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September 12th, 2014, 06:43 AM   #1
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Dimensional Analysis

Hello,

I am practicing dimensional analysis, and I wanted to make sure that I am solving this correctly. May you check my work?

A can of soda contains 335mL of liquid. How many cubic meters would be occupied by the liquid from a 12 pack of cola?

This is how I set up the problem:

$\displaystyle \frac {335ml }{can} • \frac {10^-3 L}{1mL} • \frac {10^-3m^3} {L} • \frac {can}{12 pack}$ =

Did I set up the problem correctly?

After this point, "solving" is where I am confused. I know that I must multiply and divide, but the cube and exponents confuses me.

Can you explain this to me or provide an example?

Thank you!
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September 12th, 2014, 06:47 AM   #2
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The first factor is just what you started with. The second factor, converting mL to L, is correct. The third factor, converting L to m^3, is also correct.

The fourth factor says that there's only one can to a 12-pack, which seems low to me. I'd expect, oh, about 11 more.

Once you have all the factors right, just multiply and cancel out what you can. (You might find it easier to cancel first, then multiply what's left; all but two units, one on top and one on bottom, should disappear.)
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September 12th, 2014, 07:19 AM   #3
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Okay thanks

I solved and I came with the following:

$\displaystyle \frac{335mL}{can} = 2.79 • 10^-6 m^3 pack $

Is this correct? If not, can you show me where I made the mistake?

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September 12th, 2014, 07:36 AM   #4
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You have divided by 12 instead of multiplying ($12\text {cans}}{\text{pack}}$
Your $\text{pack}$ unit should be in the denominator too.
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September 12th, 2014, 08:26 AM   #5
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Right. Because 12 cans = 1 pack you need to multiply by (12 cans)/(1 pack).

You could also multiply by (1 pack)/(12 cans) but that wouldn't help you.
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September 12th, 2014, 08:36 AM   #6
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I would work it out bit by bit like this:

1 can = 335 millilitres
335 millilitres = 335 millilitres $\displaystyle \times \frac{1 litre}{1000 millilitres}$ = 0.335 litres
0.335 litres = 0.335 litres $\displaystyle \times \frac{1 m^3}{1000 litres}$ = 0.000335 m$\displaystyle ^3$

Therefore:
12 cans = 12 cans $\displaystyle \times \frac{0.000335 m^3}{1 cans}$ = 0.00402 m$\displaystyle ^3$.

Notice that in each step the previous units cancel, leaving a new unit in its place. In practise most people don't bother with the units until the very last step, happy in the knowledge that things are consistent.

-----

Extra bit:

Dimensional analysis is actually a really powerful tool. I figured I would provide a dimensional analysis problem here that'll hopefully give you an idea of cool it is and why we do lots of checks using it in our workplace to stop silly errors

Let's say someone gives you a formula and asks you to check if it's correct. The formula is:

$\displaystyle \Delta p = \frac{G^2 v_i}{2}\left[(1 + \sigma^2)\left(\frac{v_o}{v_i}\right) + f \frac{A}{A_{ff}}v_m\right]$

and the meanings of the letters:

$\displaystyle \Delta p$ = pressure drop (Pascals)
$\displaystyle G$ = mass velocity (kg/(m$\displaystyle ^2$s))
$\displaystyle v_i$ = specific volume of inlet (m$\displaystyle ^3$ / kg)
$\displaystyle v_o$ = specific volume of outlet (m$\displaystyle ^3$ / kg)
$\displaystyle v_m$ = average specific volume (m$\displaystyle ^3$ / kg)
$\displaystyle \sigma$ = ratio of free-flow to finned passages
$\displaystyle f$ = Darcy-Weisbach friction factor (dimensionless)
$\displaystyle A$ = total area (m$\displaystyle ^2$)
$\displaystyle A_{ff}$ = frontal fin area (m$\displaystyle ^2$)

... Yuck! You might never have seen something like this and might think "yeah, screw that", but you can check whether it's at least a possible solution without ever having touched the subject material using dimensional analysis!

So, in dimensional analysis we only care about the units, so we ignore any factors that are dimensionless. Basically, we just get rid of the numbers. However, if we're adding or subtracting things we just need to make a quick check to see that the thing being added/subtracted to it has the same units. We also denote the units of something by using square brackets around the quantity, so

$\displaystyle [\Delta p] = [G^2][v_i]\left[\left(\frac{[v_o]}{[v_i]}\right) + \frac{[A]}{[A_{ff}]}[v_m]\right]$

The 2 and $\displaystyle f$ have gone because they are dimensionless factors (i.e. they are multiplied/divided). We don't care about those! The $\displaystyle [(1+[\sigma^2])$ needs a bit of care... we're adding a dimensionless number (1) to another dimensionless thing ($\displaystyle [\sigma^2]$). This is okay and because the result is dimensionless and is then multiplied to something, we can get rid of it.

Now we just put the units in:

$\displaystyle Pa = \left(\frac{kg}{m^2s}\right)^2\frac{m^3}{kg}\left[\left(\frac{m^3/kg}{m^3/kg}\right) + \frac{m^2}{m^2}m^3/kg\right]$

we cancel the stuff that we can and end up with

$\displaystyle Pa = \frac{kg}{ms^2} \left[1 + m^3/kg\right]$

we're adding a dimensionless quantity (indicated by the 1) and a quantity with units of $\displaystyle m^3/kg$. This is not allowed! See the bold bit above. Therefore, we know that this formula is absolutely wrong.

In fact, the real one (which I just copied from a text book) is this:

$\displaystyle \Delta p = \frac{G^2 v_i}{2}\left[(1 + \sigma^2)\left(\frac{v_o}{v_i}\right) + f \frac{A}{A_{ff}}\frac{v_m}{v_i}\right]$

There's an extra $\displaystyle v_i$ in it below the $\displaystyle v_m$. If we repeat what we did above, we instead get

$\displaystyle Pa = \frac{kg}{ms^2}\left[1 + 1\right] = \frac{kg}{ms^2}$

The 1+ 1 bit disappears because the result is dimensionless and is being multiplied by something.

Okay... so is the unit Pascals equal to $\displaystyle \frac{kg}{ms^2}$? We can check by taking an existing formula for pressure:

$\displaystyle Pressure = Force/Area$
$\displaystyle P = \frac{F}{A}$

We can do dimensional analysis on this too

$\displaystyle [P] = \frac{[F]}{[A]}$
$\displaystyle Pa = \frac{N}{m^2}$

and we know Newton's second law is

$\displaystyle Force = mass \times acceleration$
$\displaystyle F = ma$

so...

$\displaystyle [F] = [m][a]$
$\displaystyle N = kg m/s^2$

putting this back into what we obtained from our original formula, we get

$\displaystyle Pa = \frac{N}{m^2} = \frac{kg m/s^2}{m^2} = \frac{kg}{ms^2}$

So it matches. This means the equation is dimensionally sound. It doesn't prove it's right, but it at least shows that it's sensible.

Last edited by Benit13; September 12th, 2014 at 08:45 AM.
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September 12th, 2014, 08:42 AM   #7
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I'm not drinking a can of soda for a long while.....
Thanks from Benit13
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September 12th, 2014, 09:03 AM   #8
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Thanks guys! I see where I should have flipped the 12 cans/1 pack.

Thank you so much for explaining it to me, and I agree no more soda!
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September 12th, 2014, 10:09 AM   #9
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Quote:
Originally Posted by Denis View Post
I'm not drinking a can of soda for a long while.....
If you think that is scary, try this. According to my calculations to work off the energy of one 12 oz can of Pepsi you need to walk up about 200 flights of stairs. (Based on my weight of 250 lbs.)

-Dan
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September 12th, 2014, 10:46 AM   #10
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Originally Posted by topsquark View Post
I'm just a multitudinous profusion of oracular profundity.
-Dan
Soooo...you're a man of few words, but you use them OFTEN !!
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