Chapter 21 Simultaneous heat and mass transfer page 21-1
Chapter 21 Simultaneous Heat and Mass Transfer
1. CCl4 concentration in air
From Eq. 20.2-19, y1 = y1i – (Cp/Hvap) (T – Ti)
2. Evaporation of n-butyl alcohol
From the Chilton-Colburn Analogy Eq. 20.2-13),
2/3 = h
2/3
3. Burning meteor
For a 1st order reaction, q = N1 Hrxn
Where mass transferred N1 = kGA[cO2(bulk) – 0]
4.Antibiotic drying
For a constant drying rate, the temperature of the cake is constant. We must find this constant
temperature to calculate how fast the acetone evaporates. One constraint is given in the problem;
Chapter 21 Simultaneous heat and mass transfer page 21-2
the other can be found from the Chilton-Colburn analogy. A mass and energy balance on the
cake gives:
~
~
Applying the Chilton-Colburn analogy we have:
CTcmmolc
TC
mol
cal
Kmol
cal
cm
mol
TT
H
Cc
TT
H
C
c
TT
C
cH
C
h
k
ii
ii
vap
p
i
vap
p
i
i
p
ivap
p
863
3
1086.21014.1/
40
7800
5
22400
1
~
~
~
ˆ
ˆ
~
ˆ
Solving the equations simultaneously we have:
3
666
1045.16.101024.0100.4
cm
mol
Cc
i
i
A mass balance on the trays gives:
f
i
if
xx
Mkc
t
tAcckxx
M
0
0
Since the density is based on the crystals, we must use mass ratios instead of mass fractions to
express the initial and final concentrations of acetone.
Chapter 21 Simultaneous heat and mass transfer page 21-3
824 s13.7min
5. Water treatment
a) Total air flow
To find the minimum air flow, start at 20 °C and 0 kJ/mol and draw a line which is tangent to the
equilibrium line.
9
10
11
12
13
14
15
Equilibrium
We can now find the gas flow rate:
min
mol
,
min
mol
,.G.G
min
.
.
min
min
0002390001595151
442
442
b) The height of the column is given by:
0
H
*
H
dH
HTU HH
To simplify the integral, we assume an average driving force based on a linear equilibrium line.
The endpoint of the operating line is about 5.0 kJ/mol, so the driving force is:
2
H.
mol
The integral becomes:
0
305
H
kJ
HH .mol
6. Cooling water
a) Tower diameter
2
2
80 50
0.892
mol mol
d
dm
b) To find the minimum air flow, start at 30 °C and 1 kJ/mol and draw a line which is tangent to
the equilibrium line.
0
1
2
3
8
9
10
11
12
13
14
15
20 25 30 35 40 45 50 55 60 65 70
Water Temperature (°C)
Molar Enthalpy (kJ / mol wet air)
Equilibrium
The endpoint of this line is about 8.6 kJ/mol. The slope of this line is:
025 338
0 075
min
LkJmolC
..
GmolC.kJ
Chapter 21 Simultaneous heat and mass transfer page 21-6
For twice the minimum air flow we have:
338 2
mol / s mol
./ s
c) The height of the column is given by:
0
H
To simplify the integral, we use an average driving force. The endpoint of the operating line is
about 4.8 kJ/mol, so the driving force is:
2 1 98 48 35
2
kJ kJ
.. kJ
mol mol
H.
mol
The integral becomes:
0
048 1
35
H
H
kJ
.
dH H H mol
kJ
HH .mol
7. Countercurrent cooling tower H2O,
Air
To do it numerically, first we calculate the equilibrium line
Hi(T). Here the value of
H2O
Chapter 21 Simultaneous heat and mass transfer page 21-7
0
2000
0 20406080
T (C)
The slope of the operating line is from Eq. 20.3-14. For the water stream,
2
Hence at the top of the tower,
Hin = 702 + 98.7*(66 – 20) = 5243 J/mol
The numerical integration gives
Hout d
H
Hi –
H = 2.65
By evaluating at the average temperature, 43°C, and 1 atm,
c = 1/82.05/316 = 3.8610-5 mol/cm3 = 38.6 mol/m3
k = 80/10/63/38.6*2.655 = 0.00873 m/s = 0.873 cm/s
(b) In this case, the water flow rate is fixed at 1000 kg/min, and the air flow rate is to be
reduced. To solve this problem, we use a trial-and-error approach to get the solution:
(i) assume a value for nair,
(ii) calculate the slope of the operating line,
Hout d
H
(nair/kc)1 =
(nair/nH2O)1
8. Thermal diffusion in Two-bulb capillary
Following example 20.5-1, at steady state, eq. 20.5-2 becomes
–
j = 0 = Dc(x1 – x1x2T/T) => 0 = dx1
9. Thermal diffusion pump
The pump is illustrated on the right.
For Knudsen diffusion in the membrane, from Eq. 6.5-15, 23°C
47°C
D = 4850 dT/M
= 4850*12010-8 T/28.97
10. Effect of diffusion-engendered convection on heat transfer
dz(q + vC T) (1)
In addition, combining Fourier’s Law gives
v dT
dz = d2T
dz2 (2)
T0
where = k/Cp
(b) The general solution of Eq. (2) is
T(z) = Aevz/ + B
Applying the 1st B. B., we have T0 = A + B => T = A(evz/ -1) + T0
Applying the 2nd B. C., we have Tl = A(evl/ -1) + T0 => A = Tl – T0
evl/ -1
11. Thermal conductivity of reacting mixture
Let 1 = NO2, 2 = N2O4
(a) Mass balance shows that
Chapter 21 Simultaneous heat and mass transfer page 21-10
01
dz + y n /2 => n(1 – y/2)dz = –Dcdy
Integration gives n1z = 2Dc ln
1 – y1/2
1 – y10/2
1 – y1l/2
at the average temperature, c = 1/(82.05*333) = 3.6610-5 mol/cm3
n1 = 2*0.07*3.6610-5
0.16 ln
1 – 0.85/2
1 – 0.48/2 = -1.4310-6 mol/cm2-s
(c) Energy balance shows that
d2T
dz2 = v dT
dz (6)
12. Heat transfer across a reactive gas layer
(a) The concentration profile :
The reaction only occurs at the surface, therefore, n1 and n2 are independent of z.
Furthermore, by mass balance, we have n1 = -2n2 (1)
Chapter 21 Simultaneous heat and mass transfer page 21-11
z = l, y1 = 1
By integrating eq. (4), we have n1 = 2Dc
z ln(1 – y/2) (5)
1
Substituting boundary condition,
Thus, the concentration profile is y1 = 2 + 2ln(2) (z/l) (7)
(b) The energy balance is
The temperature profile is T = T0 + (Tl – T0) e–Dln(2)z/l – 1
e–Dln(2)/ – 1 (10)
(c) At z = 0, the heat flux is