- #1

binbagsss

- 1,254

- 11

If I, after defining my dimensionless numbers, enter the command :

simplify(*expression*)

where the expression is a function of variables that have all been used at least once in the definition of the dimensionless numbers, is it able to give this expression in terms of these dimensionless numbers?

E.g dimensionless numbers :

## Re = \frac{\rho U_0 H_0}{ \mu } ##

## Ca =\frac{ \mu U_0 }{ \sigma} ##

## St = \frac{T_0 U_0}{H_0} ##

##Fr = \frac{U_0 }{\sqrt{G_0 H_0}}##

##Oh = \frac{\mu}{\sqrt{ \rho \sigma H_0}} ##

##Bo =\frac{\rho G_0 H_0^2 }{\sigma} ##

If I do :

simplify(## \frac{G_0}{H_0 U_0}##)

Will it give the answer as ##\frac{1}{Fr^2}##

Or would I instead need to type something like :

simplify(( ##\frac{G_0}{H_0 U_0}##), )Fr=...)

So e.g will it tell me what combination of dimensionless numbers I can write ##\frac {\mu}{U_0^2 H_0 } ## as, if I write :

Simplify (##\frac {\mu}{U_0^2 H_0 } ## )

Or :

Simplify ((##\frac {\mu}{T_0 H_0 } ##) , Re=...,Fr=...,Ca=...St=..., Bo=..., Oh=...)

Many thanks !