This is probably a bit basic but I was asked a question the other day regarding the procedure for starting up a centifugal pump. The procedure usually involves shutting the discharge valve, turning on the pump, then openiong the discharge valve. I was told the reason for this procedure was to limit starting current on the motor (small motor - no starting resistors). I haven't been able find a good description for why the discharge valve being shut helps.
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Starting current for squirrel cage prime mover on a cent. pump 13
- Thread starter jlhaas
- Start date
Did every one forget to mention why the pump is started and stopped when the valve is closed? It is done to limit surges in the water system mains. It is common practice to use "booster pump control valves" in municiple water systems to keep from from blowing out the "weak" links. It has nothing to do with motor inrush, this is handled using soft starters if neccessary (>75 hp typ).
Thanks for the comment GOTWW that is a more reasonable cause for the closed valve procedure. Some times we electrical people forget the system requirements and total physical phenomena.
By the other hand the motor current from the moment that the contactors are closed to the final steady state operation is changing constantly following a performance defined by the motor reactance and slip, assuming that the voltage remains constant.
I must agree with Jbartos and Electricpete that current taken by a motor is based on the solution to the equation; v=Ldi/dt, where L is changing due to the rotor frequency and saturation.
The solution for the current “i” as function of the time “t”, has a exponential component and a variable component proportional the the inductance change until finally becomes a steady state component when the load-speed equilibrium is reached.
By the other hand the motor current from the moment that the contactors are closed to the final steady state operation is changing constantly following a performance defined by the motor reactance and slip, assuming that the voltage remains constant.
I must agree with Jbartos and Electricpete that current taken by a motor is based on the solution to the equation; v=Ldi/dt, where L is changing due to the rotor frequency and saturation.
The solution for the current “i” as function of the time “t”, has a exponential component and a variable component proportional the the inductance change until finally becomes a steady state component when the load-speed equilibrium is reached.
electricpete
Electrical
GOTTW:
I don't think the consideration of water hammer has been forgotten. My very first post:
"From motor standpoint it would be preferable to start these [axial flow pumps] with valve full open (but there may also be fluid considerations such as water hammer)."
I don't think it is reasonable to suggest that the selection of valve closed start is always driven by water hammer consideration and never driven by acceleration time/motor heating/voltage sag duration considerations. Either one may be important or unimportant depending on the application. Yes we have talked about the motor considerations more. This is after all a motor forum.
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I don't think the consideration of water hammer has been forgotten. My very first post:
"From motor standpoint it would be preferable to start these [axial flow pumps] with valve full open (but there may also be fluid considerations such as water hammer)."
I don't think it is reasonable to suggest that the selection of valve closed start is always driven by water hammer consideration and never driven by acceleration time/motor heating/voltage sag duration considerations. Either one may be important or unimportant depending on the application. Yes we have talked about the motor considerations more. This is after all a motor forum.
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electricpete
Electrical
A more concise formulation of the last 2 paragraphs of my previous post is as follows:
If theta = Pi/2,
i(t) = (1/L)*integral V0sin(w*t-Pi/2) dtau from tau=0 to t
i(t) = -V0/(L*w)*sin(w*t)=-I0*sin(w*t)
(no dc component)
If theta =0,
i(t) = (1/L)*integral V0sin(w*t) dtau from tau=0 to t
i(t) = V0/(L*w)*(1-cos(w*t))=I0*(1-cos(w*t))
(maximum dc component)
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If theta = Pi/2,
i(t) = (1/L)*integral V0sin(w*t-Pi/2) dtau from tau=0 to t
i(t) = -V0/(L*w)*sin(w*t)=-I0*sin(w*t)
(no dc component)
If theta =0,
i(t) = (1/L)*integral V0sin(w*t) dtau from tau=0 to t
i(t) = V0/(L*w)*(1-cos(w*t))=I0*(1-cos(w*t))
(maximum dc component)
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Returning to the original posting, I wonder whether we have adequately answered the question.
The original question was in regard to the effect of the discharge valve on the starting current of the motor.
Firstly, the closed discharge valve will reduce the shaft torque presented by the pump during start.
If you use a full voltage starting method, the reduction in starting torque will serve to reduce the starting time, but will not alter the start current. Under full voltage starting conditions, the start current is a function of the motor design, the applied voltage and the rotor speed.
The start current will initially be the Locked rotor current and will reduce gradually as the motor accelerates and will only fall quickly as the motor reaches full speed.
At the instance of the contactor closure, there is an inrush current that will last for a couple of cycles only.
The locked rotor current of the motor is typically in the range of 550% to 900% of the rated current of the motor. The inrush current could be as high as twice this.
Secondly, if you use a reduced voltage starting method, then the lowest start current is dependent on the motor design and the shaft load. By reducing the starting torque required by the pump, you will reduce the minimum start current required to start the pump. This can only be achieved provided that an appropriately engineered starting system is employed.
To summarise, the statement "limit starting current on the motor" could be true if an appropriately engineered reduced voltage starter is used, otherwise, the start current will be the same, but for a shorter period of time.
Best regards,
Mark Empson
The original question was in regard to the effect of the discharge valve on the starting current of the motor.
Firstly, the closed discharge valve will reduce the shaft torque presented by the pump during start.
If you use a full voltage starting method, the reduction in starting torque will serve to reduce the starting time, but will not alter the start current. Under full voltage starting conditions, the start current is a function of the motor design, the applied voltage and the rotor speed.
The start current will initially be the Locked rotor current and will reduce gradually as the motor accelerates and will only fall quickly as the motor reaches full speed.
At the instance of the contactor closure, there is an inrush current that will last for a couple of cycles only.
The locked rotor current of the motor is typically in the range of 550% to 900% of the rated current of the motor. The inrush current could be as high as twice this.
Secondly, if you use a reduced voltage starting method, then the lowest start current is dependent on the motor design and the shaft load. By reducing the starting torque required by the pump, you will reduce the minimum start current required to start the pump. This can only be achieved provided that an appropriately engineered starting system is employed.
To summarise, the statement "limit starting current on the motor" could be true if an appropriately engineered reduced voltage starter is used, otherwise, the start current will be the same, but for a shorter period of time.
Best regards,
Mark Empson
Perhaps a little of track here but I would be glad if someone could put me in the right direction. We have a problem starting 11kV 2MW pumps at the end of a long and weak distribution system. All of the usual technigues thought of and modelled but gave voltage drops outside limits. We went back to manufatturer to ask about using a donkey motor to wind pump up to speed and then apply DOL stating. His reply was for a 1M donkey motor!!!!. Has anyone a practical solution?
Hello Rodmcm
The minimum start current required, is a fucntion of the required load torque, the motor speed torque curves and the motor speed current curves. It is easy to determine what the lowest start current is provided that this information is available. At a guess, I would exect that you would need a start current in the order of 400% to start this motor unless it has particularly good high slip curves. You can reduce the minimu start current requirements by modifying the start torque requirements of the driven load. In the case of a pump, you can close the outlet valve to reduce the start torque requirements.
If the motor is not yet in existance, you do have the option of using a multistage secondary resistance starter. This will allow you to dramaticaly reduce the start current requirements with careful starter design. You are however faced with brushes and swithcgear that need continuous maintantence, and of course, a slip ring motor.
The other way to minimise the start current, is to use an inverter drive system, but in this application, the costs would be prohibitive.
Best regards,
Mark Empson
The minimum start current required, is a fucntion of the required load torque, the motor speed torque curves and the motor speed current curves. It is easy to determine what the lowest start current is provided that this information is available. At a guess, I would exect that you would need a start current in the order of 400% to start this motor unless it has particularly good high slip curves. You can reduce the minimu start current requirements by modifying the start torque requirements of the driven load. In the case of a pump, you can close the outlet valve to reduce the start torque requirements.
If the motor is not yet in existance, you do have the option of using a multistage secondary resistance starter. This will allow you to dramaticaly reduce the start current requirements with careful starter design. You are however faced with brushes and swithcgear that need continuous maintantence, and of course, a slip ring motor.
The other way to minimise the start current, is to use an inverter drive system, but in this application, the costs would be prohibitive.
Best regards,
Mark Empson
electricpete
Electrical
"We have a problem starting 11kV 2MW pumps at the end of a long and weak distribution system"
Is the problem:
A - Motor trips due to long start time
or
B - Unacceptable voltage drop in the system
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Eng-tips forums: The best place on the web for engineering discussions.
Is the problem:
A - Motor trips due to long start time
or
B - Unacceptable voltage drop in the system
=====================================
Eng-tips forums: The best place on the web for engineering discussions.
Rodmcm (Electrical)
Your option is optimize the starting Amperes vs Torque.
The best motor to convert inrush current into torque is the IWRM (induction wound rotor motor).
Vector drivers can attempt a similar performance but the most cost effective way to get torque is modifying the rotor circuit resistance.
Your option is optimize the starting Amperes vs Torque.
The best motor to convert inrush current into torque is the IWRM (induction wound rotor motor).
Vector drivers can attempt a similar performance but the most cost effective way to get torque is modifying the rotor circuit resistance.
jbartos
Electrical
- Jan 15, 2000
- 6,440
Suggestion: Try to shop around for a custom made electromagnetic coupling.
There are some readily available; however, for smaller size motors, e.g.
There are some readily available; however, for smaller size motors, e.g.
The main problem is unacceptable voltage drops due to the long weak line. Mechanical people are looking at mechanical solutions. What surprised me was the manufacturers recommendation for 1MW donkey motors to start a 2.2MW pump on closed head. I have seen 9MW refiner motors started with a 250KW donkey motor but the refiner flywheel is of course free wheeling. Does anyone know how the manufacturer got to 1MW from a pump curve against closed head.?
hello edison123 and jbartos.
The use of a fluid coupling, or a magnetic clutch coupling will not directly reduce the starting current. It may result in a reduced start time.
If you were able to decouple the load from the motor shaft and use a reduced voltage starter to start them motor under zero load conditions, and then apply the load to the motor at full speed, then there may be an opportunity. Is this what you meant??
Best regards,
Mark Empson
The use of a fluid coupling, or a magnetic clutch coupling will not directly reduce the starting current. It may result in a reduced start time.
If you were able to decouple the load from the motor shaft and use a reduced voltage starter to start them motor under zero load conditions, and then apply the load to the motor at full speed, then there may be an opportunity. Is this what you meant??
Best regards,
Mark Empson
Hi Marke,
I fully agree that use of fluid coupling wil reduce only the start up time and not the starting inrush current. I had assumed that the motor tripped due to longer start up time than due to lower voltage caused by inrush current. My mistake.
I fully agree that use of fluid coupling wil reduce only the start up time and not the starting inrush current. I had assumed that the motor tripped due to longer start up time than due to lower voltage caused by inrush current. My mistake.
jbartos
Electrical
- Jan 15, 2000
- 6,440
Comment on electricpete (Electrical) Mar 18, 2004 marked ///\\Regardless of what you do to the pump, you will initially have a starting current ~5 - 7 times full load amps (excluding initial exponentially-decaying component).
///Yes, unless there is a different motor applied with the lower NEMA Code Letter, e.g. e.g. A, which implies lower starting currents\\\
///Yes, unless there is a different motor applied with the lower NEMA Code Letter, e.g. e.g. A, which implies lower starting currents\\\
The simple answer as the the reason might be to eliminate water hammer as the result of the inrush. Did someone mention that already? The maximum inrush current will not be effected whether the valve is open or closed, and that only lasts a very short time anyway. So the more likely answer is that starting the pump with the valve open causes a hydraulic surge (water hammer). Closing the valve before starting prevents this, and then opening it slowly allows the pressure gradually into the system.
Going back to a previous statement "Firstly, the closed discharge valve will reduce the shaft torque presented by the pump during start." Mark April 19th.
How does this happen in that the pump will still lift water and impress it against a shut valve?
How does this happen in that the pump will still lift water and impress it against a shut valve?
Hello Rodmcm
The pump will cavitate. That is, it will spin the fluid within the housing, but none can exit from the housing due to the closed valve.
There will be cavitation losses in the pump, but there is no fluid moving so no "work done"
If the pump was a positive displacement pump, this would not be true.
Best regards,
Mark Empson
The pump will cavitate. That is, it will spin the fluid within the housing, but none can exit from the housing due to the closed valve.
There will be cavitation losses in the pump, but there is no fluid moving so no "work done"
If the pump was a positive displacement pump, this would not be true.
Best regards,
Mark Empson
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