How to determine voltage pull down and duration during motor start? 1

[Gwhiz]

I have a motor with the following characteristics: 2400 V, 1500 HP, 320 FLC, 1920 LRC used for a boiler feed pump in a coal fired generation planct.

How can I determine the value that the supply voltage will be pulled down to and how long it will last when the motor starts?

Thanks in advance
Gwhiz

 

[electricpete]

The minimum voltage at terminals of the motor during starting will be a function of Motor LRA, Motor starting power factor, and supply system impedance (complex).

To a fairly good approximation you can look at only the inductive portion of the system impedance, and assume the motor starting current is inductive, and solve algebraically (without complex/vector math). The tough part will be determining your system impedance. It has to take into account all cables and transformers between the point you are calculating voltage and the supply (make sure they are all expressed on the same base). Also to determine worst case, you will have to know what tolerances voltage is controlled at the source (select the minimum), and the tap settings of all intervening transformers.

Starting time is often determined using a simulation. Input to the simulation requires three things: the motor torque-speed curve, the pump torque-speed curve (can be developed from the pump dp-flow curve at full speed using pump laws if density of fluid is known), and the total inertia of rotating parts. Similar to the linear relation a = F/m, the rotational formula is dw/dt = (Telec-Tmech)/J where w is radian rotational speed, Telec is motor torque for that speed, Tmech is pump torque for that speed, and J is rotating inertia also called wk^2. Stat with w(0)=0. Then perfrom time-step simulation something like: w(i)=w(i-1)+deltaT*(Telec-Tmech)/J where Telec and Tmech are read of the Torque-speed curves for the present value of w.

I think there may be a somewhat simpler graphical method described in "Managing Motors" by Nailen, but it still requires you to have both of the Torque-speed curves and inertia available.

 

[electricpete]

The torque that you use for the starting-time study should be based on the reduced voltage that you calculated in the first part. Normally the torque-speed curve is shown at full voltage. For reduced voltage at motor terminals, the torque will reduce in proportion to voltage-squared.

The reduced voltage would last for the majority of the start (doesn't come up much until very near the end), so it is a reasonable and conservative assumption that the starting voltage lasts through the whole start. Otherwise for more precise solution, you could use your simulation to determine current as a fucntion of speed and voltage drop as function of speed, which would input into the Telec term in your simulation.

 

[electricpete]

I have seen a starting-time simulation of the above type where current was calculated vs time and then plotted against the relay curves to verify that the motor would not trip during normal start.

The author of the calculation (a registered PE who signed and sealed the calc) chose to ignore the pump torque (assume it was zero). The error from this assumption is clearly non-conservative, but I don't know how much error is introduced (it's a reasonable assumption of motor torque is much more than pump torque throughout the range). The plant is now operating and motors don't trip during starts.... I chalk it up to luck, not careful analysis.

 

[blitz]

You asked:

I have a motor with the following characteristics: 2400 V, 1500 HP, 320 FLC, 1920 LRC used for a boiler feed pump in a coal fired generation planct.

How can I determine the value that the supply voltage will be pulled down to

******
I assume you have a standard squirrel cage induction motor.
To quickly estimate the voltage dip when starting one needs to know two things:

1. The motor starting kVA
2. The fault level at the MCC bus in kVA

Step 1: Motor starting kVA

A good approximation is motor kVA x LRC/FLC

Motor kVA = (HP x 0.746)kW / (Power factor x Efficiency)
= 1500 x 0.746/0.85 x 0.9 .... assumed values
= 1463 kVA, say 1500 kVA
(note that this figure will always be approximately equal to the HP)

thus for your case motor starting kVA = 1500 x 1920/320
= 9000 kVA
= 9 MVA

Step 2: Bus fault kVA

This is calculated in the normal way, but if it is not known one can estimate it using the bus supply transformer rating. For sake of the example let's assume that the bus is supplied by a 10 MVA transformer with a 6% impedance. The maximum fault level on the secondary terminals of the transformer, assuming infinite source, will then be:

Fault Level(max) = Xmer MVA x 100/%Z
= 10 x 100/6
= 167 MVA

The connection from the xmer to the MCC puts more impedance into the circuit and therefore the actual fault level at the MCC bus will be less tha 167 MVA. To get the worst case voltage dip, one needs to know the lowest available fault level at the bus. This will have to be a personal assessment, but for the purpose of this example, assume:

Fault level (min) = 120 MVA.

Step 3: Calculating voltage dip

Without going into the derivation of the equation, which is just based on simple voltage division across series impedances, the voltage dip at the MCC bus in percent:

MSkVA = Motor start kVA = 9 000 kVA
MinFL = Minimum Fault level = 120 000 kVA

Voltage dip (%) = 100 x MSkVA / (MSkVA + MinFL)
= 100 x 9/(120+9) .. using MVA ipo kVA
= 900/129
= 6.98%, say 7%

Of course this is NOT the voltage dip at the motor terminals. To find that we have to add the effect of the motor connection as well. If the motor supply was designed to have a voltage drop of less than 2.5% when running at full load, a coomon value, the same cable will drop 6 x 2.5% = 15% when the motor is started. Therefore, the voltage dip at the MOTOR terminals will be 7%+15% = 22%. Anything more than 20% should be treated with caution, but is not necessarily fatal.

******************
and how long it will last when the motor starts?

This cannot be calculated with the information you have supplied. It is a function of not only electrical, but also mechanical parameters like motor and load inertias. Watch that motor terminal voltage, the lower it is, the less accelerating torque is available and the longer the motor will take to run up to speed.

Hope this is useful. Cheers

 

[redtrumpet]

Blitz - are you sure about the voltage drop across the cable during starting being directly proportional to the running voltage drop? As electricpete stated, the motor will draw primarily inductive current on start (low power factor), whereas the running motor current has a higher power factor. Since the resistance and inductive reactance of the cable form a complex impedance, the voltage drop is a function of power factor as well as current. The difference in voltage drop for the same current at say 0.3 and 0.8 power factor can be appreciable.

 

[electricpete]

Good point by Redtrumpet. Exact solution would definitely require complex analysis, but blitz' solution is probably fairly close since all impedances are primarily inductive.

As a learning excercize for myself I derived blitz' formula:

MinFL = Fault KVA
V = applied voltage (at infinite bus)
Zsys = System impedance (associated with fault kva)

MSkVA = Motor start kVA = 9 000 kVA
Zmot = Motor starting impedance (associated with MSkva)

MinFL=|V|^2/Zsys (by definition of complex power = kva)
Zsys = |V|^2/MinFL (rearranged above equation to solve for Zsys)

ASSUME THAT Zmot>>Zsys
MsKVA = |V|^2/Zmot (definition of kva)
Zmot~|V|^2/MsKVA (rearranged above equation).

Voltage divider gives the voltage at terminals during start as follows:
Vterminal=V*Zmot/(Zmot+Zsys)
substitute in for Zmot and Zsys....
Vterminal=V*<|V|^2/MsKVA>/(<|V|^2/MsKVA>+<|V|^2/MinFL>)
cancel out |V|^2 and multiply by (MsKVA*MinFL)/(MsKVA*MinFL)
Vterminal=V*<(MsKVA*MinFL)/MsKVA> / (<(MsKVA*MinFL)/MsKVA>+<(MsKVA*MinFL)/MinFL>)
= V*MinFL/(MinFL+MsKVA).
Voltage DROP = V-Vterminal = V*(1-Vterminal) = V * [1 - MinFL/(MinFL+MsKVA)] = V*MsKVA/(MinFL+MsKVA)

Same result as per blitz, with the assumption that Zmot>>Zsys. Also if we are to treat KVA as a scalar (not vector) quantity, then we have to assume that the power factor angle of both impedances are similar, which is reasonable.

One other thing worth pointing out is that fault level at a generating station may be expressed based on generator transient or subtransient reactance (if that quantity is intended to be used for determining fault interrupting requirements for breakers). But you probably want to use fault level and/or impedance values based on generator syncronous reactance, since you're interested in voltage a few seconds into the transient.

 

[electricpete]

In my first message I said that &quot;Starting time is often determined using a simulation.&quot; Actually it is simple integration if we have three pieces of information (and if we choose to neglect the change in Torque due to reduced voltage). Required info is: the motor torque-speed curve, the pump torque-speed curve , and the total inertia of rotating parts.

Similar to the linear relation a = F/m, the rotational formula is dw/dt = (Telec-Tmech)/J where w is radian rotational speed, Telec is motor torque for that speed, Tmech is pump torque for that speed, and J is rotating inertia also called wk^2.


Rearrange as dt = Jdw/(Telec-Tmech), then integrate both sides to give:
t = J*Integral(1/[Telec-Tmech])dw from w=0 to w=2*Pi*F_operating.

Can be easily solved by selecting perhaps 10 frequencies... reading off Telec and Tmech from the curve for each frequency, and performing numerical integration using a spreadsheet.

 

[blitz]

Thanks for the relevant comments. Of course the result of my quick and dirty will not be absolutely correct, for all the reasons stated by the other correspondents, but as a reasonable and quick estimate it probably holds. I normally do this just to see whether a problem can be expected and will then perform a full calculation if required. If it is non-critical with an estimated dip less than 15% and there is no &quot;funny&quot; load type to accelerate, I normally would not bother with a very detailed study. (A &quot;funny&quot; load would be about anything except centrifugal pumps or fans, and even then I take note of what is being pumped; slurry can be a motor killer on start-up!)

On something large and important like a boiler feed pump it would be advisable to run a full calculation or simulation for the various supply conditions. Most of the effort is in assembling the input data, not doing the calculations.