Convective Heat Transfer

• Evaluating processes where there is convective heat transferred to/from a solid surface

• External or Internal • Gas or liquid

Convection

• Two Major Types • Forced

– Flow is driven by external force • Pump • Moving surface

• Natural – Flow is driven by density difference due to

temperature gradient in the fluid

Convective Heat Transfer

( )ambsurfsurfconv TTAh q −=

Calculating “h”

• Forced Convection – Nu = ARemPrn, where A, m and n are constants

• Natural Convection – Nu = A(GrPr)m = Aram, whre A and m are

constants

Fluid of ConductionFluid of Convection

kh uN ==δ

Dimensionless Numbers

kh Nu δ

=

νδ

= ∞v eR

αν

= r P

( )2

3ambsurf TTg

r Gν

δ−β=

Natural Convection

• Buoyancy force • Requires temperature dependence of fluid density • Requires the presence of gravity • Magnitude of natural convection characterized by

Grashof or Rayleigh #

( )ForceViscousForceBuoyancyTTg

G ambsurf

r 2

3=

−=

ν

δβ

Pr a GrR =

Empirical Relations ( ) (7.23) PlateFlat flowLaminar 0.6,Pr ,Pr Re332.0 3/1,

, ≥== xlocx

locx kxh

Nu

( ) (7.30) Plate Flat flow Laminar 0.6,Pr ,Pr Re664.0k

LhNu 3/1L

LL ≥==

( ) (7.36) Plate Flat flow Turbulent ,06Pr0.6 ,Pr Re 0296.0k

xhNu 3/10.8

xloc,x

loc,x ≤≤==

[ ] ( ) (7.38) Plate Flat flow Turbulent ,06Pr0.6 ,Pr ARe037.0k

LhNu 3/18.0L

LL ≤≤−==

5.0c,x

8.0c,x Re664.0Re037.0A −= (6.24) location transition at Re ,

xURe c

c,x µρ ∞=

kLh Pr871e0.037R uN 318.0

LL =

−=

General Approach to Calculating “h”

• Determine Natural or Forced Convection • Collect the appropriate physical data of the fluid

– Thermal conductivity, Viscosity, Prandtl Number, Coefficient of Expansion

• Calculate appropriate dimensionless numbers • Use proper correlations to determine Nu • Calculate “h” from Nu

Evaluating Physical Properties

• Film Temperature – Average between Tsurf and Tamb

– Used for force convection, and external natural convection

• Average Temperature – Average between the two surface temperature

of an enclosure • Natural convection in an enclosure

Linear Interpolation

• Evaluating physical properties that are a temperature between to values in a table

( )lowhighlowhigh

lowbetweenlowbetween yy*

TTTT y y −

−−

+=

Flow over a flat plate

• Problem – P = 83.4kPa – Tair = 30C, Vair = 6m/s – Plate Area = 1.5m x 8m, Tplate = 120C

• What will result in more heat transfer?

– Flow perpendicular to the 1.5m edge – Flow perpendicular to the 8m edge

Assumptions

• Steady State • Heat transfer from only one side of the plate • Ideal gas • Neglect radiation • Thermal conductivity not a function of pressure • Characteristic length is the length of the plate • Laminar flow (ReL ≤ 5x105)

Where do we start??

• Question asks for the convection heat transfer for different orientations

( )∞−= TTAh q surfsurfconv

Calculating h

• Physical properties (at the film temperature) – k, ν, Pr

• Re • Pr • Correct correlation for Nu • Back out h

Physical Properties

K348C752

C30C120 T f ==+

=

Temperature (K)

k (W/mK) ν(m2/s) Pr

340 0.0290 1.96x10-5 0.707

348 0.0296 2.04x10-5 0.706

350 0.0297 2.06x10-5 0.706

Solution

( )6

25

10x.931

sm10x49.2

m8sm6

eR ==−

atm823.0kPa325.101

atm1kPa4.83P ==

sm10x49.2

atm823.0atm1

sm10x05.2

25

25

atm823.0P−−

= =

=ν

Solution (fully turbulent)

kh Pre0.037R u N 318.0 δ

==

KmW13

8mmKW0296.0

3517 h 2==

( ) ( ) 3517 706.010x93.10.037 u N 318.06 ==

( )( )( ) W14056C30C120m5.1m8Km

W13 q 2conv =−=

Solution (1.5m Length) ( )

52

510x.6143

sm10x49.2

5.1sm6

eR ==−

kh Pre0.664R u N 312/1 δ

==

( ) ( ) 355 706.010x614.30.664 u N 312/15 ==

Solution

CmW7

2.5mmKW0296.0

355 h 2==

( )( )( ) W7565C30C120m5.1m8Km

W7 q 2conv =−=

Solution (laminar + turbulent)

( )k

h Pr871e0.037R u N 318.0 δ=−=

KmW10.1

8mmKW0296.0

2738 h 2==

( ) ( ) 2738 706.087110x93.10.037 u N 318.06 =

−=

Solution ( )( )( ) W10908C30C120m5.1m8

KmW10.1 q 2conv =−=

q (δ=8m) Turbulent 14 kW

q (δ=1.5m) Turbulent 7.5 kW

q (δ=8m) Laminar +Turbulent

11 kW

Flow Over Cylinders and Sphere

• External flow over a fluid over a heated/cooled solid object.

• Fluid flow pattern is disturb by a solid stationary object

Characterizing Fluid Flow Over Objects

µδρ

νδ U Re U∞∞ ==

2UAC F

2FDD

∞=ρ

nm Pr Re C Nu δ=

Examples (qconv.)

• Forced convection – Calculating q • Atmospheric Air at 5C flows across a 10cm

diameter pipe at 8m/s • Pipe surface temperature is 95C • Determine qconv

Evaluating Fluid Properties

323K 50C 2

5C95C Tfilm ==+

=

Temperature (K)

k (W/mK) ν(m2/s) Pr

320 0.0275 1.77x10-5 0.710

323 0.0277 1.80x10-5 0.709

330 0.0283 1.86x10-5 0.708

Calculations

• Forced convection • Reynolds Number • Select proper correlation • Determine Nussult Number • Calculate heat transfer coefficient • Calculate qconv

Solution ( )

44,444

sm10x8.1

m1.0sm8

eR 25

==−

kh Pre0.027R u N 31805.0 δ

==

( ) ( )Km

W36.8 0.1m

mKW0277.0

709.0444,440.027 h 231805.0 ==

Solution

( ) ( ) W1040C5C95m1m1.0Km

W36.8 q 2conv =−= π

Example (Iterative Problem)

• Given – Transistor mounted on a PCB. Dissipated 0.58W of

heat. Maximum surface temperature allowed is 90C – Fan blows air across the transistor at a rate of 10m/s

• Assumptions – Steady State – All heat rejected by convection side surface only – Atmospheric Conditions

Diagram of Problem

PCB

L = 0.53cm

D = 0.44cm

Direction of air flow 10m/s

Tsurf,max = 90C

What We Know

• Dimensions of the transistor • Fluid is air • Velocity of air • Surface temperature • Dissipation of heat

– This is the amount need to be transferred by convection!!!

What We Don’t Know

• The heat transfer coefficient (h) • The ambient air temperature

• Problem

– The heat transfer coefficient is dependent on the ambient air temperature

– Need to know Tamb to calculate h…!

Solution Methodology • This is an iterative problem

– Guess an Ambient Temperature (T∞) – Calculate the film temp – Determine the physical properties at Tfilm – Calculate Re and Nu – Determine h – Solve for T∞ – Compare guess with calculated value of T∞ – If Tguess = Tcalc STOP – If Tguess ≠ Tcalc THEN

• Use Tcalc as new Tguess for ambient temperature

Iteration 1

Temperature (K)

k (W/mK) ν(m2/s) Pr

330 0.0283 1.86x10-5 0.708

340 0.0290 1.96x10-5 0.707

350 0.0297 2.06x10-5 0.706

2TC90 Tfilm∞+

= 4C4 T 1guess, =∞

Iteration 1

( )2245

sm10x96.1

m0044.0sm10

eR 25

==−

31466.0 Pre0.683R u N =

Iteration 1 ( ) ( ) 2.22707.022450.683 u N 31466.0 ==

KmW16.146

m0044.0mKW0290.0

2.22kNu h 2=

==δ

( )∞−= TTAh q surfsurfconv

Iteration 1

• Since calculated value does not equal guess – 44C ≠35.8

• Use calculated value as new guess to determine Tfilm

( )( )[ ]( )∞−= TCcmcm 900044.00053.0Km

W146.16 .58W 0 2 π

35.8C , =∞ calcT

KCCCTfilm 8.3358.622

8.3590 ==+

=

Iteration 2

Temperature (K)

k (W/mK) ν(m2/s) Pr

330 0.0283 1.86x10-5 0.708

335.8 0.0287 1.92x10-5 0.707

340 0.0290 1.96x10-5 0.707

5.7C3 T 2guess, =∞

Iteration 2 ( )

2293

sm10x92.1

m0044.0sm10

eR 25

==−

4.22Pre0.683R u N 31466.0 ==

KmW146

m0044.0mKW0287.0

4.22kNu h 2=

==δ

( )( )[ ]( )∞−= TC90cm0044.0cm0053.0Km

W146.17 .58W0 2 π

35.8C T calc, =∞

Cross Flow Over Tubes

V,T

ST

SL

A1

D

3/1mmax,D1D PrReC13.1Nu =

Internal Flow

• Flow inside pipes (round) and ducts (not round)

• Tin ≠Tout

• Laminar v. Turbulent • Velocity to be used? • Consider development of pipe flow

Flow Inside Pipe ro

( )

−−=

2

o

2o r

r1rdxdP

41ruµ

dxdP

8r

u2o

m µ−= mcuAm ρ=

µρ Du

Re m=

Mean Temperature

( )inoutp TTCmq −=

∫= cudAm ρ

∫=cA

cpmp TdAuCTCm ρ

∫

∫∫===

or

02om

cAc

p

cAcp

m uTrdrru

2m

uTdA

Cm

TdAuC

T

ρρ

Forced Internal Convection Entry Effects

• Consider pipe flow • Sieder Tate correlation for laminar flow

– ReD < ~2400

PrRePe,LDPe86.1Nu D

14.0

s

3/1

D =

=

µµ

Kay’s

3/2

D

D

D

PrReLD04.01

PrReLD0668.0

66.3Nu

+

+=

• Constant Surface Temp

Forced Internal Convection Turbulent

• Turbulent Flow (ReD >~104) – Pr range: 0.7 < Pr < 100 – L/D > 60 – All fluid properties at mean bulk temperature

• (Tinlet + Toutlet)/2 – Values of n

• Heating fluid n = 0.4 • Cooling fluid n = 0.3

nDDNu PrRe023.0 8.0=

Forced Turbulent Continued 14.0

s

3/18.0DD PrRe027.0Nu

=

µµ

( )

( )1Pr8f7.121

Pr1000Re8f

Nu3/2

D

D−+

−

=

Example Forced Internal Convection

• Pipe dimensions: D = 3cm, L = 5m • Water flows through the pipe at 10L/min. • Heating water with a resistance heater

– Inlet = 15C, Outlet = 65C • Assume outside of pipe is perfectly insulated.

• Determine the required heater power • Inner surface temperature of the pipe at the exit

Solution

• Water physical properties at 40C • Density = 992.2 kg/m3, k = 0.633W/mK, ν=0.663x10-6m2/s, Cp = 4175J/kgK, Pr = 4.33 ( ) 242 1007.703.0

4mxmAcross

−==π

( ) 2471.0503.0 mmmAsurf == π

Solution

skg

sm

mkgm 165.0

60min1

min01.02.992

3

3 ==

( ) WCCkgK

Js

kgq 3444415654175165.0 =−=

( )ambssurf TThAq −=

Solution

sm

mx

m

V 236.01007.7

sec60min1

min01.0

24

3

==−∞

)(1076010663.0

03.0236.0Re 2

6Turbulent

smx

msm

D ==−

( ) ( ) 5.6933.410760023.0 4.08.0 ==DNu

KmW

mmKW

h 214625.6903.0

633.0==

Solution

C

KmWmW

CTs 1151462

471.034444

65

2

2=+=

Buoyancy

• Archemides Principle.: “A body wholly or partly immersed in a fluid is buoyed up with a force equal to the weight of the fluid displace by the body.”

x

y

z

Buoyancy

Resultant force hold body in equilibrium If ρl > ρb then the resultant force will cause the

object to float.

( ) bodybodyfluid gVFnet ρρ −=

bodybody gVFweight ρ=

bodyfluid gVFbouyancy ρ=

Note that when a body is floating the buoyancy force is calculated using only the submerged portion of the body.

Fluid Mechanics Governing Equation

2

2

yug

dxdP1

yuV

xuu

∂

∂+−−=

∂∂

+∂∂ ∞ ν

ρ

gdx

dP∞

∞ −= ρ

2

2

yug

yuV

xuu

∂

∂+

−=

∂∂

+∂∂ ∞ ν

ρρρ

−−

−≈

∂∂

−=∞

∞TT

1T

1

P

ρρρ

ρρ

β

Natural Convection

T

ρ

( ) 2

2

yuTTg

yuV

xuu

∂

∂+−=

∂∂

+∂∂

∞ νβ

2

2p

yTk

yTV

xTuC

∂

∂=

∂∂

+∂∂ρ

( ) ( )viscousinertialTTgTTg

Gr 2

3s

2

2

3s =

−=

−= ∞∞

µ

δβρ

ν

δβ

( )nPrGrCk

hNu δδ==

Example Natural Convection

Ts=45ºC

Tamb=15ºC

0.3m 0.75m

Very Long

What is the heat loss?

Fluid Properties

K303K2732

C45C15T f =+°+°

=

ν (m2/s) α (m2/2) K (W/mK β (K-1) Pr 1.62x10-5 2.29x10-5 0.0265 0.0033 0.71

( )( )( )( ) 7

225

312

3.0 1007.771.0

1062.1

3.015450033.081.9x

smx

mCCKsm

Ra m =

°−°=

−

−

=δ

From the Sides

8.47

Pr492.01

Ra670.068.0Nu 9/416/9

25.0m3.0 =

+

+==δ

KmW23.4

m3.0mKW0265.0

8.47h 2==

( )( )( )8

25

312

m375.0 10x95.1

sm10x62.1

m375.0C15C45K0033.0sm81.9

Ra =°−°

=−

−

=δ

From the Top

( ) m375.02

m75.02

WWL2

LWPAs ==≈

+==δ

Example Internal Convection

• Double Paned Window – Tinside = 12C, Toutside = 2C – Area

• Height = 0.8m, Width = 2m, Gap = 2cm – Determine qloss

Evaluating Fluid Properties

280K 7C 2

2C12C Tfilm ==+

=

Temperature (K) k (W/mK) ν(m2/s) Pr

280 0.0246 1.40x10-5 0.717

Assume ideal gas

1-0.0036K K280

1 )K(T

1 ===β

Solution

( )( )14300

sm10x40.1

m02.0C2C12K0036.0sm81.9

rG 225

312

=

−=

−

−

( ) 10253 717.014300 GrPr Ra ===

kh HPra0.42R u N

3.0012.025.0 δ

δ=

=

−

Solution

( ) ( )Km

WmKW

mmh 2

3.0012.025.0 .71

0.02m

0246.0

02.08.0717.0102530.42 =

=

−

( ) ( ) WCCmmqconv 2721228.0Km

W1.7 2 =−=

Alternative Question (Dimension)

• Double Paned Window • Maximum allowable qloss = 35W • Tinside = 12C, Toutside = 2C • Height = 0.8m, Gap = 2cm

• How wide of a window is allowable?

Alternative Solution

( ) ( ) W35C2C12Widthm8.0Km

W1.6 q 2conv =−=

( )( )2.7m

C2C12m8.0Km

W1.6

W35Width

2

=−

=

Example 3 (Temp)

• Horizontal Plate is convecting 90W to ambient air.

• Plate Area = 0.6m x 0.6m • Ambient air temperature is 30C

• What is the surface temperature of the

plate?

Solution

• Natural Convection • Have to iterate on Tsurf. • Need to guess to determine air properties • Characteristic length δ = Area/Perimeter (A/P)

Solution • Guess a surface temperature • Calculate air properties at film temperature • Determine Ra • Select proper correlation • Determine Nu • Calculate h • Calculate Tsurf. • Compare calculated Tsurf with guess Tsurf.

– If Tguess = Tcalc STOP – If Tguess ≠ Tcalc THEN

• Use Tcalc as new Tguess for surface temperature

Evaluating Fluid Properties

2C03T

T surffilm

+=

Temperature (K) k (W/mK) ν(m2/s) Pr β(1/K)

320 0.0275 1.77x10-5 0.710 0.00312

323 0.0277 1.80x10-5 0.709 0.00310

330 0.0283 1.86x10-5 0.708 0.003

0C7 T 1guess,surf =

Solution

( )7

225

321

210x27.1

sm10x80.1

m4.2m36.0C30C70K0031.0

sm81.9

rG =

−

=

−

−

( ) 67 10x.978 709.010x2.1 GrPr Ra ===

kh a0.54R u N 25.0 δ

==

Solution

( )Km

W.465

2.4m0.36m

mKW0277.0

10x97.80.54 h 22

25.06 =

=

( ) ( ) W90C30Tm6.0m6.0Km

W5.46 q surf2conv =−=

C76Tsurf =

Evaluating Fluid Properties

2C03T

T surffilm

+=

Temperature (K) k (W/mK) ν(m2/s) Pr β(1/K)

320 0.0275 1.77x10-5 0.710 0.00312

326 0.0280 1.82x10-5 0.709 0.00306

330 0.0283 1.86x10-5 0.708 0.003

6C7 T 2guess,surf =

Solution

( )7

225

321

210x40.1

sm10x82.1

m4.2m36.0C30C76K00306.0

sm81.9

rG =

−

=

−

−

( ) 67 10x.899 709.010x41.1 GrPr Ra ===

kh a0.54R u N 25.0 δ

==

Solution

( )Km

W.675

2.4m0.36m

mKW028.0

10x9997.90.54 h 22

25.06 =

=

( ) ( ) W90C30Tm6.0m6.0Km

W5.66 q surf2conv =−=

C74Tsurf =