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5. Heat transfer
Ron ZevenhovenÅbo Akademi University
Thermal and Flow Engineering / Värme- och strömningstekniktel. 3223 ; ron.zevenhoven@abo.fi
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
∆..
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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) ]
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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
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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
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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
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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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Example: convective heat transfer /2
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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Example: convective heat transfer /2
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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Example: convective heat transfer: tube flow /2
Picture: KJ05
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Example: convective heat transfer: tube flow /3
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)
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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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Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland
H = 2.5 m
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5.3 Combined conduction andconvection heat transfer; the lumped system approximation
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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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Example: Conduction + convection /2
Pictures: KJ05
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Example: Conduction + convection /3
Pictures: KJ05
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Example: Conduction + convection /4
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
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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
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Convective heat transfer & condensation /3
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 !
.
.
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Convective heat transfer & condensation /4
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
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Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland
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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
.
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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
.
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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
Convective heat transfer & boiling /2
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Convective heat transfer & boiling /3
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
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Convective heat transfer & boiling /4
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
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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
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5.5 Radiation heat transfer
See also, for more detailÅA course 424304Process engineering thermodynamics
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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
Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland
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):
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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
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• 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 !
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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).
)(
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• For a graybody, no effect of wavelength, or direction :ε = ελ = f(T) ≠ f(λ)
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Tables and pictures: KJ05
Radiation interaction with (a) a general surface; (b) a black surface.
Gray bodies - emissivity data
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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
Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland
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
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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
.
.
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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
.
. .
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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
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Example: radiation from the sun /2
Picture: T06
Picture: KJ05
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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
. .
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Radiosity: radiation resistance
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• 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
...
.
.
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View factorsPictures: KJ05
Fi→j = 0, Fj→i =0 Fi→j = 1, Fj→i = ½Ai = πR2, Aj = 2πR2
ij
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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:
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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
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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.
.
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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.
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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
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εAε
FAεAε
)TT(σQ
and similar for Q2→1
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Radiant heat transfer /6 Gray surfaces
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
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A
A
TTAQ
42411112 TTAQ
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Radiant heat transfer /7 Gray surfaces
Radiation in a two-surface enclosurePicture: KJ05
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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)
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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
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5.6 Non-stationary (”transient”)conductive heat transfer(short introduction)
See also VST course 424508 ”Transport processes”
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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
.
.
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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”
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Transient heat conduction 1-D /2
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
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Transient heat conduction 1-D /3
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
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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)
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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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Transient heat conduction 1-D /7
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)
http
://w
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uk/b
inde
r.ht
m
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Example: transient heat conduction (1-D)
http
://w
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.rec
omac
.co.
uk/b
inde
r.ht
m
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Transient heat conduction 1-D /8
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)
Åbo Akademi University | Thermal and Flow Engineering | 20500 Turku | Finland