+ All Categories
Home > Documents > boundary layers

boundary layers

Date post: 15-Jul-2016
Category:
Upload: mohammedtalib
View: 237 times
Download: 0 times
Share this document with a friend
Description:
transport phenomena boundary layers explanation
11
Flat plat boundary layer analysis cannot be applied to curved objects i.e. cylinder and sphere Flow reversal occurs The boundary layer detaches from the wall surface 1 Boundary Layer over Curvature
Transcript

1

• Flat plat boundary layer analysis cannot be applied to curved objects i.e. cylinder and sphere– Flow reversal occurs– The boundary layer detaches from the wall surface

Boundary Layer over Curvature

2

• For a flow over a curved body, the x-coordinate is measured along the curved surface

• The y-coordinate is measured normal to the surface

3

Pressure & velocity change in a converging diverging duct

V P V

P

4

• Turbulent eddies formed due to separation and cannot convert their rotational energy back into pressure head

• To prevent, streamlining reduces adverse pressure gradient beyond the max thickness and delays separation

5

• There is also an effect on heat transfer coefficient i.e. flow across a cylinder inside heat exchanger

Nu = hd/k

High h ( turbulent)

Low h (thick boundary layer)

6

• Several dimensionless groups in boundary layer analysis which explain the overall behaviour of the flow

• Re = inertial forces/viscous forces (V∞L/)– High inertial force = transition from laminar to turbulent

• Pr = momentum diffusivity/thermal diffusivity (/)– Relative thickness of the hydrodynamic and thermal boundary

layers

• Sc = momentum diffusivity/mass diffusivity (/Dab)– Relative thickness of the hydrodynamic and concentration

layers• Nu = heat transfer by convection/conduction (hΔT/kΔTL)• St = convective/diffusive mass transport (KL/D)

7

Similar relationship can also be observed between concentration (Sc) and hydrodynamic boundary layers

hydrodynamic

Thermal

h 1/T

8

Boundary Layer Analysis• We use example of convective transport on a flat plate • By making the following assumptions:• Steady state: ( )/t = 0• Incompressible flow: ρ = constant• Constant properties: k, Dab, Cp, ρ, μ• Newtonian fluid: xy = -Vy/x, yx = -Vxy• Two dimensional flow: ( )/z = 0, Vz = 0• No energy generation: q’ = 0• No species generation: R’ = 0

9

• Continuity:

• Momentum:

• Energy:

• Mass:

0

yV

xV yx

2

2

2

21yV

xV

xP

yVV

xVV xxx

yx

x

2

2

2

21yV

xV

yP

yV

VxV

V yyyy

yx

2

2

2

2

yT

xT

yTV

xTV yx

2

2

2

2

yC

xCD

yCV

xCV aa

aba

ya

x

10

• Next, we use some dimensionless equations as defined below:

• x* = x/L, y* = y/L (L = plate length)

• U* = Vx/V∞, V* = Vy/V∞ (V∞ = free stream velocity)

• T* = (T-Ts)/(T∞-Ts) (Ts = wall surface; T = free steam T)

• C* = (Ca-Cas)/(Ca∞-Cas) (Cas = wall surface; Ca = free stream conc.)

• P* = P/ρv∞2

• We also need to derive the partial derivative terms• For example:• x*/x = 1/L, y*/y = 1/L• U*/x = 1/V(Vx/x), V*/y = 1/V(Vy/y)• U*/y = 1/V(Vx/y), V*/y = 1/V(Vy/y)

11

• The equations are transformed into dimensionless forms as below:

• Continuity:

• Momentum:

• Energy:

• Mass:

0*

*

*

*

yV

xU

2*

*2

2*

*2

*

*

*

**

*

**

Re1

yU

xU

xP

yUV

xUU

2*

*2

*2

*2

*

**

*

**

PrRe1

yT

xT

yTV

xTU

2*

*2

2*

*2

*

**

*

**

Re1

yC

xC

ScyCV

xCU

2*

*2

2*

*2

*

*

*

**

*

**

Re1

yV

xV

yP

yVV

xVU


Recommended