I'm reading a blood viscosity related theory and I have an example formula which I don't understand. It turns lt (fluid liters) or $cm^3$ into $cm^2$ and comes out with a number. I don't understand how the conversion is made from area to volume and vice-versa.
My formula is: $Q = S * v$
$S = 2,5 cm$
$Q = 5L*min^{-1}$
$v = ?$
So my simlpe equation is: ${\bf (5 cm^3*min^{-1}) = 2,5 cm * v}$
In my (friend's) notes I see the following fraction: $\frac{\frac{5*10^{-3}}{60}}{2,5*10^{-4}}$
NowI don't understand how the $5cm^3$ got turned into $\frac{5*10^{-3}}{60}$ and why the 2,5 must be multiplied with $10^{-4}$.
It's like converting the two units at once to match each other, but I'm missing something... the math between :-) (if there's any).
Thanks
Well $1L$ is not $1\ \rm{cm}^3$ but $1000\ \rm{cm}^3$ and I think that $S$ is a section so that it should be in $\rm{cm}^2$. If you work in $\rm{cm}$ and in $\rm{s}$ you'll get $$v=\frac{5000\,\rm{cm}^3/60\,\rm{s}}{2.5\,\rm{cm}^2}=\frac {100}3\frac{\rm{cm}}{\rm{s}}$$
You could too use the meter as a unit and in this case you'll get : $$v=\frac{5\cdot 10^{-3}\,\rm{m}^3/60\,\rm{s}}{2.5\cdot 10^{-4}\,\rm{m}^2}=\frac {1}3\frac{\rm{m}}{\rm{s}}\approx 0.3333 \frac{\rm{m}}{\rm{s}}$$ (so that the $10^{-4}$ is right ! :-))
For this kind of problem remember to always write the units (as I did here) and handle them as variables (FrenzY DT proposed you the interesting link to 'dimensional analysis').
Here you had : $$\rm{cm}=10^{-2}\,\rm{m}$$ so that you may use : $$2.5(\rm{cm})^2=2.5\left(10^{-2}\,\rm{m}\right)^2=2.5\cdot 10^{-4}\,\rm{m}^2$$ $$\frac {(\rm{cm})^3}{(\rm{cm})^2}=\rm{cm}$$ the same way you had : $$\rm{m}=60\,\rm{s}\quad \text{i.e.}\ \ 1\ \text{minute}=60\ \text{seconds}$$