitation is
equivalent to varying the 'ideal' voltage source inside the generator..
Seven miles for the transmission line isn't at all unusual. Our plants have
lines that run more than 20 miles to the next station, and even that is not
out of the ordinary. Our lines are slightly capacitive when unloaded, but
the substation load at the other end is dominant. But the regional operator
gets most of the voltage support he needs (MVAR generation) from other
equipment (capacitor banks, peaking units, etc [we're a base load unit]) and
we only carry enough MVAR to stay lagging by about 0.98. This is for
stability reasions mostly.
Yet another factor in deciding just how much reactive load to carry is
system stability. If the voltage at your end of the transmission line is
too low, then the power you're generating won't actually flow down the line.
So your generator doesn't apply as much counter-torque to the prime-mover
and the shaft speed picks up. You 'slip poles', drop out of synchronism
with the grid and "bad things" happen. Many generator unit voltage
regulators have "URAL" (Under Reactive Ampere Limit) circuits that cut in
automatically to prevent excitation from being reduced too far. Others have
some procedural curve the operators must follow, something like MW on the
horizontal axis and MVAR on the vertical, with shaded areas saying "Always
operate above this line..."
Such limits also consider the possibility of a line tripping (if feeding
multiple lines). When one line trips, the generating end will increase its
torque angle (higher impedance in the transmission line). If this gets too
close to 90 degrees, "bad things" happen (generator could pull out of sync
and start slipping). Some installations even consider the effects of
whether a line has 'fast reclosers' or not. If the line trips but recloses
quickly enough, the generator end will not have advanced its torque angle
very far before reclosure. This is common if the lines are subject to
lightning strikes. Lightning causes flashover, the line trips, the
generating unit starts to increase its torque angle, the line recloses,
generator returns back to original torque angle (well, there are a few
'oscillations' before it damps out).
This is the case for us, if certain transmission line protective equipment
is out of service, we must use a different curve to determine minimum MVAR
(have to carry more in case of no 'fast reclosers').
Welcome to the world of *POWER* electrical engineering :-)
P.S. Another good book on this stuff is "Electric Power Transmission
Systems", J.Robert Eaton / Edwin Cohen. Mines getting pretty old, but still
a lot of good info.
I appreciate alot of the feedback but some of the information appears
to be 2 steps ahead of where I currently am regarding theory digestion.
Plus, some of the nomenclature (and theory) you folks are using is
passing right over my head. Advanced rotating machinery and the
associated equipment typically used in power generation are not areas
covered in college. It also doesnt help that I do not have a mentor or
any senior engineering type person to keep me on the straight and
narrow. I am going to have to teach this info to myself.
I will attempt to get ahold of the recommended reading material since
seeing diagrams supported by mathematical equations have always
solidified things more in my brain. Hope it works out.
Thanks for all the help.
Well, start with AC circuit analysis of the simple scheme I mentioned. A
variable AC source to represent the 'internal voltage' of the generator, a
series inductance that represents the machine's 'synchronous impedance'
(don't worry about the name, just treat it as an inductor). Zero resistance
in the circuit (it's several orders of magnitude less than the inductance
and can be ignored with little error). And a fixed AC source (the 'infinite
Another thing to look up would be 'four-terminal networks'. Studying these
can give you insight into how real and reactive power flow between an AC
source and load over a line.
Yes, the phasor diagrams can help a lot.
Also, if you're not familiar with P.U. (per unit) calculations/terms, take a
moment with those. They're not real hard or anything, but a lot of graphs
and nomographs are based on P.U. terms (some xfmr and gen ratings such as
synch. impedance are also listed on a P.U. basis).
Good Luck, come back anytime if you get 'stuck'. Myself, or others (Don and
Charles are two that come to mind that are quite well versed in this area)
will be glad to help you get through any 'rough spots'.
.3 degrees ahead of the load voltage.
The complex power generated is 0.8 +j 0.6 +(I^2) (0+j0.1) =0.8+j0.7
P=0.8 and Q = 0.7 so the pf is 0.75 lag
Load at voltage Vl and generator at voltage Vg. Assume that the load voltage
is constant at all times and we can take it as reference (angle 0) All V and
I below are phasor quantities
Current Il = Vg-Vl/Zline =|Il| @ angle theta
To simplify take Z=jX
Sl=Pl+jQl (load complex power =(Vl)Il* (* implying conjugate)
generator Vg =Vl +jX(Il) =|Vg| @ angle delta
Sg =VgIl* =( Vg)Il* =Vg(Vg-Vl)/jX
do the manipulations and you will end up with
Sg =((|Vg||Vl|)/X )sin delta +j[ ((|Vg|^2) /X) -((|Vg|^2)/X) cos delta]
Your machines are not affecting the grid so frequency and voltage at the
load are constant
try to increase voltage and there is a local voltage increase but the grid
voltage is unchanged so that Vg in the above changes magnitude and delta
changes to compensate. This affects the reactive.
Try to change speed and the generator power increases so that there is a
change in delta (and some change in Vg resulting in a change in reactive.
If you have two machines feeding a load over separate lines and they are the
only generation sources, then you have two racoons in a bag.
I can send you a program for the latter situation but what limits are on
files that you can get?
Don Kelly firstname.lastname@example.org
remove the X to answer
For me, when you say 'brush' generators I think of Charles F. Brush. Do you
synchronous generator? (With slip rings and carbon brushes.)
What is an aeroderivative turbine?
Consider a generator (reactive) capability curve - a plot of watts vs vars
capability of a synchronous machine based on allowed temperature. Generally it
shows 2 quadrants
out of 4. The horizontal axis starts at 0 and represents watts out of the
vertical axis is along the left side and represents vars.
The power system considers inductive reactance as being positive.* (The * is a
In power system speak, inductors consume vars and capacitors supply (make) vars.
machine can do either. A more complete but slightly confusing statement is that
consume vars supplied by capacitors and capacitors consume vars supplied by
vars represent energy stored in magnetic or electric fields, vars are not
but shuttle back and forth between reactive elements.
Expand the capability curve to 4 quadrants. To the right of the var axis is
(watts 'made') and to the left is negative watts (watts 'consumed'). In a
similar fashion above
the watt axis is positive vars (vars 'made') and below is negative vars (vars
_Generators take on the pf of the load they serve._ <- Important
The generator controls the voltage and frequency but the load controls
everything else. <- Also
important (Note that I said 'The' generator which is all of the generators in a
together as one.)
If the _load_ has a lagging pf, the generator operates in the lag. If the _load_
has a leading
pf, the generator operates in the lead.
If you examine the math, you'll find that current will always be within -90d
(lagging) to +90d
(leading) of the voltage at a _load_ made up of real components like resistors
watts), inductors and capacitors.* The load, not the generator, determines
whether the current
is 'leading' or 'lagging' the voltage.
There are 4 possible conditions at any given node*** in a power system.
1. If watts and vars are flowing into a node, then that node represents an
(inductance and resistance); it consumes watts and vars and the node is a
lagging pf load. This
is the lower left quadrant (an induction motor for example). It is served by a
operating in the lag.
2. If watts are into and vars are out of a node, then that node represents a
capacitive load; it
consumes watts and supplies vars and the node is a leading pf load. This is the
quadrant (an overexcited synchronous motor or lightly loaded transmission
system**). It is
served by a generator operating in the lead.
3. If watts and vars are flowing out of a node, then that node represents a
source (an overexcited generator); it supplies watts and inductive vars and is a
source. This is the upper right quadrant ('normal' generator operation). It is a
serving a lagging pf load.
4. If watts are out and vars are in to a node, then that node represents an
source (an underexcited generator); it supplies watts and uses inductive vars
and is a leading
pf source (normal generator operation on a lightly loaded transmission system**
or a normal
induction generator). It is a generator serving a leading pf load.
What a node appears to be depends on which 'side' it is viewed from. Conditions
1 and 3 are
opposite views of the same node as are conditions 2 and 4. Conditions 1 and 3
operation at lagging pf while conditions 2 and 4 represent operation at leading
pf. What is
'in' to the node on one side is 'out' of the node on the other. If watts or vars
into a node are
considered positive then watts or vars out of the same node are considered
negative. When watts
are summed and vars are summed at a node the sums must be 0 watts and 0 vars.
What comes in,
must go out. If not, something is going to melt or at least glow or move or arc.
Of course, if
the node represents a utilization (or generating) device, this may be the
desired effect as the
energy leaves (or enters) the node in some form other than electrical.
So, the upper right and lower left quadrants are opposite sides of a lagging pf
node and the
upper left and lower right quadrants are opposite sides of a leading pf node.
The generator has to provide what the load demands and takes the name lagging
loads in the lower left quadrant where the current lags the voltage by up to
90d. A generator
operating in the lead takes the name from loads in the upper left quadrant where
leads the voltage by up to 90d.
I prefer to use overexcited and underexcited for synchronous machinery. If a
overexcited, the field current is higher than required for unity pf operation
and it raises the
system voltage. It is the same whether the machine is a motor or generator (as
opposed to the
terms leading and lagging). The volt-ampere vector will be above the watt axis
and to the right
for a generator (supplying watts) and to the left for a motor (using watts). A
in this region supplies vars to an inductive reactance and appears to be a
capacitor in parallel
with a watt source or sink.
If the machine is underexcited, the field current will be less than required for
operation and it lowers the system voltage. The va vector will be below the watt
axis. Again it
will be to the right if the machine is supplying watts (generating) and to the
left if the
machine is using watts (motoring). A machine operating in this region consumes
vars from a
capacitive reactance (or an overexcited generator) and appears to be an inductor
with a watt source or sink.
I suppose that a problem arises with synchronous motor capability curves because
I suspect that
synchronous motor capability curves are drawn with watts consumed to the right
above the watt line. From a power system standpoint, that capability curve
watts increasing to the right and is a mirror image across the vertical axis of
what is really
happening (from a power system engineer's view).
*If the voltage angle is taken as reference (0), the current angle will be
the minus (-) sign before arctan) where x is positive for inductance and
capacitance. Positive x (inductive reactance, jwl) gives a negative current
while negative x (capacitive reactance, 1/jwc) gives a positive current angle
resistance (r) is assumed positive but the math works if watt sources are
negative r. (the 'w' here is omega, the radian frequency, i.e., 2*pi*Hz) (the
comes from -j=1/j where j=sqrt(-1))
** It is possible to load a line in such a way that the vars flow into or out of
both ends and
not 'thru' the line. This is because a line's lagging var requirement
(inductance) is determined
by current (which can vary a lot) while the leading vars supplied by a line
determined by voltage (which varies relatively little). At low currents (low
can dominate while at high loads inductance can dominate. There is also a
where lagging vars are canceled by leading vars so that 0 vars are required by
one end of the
line (maybe both ends under the right conditions). You can make line vars do
almost anything if
you can swing the load far enough.
*** A node is a place where a set of conditions (watts, vars, volts, and amps)
(calculated, measured?). It is usually where elements are connected together but
can be anywhere.
If it were up to me, I'd remove all the pf meters at generating stations and
make the operators
pay attention to the var load (and capability). That would have prevented the
Eastlake-5 in Aug. of 2003.
I assumed that meant slip rings and brushes vs. a brushless excitation
A turbine derived from a airplane engine. Usually the power section
(compressor, combustor & compressor drive turbine) is nearly identical to a
turbojet power section. The turbine that spins the generator is downstream
and is customized for shaft power delivery vs. the original design that was
designed for aircraft.
Yep, absolutely. The plant I worked at years ago had frequency, MW, MVAR,
stator kA, stator kV, and in the substation kV and A for each line. I don't
remember a PF meter. Commands from dispatch were in MW and MVAR, with
voltage limit bands.
Again, I apologize for the confusion. The generator has a brushless,
perm magnet exciter utilizing a single shaft. Brush is the
manufacturer. No slip ring. No burshes.
A gas combustion turbine of the construction typically used in aricraft
engines. Very simple in desgin when compared to the massive frame
engines. Not quite as robust an engine as the frame machines.
Thanks again all. I will come back and respond when I get a chance to
review all this wonderful information. I just happen to glance and see
the above questions - which may or may not be causing problems.
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