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2011/01/18 at 10:50 pm #10353adminKeymaster
If I have a 1 HP motor that runs 1 hour, it will consume 750 watts, if it starts one time. But if the motors works one hour, and it starts 4 times, and works 15 minutes each time. What will be the difference in electricity consumption?.
Thanks.
2011/01/27 at 3:55 pm #11757adminKeymasterHello
4×15 minutes = 60 minutes the normal consumption is equal .
You must determine consumption in start time and the time .
So if the motor star one time you only a start consumption more the normal consumption.
If the motor star n times the consumption in the start is equal n x oine time + the normal consumption in each time.
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2011/01/27 at 6:58 pm #11758adminKeymasterHi
Your motor start at 4 times per hour : It consume in general 7 xIn per start during a court time +15min x4x In per hours
Determination of In:
In = P motors/ (U motors x Square(3)*x cos phi)
*only on 3P motor
For one motor on 3P
In= 750 W /( 400v? x1.732…x0.8)= 1.36 Amps
At starting during few second : Icomsume = 7xIn = 9.52 Amps : P=9.52x400vx1.732x.8=5.28 KW during a court time (between 1 and
10 second is depend of utilisation of the motor :Its very different to use a motor for pumping or for Lifting.
for pumping : 7. For lifting : 14)
The difference is the time during the motor Start and the application for the motor.
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Thierry
2011/01/27 at 10:26 pm #11761adminKeymasterHi,
Don't forget the efficiency of the motor. You are dealing with a small motor which will have en effciency probably between 70% and 80% depending on the speed and quality of the motor so the power consumed will be typically 120% to 125% of the output power. Use the motor mameplate data to determine the input (consumed) power
Pinput = 1.732 x line volts x Irated x Power factor (cos phi) – assuming a 3 phase motor
alternatively
Pinput = Shaft power / Effeiciency
2011/01/28 at 10:01 am #11774adminKeymasterWell all equations aside, can't help wondering if the origins of the original question are not those based on the old techno myth that “It’s cheaper to keep something running continuously i.e. a fluorescent tube than switching it on and off many times because of the start current, current surge.
In reality if this held true I suspect that your Hall effect electricity meter would fly off the wall.
In short a motor of that size would get up to speed that fast that the electricity meter would not have time to register the load change so you would not see any increase in energy consumption.
Curiously I once met an energy conservation consultant whom was asked a similar question in a meeting I attended by the client. He confirmed to the client that it was cheaper to keep the luminaires switched on. When challenged on this point by myself his response was “Well that certainly used to be the case”, not a lot more I can say to that.
2011/02/10 at 7:20 pm #11817adminKeymasterThanks for your opinion.
2011/02/10 at 7:21 pm #11818adminKeymasterGraham Fields said:
Well all equations aside, can't help wondering if the origins of the original question are not those based on the old techno myth that “It’s cheaper to keep something running continuously i.e. a fluorescent tube than switching it on and off many times because of the start current, current surge.
In reality if this held true I suspect that your Hall effect electricity meter would fly off the wall.
In short a motor of that size would get up to speed that fast that the electricity meter would not have time to register the load change so you would not see any increase in energy consumption.
Curiously I once met an energy conservation consultant whom was asked a similar question in a meeting I attended by the client. He confirmed to the client that it was cheaper to keep the luminaires switched on. When challenged on this point by myself his response was “Well that certainly used to be the case”, not a lot more I can say to that.
Thanks for your opinion.2011/02/11 at 1:19 pm #11820MatthieuParticipantHello,
Don't foreget that during the start of an induction motor the CosPhi is quite different from nominal speed running.
You can not multiply the rated (active) power by the starting current / nominal current ratio to get the active power consumed during the start.
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