Couette Flow

BY

VIRENDRA KUMAR

PHD PURSUING (IIT DELHI)

IntroductionIn 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.

Journal bearing

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 +𝜌𝑔𝑥+𝜇 (𝜕2𝑢𝜕𝑥2 + 𝜕

2𝑢𝜕 𝑦2 +

𝜕2𝑢𝜕 𝑧 2 )+𝜇3 𝜕

𝜕 𝑥 (𝜕𝑢𝜕𝑥 + 𝜕𝑣𝜕 𝑦 + 𝜕𝑤𝜕 𝑧 )

ρ+

ρ+

𝜕𝑢𝜕𝑥 +

𝜕𝑣𝜕 𝑦 +

𝜕𝑤𝜕 𝑧 =0 Continuity equation for study incompressible flow

Analytical solution oF Couette flow

means

Now Steady Navier-Stroke equation can be reduce to

Invoking 0 0

ρ X-momentum

𝜕𝑝𝜕 𝑥=𝜇( 𝜕

2𝑢𝜕 𝑦 2 )

We choose 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 means𝑝=𝑝 (𝑥 )𝑜𝑛𝑙𝑦

• The governing equation is :

𝑢= 12𝜇𝜕𝑝𝜕𝑥 𝑦

2+𝐶1 𝑦+𝐶2

The boundary conditions are:

After invoking boundary conditions:

Where P is non-dimensional Pressure gradient.

s𝑜 ,𝑢= 𝑦h𝑈 − h

2

2𝜇 ∙𝜕𝑝𝜕 𝑥 ∙

𝑦h (1− 𝑦h )

𝑢𝑈=

𝑦h− h2

2𝜇𝑈 ∙𝜕𝑝𝜕 𝑥 ∙

𝑦h (1− 𝑦h ) Let P

The velocity profile in non-dimensional form

• when the equation reduced to:

(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.

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

Maximum and minimum velocity and it’s location• For maximum velocity :

• 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 : For P ≥ 1The Min. velocity : For P ≤ 1

Volume flow rate and average velocity

• The volume flow rate per unit width is:

𝑢𝑎𝑣𝑔=( 12+ 𝑃

6 )𝑈

• The Average velocity:

• For P=-3, volume flow rate (Q) and average velocity uavg=0

Shear stress distribution

• By invoking Newton’s law of viscosity:

• In the dimensionless form, the shear stress distribution becomes

h𝜏𝜇𝑈=1+𝑃 (1− 2 𝑦

h )• Shear stress varies linearly with the distance from the

boundary.

• For P=0, Shear stress remains constant across the flow passage:

• At y=h/2, i.e., at the center of the flow passage, shear stress is independent of pressure gradient (P).

Force, Torque and Power

¿𝜇𝜋 𝐷𝑁

60 𝑡 ∙𝜋𝐷𝐿=𝜇𝜋 2𝐷2𝑁𝐿

60 𝑡

𝑇𝑜𝑟𝑞𝑢𝑒𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑𝑡𝑜𝑜𝑣𝑒𝑟𝑐𝑜𝑚𝑒 h𝑡 𝑒𝑣𝑖𝑠𝑐𝑜𝑢𝑠 𝑒𝑓𝑓𝑒𝑐𝑡=𝑣𝑖𝑠𝑐𝑜𝑢𝑠𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒× 𝐷2

𝑇=𝜇𝜋 2𝐷3𝑁𝐿

120 𝑡

𝑃𝑜𝑤𝑒𝑟 𝑎𝑏𝑠𝑜𝑟𝑏𝑒𝑑𝑖𝑛𝑜𝑣𝑒𝑟𝑐𝑜𝑚𝑖𝑛𝑔 h𝑡 𝑒𝑣𝑖𝑠𝑐𝑜𝑢𝑠𝑟𝑒𝑠𝑖𝑠𝑖𝑡𝑎𝑛𝑐𝑒=𝑇 ∙𝜔

𝑃𝑜𝑤𝑒𝑟=𝜇𝜋3𝐷3𝑁2 𝐿360 0 𝑡 watts

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