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Convective Heat Transfer (6) Forced Convection (8) Martin Andersson

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Convective Heat Transfer (6)Forced Convection (8)Martin Andersson

Agenda

• Convective heat transfer• Continuity eq.• Convective duct flow (introduction to ch. 8)

Convective heat transfer

Convective heat transfer

Convective heat transfer

AttQ wf )( −=α!

värmeledande kropptf

y

x

tw(y)

t(x,y)fluid

Heat conducting body

x = 0 ⇒ u, v, w = 0 ⇒ heat conduction in the fluid

AttQ wf )( −=α!

!

0/ (6 3)f

x

w f w f

tyQ A

t t t t

λ

α =

⎛ ⎞∂⎜ ⎟∂⎝ ⎠= = − −

− −

&

värmeledande kropptf

y

x

tw(y)

t(x,y)fluid

heat conducting body

Convective heat transfer

Introduction of heat transfer coeficient:

Convective heat transfer

Objective (of chapter 6-11): Determine α and the parameters influencing it for

prescribed tw(x) or qw(x) = Q/A

Order of magnitude for α

Medium α W/m²KAir (1bar); natural convection 2-20Air (1bar); forced convection 10-200Air (250 bar); forced convection 200-1000Water forced convection 500-5000Organic liquids; forced convection 100-1000Condensation (water) 2000-50000Condensation (organic vapors) 500-10000Evaporation, boiling, (water) 2000-100000Evaporation, boiling (organic liquids) 500-50000

How to do it? (to describe convective HT)

What are the tools?

Fluid motion:Mass conservation equation (Continuity eqn)Momentum equations (Newton’s second law)

Energy balance in the fluidFirst law of thermodynamics for an open system

Continuity eq.Expresses that mass is constant and not destroyable

Especially for

steady state, incompressible flow, two-dimensional case

)56(0yv

xu

−=∂∂

+∂∂

)46(0)w(z

)v(y

)u(x

−=ρ∂∂

+ρ∂∂

+ρ∂∂

+τ∂ρ∂

Resulting momentum equations – 2 dim.

⎟⎟⎠

⎞⎜⎜⎝

∂+

∂+

∂−=⎟

⎟⎠

⎞⎜⎜⎝

∂+

∂+

∂2

2

2

2

:ˆyu

xu

xpF

yuv

xuuux x µρ

τρ

⎟⎟⎠

⎞⎜⎜⎝

∂+

∂+

∂−=⎟

⎟⎠

⎞⎜⎜⎝

∂+

∂+

∂2

2

2

2

:ˆyv

xv

ypF

yvv

xvuvy y µρ

τρ

Inpossible to solve by hand, need to be simplified… (chapter 6, 7 and 8)

Temperature Equation

⎟⎟⎠

⎞⎜⎜⎝

∂+

∂+

∂=

∂+

∂+

∂2

2

2

2

2

2

zt

yt

xt

cztw

ytv

xtu

pρλ

Unpossible to solve by hand, need to be simplified… (chapter 6, 7 and 8)

Boundary layer approximations (laminar case)

u(x,y)

δ (x)

U∞

y x

δ T

y x

t wt(x,y)t∞

Why different fields for temperature and velocity ???

Boundary layer theory developed by Prandtl

vu >>

yv

xv

xu

yu

∂>>

∂ ,,

xt

yt

∂>>

If the boundary layer thickness is very small

If 2D

Boundary layer approximations – Prandtl’s theory

)(xpp =

2

2

yu

dxdpF

yuv

xuu x

∂+−=⎟

⎟⎠

⎞⎜⎜⎝

∂+

∂µρρ

2

2

yt

cytv

xtu

p ∂

∂=

∂+

ρ

λ

From Navier-Stokes eqn in the y-direction it is foundthat the pressure is independent of y

Then Navier Stokes in the x-direction is simplified to:

Also the temperature field is simplified

Boundary layer approximations – Prandtl’s theory

21 konstant2

p Uρ+ =

dxdUU

dxdp

ρ−=

λ

µ

λ

ρν pp cc==Pr

Bernoullis eqn describes the flow outside the boundary layer

The dimensionless Prandtl number is introduced

Boundary layer equations

0=∂

∂+

yv

xu

2

2

yu

dxdUU

yuv

xuu

∂+=

∂+

ρµ

Pr 2

2

yt

ytv

xtu

∂=

∂+

ρµ

Mass conservation

Momentum conservation

Energy conservation

Boundary layers

U∞

U∞

yx

laminar boundarylayer

transition turbulent boundary layer

fully turbulentlayer

buffer layerviscous sublayer

xc

ν/Re cc xU∞=

5105 ⋅

Pr)(Re,Nu 7f=

Continuity eq. (expresses that mass is constant and not destroyable)

x:

Net mass flow out in x-direction

Analogous in y- and z-directions

Net mass flow out ⇒ Reduction in mass within volume element

dzdyum 1x ρ=!

2 1

1

( )

xx x

mm m dxx

u dy dz u dxdy dzx

ρ ρ

∂= + =

∂∂

= +∂

&& &

dzdydx)u(x

mx ρ∂∂

=Δ !

dydxdzwz

mdzdxdyvy

m zy )( )( ρρ∂

∂=Δ

∂=Δ !!

zyx mmmutströmmatNetto !!! Δ+Δ+Δ:out flownet ,

y

xz dx

dy

dz

Cont. continuity eq.

Reduction per time unit:

Especially for steady state, incompressible flow, two-dimensional case

dzdydxτ∂ρ∂

! )w(z

)v(y

)u(x

ρ∂∂

+ρ∂∂

+ρ∂∂

=τ∂ρ∂

)46(0)w(z

)v(y

)u(x

−=ρ∂∂

+ρ∂∂

+ρ∂∂

+τ∂ρ∂

)56(0yv

xu

−=∂∂

+∂∂

Consider a mass balance for the volume element at the previous page

Navier – Stokes’ ekvationer (eqs.)Derived from Newton’s second law

m = ρ dx dy dz

but

u = u(x, y, z, τ), v = v(x, y, z, τ),

w = w(x, y, z, τ)

Fam!!

=⋅

⎟⎠⎞

⎜⎝⎛

τττ=

ddw,

ddv,

ddua

!

Forces

F!

a. volume forces (Fx , Fy , Fz) are calulated per unitmass, N/kg

b. stresses σij N/m2

The surface forces act on the boundary surfaces ofthe fluid element and are acting as either normal forces or shear forces

Forces The surface forces are calulated per unitarea and are called stresses

F!

⎥⎥⎥

⎢⎢⎢

σσσ

σσσ

σσσ

zzzyzx

yzyyyx

xzxyxx

ij

a. volume forces (Fx , Fy , Fz) are calulated per unitmass, N/kg

b. stresses σij N/m2

! ! !"z""y""x"

Stresses for an element dxdydz

Signs for the stresses

dyy

σxx

σyy

σxx

σyyσyx

σxy

x

.

Signs for the stresses

Resulting stresses

dyy

σxx

σyy

σxx

σyyσyx

σxy

σyy

σxx

σyxσxy

dy)(y yyyy σ∂∂

dx)( xxx

xx σ∂∂

dy)(y yxyx σ∂∂

dx)(x xyxy σ∂∂

x

dzdxdy)(z

dydxdz)(y

dxdydz)(x zxyxxx σ

∂∂

+σ∂∂

+σ∂∂

dxdydz)(x jijσ

∂∂

Net force in x-direction (for the 3D case)

Examples of stresses

xupep xxxx∂

∂+−=+−= µµσ 22

⎟⎟⎠

⎞⎜⎜⎝

∂+

∂===

xv

yuexyyxxy µµσσ 2

yvpep yyyy∂

∂+−=+−= µµσ 22

Resulting momentum equations

⎟⎟⎠

⎞⎜⎜⎝

∂+

∂+

∂−=⎟

⎟⎠

⎞⎜⎜⎝

∂+

∂+

∂2

2

2

2

:ˆyu

xu

xpF

yuv

xuuux x µρ

τρ

⎟⎟⎠

⎞⎜⎜⎝

∂+

∂+

∂−=⎟

⎟⎠

⎞⎜⎜⎝

∂+

∂+

∂2

2

2

2

:ˆyv

xv

ypF

yvv

xvuvy y µρ

τρ

6.20

6.21

Energy eq. (First law of thermodynamics of an open system), ⇒ Temperature field eq.

dHQd =!

y

xz dx

dy

dz

Net heat to element =

Change of enthalpy flow

Neglecting kinetic and potential energy

.

Heat conduction in the fluid (calulating the heat flow as in chapter 1)

Qd !

dxdydz)xt(

xdydzxt

dxxQQQ

xtdydz

xtAQ

xxdxx

x

∂∂

λ∂∂

−∂∂

λ−=

=∂∂

+=

∂∂

λ−=∂∂

λ−=

+

!!!

!

dxdydz)xt(

xQQQ xdxxx ∂

∂λ

∂∂

−=−=Δ +!!!

6.24

.Analogous in y- and z-directions

(6-27)

dydxdz)zt(

zQ

dzdxdy)yt(

yQ

z

y

∂∂

λ∂∂

−=Δ

∂∂

λ∂∂

−=Δ

!

!

{ }zyx QQQQd !!!! Δ+Δ+Δ−=↑

Qd !heatfor convention sign

dzdydx)zt(

z)yt(

y)xt(

xQd

⎭⎬⎫

⎩⎨⎧

∂∂

λ∂∂

+∂∂

λ∂∂

+∂∂

λ∂∂

=!

!

6.25

6.26

Enthalpy flows and changes

• Flow of enthalphy in the x-direction

hdzdyuhmH xx ρ== !!

dzdydxxhudzdydx

xuhHd x

∂+

∂=⇒ ρρ! 6.28

Enthalpy changes

• y- and z-directions

dzdydxyhvdzdydx

yvhHd y

∂+

∂= ρρ!

dzdydxzhwdzdydx

zwhHd z

∂+

∂= ρρ!

6.29

6.30

Total change in enthalpy

x y zdH dH dH dH

u v wh dxdy dzx y z

h h hu v w dxdy dzx y z

ρ

ρ

= + + =

⎛ ⎞∂ ∂ ∂= + + +⎜ ⎟∂ ∂ ∂⎝ ⎠

⎛ ⎞∂ ∂ ∂+ +⎜ ⎟∂ ∂ ∂⎝ ⎠

& & & &

Enthalpy vs temperature

( )

dtthdp

phdh

tphh

pt⎟⎟⎠

⎞⎜⎜⎝

∂+⎟

⎟⎠

⎞⎜⎜⎝

∂=⇒

=

,

6.33

Enthalpy vs temperature

p

p thc ⎟⎟⎠

⎞⎜⎜⎝

∂=

For ideal gases the enthalpy is independent of pressure, i.e.,

0)/( ≡∂∂ tph

.For liquids, one commonly assumes that the derivative

tph )/( ∂∂

is small and/or that the pressure variation dp is small compared to the change in temperature.

Then generally one states dtcdh p=

By definition

i.e., enthalphy is coupled to temperature via the heat capacity

Temperature Equation

⎟⎟⎠

⎞⎜⎜⎝

∂+

∂+

∂=

∂+

∂+

∂2

2

2

2

2

2

zt

yt

xt

cztw

ytv

xtu

pρλ

6.36

Rewriting eqn 6.32 as a fuction of T instead of h

.

Chapter 8 - Convective DuctFlow

.

laminar if pipe or tube ReD < 2300

ν=

DuRe mDmAum ρ=!

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎠⎞

⎜⎝⎛−=

2

m Rr12

uu

⎟⎟⎠

⎞⎜⎜⎝

⎛−= 2

2

max by1

uu

kärna, core

gränsskikt, boundarylayer

2bU0x

y

Chapter 8 - Convective DuctFlow

Parallel plate duct

Circular pipe, tube

.

Di Re0575.0DL

=

y

-y

gränsskikt, boundary layer

gränsskikt, boundarylayer

kärna, core

Chapter 8 Convective Duct Flow

kärna, core

gränsskikt, boundarylayer

2bU0x

y

Typical entrance region flow profile

Cont. duct flow

If ReD > 2300

U0

omslag, transition

laminärt gränsskikt turbulent gränsskikt

xy

Laminar boundary layer Turbulent boundary layer

Fully developed turbulent flow

. Pressure drop fully developed flow

Dh = hydraulic diameter

2u

DLfp

2m

h

ρ=Δ

ReCf =

ν= hmDuRe

4 4 x cross section area = perimeter

tvärsnittsareanmedieberörd omkrets

×=

Amum ρ

=!

.

0 1 2 3 4 5 61E-4

0.001

0.01

0.1

2/2mup

ρΔ

Pressure drop - entrance region (circular pipe)

x/DhReDh

Figure 8.4

Convective heat transfer for an isothermal tube

rdr

x dx

tw= konstant; constant

Velocity field fully developed

”Heat balance” for an element dxdr2πr

Heat conduction in radial directionEnthalpy transport in x-direction

⎟⎟

⎜⎜

⎛⎟⎠⎞

⎜⎝⎛−=

2

m Rr1u2u

rdr

x dx

tw= konstant; constant

Convective heat transfer for an isothermal tube

Boundary conditions:

x = 0 : t = t0r = R : t = twr = 0 : (Symmetry)

)"128("rtr

rr1

xtucp −⎟

⎠⎞

⎜⎝⎛∂∂

∂∂

λ=∂∂

ρ

0rt=

∂∂

Remember the figure from previous page

Commonly called Graetz problem

Introduce

Introduce u = 2um (1 - rʹ2)

or

p

,w

ca

tt,Rxx,Rrr

ρλ

=

−=ϑ=ʹ=ʹ

⎟⎠⎞

⎜⎝⎛

ʹ∂ϑ∂ʹ

ʹ∂∂

ʹ=ʹ∂ϑ∂

rRrR

rRrR1

xR1

au

⎟⎠⎞

⎜⎝⎛

ʹ∂ϑ∂ʹ

ʹ∂∂

ʹ−ʹ=ʹ∂ϑ∂

rr

r)r1(r1

xaRu2

2m

)208(r

rr)r1(r

1x

PrRe 2D −⎟⎠⎞

⎜⎝⎛

ʹ∂ϑ∂ʹ

ʹ∂∂

ʹ−ʹ=ʹ∂ϑ∂

Nusselt approach still valid

Coupling between velocity and average velocity

)208(r

rr)r1(r

1x

PrRe 2D −⎟⎠⎞

⎜⎝⎛

ʹ∂ϑ∂ʹ

ʹ∂∂

ʹ−ʹ=ʹ∂ϑ∂

⎟⎠⎞

⎜⎝⎛∂∂

∂∂

=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

∂∂

+∂

∂=

τ∂∂

rtr

rr1a

rt

r1

rtat2

2

Assume ϑ = F(xʹ) ⋅ G(rʹ).

After some (8.24-8.28) calculations one finds

β0 < β1 < β2 < β3 < β4 …

)298(e)r(GC0i

PrRe/xii

D2i −ʹ=ϑ ∑

=

ʹβ−

Inspiration for one of the teoretical home assignments

.

β0 = 2.705

β1 = 6.667

β2 = 10.67

Temperature profile in the thermal entrace region for a circular pipe with uniform wall temperature and fully developed laminar flow

Bulktemp. tB

the enthalpy flow of the mixture:

the enthalpy flow can be written

Bptcm!

!∫Δ

πρR

0 "h"p

"m"

tcurdr2 "#"\$%&

⎟⎟⎠

⎞⎜⎜⎝

⎛ πρ=πρ=

πρ=

R

0m

2

R

0B

u4Drdru2m

drurt2m1t

!

!!

)348(

urdr

urtdrt R

0

R

0B −=

um

dx

Local Nusselt number

0.001 0.01 0.1 11

5

10

50

100

3.656konstant temperatur,constant wall temperature

konstant värmeflöde, uniform heat flux

konstant värmeflöde, uniform heat fluxHastighetsfält ej fullt utbildat; velocity field not fully developed

4.364

PrRex/D

D

NuD

Average Heat Transfer Coefficient

1.

If the velocity field is fully developed ⇒

N.B.! Higher values if velocity field not fully develped. Eq. (8-38) gives the average value.

tw = constant

[ ]

14.0

3/2/PrRe04.01/PrRe0668.0656.3 ⎟⎟

⎞⎜⎜⎝

⎭⎬⎫

⎩⎨⎧

++==

w

B

D

DD

xDxDDNu

µµ

λα

)388(D/LPrRe86.1DNu

14.0

w

B3/1

DD −⎟⎟⎠

⎞⎜⎜⎝

µµ

⎭⎬⎫

⎩⎨⎧=

λα

=

1.0PrReD/LD

<

t0

x tw = konst

2.

Fully developed flow and temperature fields

NuD = 4.364 (8-50)

qw = α (tw − tB)

⇒ tw − tB = konst.If tB increases, tw must increase as much

Average value including effects of the thermal entrance length

↓↓

!!"!!#\$ konstkonst

⎪⎪

⎪⎪

>+

<⎟⎟⎠

⎞⎜⎜⎝

=λα

=

03.0PrReD/xomPrRe

D/x0722.0364.4

03.0PrReD/xomD/x

PrRe1953.1

DNu

DD

D

3/1

DD

t0

...

...

qw = konst. = constant

.

2.

Fully developed flow and temperature fields

NuD = 4.364 (8-50)

qw = α (tw − tB)

⇒ tw − tB = konst.

If tB increases, tw must increase as much

tw highest at the exit!

↓↓

!!! "!!! #\$ constconst

t0

...

...

qw = konst. = constant