PROBLEM 7.109 (Cont.)
Convection Calculations: For the prescribed conditions, the Reynolds number associated with the
dry-bulb thermometer is
Approximating the Prandtl number ratio as unity, from Eq. 7.53 and Table 7.4,
From Eq. (1) the air temperature is
()
8 24 4 44
2
0.95 5.67 10 W/m K
T 45 C 318 308 K 45 C 0.55 C 45.6 C.
120 W/m K
−
∞×× ⋅
=+ − =+=
⋅
 
<
The relative humidity may now be obtained from Eq. (2). The Reynolds number associated with the
wet-bulb thermometer is
From Eq. 7.53 and Table 7.4, it follows that
Using the mass transfer analog of Eq. 7.53, it also follows that
Also,
Hence the relative humidity is, from Eq. (2)
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 φ∞, causing Twb to be significantly less than T∞.
PROBLEM 7.110
KNOWN: Velocity, diameter and temperature of a spherical droplet. Conditions of surroundings.
FIND: (a) Expressions for droplet evaporation and cooling rates, (b) Evaporation and cooling rates
for prescribed conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Negligible temperature gradients in the drop, (2) Heat and mass transfer
analogy is applicable, (3) Perfect gas behavior for vapor.
PROPERTIES: Table A-4, Air (T∞ = 298K, 1 atm): ν = 15.71 × 10–6 m2/s, k = 0.0261 W/m⋅K, Pr
ANALYSIS: (a) The evaporation rate is given by
( )
( ) ( )
2
evap m s A,s A, m A,sat A,sat
m hA h D T T .
ρ ρ π ρ φρ
∞ ∞∞

= −= −

<
The cooling rate is obtained from an energy balance performed for a control surface about the droplet,
With As = πD2, it follows that
(b) To obtain
m
h,
the mass transfer analog of the Ranz–Marshall correlation gives
PROBLEM 7.110 (Cont.)
Hence
( ) ( )
1/ 2 1/ 3
D
Sh 2 0.6 1337 0.6 20.5=+=
evap
The evaporative heat flux is then
Using the heat transfer correlation, the Nusselt number is
and the sensible heat flux is
conv
The net radiative flux is
COMMENTS: (1) Evaporative cooling provides the dominant heat loss from the drop. (2) To test
the validity of assuming negligible temperature gradients in the drop, calculate
From Table A-6,
k 0.631 W/m K,= ⋅
hence
PROBLEM 7.111
KNOWN: Cranberries with an average diameter of 15 mm rolling over a fine screen. Thickness of
the water film is 0.2 mm.
FIND: Time required to dry the berries exposed to heated air with a velocity of 2 m/s and temperature
of 30°C.
SCHEMATIC:
Berry, D = 15 mm,
ASSUMPTIONS: (1) Steady–state conditions, (2) Air stream is dry, (3) Water film on the berries is
also at 30°C, (4) Convection process is uniform over the exposed surface, and (5) Heat-mass analogy
is applicable.
PROPERTIES: Table A-6, Water (Tf = 30°C = 303 K):
3
A,f 995.8 kg / m ,
ρ
=
ρ
A,g = 0.02985
ANALYSIS: The evaporation rate of water from the berry surface is given by the rate equation,
( )
m s A,s A,
nhA
ρρ
∞
= −
(1)
where As = πD2 and
m
h
is determined using the heat-mass analogy, Eq. 6.60,
where Le = α/DAB and for this problem, n = 0.4. The heat transfer coefficient
h
is estimated with the
Whitaker correlation, Eq. 7.56,
Substituting numerical values, find
and using the heat-mass analogy,
PROBLEM 7.111 (Cont.)
where
Using Eq. (1), the evaporation rate is
The time, to, required to evaporate the water film of thickness δ = 0.2 mm is
PROBLEM 7.112
KNOWN: Diameter, velocity and surface vapor concentration of alcohol droplet falling in quiescent air.
Latent heat of vaporization and diffusion coefficient. Air temperature.
FIND: Droplet surface temperature
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Applicability of heat and mass transfer analogy, (3)
Negligible radiation, (4) Negligible vapor concentration in air (
A,
ρ
∞
= 0).
PROPERTIES: Table A.4, air (
T
∞
= 300 K): ν = 15.89 × 10-6 m2/s, k = 0.0263 W/m⋅K, Pr = 0.707.
ANALYSIS: Application of a surface energy balance yields
With
4 62
D
Re VD 1.8 m s 5 10 m 15.89 10 m s
ν
−−
= = ×× ×
= 56.6 and Sc =
AB
D
ν
= 1.59, the
Ranz-Marshall correlation yields
With
m
hh
=
( ) ( )
DAB D
ShD DNukD
,
COMMENTS: The large vapor density,
A,s
ρ
, renders the evaporative cooling effect significant.
PROBLEM 7.113
KNOWN: Diameter and density of liver cells, diameter of droplets.
FIND: (a) Terminal velocity of the droplets when each droplet contains one liver cell, (b) Time
of flight of a droplet containing one liver cell if the distance between injector and scaffold is L =
4 mm, (c) Initial evaporation rate from the droplet, (d) Comparison of the mass variation due to
evaporation to variation due to liver cell populations ranging from one to five per droplet.
SCHEMATIC:
ASSUMPTIONS: (1) Constant properties, (2) Negligible evaporative cooling, (3) Stokes’ law is
valid, CD = 24/ReD, (4) Neglect mass of displaced air in force balance, (5) Evaporation rate is
unaffected by change in droplet diameter, (6) Negligible microscale mass transfer effects.
PROPERTIES: Table A.4, air: (T = 25 °C = 298 K): ρ = 1.171 kg/m3, ν = 15.71 × 10-6 m2/s,
Table A.6, liquid water: (T = 25 °C = 298 K): ρ = 997.4 kg/m3, Table A.6, water vapor: (T = 25
°C = 298 K): vg = 44.25 m3/kg. Table A.8, water vapor in air: (T = 25 °C = 298 K): DAB = 0.26 ×
10-4 m2/s.
ANALYSIS:
(a) At terminal velocity, the face balance is
The volume of the droplet is
-6 3 -14 3
p
4
= π × (25 × 10 m) = 6.54 × 10 m
∀
while the volume of a
The mass is therefore
-15 3 3 -14 3 -15 3 3
M = 4.12 × 10 m × 2400 kg/m + (6.54 × 10 m – 4.12 × 10 m ) × 997.4 kg/m
F
D
D
P
= 50 µm
D
lc
= 20 µm
F
D
D
P
= 50 µm
D
lc
= 20 µm
PROBLEM 7.113 (Cont.)
Combining Equations 1, 2 and 3 yields
-11 2 -6
gp
-6 2 -9 2 3
f
MD 7.10 × 10 kg × 9.8 m/s × 50 × 10 m
V = =
12νA ρ 12 × 15.71 × 10 m /s × 1.963 × 10 m × 1.171 kg/m
(b) The time of flight is
(c) With Sc = ν/DAB = 1.571 × 10-5 m2/s / 0.26 × 10-4 m2/s = 0.604, the heat and mass transfer
analogy may be applied to Whitaker’s correlation to yield
The Reynolds number is
-6 -6 2
Dp
Re = VD /ν = 0.08 m/s × 50 × 10 m/15.71 × 10 m /s = 0.255.
Hence,

(d) The sensitivity may be estimated by comparing the change in mass due to evaporation to the
difference in mass due to liver cell loading.
Evaporation
Loading
The droplet mass with 3 liver cells is
The change in mass relative to one liver cell in the droplet is
COMMENTS: (1) Inspection of Figure 7.9 shows that Stokes’ law is valid at ReD = 0.255. (2)
The Whitaker correlation is used outside of the Reynolds and Schmidt number ranges for which it
was developed. The Sherwood number is
D p AB
h D /D
= 2.2. Since this is close to the limiting
value of 2.0, it is probably reasonably accurate.
PROBLEM 7.114
KNOWN: Dimension and approximate shape of E. coli bacterium. Binary diffusivity, nutrient
value, propulsion efficiency and concentration difference from free stream water–based solution
to bacterium shell.
FIND: Estimate the maximum E. coli speed in body diameters per second.
SCHEMATIC:
Water–based solution
D
AB
= 0.7 ×10
-9
m
2
/s
Water–based solution
D
AB
= 0.7 ×10
-9
m
2
/s
ASSUMPTIONS: (1) Negligible capability of bacterium to store energy, (2) Constant properties,
(3) Steady-state, (4) Stokes’ law is valid, that is CD = 24/ReD, (5) Negligible microscale mass
transfer effects.
PROPERTIES: Table A.6, water (T = 37 °C = 310 K): ρ = 993 kg/m3, ν = 6.999 × 10-7 m2/s, Pr
= 0.701.
ANALYSIS: For the spherical bacterium shell
-6 -7 2
D
Re = VD/ν = (V × 2 × 10 m)/(6.999 × 10 m /s) = 2.858 (s/m) × V
(1)
The power required to propel the bacterium is
D
We may combine Equations 1 and 2 to yield
For η = 0.5, the energy to be delivered from the water-based solution to the bacterium is
PROBLEM 7.114 (Cont.)
-18
m
E = 43.23 × 10 Ws/m × h
(4)
Applying the heat and mass transfer analogy to the Whitaker correlation yields
Therefore
Combining Equations (3) through (5) and solving for V yields
COMMENTS: (1) The maximum Reynolds number is ReD = VD/ν = 70 × 10-6 m/s × 2 × 10-6
m/6.999 × 10-7 m2/s = 200 × 10-6. The Whitaker correlation is extrapolated outside of its range of
application and provides a Sherwood number of 2.093 and a mass transfer coefficient of 732 ×
10-6 m/s. Using a Sherwood number of two, one would calculate a mass transfer coefficient of
PROBLEM 7.115
KNOWN: Diameter and temperature of sphere wetted with kerosene. Air flow conditions.
FIND: (a) Minimum kerosene flow rate, (b) Air temperature required to maintain wetted surface at
300K.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Sphere mount has a negligible influence on the
flow field and hence on
h,
(3) Negligible kerosene vapor concentration in free stream.
PROPERTIES: Table A-4, Air (300K): ν = 15.89 × 10-6 m2/s, k = 0.0263 W/m⋅K, ρ = 1.161
kg/m3, Pr = 0.707; Kerosene (given): ρA,sat = 0.015 kg/m3, hfg = 300 kJ/kg; Kerosene vapor-air
(given): DAB = 10–5 m2/s.
ANALYSIS: (a) The kerosene flowrate is
( )
A m A,sat A,
n hA .
ρρ
∞
= −
Using the mass transfer
analog of Eq. 7.56 and neglecting the viscosity ratio,
(b) An energy balance on the sphere yields
( )
A fg s
n h hA T T .
∞
= −
Using the Whitaker
correlation and neglecting the viscosity ratio,
COMMENTS: The small temperature excess (2.3K) is due to comparatively small values of ρA,sat
and hfg for kerosene.
PROBLEM 7.116
KNOWN: Dimensions of slot jet array. Jet exit velocity and temperature. Temperature of paper.
FIND: Drying rate per unit surface area.
SCHEMATIC:
ASSUMPTIONS: (1) Applicability of heat and mass transfer analogy, (2) Paper motion has
negligible effect on convection (U << Ve).
PROPERTIES: Table A-4, Air (300 K, 1 atm): ν = 15.89 × 10-6 m2/s; Table A-6, Saturated water
ANALYSIS: The mass evaporation flux is
( )
A m A,s A,e m A,sat
nh h
ρρ ρ
′′ = −=
For an array of slot nozzles,
2/3
3/4
r,o
0.42 r r,o r,o r
Sh 2 2 Re
A
3 A /A A /A
Sc

=

+

where
15.89 10 m / s
×
Hence
COMMENTS: The mass fraction of water vapor to air leaving the sides of the dryer is
( ) ( )
4
A air e
n S L / V W L 7 10 .
ρ
−
′′ × ×=×
Hence, the assumption of dry air throughout the dryer is
reasonable.