ME 150 â€“ Heat and Mass Transfer

Chap. 12: Introduction to Convection

Principles of Convection Convective heat transfer = Heat transfer between a fluid and a surface in contact with the fluid flow

Simple case: flat plate, 1-dimensional

General case: arbitrary body, 3-dimensional

Prof. Nico Hotz

1

ME 150 – Heat and Mass Transfer

Chap. 12: Introduction to Convection

General description of convection: Goal: Calculation of convective heat transfer coefficient h [W/m2.K]

qʹ′ʹ′ = h ⋅ (T0 − T∞ )

Velocity and temperature of the fluid depend on the position: q = ∫ qʹ′ʹ′ ⋅ dA0 = (T0 − T∞ ) ⋅ ∫ h ⋅ dA0 A0

A0

Practical assumption for some problems: local h can be replaced by average h : q = h ⋅ A0 ⋅ (T0 − T∞ )

General body:

1 h= h ⋅ dA0 ∫ A0 A0 Prof. Nico Hotz

Flat plate (1D):

L

1 h = ⋅ ∫ h( x) ⋅ dx L 0 2

ME 150 – Heat and Mass Transfer

Chap. 12.1: Conservation Equations

Conservation Equations for flowing fluid:

Fluid motion is descriped by 5 variables:

U = (U x , U y , U z ) = (u, v, w)

ρ T

5 equations necessary: Continuity (mass conservation) = 1 equation Momentum conservation = 3 equations Energy conservation = 1 equation We will consider 2-dimensional problems, therefore 4 equations necessary.

Prof. Nico Hotz

3

ME 150 – Heat and Mass Transfer

Chap. 12.1.1: Mass Conservation

Continuity Equation – Steady-state Considering an infinitesimal volume element

m = Density ⋅Velocity ⋅ Area Prof. Nico Hotz

4

ME 150 – Heat and Mass Transfer

Entering mass flows: (in the following: dz = 1) Leaving mass flows:

Chap. 12.1.1: Mass Conservation

m in = ρ ⋅ u ⋅ dy ⋅ dz + ρ ⋅ v ⋅ dx ⋅ dz

⎡ ⎤ ∂ ∂ ⎡ ⎤ mout = ⎢ ρ ⋅ u + ( ρ ⋅ u ) ⋅ dx ⎥ dy + ⎢ ρ ⋅ v + ( ρ ⋅ v) ⋅ dy ⎥ dx ∂x ∂y ⎣ ⎦ ⎣ ⎦

Mass conservation: Difference between in and out = 0: ∂ (ρ ⋅ u )⋅ dx ⋅ dy + ∂ (ρ ⋅ v )⋅ dx ⋅ dy = 0 ∂x ∂y

General form in vector notation:

div( ρ ⋅u ) = 0

→

∂ ( ρ ⋅ u ) ∂ ( ρ ⋅ v) + =0 ∂x ∂y

Simplification for incompressible liquid (ρ = const.)

∂u ∂v + =0 ∂x ∂y Prof. Nico Hotz

5

ME 150 – Heat and Mass Transfer

Chap. 12.1.2: Momentum Conservation

Momentum Conservation (steady-state) Consider momentum fluxes through stationary control volume Example: x-direction, 2-D (dz = 1)

x-momentum flux in x-direction

ρ ⋅ u ⋅ dy ⋅ dz ⋅ u Velocity

x-momentum flux in y-direction

( ρ ⋅ v ⋅ dx ⋅ dz ) ⋅ u

Prof. Nico Hotz

Mass flow

6

ME 150 – Heat and Mass Transfer

Chap. 12.1.2: Momentum Conservation

Entering momentum fluxes:

(ρ ⋅ u)⋅ u ⋅ dy + (ρ ⋅ u)⋅ v ⋅ dx

Leaving momentum fluxes:

(ρ ⋅ u )⋅ u ⋅ dy +

∂ ((ρ ⋅ u )⋅ u ⋅ dy )⋅ dx + (ρ ⋅ u )⋅ v ⋅ dx + ∂ ((ρ ⋅ u )⋅ v ⋅ dx )⋅ dy ∂x ∂y

Difference between in and out = Change in momentum ∂ [(ρ ⋅ u )⋅ u ⋅ dy ]⋅ dx + ∂ [(ρ ⋅ u )⋅ v ⋅ dx ]⋅ dy = ⎡⎢ ∂ ( ρ ⋅ u ⋅ u ) + ∂ ( ρ ⋅ u ⋅ v)⎤⎥ ⋅ dx ⋅ dy ∂x ∂y ∂y ⎣ ∂x ⎦ ⎡ ∂ ( ρ ⋅ u ) ∂u ∂ ( ρ ⋅ v) ∂u ⎤ = ⎢u + ρ ⋅u +u + ρ ⋅ v ⎥ ⋅ dx ⋅ dy ∂x ∂x ∂y ∂y ⎦ ⎣ ⎧ ⎫ ⎪ ⎪ ⎡ ∂ ( ρ ⋅ u ) ∂ ( ρ ⋅ v) ⎤ ⎪ ∂u ∂u ⎪ = ⎨ ρ ⋅ u + ρ ⋅ v + u ⋅ ⎢ + ⎥ ⎬ ⋅ dx ⋅ dy ∂ x ∂ y ∂ x ∂ y ⎣ ⎪ ⎦ ⎪ ⎪ ⎪ =0 Continuity equation ⎩ ⎭ Prof. Nico Hotz

7

ME 150 – Heat and Mass Transfer

Chap. 12.1.2: Momentum Conservation

Rate of change in momentum in the control volume: ⎛ ∂u ∂u ⎞ ρ ⋅ ⎜⎜ u + v ⎟⎟ ⋅ dx ⋅ dy ∂y ⎠ ⎝ ∂x

Newton‘s Second Law: Rate of change in momentum = Force

Possible forces: Surface forces: • Viscous normal stress • Normal stress from pressure • Shear stress Volume forces (e.g. gravitation)

Prof. Nico Hotz

8

ME 150 – Heat and Mass Transfer

Chap. 12.1.2: Momentum Conservation

Net force = Difference between opposing areas Oberflächenkräfte y

[p

dy

p

xx

p

yy

+

∂ (p )dy]dx ∂y yy

dy

xy

[p

yx

ρ g dxdy1

+

∂ ( p )dy]dx ∂y yx

x

[p

dy p

yx

dx

p

xy

+

∂ (p )dx ]dy ∂x xy

xx

+

∂ (p )dx]dy ∂x xx

[p

dx yy

dx

Terminology:

x

pxx

First index: Area (perpendicular to direction) Second index: Direction of force

Prof. Nico Hotz

pxy

pyx 9

ME 150 – Heat and Mass Transfer

Chap. 12.1.2: Momentum Conservation

Net force in x - direction: ⎡ ⎤ ⎡ ⎤ ∂ ∂ p + ( p ) dx ⋅ dy + p + ( p ) dy ⎢ xx ∂ x xx ⎥ ⎢ yx ∂ y yx ⎥⋅ dx − p xx ⋅ dy − p yx ⋅ dx ⎣ ⎦ ⎣ ⎦ =

∂ p yx ∂ p xx ⋅ dx ⋅ dy + ⋅ dx ⋅ dy ∂x ∂y

From Fluid Dynamics (without derivation): normal: external pressure + viscous

⎛ ∂u ∂v ⎞ 2 ∂u pxx = − P + σ x = − P − ⋅ µ ⋅ ⎜⎜ + ⎟⎟ + 2 ⋅ µ ⋅ 3 ∂x ⎝ ∂x ∂y ⎠

tangential: viscous

⎛ ∂v ∂u ⎞ p yx = τ yx = τ xy = µ ⎜⎜ + ⎟⎟ ⎝ ∂x ∂y ⎠ Prof. Nico Hotz

10

ME 150 – Heat and Mass Transfer

Chap. 12.1.2: Momentum Conservation

Sum of surface forces (x - direction): ⎧⎪ ∂P ∂ ⎡ 2 ⎛ ∂u ∂v ⎞ ∂u ⎤ ∂ ⎡ ⎛ ∂v ∂u ⎞⎤ ⎫⎪ + ⎢− µ ⋅ ⎜⎜ + ⎟⎟ + 2 µ ⋅ ⎥ + ⎢ µ ⋅ ⎜⎜ + ⎟⎟⎥ ⎬ ⋅ dx ⋅ dy ⎨− ∂x ⎦ ∂y ⎣ ⎝ ∂x ∂y ⎠⎦ ⎪⎭ ⎪⎩ ∂x ∂x ⎣ 3 ⎝ ∂x ∂y ⎠

Volume force: gravity

ρ ⋅ g x ⋅ dx ⋅ dy

Momentum Conservation (x-direction): Navier Stokes ⎡ ∂P ∂ ⎧ 2 ⎛ ∂u ∂v ⎞ ⎡ ∂u ∂u ⎤ ∂u ⎫ ∂ ⎧ ⎛ ∂v ∂u ⎞⎫⎤ ρ ⋅ ⎢u + v ⎥ = ⎢− + ⎨− µ ⋅ ⎜⎜ + ⎟⎟ + 2 ⋅ µ ⎬ + ⎨µ ⋅ ⎜⎜ + ⎟⎟⎬⎥ + ρ ⋅ g x ∂ x ∂ y ∂ x ∂x ⎩ 3 ⎝ ∂x ∂y ⎠ ∂x ⎭ ∂y ⎩ ⎝ ∂x ∂y ⎠⎭⎦⎥ ⎢⎣ ⎣ ⎦

Momentum change

Normal viscous stress Pressure

Shear stress Gravity

Prof. Nico Hotz

11

ME 150 – Heat and Mass Transfer

for incompressible fluids:

Chap. 12.1.2: Momentum Conservation

∂u ∂v + =0 ∂x ∂y

⎛ ∂u ∂v ⎞ 2 − µ ⋅ ⎜⎜ + ⎟⎟ 3 x ∂y ⎠ ⎝∂

=0 Continuity equation

Rearranging the right-hand side of Navier Stokes: ∂P ∂ 2u ∂ ⎛ ∂v ∂u ⎞ ⎟⎟ + ρ ⋅ g x − + 2 ⋅ µ ⋅ 2 + µ ⋅ ⎜⎜ + ∂x ∂x ∂y ⎝ ∂x ∂y ⎠ ∂P ∂ 2u ∂ 2u ∂ 2v ∂ 2u = − + µ⋅ 2 + µ⋅ 2 +µ⋅ + µ ⋅ 2 + ρ ⋅ gx ∂x ∂x ∂x ∂x ⋅ ∂y ∂y ∂P ∂ 2u ∂ = − + µ⋅ 2 +µ⋅ ∂x ∂x ∂x

⎛ ∂u ∂v ⎞ ∂ 2u ⎜⎜ ⎟⎟ + µ ⋅ 2 + ρ ⋅ g x + x ∂y ⎠ ∂y ⎝∂

=0 Continuity equation

Prof. Nico Hotz

12

ME 150 – Heat and Mass Transfer

Chap. 12.1.2: Momentum Conservation

Navier Stokes in x - direction (incompressible fluid): ⎛ ∂ 2 u ∂ 2 u ⎞ ⎛ ∂u ∂u ⎞ ∂P ρ ⋅ ⎜⎜ u + v ⎟⎟ = − + µ ⋅ ⎜⎜ 2 + 2 ⎟⎟ + ρ ⋅ g x ∂y ⎠ ∂x ∂y ⎠ ⎝ ∂x ⎝ ∂x

and in y - direction, respectively: ⎛ ∂ 2 v ∂ 2 v ⎞ ⎛ ∂v ∂v ⎞ ∂P ρ ⋅ ⎜⎜ u + v ⎟⎟ = − + µ ⋅ ⎜⎜ 2 + 2 ⎟⎟ + ρ ⋅ g y ∂y ⎠ ∂y ∂y ⎠ ⎝ ∂x ⎝ ∂x

In vector notation:

2 ρ ⋅ U • ∇ U = − ∇P + µ ⋅ ∇ U + ρ ⋅ g

(

)

2-D vector operators: U = i ⋅u + j ⋅v

∂ ∂ ∇ = i ⋅ + j⋅ ∂x ∂y

Prof. Nico Hotz

2 ∂2 ∂2 ∇ = 2+ 2 ∂x ∂y 13

ME 150 – Heat and Mass Transfer

Chap. 12.1.3: Energy Conservation

Energy Conservation (steady-state) We have to condsider: • Convective heat transfer • Thermal conduction • Internal heat sources • Work due to friction and volume forces We can neglect: • Kinetic energy • Potential energy Applicable for problems with: low Mach numbers, Δh < 1000 m Prof. Nico Hotz

14

ME 150 â€“ Heat and Mass Transfer

Chap. 12.1.3: Energy Conservation

Components of Heat Transfer: D = Conduction V = Convection

Prof. Nico Hotz

15

ME 150 – Heat and Mass Transfer

Chap. 12.1.3: Energy Conservation

Balance for convection: e = specific internal energy (J/kg) EV ,tot = (EV , x − EV , x + dx ) + (EV , y − EV , y + dy ) = ⎡ ⎤ ∂ = (ρ ⋅ e ⋅ u )⋅ dy − ⎢(ρ ⋅ e ⋅ u )⋅ dy + ( ρ ⋅ e ⋅ u ) ⋅ dy ⋅ dx ⎥ + ∂x ⎣ ⎦ ⎡ ⎤ ∂ + (ρ ⋅ e ⋅ v )⋅ dx − ⎢(ρ ⋅ e ⋅ v )⋅ dx + ( ρ ⋅ e ⋅ v) ⋅ dx ⋅ dy ⎥ = ∂y ⎣ ⎦ ⎡ ∂ ⎤ ∂ = − ⎢ ( ρ ⋅ e ⋅ u ) ⋅ dx ⋅ dy + ( ρ ⋅ e ⋅ v) ⋅ dx ⋅ dy ⎥ ∂y ⎣ ∂ x ⎦

Prof. Nico Hotz

16

ME 150 – Heat and Mass Transfer

Chap. 12.1.3: Energy Conservation

Balance for conduction: E D ,tot = (E D , x − E D , x + dx )+ (E D , y − E D , y + dy ) = ⎛ ⎞ ⎛ ⎞ ∂ ∂ = E D , x − ⎜⎜ E D , x + ED , x ⋅ dx ⎟⎟ + E D , y − ⎜⎜ E D , y + ED , y ⋅ dy ⎟⎟ = ∂x ∂y ⎝ ⎠ ⎝ ⎠ ⎡ ∂ ⎤ ∂ = − ⎢ E D , x ⋅ dx + ED , y ⋅ dy ⎥ ∂y ⎣ ∂ x ⎦

Using Fourier‘s Law: ∂ ⎧ ∂T ⎫ ∂ ⎧ ∂T ⎫ E D ,tot = − ⎨− k ⋅ dy ⋅ ⎬ ⋅ dx − ⎨− k ⋅ dx ⋅ ⎬ ⋅ dy = ∂x ⎩ ∂x ⎭ ∂y ⎩ ∂y ⎭ ⎡ ∂ ⎛ ∂T ⎞ ∂ ⎛ ∂T ⎞⎤ ⎟⎟⎥ ⋅ dx ⋅ dy = ⎢ ⎜ k ⋅ ⎟ + ⎜⎜ k ⋅ ⎣ ∂x ⎝ ∂x ⎠ ∂y ⎝ ∂y ⎠⎦

Prof. Nico Hotz

17

ME 150 â€“ Heat and Mass Transfer

Chap. 12.1.3: Energy Conservation

Rate of work due to friction (surface forces) and volume forces:

Rate of work = Power = Force â—? Velocity

total of 10 force components for a 2D-volume element Prof. Nico Hotz

18

ME 150 – Heat and Mass Transfer

Chap. 12.1.3: Energy Conservation

Rate of work of these 10 force components: ∂ ∂ Wtot = (u ⋅ p xx ⋅ dy ) ⋅ dx + (v ⋅ p yy ⋅ dx ) ⋅ dy + ∂x ∂y ∂ ∂ + (u ⋅ p yx ⋅ dx ) ⋅ dy + (v ⋅ p xy ⋅ dy ) ⋅ dx + fas ∂y ∂x + u ⋅ ρ ⋅ g x ⋅ dx ⋅ dy + v ⋅ ρ ⋅ g y ⋅ dx ⋅ dy

Substitute for the surface forces pressure and viscous components:

p xx = − P + σ x p yy = − P + σ y p xy = p yx = τ xy = τ yx

Prof. Nico Hotz

19

ME 150 – Heat and Mass Transfer

Chap. 12.1.3: Energy Conservation

Using this substitution: ⎧ ∂ ⎫ ∂ ∂ Wtot = ⎨ (u ⋅ σ x ) − (u ⋅ P) + (u ⋅τ xy ) + u ⋅ ρ ⋅ g x ⎬ ⋅ dx ⋅ dy + ∂x ∂y ⎩ ∂x ⎭ ⎧ ∂ ⎫ ∂ ∂ + ⎨ (v ⋅ σ y ) − (v ⋅ P) + (v ⋅τ xy ) + v ⋅ ρ ⋅ g y ⎬ ⋅ dx ⋅ dy ∂y ∂x ⎩ ∂y ⎭

Condition for steady-state energy balance:

0 = E D,tot + EV ,tot + Wtot + qs ⋅ dV

Prof. Nico Hotz

20

ME 150 – Heat and Mass Transfer

Chap. 12.1.3: Energy Conservation

Using the individual components: ⎡ ∂ ⎤ ∂ 0 = − ⎢ (ρ ⋅ e ⋅ u ) + (ρ ⋅ e ⋅ v )⎥ + ∂y ⎣ ∂x ⎦

Convection

⎡ ∂ ⎛ ∂T ⎞ ∂ ⎛ ∂T ⎞⎤ ⎟⎟⎥ + + ⎢ ⎜ k ⋅ ⎟ + ⎜⎜ k ⋅ Conduction ∂ x ∂ x ∂ y ∂ y ⎠ ⎝ ⎠⎦ ⎣ ⎝ ∂ ∂ ⎡ ∂ ⎤ + ⎢ (u ⋅ σ x ) − (u ⋅ P ) + (u ⋅τ xy ) + u ⋅ ρ ⋅ g x ⎥ + Work due to ∂x ∂x ⎣ ∂x ⎦ ⎡ ∂ ⎤ ∂ ∂ + ⎢ (v ⋅ σ y ) − (v ⋅ P ) + (v ⋅τ xy ) + v ⋅ ρ ⋅ g y ⎥ + ∂y ∂x ⎣ ∂y ⎦ + q s

Prof. Nico Hotz

forces

Sources

21

ME 150 – Heat and Mass Transfer

Chap. 12.1.3: Energy Conservation

Next steps: - Use explicit expressions for σ und τ - Neglect volume force - Combine friction term to Φ ρ ⋅u ⋅

⎛ ∂u ∂v ⎞ ∂e ∂e ∂ ⎛ ∂T ⎞ ∂ ⎛ ∂T ⎞ + ρ ⋅ v ⋅ = ⎜ k ⋅ ⎟ + ⎜⎜ k ⋅ ⎟⎟ − P ⋅ ⎜⎜ + ⎟⎟ + µ ⋅ Φ + q s ∂x ∂y ∂x ⎝ ∂x ⎠ ∂y ⎝ ∂y ⎠ ⎝ ∂x ∂y ⎠

Convection

Conduction

Pressure

Sources

Φ = Effect of viscous friction 2 ⎡⎛ ∂u ⎞ 2 ⎛ ∂v ⎞ 2 ⎤ 2 ⎛ ∂u ∂v ⎞ 2 ⎛ ∂u ∂v ⎞ Φ = ⎜⎜ + ⎟⎟ + 2 ⋅ ⎢⎜ ⎟ + ⎜⎜ ⎟⎟ ⎥ − ⋅ ⎜⎜ + ⎟⎟ ⎢⎣⎝ ∂x ⎠ ⎝ ∂y ⎠ ⎥⎦ 3 ⎝ ∂x ∂y ⎠ ⎝ ∂y ∂x ⎠

Prof. Nico Hotz

22

ME 150 – Heat and Mass Transfer

Chap. 12.1.3: Energy Conservation

More simplifications: Fluid is incompressible

ρ = constant

∂u ∂v + =0 ∂x ∂y

Continuity

Pressure term = 0

Energy purely thermal: KE = PE = 0 de = c p ⋅ dT = cv ⋅ dT = c ⋅ dT

Energy conservation in terms of temperature: ⎛ ∂T ∂T ⎞ ∂ ⎛ ∂T ⎞ ∂ ⎛ ∂T ⎞ ⎟⎟ = ⎜ k ⎟⎟ + µ ⋅ Φ + q s +v ⎟ + ⎜⎜ k ∂y ⎠ ∂x ⎝ ∂x ⎠ ∂y ⎝ ∂y ⎠ ⎝ ∂x

ρ ⋅ c ⋅ ⎜⎜ u

2

with simplified Φ

⎛ ∂u ∂v ⎞ ⎛ ∂u ∂v ⎞ Φ = ⎜⎜ + ⎟⎟ − 4 ⋅ ⎜⎜ ⋅ ⎟⎟ ⎝ ∂y ∂x ⎠ ⎝ ∂x ∂y ⎠

Prof. Nico Hotz

23

ME 150 – Heat and Mass Transfer

Chap. 12.1.3: Energy Conservation

Neglecting friction (Φ = 0) and taking k = constant: Heat Transfer Equation for laminar, incompressible flows without friction and with constant thermal conductivity: ⎛ ∂ 2T ∂ 2T ⎞ ⎛ ∂T ∂T ⎞ ⎜⎜ u ⋅ ⎟⎟ = α ⋅ ⎜⎜ 2 + 2 ⎟⎟ +v⋅ ∂y ⎠ ∂y ⎠ ⎝ ∂x ⎝ ∂x

α=

k ρ ⋅ cp

Using vector operators:

2 U • ∇T = α ⋅ ∇ T

U = u ⋅i + v⋅ j

∂T ∂T ∇T = i +j ∂x ∂y

Prof. Nico Hotz

24

ME 150 – Heat and Mass Transfer

Chap. 12.1.3: Energy Conservation

Remark on transient problems: For transient problems, an additional term is needed

Transient change of energy content of a control volume:

Prof. Nico Hotz

ρ ⋅ c p ⋅ ∂T dt

25

ME 150 â€“ Heat and Mass Transfer

Prof. Nico Hotz

26

Advertisement