Lecture 11b

Introduction to Convection:

Flow and Thermal Considerations

Chapter Six and Appendix D

Sections 6.1 through 6.8 and D.1 through D.3

Lecture 11b

Dimensionless Parameters

ForceViscous

ForceInertia

F

F

LV

LVVL

s

IL

2

2

/

/Re

For a small Re: Viscous force is important

For a large Re: Viscous force is negligible


yDiffusivitThermal

yDiffusivitMomentum

k

cp

Pr

Pr ≈ 1 for gases, Pr<<1 for liquid metals, Pr>>1 for oils

3/1Prt


yDiffusivitMass

yDiffusivitMomentum

DDSc

ABAB

/

3/1Scc


yDiffusivitMass

yDiffusivitThermalSc

DLe

AB

Pr

3/1Lec

t


fk

hLNu Dimensionless T gradient at the surface

AB

m

D

LhSh

2/2VC s

f

Dimensionless surface shear stress

Dimensionless C gradient at the surface

Boundary Layer Equations

Normalized boundary layer equations

2

2

*

*

Re

1

*

*

*

**

*

**

y

u

dx

dp

y

uv

x

uu

L

Velocity:

Thermal: 2

2

*

*

PrRe

1

*

**

*

**

y

T

y

Tv

x

Tu

L

Concentration: 2

2

*

*

Re

1

*

**

*

**

y

C

Scy

Cv

x

Cu A

L

AA

Boundary Layer Analogies

To establish relationship between Cf, Nu and Sh Fluid Flow Heat Transfer Mass Transfer

)*

*,Re*,*,(*dx

dpyxfu L )

*

*Pr,,Re*,*,(*

dx

dpyxfT L )

*

*,,Re*,*,(*dx

dpScyxfC LA

0**

*

Re

2

yL

f y

uC

)Re*,(Re

2L

Lf xfC

0**

*

yf

f

y

T

k

LhNu

0**

*

y

A

AB

m

y

C

D

LhSh

Pr),Re*,( LxfNu ),Re*,( ScxfSh L

Pr),(ReLfNu ),(Re ScfSh L)(ReRe

2L

Lf fC

Boundary Layer Analogies

nL

f

xfk

hLNu Pr)Re*,(

nL

AB

m ScxfD

LhSh )Re*,(

nLn Sc

Shxf

Nu )Re*,(

Pr

npn

ABm

LecLeD

k

h

h 1

From Observations:

Reynolds Analogies

ShNuC Lf

2

Re

If dp*/dx*=0, Pr=Sc=1, the boundary layer equations for

fluid flow, heat transfer and mass transfer become same form.

mf StStC 2/

PrRe

Nu

Vc

hSt

p

Sc

Sh

V

hSt m

m Re

Pr = Sc =1

Chilton-Colburn Analogies

Hf jSt

C 3/2Pr

20.6 < Pr < 60

mmf jScSt

C 3/2

20.6 < Sc < 3000

For Laminar flow, need dP*/dx* =0;

For turbulent flow, doesn’t need dp*/dx*=0

Evaporative Cooling

• The term evaporative cooling originates from association of the latent energy created by evaporation at a liquid interface with a reduction in the thermal energy of the liquid. If evaporation occurs in the absence of other energy transfer processes, the thermal energy, and hence the temperature of the liquid, must decrease.

• If the liquid is to be maintained at a fixed temperature, energy loss due to evaporation must be replenished by other means. Assuming convection heat transfer at the interface to provide the only means of energy inflow to the liquid, an energy balance yields

Evaporative Cooling

conv evapq q

""""radevapaddconv qqqq

fgAsatAmfgAs hTshhnTTh *]})([*{*)( ,,"

""evapconv qq

fgAsatAm

s hTsh

hTT *])([)( ,,

With radiation from the interface and heat addition by other means,

Example 1

Analogy between heat and mass transfer

Example 1

Known: Boundary layer temperature and heat flux at a

location on a solid in an air stream given T and V

Find: Water vapor concentration and flux associate with

the same location on a larger surface

Assumptions: Steady state, 2-D, incompressible

boundary layer behavior, constant properties; boundary

layer approximations are valid; molar fraction of water

vapor is much less than unit.

Example 1

Properties: Table A.4 air (50 C) v, k, Pr, Table A.6

saturated water vapor (50C), , Table A.8 Water vapor-

air (50C), DAB =

Analysis:

T* = f(x*, y*, ReL, Pr, dp*/dx*)

CA*= f(x*, y*, ReL, Sc, dp*/dx*)

Example 1

For case 1 (L=1m):

ReL,1 = 5.5x106, Pr=0.7

For case 2 (L=2m):

ReL,2 = 5.5x106, Sc=0.7

ReL,1 =ReL,2, Pr= Sc, x1*=x2*, y1*=y2*

Example 1

T*=(T-Ts)/(T –Ts) = f(x*, y*, ReL, Pr, dp*/dx*)

CA*=(CA-CA,s)/(CA, - CA,s)=f(x*, y*, ReL, Sc, dp*/dx*)

We expect T* = CA* = f ( …. )

We can calculate CA from T*

and hm from h (Sh=Nu ?)

And NA” = hm (CA,s-C A,)

Example 2

Dry air at atmospheric pressure blows across a thermometer whose bulb

has been covered with a dampened wick. This classic “wet-bulb”

thermometer indicates a steady-state T reached by a small amount of

liquid evaporating into a large amount of unsaturated vapor-gas mixture.

The thermometer reads at 18.3 ºC. At this T, the following properties

were evaluated:

Vapor pressure of water: 0.021 bar, density of air: 1.22 kg/m3

Latent heat of water vaporization: 2458 J/kg, Pr: 0.72, Sc: 0.61

Specific heat, cp of air: 0.56 J/kg/c

What is the Temperature of dry air ?

Example 2

Known: Thermophysical properties and T of water

Find: Temperature of dry air

Schematic:

pH2O=0.021 barTs=18.3 Chfg=2458 J/kg

T =?CH2O, = 0=1.22 kg/m3

cp=0.56 J/kg/CPr=0.72 Sc=0.61

q”conv

q”evap

Example 2

Assumption: Steady-state, constant properties

Analysis:

Energy balance: q”conv = q”evap

The energy required to evaporate the water is supplied by convective heat

transfer.

q”conv = h(T – Ts) = hfg*MH2O*N”H2O

Example 2

Where N”H2O is the molar flux of water transferred from thermometer to air

N”H2O = hm*(CH2O,S-CH2O,)

)( ,2,22

OHSOHmOHfg

s CCh

hMhTT

How do we determine hm/h and CH2O,S ?

Example 2

Pr = 0.72, Sc = 0.61,

Reynolds analogy is not accurate

Chilton-Colburn analogy can be used to relate hm and h.

JH = Jm

or St*Pr2/3 = Stm*Sc2/3

Stm/St= (Pr/Sc)2/3

Example 2

St = h/(Vcp), Stm = hm/V

Stm/St= (hm/h)*(cp) = (Pr/Sc)2/3

hm/h = (Pr/Sc)2/3/(cp)

From ideal gas law: PV = nRT

CH2O,S = n/V = P/(R*Ts)

Example 2

)( ,2,22

OHSOHmOHfg

s CCh

hMhTT

)*

()(Pr/ 3/2

2

sPOHfgs TR

P

c

ScMhTT

M H2O = 18 (kg/kmol), R = 0.08314 (m3bar/kmol/K)

T = 18.3C + 2458 (J/kg)* 18 (kg/kmol) * (0.72*0.61)2/3

/(1.22 kg/m3*0.56J/kg/C)*(0.021 bar/0.08314 (m3bar/kmol/K)

/(18.3+273.15)K = 18.3 + 32.4 = 50.7 C

Example 3

As a means to prevent ice formation on the wings of small aircraft, it is

proposed that electric resistance heating elements be installed within the

wings. To determine the representative power requirements, considering

normal flying conditions for which plane moves at 100 m/s in air that is at

-23 ºC and has properties of k=0.022 W/mK, Pr=0.72, and ν=16.3x10-6 m2/s.

If the characteristic length of the airfoil is L=2m and the wind tunnel

measurements indicate an average friction coefficient of =0.0025 for the

normal conditions, what is the average heat flux needed to maintain a surface

temperature of Ts=5 C?

fC

Example 3

Known: Nominal operating conditions of aircraft,

characteristic length and average friction coefficient of wing

Find: Average heat flux needed to maintain prescribed

surface temperature of wing

Schematic:

Example 3

Assumption: Steady-state, constant properties

Analysis:

The average heat flux that must be maintained over

the surface of the air foil is

where the average convection coefficient ( ) may be

obtained from the modified Reynolds analogy

(Chilton-Colburn Analogy).

h

Example 3

Lecture 11b

Lecture 11b

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