Engineering, 24.03.2020 04:07 rwerjekrryery6750

The liquid-phase reaction follows an elementary rate law and is carried out isothermally in a flow system. The concentrations of the A and B feed streams are 2 M before mixing. The volumetric flow rate of each stream is 5 dm 3 /min, and the entering temperature is 300 K. The streams are mixed immediately before entering. Two reactors are available. One is a gray, 200.0-dm 3 CSTR that can be heated to 77 C or cooled to 0 C, and the other is a white, 800.0-dm 3 PFR operated at 300 K that cannot be heated or cooled but can be painted red or black. Note that k 0.07 dm 3 /mol min at 300 K and E 20 kcal/mol.

a) Which reactor would and what conditions do you recommend? Explaining the reason for your choice. Back up your reasoning with calculations.

b) How long would it take to achieve 90% conversion in a 200 dm3 batch reactor with initial concentrations of A and B at 1 mol/L and operated at 77 °C?

c) What would be your answer to part (b) if at 0 °C?

d) What conversion would be obtained if the CSTR and PFR were operated at 300 K and connected in series? In parallel with 5 dm min of the mixture to each?

e) What can you say about the various reactor types based on the above?

Answers: 3

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Answers

Answer from: Bri0929

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.

- 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:

$F_A_0=F_B_0=2\frac{mol}{L} *\frac{1L}{1dm^3} *5\frac{dm^3}{min} =10\frac{mol}{min}$

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:

$\frac{dX}{dV}=\frac{-r_A}{F_A_0}$

So the rate, as it is an elemental reaction, we have:

$\frac{dX}{dV}=\frac{kC_AC_B}{F_A_0}=\frac{kC_A_0^2(1-X)^2}{F_A_0}$

Integrating as shown below:

$\int\limits^X_0 { \frac{dX}{(1-X)^2}} \, dX=\frac{kC_A_0^2}{F_A_0}\int\limits^V_0 { dV}}\\\frac{X}{X-1}= \frac{kC_A_0^2}{F_A_0}V\\\frac{X}{X-1}=\frac{0.07\frac{dm^3}{mol*min}(2\frac{mol}{dm^3})^2}{10\frac{mol}{min} } (800.0dm^3)\\\frac{X}{X-1}=22.4\\X=0.957$

Thus, the achieved conversion is the 95.7% which looks promising under its operation conditions.

- CSTR: the design equation is:

$V=\frac{F_A_0X}{-r_A}=\frac{F_A_0X}{kC_A_0^2(1-X)^2}$

In this case, the greater rate constant is selected as shown below:

$k(T)=k(300K)*exp[\frac{E}{R}(\frac{1}{300K}-\frac{1}{273.15K} )]\\\\k(0^0C=273.15K)=0.07\frac{dm^3}{mol*min} *exp[\frac{20000cal/mol}{1.9872cal/mol*K}(\frac{1}{300K}-\frac{1}{273.15K} )]\\k(0^0C=273.15K)=0.00259\frac{dm^3}{mol*min}\\\\k(77^0C=350.15K)=0.07\frac{dm^3}{mol*min} *exp[\frac{20000cal/mol}{1.9872cal/mol*K}(\frac{1}{300K}-\frac{1}{350.15K} )]\\k(77^0C=350.15K)=8.55\frac{dm^3}{mol*min}$

In such a way, it is better at 77 °C

Therefore, the conversion is found by solving the design equation with solver as:

$\frac{10\frac{mol}{min} X}{(8.55\frac{dm^3}{mol*min})(2\frac{mol}{dm^3} )^2(1-X)^2}-200.0dm^3=0\\X=0.962$

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:

$\frac{dX}{dt}=\frac{-r_A}{C_A_0} =\frac{kC_A_0^2(1-X)^2}{C_A_0} =kC_A_0(1-X)^2$

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:

$\int\limits^{0.9}_0 { \frac{dX}{(1-X)^2}} \, dX=kC_A_0\int\limits^t_0 { dt}}\\\\\frac{0.9}{1-0.9}=(8.55\frac{dm^3}{mol*min} )(1\frac{mol}{dm^3} )t\\\\t=\frac{9}{8.55min^{-1}} =1.05min$

c) In this case, the rate constant is 0.00259 dm³mol⁻¹min⁻¹ as previously computed, so the time turns out:

$\int\limits^{0.9}_0 { \frac{dX}{(1-X)^2}} \, dX=kC_A_0\int\limits^t_0 { dt}}\\\\\frac{0.9}{1-0.9}=(0.00259\frac{dm^3}{mol*min} )(1\frac{mol}{dm^3} )t\\\\t=\frac{9}{0.00259min^{-1}} =3474.9min$

- Series: the first CSTR will provide the following conversion, considering that the rate constant is at 300K:

$\frac{10\frac{mol}{min} X}{(0.07\frac{dm^3}{mol*min})(2\frac{mol}{dm^3} )^2(1-X)^2}-200.0dm^3=0\\X=0.657$

Consequently, the inlet concentration and flow rate to the PFR are:

$C_A_1=2\frac{mol}{dm^3} (1-0.657)=0.686\frac{mol}{dm^3} \\F_A_1=10\frac{mol}{min} (1-0.657)=3.43\frac{mol}{min} \\$

In this manner, the new integration yields the following conversion: $\int\limits^X_0 { \frac{dX}{(1-X)^2}} \, dx=\frac{kC_A_1^2}{F_A_1}\int\limits^V_0 { dV}}\\\frac{X}{X-1}=\frac{0.07\frac{dm^3}{mol*min}(0.686\frac{mol}{dm^3})^2}{3.43\frac{mol}{min} } (800.0dm^3)\\\\X=0.885$

And the overall conversion is:

$X_{overall}=\frac{10-[3.43(1-0.885)]}{10} =0.96$

- 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:

$\frac{5\frac{mol}{min} X}{(0.07\frac{dm^3}{mol*min})(2\frac{mol}{dm^3} )^2(1-X)^2}-200.0dm^3=0\\X=0.742$

And the PFR:

$\int\limits^X_0 { \frac{dX}{(1-X)^2}} \, dx=\frac{kC_A_0^2}{F_A_0}\int\limits^V_0 { dV}}\\\frac{X}{X-1}=\frac{0.07\frac{dm^3}{mol*min}(2\frac{mol}{dm^3})^2}{5\frac{mol}{min} } (800.0dm^3)\\\\X=0.978$

Thus, the outlet A's flow rate from the CSTR and the PFR is:

$F_A_1^{CSTR}=5\frac{mol}{min} (1-0.742)=1.29\frac{mol}{min} \\F_A_1^{PFR}=5\frac{mol}{min} (1-0.978)=0.11\frac{mol}{min} \\$

Thus, the outlet flow rate is 1.4 mol/mol, so the overall conversion is:

$X_{overall}=0.86$

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.

Answer from: Quest

how is this a question? in order for me to answer it i have to understand the question

explanation:

Answer from: Quest

correct me if i’m wrong but i believe it’s false

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The liquid-phase reaction follows an elementary rate law and is carried out isothermally in a flow s...