Mechanosensory Pathways in Angiocrine Mediated Tissue Regeneration - Mechanical and Chemical Signaling in Angiogenesis - page 26

into the flow cell 24 h before the experiment to allow the cells to grow to con-
fluence. The experiment is performed at 37 C and 5 % CO 2 . The peristaltic pump
is placed inside the incubator to minimize the tubing length and as a result the flow
path is considerably reduced. At the completion of the experiment, the flow cell
can be disassembled by stopping the pump and fixing the cells for visualization or
lysing for assaying.
As fluid flows through the parallel-plate flow chamber there are several vari-
ables that are used to calculate the shear stress imparted on the cell monolayer on
the internal plate of the chamber. The magnitude of shear stress (s) can be
determined by quantifying the following variables: fluid viscosity (l), chamber
width (b), distance between plates (h), and volumetric flow rate (Q). The shear
stress acting on the cell monolayer adhering to the plate is expressed in the
following mathematical form:

s w ¼ 6lQ
bh 2

ð 1 Þ

In order to derive the shear stress model, the Navier-Stokes equations can be
used. In the following Navier-Stokes equation for a fluid, q is density, u is fluid
velocity, l is viscosity, P is pressure, and t is time.

¼r P þ l r 2 u

q o u

ot þ u r u

ð 2 Þ

Since it is assumed that the flow is fully developed, there is no change in flow
velocity with respect to time, and as a result, the entire left side of the equation is
neglected.

0 ¼r P þ l r 2 u

ð 3 Þ

Rearranging and simplifying the terms, the equation shows the relation between
the pressure gradient in the flow direction and the change in shear stress in the
direction normal to the plate:

d 2 u

dy 2 ¼ 1

dP
dx

ð 4 Þ

l

Since it is assumed that the no-slip boundary condition applies, the following
boundary conditions exist for this second order differential equation, where
h = the height of flow chamber.

du

(1)

dy ¼ 0aty ¼ 0 due to the shear rate being zero at the axis of symmetry
(2) u ¼ 0aty ¼ 2 due to the no-slip condition at the top and bottom

integrating Eq. ( 4 ) once gives:

y þ c 1

du
dy ¼

1
l

dP
dx

ð 5 Þ