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Location: hyderbad,calicut
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Inductoin Motor

09/23/2010 11:10 AM

we r using star configaration for reducing starting current,and now i want 2 know "upto 40% load star configaration is more efficient than the delta configaration"-why it's happened,,and i want mathematcal calculations.

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Anonymous Poster
#1

Re: inductoin motor

09/23/2010 11:23 AM

Can U Plz Uz Non SMS engls ?

Look at the motor characteristics curve Power/Current, think about the two types of losses.

UD15

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#2
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Re: inductoin motor

09/23/2010 11:32 AM

thank u sir,

i want some more explanation and mathematical proof

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#3
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Re: inductoin motor

09/23/2010 11:45 AM

Our Guest has very politely asked you to think after examining the pertinent documentation. Instead of demonstrating your thoughts, you insist to be shown a mathematical presentation and explanation. Well I don't know if you can think and I don't know if you understand the documentation you found. So I have no idea where to start or if any effort on my part will be properly received. If you explain what you do or don't understand then somebody may help you. Until you show me anything that you understand, I certainly will not help.

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#4
In reply to #3

Re: inductoin motor

09/23/2010 4:08 PM

Give up! It's the easy option Fred.

Rude, arrogant & demanding I think sums it up. But it won't get him anywhere.

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#5
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Re: inductoin motor

09/23/2010 4:18 PM

Sigh, at least one can try.

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#6

Re: Inductoin Motor

09/23/2010 5:39 PM

Ooooh, oooooh, let me try!

""up to 40% load star configaration is more efficient than the delta configaration""

Whoever told you this is feeding you a steaming pile of bovine excrement. This is a commonly used pile of excrement, I have no idea why it persists.

When you connect a motor in Star when it is intended to be connected in Delta, then you are reducing the effective voltage by the square root of 3, or 1.732 which equates to 58% voltage. Torque reduces by the square of the voltage, so your torque is reduced to 33% of normal. But since you have not changed the frequency or the pole configuration of the motor, your speed remains the same, so your effective power (kW or HP) at the shaft is reduced to 33% of normal as well. If your connected load is ALSO reduced, you can theoretically reduce the energy consumption. The theory on why they say 40% instead of 33% is because some of the energy losses are fixed and will not change, so they postulate that the load can be higher than 33% and it will still work. This is, of course, complete bovine excrement.

In reality, this is so far fetched that it borders on criminal. ANY AC motor is only going to draw as much power as it needs to do the work of the connected load, plus any associated losses. So if properly connected in Delta, and with a connected load of 33% of maximum, then the power consumption is still based on that 33% loading. So it's really pretty much the same either way. The only difference is that your magnetic losses are a tiny bit lower in Star than they would be in Delta because of the lower effective voltage across the windings. Magnetic losses however account for only about 25% of the LOSSES, and the total losses in a 95% efficient modern motor are of course only 5% to start with. So your savings are maybe 25% of 5% or a little over 1%. The risk however is very high that this will not work at all and you will stall the motor, cause it to fail and shut down your revenue stream, in which case your savings are immediately negative.

Hence, bovine excrement...

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#7
In reply to #6

Re: Inductoin Motor

09/24/2010 1:50 AM

Exactly on point and worthy of a GA. But I was so hoping that the OP would show a little integrity, respect, intelligence, nobility or other worthy quality before getting an accurate answer.

Even blinded idealists have dreams.

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#8
In reply to #6

Re: Inductoin Motor

09/24/2010 7:05 AM

thank you sir,

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