convective heat transfer (6) forced convection (8) heat transfer for an isothermal tube r dr x dx t...
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Convective Heat Transfer (6)Forced Convection (8)Martin Andersson
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Agenda
Convective heat transfer Continuity eq. Convective duct flow (introduction to ch. 8)
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Convective heat transfer
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Convective heat transfer
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Convective heat transfer
AttQ wf )( =
vrmeledande kropptf
y
x
tw(y)
t(x,y)fluid
Heat conducting body
-
x = 0 u, v, w = 0 heat conduction in the fluid
AttQ wf )( =
0/ (6 3)f
x
w f w f
tyQ A
t t t t
=
= =
vrmeledande kropptf
y
x
tw(y)
t(x,y)fluid
heat conducting body
Convective heat transfer
Introduction of heat transfer coeficient:
-
Convective heat transfer
Objective (of chapter 6-11): Determine and the parameters influencing it for
prescribed tw(x) or qw(x) = Q/A
-
Order of magnitude for
Medium W/mKAir (1bar); natural convection 2-20Air (1bar); forced convection 10-200Air (250 bar); forced convection 200-1000Water forced convection 500-5000Organic liquids; forced convection 100-1000Condensation (water) 2000-50000Condensation (organic vapors) 500-10000Evaporation, boiling, (water) 2000-100000Evaporation, boiling (organic liquids) 500-50000
-
How to do it? (to describe convective HT)What are the tools?
Fluid motion:Mass conservation equation (Continuity eqn)Momentum equations (Newtons second law)
Energy balance in the fluidFirst law of thermodynamics for an open system
-
Continuity eq.Expresses that mass is constant and not destroyable
Especially for
steady state, incompressible flow, two-dimensional case
)56(0yv
xu
=
+
)46(0)w(z
)v(y
)u(x
=
+
+
+
-
Resulting momentum equations 2 dim.
+
+
=
+
+
2
2
2
2
:yu
xu
xpF
yuv
xuuux x
+
+
=
+
+
2
2
2
2
:yv
xv
ypF
yvv
xvuvy y
Inpossible to solve by hand, need to be simplified (chapter 6, 7 and 8)
-
Temperature Equation
+
+
=
+
+
2
2
2
2
2
2
zt
yt
xt
cztw
ytv
xtu
p
Unpossible to solve by hand, need to be simplified (chapter 6, 7 and 8)
-
Boundary layer approximations (laminar case)
u(x,y)
(x)
U
y x
T
y x
twt(x,
y)t
Why different fields for temperature and velocity ???
-
Boundary layer theory developed by Prandtl
vu >>
yv
xv
xu
yu
>> ,,
xt
yt
>>
If the boundary layer thickness is very small
If 2D
-
Boundary layer approximations Prandtls theory
)(xpp =
2
2
yu
dxdpF
yuv
xuu x
+=
+
2
2
yt
cytv
xtu
p
=
+
From Navier-Stokes eqn in the y-direction it is found that the pressure is independent of y
Then Navier Stokes in the x-direction is simplified to:
Also the temperature field is simplified
-
Boundary layer approximations Prandtls theory
21 konstant2
p U+ =
dxdUU
dxdp =
pp cc ==Pr
Bernoullis eqn describes the flow outside the boundary layer
The dimensionless Prandtl number is introduced
-
Boundary layer equations
0=
+
yv
xu
2
2
yu
dxdUU
yuv
xuu
+=
+
Pr 2
2
yt
ytv
xtu
=
+
Mass conservation
Momentum conservation
Energy conservation
-
Boundary layers
UU
yx
laminar boundarylayer
transition turbulent boundary layer
fully turbulentlayer
buffer layerviscous sublayer
xc
/Re cc xU=
5105
Pr)(Re,Nu 7f=
-
Continuity eq. (expresses that mass is constant and not destroyable)
x:
Net mass flow out in x-direction
Analogous in y- and z-directions
Net mass flow out Reduction in mass within volume element
dzdyum 1x =
2 1
1
( )
xx x
mm m dxx
u dy dz u dx dy dzx
= + =
= +
dzdydx)u(x
mx
=
dydxdzwz
mdzdxdyvy
m zy )( )(
=
=
zyx mmm ++:out flow Net
y
xz dx
dy
dz
-
Cont. continuity eq.
Reduction per time unit:
Especially for steady state, incompressible flow, two-dimensional case
dzdydx
)w(z
)v(y
)u(x
+
+
=
)46(0)w(z
)v(y
)u(x
=
+
+
+
)56(0yv
xu
=
+
Consider a mass balance for the volume element at the previous page
-
Navier Stokes ekvationer (eqs.)Derived from Newtons second law
m = dx dy dz
but
u = u(x, y, z, ), v = v(x, y, z, ),
w = w(x, y, z, )
Fam
=
=
ddw,
ddv,
ddua
-
Forces
F
a. volume forces (Fx , Fy , Fz) are calulated per unit mass, unit ???
b. stresses ij unit ???
The surface forces act on the boundary surfaces of the fluid element and are acting as either normal forces or shear forces
-
Forces
F
a. volume forces (Fx , Fy , Fz) are calulated per unit mass, N/kg
b. stresses ij N/m2
The surface forces act on the boundary surfaces of the fluid element and are acting as either normal forces or shear forces
-
Forces The surface forces are calulated per unit area and are called stresses
F
=
zzzyzx
yzyyyx
xzxyxx
ij
a. volume forces (Fx , Fy , Fz) are calulated per unit mass, N/kg
b. stresses ij N/m2
"z""y""x"
Stresses for an element dxdydz
-
Examples of stresses
xupep xxxx
+=+= 22
+
===xv
yuexyyxxy 2
yvpep yyyy
+=+= 22
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Resulting momentum equations
6.20
6.21
+
+
=
+
+
2
2
2
2
:yu
xu
xpF
yuv
xuuux x
+
+
=
+
+
2
2
2
2
:yv
xv
ypF
yvv
xvuvy y
-
Energy eq. (First law of thermodynamics of an open system), Temperature field eq.
dHQd =
y
xz dx
dy
dz
Net heat to element =
Change of enthalpy flow
Neglecting kinetic and potential energy
-
Heat conduction in the fluid (calulating the heat flow as in chapter 1)
Qd
dxdydz)xt(
xdydz
xt
dxx
QQQ
xtdydz
xtAQ
xxdxx
x
=
=
+=
=
=
+
dxdydz)xt(
xQQQ xdxxx
== + 6.24
-
.Analogous in y- and z-directions
(6-27)
dydxdz)zt(
zQ
dzdxdy)yt(
yQ
z
y
=
=
{ }zyx QQQQd ++=
Qd heatfor convention sign
dzdydx)zt(
z)
yt(
y)
xt(
xQd
+
+
=
6.25
6.26
-
Enthalpy flows and changes
Flow of enthalphy in the x-direction
hdzdyuhmH xx ==
dzdydxxhudzdydx
xuhHd x
+
= 6.28
-
Enthalpy changes
y- and z-directions
dzdydxyhvdzdydx
yvhHd y
+
=
dzdydxzhwdzdydx
zwhHd z
+
=
6.29
6.30
-
Total change in enthalpy
x y zdH dH dH dH
u v wh dx dy dzx y z
h h hu v w dx dy dzx y z
= + + =
= + + +
+ +
-
Enthalpy vs temperature
( )
dtthdp
phdh
tphh
pt
+
=
=
,
6.33
-
Enthalpy vs temperature
p
p thc
=
For ideal gases the enthalpy is independent of pressure, i.e.,
0)/( tph
.For liquids, one commonly assumes that the derivative
tph )/( is small and/or that the pressure variation dp is small compared to the change in temperature.
Then generally one states dtcdh p=
By definition
i.e., enthalphy is coupled to temperature via the heat capacity
-
Temperature Equation
6.36
Rewriting eqn 6.32 as a fuction of t instead of h
+
+
=
+
+
2
2
2
2
2
2
zt
yt
xt
cztw
ytv
xtu
p
-
Summary
Velocity profile equals the
temperature profile when Pr=1
Rec = 2300
Rec = 500000
-
.
Chapter 8 - Convective Duct Flow
-
.
laminar if pipe or tube ReD < 2300
=
DuRe mDmAum =
=
2
m Rr12
uu
= 2
2
max by1
uu
krna, core
grnsskikt, boundarylayer
fullt utbildadstrmning, fullydeveloped flow
2bU0 xy
Chapter 8 - Convective Duct Flow
Parallel plate duct
Circular pipe, tube
-
.
Di Re0575.0
DL
=
y
-y
grnsskikt, boundary layer
grnsskikt, boundarylayer
krna, core
Chapter 8 Convective Duct Flow
krna, core
grnsskikt, boundarylayer
fullt utbildadstrmning, fullydeveloped flow
2bU0 xy
Typical entrance region flow profile
-
Cont. duct flow
If ReD > 2300
fullt utbildadturbulent strmning
U0
omslag, transition
laminrt grnsskikt turbulent grnsskikt
xy
Laminar boundary layer Turbulent boundary layer
Fully developed turbulent flow
-
. Pressure drop fully developed flow
Dh = hydraulic diameter
2u
DLfp
2m
h
=
ReCf =
= hm
DuRe
4 4 x cross section area = perimeter
tvrsnittsareanmedieberrd omkrets
=
Amum
=
-
.
0 1 2 3 4 5 61E-4
0.001
0.01
0.1
2/2mup
Pressure drop - entrance region (circular pipe)
x/DhReDh
Figure 8.4
-
Convective heat transfer for an isothermal tube
rdr
x dx
tw= konstant; constant
-
Velocity field fully developed
Heat balance for an element dxdr2r
Heat conduction in radial directionEnthalpy transport in x-direction
=
2
m Rr1u2u
rdr
x dx
tw= konstant; constant
Convective heat transfer for an isothermal tube
-
Boundary conditions:
x = 0 : t = t0r = R : t = twr = 0 : (Symmetry)
)"128("rtr
rr1
xtucp
=
0rt=
Remember the figure from previous page
Energy eqn for steady state
Commonly called Graetz problem
-
Introduce
Introduce u = 2um (1 - r2)
or
p
,w
ca
tt,Rxx,Rrr
=
===
=
rRrR
rRrR1
xR1
au
=
rr
r)r1(r1
xaRu2
2m
)208(r
rr)r1(r
1x
PrRe 2D
=
Nusselt approach still valid
Coupling between velocity and average velocity
-
)208(r
rr)r1(r
1x
PrRe 2D
=
=
+
=
rtr
rr1a
rt
r1
rtat 2
2
compared with unsteady heat conduction:
Steady heat conduction
-
Assume = F(x) G(r).
After some (8.24-8.28) calculations one finds
0 < 1 < 2 < 3 < 4
)298(e)r(GC0i
PrRe/xii
D2i =
=
Inspiration for one of the teoretical home assignments
-
.
0 = 2.705
1 = 6.667
2 = 10.67
Temperature profile in the thermal entrace region for a circular pipe with uniform wall temperature and fully developed laminar flow
-
Bulktemp. tB
the enthalpy flow of the mixture:
the enthalpy flow can be written
Bptcm
R
0 "h"p
"m"
tcurdr2
==
=
R
0m
2
R
0B
u4Drdru2m
drurt2m1t
)348(
urdr
urtdrt R
0
R
0B =
umdx
-
Local Nusselt number
0.001 0.01 0.1 11
5
10
50
100
3.656konstant temperatur,constant wall temperature
konstant vrmeflde, uniform heat flux
konstant vrmeflde, uniform heat fluxHastighetsflt ej fullt utbildat; velocity field not fully developed
4.364
PrRex/D
D
NuD
-
Average Heat Transfer Coefficient
1.
If the velocity field is fully developed
N.B.! Higher values if velocity field not fully develped. Eq. (8-38) gives the average value .
tw = constant
[ ]
14.0
3/2/PrRe04.01/PrRe0668.0656.3
++==
w
B
D
DD
xDxDDNu
)388(D/LPrRe86.1DNu
14.0
w
B3/1
DD
=
=
1.0PrRe
D/LD
+
-
.
2.
Fully developed flow and temperature fields
NuD = 4.364 (8-50)
qw = (tw tB)
tw tB = konst.
If tB increases, tw must increase as much
tw highest at the exit!
constconst
t0
...
...
qw = konst. = constant
Convective Heat Transfer (6)Forced Convection (8)Agenda Convective heat transferConvective heat transferConvective heat transferSlide Number 6Convective heat transferOrder of magnitude for How to do it? (to describe convective HT)What are the tools?Continuity eq.Expresses that mass is constant and not destroyableResulting momentum equations 2 dim.Temperature EquationBoundary layer approximations (laminar case)Boundary layer theory developed by PrandtlBoundary layer approximations Prandtls theoryBoundary layer approximations Prandtls theoryBoundary layer equationsBoundary layersContinuity eq. (expresses that mass is constant and not destroyable)Cont. continuity eq.Navier Stokes ekvationer (eqs.)Derived from Newtons second lawForcesForcesForces The surface forces are calulated per unit area and are called stressesExamples of stressesResulting momentum equationsEnergy eq. (First law of thermodynamics of an open system), Temperature field eq.Slide Number 28.Enthalpy flows and changesEnthalpy changesTotal change in enthalpyEnthalpy vs temperatureEnthalpy vs temperatureTemperature EquationSummary...Cont. duct flow..Convective heat transfer for an isothermal tubeSlide Number 44 .Bulktemp. tBLocal Nusselt numberAverage Heat Transfer CoefficientSlide Number 53.
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