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Convective Heat Transfer (6)Forced Convection (8)Martin Andersson
Agenda
• Convective heat transfer• Continuity eq.• Convective duct flow (introduction to ch. 8)
Convective heat transfer
Convective heat transfer
Convective heat transfer
AttQ wf )( −=α!
värmeledande kropptf
y
x
tw(y)
t(x,y)fluid
Heat conducting body
x = 0 ⇒ u, v, w = 0 ⇒ heat conduction in the fluid
AttQ wf )( −=α!
!
0/ (6 3)f
x
w f w f
tyQ A
t t t t
λ
α =
⎛ ⎞∂⎜ ⎟∂⎝ ⎠= = − −
− −
&
värmeledande kropptf
y
x
tw(y)
t(x,y)fluid
heat conducting body
Convective heat transfer
Introduction of heat transfer coeficient:
Convective heat transfer
Objective (of chapter 6-11): Determine α and the parameters influencing it for
prescribed tw(x) or qw(x) = Q/A
Order of magnitude for α
Medium α W/m²KAir (1bar); natural convection 2-20Air (1bar); forced convection 10-200Air (250 bar); forced convection 200-1000Water forced convection 500-5000Organic liquids; forced convection 100-1000Condensation (water) 2000-50000Condensation (organic vapors) 500-10000Evaporation, boiling, (water) 2000-100000Evaporation, boiling (organic liquids) 500-50000
How to do it? (to describe convective HT)
What are the tools?
Fluid motion:Mass conservation equation (Continuity eqn)Momentum equations (Newton’s second law)
Energy balance in the fluidFirst law of thermodynamics for an open system
Continuity eq.Expresses that mass is constant and not destroyable
Especially for
steady state, incompressible flow, two-dimensional case
⇒
)56(0yv
xu
−=∂∂
+∂∂
)46(0)w(z
)v(y
)u(x
−=ρ∂∂
+ρ∂∂
+ρ∂∂
+τ∂ρ∂
Resulting momentum equations – 2 dim.
⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂+
∂
∂+
∂
∂−=⎟
⎟⎠
⎞⎜⎜⎝
⎛
∂
∂+
∂
∂+
∂
∂2
2
2
2
:ˆyu
xu
xpF
yuv
xuuux x µρ
τρ
⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂+
∂
∂+
∂
∂−=⎟
⎟⎠
⎞⎜⎜⎝
⎛
∂
∂+
∂
∂+
∂
∂2
2
2
2
:ˆyv
xv
ypF
yvv
xvuvy y µρ
τρ
Inpossible to solve by hand, need to be simplified… (chapter 6, 7 and 8)
Temperature Equation
⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂+
∂
∂+
∂
∂=
∂
∂+
∂
∂+
∂
∂2
2
2
2
2
2
zt
yt
xt
cztw
ytv
xtu
pρλ
Unpossible to solve by hand, need to be simplified… (chapter 6, 7 and 8)
Boundary layer approximations (laminar case)
u(x,y)
δ (x)
U∞
y x
δ T
y x
t wt(x,y)t∞
Why different fields for temperature and velocity ???
Boundary layer theory developed by Prandtl
vu >>
yv
xv
xu
yu
∂
∂
∂
∂
∂
∂>>
∂
∂ ,,
xt
yt
∂
∂>>
∂
∂
If the boundary layer thickness is very small
If 2D
Boundary layer approximations – Prandtl’s theory
)(xpp =
2
2
yu
dxdpF
yuv
xuu x
∂
∂+−=⎟
⎟⎠
⎞⎜⎜⎝
⎛
∂
∂+
∂
∂µρρ
2
2
yt
cytv
xtu
p ∂
∂=
∂
∂+
∂
∂
ρ
λ
From Navier-Stokes eqn in the y-direction it is foundthat the pressure is independent of y
Then Navier Stokes in the x-direction is simplified to:
Also the temperature field is simplified
Boundary layer approximations – Prandtl’s theory
21 konstant2
p Uρ+ =
dxdUU
dxdp
ρ−=
λ
µ
λ
ρν pp cc==Pr
Bernoullis eqn describes the flow outside the boundary layer
The dimensionless Prandtl number is introduced
Boundary layer equations
0=∂
∂+
∂
∂
yv
xu
2
2
yu
dxdUU
yuv
xuu
∂
∂+=
∂
∂+
∂
∂
ρµ
Pr 2
2
yt
ytv
xtu
∂
∂=
∂
∂+
∂
∂
ρµ
Mass conservation
Momentum conservation
Energy conservation
Boundary layers
U∞
U∞
yx
laminar boundarylayer
transition turbulent boundary layer
fully turbulentlayer
buffer layerviscous sublayer
xc
ν/Re cc xU∞=
5105 ⋅
Pr)(Re,Nu 7f=
Continuity eq. (expresses that mass is constant and not destroyable)
x:
Net mass flow out in x-direction
Analogous in y- and z-directions
Net mass flow out ⇒ Reduction in mass within volume element
dzdyum 1x ρ=!
2 1
1
( )
xx x
mm m dxx
u dy dz u dxdy dzx
ρ ρ
∂= + =
∂∂
= +∂
&& &
dzdydx)u(x
mx ρ∂∂
=Δ !
dydxdzwz
mdzdxdyvy
m zy )( )( ρρ∂
∂=Δ
∂
∂=Δ !!
zyx mmmutströmmatNetto !!! Δ+Δ+Δ:out flownet ,
y
xz dx
dy
dz
Cont. continuity eq.
Reduction per time unit:
Especially for steady state, incompressible flow, two-dimensional case
⇒
dzdydxτ∂ρ∂
! )w(z
)v(y
)u(x
ρ∂∂
+ρ∂∂
+ρ∂∂
=τ∂ρ∂
−
)46(0)w(z
)v(y
)u(x
−=ρ∂∂
+ρ∂∂
+ρ∂∂
+τ∂ρ∂
)56(0yv
xu
−=∂∂
+∂∂
Consider a mass balance for the volume element at the previous page
Navier – Stokes’ ekvationer (eqs.)Derived from Newton’s second law
m = ρ dx dy dz
but
u = u(x, y, z, τ), v = v(x, y, z, τ),
w = w(x, y, z, τ)
Fam!!
=⋅
⎟⎠⎞
⎜⎝⎛
τττ=
ddw,
ddv,
ddua
!
Forces
F!
a. volume forces (Fx , Fy , Fz) are calulated per unitmass, N/kg
b. stresses σij N/m2
The surface forces act on the boundary surfaces ofthe fluid element and are acting as either normal forces or shear forces
Forces The surface forces are calulated per unitarea and are called stresses
F!
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
σσσ
σσσ
σσσ
=σ
zzzyzx
yzyyyx
xzxyxx
ij
a. volume forces (Fx , Fy , Fz) are calulated per unitmass, N/kg
b. stresses σij N/m2
! ! !"z""y""x"
Stresses for an element dxdydz
Signs for the stresses
dyy
σxx
σyy
σxx
σyyσyx
σxy
x
.
Signs for the stresses
Resulting stresses
dyy
σxx
σyy
σxx
σyyσyx
σxy
σyy
σxx
σyxσxy
dy)(y yyyy σ∂∂
+σ
dx)( xxx
xx σ∂∂
+σ
dy)(y yxyx σ∂∂
+σ
dx)(x xyxy σ∂∂
+σ
x
dzdxdy)(z
dydxdz)(y
dxdydz)(x zxyxxx σ
∂∂
+σ∂∂
+σ∂∂
dxdydz)(x jijσ
∂∂
Net force in x-direction (for the 3D case)
Examples of stresses
xupep xxxx∂
∂+−=+−= µµσ 22
⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂+
∂
∂===
xv
yuexyyxxy µµσσ 2
yvpep yyyy∂
∂+−=+−= µµσ 22
Resulting momentum equations
⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂+
∂
∂+
∂
∂−=⎟
⎟⎠
⎞⎜⎜⎝
⎛
∂
∂+
∂
∂+
∂
∂2
2
2
2
:ˆyu
xu
xpF
yuv
xuuux x µρ
τρ
⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂+
∂
∂+
∂
∂−=⎟
⎟⎠
⎞⎜⎜⎝
⎛
∂
∂+
∂
∂+
∂
∂2
2
2
2
:ˆyv
xv
ypF
yvv
xvuvy y µρ
τρ
6.20
6.21
Energy eq. (First law of thermodynamics of an open system), ⇒ Temperature field eq.
dHQd =!
y
xz dx
dy
dz
Net heat to element =
Change of enthalpy flow
Neglecting kinetic and potential energy
.
Heat conduction in the fluid (calulating the heat flow as in chapter 1)
Qd !
dxdydz)xt(
xdydzxt
dxxQQQ
xtdydz
xtAQ
xxdxx
x
∂∂
λ∂∂
−∂∂
λ−=
=∂∂
+=
∂∂
λ−=∂∂
λ−=
+
!!!
!
dxdydz)xt(
xQQQ xdxxx ∂
∂λ
∂∂
−=−=Δ +!!!
6.24
.Analogous in y- and z-directions
(6-27)
dydxdz)zt(
zQ
dzdxdy)yt(
yQ
z
y
∂∂
λ∂∂
−=Δ
∂∂
λ∂∂
−=Δ
!
!
{ }zyx QQQQd !!!! Δ+Δ+Δ−=↑
Qd !heatfor convention sign
dzdydx)zt(
z)yt(
y)xt(
xQd
⎭⎬⎫
⎩⎨⎧
∂∂
λ∂∂
+∂∂
λ∂∂
+∂∂
λ∂∂
=!
!
6.25
6.26
Enthalpy flows and changes
• Flow of enthalphy in the x-direction
hdzdyuhmH xx ρ== !!
dzdydxxhudzdydx
xuhHd x
∂
∂+
∂
∂=⇒ ρρ! 6.28
Enthalpy changes
• y- and z-directions
dzdydxyhvdzdydx
yvhHd y
∂
∂+
∂
∂= ρρ!
dzdydxzhwdzdydx
zwhHd z
∂
∂+
∂
∂= ρρ!
6.29
6.30
Total change in enthalpy
x y zdH dH dH dH
u v wh dxdy dzx y z
h h hu v w dxdy dzx y z
ρ
ρ
= + + =
⎛ ⎞∂ ∂ ∂= + + +⎜ ⎟∂ ∂ ∂⎝ ⎠
⎛ ⎞∂ ∂ ∂+ +⎜ ⎟∂ ∂ ∂⎝ ⎠
& & & &
Enthalpy vs temperature
( )
dtthdp
phdh
tphh
pt⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂+⎟
⎟⎠
⎞⎜⎜⎝
⎛
∂
∂=⇒
=
,
6.33
Enthalpy vs temperature
p
p thc ⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂=
For ideal gases the enthalpy is independent of pressure, i.e.,
0)/( ≡∂∂ tph
.For liquids, one commonly assumes that the derivative
tph )/( ∂∂
is small and/or that the pressure variation dp is small compared to the change in temperature.
Then generally one states dtcdh p=
By definition
i.e., enthalphy is coupled to temperature via the heat capacity
Temperature Equation
⎟⎟⎠
⎞⎜⎜⎝
⎛
∂
∂+
∂
∂+
∂
∂=
∂
∂+
∂
∂+
∂
∂2
2
2
2
2
2
zt
yt
xt
cztw
ytv
xtu
pρλ
6.36
Rewriting eqn 6.32 as a fuction of T instead of h
.
Chapter 8 - Convective DuctFlow
.
laminar if pipe or tube ReD < 2300
ν=
DuRe mDmAum ρ=!
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎟⎠⎞
⎜⎝⎛−=
2
m Rr12
uu
⎟⎟⎠
⎞⎜⎜⎝
⎛−= 2
2
max by1
uu
kärna, core
gränsskikt, boundarylayer
fullt utbildadströmning, fullydeveloped flow
2bU0x
y
Chapter 8 - Convective DuctFlow
Parallel plate duct
Circular pipe, tube
.
Di Re0575.0DL
=
y
-y
gränsskikt, boundary layer
gränsskikt, boundarylayer
kärna, core
Chapter 8 Convective Duct Flow
kärna, core
gränsskikt, boundarylayer
fullt utbildadströmning, fullydeveloped flow
2bU0x
y
Typical entrance region flow profile
Cont. duct flow
If ReD > 2300
fullt utbildadturbulent strömning
U0
omslag, transition
laminärt gränsskikt turbulent gränsskikt
xy
Laminar boundary layer Turbulent boundary layer
Fully developed turbulent flow
. Pressure drop fully developed flow
Dh = hydraulic diameter
2u
DLfp
2m
h
ρ=Δ
ReCf =
ν= hmDuRe
4 4 x cross section area = perimeter
tvärsnittsareanmedieberörd omkrets
×=
Amum ρ
=!
.
0 1 2 3 4 5 61E-4
0.001
0.01
0.1
2/2mup
ρΔ
Pressure drop - entrance region (circular pipe)
x/DhReDh
Figure 8.4
Convective heat transfer for an isothermal tube
rdr
x dx
tw= konstant; constant
Velocity field fully developed
”Heat balance” for an element dxdr2πr
Heat conduction in radial directionEnthalpy transport in x-direction
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛⎟⎠⎞
⎜⎝⎛−=
2
m Rr1u2u
rdr
x dx
tw= konstant; constant
Convective heat transfer for an isothermal tube
⇒
Boundary conditions:
x = 0 : t = t0r = R : t = twr = 0 : (Symmetry)
)"128("rtr
rr1
xtucp −⎟
⎠⎞
⎜⎝⎛∂∂
∂∂
λ=∂∂
ρ
0rt=
∂∂
Remember the figure from previous page
Energy eqn for steady state
Commonly called Graetz problem
Introduce
⇒
Introduce u = 2um (1 - rʹ2)
or
p
,w
ca
tt,Rxx,Rrr
ρλ
=
−=ϑ=ʹ=ʹ
⎟⎠⎞
⎜⎝⎛
ʹ∂ϑ∂ʹ
ʹ∂∂
ʹ=ʹ∂ϑ∂
rRrR
rRrR1
xR1
au
⎟⎠⎞
⎜⎝⎛
ʹ∂ϑ∂ʹ
ʹ∂∂
ʹ−ʹ=ʹ∂ϑ∂
rr
r)r1(r1
xaRu2
2m
)208(r
rr)r1(r
1x
PrRe 2D −⎟⎠⎞
⎜⎝⎛
ʹ∂ϑ∂ʹ
ʹ∂∂
ʹ−ʹ=ʹ∂ϑ∂
Nusselt approach still valid
Coupling between velocity and average velocity
)208(r
rr)r1(r
1x
PrRe 2D −⎟⎠⎞
⎜⎝⎛
ʹ∂ϑ∂ʹ
ʹ∂∂
ʹ−ʹ=ʹ∂ϑ∂
⎟⎠⎞
⎜⎝⎛∂∂
∂∂
=⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
∂∂
+∂
∂=
τ∂∂
rtr
rr1a
rt
r1
rtat2
2
compared with unsteady heat conduction:
Steady heat conduction
Assume ϑ = F(xʹ) ⋅ G(rʹ).
After some (8.24-8.28) calculations one finds
β0 < β1 < β2 < β3 < β4 …
)298(e)r(GC0i
PrRe/xii
D2i −ʹ=ϑ ∑
∞
=
ʹβ−
Inspiration for one of the teoretical home assignments
.
β0 = 2.705
β1 = 6.667
β2 = 10.67
Temperature profile in the thermal entrace region for a circular pipe with uniform wall temperature and fully developed laminar flow
Bulktemp. tB
the enthalpy flow of the mixture:
the enthalpy flow can be written
Bptcm!
!∫Δ
πρR
0 "h"p
"m"
tcurdr2 "#"$%&
∫
∫
⎟⎟⎠
⎞⎜⎜⎝
⎛ πρ=πρ=
πρ=
R
0m
2
R
0B
u4Drdru2m
drurt2m1t
!
!!
)348(
urdr
urtdrt R
0
R
0B −=
∫
∫
um
dx
Local Nusselt number
0.001 0.01 0.1 11
5
10
50
100
3.656konstant temperatur,constant wall temperature
konstant värmeflöde, uniform heat flux
konstant värmeflöde, uniform heat fluxHastighetsfält ej fullt utbildat; velocity field not fully developed
4.364
PrRex/D
D
NuD
Average Heat Transfer Coefficient
1.
If the velocity field is fully developed ⇒
N.B.! Higher values if velocity field not fully develped. Eq. (8-38) gives the average value.
tw = constant
[ ]
14.0
3/2/PrRe04.01/PrRe0668.0656.3 ⎟⎟
⎠
⎞⎜⎜⎝
⎛
⎭⎬⎫
⎩⎨⎧
++==
w
B
D
DD
xDxDDNu
µµ
λα
)388(D/LPrRe86.1DNu
14.0
w
B3/1
DD −⎟⎟⎠
⎞⎜⎜⎝
⎛
µµ
⎭⎬⎫
⎩⎨⎧=
λα
=
1.0PrReD/LD
<
t0
x tw = konst
2.
Fully developed flow and temperature fields
NuD = 4.364 (8-50)
qw = α (tw − tB)
⇒ tw − tB = konst.If tB increases, tw must increase as much
Average value including effects of the thermal entrance length
↓↓
!!"!!#$ konstkonst
⎪⎪
⎩
⎪⎪
⎨
⎧
>+
<⎟⎟⎠
⎞⎜⎜⎝
⎛
=λα
=
−
03.0PrReD/xomPrRe
D/x0722.0364.4
03.0PrReD/xomD/x
PrRe1953.1
DNu
DD
D
3/1
DD
t0
...
...
qw = konst. = constant
.
2.
Fully developed flow and temperature fields
NuD = 4.364 (8-50)
qw = α (tw − tB)
⇒ tw − tB = konst.
If tB increases, tw must increase as much
tw highest at the exit!
↓↓
!!! "!!! #$ constconst
t0
...
...
qw = konst. = constant