Nondimensionalizing and what to do

[member 428835]

Hi PF!

There is a paper I'm reading about fluid that is sloshed in a rectangular channel, where the width is $x$, the length of of the channel is $z$, and the channel height $y$. The paper reads: "All further discussion will be carried out in dimensionless variables, choosing the half-width $l$ of the channel as the characteristic dimension. Then in Cartesian coordinates the region $\Omega$ occupied by the fluid in the equilibrium state is determined by the inequalities $$-1<x<1,\,\,\,\,h\leq y \leq \Gamma(x)$$ where $h$ is the depth of the fluid and $y =\Gamma(x)$ is the equation for the surface."

My question is, is $h$ dimensionless? Later in the paper $h$ is a parameter, but when comparing to an experiment, is $h$ dimensional?

 

[Orodruin]

If they state that they will use dimensionless variables you should probably trust them. That means that all variables have been normalised by the corresponding natural unit.

Having h as a parameter does not prevent it from being dimensionless.

 

[member 428835]

If they state that they will use dimensionless variables you should probably trust them. That means that all variables have been normalised by the corresponding natural unit.

Having h as a parameter does not prevent it from being dimensionless.

So how do I get $y$ into dimensional form? This is confusing because they don't state what $h$ is beyond the lower bound. It makes sense to me that dimensionally $-l\leq x\leq l$ and after non-dimensionalizing becomes $-1\leq x \leq 1$, but how to get $y$ into dimensional form?

 

[Orodruin]

Multiply by the corresponding natural unit, just like for $x$.

 

[member 428835]

Multiply by the corresponding natural unit, just like for $x$.

Great, that's what I thought. And area $l^2$. Thanks!

 

[Orodruin]

Note that, in a sense, we are always doing this whenever we are using actual values for physical variables. The only difference here is that the length unit is chosen based on a dimension present in the problem, not on a preexisting definition (such as cm or inches).

If you are entering numbers into a calculator, those are dimensionless. It is up to you to keep track of the units. The same goes for computer code. (Unless you are like me and program your own C++ class to take care of units for you.)

For example, in uniform motion of 12 m taking 3 s. What you would put into the calculator is 12/3=4. This is dimensionless. It is up to you to take care of the units, which are 1 m and 1 s. Dividing the units gives you a speed unit (1 m)/(1 s) = 1 m/s.

 

[member 428835]

Yea, that's a good call! I like seeing things from this view.

Ummmm I don't use C actually but perhaps if computations get spendy I may switch over...

And thanks for the help!