[Solucionário - Capítulo 9] Reações Químicas e Reatores Químicas George W. Roberts (Wiley, 2009)

Prévia do material em texto

Copyright© - George W. Roberts 1 April 21, 2008
Solution to Problem 9-18(T-15)
This reaction takes place at high temperatures, ca. 800 – 1000 °C. The SiO2 phase
is essentially non-porous, so that O2 must dissolve into and diffuse through the solid
SiO2 in order to reach the interface between SiO2 and Si and then react. A sketch of the
system is shown below.
CO
Bulk
Gas
Gas
Boundary
Layer
SiO2 Si
CS
Ci
Ce
T(t)
z = 0 z = T(t)
Part 1)
Assumption 2: Let C be the concentration of O2 dissolved in the SiO2. The flux of
O2 is given by:
 
NO 2 = �DO
dC
dz
+ yO 2 Ni
i
� = �DO dCdz + yO 2NO 2
If yO 2 << 1, the second term on the right-hand side can be ignored and the flux of O2 is
purely diffusive, so that:
 
NO 2 = �DO
dC
dz
Assumption 4: If the reaction at the interface is very slow compared to the rate of
diffusion, the accumulation term in the O2 material balance for the SiO2 phase can be
neglected, i.e., the O2 flux in the SiO2 layer is independent of position at every time, t.
Therefore,
 
NO 2 (t) = �DO
dC
dz
= constan t
Integrating from z = 0 (the gas/SiO2 interface) to z = T(t) (the SiO2/Si interface),
Copyright© - George W. Roberts 2 April 21, 2008
 NO 2 (t) = DO (Ce �Ci)/T(t)
Part 2)
At any time, the flux of O2 reaching the SiO2/Si interface must be equal to the
rate at which O2 reacts with Si to form SiO2.
 
NO 2 (t) = DO (Ce �Ci)/T(t) = kCi
Ci = DOCe /(DO + kT(t))
Furthermore, at any time, the flux of O2 reaching the gas/SiO2 interface must be
equal to the flux of O2 leaving that interface.
 kc (CO �Cs) = DO (Ce �Ci)/T(t)
Since Ci = DOCe/(DO + kT(t)) and Ce = HCs,
 
Cs = CO / 1+
DOHk
kc (DO + kT(t)
� 
� 
� 
� 
� 
� 
 
NO 2 (t) = kc (CO �Cs) = kcCODOHk/[kc (DO + kT(t)) + DOHk)]
NO 2 (t) = HCODO /{A + T(t)]
Here, 
 
A �DO
1
k
+
H
kC
� 
� 
� 
� 
� 
� 
The rate of growth of the oxide layer is the rate of consumption of O2 at the
SiO2/Si interface, divided by the moles of O2 per unit volume of SiO2 (N1).
 
dT(t)
dt
=
NO 2 (t)
N1
=
HCODO
N1[A + T(t)]
=
B
2[A + T(t)]
where B = 2HCODO/N1.
Integrating from t = 0, T = T0 to t = t, T = T;
 
[A + T]dT =
B
2
dt
0
t
�
T0
T
�
A(T �T0 ) +
T2 �T02
2
=
Bt
2
Solving this quadratic equation for T(t),
 
T(t) = A 1+
T0
2
A2
+
2T0
A
+
Bt
A2
� 
� 
 
� 
	 
� 
1/2
� 1
� 
� 
� 
� 
� 
� 
 
 
Let � = (T0
2
+ 2AT0 )/B ,
 
T(t) = A 1+
B
A2
(t + �)
� 
� � 
	 
 
 
1/2
� 1
� 
� 
� 
� 
� 
� 
� 
� 
Copyright© - George W. Roberts 3 April 21, 2008
Part 3)
If the gas is 100% O2, the O2 concentration at the solid surface, Cs, will be equal to
CO. There will be no resistance to mass transfer, and no O2 concentration gradient,
through the gas boundary layer. The previously-derived expression for T(t) remains
valid, except that A = DO/k. This is the same result that would be obtained if the mass-
transfer coefficient kC were set equal to infinity to reflect the fact that there is no
resistance to mass transfer associated with the gas boundary layer.
Copyright© - George W. Roberts March 12, 2008
T-21 (Problem 9-19)
Engineer B is correct. It is true that changing the catalyst particle size will change
the external resistance to mass transfer. However, changing the particle size also
changes the internal (pore diffusion) resistance, since the Thiele modulus, �, is directly
proportional to the characteristic dimension of the particle, lc. Therefore, if the outlet
conversion changes when the catalyst particle size is varied at constant space time and
linear velocity, we do not know whether the conversion change is due to the variation
of the external resistance or the internal resistance (or both).
Varying the catalyst particle size is a good diagnostic experiment for an internal
transport resistance (� < 1) only when it has been established that the external transport
resistance is not significant.
Copyright© - George W. Roberts 1 April 11, 2008
Problem 9-20 (T-22)
1) The reaction is first-order, there is no change in moles on reaction, and the
reactor is isothermal. Therefore, density changes can be neglected and the rate of
disappearance of A is given by:
-rA = kCA = kCA0(1 – xA).
Since the reactor is an ideal PFR, the design equation is:
 
W
FA0
=
dxA
kCA0 (1� xA )
=
0
xA
� 1kCA0
dxA
(1� xA )
=
0
xA
� � ln(1� xA )kCAo
 
WCA0
FA0
= � = � ln(1� xA )/k
 � ln(1� xA ) = k�
To test whether the reaction is first-order and irreversible, plot ln(1-xA) versus �.
If the model fits the data, a straight line through the origin should result, with a slope of
-k. See the plot on the following page. There is very little scatter. The value of k is 3.85
(cm3/g-s). (Note: The “best fit” straight line was forced through the origin.)
2) Experiment 5 is at the same space time and temperature as Experiment 1.
However, the superficial velocity is a factor of 3 higher than in Experiment 5.
Experiment 6 is at the same space time and temperature as Experiment 2. Once again,
the superficial velocity is a factor of 3 higher in Experiment 6 than in Experiment 2.
At constant temperature and space time, the superficial velocity will only affect
the conversion if the external mass transfer resistance is significant, i.e., if the reaction is
controlled or influenced by external mass transfer.
The conversion increases by 50% (from 0.50 to 0.75) from Experiment 1 to
Experiment 5. The conversion increases by 25% (from 0.75 to 0.94) from Experiment 2
to Experiment 6. These increases are significant. The reaction appears to be influenced
or controlled by external mass transfer.
3) Experiment 7 is at the same conditions as Experiment 1, except that the
temperature is 25 °C higher in Experiment 7. The fractional conversion increases from
0.50 at 400 °C to 0.54 at 425 °C. This is a relatively small change, and it suggests that the
reaction is strongly influenced by some mass transfer resistance, consistent with the
conclusion in Part 2.
Let’s calculate an apparent activation energy. If the reaction remains first-order,
the rate constant is:
k(425 °C) = -ln(1 – xA)/� = -ln(1 – 0.54)/0.18 = 4.31 cm3/g-s
Arrhenius Relationship:
 
k(425)
k(400)
= exp
�E
8.314
1
(425 + 273)
� 1
(400 + 273)
� 
� 
� 
	 
 
� 
� 
� 
� 
� 
� 
 =
4.31
3.85
E � 18 kJ/mole
Copyright© - George W. Roberts 2 April 11, 2008
If the reaction were controlled by intrinsic kinetics, an activation energy of about
40 kJ/mole or higher would be expected. The low value calculated above suggests that
mass transfer of some kind is influencing the reaction rate. The results of this
calculation are consistent with the analysis in Part 2.
0.01
0.1
1
0 0.2 0.4 0.6 0.8 1 1.2
Data Analysis
(Problem 9-20, Part 1)
ln
(1
-x
A
)
Space Time, �
(g-s/cm
3
)
y = exp(m1*M0)
ErrorValue
0.021371-3.8529m1 
NA2.531e-05Chisq
NA0.99991R
	Solution 9-1(T-8).pdf
	Solution 9-2(T-23).pdf
	Solution 9-3(T-29).pdf
	Solution 9-4(T-7).pdf
	Solution 9-5(T-14b).pdf
	Solution 9-6(T-25).pdf
	Solution 9-7(T-27).pdf
	Solution 9-8(T-35).pdf
	Solution 9-9(T-10).pdf
	Solution 9-10(T-30).pdf
	Solution 9-11(T-33).pdf
	Solution 9-12(T-3(Part e)).pdf
	Solution 9-13(T-18).pdf
	Solution 9-14(T-26).pdf
	Solution 9-15(T-34).pdf
	Solution 9-16(T-12).pdf
	Solution 9-17(T-9).pdf
	Solution 9-18(T-15).pdf
	Solution 9-19(T-21).pdf
	Solution 9-20(T-22).pdf
	Solution 9-21(T-32).pdf