Math issues with converting nameplate voltage

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madpenguin

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This is the last question I had on my mind but thought it would be better off in a new thread.

Let me give you a hypothetical and tell you how I'm trying to solve it.

Say I have a nameplate that just states the unit is rated for 9.7kw @ 240v. I want to run this on 208v and I'm trying to find the amperage.

Is this right?

I = P/E = 9700/240 = 40.42A
R = E/I = 240/40.42 = 5.94ohm
I = E/R = 208/5.94 = 35.02A

Thanks for helping me out on this one. I think that's right but I'm mixing series and parallel rules which is adding some doubt... I guess I thought they were exclusive but maybe not.
 
madpenguin said:
This is the last question I had on my mind but thought it would be better off in a new thread.

Let me give you a hypothetical and tell you how I'm trying to solve it.

Say I have a nameplate that just states the unit is rated for 9.7kw @ 240v. I want to run this on 208v and I'm trying to find the amperage.

Is this right?

I = P/E = 9700/240 = 40.42A
R = E/I = 240/40.42 = 5.94ohm
I = E/R = 208/5.94 = 35.02A

Thanks for helping me out on this one. I think that's right but I'm mixing series and parallel rules which is adding some doubt... I guess I thought they were exclusive but maybe not.
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Depends on what kind of unit.
Your calculation assumes that it is resistive and that the resistance is constant.
 
madpenguin said:
I have a nameplate that just states the unit is rated for 9.7kw @ 240v. I want to run this on 208v and I'm trying to find the amperage.
.
Click to expand...
Recommend looking at your nameplate again:

Ususlly when a motor rated in kW, the kW number is a shaft power output - like horsepower, not the input power.

Second thing is 9.7kW ~ 13hp. That's really a lot for single phase. Sure it isn't 3phase?

Third, most all nameplates have the FLA, which you will need. There is efficiency and power factor involved, which will keep your formulas from being accurate.

cf
 
Um... Let's say it's a heavy duty clothes dryer... :wink:

Does that help any? Judging from your response, it seems I'm trying arrive at a solution without fully understanding the theory. Altho, you imply I'm correct, no? As long as the criteria are met?
 
I'll quit beating around the bush. It's a homework question... Sorry.

I don't know any more than what I've already said. clothes dryer, 9.7kw @ 240 rated. What's the amps @ 208?

I'm not really looking for a correct answer. I'm good at math. That's not the issue. Just need to know if my approach is right. Thing is, the instructor doesn't teach any of the homework questions until AFTER we have turned it in. Really irks me.

Sorry for not being straight and adding confusion. Sparky, that was one of the answers. Looks like I need to go back and redo the math, tho I would have picked that one just because it was so close... ;)

Again, I apologize for asking a homework question, but as I say, we are thrown to the wolves with homework and I'm not finding a solution in the curriculum....

EDIT - I also just realized this probably wasn't the proper forum for this. Sorry again!
 
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madpenguin said:
I'll quit beating around the bush. It's a homework question... Sorry.

I don't know any more than what I've already said. clothes dryer, 9.7kw @ 240 rated. What's the amps @ 208?

I'm not really looking for a correct answer. I'm good at math. That's not the issue. Just need to know if my approach is right. Thing is, the instructor doesn't teach any of the homework questions until AFTER we have turned it in. Really irks me.

Sorry for not being straight and adding confusion. Sparky, that was one of the answers. Looks like I need to go back and redo the math, tho I would have picked that one just because it was so close... ;)

Again, I apologize for asking a homework question, but as I say, we are thrown to the wolves with homework and I'm not finding a solution in the curriculum....
Click to expand...


Mad,

Why would you post a homework question with all of the calculations and not say that it's a homework question?
 
madpenguin said:
clothes dryer, 9.7kw @ 240 rated. What's the amps @ 208?...
Click to expand...
Your approach looks fine to me. Close dryer = assume mostly resistive load, zero temp coefficient on the heating element.

Again, I apologize for asking a homework question, but as I say, we are thrown to the wolves with homework and I'm not finding a solution in the curriculum....

EDIT - I also just realized this probably wasn't the proper forum for this. Sorry again!
Click to expand...
It's okay with me :)

cf
 
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First, solve for R with the two known variables:

R=E?/P : R=240?/9700 : R=57600/9700. R=5.938 Ohms.





Now solve for I with the two known variables:

I=E/R : I=208/5.938 : I=35.03.
 
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madpenguin said:
Again, I apologize for asking a homework question, but as I say, we are thrown to the wolves with homework and I'm not finding a solution in the curriculum.
Click to expand...
We generally don't mind helping with homework. It helps that you gave us your own answer, and your own approach, and not just asking for the answer.
 
infinity said:
Mad,

Why would you post a homework question with all of the calculations and not say that it's a homework question?
Click to expand...

Because I thought that was taboo and against the forum rules....Seen many people get shot down because of it.

But, was never really looking for an "answer", so should be ok.

They gave us another one only a water heater that had 2 elements, each of which supplied different wattage @ 250v. What's the total load @ 208... Fun stuff.
 
480sparky said:
Frist, solve for R with the two known variables:

R=E?/P : R=240?/9700 : R=57600/9700. R=5.938 Ohms.





Now solve for I with the two known variables:

I=E/R : I=208/5.938 : I=35.03.
Click to expand...

Thanks! I'm running late for school but wrote it down on a piece of paper, will check it out when I get there! :grin: Actually, I see what happened. I rounded to the hundredth with ohms, you did to the thousandth. ;)

Thanks again everyone for the help. I appreciate it.
 
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480sparky said:
First, solve for R with the two known variables:

R=E?/P : R=240?/9700 : R=57600/9700. R=5.938 Ohms.

Now solve for I with the two known variables:

I=E/R : I=208/5.938 : I=35.03.
Click to expand...

Too late to help the pengiun but maybe this is a simpler approach?:

First calculate the current at 240V.
9700/240 = 40.4A

Then, multiply that current by the ratio of the voltages
40.42*208/240 = 35.03A

No need to calculate resistance.
 
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