Sequence of Operation for 200 HP motor 1

[gpape]

I have a 200 HP Reliance motor, Model No. P44G5261, on a vertical sump pump that maintenance wants to operate as an on/off pump in case of heavy rain. Based on operation sequence, can anyone offer information on the pros and cons of going on/off versus VFD? Sump liquid is waste water with around 1.2 cP viscosity.

 

[itsmoked]

Frequent on/off on that size motor is not good.

Any generators involved?

VFD could allow steady state matching of pump speed to in-flow.

May also allow greater surge capacity for the system. If large sudden in-flows are possible and the standard "wait until almost full" before starting pump,(to minimize cycling), is being used, you would have less available capacity compared to a VFD keeping the sump almost empty.

 

[jraef]

A VFD is going to be an expensive option for an occasional use 200HP pump, and it will not pay for itself with maintenance savings alone. The steady-state speed matching to inflow is great if it is a sewage lift station or something where inflow is always happening, but for a storm drain it is probably adding a complexity risk that is unwarranted for the task at hand IMHO.

Consider a solid state soft starter. 1/5 to 1/10th the price of a VFD and it would eliminate the concerns over mechanical damage from frequent starting and stopping. Also, better soft starters now come with starts/hour lockout timers that would reduce the stress even further.

"Our virtues and our failings are inseparable, like force and matter. When they separate, man is no more."
Nikola Tesla

 

[davidbeach]

Also, better soft starters now come with starts/hour lockout timers that would reduce the stress even further.

Somehow, for this application, I think it might be better to allow the motor to run to destruction than to limit starts and allow flooding.

Probably better would be multiple pumps, set up such that higher liquid levels run more pumps and the lead pump alternates.

 

[rbulsara]

it also depends on what is 'frequent'?

Rule of thumb is not more than four full voltage (DOL) starts per hour, spaced not less than 15 minutes apart.

With a soft start (not necessarily VFD) you can have a few more starts than four. If you can have the soft start that you do not exceed 100% of the FLA, you can have as many starts as you like. YOu need to consult the motor mfr for any starting current more than 100%.

FYI, a DOL start will have 600%-700% of the rated full load current (FLA).

A typical soft start can limit the starting current to 150 to 200% of the FLA.

 

[rbulsara]

Does this motor have RTD's for winding temperature sensing?

 

[Marke]

Hello rbulsara

The energy disspated in the motor during starting is the same for DOL and for soft start, so if there is a thermal limit of fours starts per hour on DOL, the same limit would apply on soft start. There is however, considerable advantage in transient reduction, both electrical and mechanical and this will allow a much increased number of starts before the accumulated damage requires repair. This will enable a greater number of starts per hour provided that the thermal limitation is not exceeded.
"A typical soft start can limit the starting current to 150 to 200% of the FLA." Although a soft starter may be able to limit the start current to less than 200% current, the start torque developed by a motor at this current will not start a pump. Submersible pumps with long thin rotors will often start at 250 - 300%, but surface pumps can require up to 450% current to start them. This is a function of the motor starting characteristics and the pump torque curve.

Extended or frequent starting results in rotor damage rather than stator damage. RTDs do not help in this regard. A good thermal model designed and calibrated to match the motor characteristics is one of the best means of protection of the rotor.

Mark Empson

http://www.lmphotonics.com

 

[jraef]

Somehow, for this application, I think it might be better to allow the motor to run to destruction than to limit starts and allow flooding.

Probably better would be multiple pumps, set up such that higher liquid levels run more pumps and the lead pump alternates.

I somewhat agree with you davidbeach, but "run to destruction" is a term usually reserved for Fire Pumps, not storm water unless there is serious threat of loss of life involved. We don't really know anything like that about his application, and that is something that should have been addressed at the system design stage. gpape's post was in reference to what he already has to work with, a single 200HP pump. In that case, a soft starter is the optimum way to go IMHO.

Marke is correct, a soft starter will not extend the thermal capacity of the motor, nor will it decrease it either. It is exactly the same anount of energy as Across-the-Line, just stretched out over a longer period. There is some minor benefit to the cooling system of the motor having more time to dissipate it, but that is neglegible.

Side note: Nor will any soft starter be capable of starting a motor at 100% FLA. The physics don't work out. Some manufacturers make users to believe it will by allowing their adjustment range to go that low, but it s a false hope. I have never seen one start a motor with less than 150% current with any load connected, and that was a very light load.

"Our virtues and our failings are inseparable, like force and matter. When they separate, man is no more."
Nikola Tesla

 

[rbulsara]

Marke:

Thanks for expert input, specially for the heat dissipation. As for starting torque goes, most pump would start at reduced voltage, in fact they are the best candidates for the reduced voltage starter.

gpape:

So there is no advantege of thermal dissipation, I would say stay with what you have and need to stay with about 4 starts per hour, not less than 15 min apart.

jraef:
I agree, 100% staring current is stretching it, but some (chiller) vendors have professed that. 150% to 200% is a reasonable number for most pumps


Marke:
These questions are not to challage you, but for my benefit:

What contributes more to the heat build up? The fact that motor is at standstill at the start? The current have no effect?

Would you also say a squirrel cage rotor is more susceptible to heat, than the stator?

Thanks

 

[itsmoked]

Whoa! whoa! Why would 200% start NOT dump less energy into a motor than a 800% DOL?? We are talking about large I^2 differences here.... Also what about the "physics" F=MA? If we accelerate our car to 60MPH in 3 seconds we use a heck of a lot more fuel than if we do it over 10 seconds.

 

[Marke]

hello rbulsara

During start, there is a lot of slip energy dissipated in the motor. This slip energy is dissipated in the rotor. In addition, there is energy dissipated in the stator due to the I2R losses, but it is the rotor energy that does the damage. The slip energy dissipated in the rotor is equal to the full speed kinetic energy of the driven load and motor. Motors are rated in the maximum rotor energy either by the maximum locked rotor time or the maximum load inertia.
When you apply a soft starter, you reduce the instantaneous start current, but you prolong the start time. The area under the curve should be the same. If you try to start at too low a current, you will dissipate more heat in the motor due to the pronlonged time that the motor is doing work at a higher slip.

The number of starts is realy a thermal issue. If you have a motor rated at 4 starts per hour and a rated maximum start time of 10 seconds, you can start considerabley mor frequently on a pump because the start tiem required for the pump is significantly less. If you take the maximum rated start time of the rotor multiplied by the rated number of starts, that gives you a maximum starting time per hour. Provided that the maximum starting time is less than the rated maximum starting time and the total time is less than the product of the maximum start time and frequency, you can safely increase the starts per hour on a soft starter. For example, 10 seconds maximum start time and 4 starts per hour, is the thermal equivilent of 8 5 second starts per hour and a pump DOL wil start in less than 5 seconds.

During start, the rotor is subjected to mor heat and is most likely to be damaged due to the high slip energy. During run, the slip energy is low and the stator is the thermal weak point.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[itsmoked]

I copy that Marke. Thx.

 

[rbulsara]

Good info Mark.

 

[electricpete]

itsmoked - The result that the rotor heating during start of unloaded motor is equal to kinetic energy (regardless of voltage) is not something intuitive and not easily shown.

It is proved here (section 1n and 2 of the file)

http://www.geocities.com/pschimpf/RotorHeatingSymbols.htm

It is based on the equivalent circuit which gives us a means to determine current as a function of slip and torque as a function of slip. We know heating is integral I^2*r dt.. There is a tricky change of variables necessary to convert from integration over time to integration over slip. By the time you are done with the math there is not much intuition left, but the result is there.

The remaining sections of the document show how this result is modified when we add load. It turns out that total rotor heating during start of a loaded motor gets higher as we lower the voltage.


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[electricpete]

A more compact and understandable version of the above:

RH= Int{I^2R2 dt} =
= Int{I^2R2 / (ds/dt) ds}
What is ds/dt
dN/dt = T/J (f=ma type relationship)
N =(1-s)Nsync
dN/dt =-Nsync ds/dt = T/J
ds/dt=-T/J/Nsync
ds/dt = -1/(JNsync)*T = 1/J*[Pelec]/[N]
= -1/(JNsync)*[I^2R2*(1-s)/s]/[Nsync*(1-s)]
= -1/(JNsync)*[I^2R2/s]/[Nsync]
-I^2R2/[sJNsync^2] units Power / energy = time^-1

Substitute ds/dt into RH
RH = Int{-I^2R2 / (I^2R2/sJNsync^2) ds}
RH = Int{-sJ Nsync ds; s=0..sFinal}
= [-0.5JNsync^2*s^2@sFinal-s0]
sNsync=Nsync-N
= [-0.5J(Nsync-N)^2@sFinal-s0]
= [-0.5J(sFinalNsync)^2] - [-0.5J(s0*Nsync)^2]
= [-0.5J(s0*Nsync)^2] - [-0.5J(sFinalNsync)^2]
= 1/2*J*(N_sync*s_initial)^2 - 1/2*J*(N_sync*s_final)^2

Note – N and Nsync represent radian speed.

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[electricpete]

The analysis directly above applies to starting of UNLOADED motor.

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[electricpete]

Just to finish the problem if not obvious:
RH = 1/2*J*(N_sync*s_initial)^2 - 1/2*J*(N_sync*s_final)^2

let s_initial=1, s_final=0

RH = 1/2*J*(N_sync)^2 = KE_final


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[itsmoked]

Ugh.... Yes it does lose a bit of the intuitive. It makes sense that the rotor KE has got to be paid for each start. It's the strange twist of rotorheat=f(s)that sticks the sharp end, into the eye of intuiton.

The link helped.

Thanks electricpete.

 

[electricpete]

"It makes sense that the rotor KE has got to be paid for each start"

Note this does not follow from conservation of energy. Assuming stator winding resistance was 0, the amount of energy we would have to put into the motor from the power system would be 2*KE. (1*KE would be converted rotor heat and 1*KE would be converted to rotor KE)

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[electricpete]

In the example directly above, the amount of energy converted directly to ke can be determined by conservation of energy. The amount of energy dissipated as heat cannot be determined by conservation of energy.



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