# Navier-Stokes and Blood.

Let us assume that blood moves only along the blood vessel, by which we mean that there is no radial or angular components in the speed of the blood. If we consider a straight blood vessel, then, the speed will be written as

$\mathbf{u}(r,z) = u(r,z)\mathbf{\widehat{z}},$

where there is no angular dependence considering the symmetry of the problem. If we consider a steady flow, then Navier-Stokes equation (NVE) in cylindrical coordinates will give us

$0 = -\nabla p + \mu \nabla^2\mathbf{u} = -\nabla p + \frac{\mu}{r} \frac{\partial }{\partial r} \left ( r \frac{\partial u(r,z)}{\partial r} \right ) + \frac{\partial^2 u(r,z)}{\partial z^2},$

where $p$ is a function denoting the pressure field. Let us assume that blood pressure drops from arterial pressure $p_0$ by an amount $\Delta$ after traveling a distance $L$ in a linear fashion:

$p(z) = p_0 - \frac{\Delta}{L} z \;\;\ \Rightarrow \;\;\ \nabla p = - \frac{\Delta}{L} \equiv q.$

Therefore NVE yields:

$-q + \frac{\mu}{r} \frac{\partial }{\partial r} \left ( r \frac{\partial u(r,z)}{\partial r} \right ) + \frac{\partial^2 u(r,z)}{\partial z^2} = 0 \;\;\; \left [ 1 \right ].$

But, as we know, mass must be conserved. So

$\nabla \cdot \mathbf{u} = \frac{\partial u}{\partial z} = 0 \;\;\; \Rightarrow \;\;\; u(r,z) = C_1 + f(r),$

where $C_1$ is a constant and $f(r)$ is a function of the distance to the center of the flow, only. As there is no dependence in $z,$ the second partial derivative in relation to this component must be $0$ and equation $\left [ 1 \right ]$ reduces to

$-q + \frac{\mu}{r} \frac{\partial }{\partial r} \left ( r \frac{\partial f}{\partial r} \right ) = 0,$

which easily solves to

$f(r) = \frac{qr^2}{4\mu} + C_2 \ln(r) + C_3 \;\;\; \therefore \;\;\; u(r,z) \equiv u(r) = \frac{qr^2}{4\mu} + C_2 \ln(r) + C_3 + C_1 \equiv \frac{qr^2}{4\mu} + C,$

where we've taken $C_2 = 0$ because we want the function to not diverge as $r \rightarrow 0$ and $C_1 + C_3 = C.$ If the speed function is required to be smooth everywhere, one would like that $u(R) = 0,$ where $R$ is the radius of the blood vessel. This condition settles a value for $C$ and one can find the unique solution:

$u(r) = \frac{qR^2}{4\mu} \left ( \frac{r^2}{R^2} - 1 \right ).$

As one can see, even though it might seem counterintuitive, in a steady flow regime the speed of the flow in a linear-gradient pressure drop does not depend on the distance from the beginning of the pressure drop. Note by Lucas Tell Marchi
3 months, 1 week ago

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