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Q1

03/19/2012 10:55 PM

What is the Full Load Current (FLC) for a 12.5 horsepower squirrel cage induction motor operated at a voltage of 230 volts?

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Guru
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#1

Re: Q1

03/19/2012 11:28 PM

Page 141 of your text book. It's that bound paper thing you've been using to prop up the broken table leg in your kitchen.

The kitchen is the part of your apartment where the refrigerator is.

The refrigerator is the thing that keeps the beer cold.

If I have to explain what beer is to you, you need to move back in with Mom and Dad...

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#2
In reply to #1

Re: Q1

03/19/2012 11:50 PM

I don't know what text book you're referring to because page 141 of my text book has nothing to do with Full Load Currents.......sir.

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#3

Re: Q1

03/20/2012 12:22 AM

Go for single phase power equation...........

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#4

Re: Q1

03/20/2012 12:34 AM

Single-phase or three-phase? What frequency? (12.5 does not sound like a standard HP.)

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#5

Re: Q1

03/20/2012 12:48 AM

My dear first you convert 12.5HP to watts.

Find the power factor of motor (if it is not given take as 0.8).

Then apply in P = V X I X pf

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#6
In reply to #5

Re: Q1

03/20/2012 1:16 AM

Thanks for the serious answer. Here's what I came up with: 12.5*746=9325w Where 230v Pf=1.25 Eff=0.89 I=9325/(sqrt(3)*230*1.25*0.89) Should that do it? I just assumed the pf was 125% an that the eff was 0.89 because it wasn't given.

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#8
In reply to #6

Re: Q1

03/20/2012 2:13 AM

Sorry to say this, but if you think PF (Cos Φ) is >1, you are in serious trouble. We learnt in school that Sin and Cos can never exceed 1 didn't we? So please do your basics again.

On a more advanced note, a 10kW machine is more likely to be 3 Phase. Please check and let us know.

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#9
In reply to #6

Re: Q1

03/20/2012 2:31 AM

sqrt(3) ?

P.F 125% ?

Apart from all tell me where have you seen 12.5 h.p single phase motor?

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#7

Re: Q1

03/20/2012 1:28 AM

1 HP = 0.746 KW, 12HP is Pout

Pin= V.I. cosΦ for single phase i.e, I=P/V cosΦ....

Pin= √3. (VL). (IL). cosΦ... VL and IL are line voltage and line currents- 3 phase

Pin= 3 .(Vph).(Iph). cosΦ... Vph and Iph are phase voltage and phase currents- 3 phase

Take cosΦ=0.8 to 0.85

eff = Pout/Pin... assume eff=0.9

Its enough for you to calculate now for any type of such problems.

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#10

Re: Q1

03/20/2012 3:26 AM
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#11

Re: Q1

03/20/2012 6:32 AM

Just look on the motor nameplate, it'll be on there.

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#12
In reply to #11

Re: Q1

03/20/2012 7:18 AM

HP = 12.5 Power

V = 230 V

HP tobe convert to kW: 12.5 x 0.746 = 9.3 kW

For Single phase P = V x I x PF

9.3 x 1000 = 230 x I x 0.80

I = 50.5 Ampere

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#13

Re: Q1

03/20/2012 7:26 AM

HP = 12.5 Power

V = 230 V

HP tobe convert to kW: 12.5 x 0.746 = 9.3 kW

For Single phase P = V x I x PF

9.3 x 1000 = 230 x I x 0.80

I = 50.5 Ampere

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#14

Re: Q1

03/20/2012 7:42 AM

"Q1" is a dead giveaway that this is homework or a test question.

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#15
In reply to #14

Re: Q1

03/20/2012 10:37 AM

Not to mention the name.

Goes to show there's no creative juices flow here.

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#16
In reply to #15

Re: Q1

03/20/2012 1:13 PM

Hence the sacrastic tome of my response. The poster is not really an electrical student, he is a management student because all he is learning is how to get other smarter people to do his work for him.

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#17
In reply to #16

Re: Q1

03/20/2012 1:26 PM

I like the way you phrased your (appropriate) comment so it was in a context to which the OP could relate.

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#18
In reply to #17

Re: Q1

03/21/2012 10:55 AM

Actually to let you all know. I AM a student an my instructor gave us homework an never told us to or that we even could take a copy of the NEC home to do this homework. So there was information missing inwhich I decided to just assume what I thought the info was. But I've talked to my instructor an have a copy of the NEC now to finish the work. I just wanted to see what other people thought about the question an see how they would handle it. Thank you

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