External Flow:Mass Transfer

Chapter 7

Sections 7.2 - 7.5, 7.7, 7.8

Mass Transfer

Mass Transfer• Correlations for convection mass transfer coefficients associated with evaporation or sublimation from liquid or solid surfaces in external flow may be inferred from the corresponding heat transfer correlations for an isothermal surface by invoking the heat-and-mass transfer analogy:

Pr

Nu Sh

Sc

• Example: Flat Plate in Parallel Flow

1/ 2 1/ 30.332 Re mx x

AB

h xSh Sc

D

1/ 2 1/ 30.664 Re mL L

AB

h LSh Sc

D

– Laminar-to-Turbulent Flow:4 / 5 1/ 3(0.037 Re ) L LSh A Sc

4 / 5 1/ 2, ,0.037Re 0.664 Re x c x cA

– Turbulent Flow: 4 / 5 1/ 30.037 ReL LSh Sc

– Laminar Flow:

Mass Transfer

• Fluid Properties:

– Assume a dilute, binary mixture of species A (the volatile species) and species B (the free stream fluid).

Re / /L B B ABVL Sc D

Table A.8ABD

– Evaluate the properties of species B at the reference temperature specified for the correlation. If the correlation involves a property ratio, approximate

the ratio as unity.

How would you characterize the relative effectiveness of diffusion in gases, liquidsand solids?

– Evaluate the surface vapor concentration at the surface temperature (assuming saturated conditions).

, ,orA s A sC

Hence, with , the properties of the mixture may be approximated by those of species B.

1Ax

Problem: Paper Drying

Problem 7.119 Use of radiant energy to dry slurry in a paper production process.

KNOWN: Paper mill process using radiant heat for drying.

FIND: (a) Evaporative flux at a distance 1 m from roll edge and corresponding irradiation, G (W/m2), required to maintain surface at Ts = 300 K; (b) Compute and plot variations of hm,x(x), AN (x), and G(x)

for the range 0 x 1 m when the velocity and temperature are increased to 10 m/s and 340 K, respectively.

Problem: Paper Drying (cont).

ASSUMPTIONS: (1) Steady-state conditions, (2) Heat-mass transfer analogy, (3) Paper slurry (water-fiber mixture) has properties of water, (4) Water vapor behaves as a perfect gas, (5) All irradiation is absorbed by slurry, (6) Negligible emission from the slurry, (7) Rex,c = 5 105.

PROPERTIES: Table A.4, Air (Tf = 315 K, 1 atm): = 17.40 10-6 m2/s, k = 0.0274 W/mK, Pr = 0.705; Table A.8, Water vapor-air (Tf = 315 K): DAB = 0.26 10-4 m2/s (315/298)3/2 = 0.28 10-4 m2/s, Sc = B/DAB = 0.616; Table A.6, Saturated water vapor (Ts = 330 K): A,sat = 1/vg = 0.1134 kg/m3, hfg = 2366 kJ/kg.

ANALYSIS: (a) Modeling the drying process as flow over a flat plate with heat and mass transfer, the local mass flux is

6 2 5 5x BRe u x 5m s 1m 17.40 10 m s 2.874 10 5 10

A,x m,x A,s A, m,x A,sat s A,satn h h T T (1)

SCHEMATIC:

Problem: Paper Drying (cont.)

1/ 2 1/ 34 2 5 3

m,xh 0.28 10 m s 1m 0.332 2.874 10 0.616 4.24 10 m s

Estimating hx from the heat-mass transfer analogy,

1/ 3 1/ 33 2

x m,x 4 2AB

k Pr 0.0274 W m K 0.705h h 4.24 10 m s 4.34 W m K

D Sc 0.6160.28 10 m s

(4)

The flow is laminar, and invoking the heat-mass analogy,

m,x 1/ 2 1/ 3

x xAB

h xSh 0.332Re Sc

D (2)

Hence, the evaporative flux at x = 1 m is

3 3 4 2A,xn 4.24 10 m s 0.1134kg m 0 4.81 10 kg s m <

From an energy balance on the differential element at x = 1 m,

conv evap x s A,x fgG q q h T T n h . (3)

Hence, from Eq. (3), the radiant power required to maintain the slurry at Ts = 330 K is

2 4 2 3G 4.34 W m K 330 300 K 4.81 10 kg s m 2366 10 J kg

2 2G 130 1138 W m 1268 W m . <

Problem: Paper Drying (cont.)

Parametric calculations reveal the following results for Ts = 340 K and u = 10 m/s.

0 0.2 0.4 0.6 0.8 1

Distance from the leading edge, x (m)

0

0.02

0.04

0.06

0.08

Lo

cal c

oe

ff,

hm

x (m

/s)

0 0.2 0.4 0.6 0.8 1

Distance from the leading edge, x (m)

0

0.005

0.01

0.015

0.02

Eva

po

rativ

e f

lux,

n''A

x (k

g/s

.m^2

)

0 0.2 0.4 0.6 0.8 1

Distance from the leading edge, x(m)

0

10000

20000

30000

Irra

dia

tio

n,

G (

W/m

^2

)

Problem: Wet-and Dry-Bulb Thermometers

Problem 7.128 Use of wet- and dry-bulb thermometers to determinetemperature and relative humidity of air flow in a duct.

KNOWN: Dry-and wet-bulb temperatures associated with a moist airflow through a large diameter duct of prescribed surface temperature.

FIND: Temperature and relative humidity of airflow.

SCHEMATIC:

Problem: Wet-and Dry-Bulb Thermometers (cont.)

ASSUMPTIONS: (1) Steady-state conditions, (2) Conduction along the thermometers is negligible, (3) Duct wall forms a large enclosure about the thermometers.

PROPERTIES: Table A-4, Air (318K, 1 atm): = 17.7 10-6

m2/s, k = 0.0276 W/mK, Pr = 0.70;

Table A-4, Air (298K, 1 atm): = 15.7 10-6

m2/s, k = 0.0261 W/mK, Pr = 0.71; Table A-6,

Saturated water vapor (298K): vg = 44.3 m3/kg, hfg = 2442 kJ/kg; Saturated water vapor (318.5K):

vg = 15.5 m3/kg; Table A-8, Water vapor-air (298K): DAB = 0.26 10

-4 m

2/s, Sc = 0.60.

ANALYSIS: Dry-bulb Thermometer: Since Tdb > Ts, there is net radiation transfer from the surface of the dry-bulb thermometer to the duct wall. Hence to maintain steady-state conditions, the

thermometer temperature must be less than that of the air (Tdb < T) to allow for convection heat transfer from the air.

From application of a surface energy balance to the thermometer, qconv = qrad, or,

4 4db db g db sdbhA T T A T T .

The air temperature is then

4 4db g sdbT T / h T T (1)

Problem: Wet-and Dry-Bulb Thermometers (cont.)

Wet-bulb Temperature: The relative humidity may be obtained by performing an energy balance on the wet-bulb thermometer. In this case convection heat transfer to the wick is balanced by evaporative and radiative heat losses from the wick,

conv evap radq q q

evap A wb fg m A,sat wb A,sat wb fgq =n A h h T T A h .

4 4wb wb m A,sat wb A,sat wb fg w wb swbhA T T h T T A h A T T

4 4A,sat wb w s wb fg m A,satwbT T T h T T / h h / T

Convection Calculations: For the prescribed conditions, the Reynolds number associated with the dry-bulb thermometer is

-6 2

bdD dbRe VD / 5 m/s 0.003 m/17.7 10 m / s 847.

Approximating the Prandtl number ratio as unity, the Zhukauskas correlation and Table 7.4 yield

0.5 0.37m nD db D db

Nu CRe Pr 0.51 847 0.70 13.01

2

db

k 0.0276 W/m Kh 13.01 13.01 120 W/m K.

D 0.003 m

From Eq. (1) the air temperature is

8 2 4

4 4 42

0.95 5.67 10 W/m KT 45 C 318 308 K 45 C 0.55 C 45.6 C.

120 W/m K

<

Problem: Wet-and Dry-Bulb Thermometers (cont.)

The Reynolds number associated with the wet-bulb thermometer is

-6 2

wbD wbRe VD / 5 m/s 0.004 m/15.7 10 m / s 1274.

From the Zhukauskas correlation and Table 7.4, it follows that

0.6 0.37D wbNu 0.26 1274 0.71 16.71

2

wb

k 0.0261 W/m Kh 16.71 16.71 109 W/m K.

D 0.004 m

Using the heat-mass transfer analog , it also follows that

0.6 0.370.6 0.37D wb D wb

Sh 0.26Re Sc 0.26 1274 0.6 15.7

4 2

ABm

wb

D 15.7 0.26 10 m / sh 15.7 0.102 m/s.

D 0.004 m

11 3 3A,sat wb gT v 298 K 44.3 m / kg 0.0226 kg/m

11 3 3A,sat gT v 318.5 K 15.5 m / kg 0.0645 kg/m .

Problem: Wet-and Dry-Bulb Thermometers (cont.)

Hence, from Eq. (2), the relative humidity is,

8 2 4 4 4 4 23 3

6

0.95 5.67 10 W/m K 298 308 K 109W/m K 45.55 25 K0.0226 kg/m / 0.0645 kg/m

2.442 10 J/kg 0.102 m/s

0.21

COMMENTS: (1) The effect of radiation exchange between the duct wall and the thermometers is small. For this reason db.T T

(2) The evaporative heat loss is significant due to the small value of , causing Twb to be significantly

less than T.