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Large Induction motor - Capacitor start vs Reactor start 4

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RRaghunath

Electrical
Aug 19, 2002
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We have a large motor of 15MW/11kV.

When should we prefer capacitor start over Reactor start and how do these compare if we consider that running power factor is not an issue. Any help.

Thanks in anticipation.
 
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Suggestion: The capacitor in conjunction with the motor may monotonously be increasing voltage during the motor start.
The reactor in conjunction with the motor may monotonously be increasing current during the motor start.
 
15 MW/ 11 kV motor does not specify anything. When asking for suggestion minimum details are to be given. Such queries deserves to be Red Flagged.
 
Suggestion: Visit
for: Monotonic function
for: Completely monotonic function
for: Monotone decreasing
for: Monotone increasing
for: Nondecreasing function
for: Nonincreasing function
Simply (Reference:
Granino A. Korn, Ph.D., Theresa M. Korn, M.S. "Mathematical Handbook for Scientists and Engineers," McGraw-Hill Book Company,
page 94, par. 4.4-8 Monotonic Functions of Bounded Variation.):
A real function f(x) of real variable x is strongly monotonic in (a,b) if f(x) increases as x increases in (a,b) (increasing or positively monotonic function), of if f(x) decreases as x increases in (a,b) (decreasing or negative monotonic function).
f(x) is weakly monotonic in (a,b) if f(x) does not decrease in (a,b) (nondecreasing function), or if f(x) does not increase in (a,b) (nonincreasing function). Analogous definitions apply to monotonic sequences.
 
Suggestion: Visit
for: Monotonous Function
for: Monotonous Function and Monotonous Integral
for: Interesting aspects of monotonous functions
etc. for more info on monotonous as an alternative to the above posting:

electricpete (Electrical) Nov 27, 2003
mo·not·o·nous
1. Sounded or spoken in an unvarying tone.
2. Tediously repetitious or lacking in variety. See Synonyms at boring

that uses somewhat unrelated definition.
 
rraghunath,
Please do not be discouraged by the ongoing chatter expressed above. It is not germain to your question and is less than usefull in my opinion.

gsimon has a point about your question not providing enough information to facilitate an intellegent answer, but you are welcomed to my educated rambling, for whatever it is worth.

Starting capacitors provide a voltage "boost" right at the moment the motor contactor closes (assuming DOL start), which serves to dampen the probable drop that will accompany the motor inrush current. This is a very short term effect, usefull only in applications where the motor is lightly loaded at initial connection, such as centrifugal pumps. The inrush current will still be the same as DOL, just the effect that it has on the power system is affected. This subject has been discussed in this forumn in the past.

Reactor starting will drop the motor terminal voltage and thus reduce both the starting current and the output torque of the motor, which will increase the starting time. This is more usefull if your load will need some time to accellerate to full speed. The drawback is that the reactor is not adjustable and if your load conditions change, you have no flexibility.

Better yet is a solid state starter. Infinitely adjustable to whatever needs to be done. Have you considered this?

By the way, is this still the same application you posted on thread237-74451?

Quando Omni Flunkus Moritati

 
Nicely put your views jraef. I would like to substantiate.
- During dtarting of a coupled motor, the starting characteristiscs of the motor, starting characteristiscs of the load, the GD2 of the motor and GD2 of the load referred to the motor shaft and the motor speed are to be considered.
- The starting characteristiscs of the load cannot be normally changed except closing of damper/inlet valve in a fan/pump.
- The inherent starting characteristics varies as per the type of motor viz. Sq.cage, Slip Ring & synchronous.
- The inherent starting characteristiscs of the motor can be changed ( as applicable) to suit our needs by varying the supply voltage, varying the frequency and by inclusion of rotor resistance.
-The starting time is directly proportional to GD2 & speed and inversly proportional to Motor torque - Load torque.
- For any motor to start and accelerate faster the Motor torque should be substantially higher during the entire starting period.
To identify suitable type of starter, application along with available data should be posted.

 
rraghunath,
Have you considered using an Automatic-Load-Tap-Changer in the step-down transformer? Or, how about a Line-Type, Voltage-Step-Regulator in the motor's feeder. Both methods have the advantage of "closed-transition!" In addition an LTC in the transformerd primary will be far less costly then a reduced voltage starter.

Following are severaL IEEE Papers covering Large Motor Starting on Limited Capacity Systems:

1) Vol IA-14, No. 3, May/Jun 1978, PID 77-19.
"Large Motor Starting on Limited Capacity Transmission Lines", H.W. Lewis and F.A. Woodbury.

2) Vol IA-14, No. 3, May/Jun 1978, PID 77-20.
"Capacitor Starting of Large Motors", J. Stout

3) Vol IA-15, No. 6, Nov/Dec 1979, PID 78-55.
"Electric Utility Flicker Limitations", M.E. Walker.



 
ShortStub:

I think you came up just a bit short on your IEEE references:

From what Journal... or IEEE publication are your references taken from ?

Tnx

j
Code:
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Suggestions to rraghunath (Electrical) Nov 6, 2003 marked ///\\
We have a large motor of 15MW/11kV.

When should we prefer capacitor start over Reactor start

///It depends. One possible way to approach to this preferential choice is to mathematically model it. e.g.

1. Capacitor start:
Simply, the capacitor will be in parallel with the motor.
Set up a current Kirchhoff Law mathematical model.
Is(t)=Ic(t) + Im(t).....Eg1
where
Is(t)=Vs x Ys x sin(wt)....power supply
Ic(t)=C x dv(t)/dt....capacitor branch
Im(t)=(1/Lm)x Integral of [v(t') x dt'] ... simplified motor branch
Solve Eq1 by a suitable transform, perhaps Laplace transform to obtain the voltage function v(t) versus time.

2. Reactor start:
Simply, the reactor will be in series with the motor.
Set up a voltage Kirchhoff Law mathematical model.
Vs(t)=Vr(t) x Vm(t)....Eq2
where
Vs(t)=Is x sin(wt) .... power supply
Vr(t)=Lr x di(t)/dt .... reactor load
Vm(t)=Lm x di(t)/dt .... motor load
Solve Eq2 by a suitable transform, perhaps Laplace transform to obtain the current function i(t) versus time.\\
and how do these compare if we consider that running power factor is not an issue. Any help.

///Visit
for:
Extending DER Transient
Loadability Using
Electrochemical Capacitors\\
 
IA = Transactions on Industrial Applications!

BTW, Rraghunath, you realize that the capacitor method requires that the capacitor be quickly de-energized upon reaching full speed.
 
Suggestion: Visit
for:
Capacitor starting involves switching a large capacitance in parallel with the motor during acceleration. The effect is power factor correction on a grand scale; the capacitance supplies the reactive current requirement, leaving only the current associated with the actual shaft horsepower and losses supplied by the line. A speed switch or current sensor is used to disconnect the capacitance in steps as the motor reaches rated speed. The switching equipment makes this a costly method limited to very large motors or motors that cannot accept any reduction of terminal voltage.
 
Suggestion: Visit
for more recent article abstract on this topic.
Also, visit
for: Load Flow Analysis module that includes Capacitor Optimization
for: monotonously increasing voltage across the R resistor, representing monotonously increasing current across the resistor, in LR circuit. In case of ac source, the increasing function will become an envelope to the ac source.
 
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