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
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8
- #2
electricpete
Electrical
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).
The pump will affect how long that starting current lasts. The more torque the pump draws during startup, the longer it takes the unit to get up to speed and longer the starting current lasts.
For radial flow centrifugal pumps, the least torque is drawn at the lowest flow (valve near closed). That will give the shortest starting time (ensure minimum flow required for cooling/lub is available).
Axial flow pumps draw least torque at high flow (valve full open). From motor standpoint it would be preferable to start these with valve full open (but there may also be fluid considerations such as water hammer).
Mixed flow falls in between these two.
You can see this behavior from the shape of the bhp vs flow curves at:
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The pump will affect how long that starting current lasts. The more torque the pump draws during startup, the longer it takes the unit to get up to speed and longer the starting current lasts.
For radial flow centrifugal pumps, the least torque is drawn at the lowest flow (valve near closed). That will give the shortest starting time (ensure minimum flow required for cooling/lub is available).
Axial flow pumps draw least torque at high flow (valve full open). From motor standpoint it would be preferable to start these with valve full open (but there may also be fluid considerations such as water hammer).
Mixed flow falls in between these two.
You can see this behavior from the shape of the bhp vs flow curves at:
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1
- #4
PUMPDESIGNER
Mechanical
electricpete indeed stated it well, fast too, probably off the top of his little head.
So I am asking a more difficult question.
How much affect will the type of pump, high Specific Speed (Axial Flow) vs. Low Specific Speed (Radial Flow), really have on starting current?
Assuming Across the Line starting, the radial vs. axial flow affect will not enter until the pump is close to speed.
In theory, the "type of pump effect" could extend the time of higher than operating current, but probably would not increase that current.
PUMPDESIGNER
So I am asking a more difficult question.
How much affect will the type of pump, high Specific Speed (Axial Flow) vs. Low Specific Speed (Radial Flow), really have on starting current?
Assuming Across the Line starting, the radial vs. axial flow affect will not enter until the pump is close to speed.
In theory, the "type of pump effect" could extend the time of higher than operating current, but probably would not increase that current.
PUMPDESIGNER
electricpete
Electrical
I agree with your last paragraph: the type of pump could affect the duration of the starting current, but not the magnitude.
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electricpete
Electrical
NEMG MG-1:
"12.36 INSTANTANEOUS PEAK VALUE OF INRUSH CURRENT
The values in the previous tables are rms symmetrical values, i.e. average of the three phases. There will
be a one-half cycle instantaneous peak value which may range from 1.8 to 2.8 times the above values as
a function of the motor design and switching angle. This is based upon an ambient temperature of 25°C."
The above is easily understood by modeling the motor (per-phase) as a constant impedance ZLRC= R+jwL, with suddenly-applied sinusoidal component.
In the worst case scenario, ZLRC is very close to inductive and closing occurs at the zero of the voltage waveform.
There are two components of the solution:
iSteadyState = sqrt(2)*ILRC*cos(2*Pi*FL*t)
iTransient = sqrt(2)*exp(-t*R/L)
Add them together to get the worst case 2*sqrt(2) which is where the 2.8 in the NEMA quote comes from.
If I have a chance I will try to provide more detailed discussion later.
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"12.36 INSTANTANEOUS PEAK VALUE OF INRUSH CURRENT
The values in the previous tables are rms symmetrical values, i.e. average of the three phases. There will
be a one-half cycle instantaneous peak value which may range from 1.8 to 2.8 times the above values as
a function of the motor design and switching angle. This is based upon an ambient temperature of 25°C."
The above is easily understood by modeling the motor (per-phase) as a constant impedance ZLRC= R+jwL, with suddenly-applied sinusoidal component.
In the worst case scenario, ZLRC is very close to inductive and closing occurs at the zero of the voltage waveform.
There are two components of the solution:
iSteadyState = sqrt(2)*ILRC*cos(2*Pi*FL*t)
iTransient = sqrt(2)*exp(-t*R/L)
Add them together to get the worst case 2*sqrt(2) which is where the 2.8 in the NEMA quote comes from.
If I have a chance I will try to provide more detailed discussion later.
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electricpete
Electrical
On the thread below I have attached starting current waveforms and the decaying dc component is readily apparent:
thread237-9886
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thread237-9886
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electricpete
Electrical
should have been
iTransient = sqrt(2)*ILRC*exp(-t*R/L)
and ILRC = VLGrms/sqrt(R^2+<2*Pi*L)>2)
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iTransient = sqrt(2)*ILRC*exp(-t*R/L)
and ILRC = VLGrms/sqrt(R^2+<2*Pi*L)>2)
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electricpete
Electrical
You are welcome. Even though the question is answered, I never miss a chance to waste time on an academic excercize ;-) Here is another explanation of the same thing. Unfortunately I think I succeeded in making it more complicated than it really is. Sorry for the computer-algebra.
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electricpete
Electrical
One more version, a little better than the last.
ok, I'm done now.
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ok, I'm done now.
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Hello jlhaas
As Pete has very correctly stated, if you are using a full voltage start or across the line start, the start current is independent of the shaft loading on the motor. The start current under full voltage starting is a function of the motor design and the terminal voltage.
If you reduce the voltage during start, then the start current can also be reduced. The minimum start current that can be achieved by using a reduced voltage starter is a function of the motor design and the shaft torque. The statement about controlling the discharge valve to reduce the start current is only true if you are applying a reduced voltage starter.
Best regards,
Mark Empson
As Pete has very correctly stated, if you are using a full voltage start or across the line start, the start current is independent of the shaft loading on the motor. The start current under full voltage starting is a function of the motor design and the terminal voltage.
If you reduce the voltage during start, then the start current can also be reduced. The minimum start current that can be achieved by using a reduced voltage starter is a function of the motor design and the shaft torque. The statement about controlling the discharge valve to reduce the start current is only true if you are applying a reduced voltage starter.
Best regards,
Mark Empson
jbartos
Electrical
- Jan 15, 2000
- 6,440
Suggestion: A distinction is supposed to be made among:
1. Motor starting current, namely:
The current drawn by the motor during the starting period (a function of speed or slip).
2. Motor locked rotor current, namely:
Steady-state current taken from the line with the rotor locked and with rated voltage (and rated frequency in the case of alternating-current motors) applied to the motor.
3. Motor inrush current, namely:
The rapid change of current with respect to time upon motor energization. Inrush current is dependent upon the voltage impressed across the motor terminals and the motor inductance by the relationship V=L*di/dt.
The original posting refers to the starting current as defined in item 1.
The starting current includes inrush current and if the motor happens to stall, the motor locked-rotor current, as defined in item 2, materializes.
If the discharge valve is used properly, there will be two modes of operation of the motor with one motor starting current, namely:
1. The motor starts with relatively low shaft load when the discharge valve is closed. This implies that the motor starting current will be converging faster to the motor steady state current that is lower than the motor running load current (usually designed lower than the motor rated current).
See Reference: IEEE Std 399-1997 Chapter 9 "Motor Starting Studies"
for motor starting time calculations and equations, e.g. Equation 9-18.
2. Discharge valve becomes opened. The motor is loaded as designed by the motor-pump set integrator.
The motor experiences the second part of its motor starting current in terms of its adjustment or increase to its designed motor current value. There may be a motor current transient depending how fast the discharge valve is open. This current increase is not normally referred to as an inrush current even if it may have some parts of inrush current resemblance.
To summarize the advantage. The motor starts faster with the same inrush current, however, the current levels off at lower than the designed current. When discharge valve opens, the motor current adjusts to its running current as designed by the motor-pump set integrator.
Advantage of this concept is in the smaller size of the motor that accomplishes the same function as the larger size of the motor that may be required for the discharge valve kept in open position, and in the starting current that converges to the lower steady state current initially when the motor shaft load is lower.
1. Motor starting current, namely:
The current drawn by the motor during the starting period (a function of speed or slip).
2. Motor locked rotor current, namely:
Steady-state current taken from the line with the rotor locked and with rated voltage (and rated frequency in the case of alternating-current motors) applied to the motor.
3. Motor inrush current, namely:
The rapid change of current with respect to time upon motor energization. Inrush current is dependent upon the voltage impressed across the motor terminals and the motor inductance by the relationship V=L*di/dt.
The original posting refers to the starting current as defined in item 1.
The starting current includes inrush current and if the motor happens to stall, the motor locked-rotor current, as defined in item 2, materializes.
If the discharge valve is used properly, there will be two modes of operation of the motor with one motor starting current, namely:
1. The motor starts with relatively low shaft load when the discharge valve is closed. This implies that the motor starting current will be converging faster to the motor steady state current that is lower than the motor running load current (usually designed lower than the motor rated current).
See Reference: IEEE Std 399-1997 Chapter 9 "Motor Starting Studies"
for motor starting time calculations and equations, e.g. Equation 9-18.
2. Discharge valve becomes opened. The motor is loaded as designed by the motor-pump set integrator.
The motor experiences the second part of its motor starting current in terms of its adjustment or increase to its designed motor current value. There may be a motor current transient depending how fast the discharge valve is open. This current increase is not normally referred to as an inrush current even if it may have some parts of inrush current resemblance.
To summarize the advantage. The motor starts faster with the same inrush current, however, the current levels off at lower than the designed current. When discharge valve opens, the motor current adjusts to its running current as designed by the motor-pump set integrator.
Advantage of this concept is in the smaller size of the motor that accomplishes the same function as the larger size of the motor that may be required for the discharge valve kept in open position, and in the starting current that converges to the lower steady state current initially when the motor shaft load is lower.
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1
- #15
Hi all,
Under full voltage starting, motor locked rotor current and motor starting current are one and the same.
As has already been stated by others closing the discharge valve during starting;
a) will not reduce the starting current if accross the line or full voltage start techniques are adopted. They will however reduce the over-current period.
b) will reduce the starting current if reduced voltage start techniques are adopted eg star/delta, auto-transformer, primary resistance, soft start etc.
Regards,
GGOSS
Under full voltage starting, motor locked rotor current and motor starting current are one and the same.
As has already been stated by others closing the discharge valve during starting;
a) will not reduce the starting current if accross the line or full voltage start techniques are adopted. They will however reduce the over-current period.
b) will reduce the starting current if reduced voltage start techniques are adopted eg star/delta, auto-transformer, primary resistance, soft start etc.
Regards,
GGOSS
jbartos
Electrical
- Jan 15, 2000
- 6,440
Comment on GGOSS (Electrical) Mar 22, 2004 marked ///\\Hi all,
Under full voltage starting, motor locked rotor current and motor starting current are one and the same.
///Please, notice the definition of the locked-rotor current, which refers to steady state current. The motor starting current includes transients. True locked-rotor current is when the motor stalls, e.g. due to high shaft load.\\As has already been stated by others closing the discharge valve during starting;
a) will not reduce the starting current if accross the line or full voltage start techniques are adopted. They will however reduce the over-current period.
///Please, notice the difference in definitions of motor starting current and motor inrush current. There is a clear difference between those two according to IEEE Std 100 "Dictionary"\\b) will reduce the starting current if reduced voltage start techniques are adopted eg star/delta, auto-transformer, primary resistance, soft start etc.
///Agreed.\\Regards,
GGOSS
Under full voltage starting, motor locked rotor current and motor starting current are one and the same.
///Please, notice the definition of the locked-rotor current, which refers to steady state current. The motor starting current includes transients. True locked-rotor current is when the motor stalls, e.g. due to high shaft load.\\As has already been stated by others closing the discharge valve during starting;
a) will not reduce the starting current if accross the line or full voltage start techniques are adopted. They will however reduce the over-current period.
///Please, notice the difference in definitions of motor starting current and motor inrush current. There is a clear difference between those two according to IEEE Std 100 "Dictionary"\\b) will reduce the starting current if reduced voltage start techniques are adopted eg star/delta, auto-transformer, primary resistance, soft start etc.
///Agreed.\\Regards,
GGOSS
jb,
I don't have the time or the patience to argue with you at the momment, so I won't.
Please understand that different terminology is used in different parts of the globe and posts on eng-tips (which has global membership) are likely to reflect that. Also the IEEE dictionary to which you most often refer is only one of many, many references. Are you suggesting all others are incorrect?
Regards,
GGOSS
I don't have the time or the patience to argue with you at the momment, so I won't.
Please understand that different terminology is used in different parts of the globe and posts on eng-tips (which has global membership) are likely to reflect that. Also the IEEE dictionary to which you most often refer is only one of many, many references. Are you suggesting all others are incorrect?
Regards,
GGOSS
electricpete
Electrical
Good comments GGOSS.
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I have to somewhat agree with jbartos on inrush and starting current. I have seen current (whether you call it inrush or starting) at energization, well beyond the locked rotor current. I reason (not an expert) that this is because the motor is completely de-energized and acts, momentarily, as a short circuit. The duration of this current is very small compared to the "starting current".
Generally, no one is much concerned with this "inrush" current as they are with the "starting" current. However, technically, there is a difference.
Generally, no one is much concerned with this "inrush" current as they are with the "starting" current. However, technically, there is a difference.
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