Understanding Units and Dimensions: Solving Confusion in Basic Math Homework

[SherlockOhms]

Homework Statement

Just a quick question. This is fairly terrible that I'm blanking on something as fundamental as this. It's to do with units and dimensions. So, Area, for example, = L * L = L^2. If I have a quantity, R, for example and let's say it's units are L. If I use R in some arbitrary equation and square it does it then become the |R|^2 L or |R|^2 L^2 (where |R| is the scalar component of R)?

Homework Equations

The following is what got me confused about this. u(the average velocity) = 14/R² * u_max(The max velocity) * (R^(1/4) - R^(1/15).

The Attempt at a Solution

So, the units don't check out in the above equation for velocity, do they?

 

[Pythagorean]

Yes, units suffer the same operation as their values.

 

[arildno]

Well, if "R" in your problem is already NON-dimensional (say a ratio between two radii), it is wholly unproblematic.

 

[HallsofIvy]

Certainly R^{1/4}- R^{1/15} would make no sense if R is a "distance". Perhaps, as arildno suggests, R is really a ratio of distances. Could you give us more information about that formula?

 

[SherlockOhms]

Well, if "R" in your problem is already NON-dimensional (say a ratio between two radii), it is wholly unproblematic.

My apologies, R is a radius, therefore it has units of metres.

 

[SherlockOhms]

Thanks for clearing that up.

 

[Borek]

Could be it is an approximate formula that doesn't follow strict derivation, in such case units can be off.

 

[SherlockOhms]

Could be it is an approximate formula that doesn't follow strict derivation, in such case units can be off.

It's only for calculating the average fluid flow in a pipe, nothing overly complicated. We weren't told that it was an approximate solution so I couldn't say if it is or isn't.

 

[pasmith]

Homework Statement

Just a quick question. This is fairly terrible that I'm blanking on something as fundamental as this. It's to do with units and dimensions. So, Area, for example, = L * L = L^2. If I have a quantity, R, for example and let's say it's units are L. If I use R in some arbitrary equation and square it does it then become the |R|^2 L or |R|^2 L^2 (where |R| is the scalar component of R)?

Homework Equations

The following is what got me confused about this. u(the average velocity) = 14/R² * u_max(The max velocity) * (R^(1/4) - R^(1/15).

The Attempt at a Solution

So, the units don't check out in the above equation for velocity, do they?

No. Most likely the formula is valid only if length is measured in metres (and only for a particular fluid), and that the necessary constants of proportionality would have to be put back in if other length units were to be used.

 

[SherlockOhms]

No. Most likely the formula is valid only if length is measured in metres (and only for a particular fluid), and that the necessary constants of proportionality would have to be put back in if other length units were to be used.

Could you explain that? The length would be measured in metres and there are no constants of proportionality evident in the equation, so the units still wouldn't check out?

 

[pasmith]

Could you explain that? The length would be measured in metres and there are no constants of proportionality evident in the equation, so the units still wouldn't check out?

For the dimensions to be consistent, you would need
<br /> u_{avg} = u_{max}\frac{k}{R^2} (R^{1/4} - \alpha R^{1/15}) = u_{max} \frac{k}{R^{7/4}}(1 - \alpha R^{-11/60})<br />
where [k] = L^{7/4} and [\alpha] = L^{11/60}.

It may be that for the fluid in question k = 14\,\mathrm{m}^{7/4} and \alpha = 1\,\mathrm{m}^{11/60}.

 

[SherlockOhms]

For the dimensions to be consistent, you would need
<br /> u_{avg} = u_{max}\frac{k}{R^2} (R^{1/4} - \alpha R^{1/15}) = u_{max} \frac{k}{R^{7/4}}(1 - \alpha R^{-11/60})<br />
where [k] = L^{7/4} and [\alpha] = L^{11/60}.

It may be that for the fluid in question k = 14\,\mathrm{m}^{7/4} and \alpha = 1\,\mathrm{m}^{11/60}.

Gotcha. Thanks for that.

 

[Ray Vickson]

Homework Statement

Just a quick question. This is fairly terrible that I'm blanking on something as fundamental as this. It's to do with units and dimensions. So, Area, for example, = L * L = L^2. If I have a quantity, R, for example and let's say it's units are L. If I use R in some arbitrary equation and square it does it then become the |R|^2 L or |R|^2 L^2 (where |R| is the scalar component of R)?

Homework Equations

The following is what got me confused about this. u(the average velocity) = 14/R² * u_max(The max velocity) * (R^(1/4) - R^(1/15).

The Attempt at a Solution

So, the units don't check out in the above equation for velocity, do they?

This can get confusing; it all depends on where the units are "attached". If you say "radius = R meters", the quantity R is dimensionless and you can make immediate sense of things like sin(πR) or log(R), etc. However, if you say that "R = 2 meters", for example, then R has dimensions and you need to be a lot more careful.

 

[SherlockOhms]

That's a great explanation. Thanks. I find that the more you think about it, the more muddled up you can get. So, basically, in your example, R is essentially some arbitrary scalar?