Respuesta :
Answer:
a) The CSTR operating at 77 °C is recommended since it attain a higher conversion (96.2%).
b) 1.05 minutes.
c) 3474.9 minutes.
d)
- Series: 96%
- Parallel: 86%
e) The highest conversion was obtained in the first single 200-dm³ CSTR.
Explanation:
Hello,
At first, it is convenient to know that the initial flow rates of both A and B are equal since their volumetric rates and concentrations are equal:
[tex]F_A_0=F_B_0=2\frac{mol}{L} *\frac{1L}{1dm^3} *5\frac{dm^3}{min} =10\frac{mol}{min}[/tex]
Hence, we proceed.
a) In this case, one could select the answer by comparing the obtained conversion of A in the PFR and in the CSTR as shown below:
- PFR: the design equation is:
[tex]\frac{dX}{dV}=\frac{-r_A}{F_A_0}[/tex]
So the rate, as it is an elemental reaction, we have:
[tex]\frac{dX}{dV}=\frac{kC_AC_B}{F_A_0}=\frac{kC_A_0^2(1-X)^2}{F_A_0}[/tex]
Integrating as shown below:
[tex]\int\limits^X_0 { \frac{dX}{(1-X)^2}} \, dX=\frac{kC_A_0^2}{F_A_0}\int\limits^V_0 { dV}}\\\frac{X}{X-1}= \frac{kC_A_0^2}{F_A_0}V\\\frac{X}{X-1}=\frac{0.07\frac{dm^3}{mol*min}(2\frac{mol}{dm^3})^2}{10\frac{mol}{min} } (800.0dm^3)\\\frac{X}{X-1}=22.4\\X=0.957[/tex]
Thus, the achieved conversion is the 95.7% which looks promising under its operation conditions.
- CSTR: the design equation is:
[tex]V=\frac{F_A_0X}{-r_A}=\frac{F_A_0X}{kC_A_0^2(1-X)^2}[/tex]
In this case, the greater rate constant is selected as shown below:
[tex]k(T)=k(300K)*exp[\frac{E}{R}(\frac{1}{300K}-\frac{1}{273.15K} )]\\\\k(0^0C=273.15K)=0.07\frac{dm^3}{mol*min} *exp[\frac{20000cal/mol}{1.9872cal/mol*K}(\frac{1}{300K}-\frac{1}{273.15K} )]\\k(0^0C=273.15K)=0.00259\frac{dm^3}{mol*min}\\\\k(77^0C=350.15K)=0.07\frac{dm^3}{mol*min} *exp[\frac{20000cal/mol}{1.9872cal/mol*K}(\frac{1}{300K}-\frac{1}{350.15K} )]\\k(77^0C=350.15K)=8.55\frac{dm^3}{mol*min}[/tex]
In such a way, it is better at 77 °C
Therefore, the conversion is found by solving the design equation with solver as:
[tex]\frac{10\frac{mol}{min} X}{(8.55\frac{dm^3}{mol*min})(2\frac{mol}{dm^3} )^2(1-X)^2}-200.0dm^3=0\\X=0.962[/tex]
Thus, the achieved conversion is the 96.2% which looks promising under its operation conditions.
Therefore, the CSTR operating at 77 °C is recommended since it attain a higher conversion.
b) In this case, the rate constant is 8.55 dm³mol⁻¹min⁻¹ as previously computed, so the design equation is:
[tex]\frac{dX}{dt}=\frac{-r_A}{C_A_0} =\frac{kC_A_0^2(1-X)^2}{C_A_0} =kC_A_0(1-X)^2[/tex]
As long as there is no change in volume by cause of the state at which the reaction is carried out, that is liquid. Thereby, integrating and solving for time for a 90% conversion we obtain:
[tex]\int\limits^{0.9}_0 { \frac{dX}{(1-X)^2}} \, dX=kC_A_0\int\limits^t_0 { dt}}\\\\\frac{0.9}{1-0.9}=(8.55\frac{dm^3}{mol*min} )(1\frac{mol}{dm^3} )t\\\\t=\frac{9}{8.55min^{-1}} =1.05min[/tex]
c) In this case, the rate constant is 0.00259 dm³mol⁻¹min⁻¹ as previously computed, so the time turns out:
[tex]\int\limits^{0.9}_0 { \frac{dX}{(1-X)^2}} \, dX=kC_A_0\int\limits^t_0 { dt}}\\\\\frac{0.9}{1-0.9}=(0.00259\frac{dm^3}{mol*min} )(1\frac{mol}{dm^3} )t\\\\t=\frac{9}{0.00259min^{-1}} =3474.9min[/tex]
d)
- Series: the first CSTR will provide the following conversion, considering that the rate constant is at 300K:
[tex]\frac{10\frac{mol}{min} X}{(0.07\frac{dm^3}{mol*min})(2\frac{mol}{dm^3} )^2(1-X)^2}-200.0dm^3=0\\X=0.657[/tex]
Consequently, the inlet concentration and flow rate to the PFR are:
[tex]C_A_1=2\frac{mol}{dm^3} (1-0.657)=0.686\frac{mol}{dm^3} \\F_A_1=10\frac{mol}{min} (1-0.657)=3.43\frac{mol}{min} \\[/tex]
In this manner, the new integration yields the following conversion:[tex]\int\limits^X_0 { \frac{dX}{(1-X)^2}} \, dx=\frac{kC_A_1^2}{F_A_1}\int\limits^V_0 { dV}}\\\frac{X}{X-1}=\frac{0.07\frac{dm^3}{mol*min}(0.686\frac{mol}{dm^3})^2}{3.43\frac{mol}{min} } (800.0dm^3)\\\\X=0.885[/tex]
And the overall conversion is:
[tex]X_{overall}=\frac{10-[3.43(1-0.885)]}{10} =0.96[/tex]
- Parallel:
In this case, the initial flow rate to each reactor is 5 mol/min as the feed is divided, thus, the CSTR will provide the following conversion:
[tex]\frac{5\frac{mol}{min} X}{(0.07\frac{dm^3}{mol*min})(2\frac{mol}{dm^3} )^2(1-X)^2}-200.0dm^3=0\\X=0.742[/tex]
And the PFR:
[tex]\int\limits^X_0 { \frac{dX}{(1-X)^2}} \, dx=\frac{kC_A_0^2}{F_A_0}\int\limits^V_0 { dV}}\\\frac{X}{X-1}=\frac{0.07\frac{dm^3}{mol*min}(2\frac{mol}{dm^3})^2}{5\frac{mol}{min} } (800.0dm^3)\\\\X=0.978[/tex]
Thus, the outlet A's flow rate from the CSTR and the PFR is:
[tex]F_A_1^{CSTR}=5\frac{mol}{min} (1-0.742)=1.29\frac{mol}{min} \\F_A_1^{PFR}=5\frac{mol}{min} (1-0.978)=0.11\frac{mol}{min} \\[/tex]
Thus, the outlet flow rate is 1.4 mol/mol, so the overall conversion is:
[tex]X_{overall}=0.86[/tex]
e) The results showed that a simple CSTR is more than enough to attain a high conversion since the reaction is second-order, because if it would be first order a higher change in the conversion will be related with a higher volume for both CSTR and PFR, as well as it is the smallest one (cheaper one).
Best regards.
