Is this formula right?

[Jacobbus]

I 've got the formula:

t/V=((α*ω*η)/(2*A^2*ΔP))*V+((η*β)/(A*ΔP))

In which are:
t = Time [s]
V = Volume Filtraat [m^3]
α = Specific resistance of filtered solids [m/kg]
ω = density solids [kg/m^3]
η = Viscositie suspension [Pa*s]
A = Filterarea [m^2]
ΔP = presure [Pa]
β = resistance filter [1/m]

And I'm wondering if its right
the piece before the equal sign devides time in seconds over volume in m^3
so the formula must be in s/m^3
but I can't realy find out if it is, and I thougt you are better in math then I am so why not ask.
I thougt it was different because the piece before the + sign gives me:
((m/kg*kg/m^3*Pa*s)/(2*(A^2)^2*Pa))*m^3
simplify: (Pa's away kg's away A^2^2 =A^4 and m*m^-3=m^-2)
((m^-2*s)/(2*m^4))*m^3
simplify further:
s/(2*m^3)
do i do something wrong there?????????????

the piece after the + sign gives its number right in s/m^3 so if the fist part was also in s/m^3 it the 2 of em added would also be in s/m^3 right

but i cant get the fitst to be in s/m^3 so do i do something wrong or should the formula not have the times 2 in the first part?

by the way it looks a lot simpler if you write it down on paper

[WikiGuru]

Just glancing over it, but maybe it's because you need a constant, preferably one that have units to change the units you have into a correct unit (This is my chemistry for the day :biggrin: )?

I don't know if you know about the reaction rate constant, but you might and it adjusts the equation accordingly to get the units to work out.

r=k[A]^m[B]ⁿ

r is the rate (mol*L^-1*sec^-1)
k is the constant
[A] is the concentration of substance A (mol*L^-1)
[B], same as above except it's substance B
m and n are the orders of the rate in proportion to both [A] and [B], m for [A] and n for [B] (unit less)

If m+n=0: units of k are mol*L^-1*sec^-1
If m+n=1: units of k are sec^-1
If m+n=2: units of k are L*mol^-1*sec^-1
... so on and so forth.

I know all this jabbering wasn't relevant to the question (which I have very little knowledge of) but hopefully this may jar your memory into remembering some constant with such a property?

[DigiTan]

You're right about it being s/m³. I left the "2" before the plus sign out completely since it's unit-less and would probably distract me at some point. Breaking Pascals down to "kg/m²" I got...

[font="Courier"]
(m * kg * kg * s * m³ / kg * m³ * m²) / (m^4 * (kg / m²) )

[Jacobbus]

Ok thanks DigiTan, I've got it now, it was my fault to keep the 2 in.

[DigiTan]

For what it's worth, it took me a couple tries to keep from getting seconds/meter right of the "+" sign.

This kind of reminds me, there's this program for making math formula jpegs but I forgot the name. It's like "Math Print" or "Math Scribe" or something.

[DarkerLine]

[JoeImp]