440 v 3 phase 110 KW Motor

[sidsingh]

we have a 110 KW power pack going off shore on a vessel. the vessel can provide 380 V Supply volatege.
i wanted to kow will it be possible to run this motor on 3800 v supply?

what do i have to do with start up amp coz it needs 200 a for full load running...??? will it be 1200 A for start up....
what can i do with the breaker??
plz i need some help here.cheers

 

[waross]

Hire an electrical engineer or an electrician.
I turn, I'll look for a naval architect if I want to move bulkheads around.

Bill
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"Why not the best?"
Jimmy Carter

 

[davidbeach]

What Bill said.

 

[ScottyUK]

Please use the 'Preview' function to proof read your post and correct the spelling errors in future. The odd one is fair enough, but that mess is awful. The errors in the numbers are even worse: do you really have a 3.8kV supply?

It gives a very bad impression of you: the assumption made by a reader is that you are disorganised and don't care about technical accuracy. Members don't tend to bother trying to help someone that appears undeserving and, rightly or wrongly, that is a conclusion that many people reading your post will reach. Have another go at asking your question, this time using some proper grammar and taking the time to define what you are trying to achieve, and what you trying to achieve it with.




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If we learn from our mistakes I'm getting a great education!

 

[waross]

Typo, Gosh. I just though that he had an old 4160 supply with a lot of voltage drop.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[7anoter4]

Hi sidsingh
I agree with all above posts you have to be more accurate in your post [at least with the data!].That may be also due to a lack of your electric vocabulary. As I understood there is a 400 V 110 KW induction motor [60 c/s] you intend to use on 380 V [60 Hz].
The total voltage drop does not be more then 5% [if the current frequency will not be less then 57 c/s ([Hz)]- 5% it means 20 V and then the supply shall not be less then 400-20=380 V.
On steady state usually will be also a voltage drop on generator impedance and also on the cables. Then you could not maintain 5% maximum drop.
But, if you could maintain the frequency most of the time at 60 c/s you could use up to 10% total voltage drop. That means the motor terminal voltage could be 400-400*10/100=360 V [including generator and cable voltage drop].
One has to calculate the total drop for the required load.
In the starting process the voltage drop on generator and cable impedance may be up to 20% [80 V] then the voltage on motor terminal may be 400-80=320 V. The starting current will be less then 6*200=1200 A since the starting current is approximately proportional with the voltage [1200*320/400= 960 A].The starting process duration–depending on the load wk^2 and load torque – may be between 2 up 10 seconds.
So you have to set the breaker protection so, that the breaker will not trip for this time.
As a thumb rule, for a 110 kW [140 kVA] motor starting the vessel generator has to be 400-450 kVA rated.
So, as you see the above recommendation to consult an electrical engineer priorly is compulsory.
Regards