# ChE306: Heat and Mass Transfer - PowerPoint - PowerPoint

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```					External Flow

Chapter 7
Section 7.1 through 7.8

Lecture 12
Boundary Layer Analogies
To establish relationship between Cf, Nu and Sh
Fluid Flow                       Heat Transfer                            Mass Transfer
dp *     T *  f ( x*, y*, Re L , Pr,
dp *                                      dp *
u*  f ( x*, y*, Re L ,        )                                       )   C A *  f ( x*, y*, Re L , Sc,        )
dx *                                    dx *                                      dx *

2 u *                            hL    T *                             hm L   C *
Cf                               Nu                                  Sh           A
Re L y * y*0                      k    y * y*0                        DAB     y * y*0

2
Cf       f ( x*, Re L )         Nu  f ( x*, Re L , Pr)                   Sh  f ( x*, Re L , Sc)
Re L

2
Cf         f (Re L )              Nu  f (Re L , Pr)                      Sh  f (Re L , Sc)
Re L
1. Empirical Method
Nu L  f (Re L , Pr)
Nu L  C Re Pr
m
L
n
Empirical Correlations
ln( Nu L )  [ln C  m ln Re L ]  n ln Pr
ln( Nu L / Pr )  ln C  m ln Re L
n
Empirical Correlation

Nu L  C Re Pr      m
L
n

Sh L  C Re Sc     m
L
n

Pr and Sc are properties of fluid at film temperature Tf

Tf =(Ts+T)/2
2. Flat Plate in Parallel Flow
Laminar Flow:
Physical Features

Physical Features

• As with all external flows, the boundary layers develop freely without constraint.
• Boundary layer conditions may be entirely laminar, laminar and turbulent,
or entirely turbulent.
• To determine the conditions, compute
 u L u L
ReL          
     
and compare with the critical Reynolds number for transition to turbulence, Rex ,c .
ReL  Rex ,c  laminar flow throughout

ReL  Rex ,c  transition to turbulent flow at xc / L  Rex ,c / ReL
Physical Features (cont.)

• Value of Rex ,c depends on free stream turbulence and surface roughness.
Nominally,
Rex ,c  5  105.

• If boundary layer is tripped at the leading edge
Rex ,c  0
and the flow is turbulent throughout.

• Surface thermal conditions are commonly idealized as being of uniform

temperature Ts or uniform heat flux qs . Is it possible for a surface to be
concurrently characterized by uniform temperature and uniform heat flux?

• Thermal boundary layer development may be delayed by an unheated
starting length.

Equivalent surface and free stream temperatures for x   and uniform Ts
(or qs ) for x   .

Similarity Solution

Similarity Solution for Laminar,
Constant-Property Flow over an Isothermal Plate
• Based on premise that the dimensionless x-velocity component, u / u ,
and temperature, T *  T  Ts  / T  Ts , can be represented exclusively in
                      
terms of a dimensionless similarity parameter
  y  u /  x 
1/2

• Similarity permits transformation of the partial differential equations associated
with the transfer of x-momentum and thermal energy to ordinary differential
equations of the form
d3 f d2 f
2 3f      0
d   d 2

where  u / u   df / d , and

d 2T * Pr dT *
   f    0
d  2
2   d
Similarity Solution (cont.)

• Subject to prescribed boundary conditions, numerical solutions to the momentum
and energy equations yield the following results for the x-component velocity
distribution and the temperature distribution in the boundary layer:

What value of  corresponds to the edge of the velocity boundary layer?

How thick is the thermal boundary layer, relative to the velocity boundary
layer, for Pr = 1? Pr > 1? Pr < 1?
Similarity Solution (cont.)

With u / u  0.99 at   5.0 at the edge of the velocity boundary layer,
5.0                    5x
                          
 u / vx                  Rex 
1/2               1/2

u                                   d2 f
With  s                     u       u / vx
y       y 0
d 2         0

and d 2 f / d 2           0.332,
 0

 s,x
C f ,x                    0.664 Re 1/2
u / 2 2

x

With hx  qs / Ts  T   k T * / y
                                                 k  u / vx  dT * / d
1/2
y 0                                0

and dT * / d           0.332 Pr1/3 for Pr  0.6,
 0

hx x
Nux                0.332 Re1/2 Pr1/3
x
k

and                  Pr1/3
t
Similarity Solution (cont.)

• How would you characterize relative laminar velocity and thermal boundary layer
growth for a gas? An oil? A liquid metal?

• How do the local shear stress and convection coefficient vary with distance from
• Average Boundary Layer Parameters:
1
 s , x  0x  s dx
x
C f , x  1.328 Re1/2
x

1 x
hx      0 hx dx
x
Nu x  0.664 Re1/2 Pr1/3
x

• The effect of variable properties may be considered by evaluating all properties
at the film temperature.
Ts  T
Tf 
2
Turbulent Flow

Turbulent Flow
• Local Parameters:

Empirical
C f , x  0.0592 Rex 1/5
Correlations
Nux  0.0296 Rex Pr1/3
4/5

How do variations of the local shear stress and convection coefficient with
distance from the leading edge for turbulent flow differ from those for laminar flow?

• Average Parameters:
1
L

hL  0xc hlam dx  xLc hturb dx    
Substituting expressions for the local coefficients and assuming Rex ,c  5  105 ,
0.074 1742
C f ,L     1/5

ReL      ReL

Nu L  0.037 ReL  871 Pr1/3
4/5

For Rex ,c  0 or L  xc  ReL  Rex ,c  ,

C f ,L  0.074 ReL 1/5
Nu L  0.037 ReL Pr1/3
4/5
Special Cases

Special Cases: Unheated Starting Length (USL)
and/or Uniform Heat Flux

For both uniform surface temperature (UST) and uniform surface heat flux (USF),
the effect of the USL on the local Nusselt number may be represented as follows:
Laminar            Turbulent
Nu x  0                      UST         USF     UST          USF
Nu x                   b
1   / x a          a         3/4         3/4     9/10        9/10
                       b         1/3         1/3     1/9         1/9
Nu x  0  C Rex Pr1/3
m
C        0.332       0.453   0.0296   0.0308

m         1/2         1/2     4/5         4/5

Sketch the variation of hx versus  x    for two conditions:   0 and   0.
What effect does an USL have on the local convection coefficient?
Special Cases (cont.)

• UST:
qs  hx Ts  T  q  hL As Ts  T 


L                   p 1 /  p  2   p /  p 1
Nu L  Nu L                 1   / L 
 0  L                                     
p  2 for laminar flow throughout
p = 8 for turbulent flow throughout

hL  numerical integration for laminar/turbulent flow
1  xc
hL         hlam dx  xc hturb dx 
L

L                           
• USF:
q
Ts  T  s                         
q  qs As
hx

• Treatment of Non-Constant Property Effects:
Evaluate properties at the film temperature.
T T
Tf  s 
2
Flat Plate in Parallel Flow
Laminar Flow:
5x
Boundary layer thickness:                                   
Re x
 s, x
0.664
Local fraction coefficient:
Cf  2      
u / 2     Re x
hx x
Nu x        0.332 Re x Pr
1/ 2 1/ 3
Local heat transfer:
Pr ≥ 0.6                       k
hm , x x
Shx                0.332 Re x
1/ 2
Local mass transfer:                                                   Sc1/ 3
Sc ≥ 0.6                           DAB
Flat Plate in Parallel Flow
Laminar Flow:
 s,x1.328
Average fraction   Cf  2      
coefficient:           u / 2     Re x

hx x
Average heat
transfer:         Nu x        0.664 Re 1/ 2 Pr 1/ 3
Pr ≥ 0.6                  k

Average mass              hm, x x
transfer:         Shx               0.664 Re1/ 2 Sc1/ 3
Sc ≥ 0.6                  DAB
Flat Plate in Parallel Flow
Turbulent Flow:
Boundary layer thickness:               0.37 x Re 1/ 5
x

C f  0.0592 Re1/ 5
Local fraction coefficient:
Rex ≤ 107                                          x

Local heat transfer:          Nu x  0.0296 Re 4 / 5 Pr 1/ 3
x
0.6<Pr<60

Local mass transfer:            Shx  0.0296 Re 4 / 5 Sc1/ 3
x
0.6<Sc<3000
Flat Plate in Parallel Flow
Mixed Boundary Layer Conditions:

hL    hlam dx   hturb dx 
1  Xc          L

L 0            Xc        

k         u 1/ 2 X c dx              u 4 / 5 L dx  1/ 3
hL  ( ) 0.332 ( )                0.0296 ( )  1/ 5  Pr
L          v      0   x1/ 2
v       Xc x

Flat Plate in Parallel Flow
Mixed Boundary Layer Conditions:
Average fraction
0.074 1742
coefficient:
5x105<ReL≤108
Cf     1/ 5

Rex,c=5x105
Re L      Re L
Average heat
transfer:
0.6<Pr<60
Nux  (0.037 Re4 / 5  871) Pr1/ 3
L
5x105<ReL≤108
Rex,c=5x105

Average mass
transfer:
0.6<Sc<3000         Shx  (0.037 Re4 / 5  871)Sc1/ 3
L
5x105<ReL≤108
Rex,c=5x105
3. Cylinder in Cross Flow
Boundary layer formation and separation

ReD = VD/ν,   CD ≡ FD/(Af*(*V2/2)
Cylinder in Cross Flow
Boundary layer formation and separation
Cylinder in Cross Flow
Hilpert Correlation: (Pr>0.6)
hD
Nu x      C Re D Pr
m    1/ 3

k
C and m depend on ReD and geometry of the cylinder, they can be found
in Tables 7.2 & 7.3 (pages 458-459), where all properties are
evaluated at film temperature Tf

ReD = VD/ν
Cylinder in Cross Flow
Zhukauskas Correlation (circular cylinders):
hD            n Pr 1/ 4
Nu x      C Re D Pr ( )
m

k              Prs
C and m depend on ReD, they can be found in Tables 7.4 (pages 459)

where all properties are evaluated at T, except Prs at Ts

n=0.37, Pr ≤10

n=0.36, Pr>10
Cylinder in Cross Flow
Chuchill-Bernstein Correlation:
0.62 Re 1/ 2 Pr 1/ 3             Re D 5 / 8 4 / 5
Nu x  0.3              D
[1  (          ) ]
[1  (0.4 / Pr) ]
2 / 3 1/ 4
282 ,000

Pr > 0.2 and all ReD

all properties are evaluated at film temperature Tf
4. Flow Cross Sphere
24
CD       , Re D  0.5                ReD = VD/ν
Re D
 1/ 4
Nu D  2  (0.4 Re   1/ 2
 0.06 Re   2/3      0.4
) Pr ( )
s
D                  D

0.71<Pr<380,            3.5<ReD<7.6x104

All properties, except μs, are evaluated at T , μs
is at surface temperature Ts
5. Flow Across Banks of Tubes

Aligned          Staggered
Flow Across Banks of Tubes
Zhukauskas Correlation (Table 7.5, page 470):

hD                0.36 Pr 1/ 4
Nu x      C Re D,max Pr ( )
m

k                     Prs
NL ≥ 20
0.7 < Pr < 500
1000 < ReD,max < 2x106
where all properties, except Prs, are evaluated at average fluid T (Tf)
ReD,max = VmaxD/ν
Vmax = ST/(ST-D)*V for aligned or staggered arrangement
Vmax = ST/(SD-D)*V for staggered arrangement
Flow Across Banks of Tubes
Log-mean temperature difference:
(Ts  Ti )  (Ts  To )
Tlm 
Ts  Ti
ln(          )
Ts  To

Heat transfer rate per unit length:

q'  N (hDTlm )
6. Flow Through Packed Bed
6. Flow Through Packed Bed

 J H   J m  2.06 Re                 0.575
D

St Pr   2/3
 jH  jm  Stm Sc          2/3

Pr ≈ 0.7 (gas)
90 ≤ ReD ≤4000
properties at average fluid T
7. Methodology for Convection Calculation

•   Identify the flow geometry

•   Specify reference T for fluid properties

•   Use dilute species to calculate Sc

•   Calculate Re

•   Local or average surface coefficient ?

•   Select an appropriate correlation
Example 1
Air at pressure of 6 kN/m2 and a temperature of

300 ºC flows with a velocity of 10m/s over a flat

plate of 0.5m long. Estimate the cooling rate per

unit width of the plate needed to maintain it at a

surface T of 27 ºC?
Example 1

Known: Air flow over isothermal flat plate

Find:    Cooling rate per unit width, q’ (W/m)

Schematic:
Example 1

Properties: At (300+27)/2 = 163.5 ºC, from Table A.4, for air

v=30.84x10-6 m2/s, k=36.4x10-3W/mK, Pr=0.687

Analysis:
q'  hL(T  Ts )

u L 10m / s * 0.5m
Re L                      9597
            4
5.21x10 m / s2
Example 1
The flow is laminar over the entire plate, the corresponding correlation is:

hL
Nu x             0.664 Re       1/ 2
Pr   1/ 3
 57 .4
k

h  57 .4 * k / L  57 .4 * 0.0364 W / mK /(0.5m)  4.18W / m 2 K

q'  hL(T  Ts )  4.18W / m2 K * 0.5m * (300  27)C  570W / m
Example 2
The decorative plastic film on a copper sphere
of 10-mm diameter is cured in an oven at 75ºC.
Upon removal from the oven, the sphere is
subjected to an air stream at 1 atm and 23 ºC
having a velocity of 10m/s. Estimate how long
it will take to cool the sphere to 35 ºC.
Example 2
Known: Sphere cooling in air stream
Find:     Time required to cool from 75ºC to 35ºC

Schematic:                                          Copper
Sphere
D=10 mm

Air
p=1 atm
V=10 m/s
T=23 C

Ti =75 C
T(t)=35 C
Example 2
Assumptions:
1. Negligible thermal resistance and capacitance for the plastic film;

2. Spatially isothermal sphere;

Properties:

Copper: =8933 kg/m3, k=399 W/mK, cp=387 J/kgK,

Air: At T (296K), μ=181.6x10-7Ns/m2, ν=15.36x10-6m2/s,

k=0.0258 W/mK, Pr=0.709. At Ts, μ=197.8x10-7Ns/m2
Example 2
Analysis:
This is a transient problem, check Bi first
Bi = hL/k=?,
Assume Bi < 0.1, apply lumped capacitance method

Vcp Ti  T Dc p Ti  T
t     ln             ln
hAs    T  T   6h     T  T

VD10m / s * 0.01m
Re D                      6510
 15.36x10 m / s
5  2
Example 2
Analysis:
Re D  6510               From Eqn. 7.59:

 1/ 4
Nu D  2  (0.4 Re   1/ 2
 0.06 Re   2/3      0.4
) Pr ( )
s
D                  D

Nu D  47 .4

k        0.0258W / mK
h  Nu D *  47.48               122W / m 2 K
D            0.01m
Example 2
Analysis:

h  122W / m 2 K

Dc p Ti  T 8933kg / m3 * 0.01m * 387J / kgK 75  23
t    ln                                        ln          69.2s
6h    T  T          6 *122W / m K2
35  23

Check if Bi < 0.1 ?
Bi = hL/k= 122 W/m2K*(0.01m/6)/399W/mK = 5.1x10-4 <0.1
Where L = D/6
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:
Nu  Sh
Pr  Sc
• Example: Flat Plate in Parallel Flow
– Laminar Flow:
h x
Shx  m  0.332 Re1/2 Sc1/3
x
DAB
hm L
Sh L          0.664 Re1/2 Sc1/3
L
DAB

– Laminar-to-Turbulent Flow:
Sh L  (0.037 ReL  A) Sc1/3
4/5

A  0.037 Rex ,c  0.664 Re1/2
4/5
x ,c

– Turbulent Flow:
Sh L  0.037 ReL Sc1/3
4/5
Mass Transfer (cont.)

• Fluid Properties:
– Assume a dilute, binary mixture of species A (the volatile species) and
species B (the free stream fluid). Hence, with xA  1 , the properties of the
mixture may be approximated by those of species B.
ReL  VL /  B            Sc   B / DAB

DAB  Table A.8

– 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, liquids
and solids?

– Evaluate the surface vapor concentration   A,s or CA,s  at the surface temperature
(assuming saturated conditions).
Problem: Paper Drying

Problem 7.130 Use of radiant energy to dry slurry in a paper production process.

In a paper mill drying process, a sheet of paper slurry (water-fiber mixture) has a
linear velocity of 5 m/s as it is rolled. Radiant heaters maintain a sheet temperature
of Ts = 330 K, as evaporation occurs to dry, ambient air at 300K above and below
the sheet.

(a). What is the evaporation flux at a distance of x = 1m from the leading edge of the
roll? What is the corresponding value of the radiant flux (irradiation, G) that must
be supplied to the sheet to maintain its temperature at 330 K? The sheet has an
absorptivity of α=1.
(b). To accelerate the drying and paper production processes, the velocity and
temperature of the strip are increased to 10 m/s and 340K, respectively. To maintain
a uniform strip temperature, the irradiation G must be varied with x along the strip.
For 0≤ x ≤1 m, compute and plot the variation hm,x(x), N”A(x) and G(x).
Problem: Paper Drying

Problem 7.130 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), N A (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).

SCHEMATIC:

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 an ideal 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/mK, 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
            
nA, x  hm, x A,s  A,  hm, x  A,sat Ts   A,sat T 
                                                                                         (1)

Rex  u x B   5m s×1m  17.40  106 m2 s  2.874  105  5 105
Problem: Paper Drying (cont.)

The flow is laminar, and invoking the heat-mass analogy,
hm, x x
Shx               0.332 Re1/2Sc1/3
x                                                                   (2)
DAB

                                             
1/2
hm,x  0.28  104 m2 s 1m  0.332 2.874  105                  0.616 1/3  4.24  103 m s

Hence, the evaporative flux at x = 1 m is

                        
nA, x  4.24  103 m s 0.1134 kg m3  0  4.81 104 kg s  m2
                                                                                             <
From an energy balance on the differential element at x = 1 m,

G  qconv  qevap  hx Ts  T   nA, xh fg .
                                                                                     (3)

Estimating hx from the heat-mass transfer analogy,

 Pr 
k
1/3
3       0.0274 W m  K   0.705 1/3           2
hx  hm,x                       4.24  10        m s                             4.34 W m  K        (4)
DAB  Sc                                     0.28  104 m 2 s   0.616 
                   

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

G  4.34 W m2  K  330  300  K  4.81  104 kg s  m2  2366  103 J kg

G  130  1138 W m2  1268 W m2 .                                                             <
Problem: Paper Drying (cont.)

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

0.08                                                                                                                                  0.02

Evaporative flux, n''Ax (kg/s.m^2)
0.06                                                                                                                                 0.015
Local coeff, hmx (m/s)

0.04                                                                                                                                  0.01

0.02                                                                                                                                 0.005

0                                                                                                                                        0
0   0.2             0.4                          0.6        0.8          1                                                              0         0.2               0.4        0.6              0.8   1

Distance from the leading edge, x (m)                                                                                                         Distance from the leading edge, x (m)

30000

20000

10000

0
0         0.2            0.4                                          0.6           0.8                 1

Distance from the leading edge, x(m)
Problem: Wet-and Dry-Bulb Thermometers

Problem 7.139 Use of wet- and dry-bulb thermometers to determine
temperature 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.

-6 2
PROPERTIES: Table A-4, Air (318K, 1 atm):  = 17.7  10 m /s, k = 0.0276 W/mK, Pr = 0.70;
-6 2
Table A-4, Air (298K, 1 atm):  = 15.7  10 m /s, k = 0.0261 W/mK, Pr = 0.71; Table A-6,
3
Saturated water vapor (298K): vg = 44.3 m /kg, hfg = 2442 kJ/kg; Saturated water vapor (318.5K):
3                                                         -4 2
vg = 15.5 m /kg; Table A-8, Water vapor-air (298K): DAB = 0.26  10 m /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,


hAdb T  Tdb    g Adb Tdb  Ts4 .
4

The air temperature is then
  4

T  Tdb   g / h Tdb  Ts4                                                     (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,

qevap =nA Awbh fg  hm  A,sat Twb   A,sat T  Awbh fg .
                                            

                          
hAwb T  Twb   hm  A,sat Twb   A,sat T  Awbh fg   w Awb Twb  Ts4

4


   A,sat Twb    w Twb  Ts4  h T  Twb  / h fg hm /  A,sat T 


4
            
                      
Convection Calculations: For the prescribed conditions, the Reynolds number associated with the
dry-bulb thermometer is
ReD  db   VDdb /   5 m/s  0.003 m/17.7  10-6 m2 / s  847.

Approximating the Prandtl number ratio as unity, the Zukauskas correlation and Table 7.4 yield
Pr n  0.51847   0.70 
0.5      0.37
Nu D db   CRem                                         13.01
D db 

k          0.0276 W/m  K
h  13.01             13.01                 120 W/m 2  K.
Ddb             0.003 m

From Eq. (1) the air temperature is
0.95  5.67  108W/m2  K 4
T  45 C 
120 W/m2  K
3184  3084  K4  45 C  0.55 C  45.6 C. <
Problem: Wet-and Dry-Bulb Thermometers (cont.)

The Reynolds number associated with the wet-bulb thermometer is

ReD wb   VDwb /   5 m/s  0.004 m/15.7  10-6 m2 / s  1274.

From the Zukauskas correlation and Table 7.4, it follows that
Nu D wb   0.26 1274                0.710.37  16.71
0.6

k              0.0261 W/m  K
h  16.71                16.71                       109 W/m 2  K.
Dwb                   0.004 m

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

Sc0.37  0.26 1274          0.60.37  15.7
0.6
Sh D wb  0.26Re0.6
D wb

DAB       15.7  0.26  104 m2 / s
hm  15.7                                               0.102 m/s.
Dwb                 0.004 m

               0.0226 kg/m3
1
 A,sat Twb   v g  298 K 1  44.3 m3 / kg

 A,sat T   v g  318.5 K 1  15.5 m3 / kg   0.0645 kg/m3.
1
Problem: Wet-and Dry-Bulb Thermometers (cont.)

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

 

              
0.95  5.67  108 W/m2  K 4 2984  3084 K 4  109W/m2  K  45.55  25 K  
 0.0226 kg/m3                                                                                / 0.0645 kg/m3


                                                  6

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 T  Tdb.

(2) The evaporative heat loss is significant due to the small value of , causing Twb to be significantly
less than T.

Lecture 12

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