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Staring power of motor
- Thread starter omime
- Start date
- Thread starter
- #3
jbartos
Electrical
- Jan 15, 2000
- 6,440
Suggestion: There is one little thing still needed. Namely, some motor start at 85% of their nameplate rated voltage. These are typical motors. Other motors, special ones, can start with as low as 75% of their rated terminal voltage. This means the starting power can be anywhere from:
Pwatt(t)=sqrt3 x Vstarting(t) x Istarting(t) x PFstarting(t)
where Vstarting1=0.75 x Vrated<= Vstarting(t) <=Vstarting2=Vrated (or higher),
for:
t=>0, and
three phase motor. This Pwatt will vary in time, t. Perhaps, an oscilloscope could be used to obtain info for the Pwatt versus time curve.
Pwatt(t)=sqrt3 x Vstarting(t) x Istarting(t) x PFstarting(t)
where Vstarting1=0.75 x Vrated<= Vstarting(t) <=Vstarting2=Vrated (or higher),
for:
t=>0, and
three phase motor. This Pwatt will vary in time, t. Perhaps, an oscilloscope could be used to obtain info for the Pwatt versus time curve.
An effective (and stimulating!) method to calculate the motor starting parameters (torque, current, PF, power) is to use the equivalent circuit for an asynchronous motor. You will need to know the following about the motor:
- stator resistance and reactance.
- rotor resistance and reactance.
- air gap reacactance and loss resistance
- full load slip value of motor.
- number of motor poles
With the equivalent circuit values in hand (above) and some careful pencil work you can find all that you want to know about the motor through starting (transient) to rated RPM.
More Pratically:
You may also compare the torque-speed curve of the motor to the known mechanical characteristics of the load. IFF the torque available (at motor) is greater than the torque required by the load, the motor can start the load.
Regarding your power available from the line, you may also review the current-speed curve of the motor. This will give you an indication of what the current draw from the line will be through the starting phase.
Lastly, knowing current based on the current-speed curve, and the PF rating at start, you can determine the KVA demand from the line. Most motor manufacturers will provide a PF rating at start (0 speed) and at different mechanical loads (50%, 75%, etc.). Torque-speed and current-speed curves are also typically available from the manufacturer.
Be Careful:
If you use some sort of reduced voltage starting than the manufacturer's torque-speed and current-speed curves go out the window.
- stator resistance and reactance.
- rotor resistance and reactance.
- air gap reacactance and loss resistance
- full load slip value of motor.
- number of motor poles
With the equivalent circuit values in hand (above) and some careful pencil work you can find all that you want to know about the motor through starting (transient) to rated RPM.
More Pratically:
You may also compare the torque-speed curve of the motor to the known mechanical characteristics of the load. IFF the torque available (at motor) is greater than the torque required by the load, the motor can start the load.
Regarding your power available from the line, you may also review the current-speed curve of the motor. This will give you an indication of what the current draw from the line will be through the starting phase.
Lastly, knowing current based on the current-speed curve, and the PF rating at start, you can determine the KVA demand from the line. Most motor manufacturers will provide a PF rating at start (0 speed) and at different mechanical loads (50%, 75%, etc.). Torque-speed and current-speed curves are also typically available from the manufacturer.
Be Careful:
If you use some sort of reduced voltage starting than the manufacturer's torque-speed and current-speed curves go out the window.
- Thread starter
- #6
Omime:
If you have enough back power to run your motor at full load, you should not be concerned with 'starting power'. This is a fraction of full rated power (about 15%). You really need to be looking for reactive power, where will it come from, and will its demand by the motor cause an excessive voltage drop. If you're generating in island mode check to ensure that your gen set can deliver the balance 85% reactive current.
If you have enough back power to run your motor at full load, you should not be concerned with 'starting power'. This is a fraction of full rated power (about 15%). You really need to be looking for reactive power, where will it come from, and will its demand by the motor cause an excessive voltage drop. If you're generating in island mode check to ensure that your gen set can deliver the balance 85% reactive current.
swgrmfg:
I generally agree with you, except for one correction:
"This is a **multiple** of full rated power (about **600%**)"
Starting current is usually about 6-10x full rated current, so starting kVA should equal about 6-10x full-load kVA, no?
I generally agree with you, except for one correction:
"This is a **multiple** of full rated power (about **600%**)"
Starting current is usually about 6-10x full rated current, so starting kVA should equal about 6-10x full-load kVA, no?
Peebee:
It is generally understood that, when one says 'power' without qualifying the term, that usually means active power i.e. the power component of 'power' that produces work of some sort - moving something or heating something. Of course, you're correct about KVA which is composed of active power for moving, and reactive power for exciting (magnetizing). In the case of a motor the starting KVA, also called 'apparent power' is 6 times that of rated KVA.
It is generally understood that, when one says 'power' without qualifying the term, that usually means active power i.e. the power component of 'power' that produces work of some sort - moving something or heating something. Of course, you're correct about KVA which is composed of active power for moving, and reactive power for exciting (magnetizing). In the case of a motor the starting KVA, also called 'apparent power' is 6 times that of rated KVA.
electricpete
Electrical
swgmfg - I also agree with the spirit of your first message, which is that it is reactive power and associated voltage drop which is usually of concern during a start.
However your statement that "This [starting power] is a fraction of full rated power (about 15%)" seemed wrong to me.
In round numbers if starting power factor is 0.2 and starting current is 5x FLA, then the input power during starting would be approximately the same order of magnitude (or more) as full load motor input power.
I'm not sure if that's what peebee was pointing out.
However your statement that "This [starting power] is a fraction of full rated power (about 15%)" seemed wrong to me.
In round numbers if starting power factor is 0.2 and starting current is 5x FLA, then the input power during starting would be approximately the same order of magnitude (or more) as full load motor input power.
I'm not sure if that's what peebee was pointing out.
jbartos
Electrical
- Jan 15, 2000
- 6,440
Suggestion: The current transient on R-L circuits, which the starting motor essentially is (by neglecting the magnetization branch), has the following mathematical relationships:
i(t)=(V/R'(s))[1-exp(-R'(s)t/L')]
Therefore, the current is zero at a time equal to zero and then it fast increases in form of the motor inrush current. The resistance R(s) is a function of the slip, s. The slip s is equal to one (1) at the time equal to zero.
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i(t)=(V/R'(s))[1-exp(-R'(s)t/L')]
Therefore, the current is zero at a time equal to zero and then it fast increases in form of the motor inrush current. The resistance R(s) is a function of the slip, s. The slip s is equal to one (1) at the time equal to zero.
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All calculations aside, I think the point that gcaudill made about the load torque requirements is paramount in determining starting power requirements. What the motor draws is only part of the picture. Omime only provided the motor output capability. Without the load torque requirements there is no way to determine the required starting power unless you have an unlimited bus supplying it, and the very nature of his question states that he does not.
Omine, if you are serious about predetermining your starting power reqirements, check out a software package called SKM It is expensive, but very usefull if you do this a lot. We use it all the time to predetermine if a motor will start given a set of restrictive parameters. A lower cost alternative is I think available from LM Photonics, but I have not had occasion to use it so I can't vouch for it yet. Quando Omni Flunkus Moritati
Omine, if you are serious about predetermining your starting power reqirements, check out a software package called SKM It is expensive, but very usefull if you do this a lot. We use it all the time to predetermine if a motor will start given a set of restrictive parameters. A lower cost alternative is I think available from LM Photonics, but I have not had occasion to use it so I can't vouch for it yet. Quando Omni Flunkus Moritati
electricpete
Electrical
I would say that using the equivalent circuit, we can plot torque and input power (both real and reactive) vs speed, without knowing the mechanical load. The mechanical load affects how fast the speed changes.
To my knowledge the most common approach is to check the minimum voltage at the motor terminals during start. That part of the process doesn't require any knowledge of the mechanical load.
Knowledge of the mechanical load is required to answer the question of whether a given motor will accelerate a given load (under assumed minimum voltage at motor terminals). It is not so simple as comparing the motor torque speed curve to the load torque speed curve as was suggested earlier in this thread. We can have a motor/load combination which cannot be safely started even though motor torque exceeds the load torque at all speeds.
To my knowledge the most common approach is to check the minimum voltage at the motor terminals during start. That part of the process doesn't require any knowledge of the mechanical load.
Knowledge of the mechanical load is required to answer the question of whether a given motor will accelerate a given load (under assumed minimum voltage at motor terminals). It is not so simple as comparing the motor torque speed curve to the load torque speed curve as was suggested earlier in this thread. We can have a motor/load combination which cannot be safely started even though motor torque exceeds the load torque at all speeds.
electricpete
Electrical
Sorry I didn't mean to suggest that the earlier comment was wrong. That thread mentioned that a full analysis can be done with the equivalent circuit.
electricpete
No to be picky, but I have an issue here with your statement.
"I would say that using the equivalent circuit, we can plot torque and input power (both real and reactive) vs speed, without knowing the mechanical load. The mechanical load affects how fast the speed changes."
While true, this does not fully cover the needs expressed in the original post, that being his required generator size (paraphrased). Assesing the motor power requirements goes hand-in-hand with the load requirements, because how fast the speed changes determines the ability of the motor to accelerate the load without overloading. For instance, say that your calcs come up with a theoretical 100HP motor, but the load acceleration rate with a 100HP motor takes 2 minutes from start to full speed. If the motor is tripped out on overload, it was not a successful motor size application. He may need a 150HP motor to accelerate that load in 30 seconds and thus stay under the overload curve. That would require a larger generator n'est pas?
Quando Omni Flunkus Moritati
No to be picky, but I have an issue here with your statement.
"I would say that using the equivalent circuit, we can plot torque and input power (both real and reactive) vs speed, without knowing the mechanical load. The mechanical load affects how fast the speed changes."
While true, this does not fully cover the needs expressed in the original post, that being his required generator size (paraphrased). Assesing the motor power requirements goes hand-in-hand with the load requirements, because how fast the speed changes determines the ability of the motor to accelerate the load without overloading. For instance, say that your calcs come up with a theoretical 100HP motor, but the load acceleration rate with a 100HP motor takes 2 minutes from start to full speed. If the motor is tripped out on overload, it was not a successful motor size application. He may need a 150HP motor to accelerate that load in 30 seconds and thus stay under the overload curve. That would require a larger generator n'est pas?
Quando Omni Flunkus Moritati
electricpete
Electrical
No, not picky at all. I think if you look at my two paragraphs following the one you snipped you'll find that I have made exactly the same distinctionL:
- determining power required for a start does not require knowledge of mechanical load,
- determining the capability of a motor to accelerate a load does require knowledge of the mechanical load.
Your clarificaton is welcome.
Quando Omni Flunkus Puntus
- determining power required for a start does not require knowledge of mechanical load,
- determining the capability of a motor to accelerate a load does require knowledge of the mechanical load.
Your clarificaton is welcome.
Quando Omni Flunkus Puntus
jbartos
Electrical
- Jan 15, 2000
- 6,440
Suggestion to electricpete (Electrical) Mar 7, 2003 marked ///\\No, not picky at all. I think if you look at my two paragraphs following the one you snipped you'll find that I have made exactly the same distinctionL:
- determining power required for a start does not require knowledge of mechanical load,
///It appears that the mechanical load size is needed since the shaft HP may be so high, e.g. by a mechanical malfunction or friction so that the motor will even not start, it will stay at the standstill. Then, the starting power will be reduced to continuous locked-rotor power. To start the motor, it is necessary to have the mechanical load low enough for the motor not to stall, i.e. allow the motor to start turning\\- determining the capability of a motor to accelerate a load does require knowledge of the mechanical load.
///Self-evident\\
- determining power required for a start does not require knowledge of mechanical load,
///It appears that the mechanical load size is needed since the shaft HP may be so high, e.g. by a mechanical malfunction or friction so that the motor will even not start, it will stay at the standstill. Then, the starting power will be reduced to continuous locked-rotor power. To start the motor, it is necessary to have the mechanical load low enough for the motor not to stall, i.e. allow the motor to start turning\\- determining the capability of a motor to accelerate a load does require knowledge of the mechanical load.
///Self-evident\\
electricpete
Electrical
hi jraef - I see now you did say something different from me. Thx.
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