WTA power consumption of motor

Hi, everyone, i am designing a battery to supply a motor. (through inverter and transformer)
my motor rating is 8kW and supply with a 240V, is that mean that the ampere require is 33.33A ?
in this case, if i want to supply the motor for 2 seconds only, how to calculate the consumption?

Hi,

8000W/240V = 33.33A. so far you are right...and not.
(for a DC motor the 33Aeets the best, but with AC it´s somhow different...)

most probably the motor is a three phase AC motor, then it is 33.33A / 3 = 11.11A, but this only true with cos-phi = 1.
Maybe cos-phi = 0.8 then a realistic current is about 13.9A.


2 second is 2/60 min = 2/3600h
consumption = 8000W * 2/3600h = 4.4Wh

Klaus

Hi, thanks for reply.
is that possible that it is a single phase? this motor is actually the hoist motor of elevator. but i can't find the detail of the motor. it just mentioned in the Elevator catalogue.

I never thought about powering an elevator hoist motor from a battery so it still works if the mains power fails. Maybe the same high voltage DC motor is used as my furnace variable speed blower motor and is also used in hybrid cars.

Hi, sorry that i forgot to mention that i will use an inverter and a transformer in between.

Don't forget to allow for a 5X current inrush to start the motor! So your inverter design is very difficult. The power consumption is 8KW X 2 secs = 16000 Wsecs or 16000/60 X 60 W Hrs = 160/36 ~ 5 W/Hrs. A very small motor bike battery would be 12 V X 7 A/Hrs = 84 W/Hrs. But the real problem would be that you will need 700 A which a small battery would not deliver. Going to a 50V battery would help, only 175A.
It would seem to be better to use a car starter motor and mechanicaly couple it to the existing motor and use 12V. car starter motors are rated at up to 5HP ~ 4KW.
Frank

chuckey said:

Don't forget to allow for a 5X current inrush to start the motor! So your inverter design is very difficult. The power consumption is 8KW X 2 secs = 16000 Wsecs or 16000/60 X 60 W Hrs = 160/36 ~ 5 W/Hrs. A very small motor bike battery would be 12 V X 7 A/Hrs = 84 W/Hrs. But the real problem would be that you will need 700 A which a small battery would not deliver. Going to a 50V battery would help, only 175A.
It would seem to be better to use a car starter motor and mechanicaly couple it to the existing motor and use 12V. car starter motors are rated at up to 5HP ~ 4KW.
Frank

Click to expand...

hi, thanks for reply, what if i add a step up transformer at between the motor and the inverter to step up the voltage to 240V. isn't that will only require 33.33A ?

Hi,

hi, thanks for reply, what if i add a step up transformer at between the motor and the inverter to step up the voltage to 240V. isn't that will only require 33.33A ?

Click to expand...

if you find or develop a transformer that ouputs more power than you input, then please don´t tell anybody but me.

Klaus

KlausST said:

Hi,



if you find or develop a transformer that ouputs more power than you input, then please don´t tell anybody but me.

Klaus

Click to expand...

hi, very sorry that i didnt mention it in my thread. because i need to design the battery for that motor and i wondering my calculation correct or not. but then just now only i realize i forgot to mention that the output of the battery need to be convert to AC and step up to 240V.

Hi,

first check what motor is used.

Usually you may find the information on the motor itself. Tell us _all_ the information you find. Or take a picture of the data.

Without knowing that any further discussion is useless, or at least guessing.


Klaus

As my calculations show (roughly) for a 8 KW motor your battery has to deliver at least 8 KW for two seconds. So at 8 KW a 12V battery has to deliver 8000/12 ~ 660 A, which is a LARGE value. Also your transformer will need to step the voltage up to 230 V, so this has to be large and tailored made - very expensive!
You do not say how often this two second running is going to occur - once a minute? or once a month?
The way I would do it, its the cheapest BUT it is extremely dangerous. Get 31 12V batteries. this will give you 372V DC. Now with electronics you can synthesize a half sine wave of 230V RMS, which you apply to one side of the motor, the other going to the battery return. Then electronically you switch over the terminals of the motor and re-apply your half sine wave, so the motor sees half sinewaves being applied to one terminal then the other so the motor is fooled into thinking its getting an ordinary 230V sine wave. No transformer, the semi conductors only have to handle the motors 30A, A 370V battery charger will be required , but as the insulation and security devices must be of the highest order for this circuit, this could be done directly from the mains at minimal cost.
Frank