A 4160 VAC pump 1000HP motor at medium voltage Switchgear 3

[Melco001]

I have a full voltage starting induction pump motor 1000HP running on a 4160 VAC bus, can someone tell me how long it can last without power while a feeder bus transfer is done? to continue operating without having to carry out a new boot process. This is to set the motor undervoltage trip protection function.
I appreciate any help on this.

 

[dpc]

This will depend on the motor speed and total inertia, as well as the settings on your UV protection. In general, not very long. Try searching for Fast Bus Transfer or similar. This is a common issue when power plant auxiliary loads must transfer from one source to another. Also see IEEE/ANSI Standard C50.41-2000.

Dave

 

[RRaghunath]

HV motor UV protection is typically set at 75% V & 2 or 3s. This is not exactly protection for motors, rather it is a system requirement.
This is more to enable smooth transfer of power or restoration of power, without the large inrush current of all connected motors combined depressing the bus voltage to unacceptable levels.
Fast or high speed transfer schemes are designed to allow power transfer without need to trip the connected motors.
In case of normal or slow bus transfer, the motors are tripped before transfer is initiated.

 

[LionelHutz]

Stop the pump and watch how it coasts down. From my observations, pumps lose most of their speed within a second or so of stopping. They don't have much inertia and are typically running at or near full load which both work to cause a quick deceleration.

 

[panter]

Melco ,
it depends of technological process and how important is the operation of the motor i.e. the pump, if voltage drops on High voltage sistem . In this way, you determine which of the pumps must work for example at 80% of the voltage and which should be turned off and save proces through some short-term voltage drops .

 

[che12345]

Dear Mr. Melco001
I am looking at it in a different angle, for your consideration.
1. At the instant when the power failed with voltage and current dropped to 0V 0A; the motor generates a [back emf] with [decreasing voltage and frequency], independent of the load/the costing speed. This back emf decrease to 0V after say 3 cycles. If the power supply is switched back randomly, with source voltage and frequency, which if are NOT in phase with the motor generated back emf; an extremely [high current >LRA] and extremely high [mechanically torque] to pull the rotor back to the original running speed. These are detrimental to the breaker, contactor, coupling and motor.
2. When the power [voltage dropped] is <15% , no action need to be taken.
3. When the power [voltage dropped] is >15% but NOT zero, the motor may run into over-load (over-current). Tripping off the motor within say 3s should be fine.
Che Kuan Yau (Singapore)

 

[Gr8blu]

Melco001: I have a simplified Fast-Bus transfer calculation that works from first principles (motor power and speed, line voltage, slip speed, and the connected inertias of motor and load. For a 1000 HP, 4160 V machine with 1% slip, connected to a load inertia that is 10x the motor inertia, we get the following answers (which are speed-dependent).

In all cases below, please note that line voltage drops below 0.75 PU (below 3120 V) in 0.167 seconds.
Nameplate speed 1800 rpm: reclose in the 0 to 1.35 seconds range
Nameplate speed 1200 rpm: reclose in the 0 to 0.90 seconds range
Nameplate speed 900 rpm: reclose in the 0 to 0.65 seconds range

If you miss the reclose timing, wait until rotor comes to a complete stop before reclosing.

Converting energy to motion for more than half a century

 

[electricpete]

You can find a ton of info here
EPRI Report EL-4286-V2: Improved Motors for Utility Applications, Volume 2: Bus Transfer Studies
Those EPRI reports used to cost a ton (to nonmembers) but most of them (including that one linked above) are now available for free (the cost of providing your email address)

I'll throw in what I remember about it. As I recall you have two options

1. Fast transfer - Short enough denergized period that the the residual voltage doesn't drift too far out of phase too far during the deenergized period. This assumes your new source of power is in phase with your old source of power. Some relaying schemes may monitor the phase difference accross the incoming breaker before closing.

2. Slow transfer - Long enough deenergized period that the residual voltage drifts down far enough to prevent damaging transients regardless of phase angle difference upon connecting the new source. As I recall they're looking for max 133% voltage so residual voltage has to decay to 33% or less. Residual voltage is proportional to product of flux and speed, both of which decay (at different time constants). You can estimate the flux decay from the equivalent circuit, I think it's Tau_open_circuit = (Lm+L2) / (R2). After about 1.5 time constants you will be down to 33% or lower


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(2B)+(2B)' ?

 

[LionelHutz]

Pumps don't have 10X the motor inertia. They usually have under 1X the motor inertia. Unless you're going to add a big flywheel to it, you won't even get those short times given.

 

[Gr8blu]

Lionel: True - the pump assembly (shaft plus impeller) itself may not be a high-inertia load. However, if the machine has tripped chances are pretty good the system remains fully charged with fluid - which WOULD exhibit as an inertia much higher than the machine's. If the user waits for valves to close and the system to "empty" (thereby reducing the load inertia to a fraction of the machine) ... the fast-bus transfer window has already closed.

Converting energy to motion for more than half a century

 

[LionelHutz]

Sure, but go back to my first comment again. Stop it and time the deceleration then you'll know how quickly you need to fast transfer.