Couette flow, N-S Equation

1. Couette Flow
BY
VIRENDRA KUMAR
PHD PURSUING (IIT DELHI)

2. Introduction
In fluid dynamics, Couette flow is the laminar flow of a viscous fluid in
the space between two parallel plates, one of which is moving relative
to the other.
The flow is driven by virtue of viscous drag force acting on the fluid
and the applied pressure gradient parallel to the plates.
This kind of flow has application in hydro-static lubrication, viscosity
pumps and turbine.
The present analysis can be applied to journal bearings, which are
widely used in mechanical systems.
When the bearing is subjected to a small load, such that the rotating
shaft and bearing remain concentric, the flow characteristic of the
lubricant can be modeled as flow between parallel plates where the top
plate moves at a constant velocity.

3. Journal bearing

4. Navier-Stokes Equation: Cartesian Coordinates
Continuity equation for 3-D flow
X-momentum
Y-momentum
Z-momentum
𝜕𝜌
𝜕𝑡
+
𝜕
𝜕𝑥
(ρ𝑢) +
𝜕
𝜕𝑦
(ρ𝑣) +
𝜕
𝜕𝑧
(ρ𝑤) = 0
𝜌
𝜕𝑢
𝜕𝑡
+ 𝑢
𝜕𝑢
𝜕𝑥
+ 𝑣
𝜕𝑢
𝜕𝑦
+ 𝑤
𝜕𝑢
𝜕𝑧
= −
𝜕𝑃
𝜕𝑥
+ 𝜌𝑔 𝑥 + 𝜇
𝜕2
𝑢
𝜕𝑥2
+
𝜕2
𝑢
𝜕𝑦2
+
𝜕2
𝑢
𝜕𝑧2
+
𝜇
3
𝜕
𝜕𝑥
𝜕𝑢
𝜕𝑥
+
𝜕𝑣
𝜕𝑦
+
𝜕𝑤
𝜕𝑧
ρ
𝜕𝑣
𝜕𝑡
+ 𝑢
𝜕𝑣
𝜕𝑥
+ 𝑣
𝜕𝑣
𝜕𝑦
+ 𝑤
𝜕𝑣
𝜕𝑧
= −
𝜕𝑃
𝜕𝑦
+ 𝜌𝑔 𝑦 + 𝜇
𝜕2 𝑣
𝜕𝑥2 +
𝜕2 𝑣
𝜕𝑦2 +
𝜕2 𝑣
𝜕𝑧2 +
𝜇
3
𝜕
𝜕𝑦
𝜕𝑢
𝜕𝑥
+
𝜕𝑣
𝜕𝑦
+
𝜕𝑤
𝜕𝑧
ρ
𝜕𝑤
𝜕𝑡
+ 𝑢
𝜕𝑤
𝜕𝑥
+ 𝑣
𝜕𝑤
𝜕𝑦
+ 𝑤
𝜕𝑤
𝜕𝑧
= −
𝜕𝑃
𝜕𝑧
+ 𝜌𝑔 𝑧 + 𝜇
𝜕2 𝑤
𝜕𝑥2 +
𝜕2 𝑤
𝜕𝑦2 +
𝜕2 𝑤
𝜕𝑧2 +
𝜇
3
𝜕
𝜕𝑧
𝜕𝑢
𝜕𝑥
+
𝜕𝑣
𝜕𝑦
+
𝜕𝑤
𝜕𝑧
𝜕𝑢
𝜕𝑥
+
𝜕𝑣
𝜕𝑦
+
𝜕𝑤
𝜕𝑧
= 0 Continuity equation for study incompressible flow

5. Analytical solution oF Couette flow
• 𝑢 ≠ 0, 𝑣 = 𝑤 = 0
means 𝑢 = 𝑢 𝑦, 𝑧
Now Steady Navier-Stroke equation can be
reduce to
Invoking 𝜕𝑢
𝜕𝑥
+
𝜕𝑣
𝜕𝑥
+
𝜕𝑤
𝜕𝑥
= 0
0 0
ρ
𝜕𝑢
𝜕𝑡
+ 𝑢
𝜕𝑢
𝜕𝑥
+ 𝑣
𝜕𝑢
𝜕𝑦
+ 𝑤
𝜕𝑢
𝜕𝑧
= −
𝜕𝑃
𝜕𝑥
+ 𝜌𝑔 𝑥 + 𝜇
𝜕2 𝑢
𝜕𝑥2 +
𝜕2 𝑢
𝜕𝑦2 +
𝜕2 𝑢
𝜕𝑧2
X-momentum
𝜕𝑝
𝜕𝑥
= 𝜇
𝜕2 𝑢
𝜕𝑦2
We choose 𝑥 to be the direction along which all fluid particles travel, and assume the plates are
infinitely large in z-direction, so the z-dependence is not there.
0 0 0 0 00 0
𝜕𝑝
𝜕𝑦
=
𝜕𝑝
𝜕𝑧
= 0 means 𝑝 = 𝑝 𝑥 𝑜𝑛𝑙𝑦

6. • The governing equation is:
𝜕𝑝
𝜕𝑥
= 𝜇
𝜕2 𝑢
𝜕𝑦2
𝑢 =
1
2𝜇
𝜕𝑝
𝜕𝑥
𝑦2 + 𝐶1 𝑦 + 𝐶2
The boundary conditions are:
i 𝑦 = 0, 𝑢 = 0 𝑖𝑖 𝑦 = ℎ, 𝑢 = 𝑈
After invoking boundary conditions: 𝐶2 = 0 𝑎𝑛𝑑
𝐶1 =
𝑈
ℎ
−
1
2𝜇
𝜕𝑝
𝜕𝑥
∙ ℎ
Where P is non-dimensional Pressure gradient.
s𝑜, 𝑢 =
𝑦
ℎ
𝑈 −
ℎ2
2𝜇
∙
𝜕𝑝
𝜕𝑥
∙
𝑦
ℎ
1 −
𝑦
ℎ
𝑢
𝑈
=
𝑦
ℎ
−
ℎ2
2𝜇𝑈
∙
𝜕𝑝
𝜕𝑥
∙
𝑦
ℎ
1 −
𝑦
ℎ
Let P= −
ℎ2
2𝜇
∙
𝜕𝑝
𝜕𝑥

7. The velocity profile in non-dimensional form
•
𝑢
𝑈
=
𝑦
ℎ
+ 𝑃 ∙
𝑦
ℎ
1 −
𝑦
ℎ
• when 𝑃 = 0 the equation reduced to:
𝑢
𝑈
=
𝑦
ℎ
(simple couette flow )
• It can be produced by sliding a parallel plate at constant
speed relative to a stationary wall.
Fig. Simple couette flow
• For simple shear flow, there is no pressure gradient in
the direction of the flow.

8. The velocity profiles for various P
• For P < 0, the fluid motion created by the
top plate is not strong enough to
overcome the adverse pressure gradient,
hence backflow (i.e., u/U is negative)
occurs at the lower-half region.
• For P>0, the fluid motion created by top
plate is enough strong to overcome the
adverse pressure gradient, hence u/U is
+ve over the whole gap.
Velocity Profiles

9. Maximum and minimum velocity and it’s location
• For maximum velocity :
𝜕𝑢
𝜕𝑦
= 0
•
𝜕𝑢
𝜕𝑦
=
𝑈
ℎ
+
𝑃𝑈
ℎ
1 − 2
𝑦
ℎ
= 0
• It is interesting to note that maximum velocity for P=1 occurs at y/h
=1 and equals to U. For P>1, the maximum velocity occurs at a
location y/h<1.
• This means that with P>1, the fluid particles attain a velocity higher than
that of the moving plate at a location somewhere below the moving plate.
• For P=-1 the minimum velocity occurs, at y/h=0. For P<-1, the minimum
velocity occurs at allocation y/h>1, means occurrence of back flow near the
fixed plate.
The Max. velocity : 𝑢 𝑚𝑎𝑥 =
𝑈(1+𝑃)2
4𝑃
For P ≥ 1
The Min. velocity : 𝑢 𝑚𝑖𝑛 =
𝑈(1+𝑃)2
4𝑃
For P ≤ 1

10. Volume flow rate and average velocity
• The volume flow rate per unit width is:
𝑄 =
0
ℎ
𝑢 𝑑𝑦 = 𝑈
0
ℎ
𝑦
ℎ
+ 𝑃
𝑦
ℎ
1 −
𝑦
ℎ
𝑑𝑦
𝑢 𝑎𝑣𝑔 =
1
2
+
𝑃
6
𝑈
𝑄 =
1
2
+
𝑃
6
𝑈 ∙ ℎ
• The Average velocity:
𝑢 𝑎𝑣𝑔 =
𝑣𝑜𝑙𝑢𝑚𝑒 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 (𝑄)
𝑎𝑟𝑒𝑎 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑤𝑖𝑑𝑡ℎ (ℎ×1))
• For P=-3, volume flow rate (Q) and average velocity uavg=0

11. Shear stress distribution
• By invoking Newton’s law of viscosity:
𝜏 = 𝜇
𝑑𝑢
𝑑𝑦
= 𝜇
𝑑
𝑑𝑦
U
𝑦
ℎ
+ 𝑃
𝑦
ℎ
1 −
𝑦
ℎ
• In the dimensionless form, the shear stress distribution becomes
ℎ 𝜏
𝜇 𝑈
= 1 + 𝑃 1 −
2𝑦
ℎ
• Shear stress varies linearly with the distance from the boundary.
• For P=0, Shear stress remains constant across the flow passage: 𝜏 =
𝜇𝑈
ℎ
• At y=h/2, i.e., at the center of the flow passage, shear stress is
independent of pressure gradient (P).

12. Force, Torque and Power
• 𝑆ℎ𝑒𝑎𝑟 𝐹𝑜𝑟𝑐𝑒 𝑜𝑟 𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 ×
𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠ℎ𝑎𝑓𝑡
=
𝜇𝜋𝐷𝑁
60𝑡
∙ 𝜋𝐷𝐿 =
𝜇𝜋2
𝐷2
𝑁𝐿
60𝑡
• 𝑇𝑜𝑟𝑞𝑢𝑒 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑡𝑜 𝑜𝑣𝑒𝑟𝑐𝑜𝑚𝑒 𝑡ℎ𝑒 𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑒𝑓𝑓𝑒𝑐𝑡 = 𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 ×
𝐷
2
𝑇 =
𝜇𝜋2
𝐷3
𝑁𝐿
120𝑡
• 𝑃𝑜𝑤𝑒𝑟 𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑 𝑖𝑛 𝑜𝑣𝑒𝑟𝑐𝑜𝑚𝑖𝑛𝑔 𝑡ℎ𝑒 𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑟𝑒𝑠𝑖𝑠𝑖𝑡𝑎𝑛𝑐𝑒 = 𝑇 ∙ 𝜔
𝑃𝑜𝑤𝑒𝑟 =
𝜇𝜋3 𝐷3 𝑁2 𝐿
3600𝑡
watts