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I am trying to understand a paper in which they mention;

" We non-dimentionalize the problem hereafter by selecting mass, length and time units such that the fluid density $\rho\equiv1$ the gravitational acceleration g $\equiv1$ and initial tank depth $h_0\equiv1$ ".

The reson for doing this is to calculate a pressure from a plot of y vs $P/\rho g h_0$ .

So if the value in the plot is 20, what is the value in Pascal for a $\rho = 1025 kg/m^3$ , $g = 9.81 m/s^2$ and $h = 50 m$ .

Thanks in advance.

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Is the density in kg/m$^3$, rather than kg/m$^2$? –  Ted Bunn Jun 20 '11 at 19:31
Sorry about that yes it is in kg/m^3 –  GGrewal Jun 21 '11 at 10:14

1 Answer 1

As you note, the quantity on the vertical axis of the plot is $P/\rho g h_0$. So if this quantity equals 20, then $$ P/\rho g h_0=20, $$ or $$ P=20\rho g h_0. $$ The units on both sides of this expression match, so this expression is correct in any system of units. If you want to know the pressure in SI units (pascals), just use SI units for the stuff on the right. If I'm not mistaken, $\rho gh_0=5.03 \times 10^5$ Pa, so $20\rho g h_0=1.01\times 10^7$ Pa.

Another way to think about it: If you want, you can actually work out the sizes of the various units required to make the various given values equal 1. That is, you can assume that lengths, masses, and times are measured in units called L, M, T. Then you want to know how many meters there are in an L, how many kg in an M, and how many seconds in a T. You know that $$ 1025 {\rm kg/m^3}=1 {\rm M/L^3}, $$ and similarly for the other two quantities to be set equal to 1. Solve these three equations for M,L,T, and you'll know the size of the various "new" units in SI units (kg,m,s). The pressure is 20 of the new units, and now that you know what the new units are you can convert back to SI.

(Note: I'd originally, incorrectly written M/T$^3$ instead of M/L$^3$ in the above equation. Someone edited it to Tonnes/m$^3$, which is not what I meant at all. In that expression, the right-hand side is supposed to be in the "made-up" units M,L,T, which don't correspond to any standard units. M/L$^3$ is correct.)

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