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

2

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

3

• 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

6

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

8

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

9

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

− −−= − − → → =

−−

=− → → =−−−−

& &

&&

11

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

13

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

14

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

18

• Parallel – double the flow rate

• Series – double the pressure difference

P1 p2

P1 P2 p3

19

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