- #1
JohnnyGui
- 796
- 51
Hello,
I’ve been trying to derive Posseuille’s Law and I’m very close. However there seems to be a small difference between my concluded formula and the real one but I don’t know why.
Here’s what I get:
Picture a cilindrical tube with a radius ##R## in which a fluid is flowing. From what I know, there are 2 forces that play a role here. There’s the force ##F_P## that causes the flow itself and is proportional to the pressure difference ##ΔP##; it’s equal to ##F_P = ΔP \cdot π \cdot R^2##. And there’s a force ##F_V## that counteracts ##F_P## because of the fluid’s viscosity ## ɳ ## which causes friction at the walls of the tube; it’s equal to ##F_V = ɳ \cdot 2 \cdot R \cdot L \cdot \frac{v}{r}##. The ##r## is a chosen radius where you want to know ## F_V ##.
From what I understand, the ##F_P## is equal to ##F_V## at every chosen radius ##r## within radius ##R## of the tube. So we can say that:
$$ \frac{ΔP \cdot π \cdot R^2}{ɳ \cdot 2 \cdot \cdot π \cdot R \cdot L} = \frac{v}{r} $$
To calculate ##v## here, one would have to rewrite this as:
$$ \frac{ΔP \cdot R \cdot r}{ɳ \cdot 2 \cdot L} = v $$
We here now have written ##v## as a function of a chosen ##r## within the tube. If we now choose very small ##r##’s, we would get more or less the velocity of the portion of the fluid within that chosen ##r##. That velocity times the surface ##π \cdot r^2## would give the flow rate ##Q## through that chosen ##r##. We’d thus have to integrate the formula:
$$ \frac{ΔP \cdot R \cdot r}{ɳ \cdot 2 \cdot L} \cdot π \cdot r^2 = \frac{ΔP \cdot R \cdot r^3 \cdot π}{ɳ \cdot 2 \cdot L} $$
Integrating this would finally give the formula:
$$ \frac{ΔP \cdot R \cdot π \cdot r^4}{8 \cdot L \cdot ɳ} = Q $$
The correct formula is missing the ##R## parameter in the numerator, which is the radius of the tube. I don’t get why that is missing since that’s the determining factor in the formulae for the ##F_P## and ##F_V##.
I’ve been trying to derive Posseuille’s Law and I’m very close. However there seems to be a small difference between my concluded formula and the real one but I don’t know why.
Here’s what I get:
Picture a cilindrical tube with a radius ##R## in which a fluid is flowing. From what I know, there are 2 forces that play a role here. There’s the force ##F_P## that causes the flow itself and is proportional to the pressure difference ##ΔP##; it’s equal to ##F_P = ΔP \cdot π \cdot R^2##. And there’s a force ##F_V## that counteracts ##F_P## because of the fluid’s viscosity ## ɳ ## which causes friction at the walls of the tube; it’s equal to ##F_V = ɳ \cdot 2 \cdot R \cdot L \cdot \frac{v}{r}##. The ##r## is a chosen radius where you want to know ## F_V ##.
From what I understand, the ##F_P## is equal to ##F_V## at every chosen radius ##r## within radius ##R## of the tube. So we can say that:
$$ \frac{ΔP \cdot π \cdot R^2}{ɳ \cdot 2 \cdot \cdot π \cdot R \cdot L} = \frac{v}{r} $$
To calculate ##v## here, one would have to rewrite this as:
$$ \frac{ΔP \cdot R \cdot r}{ɳ \cdot 2 \cdot L} = v $$
We here now have written ##v## as a function of a chosen ##r## within the tube. If we now choose very small ##r##’s, we would get more or less the velocity of the portion of the fluid within that chosen ##r##. That velocity times the surface ##π \cdot r^2## would give the flow rate ##Q## through that chosen ##r##. We’d thus have to integrate the formula:
$$ \frac{ΔP \cdot R \cdot r}{ɳ \cdot 2 \cdot L} \cdot π \cdot r^2 = \frac{ΔP \cdot R \cdot r^3 \cdot π}{ɳ \cdot 2 \cdot L} $$
Integrating this would finally give the formula:
$$ \frac{ΔP \cdot R \cdot π \cdot r^4}{8 \cdot L \cdot ɳ} = Q $$
The correct formula is missing the ##R## parameter in the numerator, which is the radius of the tube. I don’t get why that is missing since that’s the determining factor in the formulae for the ##F_P## and ##F_V##.
Last edited: