# Math Help - weber number - dimensional analysis

1. ## weber number - dimensional analysis

can someone check this for me?

weber number = density*velocity^2*length/surface tension

find dimensions of surface tension.

density= ML-3
speed = [LT-1]^2
Length = L

so

[ML^-3]*[LT^-1]^2*[L]

= MLT

So for weber number to be dimensionless the surface tension has to cancel out MLT and therefore also has dimensions MLT.

2. Originally Posted by thermalwarrior
can someone check this for me?

weber number = density*velocity^2*length/surface tension

find dimensions of surface tension.

density= ML-3
speed = [LT-1]^2
Length = L

so

[ML^-3]*[LT^-1]^2*[L]

= MLT

So for weber number to be dimensionless the surface tension has to cancel out MLT and therefore also has dimensions MLT.
$[ ML^{-3} ] [ L T^{-1} ]^2 [ L ]$

$= [ M ] [ L^{-3} L^2 L ] [ T^{-2} ] = [M T^{-2} ]$

-Dan