4-10 analogy and differences in different transport

Advanced Heat and Mass Transfer by Amir Faghri, Yuwen Zhang, and John R. Howell

4.10 Analogy and Differences in Different Transport Phenomenon Processes


Most early work in predicting theoretically the heat and/or mass transfer in both laminar and turbulent flow cases were done using the analogy between moment m heat and mass and predictingmomentum, heat, and mass and predicting the approximate results for heat and/or mass transfer coefficient from momentummass transfer coefficient from momentum transfer or friction coefficient.

Chapter 4: External Convective Heat and Mass Transfer


1


Cl l th li it ti i i thi Clearly there are severe limitations in using this simple approach, however, it is beneficial to understand the advantages and similarities forunderstand the advantages and similarities for physical and mathematical modeling as well as the constraints involving this approach.

We present this analogy for the classical problem of heat and mass transfer over a flat plate in this section It’s applications to moreplate in this section. It s applications to more coupled geometries and boundary conditions as well as turbulent flow is not proven and caution h ld b t k i l i thi h tshould be taken in applying this approach to

other cases.



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As presented before a flat plate at constant wall temperature T is exposed to As presented before a flat plate at constant wall temperature Tw is exposed to free stream of constant velocity U∞, temperature T∞ and mass fraction ω1,∞, due to binary diffusion can be presented with the following dimensionless conservation boundary layer equations and boundary conditions corresponding to Figure 4.31.

U∞T∞

,y v

x,u

T∞ω1, w

Figure 4.31 Mass, momentum and heat transfer in laminar boundary layer

Tw and ω1, w



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ContinuityContinuity(4.348)

Momentum(4 349)

0u vx y

2

u u uu v (4.349)Energy

(4.350)S i

2

u vx y y

2

2Pru v

x y y

Species(4.351)

at y+ = 0 u+ = 0 θ = Φ = 0 (4.352)

Prx y y 2

2

u v

x y Sc x

y ( )y+ = 0 (4.353)y+ = ∞ u+ = 1 θ = Φ = 1 (4.354)

where the dimensionless variables are defined as:

wv v

, , ,(4.355)

, , ,

1 1,

1, 1,

w

w

w

w

T TT T

uu

U

vv xx Re U Lyy



4

, , ,vU

xL y

L


Let’s first consider the analogy between momentum and heat transfer. Equations (4.349) and (4.350) and appropriate boundary conditions are the same if Pr = 1. pp p yTherefore the solutions for u+ and θ are exactly the same if Pr = 1 and one expects to have a relation between friction coefficient C and heat transfer coefficient hfriction coefficient Cf and heat transfer coefficient h.

(4.356)0 0 0

f y y yw

u uuy yyC ( 356)

2 22 Re

U U U L

Tk LyhL

(4.357) 0

0

y

y

yhLNuk k T T y



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Si θ + f P 1 l l d th tSince θ = u+ for Pr = 1, one also concludes that

(4.358)0 0y y

uy y

therefore combining (4.356) and (4.357) and using (4.358)

0 0y yy y

fC Nu

(4.359)This relation between the friction coefficient and Nu is referred to as Reynolds analogy and is appropriate for Pr = 1. If Pr ≠ 1 we already concluded that

2 Re

concluded thatfor 0.5 ≤ Pr ≤ 10 from the similarity solution

presented in Section 4 6 Using this information one can generali e

1/3PrT

presented in Section 4.6. Using this information, one can generalize the result of Reynolds analogy to Pr ≠ 1 by

(4.360)1/3Pr2 R

fC Nu



6

2 Re


N f th tt ti t i il iti b t h t d t fNow focus the attention to similarities between heat and mass transfer or comparison of equation (4.350) and (4.351) with their appropriate boundary conditions. It is clear that the solution for differential equation (4.349) and (5.350) for θ and Φ are same if Sc and Pr are(4.349) and (5.350) for θ and Φ are same if Sc and Pr are interchanged appropriately.We already know the solution for equation (4.350) for vw = 0 from similarity solution for 0.5 ≤ Pr ≤ 10

1/ 2 1/3 (4.361)Therefore it can also be assumed that the solution of equation (4.351) for

1/ 2 1/30.332 Re Prx xNu

0wv

(4.362)Combining equations (4.361) and (4.362) gives

1/ 2 1/30.332 Rex xSh Sc

(4.363)1/3

1/3

PrNuSh Sc



7


It h ld b t d th t th ti ff t d t ti lIt should be noted that the convective effect due to vertical velocity at the surface in predicting h and hm are neglected and therefore the analogy presented in (4.363) is for very

flow mass transfer at the wall. This can be easily seen since the contribution of in heat flux and mass flux were primarily due to diffusion.

wv

(4.364) 0

" Cp w wy

Tq v T T ky

(4.365)

0y

11 12" wm v D

( )1, 12

0w

yy



8


Analogy in momentum, heat, and mass transfer can also be applied to complex, coupled transport phenomenon problems including phase changephenomenon problems including phase change and chemical reactions.

Obviously in these circumstances the simple Obviously, in these circumstances, the simple relation developed in equation (4.363) is not applicable.pp

To show the appropriate usefulness of this analogy of transport phenomenon we will apply it gy p p pp yto sublimation with chemical reaction for forced convection over a flat plate.



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During combustion involving a solid fuel, the solid fuel may burn directly or it may be sublimated before combustion In the latter case – which willbefore combustion. In the latter case which will be discussed in this subsection – gaseous fuel diffuses away from the solid-vapor surface. M hil th id t diff t dMeanwhile, the gaseous oxidant diffuses toward the solid-vapor interface. Under the right conditions, the mass flux of vapor fuel and theconditions, the mass flux of vapor fuel and the gaseous oxidant meet and the chemical reaction occurs at a certain zone known as the flame. The fl i ll thi i ith lflame is usually a very thin region with a color dictated by the temperature of combustion.



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Figure 4.32 shows the physical model of the problem under consideration The concentration of the fuel is highest at the solid

Figure 4.32 Sublimation with chemical reaction (Kaviany, 2001).

consideration. The concentration of the fuel is highest at the solid fuel surface, and decreases as the location of the flame is approached. The gaseous fuel diffuses away from the solid fuel surface and meets the oxidant as it flows parallel to the solid fuel surface Combustion occurs in a thin reaction zone wheresurface. Combustion occurs in a thin reaction zone where temperature is the highest, and the latent heat of sublimation is supplied by combustion. The combustion of solid fuel through sublimation can be modeled as a steady-state boundary layer type flow with sublimation and chemical reaction



11

flow with sublimation and chemical reaction.


To model the problem, the following assumptions are made:assumptions are made: The fuel is supplied by sublimation at a

steady rate.steady rate. The Lewis number is unity, so the thermal

and concentration boundary layers have the y ysame thickness.

The buoyancy force is negligible. y y g g



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The conservations of mass momentum energy and species The conservations of mass, momentum, energy and species of mass in the boundary layer are

(4.366)( ) ( ) 0u vx y

(4.367)

x y u u uu vx y y y

T (4.368)

(4 369)

,( ) ( )p p o c oTc uT c vT k m h

x y y y

( ) ( ) oo o ou v D m

(4.369)where is rate of oxidant consumption (kg/m3-s). is the heat

released by combustion per unit mass consumption of the oxidant (J/kg) which is different from the combustion heat

o o ox y y y

oxidant (J/kg), which is different from the combustion heat defined in Chapter 2. is mass fraction of the oxidant in the gaseous mixture.



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The corresponding boundary conditions of eqs. (4.366) – (4.369) are(4.366) (4.369) are

at (4.370),, , o ou U T T y

at (4.371)0, , 0f omu v

y

0y

where is the rate of solid fuel sublimation permwhere is the rate of solid fuel sublimation per unit area (kg/m2-s) and ρ is the density of the mixture

fm



14

mixture.


The shear stress at the solid fuel surface is

(4.372), 0wu yy

The heat flux at the solid fuel surface is

(4 373)

y

0Tq k y (4.373)

The exact solution of the heat and mass problem described by eqs. (4.366) – (4.369) can be obtained using conventional numerical

, 0wq k yy

( ) ( ) gsimulation, which is very complex. However, it is useful here to introduce the results obtained by Kaviany (2001) using analogy between momentum and heat transfer. Multiplying eq. (4.369) by hc,0and adding the result to eq. (4.368) , one obtains g q ( ) ,

(4.374), , ,( ) ( ) op o c o p o c o c o

Tu c T h v c T h k Dhx y y y y



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Considering the assumption that Lewis number is unity, i.e. , eq. (4.374) can be rewritten as/ 1Le D

(4.375), ,( ) ( )p o c o p o c ou c T h v c T hx y

which can be viewed as an energy equation with quantity

,( )p o c oc T hy y

which can be viewed as an energy equation with quantityas a dependent variable.

Since at , i.e., the solid fuel surface is not ,p o c oc T h

/ 0o y 0y permeable for the oxidant, eq. (4.373) can be rewritten as

(4.376),( ) , 0w p o c oq c T h yy



16

y


Analogy between surface shear stress and the surface Analogy between surface shear stress and the surface energy flux yields

, ,( ) ( )ww p o c o w p o c oq c T h c T h

u

(4.377)

The energy balance at the surface of the solid fuel is, , ,( ) ( )w

p w c o o w oc T T hu

The energy balance at the surface of the solid fuel is (4.378)w f svq m h q

where the two terms on the right-hand side of eq. (4.378) represent the latent heat of sublimation, and the sensible heat required to raise the surface temperature of theheat required to raise the surface temperature of the solid fuel to sublimation temperature and heat loss to the solid fuel.



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Combining eqs. (4.377) and (4.378) yields the rate of sublimation on the solid fuel surfacesub a o o e so d ue su ace

(4.379)where Z is transfer driving force or transfer number defined

wfm Z

U

as(4.380)

, , ,( ) ( )/

p w c o o o w

sg f

c T T hZ

h q m

Using the friction coefficient gives (4 381)

g f

wC Using the friction coefficient gives (4.381)

eq (4.379) becomes (4.382)

2 / 2fCU

2f

f

Cm U Z



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q ( ) ( )2f


The surface blowing velocity of the gaseous fuel is then

(4 383)f fm C

v U Z

(4.383)

where the friction coefficient Cf can be obtained from the

2f f

wv U Z

where the friction coefficient Cf can be obtained from the solution of boundary layer flow over a flat plate with blowing on the surface. The similarity solution of the b d l fl bl i t l if bl iboundary layer flow problem exists only if blowing velocity satisfies . In this case, one can define a blowing parameter as

1/ 2wv x

g

(4.384) 1/ 2( ) Re

( )w

xvBu



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Combination of eqs. (4.383) and (4.384) yieldsyields

(4.385)1/ 2Re2 x fZB C

Glassman (1987) recommended an f f ( )empirical form of eq. (4.385) based on

numerical and experimental results:(4.386)

0.15

ln(1 )2.6

ZBZ



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Example 4.4Example 4.4 Air with a temperature of 27 °C flows at 1 m/s over a 1-m

long solid fuel surface with a temperature of 727 °C. The concentration of the oxidant at the solid fuel surface isconcentration of the oxidant at the solid fuel surface is 0.1, and the heat released per unit mass of the oxidant consumed is 12000 kJ/kg. The latent heat of sublimation for the solid fuel is 1500 kJ/kg. Neglect the sensible heatfor the solid fuel is 1500 kJ/kg. Neglect the sensible heat required to raise the surface temperature of the solid fuel to sublimation temperature, and heat loss to the solid fuel. Estimate the average blowing velocity due tofuel. Estimate the average blowing velocity due to sublimation on the fuel surface.



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SolutionSolution The mass fractions of the oxygen at the solid

fuel surface and in the incoming air are, respectively, and . The specific0 0 1 0 21 respectively, and . The specific heat of gas, approximately taken as the specific heat of air at Tave=(Tw+T∞)/2=377 °C, is cp=1.063 kJ/kg-K The combustion heat per unit oxidant

0, 0.1w , 0.21o

kJ/kg-K. The combustion heat per unit oxidant consumed is . The latent heat of sublimation is . The density at the wall and the incoming temperatures are

, 12000 /c oh kJ kg1500 /svh kJ kg

wall and the incoming temperatures are, respectively, and . The viscosity at Tave is =60.21×10-6m2/s

31.1614 /kg m 30.3482 /w kg m



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Th t f d i i f b bt i d f i The transfer driving force can be obtained from eq. , i.e., , , ,( ) ( )

1 063 (27 727) 12000 (0 21 0 1)

p w c o o o w

sv

c T T hZ

h

The blowing parameter obtained from eq (4 386) is

1.063 (27 727) 12000 (0.21 0.1) 1500

0.3839

The blowing parameter obtained from eq. (4.386) is

Th bl i l it t th f i bt i d f (4 383)

0.15 0.15

ln(1 ) ln(1 0.3839) 0.14432.6 2.6 0.3839

ZBZ

The blowing velocity at the surface is obtained from eq. (4.383)

1/21/2 1/2Rew xw w

v BU B U x

which can be integrated to yield the average blowing velocity



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1/22w

w

v B U L

1/262 1.1614 0.1443 1 60.21 10 10.3482

0 007469 /

0.007469 m/s



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4-10 analogy and differences in different transport

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