Pe2 .(1-14)
Substitute the charge density term in Equation 1-9 with Equation 1-14. The momentum
equation in x direction yields
_u2 1 dp e 2 surf
+-E
8y2 pv dx pv X 82
2 (1-15)
where Ex = -Vx, is a constant in a uniform electric field along the channel surface.
To solve Equation 1-15, the following boundary conditions are applied:
* The electric potential is known at the Stem plane
u"rf y=0,h =po't (1-16)
* The geometric symmetry of flow field implies that the velocity gradient is zero at the
centerline of the channel, i.e.,
au
C-" =0
y y=h/2 (1-17)
* The flow is non-slipping at the stern layer, namely,
Syo-,h = 0.
y=0,h (1-18)
Integrate twice on the both sides of Equation 1-15, and make use of Equations 1-16 to 1-
18. The velocity is solved as
u(y) = -2 ) sE
4pv dxL[\ 2, pv (1-19)
Recall the expression of ,,f in Equation 1-6, the result above can be rewritten as
u(y) 1 d/p (h/ -h)2 + Ept E (e -Yl1)
4pv Ldx 2 pv (1-20)
The resulting flow, as expressed in Equation 1-20, is the combination of a pressure driven
flow and an EOF. The first term in Equation 1-20 represents the flow in the channel due to the
presence of pressure drop, while the second term stands for the electroosmosis. By introducing
an electroosmotic mobility term