Question

Consider the elementary irreversible reaction between benzoquinone (A) and cyclopentathiene (B), A + B --> Products.

Calculate the Tubular Reactor Size (PFR) required for an 87.5% conversion. Suppose isothermal operation at 25°C, where the reaction rate constant is equal to 9.92 m³/(kgmol.ks). The concentrations of the reactants in the feed are both equal to 0.08 kgmol/m³ and the liquid feed rate is equal to 0.278 m³/ks. Feed is composed of equimolar concentrations of reagents.

Answer 1

Resposta:  

O volume do reator PFR deve ser de 2,45 m³.

Explicação:  

Primeiramente, se as concentrações dos reagentes são equimolares, isto significa que Ca = Cb.

Então, a velocidade da reação será expressa por:

-ra = k*Ca*Cb = k*Ca² = k*Cb²

Obtemos então os dados pela questão:

Xa = 87,5% = 0,875

k = 9,92 m³/(kgmol.ks)

Ca = Cb = 0,08 kgmol/m³

Q = 0,278 m³/ks

Aplicando a equação do reator de fluxo tubular (PFR - Plug Flow Reactor):

$T = C_a * (-\int\limits^{X_a}_{X_{a0}} {\frac{dX_a}{-ra} })$

$\frac{V}{Q} = C_a * (-\int\limits^{X_a}_{X_{a0}} {\frac{dX_a}{k{C_a}^2(1-X_a)^2} })$

$V = Q * C_a * \frac{1}{k{C_a}^2}* (-\int\limits^{X_a}_{X_{a0}} {\frac{dX_a}{(1-X_a)^2} })$

$V = Q * \frac{1}{kC_a}* (-\int\limits^{X_a}_{X_{a0}} {\frac{dX_a}{(1-X_a)^2} })$

$V = \frac{0,278m^3.ks^{-1}}{9,92m^3.{(kgmol.ks)^{-1}*0,08kgmol/m^{3}}}* (-\int\limits^{X_a}_{X_{a0}} {\frac{dX_a}{(1-X_a)^2} })$

$V = 0,35 m^3 * (-\int\limits^{X_a}_{X_{a0}} {\frac{dX_a}{(1-X_a)^2} })$

$V = 0,35 m^3 * ( -(-{\frac{1}{(1-X_a)}}))|_{X_{a0}}^{X_a}$

$V = 0,35 m^3 * [\frac{1}{1-X_a} - \frac{1}{1-X_{a0}} ]$

$V = 0,35 m^3 * [\frac{1}{1-0,875} - \frac{1}{1-0}]$

$V = 0,35 m^3 * [\frac{1}{0,125} - 1]$

V = 2,45 m³