Convective Heat Transfer L5 (MMV031) – Chapter 6 · Especially for steady state, incompressible...
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Convective Heat Transfer L5 (MMV031) – Chapter 6 + 8 Martin Andersson
Transcript of Convective Heat Transfer L5 (MMV031) – Chapter 6 · Especially for steady state, incompressible...
Convective Heat Transfer L5 (MMV031) – Chapter 6 + 8 Martin Andersson
Agenda
• Convective heat transfer • Continuity eq. • Convective duct flow (introduction to ch. 8)
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
fx
in the fluid
tQ Ay
λ=
∂= − ∂
0/ (6 3)f
x
w f w f
tyQ A
t t t t
λα =
∂ ∂ = = − −
− −
värmeledande kropptf
y
x
tw(y)
t(x,y)fluid
heat conducting body
Convective heat transfer
Convective heat transfer
Objective: Determine α and the parameters influencing it for
prescribed tw(x) or qw(x) = Q/A
Order of magnitude for α
Medium α W/m²K Air (1bar); natural convection 2-20 Air (1bar); forced convection 10-200 Air (250 bar); forced convection 200-1000 Water forced convection 500-5000 Organic liquids; forced convection 100-1000 Condensation (water) 2000-50000 Condensation (organic vapors) 500-10000 Evaporation, boiling, (water) 2000-100000 Evaporation, boiling (organic liquids) 500-50000
How to do it? What are the tools?
Fluid motion: Mass conservation equation Momentum equations (Newton’s second law) Energy balance in the fluid First law of thermodynamics for an open system
Continuity eq.
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
Temperature Equation
∂∂
+∂∂
+∂∂
=∂∂
+∂∂
+∂∂
2
2
2
2
2
2
zt
yt
xt
cztw
ytv
xtu
pρλ
Boundary layer approximations
u(x,y)
δ(x)
U∞
y x
δ T
y x
twt(x,y)t∞
Boundary layer approximations – Prandtl’s theory
vu >>
yv
xv
xu
yu
∂∂
∂∂
∂∂
>>∂∂ ,,
xt
yt
∂∂
>>∂∂
Boundary layer approximations – Prandtl’s theory
)(xpp =
2
2
yu
dxdpF
yuv
xuu x
∂∂
+−=
∂∂
+∂∂ µρρ
2
2
yt
cytv
xtu
p ∂∂
=∂∂
+∂∂
ρλ
Boundary layer approximations – Prandtl’s theory
21 konstant2
p Uρ+ =
dxdUU
dxdp ρ−=
λµ
λρν pp cc
==Pr
Boundary layer equations
0=∂∂
+∂∂
yv
xu
2
2
yu
dxdUU
yuv
xuu
∂∂
+=∂∂
+∂∂
ρµ
Pr 2
2
yt
ytv
xtu
∂∂
=∂∂
+∂∂
ρµ
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.
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 dx dy dzx
ρ ρ
∂= + =
∂∂
= +∂
dzdydx)u(x
mx ρ∂∂
=∆
dydxdzwz
mdzdxdyvy
m zy )( )( ρρ∂∂
=∆∂∂
=∆
zyx mmmutströmmatNetto ∆+∆+∆: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
−=∂∂
+∂∂
Navier – Stokes’ ekvationer (eqs.)
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 unit mass, N/kg b. stresses σij N/m2
Forces
F
σσσ
σσσ
σσσ
=σ
zzzyzx
yzyyyx
xzxyxx
ij
a. volume forces (Fx , Fy , Fz) are calulated per unit mass, N/kg b. stresses σij N/m2
"z""y""x"
Signs for the stresses
dy y
σxx
σyy
σxx
σyy σyx σxy
x
.
Signs for the stresses
Resulting stresses
dy y
σxx
σyy
σxx
σyy σyx σxy
σyy
σxx
σyx σxy
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 ji
jσ
∂∂
Net force in x-direction
Have a look at stress-strain in solids
Stress-strain fluids
. Generally
+
−−
−=
=+δ−=σ
zzzyzx
yzyyyx
xzxyxx
ijijij
ddd
ddd
ddd
p000p000p
dp
)146()31e(2d ijijij −δ∆−µ=
µ=
zzzyzx
yzyyyx
xzxyxx
zzzyzx
yzyyyx
xzxyxx
eee
eee
eee
2
ddd
ddd
ddd
∆∆
∆µ−
3/0003/0003/
2
)156(xu
xu
21e
i
i
j
iij −
∂∂
+∂∂
=
)166(zw
yv
xueii −
∂∂
+∂∂
+∂∂
==∆
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
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
.
Värmeledning i fluiden, heat conduction in the fluid
Qd
dxdydz)xt(
xdydz
xt
dxx
QQQ
xtdydz
xtAQ
xxdxx
x
∂∂
λ∂∂
−∂∂
λ−=
=∂∂
+=
∂∂
λ−=∂∂
λ−=
+
dxdydz)xt(
xQQQ xdxxx ∂
∂λ
∂∂
−=−=∆ +
. Analogous in y- and z-directions (6-27)
dydxdz)zt(
zQ
dzdxdy)yt(
yQ
z
y
∂∂
λ∂∂
−=∆
∂∂
λ∂∂
−=∆
{ }zyx QQQQd ∆+∆+∆−=
↑Qd heatfor convention sign
dzdydx)zt(
z)
yt(
y)
xt(
xQd
∂∂
λ∂∂
+∂∂
λ∂∂
+∂∂
λ∂∂
=
Enthalpy flows and changes
• x-direction
hdzdyuhmH xx ρ==
dzdydxxhudzdydx
xuhHd x
∂∂
+∂∂
=⇒ ρρ
Enthalpy changes
• y- and z-directions
dzdydxyhvdzdydx
yvhHd y
∂∂
+∂∂
= ρρ
dzdydxzhwdzdydx
zwhHd z
∂∂
+∂∂
= ρρ
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
ρ
ρ
= + + =
∂ ∂ ∂= + + + ∂ ∂ ∂ ∂ ∂ ∂
+ + ∂ ∂ ∂
Energy equation, intermediate step
t t tx x y y z z
h h hu v wx y z
λ λ λ
ρ
∂ ∂ ∂ ∂ ∂ ∂+ + = ∂ ∂ ∂ ∂ ∂ ∂
∂ ∂ ∂+ + ∂ ∂ ∂
Enthalpy vs temperature
( )
dtthdp
phdh
tphh
pt
∂∂
+
∂∂
=⇒
=
,
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=
Temperature Equation
∂∂
+∂∂
+∂∂
=∂∂
+∂∂
+∂∂
2
2
2
2
2
2
zt
yt
xt
cztw
ytv
xtu
pρλ
Boundary layer approximations
u(x,y)
δ(x)
U∞
y x
δ T
y x
twt(x,y)t∞
Boundary layer approximations – Prandtl’s theory
vu >>
yv
xv
xu
yu
∂∂
∂∂
∂∂
>>∂∂ ,,
xt
yt
∂∂
>>∂∂
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=
.
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 Duct Flow
Parallel plate duct
Circular pipe, tube
.
Di Re0575.0
DL
=
y
-y
gränsskikt, boundary layer
gränsskikt, boundarylayer
kärna, core
Chapter 8 Convective Duct Flow
kärna, core
gränsskikt, boundarylayer
fullt utbildadströmning, fullydeveloped flow
2bU0x
y
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 = hydraulisk diameter = hydraulic diameter
2u
DLfp
2m
h
ρ=∆
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
ρ∆
Pressure drop - entrance region
x/DhReDh
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 direction Enthalpy transport in x-direction
−=
2
m Rr1u2u
rdr
x dx
tw= konstant; constant
Convective heat transfer for an isothermal tube
⇒
Randvillkor; Boundary conditions:
x = 0 : t = t0
r = R : t = tw
r = 0 : (symmetri, symmetry)
drrtrdx2
rQQQ rdrrr
∂∂
πλ∂∂
−=−=∆ +
)"128("rtr
rr1
xtucp −
∂∂
∂∂
λ=∂∂
ρ
0rt=
∂∂
Introduce
⇒
introduce u = 2um (1 - r′2)
eller, or
p
,w
ca
tt,Rxx,Rrr
ρλ
=
−=ϑ=′=′
′∂ϑ∂′
′∂∂
′=′∂ϑ∂
rRrR
rRrR1
xR1
au
′∂ϑ∂′
′∂∂
′−′=′∂ϑ∂
rr
r)r1(r1
xaRu2
2m
)208(r
rr)r1(r
1x
PrRe 2D −
′∂ϑ∂′
′∂∂
′−′=′∂ϑ∂
)208(r
rr)r1(r
1x
PrRe 2D −
′∂ϑ∂′
′∂∂
′−′=′∂ϑ∂
∂∂
∂∂
=
∂∂
+∂
∂=
τ∂∂
rtr
rr1a
rt
r1
rtat2
2
compare with unsteady heat conduction:
Assume ϑ = F(x′) ⋅ G(r′).
After some calculations one finds
β0 < β1 < β2 < β3 < β4 …
)298(e)r(GC0i
PrRe/xii
D2i −′=ϑ ∑
∞
=
′β−
.
β0 = 2.705
β1 = 6.667
β2 = 10.67
Bulktemp. tB
the enthalpy flow of the mixture:
the enthalpy flow can be written
Bptcm
∫∆
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
xD
xDDNu
µµ
λα
)388(D/LPrRe86.1DNu
14.0
w
B3/1
DD −
µµ
=
λα
=
1.0PrRe
D/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
=
−
03.0PrRe
D/xomPrReD/x
0722.0364.4
03.0PrRe
D/xomD/xPrRe
1953.1DNu
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
