# How to calculate blood viscosity formula using $cm^3$ (lt) and $cm^2$ (area) units

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

-
When you divide physical variables, their dimensions get divided too. Examples: $\text{[L]}^3$ is for volume, $\frac{\text{[L]}}{\text{[T]}}$ is for linear velocity, $\frac{\text{[L]}^3}{\text{[T]}}$ is for flow rate. $\text{[L]}^2$ is for area, or permeability. You have to convert units to make them cancel correctly, e.g. 1 min = 60 sec, though both of them are of dimension $\text{[T]}$. –  FrenzY DT. Aug 21 '12 at 8:43
Thanks for the hint, I'll try to check it out on paper! –  atmosx Aug 21 '12 at 8:50

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 : $$\rm{m}=60\,\rm{s}\quad \text{i.e.}\ \ 1\ \text{minute}=60\ \text{seconds}$$

-
Your result though seems quite close. In my notes the result is $0,33 \frac{m^2}{s}$ –  atmosx Aug 21 '12 at 8:56
@atmosx: yes you are right (I corrected my second computation !) –  Raymond Manzoni Aug 21 '12 at 9:03
Thanks for clarifying the unit conversion! I was trying to figure it out myself right now :-) –  atmosx Aug 21 '12 at 9:32
@atmosx: please continue it's an excellent exercise ! :-) –  Raymond Manzoni Aug 21 '12 at 9:34
@atmosx: a late thought : if the answer is really in $\frac {\rm{m}^2}{\rm{s}}$ as you wrote and not in $\frac {\rm{m}}{\rm{s}}$ then $S$ could be in $\rm{cm}$ as you wrote but in this case the denominator will be $2.5\cdot 10^{-2}\rm{m}$ (this was my initial answer) and not $2.5\cdot 10^{-4}\rm{m}^2$. I fear you'll have to verify that... –  Raymond Manzoni Aug 21 '12 at 9:55