Basics of Fluid Mechanics, 2014a

260 CHAPTER 8. DIFFERENTIAL ANALYSIS
y
U l
dy
z
x
flow direction
Fig. -8.16. Flow between two plates, top plate is moving at speed of U l to the right (as
positive). The control volume shown in darker colors.
length is given as ΔP 23 . The upper surface is moving in velocity, U l (The right side is
defined as positive).
Solution
In this example, the mass conservation yields
=0
{ ∫}} { ∫
d
ρdV = −
dt
cv
cv
ρU rn dA =0 (8.126)
The momentum is not accumulated (steady state and constant density). Further because
no change of the momentum thus
∫
ρU x U rn dA =0 (8.127)
A
Thus, the flow in and the flow out are equal. It can concluded that the velocity
in and out are the same (for constant density). The momentum conservation leads
∫ ∫
− PdA+ τ xy dA =0 (8.128)
cv
The reaction of the shear stress on the lower surface of control volume based on Newtonian
fluid is
τ xy = −μ dU
(8.129)
dy
On the upper surface is different by Taylor explanation as
cv
⎛
∼ =0
⎞
{ }} {
τ xy = μ ⎜
dU
⎝ dy + d2 U
dy 2 dy + d 3 U
dy 3 dy2 + ··· ⎟
⎠ (8.130)
23 The difference is measured at the bottom point of the plate.