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 )( =

vrmeledande 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 )( =

0/ (6 3)f

x

w f w f

tyQ A

t t t t

=

= =

vrmeledande 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 and the parameters influencing it for

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

Order of magnitude for

Medium W/mKAir (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 (Newtons 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

=

+

+

+

Resulting momentum equations 2 dim.

+

+

=

+

+

2

2

2

2

:yu

xu

xpF

yuv

xuuux x

+

+

=

+

+

2

2

2

2

:yv

xv

ypF

yvv

xvuvy y

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

p

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

Boundary layer approximations (laminar case)

u(x,y)

(x)

U

y x

T

y x

twt(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 Prandtls theory

)(xpp =

2

2

yu

dxdpF

yuv

xuu x

+=

+

2

2

yt

cytv

xtu

p

=

+

From Navier-Stokes eqn in the y-direction it is found that 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 Prandtls theory

21 konstant2

p U+ =

dxdUU

dxdp =

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

+=

+

Pr 2

2

yt

ytv

xtu

=

+

Mass conservation

Momentum conservation

Energy conservation

Boundary layers

UU

yx

laminar boundarylayer

transition turbulent boundary layer

fully turbulentlayer

buffer layerviscous sublayer

xc

/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 =

2 1

1

( )

xx x

mm m dxx

u dy dz u dx dy dzx

= + =

= +

dzdydx)u(x

mx

=

dydxdzwz

mdzdxdyvy

m zy )( )(

=

=

zyx mmm ++:out flow Net

y

xz dx

dy

dz

Cont. continuity eq.

Reduction per time unit:

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

dzdydx

)w(z

)v(y

)u(x

+

+

=

)46(0)w(z

)v(y

)u(x

=

+

+

+

)56(0yv

xu

=

+

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

Navier Stokes ekvationer (eqs.)Derived from Newtons second law

m = dx dy dz

but

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

w = w(x, y, z, )

Fam

=

=

ddw,

ddv,

ddua

Forces

F

a. volume forces (Fx , Fy , Fz) are calulated per unit mass, unit ???

b. stresses ij unit ???

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

Forces

F

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

b. stresses ij N/m2

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

Forces The surface forces are calulated per unit area and are called stresses

F

=

zzzyzx

yzyyyx

xzxyxx

ij

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

b. stresses ij N/m2

"z""y""x"

Stresses for an element dxdydz

Examples of stresses

xupep xxxx

+=+= 22

+

===xv

yuexyyxxy 2

yvpep yyyy

+=+= 22

Resulting momentum equations

6.20

6.21

+

+

=

+

+

2

2

2

2

:yu

xu

xpF

yuv

xuuux x

+

+

=

+

+

2

2

2

2

:yv

xv

ypF

yvv

xvuvy y

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(

xdydz

xt

dxx

QQQ

xtdydz

xtAQ

xxdxx

x

=

=

+=

=

=

+

dxdydz)xt(

xQQQ xdxxx

== + 6.24

.Analogous in y- and z-directions

(6-27)

dydxdz)zt(

zQ

dzdxdy)yt(

yQ

z

y

=

=

{ }zyx QQQQd ++=

Qd heatfor convention sign

dzdydx)zt(

z)

yt(

y)

xt(

xQd

+

+

=

6.25

6.26

Enthalpy flows and changes

Flow of enthalphy in the x-direction

hdzdyuhmH xx ==

dzdydxxhudzdydx

xuhHd x

+

= 6.28

Enthalpy changes

y- and z-directions

dzdydxyhvdzdydx

yvhHd y

+

=

dzdydxzhwdzdydx

zwhHd z

+

=

6.29

6.30

Total change in enthalpy

x y zdH dH dH dH

u v wh dx dy dzx y z

h h hu v w dx dy dzx y z

= + + =

= + + +

+ +

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

6.36

Rewriting eqn 6.32 as a fuction of t instead of h

+

+

=

+

+

2

2

2

2

2

2

zt

yt

xt

cztw

ytv

xtu

p

Summary

Velocity profile equals the

temperature profile when Pr=1

Rec = 2300

Rec = 500000

.

Chapter 8 - Convective Duct Flow

.

laminar if pipe or tube ReD < 2300

=

DuRe mDmAum =

=

2

m Rr12

uu

= 2

2

max by1

uu

krna, core

grnsskikt, boundarylayer

fullt utbildadstrmning, fullydeveloped flow

2bU0 xy

Chapter 8 - Convective Duct Flow

Parallel plate duct

Circular pipe, tube

.

Di Re0575.0

DL

=

y

-y

grnsskikt, boundary layer

grnsskikt, boundarylayer

krna, core

Chapter 8 Convective Duct Flow

krna, core

grnsskikt, boundarylayer

fullt utbildadstrmning, fullydeveloped flow

2bU0 xy

Typical entrance region flow profile

Cont. duct flow

If ReD > 2300

fullt utbildadturbulent strmning

U0

omslag, transition

laminrt grnsskikt turbulent grnsskikt

xy

Laminar boundary layer Turbulent boundary layer

Fully developed turbulent flow

. Pressure drop fully developed flow

Dh = hydraulic diameter

2u

DLfp

2m

h

=

ReCf =

= hm

DuRe

4 4 x cross section area = perimeter

tvrsnittsareanmedieberrd omkrets

=

Amum

=

.

0 1 2 3 4 5 61E-4

0.001

0.01

0.1

2/2mup

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 dxdr2r

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

=

0rt=

Remember the figure from previous page

Energy eqn for steady state

Commonly called Graetz problem

Introduce

Introduce u = 2um (1 - r2)

or

p

,w

ca

tt,Rxx,Rrr

=

===

=

rRrR

rRrR1

xR1

au

=

rr

r)r1(r1

xaRu2

2m

)208(r

rr)r1(r

1x

PrRe 2D

=

Nusselt approach still valid

Coupling between velocity and average velocity

)208(r

rr)r1(r

1x

PrRe 2D

=

=

+

=

rtr

rr1a

rt

r1

rtat 2

2

compared with unsteady heat conduction:

Steady heat conduction

Assume = F(x) G(r).

After some (8.24-8.28) calculations one finds

0 < 1 < 2 < 3 < 4

)298(e)r(GC0i

PrRe/xii

D2i =

=

Inspiration for one of the teoretical home assignments

.

0 = 2.705

1 = 6.667

2 = 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

R

0 "h"p

"m"

tcurdr2

==

=

R

0m

2

R

0B

u4Drdru2m

drurt2m1t

)348(

urdr

urtdrt R

0

R

0B =

umdx

Local Nusselt number

0.001 0.01 0.1 11

5

10

50

100

3.656konstant temperatur,constant wall temperature

konstant vrmeflde, uniform heat flux

konstant vrmeflde, uniform heat fluxHastighetsflt 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

)388(D/LPrRe86.1DNu

14.0

w

B3/1

DD

=

=

1.0PrRe

D/LD

+

.

2.

Fully developed flow and temperature fields

NuD = 4.364 (8-50)

qw = (tw tB)

tw tB = konst.

If tB increases, tw must increase as much

tw highest at the exit!

constconst

t0

...

...

qw = konst. = constant

Convective Heat Transfer (6)Forced Convection (8)Agenda Convective heat transferConvective heat transferConvective heat transferSlide Number 6Convective heat transferOrder of magnitude for How to do it? (to describe convective HT)What are the tools?Continuity eq.Expresses that mass is constant and not destroyableResulting momentum equations 2 dim.Temperature EquationBoundary layer approximations (laminar case)Boundary layer theory developed by PrandtlBoundary layer approximations Prandtls theoryBoundary layer approximations Prandtls theoryBoundary layer equationsBoundary layersContinuity eq. (expresses that mass is constant and not destroyable)Cont. continuity eq.Navier Stokes ekvationer (eqs.)Derived from Newtons second lawForcesForcesForces The surface forces are calulated per unit area and are called stressesExamples of stressesResulting momentum equationsEnergy eq. (First law of thermodynamics of an open system), Temperature field eq.Slide Number 28.Enthalpy flows and changesEnthalpy changesTotal change in enthalpyEnthalpy vs temperatureEnthalpy vs temperatureTemperature EquationSummary...Cont. duct flow..Convective heat transfer for an isothermal tubeSlide Number 44 .Bulktemp. tBLocal Nusselt numberAverage Heat Transfer CoefficientSlide Number 53.