Need to find the btu/h of electric heater.A single phase electric heater has an output equal to 40,480 btuh an impressed voltage at 240 what will the heating imput be with impressed voltage at 230?

Here is my work 40480/3.412=11864
Ohms= volts squared / watts (240*240)/11864=4.8OHMS

Watts= volts squared * ohms ........... (230*230)/4.8ohms== 10,802.532 watts
Btu's at 230v = 10,802.532*3.413 = 36,869 btu's rounded off

But this is not correct! Why?

Its one of these answers and I would like to know what im doing wrong
36850---, 37160---,37480---,or 38790 btuh

 

You used 3.412 in the first solution and 3.413 in the second solution.

 

You used 3.412 in the first solution and 3.413 in the second solution.

Meant 3.412 but still not close to answer

 

Odd...I get very close to the answers provided by just using your formulas with the right btu/watt number (3.413).

 

Using your equations and algebra, I get:

P2 = P1 * (V2 / V1)^2
which gives me an answer that isn't close to any of the choices. Are you sure your equations are correct?

 

Did the question state "choose the most correct answer"?

You know you can work the formula to find R for each of the answers. A spreadsheet could do that very quickly.

There is only 2 milli ohms difference in the load for the "correct" answer and what was calculated using the original btu/h and 240 V.

A web search revealed the conversion factor is 3.412141633

I calculated everything to three significant digits. Well, excel did, I only entered the data and the formulae.

 

Bypassing all your formulae and doing a simple proportion (40480 * 230) / 240 I get 38793. Only 3 btu/hr away from answer D. But that could be a trap.

 

Power should go with the square of voltage, not linearly. Otherwise it's just the simple proportion. Oh, I see LDC3 already showed that.

You're not supposed to estimate a change in resistance due to a lower temperature are you?

The trap was using all the time spent on those unit conversions. Knowing that power goes with the square of voltage is all you need.

 

Need to find the btu/h of electric heater.A single phase electric heater has an output equal to 40,480 btuh an impressed voltage at 240 what will the heating imput be with impressed voltage at 230?

Here is my work 40480/3.412=11864
Ohms= volts squared / watts (240*240)/11864=4.8OHMS

Watts= volts squared * ohms ........... (230*230)/4.8ohms== 10,802.532 watts
Btu's at 230v = 10,802.532*3.413 = 36,869 btu's rounded off

But this is not correct! Why?

Its one of these answers and I would like to know what im doing wrong
36850---, 37160---,37480---,or 38790 btuh

So you got an answer of 36869 BTU/h and don't think that that is close enough to 36850 BTU/h to call it a match? Look at all four answers and notice that they all end with a 0, which is a strong indication that the answers are to the nearest 10 BTU, which means that there are four sig figs.

Let's write the possible choices as kBTU/h.

36.85 kBTU/h
37.16 kBTU/h
37.48 kBTU/h
38.79 kBTU/h

Your answer was 36.87 kBTU/h and one of the possible answers was 36.85 kBTU/h. That's a difference of less than 0.05%. How close do you need it to be?

But before you go rushing off and pick the first one as your answer, be sure to consider the rest of the story.

If you are trying to get a match to four or five sig figs, then all of your work has to be done to at least that much, and more like six sig figs in order to avoid roundoff errors. But your intermediate answer is 4.8Ω, which is only two sig figs.

Look at your computations (and I've added in the units that you still insist on ignoring).

(240V*240V)/11864W = 4.8Ω

Yet if you do the calculation to six sig figs you get 4.85502Ω. Notice that you didn't even round it properly to your two sig figs! The difference between this answer and your 4.8Ω is more than 1.1%. So you've introduced an error that will result in your answer being off from the correct answer by more than 400 BTU!

If the possible answers all differed by at least twice this amount, then this might be good enough to have confidence that you could get away with such a sloppy calculation. But let's look at the differences between answers:

36850 BTU/h
37160 BTU/h, Δ = 310 BTU/h (0.84%)
37480 BTU/h, Δ = 310 BTU/h (0.86%)
38790 BTU/h, Δ = 1310 BTU/h (3.5%)

So the first three are all within your margin of error!

If you are more careful, you should be able to get close enough to the correct answer to identify it.

Once you have done that, we can walk through how the author of the problem didn't bother to maintain adequate sig figs and so ended up reporting an answer to four sig figs that didn't deserve it. Hint: They rounded the intermediate results to four sig figs.

 

Bypassing all your formulae and doing a simple proportion (40480 * 230) / 240 I get 38793. Only 3 btu/hr away from answer D. But that could be a trap.

It's a trap.

You have the right idea -- namely setting up a proportionality -- but you are setting it up as though power is proportional to voltage, when it is proportional to the square of the voltage.

 

The trap was using all the time spent on those unit conversions. Knowing that power goes with the square of voltage is all you need.

Agreed -- though you can strike a middle ground and do it symbolically and catch the mistake that strantor made while still only ending up with just the simple proportionality that LDC3 showed.

P = V^2 / R

P is power. It can have units of J/s or BTU/h. Leave it as P and deal with converting units if and when it becomes necessary.

So you have two powers, P1 and P2

P1 = V1^2 / R
P2 = V2^2 / R

Divide the second by the first and you have

P2/P1 = V2^2 / V2^2

P2 = P1 * (V2/V1)^2

 

Are we missing anything in the question? With three of the four answers under 1% error, what did the directions state?

I do realize some tests are written with a "best choice" directive where the test taker must make a determination.

 

It's hard to say what the point was. If I were asking this question and used those answers, it would be because the point of the question was to verify that computations were done properly, avoiding excessive roundoff error. But I would argue that whoever wrote the question didn't do them properly, either, although what I am sure is the "correct" answer is only off by 0.046%, so it is close enough to find if you don't accumulate too much roundoff error.

If I play just a bit with the sig figs on the intermediate results, I can get answers of

{36850, 37470, 37160, 38760} BTU/h

One of those is falling into the trap that's already been identified.

 

Need to find the btu/h of electric heater.A single phase electric heater has an output equal to 40,480 btuh an impressed voltage at 240 what will the heating imput be with impressed voltage at 230?

Here is my work 40480/3.412=11864
Ohms= volts squared / watts (240*240)/11864=4.8OHMS

Watts= volts squared * ohms ........... (230*230)/4.8ohms== 10,802.532 watts
Btu's at 230v = 10,802.532*3.413 = 36,869 btu's rounded off

But this is not correct! Why?

Its one of these answers and I would like to know what im doing wrong
36850---, 37160---,37480---,or 38790 btuh

I think I misread your point. I thought you were asking why you weren't getting one of those answers exactly. But now I think you are saying that you answered it a 36850 BTU/h and were told that that was not the correct answer. Do I understand it correctly, now?

 

That was my understanding which is why I suggested to work each answer to the heater's resistance and choose the best answer. I did that when I saw there was 2 milliohms difference at three significant places.

With the man-hours expended on this, every answer could have been cross-checked with an abacus.