boundary layes by Bradley Jacobs

ME 150 – Heat and Mass Transfer

Chap. 12.2: Laminar Boundary Layers

Laminar Boundary Layers To calculate the convective heat transfer coefficient, we have to analyze the boundary layer close to a rigid boundary: Conservation Equations Specific boundary conditions are necessary to simplify these equations. Within boundary layer: low velocities normal to wall Outside of boundary layer: Potential flow with u∞ und T∞, no effect of viscosity

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2: Laminar Boundary Layers

2D Boundary Layer: Forced convection over plate Potential flow

Flow in xdirection, y-direction ┴ to surface, z-direction is ∞

Boundary layers: Velocity: δ, Temperature δT u ( y = δ ) = 0.99 ⋅ u∞

T ( y = δ T ) − TW = 0.99 ⋅ (T∞ − TW )

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2: Laminar Boundary Layers

Goal of this analysis: Friction force on the surface (Newtonian fluid) L

F = ∫ τ w ⋅W ⋅ dx 0

⎛ ∂u ⎞ τ w = µ ⋅ ⎜⎜ ⎟⎟ ⎝ ∂y ⎠ y =0

→

⎛ ∂u ⎞ F = µ ⋅W ∫ ⎜⎜ ⎟⎟ ⋅ dx ∂y ⎠ y =0 0 ⎝

Heat transfer on the surface L

q = ∫ qʹ′wʹ′ ⋅W ⋅ dx 0

⎛ ∂T ⎞ qʹ′ʹ′ = −k ⋅ ⎜⎜ ⎟⎟ ⎝ ∂y ⎠ y =0

⎛ ∂T ⎞ q = −k ⋅W ∫ ⎜⎜ ⎟⎟ ⋅ dx ∂y ⎠ y =0 0 ⎝ Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2: Laminar Boundary Layers

Laminar Boundary Layers: Solving the governing equations with three different methods: -  Order of Magnitude Method (Approximation) -  Approximate Integral Method -  Exact Similarity Method

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2: Laminar Boundary Layers

Governing equations: KContinuity

∂u ∂ v + =0 ∂x ∂y

IMomentum

⎧ ∂u ⎛ ∂ 2u ∂ 2u ⎞ ∂u 1 ∂P +v =− + ν ⋅ ⎜⎜ 2 + 2 ⎟⎟ ⎪u ∂y ρ ∂x ∂y ⎠ ⎪ ∂x ⎝ ∂x ⎨ 2 2 ⎪u ∂v + v ∂v = − 1 ∂P + ν ⋅ ⎛⎜ ∂ v + ∂ v ⎞⎟ ⎜ ∂x 2 ∂y 2 ⎟ ⎪ ∂x ∂ y ρ ∂ y ⎝ ⎠ ⎩

ν =

EEnergy

⎛ ∂ 2T ∂ 2T ⎞ ∂T ∂T u + v ==αa⎜⎜ 2 + 2 ⎟⎟ ∂x ∂y ∂y ⎠ ⎝ ∂x

αa =

Prof. Nico Hotz

µ ρ

λk ρ ⋅ cp

ME 150 – Heat and Mass Transfer

Chap. 12.2: Laminar Boundary Layers

Boundary conditions for laminar boundary layers: y=0

u(0) = 0, v (0) = 0, T (0) = TW

y→∞

u( ∞) = u∞ , T ( ∞) = T∞ , P( ∞) = P∞

In the following, an estimation of phenomena in the boundary layer by analyzing orders of magnitude

Order of Magnitude Solution by Prandtl (1904)

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.1: Velocity Boundary Layers

Velocity Boundary Layer Variables and orders of magnitudes: x : 0 ........ L

→

Δx ≈ L

y : 0 ........ δ

→

Δy ≈ δ

u : 0 ........U ∞

→

Δu ≈ U ∞

v : 0 .........v ?

→

Δv

Approximation of order of magnitude: v ∂u ∂v + =0 ∂x ∂y

→

U ∞ Δv + ≈0 → L δ

Δv = v ≈ U ∞

δ L

Signs are irrelevant ! Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.1: Velocity Boundary Layers

Rearranging of pressure terms (still unknown)

(dP )x

(dP )y

≈ ΔP x

≈ ΔP y

x - momentum equation: ⎛ ∂ 2 u ∂ 2 u ⎞ ∂u ∂u 1 ∂P u +v =− + ν ⋅ ⎜⎜ 2 + 2 ⎟⎟ ∂x ∂y ρ ∂x ⎝ ∂ x ∂ y ⎠

L>>δ

⎛ ⎜ U ∞ U ∞ U∞ δ U ∞ 1 ΔPx U∞ ⋅ + U∞ ⋅ ⋅ ≈ ⋅ + ν ⋅ ⎜ 2 + 2 L L δ ρ L L δ ⎜  ⎝ ≈ 0

Prof. Nico Hotz

⎞ ⎟ ⎟ ⎟ ⎠

ME 150 – Heat and Mass Transfer

Chap. 12.2.1: Velocity Boundary Layers

Significance of terms: 2

U 1 ΔPx U 2 ⋅ ∞ ≈ ⋅ + ν ⋅ ∞2 L ρ L δ ≈1     Reibung Druck Friction Trägheit Inertia Pressure

Similarly for y – momentum equation: 2

2 U ∞ ⋅ δ 1 ΔPy U∞ ≈ ⋅ + ν ⋅ 2 L ρ δ  δ ⋅ L  

U ∞ δ 1 ΔPy U δ ⋅ ≈ ⋅ + ν ⋅ ∞2 ⋅ → L L ρ δ δ L

Inertia

Pressure

Friction

Pressure terms: ∂P ∂P dP = ⋅ dx + ⋅ dy ∂x ∂y

→

ΔPy ΔPx ΔP = ⋅L + ⋅ δ = ΔPx + ΔPy L δ

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.1: Velocity Boundary Layers

Pressure drop is caused by friction forces

→

Pressure and friction terms have the same oder of magnitude 1 ΔPx U ⋅ ≈ν ⋅ ∞2 ρ L δ

ΔPx ≈ L ⋅ ρ ⋅ν ⋅

→

U∞

δ2

= L⋅µ ⋅

U∞

δ2

Similarly for y – component: U 1 ΔPy ⋅ ≈ ν⋅ ∞ ρ δ δ ⋅L

ΔPy ≈ ν ⋅ ρ ⋅

→

U∞ U = µ⋅ ∞ L L

Total pressure difference: L>>δ

U ΔP ≈ µ ⋅ ∞ L

U ⎛ L ⎞ ⎜ ⎟ + µ ⋅ ∞ L ⎝ δ ⎠

Prof. Nico Hotz

ΔP y << ΔP x

ME 150 – Heat and Mass Transfer

Chap. 12.2.1: Velocity Boundary Layers

Conclusion: P = P( x) !

and

Therefore: x – momentum equation simplified:

dP =

∂P ⋅ dx ∂x

P( x) = P∞ ( x)

∂u ∂u 1 dP∞ ∂ 2u u⋅ + v⋅ =− ⋅ +ν⋅ 2 ∂x ∂y ρ dx ∂y

very often: P1(∞) = P2(∞) since: and →

P1(0) = P1(∞) P2(0) = P2(∞) P1(0) = P2(0)

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.1: Velocity Boundary Layers

Summary: Results of “Order of magnitude analysis “ so far: •  Pressure term in x – momentum equation = 0 •  y – momentum equation is negligible Relevant equations for velocity boudary layer: ∂u ∂v + =0 ∂x ∂y

Continuity

∂u ∂u ∂ 2u u⋅ +v⋅ =ν⋅ 2 ∂x ∂y ∂y

x − momentum equation

with boundary conditions: y=0 y→∞

u=v=0 u = U∞

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.2: Temperature Boundary Layers

Temperature Boundary Layer Thickness of velocity and temperature boundary layers are not identical !

(1) Analysis with assumption:

δ T >> δ Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.2: Temperature Boundary Layers

Order of magnitudes for energy equation: x : 0 ........ L

→

Δx ≈ L

y : 0 ........ δ T

→

Δy ≈ δ T

u : 0 ........U ∞

→

Δu ≈ U ∞

→

Δv ≈ U ∞ ⋅

v : 0 ........U ∞ ⋅ T : TW .....T∞

∂T u ∂x

δT L

→

∂T +v ∂y

δT

L ΔT ≈ TW − T∞

⎛ ∂ 2 T ∂ 2 T ⎞ = α ⎜⎜ 2 + 2 ⎟⎟ ∂ y ⎠ ⎝ ∂ x

⎛ ⎞ ⎜ ⎟ ⎜ T − T TW − T∞ ⎛ TW − T∞ ⎟ δ T ⎞ TW − T∞ W ∞ U∞ ⋅ + ⎜U ∞ ⋅ ⎟ ≈ α ⋅ ⎜ + ⎟ 2 2 L L δ L δ ⎝ ⎠ ⎜  ⎟ T    T     in ⎜ Conduction Conduction in ⎟ x − direction y − direction ⎠ ⎝

L >> δT Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.2: Temperature Boundary Layers

Simplified energy equation with boundary conditions: ∂T ∂T ∂ 2T u⋅ +v⋅ =α ⋅ 2 ∂x ∂y ∂y

with:

y =0 :

T = TW

y→ ∞ :

T = T∞

meaning:

δT >> δ

u(δ T ) = U ∞

(2) Analysis with assumption: Temperature boundary layer is smaller than velocity boundary layer

δ T << δ

u(δ T ) < U ∞

Orders of magnitude are different !

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.2: Temperature Boundary Layers

For order of magnitude, we can assume a linear velocity profile U∞

≈

Δu

δT

Δu ≈ U ∞ ⋅

δT δ

Velocity component v from continuity: ∂u ∂v + =0 ∂x ∂y

→

Δu Δv ≈ L δT

→

Prof. Nico Hotz

Δv ≈ U ∞ ⋅

δ T ⎛ δ T ⎞ ⋅ ⎜ ⎟ δ ⎝ L ⎠

ME 150 – Heat and Mass Transfer

Chap. 12.2.2: Temperature Boundary Layers

Energy equation for δT < δ (with orders of magnitude) ⎛ ⎞ ⎜ ⎟ ⎜ T − T TW − T∞ ⎟ δ T ⎞ ⎛ δ T ⎞ TW − T∞ ⎛ δ T ⎞ TW − T∞ ⎛ W ∞ U ∞ ⋅ ⎜ ⎟ ⋅ + ⎜U ∞ ⋅ ⎟ ⋅ ⎜ ⎟ ⋅ ≈ α ⋅ ⎜ + ⎟ 2 2 δ L L δ δ L δT ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎜  ⎟ T        in ⎜ Conduction Conduction in ⎟ x − direction y − direction ⎠ ⎝

additional term An additional term appears in both inertial terms. Summary of all findings: - no effect of pressure - no y-momentum equation - no thermal diffusion in x-direction Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.2: Temperature Boundary Layers

Complete simplified equation system for boundary layer: ∂u ∂v + =0 ∂x ∂y

Continuity

∂u ∂u ∂2 u u⋅ +v⋅ =ν⋅ 2 ∂x ∂y ∂y

x − Momentum

∂T ∂T ∂2 T u⋅ +v⋅ =α ⋅ 2 ∂x ∂y ∂y

Energy

With boundary conditions: y = 0:

u (0) = v(0) = 0

T (0) = TW

y →∞:

u (∞ ) = U ∞

T (∞) = T∞

x=0

u = U∞

T = T∞ for all y

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.3: Surface Friction

Surface friction from order of magnitude solution Momentum equation: ∂u ∂u ∂ 2u u⋅ +v⋅ = ν ⋅ 2 → ∂x ∂y ∂y

U ∞2 ⎛ δ ⎞ U U + ⎜U ∞ ⋅ ⎟ ⋅ ∞ ≈ ν ⋅ ∞2 L ⎝ L ⎠ δ δ

Factor 2 negligible: 2

U U 2 ⋅ ∞ ≈ ν ⋅ ∞2 L δ

→

U∞ 1 ≈ν⋅ 2 L δ

Solving for δ: ⎛ ν ⋅ L ⎞ ⎟⎟ δ ≈ ⎜⎜ ⎝ U ∞ ⎠

1 2

→

δ L

≈ (Re L )

−

1 2

Prof. Nico Hotz

with

Re L =

U∞ ⋅ L ρ ⋅U∞ ⋅ L = ν µ

ME 150 – Heat and Mass Transfer

Chap. 12.2.3: Surface Friction

Wall shear stress from velocity gradient: ⎛ ∂ u ⎞ τ W = µ ⋅ ⎜⎜ ⎟⎟ ⎝ ∂ y ⎠ y =0

→

U∞

1 U∞ τW ≈ µ ⋅ ≈ µ⋅ ⋅ (Re L )2 δ L 1 2

1 U ∞ ⎛ U ∞ ⋅ L ⎞ U ∞ ⋅ ρ 2 − ( ) ≈ µ⋅ ⋅ ⎜ ≈ ρ ⋅U ∞ ⋅ Re L 2 ⎟ ⋅ L ⎝ ν ⎠ U ∞ ⋅ ρ

−

τ W ≈ ρ ⋅ U ∞ ⋅ (Re L )

1 2

Friction coefficient: ρ

τ W ≈ c f ⋅ ⋅U ∞ 2

with

Prof. Nico Hotz

c f = 2 ⋅ (Re L )

−

1 2

ME 150 – Heat and Mass Transfer

Chap. 12.2.3: Surface Friction

Friction force = Wall shear stress integrated over the entire area L

F = ∫ τ W ⋅ W ⋅ dx 0

→

−

F ~ ρ ⋅ U ∞ ⋅ (Re L )

Prof. Nico Hotz

1 2

⋅W ⋅ L

ME 150 – Heat and Mass Transfer

Chap. 12.2.4: Convective Heat Transfer

Heat Transfer from order of magnitude solution (δ << δT) Energy equation: ∂T ∂T ∂2 T u⋅ +v⋅ =α ⋅ 2 ∂x ∂y ∂y

→

U∞

(TW − T∞ ) + U L

∞

⋅

δT L

⋅

(TW − T∞ ) ≈ α ⋅ TW − T∞ δT 2

δT

Simplified and solved for δT/L: U∞ α ≈ 2 L δT

Prandtl Number Pr

⎛ α ⎞ ⎟⎟ ≈ ⎜⎜ L ⎝ U ∞ ⋅ L ⎠

δT

→

Pr =

≈ Re

−1

⎛ α ⎞

⋅ ⎜ ⎟ ⎝ ν ⎠

Peclet Number Pe Pe = Re⋅ Pr

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.4: Convective Heat Transfer

Temperature boundary layer thickness for δ << δT δT −1 −1 −1 2 ≈ Pr ⋅ Re L 2 ≈ Pe 2 L Material property

Flow property

Precondition for δ << δT −1 δ ≈ Re L 2 L

Pr << 1

→

δT −1 ≈ Pr 2 >> 1 δ

True for low viscosity and/or high thermal conductivity, e.g.liquid metals, Hg: Pr < 0.03 at room temperature Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.4: Convective Heat Transfer

Summary of dimensionless parameters:

U ∞ ⋅ L Inertial Forces = ν Viscous Forces

Reynolds Number

Re =

Prandtl Number

Pr =

Peclet Number

Pe = Re⋅ Pr =

Momentum Diffusivity Thermal Diffusivity

Prof. Nico Hotz

U∞ ⋅ L

Mass Transfer Heat Transfer

ME 150 – Heat and Mass Transfer

Chap. 12.2.4: Convective Heat Transfer

Calculation of convective heat transfer coefficient from Fourier‘s Law: ⎛ ∂ T ⎞ ⎟⎟ − k ⋅ ⎜⎜ qʹ′ʹ′ ⎝ ∂ y ⎠ y =0 h= = ΔT TW − T∞

k⋅ →

h≈

TW − T∞

δT

TW − T∞

≈

δT

Using the result for the temperature boundary layer: h ≈

1 1 k ⋅ Pr 2 ⋅ Re L 2 L

Nusselt Number Nu as basis for calculating h: Nu L ≡

1 1 h⋅ L ≈ Pr 2 ⋅ Re L 2 k

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.4: Convective Heat Transfer

Heat flux through a rectangular area (W = width, L = length) 1

q = L ⋅W ⋅ h ⋅ (TW − T∞ ) ≈ W ⋅ (TW − T∞ )⋅ k ⋅ Pr 2 ⋅ Re L 2

Note: Nusselt Number NuL leads to h averaged over 0 …. L: h !

Second case: δ >> δT: Energy equation (with ‘additional terms‘): ∂T ∂T ∂2 T u⋅ +v⋅ =α ⋅ 2 ∂x ∂y ∂y

→ U ∞⋅

T −T δ T (TW − T∞ ) δ δ (T − T∞ ) ⋅ + U ∞⋅ T ⋅ T ⋅ W ≈ α⋅ W 2 ∞ δ L δ L δT δT

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.4: Convective Heat Transfer

Energy equation simplified: U∞ ⋅

δT 1 ≈α⋅ 2 δ ⋅L δT

Leading to:

where

δ T3 3

δ ≈ L ⋅ (Re L −1

≈

α ⋅ Re L

)

L ⋅U ∞

≈

and results in:

δT −1 −1 ≈ Pr 3 ⋅ Re L 2 L

Range of validity:

− δT ≈ Pr 3 << 1 δ

Prof. Nico Hotz

−

1 2

α ν

−3

⋅ Re L

Pr > 1

ME 150 – Heat and Mass Transfer

Chap. 12.2.4: Convective Heat Transfer

Summary for case 2: δ > δT Derivation analogously to case 1: δT L

≈ Pr

−1

⋅ Re L

−1

1 1 k h ≈ ⋅ Pr 3 ⋅ Re L 2 L 1 1 h⋅L 3 Nu ≡ ≈ Pr ⋅ Re L 2 k 1

q ≈ W ⋅ (TW − T∞ )⋅ k ⋅ Pr 3 ⋅ Re L 2

Difference between both cases:

δ < δT

∝ Pr

1 2

δ > δT

Prof. Nico Hotz

∝ Pr

1 3

ME 150 – Heat and Mass Transfer

Chap. 12.2.4: Convective Heat Transfer

Summary for velocity boundary layer (independent of δT) Thickness of velocity boundary layer Wall shear stress:

Friction coefficient:

Friction force:

δ L

−1

≈ Re L

τW ≈ ρ ⋅U ∞

cf ≈

−1 ⋅ Re L 2

−1 Re L 2

F ≈ ρ ⋅U ∞ ⋅W

Prof. Nico Hotz

−1 ⋅ L ⋅ Re L 2

ME 150 – Heat and Mass Transfer

Chap. 12.2: Laminar Boundary Layers

Laminar Boundary Layers: Solving the governing equations with three different methods: -  Order of Magnitude Method (Approximation) -  Approximate Integral Method -  Exact Similarity Method

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.5: Approximate Integral Method

Approximate Integral Method Principle Idea: - Consider boundary layer equations in integral form - Integration over height H > δ and δT - Approximation of profiles within the boundary layer

Results:

Velocity

- determine surface gradients more accurately,

Temperature

- determine h and τ more accurately

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.5: Approximate Integral Method

Derivation of Approximate Integral Method Momentum equation: ∂u ∂u ∂2 u u⋅ +v⋅ =ν⋅ 2 ∂x ∂y ∂y

→

∂ 2 ∂ ∂2 u (u ) + (u ⋅ v) = ν ⋅ 2 ∂x ∂y ∂y

Equations are identical for incompressible fluids: ⎛ ∂ u ∂ v ⎞ ∂ 2 ∂ ∂u ∂u ∂v ∂u ∂u ⎟⎟ (u ) + (u ⋅ v) = 2u ⋅ +v⋅ +u⋅ = u⋅ +v⋅ + u ⋅ ⎜⎜ + ∂x ∂y ∂x ∂y ∂y ∂x ∂y ∂ x ∂ y ⎝⎠ =0 due to continuity

Similarly for energy equation: ∂T ∂T ∂2 T u⋅ +v⋅ =α ⋅ 2 ∂x ∂y ∂y

→

∂ ∂ ∂2 T (u ⋅ T ) + (v ⋅ T ) = α ⋅ 2 ∂x ∂y ∂y

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.5: Approximate Integral Method

Integration over the boundary layer (H >> δ, δT) Momentum equation H

d ∂u y=H 2 ( ) u ⋅ dy + u ⋅ v = ν ⋅ y =0 d x ∫0 ∂y

Energy equation y=H

y =0

d ∂T y=H ( ) u ⋅ T ⋅ dy + v ⋅ T = α ⋅ y =0 d x ∫0 ∂y

y=H

y =0

Using known values at the boundary: H

Momentum equation

d u 2 ⋅ dy + U ∞ ⋅ v y = H ∫ dx0

⎧ ⎫ ⎪⎛ ⎛ ∂ u ⎞ ⎪⎪ ⎪ ∂ u ⎞ ⎟⎟ ⎟⎟ ⎬ − (u ⋅ v ) y = 0 = ν ⋅ ⎨⎜⎜ − ⎜⎜    ⎝ ∂ y ⎠ y = H ⎝ ∂ y ⎠ y = 0 ⎪ ⎪ =0   ⎪⎩ ⎪⎭ =0

⎧ ⎫ ⎪⎛ H ⎛ ∂ T ⎞ ⎪⎪ d ⎪ ∂ T ⎞ Energy equation ⎟⎟ ⎟⎟ ⎬ u ⋅ T ⋅ dy + v y = H ⋅ T∞ − v y = 0 ⋅ TW = α ⋅ ⎨⎜⎜ − ⎜⎜ ∫ dx0    ⎝ ∂ y ⎠ y = 0 ⎪ ⎪⎝∂y⎠ y=H =0  ⎪⎩ ⎪⎭ =0 Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.5: Approximate Integral Method

Leading to simplified equations: Momentum equation

Energy equation

H ⎛ ∂ u ⎞ d 2 ⎜⎜ ⎟⎟ u ⋅ dy + U ⋅ v = − ν ⋅ ∞ ∫ y=H dx 0 ⎝ ∂ y ⎠ y = 0

next step: determining v

H ⎛ ∂ T ⎞ d ⎜⎜ ⎟⎟ u ⋅ T ⋅ dy + T ⋅ v = − α ⋅ ∞ ∫ y=H dx 0 ⎝ ∂ y ⎠ y = 0

y=H

using continuity: ∂u ∂v + =0 ∂x ∂y

∂v ∂u =− ∂y ∂x

→

and integrating over boundary layer: H

d du dv − ∫ u ⋅ dy = − ∫ ⋅ dy = ∫ ⋅ dy = v y = H − v y =0  dx 0 dx dy 0 0 =0

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.5: Approximate Integral Method

Combining integrals:

M E

H H H ⎛ ∂ u ⎞ d d d 2 ⎜⎜ ⎟⎟ u ⋅ dy − U ⋅ u ⋅ dy = u ⋅ ( u − U ) ⋅ dy = − ν ⋅ ∞ ∞ ∫ ∫ ∫ dx 0 dx 0 dx 0 ⎝ ∂ y ⎠ y = 0 H H H ⎛ ∂ T ⎞ d d d ⎜⎜ ⎟⎟ u ⋅ T ⋅ dy − T ⋅ u ⋅ dy = u ⋅ ( T − T ) ⋅ dy = − α ⋅ ∞ ∞ dx ∫0 dx ∫0 dx ∫0 ∂ y ⎝ ⎠ y = 0

Dividing into two parts: inside + outside of boundary layer H

∫ u ⋅ (u −U

δ ∞

) ⋅ dy = ∫ u ⋅ (u − U ∞ ) ⋅ dy + ∫ u ⋅ (u − U ∞ ) ⋅ dy 0 δ   = 0 , since u (δ ) = u ( H ) =U ∞

δT

∫ u ⋅ (T − T

∞

) ⋅ dy = ∫ u ⋅ (T − T∞ ) ⋅ dy + ∫ u ⋅ (T − T∞ ) ⋅ dy 0

δT   = 0 , since T (δ T ) = T ( H ) = T∞

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.5: Approximate Integral Method

Boundary Layer Equations in Integral Form δ ⎛ ∂u ⎞ d u ⋅ (u − U ∞ ) ⋅ dy = −ν ⋅ ⎜⎜ ⎟⎟ ∫ dx 0 ⎝ ∂y ⎠ y = 0 δ

⎛ ∂T ⎞ d T α ⎟⎟ u ⋅ (T − T∞ ) ⋅ dy = − a ⋅ ⎜⎜ ∫ dx 0 ⎝ ∂y ⎠ y = 0

Variables of interest

Integral equations are exact, valid as the differential equations Still unknown: u(y), T(y)

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.5: Approximate Integral Method

Assumption of linear profiles inside the boundary layer u( y) ⎛ y ⎞ = A1 ⋅ ⎜ ⎟ + A2 U∞ ⎝ δ ⎠ T

Boundary conditions to determine constants Ai, Bi Leading to:

⎛ y T ( y) − Tw = B1 ⋅ ⎜⎜ T∞ − Tw ⎝ δ T

y <δ

⎞ ⎟⎟ + B2 ⎠

y = 0:

u (0) = 0

y =δ :

u (δ ) = U ∞

y = δT :

T (δ T ) = T∞

A2 = 0

A1 = 1

B2 = 0

B1 = 1

Prof. Nico Hotz

y < δT

T (0) = TW

ME 150 – Heat and Mass Transfer

Chap. 12.2.5: Approximate Integral Method

Momentum equation: Using assumed linear profile δ

U∞ d y ⎛ y ⎞ U ⋅ ⋅ U − U ⋅ dy = − ν ⋅ ⎜ ⎟ ∞ ∞ dx ∫0 δ ⎝ ∞ δ δ ⎠

Dividing by

U∞2,

substituting with new variable: ζ =

⎧ ⎫ ⎪ 1 ⎪ d ⎪ ν ⎪ ( ) δ ζ ⋅ ζ − 1 ⋅ d ζ = − ⎨ ⎬ dx ⎪ ∫0 U ∞δ   ⎪ ⎪⎩ ⎪⎭ =− 1 6

Integration over x: dδ 6 ⋅ ν δ⋅ = dx U ∞

→

6⋅ν δ ⋅ dδ = ⋅ dx U∞

Prof. Nico Hotz

→

δ (x )2 2

6⋅ν ⋅x U∞

ME 150 – Heat and Mass Transfer

Chap. 12.2.5: Approximate Integral Method

Result: δ = f( x ) δ=

12 ⋅ ν ⋅ x U∞

with Re x =

x ⋅U ∞

δ x

12 −1 = 3.464 ⋅ Re x 2 Re x

Wall shear stress: ⎛ du ⎞ U∞ ρ ⋅ U ∞2 τ w = µ ⋅ ⎜⎜ ⎟⎟ = µ ⋅ = δ (12 ⋅ Re x )1 2 ⎝ dy ⎠ y =0

Wall shear stress coefficient: cf =

τw 1 ⋅ ρ ⋅U ∞ 2 2

= 0.577 ⋅ Re x −1 2

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.5: Approximate Integral Method

Energy equation: Using assumed linear profiles δ

⎞ T∞ − Tw d T y ⎛ y ⎜ ⎟ [ ] [ ] U ⋅ ⋅ T − T ⋅ − T − T ⋅ dy = − α ⋅ ∞ ∞ w ∞ w ⎟ dx ∫0 δ ⎜⎝ δT δT ⎠

Dividing by U∞.(T∞ -TW), using new variable: ζ T =

δT

⎧ ⎫ ⎪ 2 1 ⎪ d ⎪δ T α ⎪ ζ T ⋅(ζ T − 1)⋅ dζ T ⎬ = − ⎨ ∫ dx ⎪ δ 0 U ∞ ⋅ δT    ⎪ ⎪⎩ ⎪⎭ =− 1 6

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.5: Approximate Integral Method

Integration over x: Assumption: δT/δ = constant d ⎛ δ δ T ⎜⎜ T dx ⎝ δ

⎞ ⎛ δ T ⎞ d δ 1 d 2 6 ⋅α ⎟ = ⎜ ⎟ ⋅ δ T ⋅ (δ T ) = ⎛⎜ T ⎞⎟ ⋅ ⋅ δ = T ⎟ ⎝ δ ⎠ dx U∞ ⎝ δ ⎠ 2 dx ⎠ δ T 2 12 ⋅ α ⋅ x → ⋅δ T = δ U∞

( )

Using earlier result for δ: δ T3 12 ⋅ a ⋅ x 1 a = ⋅ = δ3 U∞ δ2  ν 1

→

−1 ⎛ δ T ⎞ ⎜ ⎟ = Pr 3 ⎝ δ ⎠

Result for δT : δT −1 −1 δ −1 3 2 = ⋅ Pr = 3.464 ⋅ Re x ⋅ Pr 3 x x Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.5: Approximate Integral Method

Heat flux from Fourier‘s Law: qWʹ′ʹ′ = − k ⋅

Using solution of δT :

dT dy

= −k⋅

TW − T∞

δT

y =0

1 1 k 2 qʹ′wʹ′ = ⋅ 0.289 ⋅ Re x ⋅ Pr 3 ⋅ (Tw − T∞ ) x

Calculating convective heat transfer coefficient: q ʹ′wʹ′ = h ⋅ (Tw − T∞ )

→

1 1 q ʹ′wʹ′ k = ⋅ 0.289 ⋅ Re x 2 ⋅ Pr 3 Tw − T∞ x

Dimensionless local Nusselt number Nu(x): Nu =

α ⋅x k

= 0.289 ⋅ Re x ⋅ Pr 2

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2: Laminar Boundary Layers

Laminar Boundary Layers: Solving the governing equations with three different methods: -  Order of Magnitude Method (Approximation) -  Approximate Integral Method -  Exact Similarity Method

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.6: Exact Solution

Exakt Solution using Similarity Solution (according to Blasius)

Governing equations:

∂u ∂v + =0 ∂x ∂y

Continuity

∂u ∂u ∂2 u u⋅ +v⋅ =ν⋅ 2 ∂x ∂y ∂y

x − Momentum

∂T ∂T ∂2 T u⋅ +v⋅ =α ⋅ 2 ∂x ∂y ∂y

Energy

Analytical solution exists for potential flows, i.e. a potential function Ψ exists with: u=

∂ψ ∂y

Prof. Nico Hotz

v=−

∂ψ ∂x

ME 150 – Heat and Mass Transfer

Chap. 12.2.6: Exact Solution

Similarity variable to transform governing equations to ODEs: 1 U∞ y η = y⋅ = ⋅ Re x 2 ν⋅x x

with Re x =

x ⋅U ∞ ν

Instead of potential function, we use a function f(η): u = U ∞

Correlation to potential function Ψ:

df dη

∂ Ψ ∂ Ψ ∂η ∂f = ⋅ = U∞ ∂y ∂η ∂ y ∂η

∂η ⋅ Ψ = U ∞ ⋅ f (η ) ∂y f (η ) =

Prof. Nico Hotz

1 ⋅Ψ U∞ ⋅ν ⋅ x

ME 150 – Heat and Mass Transfer

Chap. 12.2.6: Exact Solution

Calculation of v from the definition of potential flow: ∂Ψ ∂ v=− =− ∂x ∂x

(

⎛ ⎞ ∂ f (η ) 1 U ∞ ⋅ ν ⎜ U ∞ ⋅ ν ⋅ x ⋅ f (η ) = − ⎜ U ∞ ⋅ ν ⋅ x ⋅ + ⋅ f (η ) ⎟⎟ ∂x 2 x ⎝ ⎠

)

Derivations: ∂ f (η ) ∂ f (η ) ∂η = ⋅ ∂x ∂η ∂ x

∂η y U∞ η =− =− ∂x 2⋅ x ν ⋅ x 2⋅ x

v as a function of f(η) and η: v=

1 ν ⋅U ∞ 2 x

⎛ ∂ f (η ) ⎞ ⋅ ⎜⎜η ⋅ − f (η )⎟⎟ ∂η ⎝ ⎠

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.6: Exact Solution

Derivations of u necessary for momentum equation: ∂u U∞ ∂2 f =− ⋅η ⋅ 2 ∂x 2⋅ x ∂η

df u = U∞ dη

∂u U∞ ∂2 f = U∞ ⋅ ⋅ ∂y ν ⋅ x ∂η 2 U ∞2 ∂ 3 f ∂ 2u = ⋅ 3 2 ∂y ν ⋅ x ∂η

Momentum equation with similarity variable: ODE

∂3 f ∂2 f 2⋅ 3 + f ⋅ 2 = 0 ∂η ∂η

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.6: Exact Solution

Boundary conditions for similarity variable: y = 0:

η∝ y

y →∞

η=0 η →∞

Boundary conditions for u: u( x,0) = v( x,0) = 0

u( x, ∞) = U ∞

Boundary conditions for potential function f(η) : u U∞

= y =0

df =0 dη η =0

Prof. Nico Hotz

u U∞

= y =∞

df =1 dη η =∞

ME 150 – Heat and Mass Transfer

Chap. 12.2.6: Exact Solution

Numerical solution (using series expansion or numerical integration) u η=y ∞ νx 0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0 4.4 4.8 5.2 5.6 6.0 6.4 6.8

4.92

df u = dη u∞

d2 f dη 2

0 0.027 0.106 0.238 0.420 0.650 0.922 1.231 1.569 1.930 2.306 2.692 3.085 3.482 3.880 4.280 4.679 5.079

0 0.133 0.265 0.394 0.517 0.630 0.729 0.812 0.876 0.923 0.956 0.976 0.988 0.994 0.997 0.999 1.000 1.000

0.332 0.331 0.327 0.317 0.297 0.267 0.228 0.184 0.139 0.098 0.064 0.039 0.022 0.011 0.005 0.002 0.001 0.000

Prof. Nico Hotz

0.99

See later!

ME 150 – Heat and Mass Transfer

Chap. 12.2.6: Exact Solution

Determining boundary layer thickness δ: u df = = 0.99 U∞ dη

→

η99 = 4.92 =

δ x

6 η = 4.92

U∞

= 0.99

⋅ Re x 2

y x

η = ⋅ Re x 2

Back in x – coordinates: δ = 4.92 ⋅

−1 x ⋅ Re x 2

δ Compared with ≈ Re −L1 2 L ‘Order of Magnitude‘ and δ −1 ‘Integral‘ = 3.464 ⋅ Re x 2 x Solutions: Prof. Nico Hotz

0 0.5

1.0

df = U ∞ dη

ME 150 – Heat and Mass Transfer

Chap. 12.2.6: Exact Solution

Wall shear stress coefficient: local value ⎛ ∂ u ⎞ ⎟⎟ ν ⋅ ρ ⋅ ⎜⎜ ⎝ ∂ y ⎠ y =0 2 ⋅ ν U ∞ ⎛ d 2 f ⎞ −1 ⎜ ⎟ c fx = = ⋅ ⋅ ⎜ 2 ⎟ = 0.664 ⋅ Re x 2 1 U∞ ν ⋅ x ⎝ dη ⎠η =0 ρ ⋅U ∞2   2 = 0.332 from table

Average over entire length L: L

1 1 ν −1 c fL = ⋅ ∫ c fx ⋅ dx = 0.664 ⋅ ⋅ ∫ ⋅ dx = 1.328 ⋅ Re L 2 L 0 L 0 U∞ ⋅ x

Total force on surface: FW = c fL ⋅

ρ 2

⋅ U ∞2 ⋅ W ⋅ L = τ L ⋅ W ⋅ L

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Chap. 12.2.6: Exact Solution

Temperature boundary layer Definition excess temperature: Equation to be solved (numerically): Boundary conditions:

θ= d 2θ dη 2

T − TW T∞ − TW +

Pr =

ν a

Pr dθ ⋅f⋅ = 0 2 dη

η = 0: θ = 0 η → ∞: θ = 1

Results (without derivation): 12

1 ⎛ dθ ⎞ ⎛ dθ ⎞ qWʹ′ʹ′ k ⎛ U ∞ ⎞ ⎛ dθ ⎞ ⎟⎟ = ⋅ Re x 2 ⋅ ⎜⎜ ⎟⎟ h= = k ⋅ ⎜⎜ ⎟⎟ = k ⋅ ⎜ ⎟ ⋅ ⎜⎜ TW − T∞ dy ν ⋅ x d η x d η ⎝ ⎠ ⎝ ⎠η =0 ⎝ ⎠η =0 ⎝ ⎠ y =0

Prof. Nico Hotz

ME 150 – Heat and Mass Transfer

Results from numerical solution:

Chap. 12.2.6: Exact Solution

For Pr ≥ 0.6 :

⎛ dθ ⎞ ⎜⎜ ⎟⎟ = 0.332 ⋅ Pr1 3 ⎝ dη ⎠η =0

Local h:

1 k h = ⋅ Re x 2 ⋅ 0.332 ⋅ Pr 1 3 x

Pr ≥ 0.6

Nusselt number:

1 1 h⋅ x 2 Nu x = = 0.332 ⋅ Re x ⋅ Pr 3 k

Pr ≥ 0.6

Averaged over the range x = 0 ….. L L

1 1 ⎛ U ⎞ h = ⋅ ∫ h( x) ⋅ dx = ⋅ ∫ k ⋅ ⎜ ∞ ⎟ ⋅ 0.332 ⋅ Pr 1 3 ⋅ dx = 2 ⋅ h( L) L 0 L 0 ⎝ ν ⋅ x ⎠

Prof. Nico Hotz

Pr ≥ 0.6

ME 150 – Heat and Mass Transfer

Chap. 12.2.6: Exact Solution

Nusselt number averaged over range x = 0 …. L Nu =

h ⋅L = 2 ⋅ Nu L = 0.664 ⋅ Re1L 2 ⋅ Pr1 3 k

Pr ≥ 0.6

Compared with ‘Order of Magnitude‘ and ‘Integral‘ Solutions: Nu ≈ Re1L 2 ⋅ Pr 1 3

Nu x = 0.289 ⋅ Re x 2 ⋅ Pr

Numerical solution for small Prandtl numbers: Nu x = 0.565 ⋅ (Re x ⋅ Pr)

Nu = 2 ⋅ Nu L

Prof. Nico Hotz

Pr ≤ 0.6

ME 150 – Heat and Mass Transfer

Chap. 12.2.6: Exact Solution

Semi-empirical solution (curve fitting) for any Prandtl number (Churchill and Ozoe 1973): 1

Pe x = Re x ⋅ Pr > 100 : →

Nu x =

0.3387 ⋅ Re x ⋅ Pr

(

⎡ 0.0468 ⎢⎣1 + Pr

)

⎤ ⎥⎦

3 1

Averaged values: 1

h( x ) ≈

Re x 2 x

≈ x

−1

→ h = 2 ⋅ h( L )

Prof. Nico Hotz

Nu = 2 ⋅ Nu L

ME 150 – Heat and Mass Transfer

Chap. 12.2.6: Exact Solution

Summary Results: Exact solution δ

Boundary layer thickness:

Friction coefficient:

Heat transfer

c fx =

= 4.92 ⋅ Re

−1 2 x

−1 0.664 ⋅ Re x 2

1 δ = Pr 3 δT

−1 c f x = 1.328 ⋅ Re x 2

Pr > 0.6

Nu x =

1 1 h⋅ x = 0.332 ⋅ Re x 2 ⋅ Pr 3 k

Pr < 0.6

Nu x =

1 h⋅ x = 0.565 ⋅ (Re x ⋅ Pr) 2 k

Pe x = Re x ⋅ Pr > 100

Prof. Nico Hotz

Nu =

0.3387 ⋅ Re x

(

1 2 ⋅ Pr 3

⎡ 0.0468 ⎢1 + Pr ⎣

)

3 ⎤

⎥ ⎦ 56

ME 150 â&#x20AC;&#x201C; Heat and Mass Transfer

Prof. Nico Hotz