Mass, volume, density of chromimium-nickel alloy, "Nichrome"

richiesmasher

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This is.. Vexing me.

I'm on a physics forum, and I have a question about the density of Chromium,

Basically, they give you Nichrome, which consists of 20% Chromium and 80% Nickel.

They want you to find the density of pure chromium.

I know Density (D)= Mass(M) / Volume (V)

and I know that M= V x D



https://www.physicsforums.com/threads/density-of-pure-chromium.936452/

Here is the thread, I recommend you view it to clear confusion.
 
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If you multiply both sides by V2, then: B = V2 + V1
Do you understand that?

B/V2 =1 +V1/V2

B/V2 x V2 = V2(1+V1/V2)
I seem to be stuck here.


Edit, I got it, you have to break it apart like 1 x v2, and V1/V2 x V2

yes I see now that B= V2+V1
 
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If you multiply both sides by V2, then: B = V2 + V1
Do you understand that?

What is it that you still need help with, since you said, "yes, I see now"?

Well let's just say I reached the point in the overall equation where I've simplified to

D3= (5D1V1)/(V1+V2)
Both numerator and denominator here can be represented as A and B.

The ratio he gave me was r = V1/V2 = D2/(4D1)

He then said what I have to do now is divide both numerator and denominator by V2, and plug in the D2/(4D1) = r



So I attempted to divide both by V2, I got both wrong.

He gave me a hint B/V2 = 1 + V1/V2

I have no idea why that hint works, where the one came from, why he put the ratio in.

But Denis asked me, if I multiply both sides of B/V2 = 1+V1/V2 by V2

I would get B = V2+V1, this I understand, but I'm still now sure what the mentor on the other site was trying to do, in short I'm trying to divide both by V2 and im kinda lost here, is what Denis did a hint?

Edit: I'm thinking, if B = V1+V2, like that's one term, I can't just divide it by V2 and cross out both V2's and be left with V1 alone...
SO I thought I have to add them first, and I know that because of the equation D2V2 = 4 D1V1, then a value for V1 is : (D2V2)/(4D1)
and a value for V2 is : (4D1V1)/D2 so I added those two terms and I got (16D1^2 *V1 + D2^2 *V2)/(4D1D2)... now I have to divide by V2, like I'm supposed to?
And then I have to do the same for A?...
 
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He gave me a hint B/V2 = 1 + V1/V2

I have no idea why that hint works, where the one came from, why he put the ratio in.

But Denis asked me, if I multiply both sides of B/V2 = 1+V1/V2 by V2

I would get B = V2+V1, this I understand, but I'm still now sure what the mentor on the other site was trying to do, in short I'm trying to divide both by V2 and im kinda lost here, is what Denis did a hint?

Edit: I'm thinking, if B = V1+V2, like that's one term, I can't just divide it by V2 and cross out both V2's and be left with V1 alone...
SO I thought I have to add them first, and I know that because of the equation D2V2 = 4 D1V1, then a value for V1 is : (D2V2)/(4D1)
and a value for V2 is : (4D1V1)/D2 so I added those two terms and I got (16D1^2 *V1 + D2^2 *V2)/(4D1D2)... now I have to divide by V2, like I'm supposed to?
And then I have to do the same for A?...

You need to learn better how to simplify fractions. "Canceling" doesn't just mean "cross the same thing off in two places and they disappear"; it means dividing the (entire) numerator and denominator of a fraction by the same thing, or equivalently removing a common factor, to get an equivalent fraction.

Here is the correct work for dividing V1+V2 by V2 and simplifying, which I do by distributing the division (dividing each term separately):

\(\displaystyle \dfrac{V_1+V_2}{V_2} = \dfrac{V_1}{V_2} + \dfrac{V_2}{V_2} = \dfrac{V_1}{V_2} + 1\)

Here I simplified the second fraction by dividing the numerator and denominator by V2 to get 1; I couldn't do that in the original form because V2 was not a factor of the entire numerator.

As for your work on the bigger problem of which this is a tiny part, you seem to be blindly combining things without a goal. The reason for using V1 = (D2V2)/(4D1) is to eliminate a variable so you can simplify; if you also use V2 = (4D1V1)/D2, then you still have both variables, and all you've done is to make things more complicated.

I'm not sure the method you are being shown elsewhere is correct, but if you want to follow it, try to think about why each step is being done so you can do it rationally. My goal at the moment is just to help you handle canceling correctly.

But perhaps you need to quote for us the entire original problem, so we can check whether this person is misleading you. One trouble I see is that he told you volumes don't add up, but then he used V1 + V2 anyway!
 
You need to learn better how to simplify fractions. "Canceling" doesn't just mean "cross the same thing off in two places and they disappear"; it means dividing the (entire) numerator and denominator of a fraction by the same thing, or equivalently removing a common factor, to get an equivalent fraction.

Here is the correct work for dividing V1+V2 by V2 and simplifying, which I do by distributing the division (dividing each term separately):

\(\displaystyle \dfrac{V_1+V_2}{V_2} = \dfrac{V_1}{V_2} + \dfrac{V_2}{V_2} = \dfrac{V_1}{V_2} + 1\)

Here I simplified the second fraction by dividing the numerator and denominator by V2 to get 1; I couldn't do that in the original form because V2 was not a factor of the entire numerator.

As for your work on the bigger problem of which this is a tiny part, you seem to be blindly combining things without a goal. The reason for using V1 = (D2V2)/(4D1) is to eliminate a variable so you can simplify; if you also use V2 = (4D1V1)/D2, then you still have both variables, and all you've done is to make things more complicated.

I'm not sure the method you are being shown elsewhere is correct, but if you want to follow it, try to think about why each step is being done so you can do it rationally. My goal at the moment is just to help you handle canceling correctly.

But perhaps you need to quote for us the entire original problem, so we can check whether this person is misleading you. One trouble I see is that he told you volumes don't add up, but then he used V1 + V2 anyway!

OK, I'm quoting everything here.

''Let me then give an alternative solution, if I may, that would be the preferred way of computing this quantity: D3=(D1V1+D2V2)/(V1+V2). In addition M2=4M1 so that D2V2=4D1V1.Numerator and denominator of the first equation can be divided by V2, and this can be readily solved for D1, (by also using the second equation). It is difficult sometimes to assess exactly what level the student is at, but this second solution would be a much improved approach, as its correct that these are normally specified 20% vs. 80% by mass.

Suggestion for the OP @Richie Smash : See if you can follow what I did in post 17, and see if you can come up with the answer for D1 that you get from those two equations. (Do not use V1/(V1+V2)=.2 and V2/(V1+V2)=.8, because we have determined that that is not what a 20%- 80% composition refers to). If my arithmetic is correct, I get a slightly different answer than 7.2 gm/cm^3. I'd be interested in seeing if you get the same thing I did.

You have two equations and two unknowns: D1 and r=V1/V2. It may look like 2 equations and 3 unknowns: D1, V1 and V2, but the ratio of V1 to V2 shows up in both equations, so it simplifies. It takes a little more algebra to solve it this way, than with the previous assumption of 20-80 by volume, but the previous assumption is really a false assumption that lacks validity. The 20-80 by mass is really the way this problem should be worked.''

Then I did this
D3 = (D1V1 +D2V2)/(V1+V2)
D3 = (D1V1 +4 D1 V1)/(V1+V2)
D3 = (D1+4D1V1)/V2
D3 = D1

And he said resuming quote.

''Your 3rd line is incorrect. Let me show you what you should get when you divide numerator and denoninator by V2: D3={5D1 (V1V2)}/{(V1V2)+1}. Also hang on to the second equation D2V2=4D1V1, and write it as r=(V1V2)=D2/(4D1) and substitute it into the equation that I just wrote out for you. You got to work at getting the algebra correct in these problems. Much of the introductory physics is algebra.


Read post 26 again please. I may have edited it since you last looked at it. Simply substitute r=V1/V2=D2/4D1 back into the first equation: D3=(5D1r)/(r+1). The result that you get will be a somewhat clumsy expression which can then be simplified somewhat, and solved for D1. And the algebra in this one is non-trivial as algebra goes, but you need to stick with it, if you want to get the correct answer.

The first thing you need is to take the equation that is of the form D3=A/B and divide numerator and denominator by V2: D3=(A/V2)/(B/V2). Then process each part A/V2 and B/V2 separately. You should also have recognized that D1V1+4D1V1=5D1V1. Doing the complete algebra: (1)(D1V1)+(4)(D1V1)=(4+1)(D1V1)=5D1V1 . For the above A=5D1V1, and B=V1+V2. It should be a simple matter to divide A by V2, and B by V2, and then substitute in for r=V1/V2=D2/(4D1)''

END QUOTE

I hope this gives more insight, because I've been struggling.
.





 
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Actually, I suggested showing us the original problem, not the solution you are trying to understand. I'll look at what you sent, but I can't tell whether it is appropriate without seeing the problem that is being solved. For example, I would expect it to say whether the alloy is 20% by weight or by volume. And where does the problem come from, and were there examples of methods to use for working with these densities?
 
Actually, I suggested showing us the original problem, not the solution you are trying to understand. I'll look at what you sent, but I can't tell whether it is appropriate without seeing the problem that is being solved. For example, I would expect it to say whether the alloy is 20% by weight or by volume. And where does the problem come from, and were there examples of methods to use for working with these densities?

This is the full question :

''This question concerns the resistance wire Nichrome.
''Nichrome is an alloy of nickel (80%) and chromium (20%).
The density of Nichrome is 8.56 g cm^-3.

Determine a likely value for the density of pure chromium in g cm^-3.
Show all steps in your calculation (the density of nickel 8.9 g cm ^ -3).''

The question comes from A book called ''Heneimann, Physics for CXC, and no no examples, just the values given and my knowledge of the density formula.
 
''This question concerns the resistance wire Nichrome. ''Nichrome is an alloy of nickel (80%) and chromium (20%). The density of Nichrome is 8.56 g cm^-3.

Determine a likely value for the density of pure chromium in g cm^-3.
Show all steps in your calculation (the density of nickel 8.9 g cm ^ -3).''

...no no examples, just the values given and my knowledge of the density formula.
Hmm... So your book mentions density, but provides absolutely zero examples of what they're talking about. That is extremely odd. Unfortunately, since your book does not define what it means by percentage breakdown (in terms of volume, or of mass?) and shows no examples from which we could discern what they intend, it is difficult, if not impossible, to advise with specificity. (The answer will vary with the definition, I suspect.)

Perhaps your instructor can provide this missing information...? Thank you! ;)
 
Hmm... So your book mentions density, but provides absolutely zero examples of what they're talking about. That is extremely odd. Unfortunately, since your book does not define what it means by percentage breakdown (in terms of volume, or of mass?) and shows no examples from which we could discern what they intend, it is difficult, if not impossible, to advise with specificity. (The answer will vary with the definition, I suspect.)

Perhaps your instructor can provide this missing information...? Thank you! ;)

I think he wants me to solve the algebra problem first then ask him those questions, and that was my plan all along xD.

If this gives some insight,
D3 = (D1 V1+ D2 V2)/V3?

Was the original equation, and we solved for this, like so:

Using the density of Nichrome given by the book, I found 20% of 8.56, and 80% of 8.56

The two values I got were then used for V1 and V2, those values being V1 = 1.712 and V2 = 6.848
Of course V1+V2 = V3.

If you substitute those along with D3 = 8.56 and D2 = 8.9

into that formula, you'll get D1 = 7.2 g cm cubed.

Which is exactly the answer in the book... But then my instructor said, yes we found the answer, but it's not a good method to arbitrarily assume that those values would be associated with volume.

So then he said its actually better to assume that for mass, even if the book doesnt say, and proceeded to give me that equation where I ended up being stuck because my algebra skills aren't up to par.

If you look at the post where I quoted him you shall see, I hope this really and truly clarifies everything, I came here because I left that thread hanging, feeling kind of shameful I couldnt figure his proper method ... so I want to learn this algebra and go back to him with newfound insight.

https://www.physicsforums.com/threads/density-of-pure-chromium.936452/
Here's the thread if you want to see.
 
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I think he wants me to solve the algebra problem first then ask him those questions, and that was my plan all along.
Okay; so, despite the fact that we don't know what the question is saying, specifically, we're supposed to guess and hope. Well, if that's how he's running his class.... :shock:

If this gives some insight, D3 = (D1 V1+ D2 V2)/V3? was the original equation....
See, we didn't know that. This equation has not been displayed in any of the previous posts, and the closest you got to this was what somebody else had "given" you. It really is helpful for us the receive all of the available information at the start! (In particular, if you were given other additional info in the original exercise, please include that in your next reply. Also, did the exercise tell you why you were being given this equation?)

...and we solved for this, like so:

Using the two densities given by the book, I found 20% of 8.56, and 80% of 8.56
So your first step was to make the assumption that the chromium has a density which is twenty percent of the density of the nichrome, and that the nickel has a density which is eighty percent of the density of the nichrome.

On what basis did you make this assumption? (Even the helper earlier suggested that surely the percentages were a matter of mass or volume, not of density!) And since you've already determined the density of the chromium, what is left to do? I mean, this is what the exercise requested, isn't it?

(By the way, I do not at all expect that "1.712" is the correct value for the density of chromium, and your value of "6.848" for the density of nickel is contradicted by what the exercise gave you.) :shock:
 
So your first step was to make the assumption that the chromium has a density which is twenty percent of the density of the nichrome, and that the nickel has a density which is eighty percent of the density of the nichrome.

On what basis did you make this assumption? (Even the helper earlier suggested that surely the percentages were a matter of mass or volume, not of density!) And since you've already determined the density of the chromium, what is left to do? I mean, this is what the exercise requested, isn't it?

(By the way, I do not at all expect that "1.712" is the correct value for the density of chromium, and your value of "6.848" for the density of nickel is contradicted by what the exercise gave you.) :shock:

No no, Those are what the assumptions of the volumes are, I simply used the number they gave me, in the question, 8.56 for the density.
So I assumed that the volume of nickel and chromium had to be 80 and 20 percent of that number respectively.
And that's exactly what we did.

And from there we solved for D1.

No my book has no extra information, and yes next time I'll post everything, even if I think it's not relevant, from the start.
 
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In that thread, you state that the "book randomly has that table of diameter versus current carrying capacity/A, so my first thought was to substitute the diameter of one of the values, into the volume of a cylinder formula, from this I would have a Volume....". In fact, that table is for the first half of the exercise, related to the resistance of constantan wire. It does not at all relate to the second half of the exercise.

The person in the thread to which you've linked (above) made reference to some other (third?) thread you're running for this exercise; this person made the explicit assumption that the percentages in the alloy referred to percentages by mass (rather than by volume). You have specified that you do not have this information, and that the instructor will not provide it to you until you've completed the exercise. Did you provide this information to this person in a private message or in the third or some other thread? Or are you assuming that his assumption was correct? (FYI for other readers: This specification is not apparent in the exercise itself, shown here in the textbook.)

No no, Those are what the assumptions of the volumes are, I simply used the number they gave me, in the question, 8.56 for the density.
So I assumed that the volume of nickel and chromium had to be 80 and 20 percent of that number respectively.
Why would the volumes of the inputs (which can vary, from one smelting to another) always be fixed at 1/5 and 4/5 of the density of the resulting alloy, and be utterly unrelated to the actual volume of the alloy? :shock:

Somehow, I'm getting the impression that the reason your working is so incomplete and confusing on this forum is that, by feeding various bits of progress onward to various help resources, you've gotten enough that you feel you can hand something in. But what you've proposed as "exactly what [they] did" to get a solution seems highly unlikely to be what was intended.
 
In that thread, you state that the "book randomly has that table of diameter versus current carrying capacity/A, so my first thought was to substitute the diameter of one of the values, into the volume of a cylinder formula, from this I would have a Volume....". In fact, that table is for the first half of the exercise, related to the resistance of constantan wire. It does not at all relate to the second half of the exercise.

The person in the thread to which you've linked (above) made reference to some other (third?) thread you're running for this exercise; this person made the explicit assumption that the percentages in the alloy referred to percentages by mass (rather than by volume). You have specified that you do not have this information, and that the instructor will not provide it to you until you've completed the exercise. Did you provide this information to this person in a private message or in the third or some other thread? Or are you assuming that his assumption was correct? (FYI for other readers: This specification is not apparent in the exercise itself, shown here in the textbook.)


Why would the volumes of the inputs (which can vary, from one smelting to another) always be fixed at 1/5 and 4/5 of the density of the resulting alloy, and be utterly unrelated to the actual volume of the alloy? :shock:


Somehow, I'm getting the impression that the reason your working is so incomplete and confusing on this forum is that, by feeding various bits of progress onward to various help resources, you've gotten enough that you feel you can hand something in. But what you've proposed as "exactly what [they] did" to get a solution seems highly unlikely to be what was intended.

No he didn't specify that, or say he will provide it to me, I didn't provide in private, I just assume his assumption was correct, also No I'm not running a third thread, just this one here, and the one on that site.

I suppose you're right, I just want to figure out his algebra at the very least... and call it a day, I apologize for the inconvenience. You may delete any confusing post in this thread or edit to suit.

Thanks for taking the time to help me :)
 
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