chapter 8 internal forced convection
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1
Chapter 8 Internal Forced convection 8.1 Introduction
• The laminar & turbulent flows in channels
• The developing & fully developed regions
• The heat transfer rate under two boundary conditions:
- Constant heat flux
- Constant surface temperature
• Method to select a cooling fan
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The laminar & turbulent in ducts
• The critical Reynolds number
Similar to the case of external flow, the flow in a duct can be laminar or turbulent. The critical Reynolds number is
Dh is called hydraulic diameter which is defined as 4A/P, P is the wetted perimeter & A is the cross-sectional area of the duct, and ּט is the kinematic viscosity of the coolant.
For a duct of rectangular cross-sectional area
Re 2,300h hcr
uD uD
Re 2300...
2300 Re 10,000...
10000 Re...
Lanimar
Transitional
Turbulent
244 4 4 4,, , , , .2( )h h
A ab
DA ab AD circular channel D D
P a b P D
a bL
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• The developing and fully developed regions
The flow in a duct can be divided into two regions
Near the inlet of the duct, the boundary layer starts developing from both sides of the channel & increases along the flow direction. At the point x* where the two boundaries meet at the center and they cannot increase anymore. From the inlet to the meeting point is called developing region. Down stream of the developing region is called developed or fully developed region.
- For laminar flow, the thermal boundary layer developing length is
and the velocity boundary developing length is
- For turbulent flow the developing length for both thermal and velocity
boundary layers is
0.05Re Pr hx D
0.05Re hx D
10 hx D
h
x* x
4
• The Nussult number and heat transfer coefficient
At any location in the duct, the Nussult number & heat flux can be expressed
h = convection heat transfer coefficient
Tsx = Local surface temperature of the duct
Tmx = local mean fluid temperature
• It was found experimentally as well as theoretically that, for a given channel, the convection heat transfer coefficient is constant in the developed region. Near the inlet, where the boundary layer thickness increases from zero to half height of the duct. Both the temperature gradient and the heat transfer coefficient are very large at the leading edge and decrease along the flow direction and then meet the constant values in the developed region, as shown in figure above and values of heat transfer coefficients for various forms of channels are shown in Table 8-3
• The channel surface temperature is,
........... ( )h
hD sx mx
hDNu q h T T
k
h
x* x
sx mx
qT T
h
5
8-3 the Nusselt correlation equations
Nusselt number correlation equations for laminar flow
h
hD
hDNu
k
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8-3 the Nusselt correlation equations
• Nusselt number for turbulent flow and the transitional region
(for Re is greater than 2300)
• The average bulk mean temperature is used to get the properties of the coolant.
• The total heat transfer rate
• The distribution of the surface temperature of the duct depends on the boundary conditions.
2mi mo
ave
T TT
( )p mo miQ mc T T
Tmi Tmo
Tsi Tso
0.8 0.40.023( ) Prh
h hD
hD uDNu
k
7
8-4 Internal forced convection with constant heat flux • The distribution of surface and bulk mean fluid temperatures
• The distribution of buck mean
fluid temperature is linear
Ts
Tm
x
Tmi
Tmo
Tso
0
[( ) ]
( )
( )
mx
mi
p mx mx m x p m x
p mx
x x T
p mx p mxo T
p mx mi
mx mip
qdA mc T dT T mc dT
q pdx mc dT
qp dx mc dT mc dT
qpx mc T T
qpT T x
mc
dATmx
Tmx+ dTmx
Tmi
0 x x+dx
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8-4 Internal forced convection with constant heat flux
• The variation of Tsx is also linear in the fully developed region
• The rate of heat transfer is equal to the rate of heat absorbed by the coolant
• The maximum surface temperature
( )sx mx sx mx sx mip
q qp qq h T T T T T T x
h mc h
( )
( ) ( )
p mo mi
sx mx so mo
Q mc T T
or
Q hA T T hA T T
so mo mo
Q qT T T
hA h
Ts
Tm
x
Tmi
Tmo
Tso
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8-5 Internal convection heat transfer--Constant surface temperature
• The energy balance on a elemental control volume (Ts is larger than Tm)
Integrating from the entrance (x = 0), where the inlet fluid mean temperature is Tmi, to any point x along the duct, where the mean fluid temperature is equal to Tmx
• The fluid temperature at any point x is
( )s mxdQ h T T dA= −&
Tmx Tmx+dTmx
Δx
( ) p
hpx
mcmx s s miT T T T e
−
= − −
dA
( )( )........ s mx
mx s mxs mx p
d T T hpdT d T T dx
T T mC
−− − −
− &
ln , ( ) p
hpx
mcs mxs mx s mi
s mi p
T T hpx T T T T e
T T mc
−−=− → → − = −
−&
&
10
8-5 constant surface temperature
• The maximum fluid temperature is at x = L or the outlet fluid temperature
• The heat transfer rate
• logarithmic mean temperature difference
T
Tmi
Ts
Tmo
x
ln
( ) ( )( )
ln ln
mo mi s mi mop mo mi s
s mo s mo
s mi s mi
hPL T T hA T TQ mc T T hA T
T T T T
T T T T
− −= − =− = = Δ
− −− −
& &
ln
ln
mo mi
s mi
s mo
T TT
T T
T T
−Δ =
−−
( ) ,
lnln
p p
hPL hPL
mc mcs momo s s mi
s mi
s mop
s mos mi p
s mi
T TT T T T e e
T T
T T hpL hpLmc
T TT T mcT T
− −−= − − → → =
−−
=− → → =−−−−
& &
&&
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12
Cooling of a hollow PCB
• Given : Hollow PCB 12cm x 18cm, total heat dissipation = 40W
Tmi = 20oC, air volume flow rate at inlet section = 0.72litre/s
channel cross-sectional area = 0.3cm x12cm
• Find (a) Tmo, (b) Tsmax
• Solution:
- Assumptions:
1, Pressure at 1 atm.
2. Smooth inner surface
3. Steady state operation
- the inlet condition
18cm
3
3 3
3
20 , 1.204 / , 1007 / , 0.72( / )
1.294 0.72 10 0.86 10 /
( )
4020 66
0.86 10 1007
omi p
i i
p mo mi
omo mi
p
T C kg m c J kgK V litle s
m V x x x kg s
Q mc T T
QT T C
mc x x
− −
−
= = = =
= = =
= −
= + = + =
&
&&& &
&
&
12cm
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Cooling of a hollow PCB
3 5 2
33 2
20 661, 43
2 2
2, 1.12 / , 1007 / , 0.0267 / , 1.72 10 / ,Pr 0.725
4 4 3.6 103,Re 0.12 0.003 3.6 10 , 0.00585
2(0.12 0.003)
omo miave
p
cc h
T TT C
kg m c J kgK k W mK v x m s
A x xA x x m D m
P
3
3
0.86 102 /
1.12 3.6 10
Re 730 2300
c
h
m xu m s
A x x
uD
T
xTmi
Tmo
Tso
2
max
max
max
0.124, . 40 8.24
0.0030.0267
5, 8.24 36 /0.00585
6,
.... ( ) ( )
40.... 66 92
36 2(0.12 0.18)
h
so s
s sx mx s s mo
os mo
s
Nu aspect ratio Nu
kh Nu W m K
D
T T
Q hA T T hA T T
QT T C
hA x x
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Cooling of a constant surface temperature hollow PCB
• Given : Board 12cm x 20cm, flow rate = 0.72x10-3m3/s, channel 0.3cm x 12cm, Tmi= 20oC, Ts = 60oC
• Find: Tmo , Heat dissipation rate
• Solution : Assumptions 1. Steady state operation
2. Air behaves as ideal gas
3. pressure is equal to 1 atm.
4. Assume Tmo to be 50oC3
5 2
3
20 5035
2
1.145 / , 1007 / , 0.02625 /
1.655 10 / , Pr 0.7268
0.72 102 /
0.003 0.12
0.00585
Re 727
oave
p
c
h
h
T C
kg m c J kgK k W mK
x m s
V xu m s
A x
D m
uD
−
−
+= =
= = =
= =
= = =
=
= =
&20cm
12cm
15
20.0251440 7.54 7.54 32.4 /
0.00585h
a kNu h Nu W m K
b D → →
2
32.4 0.0432
0.0008244 1007
ln
2(0.003 0.12) 0.2 0.0432
1.145 0.00072 0.0008244 /
( ) 60 (60 20) 53
20 5332.4 0.0432
7lnln
40
32.4 0,0432
s
p
s
hA xmc ox
mo s s mi
mi mos
s mo
s mi
A PL x m
m V x kg s
T T T T e e C
T TQ hA T x
T T
T T
x x
− −
= = + =
= = =
= − − = − − =− −
= Δ = =−−
=
&
&&
&
19.4 26.5
( ) 0.0008244 1007 (53 20) 27.4p mo mi
W
or
Q mc T T x x W
=
= − = − =& &
&
&
16
Method to select a cooling fan
• Characteristic curves
- The static pressure developed by a given fan depends on its rpm and the flow rate of the fluid which it propels. The fan curve is usually provided by the manufacturers.
- The system curve is the total pressure loss verses flow rate or velocity of
a given flow system
- The intersection of the two curves is the operation point of the fan
4
2
0
Δp
m3/s
rpmP1 p2
P1 p2P1 p2
17
Installation considerations • Inlet or outlet of the duct - Preferred position is at the inlet : positive pressure inside the cabinet to prevent air infiltration into the box from cracks or other openings and the air is denser and cooler at the position of the fan. - Heat generated by the motor is forced into the system. The inlet air
temperature is higher.• Do not used forced convection if nature convection is adequate • Critical electronics should be mounted near the inlet where the coolant• temperature is lower• Air velocity should be less than 7m/s, otherwise noise will be too large.• Arrange the system to use nature convection to help forced convection• Series operation helps to increase the pressure head and parallel
operation helps to increase the flow rate. • Arrange the openings on the side surfaces, not on the top surface • The maximum air temperature at the exit port should be less than 70oC• Make a good arrangement of the boards for small flow resistance • Consider the effect of air pressure change due to altitude effect
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• Parallel – double the flow rate
• Series – double the pressure difference
P1 p2
P1 P2 p3
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Heat transfer coefficient
The actual average heat transfer coefficient is larger than the following
developed value. For a given inlet coolant temperature, the surface
temperature is smaller. Then the device temperature mounted on the surface is
Smaller than that calculated by following developed h value.
h
x* x
h
x* x
20
• The following pages will not be taught
21
8-5 Internal convection heat transfer--Constant surface temperature
• The energy balance on a elemental control volume (Ts is larger than Tm)
Integrating from the entrance (x = 0), where the inlet fluid mean temperature is Tmi, to any point x along the duct, where the mean fluid temperature is equal to Tmx
• The fluid temperature at any point x is
( )s mdQ h T T dA
Tm Tm+dTm
Δx
( ) p
hpx
mcmx s s miT T T T e
−
= − −
dA
d &Q &mc
pdT
mxh(T
s T
mx)dA &mc
pdT
mxh(T
s T
mx) pdx
dT
mx d(T
s T
mx)
d(Ts T
mx)
Ts T
mx
hp&mC
p
dx
ln , ( ) p
hpx
mcs mxs mx s mi
s mi p
T T hpx T T T T e
T T mc
−−=− → → − = −
−&
&
22
Forced convection- internal flow
• The mean film temperature, if one of the temperature is unknown, assume one.
• Properties of the coolant• Calculate the Reynolds number - Re ≤ 2300, the flow is laminar - Re > 2300, treat the flow as turbulent • Select the Nusselt correlation equation - The boundary conditions ; constant surface or constant surface heat flux - Flow conditions ; laminar or turbulent flow• Calculate the Nusselt number and heat transfer coefficient• Calculate the heat transfer rate or the unknown temperature or both or the
area
• Compare the assumed temperature and the calculated one. If the different is large, re-assume a temperature and repeat the process.
ln
( )
( ) ( )
( )
p mo mi
s s m p mo mi
s s p mo mi
Q mc T T
or
q const Q hA T T mc T T
T const Q hA T mc T T
= −
= → = − = −
= → = Δ = −
& &
&& &
& &
23
The pressure loss in a flow system
The total pressure in a flow system is represented by
Pressure loss can be written in terms of loss of velocity head
pressure loss =
k is called loss factor and it is dimensionless. Its value depends on the type of obstructions
Type of obstruction k Inlet loss 0.5 outlet loss 1.0
channel
L : length of the channel and Dh is the hydraulic diameter f : friction factor: For fully developed flow is shown in Table 8 – 1
t iP pΔ = Δ∑
22, ( / )
2
uk N m
h
Lf
D
24
8-2 The velocity and temperature fields in a duct
• The mass flow rate
um is the average velocity in the duct.
• The bulk mean fluid temperature
The energy transport rate of the fluid in the duct
Tm is the average or mean temperature in the duct
p p p m mpA A
EE c Tdm c T udA mc T T
mc
m
mu
A
25
Example: Hollow PCB- continue
• Find the system or pressure loss curve of example .
u (m/s)
0 0
0.5 0.246
1.0 0.94
1.5 2.07
2.0 3.65
3 5 2
.
2 2 2 2
2
1.120 / , 1.72 10 / , 2 / , 0.005854 ,
12 960.18 ,Re 730, 40 96
0.3 Re
(0.5 1) (0.5 1) 962 2 2 2
0.0485 0.888
h
h
t i inlet outlet ch
th h h
t
kg m x m s u m s D m
L m AR fuD
p p p p p
u L u u L up f
D uD D
p u u
Δ Δ Δ Δ Δ
Δ
Δ
tpΔ
0 1 u 2
4
2
0
Δpt
26
Example 15-13 continue
• Plot the system curve The characteristic of the fan
• Superimpose the two curves
For u = 2m/s, the rpm of the fan is selected
0 1 u 2
4
2
0
Δpt
0 1 u 2
4
2
0
Δpt
rpm
0 1 u 2
4
2
0
Δpt
m/s