Question 10.8: Determination of Optimum Temperature for Operation of a Sing...

Determination of Optimum Temperature for Operation of a Single CSTR in Which a Reversible Exothermic Reaction is Being Carried Out

The following reversible reaction takes place in a CSTR:

C \underset{k_2}{\stackrel{k_1}{\leftrightarrow}} B

where both the forward and the reverse reaction obey first-order kinetics. The rate constants may be written in the Arrhenius form as

k_1=A_1 e^{-E_1 / R T}

k_2=A_2 e^{-E_2 / R T}

 Determine the minimum reactor volume that will be required to obtain a fraction conversion f_{C} if the feed is pure C and the volumetric feed rate is V_{0}. What will be the temperature of the effluent stream?

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For this CSTR the appropriate design equation is

\tau=\frac{V_R}{V_0}=\frac{C_{ C 0} \int_0^{f_e} d f_c}{-r_{ CF}}=\frac{C_{ C 0} f_c}{-r_{ C F}}                  (A)

with

-r_{ C F}=k_1 C_{ C F}-k_2 C_{ B F}                 (B)

From stoichiometric considerations

C_{ C F}=C_{ C 0}\left(1-f_c\right)                 (C)

C_{ B F}=C_{ C 0} f_c                    (D)

Combining equations (A) to (D) gives

\frac{V_R}{ V _0}=\frac{f_c}{k_1\left(1-f_c\right)-k_2 f_c}=\frac{f_c}{k_1-f_c\left(k_1+k_2\right)}

To minimize the required reactor volume, one may set the temperature derivative of V_{R} equal to zero:

\left(\frac{\partial V_R}{\partial T}\right)_{f_c}=\frac{-V_0 f_c \frac{\partial}{\partial T}\left[k_1-f_c\left(k_1+k_2\right)\right]}{\left[k_1-f_c\left(k_1+k_2\right)\right]^2}=0

or

\frac{\partial k_1}{\partial T}-f_c\left(\frac{\partial k_1}{\partial T}+\frac{\partial k_2}{\partial T}\right)=0

However, from the Arrhenius relation,

\frac{d \ln k}{d T}=\frac{E_A}{R T^2} \quad \text { or } \quad \frac{d k}{d T}=\frac{k E_A}{R T^2}

Thus, the optimum temperature will be that at which

\frac{k_1 E_1}{R T^2}\left(1-f_c\right)-f_c \frac{k_2 E_2}{R T^2}=0

or

\frac{k_1}{k_2}=\frac{E_2}{E_1}\left(\frac{f_c}{1-f_c}\right)

Substitution of the Arrhenius form of the rate constants yields

\frac{A_1 e^{-E_1 / R T}}{A_2 e^{-E_2 / R T}}=\frac{E_2}{E_1}\left(\frac{f_c}{1-f_c}\right)

One can then solve for the optimum temperature:

\frac{1}{T}\left(-\frac{E_1}{R}+\frac{E_2}{R}\right)=\ln \left[\left(\frac{E_2 A_2}{E_1 A_1}\right)\left(\frac{f_c}{1-f_c}\right)\right]\equiv \ln \beta

or

T=\frac{E_2-E_1}{R \ln \left\{\left[\left(E_2 A_2\right) /\left(E_1 A_1\right)\right]\left[f_c /\left(1-f_c\right)\right]\right\}}=\frac{E_2-E_1}{R \ln\beta}

At this temperature

k_1=A_1 e^{-\left(E_1 \ln \beta\right) /\left(E_2-E_1\right)}=A_1 \beta^{-E_1 /\left(E_2-E_1\right)}

and

k_2=A_2 \beta^{-E_2 /\left(E_2-E_1\right)}

Thus, the minimum reactor volume is given by

V_R=\frac{ V _0 f_c}{A_1 \beta^{-E_1 /\left(E_2-E_1\right)}\left(1-f_c\right)-A_2 \beta^{-E_2 /\left(E_2-E_1\right)} f_c}

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