Three heat transfer mechanisms - Åbo Akademi · and heat transfer coefficient h, unit: W/(m2·K)...

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5. Heat transfer

Ron ZevenhovenÅbo Akademi University

Thermal and Flow Engineering / Värme- och strömningstekniktel. 3223 ; [email protected]

Processteknikens grunder (”PTG”) Introduction to Process Engineering

v.2014

Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland

Three heat transfer mechanismsConduction Convection Radiation

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Pic: BÖ88


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5.1 Conductive heat transfer


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Fourier’s Law /1

In a non-moving medium (i.e. a solid, or stagnant fluid) in the presence of a temperature gradient, heat is transferred from high to low temperature as a result of molecular movement: heat conduction(sv: värmeledning)

For a one-dimensional temperature gradient ∆T/∆x or dT/dx, Fourier’s Law gives the conductive heat transfer rate Q through a cross-sectional area A (m2). If λ is a constant:

with thermal conductivity λ, unit: W/(mK)

(sv: termisk konduktivitet eller värmeledningsförmåga)

)(W/m (W) 2

dx

dT

A

Q"Q

dx

dTAQ

Pictures: T06

.

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For a general case with a 3-dimensional temperaturegradient T = (∂T/∂x, ∂T/∂y, ∂T/∂z), Fourier’s Law gives (for constant λ) a heat flux vector Q” = - λ T

The temperature field inside the conductingmedium can be written as T = T(t, x) with time t and three-dimensional locationvector x

For stationary (sv: stationärt, tidsinvariant) heat transfer ∂T/∂t = 0 at each position x

The heat transfer vector is perpendicular(sv: vinkelrätt) to the isothermal surfaces

Note that material property λ is, in fact, a function of temperature:

more accurately Q” = - (λ(T)·T)

∆∆.

Figure: KJ05

∆

Q is a vector with direction - T

∆..


Fourier’s Law /2

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Material conductivity data

r

Pictures: T06

Thermalconductivity is temperature-dependent!

Table: KJ05

λ


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Fourier’s Law /3

Alternative notation based on energy instead of temperature: use the heat concentration (heat per volume) ρ· cp·T(with density ρ, specific heat cp), with unit J/m3

This gives (for 1-dimensional conduction):

/s)(m cρ

λ ay diffusivit heat with

)(W/m dx

)Tcρ(da

dx

)Tcρ(d

cρ

λ"Q

2

p

2pp

p

Typical values λ (W/mK) a (m2/s)

Gases ~ 0.02 ~ 20×10-6

Liquids ~ 0.2 (water ~ 0.6) ~ 0.1×10-6

Non-metallic solids ~ 2 ~ 1×10-6

Metals 20 – 200 5 - 50×10-6

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For stationary heat conduction through a plane wall whereλ = constant, Fourier’s Law Q” = -λ· dT/dx integrates to

(see ÖS96 p.55)

Q” x1∫x2 dx = - λ T1∫T2 dT → Q” = - λ· (T2-T1)/(x2-x1)

or if λ = λ(T) it integrates to

Q” x1∫x2 dx = - T1∫T2 λ(T)dT → Q” = -(T1∫T2 λ(T)dT )/(x2-x1)

For example, with λ(T)=λ0· (1+αT), the heat flux Q” for T=T0 @ x=0 and T=T1 @ x=L integrates to

Q” = - (T0∫T1 λ0· (1+α·T)dT )/L = - (λ0/L)·[T + ½· α·T2]T0T1

= - (λ0/L)· [T1-T0+ ½· α· (T1²-T0²) ]


Fourier’s Law /4

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Conductance and resistance Fourier’s Law Q” = -λ·dT/dx can

be interpreted as a general physicallaw of the type:

flow , heat, current =

driving force / resistance;

current =

voltage difference /

resistance;

heat transfer =

temperature difference /

resistance;

fluid flow = pressure drop /

flow resistance;

Heat resistance can be formulated as

Rheat = ∆T/Q

(unit: K/W or °C/W)

Heat conductance as

Gheat =1/Rheat = Q/∆T

(unit: W/K or W/°C)

For a plane material with thickness L (m) and conductivity λ (W/mK):

Gheat = λ·A/L

Rheat = L/(λ·A)

.

.

.


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Heat conduction: plane surfaces /1/2

The heat flux Q” through a layered planar wall of materials with thickness di

and conductivity λi is foundby considering the material as a series of heat resistances di/λi

Note that Q” is the same for every location, so λ·∆T/∆x is constant: Q” = -λ1·∆T1/∆x1 = -λ2·∆T2/∆x2 for a two-layermaterial

Thus, for each layer Q”· (di/λi) = - ∆Ti = (Ti-1 – Ti)

T1T0=Tin T2=Tout

λ1 λ2Tin Tout

Q”

d1 d2

x

.

.

.

.

.


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Heat conduction: plane surfaces /2/2

Thus, for each layer

Q”· (di/λi) = - ∆Ti = (Ti-1 – Ti)

For two layers ”1” and ”2”, the (unknown?) T1 can be eliminated: 1) Q”· (d1/λ1) = (T0 – T1) and 2) Q”· (d2/λ2) = (T1 – T2) 1) into 2) gives Q”·(d2/λ2) = (T0 – Q”· (d1/λ1) –T2)

→ Q”·(d2/λ2+ d1/λ1) = (T0 – T2) = Q”· d1+2/λ1+2

Similarly for N layers: Q”· (d1/λ1+ d2/λ2+ d3/λ3+ ... dN/λN) = ∆Ttotal

T1T0=Tin T2=Tout

λ1 λ2Tin Tout

Q”

d1 d2

x

.

.

.

.

.

..

.

.

.

.


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Example: a brick wall /1

An oven wall consists in succession of – a layer of firebricks λf = 1.21 W/(m· K), – a layer of insulation material λi = 0.08 W/(m· K) and– a layer of outside bricks λb = 0.69 W/(m· K).

Each layer is 10 cm thick. Inside the oven the temperature Tin = 872°C, outside the temperature is Tout = 32°C. The surface area of the wall is 42 m2.

a. Calculate the heat loss by conduction during 24 h.b. Calculate the centretemperature Tm in the middleof the layer of insulation

Pic

ture

: http

://w

ww

.wes

t-no

rfol

k.go

v.uk

/Def

ault.

aspx

?pag

e=22

400


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Example: a brick wall /2

T1 T2T0 T3

λf λbλiTin Tout

Q”


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Cylindrical, spherical geometry /1

For stationary conduction in radial direction in cylindrical or spherical geometries, Fourier’s Law becomes:

) sphericalD-(1 )(m

l)cylindrica D-(1 )(m

with(W)

2

2

2r4A(r)

rL2A(r)

dr

dT)r(A)r(Q

Picture: KJ05

Picture: T06

see the literature for cases with angulargradients dT/dθ, dT/dφ etc.


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Cylindrical, spherical geometry /2

For the heat resistance of a cylindrical or sphericalsection [din, dout] with thickness din-dout it is foundfor the heat resistanceRheat = ∆T/Q :

Picture: KJ05

Picture: T06

mean) (geometric with

mean) ic(logarithm with

inoutgeom

outin

inout

geom

inoutspherical

in

out

inoutln

in

out

ln

inoutlcylindrica

ddd

dλπd

-dd

λπd

-dd R

dd

ln

-ddd

λπL

dd

ln

λπLd

-dd R

.

(see ÖS96 p.58)Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland

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Example: insulated hot water tube /1

A 19 mm outside diameter copper tube for hot water is insulated with a 18 mm thicklayer of insulation material with thermal conductivity λ=0.067 W/(mK)

The temperature profilethrough the insulation layer is given by the equation

Derive this expression; and) For T(ri) = 360 K and T(r0) = 315 K Calculate the heat transfer rate per unit length Q’=Q/L(W/m), and Calculate the heat fluxes Q” at r = ri and at r = r0 (W/m2)

Pictures: T06

.

.

.


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(Example: insulated hot water tube /2) Derive:

)r(T)r

rln(

)rr

ln(

)r(T)r(T)r(T)rln(

)rr

ln(

)r(T)r(T)r(T)rln(

)rr

ln(

)r(T)r(T)r(T

)rln()

rr

ln(

)r(T)r(T)r(T'c'c)rln(

)rr

ln(

)r(T)r(T)r(T'c)rln(

)rr

ln(

)r(T)r(T)r(T

)rr

ln(

)r(T)r(TLπλc)

r

rln(

Lπλ

c)rln()rln(

Lπλ

c)r(T)r(T

:known are )T(r and )T(r that fact use ; 'c)rln(Lπλ

c)r(T:egrateint

r

dr

Lπλ

c)r(dTcconstant

dr

)r(dTrLπλ)r(Q

:scoordinate lcylindrica for direction)-r D,-(1Law sFourier'

10

Picture: T06


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Example: insulated hot water tube /3

Picture: T06


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5.2 Convective heat transferforced convection, natural convection


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Heat convection /1 In convection (sv: konvektion)

heat transfer, heat is entrained with a movingconducting medium.

The medium flow may be the result of external forces: forced convection(sv: påtvungen konvektion); or is the result of densitydifferences caused by temperature differences: free, or natural convection(sv: fri, eller naturlig konvektion)

Forced convection is usuallymuch more important thannatural convection Pictures: T06

Forced convection Free convection

wall

medium dy

dTλconv"Q :Law sFourier'

to according medium flowing

the into surface the from away

conducted is heat the:Note

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The general expression for heat transfer by convection is

Q = h· A· ∆T (W) *

for heat exchange surface A (m2), temperature difference ∆T (K, °C) between the media or materials, and heat transfer coefficient h, unit: W/(m2·K)

The heat transfer coefficientdepends on

– Geometry– Flow velocity (or velocities, if

both media are fluids)– Type of flowing media (gas,

liquid)– Temperature

The heat transfer coefficient(sv: värmeövergångs-koefficient) is a purely emperical and phenomeno-logical quantitythat contains ”all we do not know” about the transfer process

Picture: T06

.

* Using the concepts of resistance & conductance gives Rheat = 1 / (h·A) ; Gheat = h·A


Heat convection /2

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Heat transfer coefficient

Some typical values:

Table & pictures: KJ05

(a) Forced and (b) Natural convectionover a cylinder.

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Boundary layers

Growth of the velocity boundary layeron a flat surface.* This can be a solid surface or

another flowing medium

At the interface of a surface* and a flowing medium, a thinlayer (0.01 – 1 mm) of fluid is created in which the velocityincreases from v = 0 at the interface to the free-flowvelocity v = v∞ (or 0.99·v∞ ) In this boundary layer

(sv: gränsskikt) all the thermaland/or viscous effects of the surface are concentrated The boundary layer can

develop from laminar to turbulent flow

Pictures: KJ05

See section 6.2


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Laminar ↔ turbulent fluid flow

Pictures: T06

Osborne Reynolds’s dye-streakexperiment (1883) for measuring laminar → turbulent flow transition

laminar: Re < 2100

laminar → turbulent

turbulent: Re > 4000

See section 6.2


For circular tube flow, the laminar → turbulent flow transition occurs at Reynolds number Re = 2100 – 2300 (fully turbulent at Re=4000) with dimensionless Re defined as Re = ρ··<v>· d/ηfor ρ = fluid’s density (kg/m3), <v> = fluid’s average velocity (m/s), d = tube diameter (m) and η = fluid’s dynamic viscosity (Pa· s)

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Viscosity Viscosity (sv: viskositet) is a

measure of a fluid's resistance to flow; it describes the internal friction of a moving fluid. More specifically, it defines the

rate of momentum transfer in a fluid as a result of a velocitygradient. Dynamic viscosity η

(unit: Pa.s) is related to a kinematic viscosity, ν(unit: m2/s) via fluid density ρ (kg/m3)

as: ν = η/ρη

See section 6.2

Picture T06 Picture: KJ05

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Thermal boundary layers /1

Growth of the thermal boundary layeron a flat surface.

The boundary layer is a thinmore or less stagnant region where the fluid flow velocitydecreases until reachingv = 0 (”no slip”) at the surface The heat transfer resistance on

the flow side can be assumedconcentrated in and limited to the boundary layer A thermal boundary layer

is also formed as a result of the temperature difference BUT: the thicknesses the of the

hydrodynamic (δH) and thermalboundary (δT) layer are not the same! Pictures: KJ05

Growth of the velocity boundary layeron a flat surface.


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Thermal boundary layers /2 Typically the heat conductivity of the

(flowing) fluid medium is (much) lower than that of the surfacematerial, and the very thin fluid layer may be the largest heat transfer resistance

For the boundary layer, two heat balances hold, which gives someinsight into the heat transfer coefficient:

Q” = Q/A = h·(Ts-Tf) = -λfluid·dT/dy│surface ≈ -λ fluid·∆T/∆y│boundary layer

with ∆T = (Tf-Ts) and ∆y = δthermal boundary layer = δT this gives

h·∆T = λfluid·∆T/δT → h ≈ λfluid/δT

This shows that thin boundary layers promote (heat) transfer !

Pictures: KJ05

. .


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Prandtl number The relative thicknesses of the thermal and hydrodynamic boundary

layers depends on: - Thermal conductivity λ- Specific heat cp

- Dynamic viscosity η(or kinematic viscosity v = η/ρ with density ρ )

defining the dimensionless Prandtl number Pr

Typical Prandtl numbers

for common fluids.

For a typical (di-atomic) gas

Pr ~ 0.7; Pr⅓ ~ 0.9

3

13

13

1

3

1

Pr)/(

a

cc pp

T

H

Picture: KJ05

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Nusselt number /1 Consider convective heat transfer Qconv = h·A·∆T

through the boundary layer with thickness δT at surfaceA with h = λfluid/δT

Can be compared with the conduction of heat througha stagnant layer of the same thickness of the same material Qconductive = - λfluid·A·∆T/δT

→ the ratio of convective and conductive heat transfer: Qconvective / Qconductive = h·δT/λfluid = Nu

This defines the dimensionless Nusselt number Nu(for a boundary layer Nu = 1)

For a more general geometry with acharacteristic size Lchar (for example, Lchar

= volume / surface = V/A), Nu is defined as: Pr)(Re,f

LhNu

fluid

char

.

.

.

.

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(Nusselt number /2 dimensional analysis) Dimensional analysis can be used to determine an expression for

Nusselt number Nu; starting with considering the relevant parameters for heat transfer coefficient h: h = f (v, L, η, ρ, λ, cp) for – flow with velocity v (m/s) – around or in a geometry with characteristic length L (m), – fluid viscosity η (Pa.s), (Pa.s = kg m-1 s-1)– fluid density ρ (kg m-3), – fluid conductivity λ (W m-1 K-1), (W m-1 K-1 = kg m s-3 K-1) and – fluid specific heat cp (J kg-1 K-1), (J kg-1 K-1 = m2 s-2 K-1).

This defines a problem with 7 variables, 4 base units (m, s, kg, K) Buckingham’s Pi theorem : a dimensionless expression with

7 – 4 = 3 dimensionless groups can be derived. Express the 4 units as function of the variables, for example:

m → L kg → ρL3 K → v2cp-1 s → L v-1


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Using these new dimension variables, the expression for h can be derived:

Pr)f(Re,Nuλ

hL

RePr,

Ref

λ

L

PrRe

h

λ

λ

cη

L

Re

h

cη

L

Re

h

ηvcρ

vLρ

Re

h

vcρ

Re

Re

h

vcρ

h

RePr,

Ref

vLρ

η

cη

λ,

Ref

vLcρ

λ,

Ref,

vLcρ

λ,,

Lvρ

η,,f

vcρ

h

vcLvL

c,

vcLLvLρ

λ,

LLρ

ρ,

vLLLρ

η,

L

L,

LvL

vf

vcvLLρ

h

Lvs ,cv K ,ρL kg L,m using

Ksm

c,

Kkgms

λ,

kgm

ρ,

skgm

η,

m

L,

ms

vf

Kkgs

h

ppppp

pppp

p

p

pp

1-1-p

23

p

See also VST course 424302 (7 sp) (rz13)Massöverföring & separationstechnik

(Nusselt number /3 dimensional analysis)

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Heat convection: isothermal flat plate

a) The boundary layer is laminar over the entire plate.

b) The laminar and turbulent boundary layers are of comparable extent.

c) The turbulent boundary layer extends over almost the entire plate.

Picture: KJ05

>

For flow over an isothermalflat plate, the laminar →turbulent transition occursat Rex = v·ρ·x / η ≈ 5×105

Fluid properties for the boundary layer are taken at temperature average

Tfilm = ½· (Tfluid flow + Tsurface)

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Nusselt number: isothermal flat plate

60 Pr 0.6 10Re for

60 Pr 0.6 10 Re 105 for

0.6 Pr 105Re for

:numbers Nusselt L length over averaged

60 Pr 0.6 105Re for

0.6 Pr 105Re for

:numbers Nusselt local

8L

8L

5

5L

5x

5x

3154

3154

3121

3154

3121

0370λ

8710370λ

6640λ

02960λ

3320λ

//

//

//

//

//

PrRe.

Pr)Re.(

PrRe.

PrRe.

PrRe.

LL

LL

LL

xx

x

xx

x

LhNu

LhNu

LhNu

xhNu

xhNu

Picture: KJ05

For constant heat flux plates,the first coefficients in Nux become0.332 → 0.453; 0.0296 → 0.0308;for the averaged numbers no significantdifference with isothermal case.

>

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Example: convective heat transfer /1

A computer chip is cooled by a flow of air at T= 20°C, p = 1 bar, with velocity 3.4 m/s. If the maximum allowable temperature of the chip is 65°C, how much heat (in W) must then be removed by forcedconvection ?

Assume Pr = 0.71; η = 2·10-5 Pa·sand λ = 0.027 W/mK for air

Picture: KJ05


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Picture: KJ05


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Nusselt number: different geometries

Flow situations can be divided into two important geometries or situation types:– Flow along an (outside) surface or

around an obstable – Flow inside a channel (most importantly

tube flow)

A generalised expression for flow around obstacles(with characteristic diameter D) is

Nu = h· D/λ = C·ReDm· Prn

with values for C and m (as function of ReD) and n ~ ⅓ (as function of Pr) taken from tables

A correction factor for Tsurface « Tflow or Tsurface»Tflow can be added, since ρ=ρ(T), η=η(T), cp=cp(T)


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Nusselt number: flow around obstacles

Convection over a sphere

term last via T T or T T for allowing

380Pr0.7 and 80000 Re3.5 for

flowsurfaceflowsurface

D

41403221 060402 /.// )(Pr)Re.Re.(surface

DDDhD

Nu

Picture T06


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Nusselt number: flow around obstacles

Convection for crossflow over a cylinder

Equation and table T06

term last via T T or T T for allowing flowsurfaceflowsurface

/

surface

nmDD )

Pr

Pr(PrReC

λ

hDNu

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Example: convective heat transfer /1Birds fluff their feathers to stay warm

during winter. Estimate the convective heat transfer from a small bird in a v = 9 m/s wind, treating the bird as a d = 6 cm diameter sphere.

Data: the surface temperature at the feathers is -7°C, the temperature of the surrounding air is -10°C, pressure is 1 bar.The convective heat transfer coefficient h is related to windvelocity v by:

for air viscosity η = 1.68×10-5 Pa·s; air density ρ = 1.33 kg/m3; air thermal conductivity λ = 0.024 W/(m·K)

0.50.53Re Nu :or ;

50

530.

vd.

hd

Picture: T06


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Pictures: T06


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Nusselt number: tube flow /1 For laminar flow in a tube (ReD < 2100) with diameter D,

for constant wall temperature: NuD = h· D/λ = 3.66 For laminar flow in a tube (ReD < 2100) with diameter D,

for constant heat flux: NuD = h· D/λ = 4.36

R

Laminar tube flow:vx(r) = (1 - r²/R²)·vmax

cross-sectional averagevelocity <v> = ½·vmax

Turbulent tube flow:vx(r) ≈ (1 - r/R)1/7·vmax

cross-sectional averagevelocity <v> = 0.875·vmax

Laminar Turbulent

Picture: BMH99


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Nusselt number: tube flow /2

For turbulent flows in pipes (with diameter D) the following expression are used:

NuD = 0.023· ReD0.8· Prn

with ReD > 10000; 0.7 < Pr ≤ 160;

L/D ≥ 10 and

n=0.3 for cooling (Twall < Tflow) or

n=0.4 for heating (Twall > Tflow)

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Nusselt number: tube flow /3

For turbulent flows in pipes (with diameter D) with large (or at least significant) temperature differencesbetween wall and flow – and this is usually the case - the following expression is usedwith a viscosity correction for near-wall fluid viscosity

ηwall = η at wall temperature:

NuD = 0.027· ReD0.8· Prn· (η/ηwall)0.14

with ReD > 10000; 0.7 < Pr ≤ 16700; L/D ≥ 10 and

n=0.3 for cooling (Twall < Tflow) or n=0.4 for heating (Twall > Tflow)

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Nusselt number: entrance effects In the entrance region of a tube flow

the boundary layer must develop, giving a higher heat transfer coefficient than in the fully developedflow downstream.

For laminar flow this can be accounted for by the dimensionlessGraetz number, Gz = Re· Pr· D/x for length section x in a tube with diameter DFor the entrance region Gz > 20, for developed flow Gz < 5

For turbulent flow this can be corrected for by a factor 1+ (D/L)0.7

for length L, diameter D

(a) Simultaneously developing velocity and thermal profiles.

(b) Development of boundary layers with an unheated starting length.Gz ↔ 1/Gz are used mixed in the literature !

Picture: KJ05

See section 6.3

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Tube flow: constant wall temperature

For a tube flow situation where the tube wall is held at a constanttemperature, the expression for a heat exchanger can be used: *)

with temperature differences ∆Ti at position ”i” (in) and ∆Te at position ”e” (exit), heat exchange surface A (m2) and total heat transfer coefficient U (W/m2K)

Heat transfer coefficients h in U can be found usingexpressions for Nu for flow inside and around tubes

Picture: KJ05 lm

e

i

ei T∆A

T∆T∆

ln

T∆T∆AQ

UU

*) see section 4.1


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Total heat transfer coefficient A lumped or total heat transfer

coefficient U (sv: värmegenom-gångskoefficient) can be defined based on partial heat transfer resistances.

For example a heat flux Q” (W/m2) through from fluid (flow) 1 through a wall (& dirt layer) to fluid (flow) 2:

Q” = U·(T1-T2) with (average) temperatures T1 and T2

The resistance 1/U equals the sum of resistances:

If the interfaces are curved, A1 ≠ A2 ≠ A3, then this must be taken intoaccount:

ARA

A

hA

A

λ

d

A

A

λ

d

A

A

hU total heatd,avgd

d

w,avgw

w

Picture: BMH99

total heattotal heat

d

d

w

w

G

AAR

hλ

d

λ

d

hU

.

.A1 A2A3


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Example: convective heat transfer: tube flow /1

A solar collector is used to supply hot water. A d = 1.2 cm inner diameter copper tube is fixed to the back of a collector plate, which is held at 75°C by incident sunlight.

Water enters the tube at Tin = 25°C at a mass flow rate m = 0.0122 kg/s. Assuming that Ttube wall = 75°C also: calculate the length L of tube needed so that the water exittemperature will be 55°C. Neglect the heat transfer resistance in tubewall and on the outside.

Data: water 25°C: ρ = 997 kg/m3; η = 8.72·10-4 Pa· s, λ = 0.611 W/m· K, cp = 4.178 kJ/kg·K;

water 55°C: ρ = 986 kg/m3; η = 4.21·10-4 Pa· s, λ = 0.648 W/m· K, cp = 4.179 kJ/kg·K;

.

Picture: KJ05


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Picture: KJ05


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Picture: KJ05


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A diagram for ∆Tlm

lm

L

L T∆

T∆T∆

ln

T∆T∆

Picture: BMH99

This diagram gives the logarithmic mean temperature ∆Tlm as function of two temperature differences ∆TL and ∆T0:

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Natural convection /1

• A heated surface can cause a temperature gradient in a surrounding medium, and the resulting density differences ∆ρ induce (buoyancy driven) convection. Can be important in situations without forced flow or mixing.

• Natural convection (or ”free convection”) can also be induced by concentration differences ∆c, and sometimes temperature and concentrationeffects must both be considered.

Picture: BMH99


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Natural convection /2 : Nusselt number

For natural convection (laminar is most common) along a surface, as a result of a temperature difference ∆T = Twall (at the surface) - T∞ (in the undisturbed medium), the Nu numberfor heat transfer becomes a function of

– Gravity, g– Coefficient of expansion of the medium β = (1/V)·(∂V/∂T)p=const

taken for T=T∞ (for an ideal gas at temperature T, β = 1/T !!)

The effect of ∆T and β on density is as β·∆T = ∆ρ/ρwall

For heat transfer it is found that Nu = f(d3· β·∆T· g/(a.ν)) with a = λ/(ρ· cp) and ν = η/ρ, both for T = ½ (Twall+T∞)

→ Nu = f (Gr· Pr) with dimensionless numbers Pr = ν/a and Grashof number Gr = L3· β·∆T· g· (ρ/η)².


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Natural convection /3 Nusselt number

Vertical plates or cylinders Nu = 0.59· (Gr· Pr)0.25 103<Gr· Pr<108

Nu = 0.13· (Gr· Pr)1/3 Gr· Pr>108

Horizontal cylinders Nu = 0.53· (Gr· Pr)0.25 103<Gr· Pr<108

Nu = 0.13· (Gr· Pr)1/3 Gr· Pr>108

Horizontal plate Nu = 0.54· (Gr· Pr)0.25 105<Gr· Pr<2·107

(above heated or below cooled) Nu = 0.17· (Gr· Pr)1/3 2·107<Gr· Pr Horizontal plate Nu = 0.27· (Gr· Pr)0.25 Gr· Pr>3·105

(below heated or above cooled)

Note: natural convection is more important than forcedconvection if Re2 << Gr (and vice versa). If Re² ~ Gr thenconsider both natural and free convection.When Nuforced convection and Nunatural convection are known, theycan be combined (as parallel resistances ~ 1/Nu) 1/Nutotal = 1/Nuforced convection + 1/Nunatural convection

Note: Gr ·Pr = Ra (Rayleigh number)


54/120

Compare the natural convectionheat transfer coefficients, h, for a 2.5 m high wall for two situations:1. Air 22°C, wall 10°C2. Air 22°C, wall 17°C

Data for air (p = 1 bar) at 15 ~ 20°C: ρ= 1.21 kg/m3; η = 1.8·10-5 Pa· s; λ = 0.025 W/m· K; Pr = 0.71. Ideal gas

Example: Natural convectionpi

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Example: Natural convection

pict

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H = 2.5 m

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5.3 Combined conduction andconvection heat transfer; the lumped system approximation


57/120

Example: Conduction + convection /1

A 2 cm×2 cm×0.3 cm computer chip is cooled by a forced air flow with heat transfer coefficient h = 152 W/(m2· K).

The electronic component is located as a thin layer at the bottom surface of the chip, generating 1.6 W heat.

Air temperature Tf is 20°C Calculate the top and bottom

surface temperatures of the chip T2 and T1

For silicon, λ= 148 W/(m·K)Pictures: KJ05

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Pictures: KJ05


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Pictures: KJ05


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Pictures: KJ05


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Lumped system approximation /1

In circumstances where one thermalresistance is much larger than another(or more others), simplifications can be made.

For example, a two-layer structure of rubber (λ = 0.13 W/m·K) and copper(λ = 400 W/m·K), both with thickness2 cm, the thermal resistances (per m2) are 0.154 K/W and 0.00005 K/W, respectively

Very often conductive heat resistances are much smaller thanconvective heat resistances

Pictures: KJ05

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Lumped system approximation /2

The cool down (or heat-up ) of a body that is suddenly immersed in a colder (or hotter) fluid depends on how fast heat is transferred to/from the interface surface at both sides of the interface

The time it takes for the centre of the immersed body to reach the same temperature as its surface depends on the ratio between the resistances for convective and conductive heat transfer, Rheat conduction and Rheat convection

In these transient (sv: övergående) situations temperatures are a functionof time and position, but the lumped system approach can be used when variations with position (inside a material) can be ignored

Picture: KJ05


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Biot number For a body with a characteristic

size Lchar = volume / surface = V / A, thermal conductivity λ and interface heat transfer coefficienth, these resistances and their ratioare equal to

Dimensionless ratio hLchar/λ is known as dimension-lessBiot number Bi = internal restance / boundary resistance

In general, if Bi < 0.1 there is little temperature variation inside the immersed body; it’s ~ isothermal

Picture: KJ05

BihL

R

R;

A

LR;

hAR charchar

conv heat

cond heatcond heatconv heat

1


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Example: cooling of a steel ball /1 A hot steel ball is annealed by dropping it in a

cool oil bath. The ball is d = 0.5” in diameter, mass m = 0.018 lbm and itsinitial temperature is Ti = 400°F. If the heat transfer coefficient betweenball and oil is h= 16 BTU/(h· ft²·°F),

calculate how long it will take for the ball to cool to 150°F. Assume the oil bath temperature constant at Tf = 70°F. Use λsteel = 32.8 BTU/(h· ft· R); cp steel = 0.12 BTU/(lbm· R)

First, convert the data to SI units: d = 0.5” = 1.27 cm; m = 8.2 g;Tf = 70°F = 21°C; Ti = 400°F = 204°C; Tend = 150°F = 66°Cwith 1 BTU/(h· ft²·°F) = 5.678 W/(m2K) → h = 91 W/(m2· K) ;with 1 BTU/(h· ft·R) = 1.731 W/mK → λsteel = 56.8 W/m·Kwith 1 BTU/(lbm· R) = 4.188 kJ/kgK → cpsteel = 0.503 kJ/kg·K

Picture: KJ05

note: ∆T(R) = ∆T(°F)


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Example: cooling of a steel ball /2

Picture: KJ05


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5.4 Convective heat transfer with condensation or boiling


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Convective heat transfer & condensation /1

Phase transitions such as boiling (sv: kokning) and condensation(sv: kondensation) involve rather large amounts of heat, large densitychanges and high heat transfer rates

Condensation occurs when the temperature of a surface is lower than the condensation(i.e. saturation) temperature of a vapour, or Tsurf < Tcondens = Tsat,vapour (for a certainpressure)

Depending on if the liquid can wet the surface(depending on surface tension, (sv: ytspänning) either film condensation(sv: filmkondensation) or dropletcondensation (sv: dropp-kondensation)occurs

Film condensation covers the whole surfacewhiledroplet condensation leaves much of the surface exposed to vapour

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Convective heat transfer & condensation /2 Condensation (and also boiling) can occur

– On horizontal or vertical surfaces, which has an effect on how liquid is removed (by gravity);

– On the inside or outside of tubes Heat transfer rates for droplet

condensation are ~10x higher thanfor film condensation. For droplet condensation, heat transfer coefficients as high as 170 – 290 kW/m2· Khave been reported

For steam, typical heat transfer coefficients are (ÖS96)hcondensation ~ 11000 W/(m2· K)

for film condensationhcondensation ~ 45000 W/(m2· K)

for droplet condensation

Droplet condensation (left) and film condenation (right) of steam on a copper plate(also shown is a 1.7 mm thermo-couple)

source: H89 Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland

69/120


Approximate values for condensation heat transfer coefficients for vapours at 100 kPa

source: H89

Note: the heat transfer rate Q for surface area A (m2) is equal to

Q = h·A·(Tsat,vapour – Tsurf)

≠ h·A·(Tvapour – Tsurf)

If the temperature difference

(Tsat,vapour – Tsurf) is small compared to

(Tvapour – Tsurf) then ”dry” heat convection may be more important, note that h”dry” convection ≠ hcondensation !

.

.


70/120


Non-condensable gases in the vapour greatly reduce the rate of condensation. These gases accumulate at the condenser surface and lower the vapour pressure of the condensing vapour.

When designing a condenser, arrangements for purging inert gases should be included.

If 0.01 < pinert / ptotal < 0.4 this can be taken into account by the emperical correction factor

(for example, 20% air in a condensing vapour decreases the heat transfer by a factor of 4 !)

If pinert / ptotal > 0.4 the heat transfer coefficient can be based on gas cooling without condensation

inert

total

p

p

h

h10.

vapour pure

gases econdensabl-non with mixture

0

0.2

0.4

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Example: condensation & convection (ÖS96-6.5) 1/2

Steam at 180°C, 280 kPa and velocity 15 m/s hits (in cross-flow) a 30 mm diameter round tube with temperature a) 140°C or b) 100°C. Calculate the heat transfer density Q” (W/m2) for these two cases.

Steam data: viscosity η ≈ 1.45·10-5 Pa· s, conductivity λ ≈ 0.028 W/(m· K), Pr ≈ 1.0

.


72/120

Example: condensation & convection (ÖS96-6.5) 2/2

Steam at 180°C, 280 kPa and velocity 15 m/s hits (in cross-flow) a 30 mm diameter round tube with temperature a) 140°C or b) 100°C. Calculate the heat transfer density Q” (W/m2) for these two cases.

Steam data: viscosity η ≈ 1.45·10-5 Pa· s, conductivity λ ≈ 0.028 W/(m· K), Pr ≈ 1.0

.


73/120

Convective heat transfer & boiling /1

Boiling occurs when the temperature of a surface is higher than the condensation (i.e. saturation) temperature of a liquid, or Tsurf > Tcondens = Tsat,liquid (for a certainpressure)

Eight situations can be distinghuished, based on the following possibilities:

– The hot surface is horizontal, or vertical

– The liquid is boiling, or undercooled

– The boiling liquid is flowing, or stagnant

The heat transfer rate Q for surface area A (m2) is equal to Q = h·A·(Tsurf - Tsat,liquid)

≠ h·A·(Tsurf - Tliquid)

A water kettle is a typical horizontal wall, stagnant medium boiler

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The boiling process of a liquid depends primarily on the temperature difference ∆T = Tsurface – Tsat, liquid

Different boiling regimes can be distinguished: (I) free convection, overheating; (II) & (III) nucleation boiling (bubbles formation); (IV) & (V) film boiling (closed vapour blanket); (VI) film boiling with thermal radiation picture: H89



75/120


Boiling (of methanol) on a copper tube: (a) nucleate boiling 243 kW/m2, ∆T = 37°C; (b) transition boling 218 kW/m2, ∆T = 62°C; (c) film boling 41 kW/m2, ∆T = 82°C;

pictures: H89


76/120


Typical numbers for heat transfer coefficients in pipeevaporators (typically vertical heated by condensing steam on the outside) are h = 2000-5000 W/m2·K for dilute aqueous solutions, h = 500-1000 W/m2· K for hydrocarbons.

For boiling water: h = 5000 ~ 17000 W/m2·K (ÖS96)

Simple relations for boiling water heat transfer coefficients at atmospheric pressure

source: H89


77/120

Heat transfer coefficients: overview

Source: BMH99

Unit h and U: W/m2·K

Final comment: often symbol α (see e.g.ÖS96) is used for heat transfer coefficientinstead of h, to avoid confusion or mixing with enthalpy h


78/120

5.5 Radiation heat transfer

See also, for more detailÅA course 424304Process engineering thermodynamics


79/120

Radiation heat transfer /1

Radiative heat transfer (sv: värme-överföringgenom strålning) involves the transfer of heat between surfaces of different temperature separated by a transparent (”diathermal”) medium, by electromagnetic waves

Radiant energy can be exchangedwithout any intervening medium and across (very) long distances.

For heat transfer, most important is the wavelength range 0.3 ≤ λ≤ 100 µm (300 ≤ λ≤ 105 nm), primarily the infrared region

A complicating factor is that radiant energyand also radiant properties of materials are dependent on wavelength λ

Picture: T06


The own* radiation QR (W) from a surface with emissivity ε (-) (sv: emissivitet), surface A (m2) and temperature T(K) equals

QR = ε· σ·A·T4 = ε·A· Eb

with Stefan-Boltzmann coefficientσ = 5.67×10-8 W/m2K4

For a blackbody surface– ε =1 in the Stefan-Boltzmann Law– all incident radiation is absorbed– radiation is maximum for its temperature

at each wavelength– the intensity of the emitted radiation is

independent of direction: it is a diffuse emitter

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Radiation heat transfer /2

Pictures: T06

.

.

* besides reflectedincoming radiation– see below

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0

4TdEb ħ = h/2π, h = Planck’s constant 6.626×10-34 J.s c0= vacuum speed of light 2.998×108 m/skB = Boltzmann’s constant 1.381 ×10-23 J/KT = temperature K, λ = wavelength m

Blackbody radiation Planck’s

radiation Lawgives the spectraldistributionof the radiationemitted by a blackbody(sv: svart kropp, svart strålning)

The area under the curve equalsthe radiation(W/m2):

82/120

Emissivity, gray surfaces

)(W/m 2

0

4Td)T,(E

Note:radiation heat transfer can be important also at lowtemperatures such as roomtemperature !

For a blackbody, the wavelength λmax for whichthe intensity is maximal is related to temperature T byλmax·T = 2898 µm· Kλmax = 10 µm @ T = 300 K

λmax = 0.5 µm @ T = 6000 K

Emissivity of a gray (sv: grå) surface is defined as ε = Eλ / Eλb = constant

Wien’s displacement law

83/120

• For a blackbody, the wavelength λmax for whichthe intensity is maximal is related to temperature T by

λmax· T = 2898 µm· K

Wien’s displacement law

λmax = 10 µm @ T = 300 K

λmax = 0.5 µm @ T = 6000 K

(Tsun ~ 5800 K)

Note: radiation heat transfer can be important also at low temperaturessuch as room temperature !


Emissivity, non-black surfaces

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• Emissivity of a real surface is defined by

for so-called hemi-sphericalemissivity ε, or ε(T).

• For a specific wavelength, the spectral hemi-sphericalemissivity is ελ or ελ(T).

)(

),(),(

)(

),(),(),(

4

0 0

TE

TET

TT

dTETdTE

b

b

• For a graybody, no effect of wavelength, or direction :ε = ελ = f(T) ≠ f(λ)


85/120

Tables and pictures: KJ05

Radiation interaction with (a) a general surface; (b) a black surface.

Gray bodies - emissivity data

86/120

Emission, reflection, transmission An energy balance for a

surface depends on reflected, absorbed, transmitted and of course emitted energy

Reflectivity ρ (-), absorptivity α (-) and transmittivity τ (-) are defined as

Energy balance → ρ + α + τ = 1For an opaque (sv: opak, ogenomskinlig) material τ = 0 → ρ + α = 1More exact, for incident angle θ, temperature T and wavelength λρ = ρ(λ,θ,T), α = α(λ,θ,T), τ = τ(λ,θ,T), as also ε = ε(λ,θ,T)

Picture: KJ05

energy incident

energy dtransmitte ;

energy incident

energy absorbed ;

energy incident

energy reflected

Incident angleθ with respectto normal


For all gray surfaces:

absorptivity = emissivity α(λ,θ,T) = ε(λ,θ,T), or α = ε

An exception to this law would mean a violation of the Second Law of Thermodynamics, since it would allowheat transfer from cold to hot surfaces

87/120

Kirchhoff’s Law

(Consider the situation in the Figure: a small, gray, diffuse body with surface A at temperature T1 in an evacuated oven with black walls at temperature T2, and T2 > T1. The object receives radiation from the walls: Q = α·σ·T2

4 (W) and after some time itstemperature has risen to T2. It will emit radiation whichis then at a rate Q = ε·σ·T2

4 (W) Thus α = ε, but strictly speaking only if the twosurfaces that ”exchange” radiation are at the same temperature ! In practice OK for ∆T up to a few 100 K.)

Picture: KJ05

.

.


88/120

The radiant heat transfer of a surfacewith the surroundings can be expressed as

Qrad = ε·A·σ·(T4 – T4surroundings)

assuming that– the surface and the surroundings are isothermal– the whole surface A is incident with the

surroundings– the surroundings behave as a blackbody or

the area of the surroundings >> A– the surface is a diffusive emitter and reflector

The effect of an incident angle θ with respect to the normal to the surface:

Lambert’s cosine Law : Q(θ) = Q(0)·cos(θ)

Radiant heat transfer /3

Picture: T06

.

. .

89/120

Example: radiation from the sun /1

Consider the sun as a blackbody radiator at 5800 K. The diameter of the sun is ~ 1.39×109 m, the diameter of earth is ~ 1.29×107 m; the distance between earth and sun is ~ 1.5×1011 m.

Calculate:1. The total energy emitted by the sun (W)2. The amount of the sun’s energy

intercepted by earth (W)3. The solar radiation flux (W/m2) that

strikes reaches the earth’s atmosphereat a right angle. Compare it with a measured solar flux of 720 W/m2

measured on a clear day in Colorado, USA. Pictures: T06


90/120

Example: radiation from the sun /2

Picture: T06

Picture: KJ05


91/120

Radiosity: opaque, gray surfaces• For non-black surfaces the

radiation flux Q”R,out leaving a surface equals the sum of emittedown radiationε· Eb = ε· σ·T4 plus (partially) reflected incoming radiation flux ρ· Q”R,in = ρ·G

• Thus

Q”R,out = Q”R,own,out + ρ· Q”R,in or J = ε · Eb + (1-ε)·G

for an opaque gray medium: τ = 0, so α = ε = 1-ρ

• For this, the term radiosity, J, is defined as all the radiationleaving a surface including emitted and reflected radiation

.

.

.

Pic: KJ05

. .


Radiosity: radiation resistance

92/120

• The energy balance for the surfacestates that the net leavingradiation flux Q”R,out,net equals the own emission minus the absorbed incoming radiation:

Q”R,out, net = Q”R, out - Q”R,in = J – G = ε·Eb + (1-ε)·G - G = ε·Eb – ε·G which gives after elimination of G: Q”R,out, net = ε · (Eb - J)/(1 - ε)

• The net heat flow leaving the surface A (m2) is thenQR,out,net = A· ε· (Eb - J)/(1 - ε) = (Eb - J)/RR, with surfaceresistance to radiation RR = (1-ε)/(A·ε)

• For a blackbody surface, resistance RR = 0.

.

Pic: KJ05

...

.

.


93/120

View factorsPictures: KJ05

Fi→j = 0, Fj→i =0 Fi→j = 1, Fj→i = ½Ai = πR2, Aj = 2πR2

ij

Ai Aj

ji

iji dAdA

Sπ

θcosθcos

AF

:geometry any for ,Definition

N

jjiF

The view factors (sv: vinkelfaktorer) Fi→j (also referredto as configuration factors, shape factors, angle factors)quantify how much (i.e. what fraction) of the radiation from surface ”i” reaches other surface”j”, (by a straight-line route) as determined by Lambert’s cosine law, and vice versa.

For any two surfaces Ai and Aj in any geometricalarrangement: Fi→j·Ai = Fj→i·Aj

0 ≤ Fi→j ≤ 1, and for many surfaces:

94/120

Example: view factors

Radiation occurs between floor, walls and ceilingof a room as shown in the Figure. Calculate the view factors from the end wall (1) to the otherfive surfaces (2...6) using the diagram given.

Pictures: KJ05


95/120

Radiant heat transfer /4 Black surfaces

For two black surfaces, some of the radiation form surface ”1” strikes surface ”2” and vice versa

For two black surfaces with different temperatures, the net radiative heat transfer between them equals

Q12 = radiation leaving ”1” and arriving at ”2” minus radiation leaving ”2” and arriving at ”1”= Eb1A1F1→2 - Eb2A2F2→1 ,

and with A1F1→2 = A2F2→1 this gives

Q12 = A1F1→2 (Eb1 – Eb2) = A1F1→2 σ(T14 – T2

4)

Picture: KJ05.

.


96/120

Radiant heat transfer /5 Gray surfaces

with radiation resistance RR= 1/(A1F1→2)

• Thus – see next page and p. 91 above – there are three resistances: 1-ε/A1·ε1, 1-ε2/A2·ε2, and 1/(A1F1→2) = 1/(A2F2→1)

In analogy with radiation betweenblackbodies, with J instead of Eb:Q12 = radiation leaving ”1” and arriving at ”2” minus radiationleaving ”2” and arriving at ”1”

= J1A1F1→2 – J2A2F2→1 ,

and with A1F1→2 = A2F2→1 this gives

Q12 = A1F1→2 (J1 – J2) = (J1 – J2) / R1→2

Pic: KJ05

.

. Two isothermal, gray, diffuse surfaces exchanging heat by radiation.


Heat radiation: two general surfaces

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Surface A1

at temperature T1

with emissivity ε1

Surface A2

at temperature T2

with emissivity ε2

F1→2 = view factor A1 → A2

Radiative heat transferSurface A1 → surface A2

εε

AA

Fεε

)TT(Aσ

εAε

FAεAε

)TT(σQ

and similar for Q2→1


98/120


For radiation between two parallel plates ”1” and ”2” with temperaturesT1 and T2, emissivities ε1 and ε2, and noting that F1→2 = 1 = F2→1, and surfaces A1 = A2 = A this gives the result

For the special case of surface A2 completely surrounding A1, noting that F1→2 =1 and F1→2·A1 = F2→1· A2 , the result is

This simplifies for A2 >> A1 (for example earth surrounded by space) to

1

11

21

42

41

12

TTAQ

2

1

21

42

411

12

111

A

A

TTAQ

42411112 TTAQ

99/120


Radiation in a two-surface enclosurePicture: KJ05


100/120

Example: radiation between gray surfaces

A tube that transports steam at 150°C runsalong a long corridor as illustrated in the Figure. The temperature of all walls is 10°C and wall emissivity is 0.85.The temperature of the tube on the outsideof the isolation at diameter 0.15 m is measured to be 25°C, at an emissivity of 0.85.

a. Calculate the heat losses per meter tube length as a result of heat radiation, and

b. By how much would the heat losses decrease if the isolation is painted with Al-paint with emissivity 0.42

Source: ÖS96-7.4


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Example: radiation between gray surfaces


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Radiative heat transfer coefficient

_

_ _

If the temperature difference ∆T = T1 – T2 of a radiative heat transfer process is small compared to T1 and T2, a radiative heat transfer coefficient hR can be defined by (for the radiation from surface ”1” to ”2”)

Q”R = hR· ∆T = QR / A1 = ε1· σ· (T14 - T2

4) (W/m2)

→ hR = ε1· σ· (T14-T2

4)/(T1-T2) = ε1· σ· (T12 + T2

2)· (T1+T2)

using (x4-y4) = (x2-y2)·(x2+y2) = (x-y)·(x+y)·(x2+y2)

This can also be simplified to hR = ε1· σ· 4·T3 where

T = ½· (T1+T2) → 4T3 ≈ (T12 + T2

2)· (T1+T2)

103/120

Example: convection and radiationfrom an outdoor grill The outside surface temperature of an

outdoor grill is Ts = 50°C. It looses heat to the surroundings by natural convection withh = hconv= 5.4 W/m2K and by heat radiation with emissivity = 0.87.The grills surface area is A = 0.63 m2. For surroundings temperature Tsurr = 20°C, calculate the heat transfer from the grill to the environment.

Picture: KJ05


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Radiative heat transfer resistance

Radiative heat transfer resistance is defined as Rheat, R = ∆T/QR = ∆T/(Q”R·A)

similar to conductive and convective heat transfer

With Q”R /∆T = hR = ε1· σ· (T12 + T2

2)·(T1+T2)

≈ ε1· σ· 4·T3

this gives

Rheat,R = 1 / (hR·A)

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More on heat radiation VST course 424304.Useful literature on Radition Heat Transfer: IdWBL07, BS06

105/120

5.6 Non-stationary (”transient”)conductive heat transfer(short introduction)

See also VST course 424508 ”Transport processes”


106/120

Non-steady heat conduction

where in principle heat Q is a 3-dimensional vector Q that creates (or is the result of !) a vector temperature gradient:(in Cartesian coordinates)

Non-steady or transient (sv: övergående) heat conductionthrough a stagnant medium depends not only on heat conductivity λ but also on heat capacity c (or cp, cv). A general energy balance for mass m gives

t

TcmQQ outin

z

T,

y

T,

x

TT

Picture: ÖS96

.

.


107/120

Transient heat conduction 1-D /1

For 1-dimensional transient heat conduction in a balance volume dVwith mass dm = ρ· dV = ρ·A·dx :

2

2

2

2

x

Ta

t

T

x

TA

x

xT

A

t

TAc

x

TA

x

Q

t

TcA

dxx

Q

t

TcdmQQ outin

-

-Q Laws Fourier' with

Aρdm/dx with

x

w

L

dx

Q.

A = L·w

”Diffusion equation”

108/120


The initial and boundary conditions(sv: start- och randvillkor) determine a heat transfer process

The three most important cases are:

0 t for t))T(0,- (Th x

t)T(0,- and0 t for T T(x,0)

:0t at h0 convection surface of change Sudden 3.

0t for Q x

t)T(0,- and0 t for T T(x,0)

:0t at Q0 flux heat surface of change Sudden 2.

0t for Tt)T(0, and0 t for T T(x,0)

:0t at T T etemperatur surface of change Sudden 1.

surr0

"0x0

"0x

10

10

x

w

L

dx

Q.

A = L·w

x

w

L

dx

Q.

Q.

A = L·w

109/120


Case 1: Assume a material with flat boundary at x=0, infinite length in x-direction, with T=T0 at all x

At time t≥0 the temperature at x=0 is increased to T=T1

and heat starts to enter (diffuse into) the material. At x→∞, T stays at T0.

conditions initial

and boundary&

2

2

x

Ta

t

T

Picture: BMH99

110/120

Transient heat conduction 1-D /4 With dimensionless variables

θ = (T-T0)/(T1-T0)

and

ξ = x / (4at)½

this gives the following solution:

)(yerfde

deTT

TT

y

at

x

0

4

001

1

2

2

2

21

with

ÖS96: erf(x) ≈ 1 - exp(- 1.128x - 0.655x2 - 0.063x3)

111/120

Transient heat conduction 1-D /5 At x = 0 the slope of the

penetration profile linesequals

∂T/∂x = -(T1-T0)/(πat)½

where x = (πat)½

is referred to as penetration depth.

Fourier number Fo is (for heat transfer) defined asFo = at/d2 = t /(d2/a)) for a medium with thickness d Fo gives the ratio between time t and the penetration time d2/a

The penetration depth concept is valid for Fo < 0.1Picture: BMH99


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Transient heat conduction 1-D /6 Depending on how long the conduction process goes on,

the medium can be considered semi-infinite if the otherside doesn’t notice any change: Fo < 0.1

If the penetration depth ~ medium dimensions, i.e. for Fo > 0.1 the medium is finite and the penetration theory cannot be used

For finite media, temperature profilesymmetry gives results of the type(T-T0)/(T1-T0) = f1(t)· f2(x/R) *)for half-size R = ½· d

For simple geometries, diagrams can be used to find average or centretemperatures – see next page

*) ”Separation of variables”- see course 424508 Picture: BMH99

d


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Mean (left) and centre (right) temperature of a body duringnon-stationary heat conduction Pictures: BMH99


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Example: transient heat conduction (1-D)

An asphalt layer of 6 cm is put on a road at temperatureT = 35°C in stagnant air of 10°C. The temperature of the road the asphalt is put on is also 10°C.

How long does it take for the asphalt to reach an averagetemperature <T> of 15°C and what is the centretemperature Tm of the asphalt layer then?

Assume that the temperature profilein the asphalt is symmetric. Data for the asphalt: ρ = 1000 kg/m3, cp=920 J/(kg· K), λ=0.17 W/(m· K)

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inde

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m


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Example: transient heat conduction (1-D)

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://w

ww

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m


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For a plane wall at initial temperature Ti, a sudden convective heat transfer with surrounding temperature Tf and heat transfer coefficient h the problem can be described as

When heat transfer at the medium surface is slow compared to conduction inside the medium, Bi << 1 and the temperature profilein the medium is uniform

Lx for T)hA(Tx

TλA- L;x for )ThA(T

x

TλA-

0;T for TT(x,0) conditions boundary & initial with

x

Ta

x

T

ff

i

2

2

Picture: KJ05


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(Transient heat conduction 1-D /9)

The solution is in the form of an infinite series, which for Bi < 0.1 gives the same result as the lumped system approximation discussed earlier:

1X for θBiX

θ 0,τ for 1θ

with X

θ

τ

θ gives

λ

hL Bi Fo,

L

atτ ,

L

xX ,

T-T

T-Tθ

variables ess dimensionl Using

2

2

2fi

f

10. for Bi

FoBitcm

Ahθ

p

expexpPictures: KJ05

See also section 5.3

λ

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(Transient heat conduction 3-D) For 3-dimensional transient

heat conduction in a balancevolume dV with massdm =ρ· dV = ρ· dx· dy· dz the temperature gradient vector is

and the energy balance gives

(in Cartesian coordinates)

With the appropriateinitial- and boundaryconditions this can be solved, usuallyrequiring a numericalmethod→ VST course 424508

Transport processes

z

T

y

T

x

TT ,,

Taz

T

y

T

x

Ta

t

T 22

2

2

2

2

2

Picture: ÖS96


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Heat transfer: final remarks During / at the same time as a heat transfer process, heat may be

generated (or taken up), for example as a result of an exothermic (or endothermic) chemical reaction. This can be taken into account by including it into the balance equations:

Also, the contact between two media or materials is typically not perfect and a contact resistanceshould be taken into account based on surface roughness etc., giving small gaps of typically 0.5 – 50 µm

Tabelised data is used.

Picture: KJ05

)(W/mq production heat for 3qz

T

y

T

x

T

t

Tc

2

2

2

2

2

2


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Sources #5 BMH99: Beek, W.J., Muttzall, K.M.K., van Heuven, J.W. ”Transport phenomena” Wiley,

2nd ed. (1999) BS06: H.D. Baehr, K. Stephan ”Wärme- und Stoffübertragung”, 5. ed., Springer (2006)

Chapter 5 BÖ88: Y. Bayazitoglu, M. Necati Özisik “Elements of heat transfer“ McGraw-Hill (1988) H89: J.P. Holman ”Heat transfer” McGraw-Hill (1989) Chapter 9 IdWBL07: F.P. Incropera, D.P. DeWitt, T.L. Bergman, A.S. Lavine ”Funda-mentals of

heat and mass transfer”, Wiley, 6th ed. (2007) Chapters 12 & 13 KJ05: D. Kaminski, M. Jensen ”Introduction to Thermal and Fluids

Engineering”, Wiley (2005) S04: B. Sörensen, “Renewable energy” 3rd ed. Elsevier Academic Press (2004) SSJ84: J.M. Smith, E. Stammers, L.P.B.M Janssen ”Fysische Transportverschijnselen I” TU

Delft / D.U.M. (1984) (in Dutch) T06: S.R. Turns ”Thermal – Fluid Sciences”, Cambridge

Univ. Press (2006) rz13: R. Zevenhoven, ”Massöverföring & separations- teknik”

kursmaterial 424302 Åbo Akademi University (2013) ÖS96: G. Öhman, H. Saxén ”Värmeteknikens grunder”, Åbo Akademi

University (1996)


Three heat transfer mechanisms - Åbo Akademi · and heat transfer coefficient h, unit: W/(m2·K)...

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Transcript of Three heat transfer mechanisms - Åbo Akademi · and heat transfer coefficient h, unit: W/(m2·K)...