A liquid-phase reaction A + B → 2C is conducted in a nonisothermal multitubular PFR.The reactor tubes (7 m long, 2 cm in diameter) are surrounded by a coolant which maintains a constant wall temperature. The reaction is pseudo-first-order with respect to A, with kA = 4.03 × lo^5 e^-5624/T s-¹. The

Question

A liquid-phase reaction A + B → 2C is conducted in a nonisothermal multitubular PFR.The reactor tubes (7 m long, 2 cm in diameter) are surrounded by a coolant which maintains a constant wall temperature. The reaction is pseudo-first-order with respect to A, with $k_A = 4.03 \times 10^5 e^{- 5624/T} s^{-1}$ . The mass flow rate is constant at $0.06 kg s^{-1}$ , the density is constant at $1.025 g cm^{-3}$ , and the temperature at the inlet of the reactor (T,) is 350 K.

(a) Develop expressions for $df_A/dx$ and $dT/dx$ .
(b) Plot $f_A(x)$ profiles for the following wall temperatures ( $T_s$ ): 350 K, 365 K, 400 K, and 425 K.

Data: $c_{Ao} = 0.50 mol L^{-1} ; c_p = 4.2 J g^{-1} K^{-1} ; \Delta H_{RA} = -210 KJ mol^{-1} ; U = 1.59 kW m^{-2} K^{-1}$ .

(a) Develop expressions for [latex]df_A/dx[/latex] and [latex]dT/dx[/latex].
(b) Plot [latex]f_A(x)[/latex] profiles for the following wall temperatures ([latex]T_s[/latex]): 350 K, 365 K, 400 K, and 425 K.

Data: [latex]c_{Ao} = 0.50 mol L^{-1} ; c_p = 4.2 J g^{-1} K^{-1} ; \Delta H_{RA} = -210 KJ mol^{-1} ; U = 1.59 kW m^{-2} K^{-1} [/latex] .

Accepted Answer

(a) The rate law is :

[latex] (- r_A) = k_Ac_A = k_Ac_{Ao} (1 - f_A) [/latex] (A)

where [latex] k_A [/latex] is given in Arrhenius form above. Substitution of equation (A) in the material- balance equation 15.2-4, [latex] (-r_A) = (F_{Ao}/ \pi r^2)df_A/dx [/latex] , results in (with R = D/2 and [latex]F_{Ao}/c_{Ao} = q_o [/latex] ) :

[latex] \frac {df_A}{dx} - \frac{\pi R^2( -r_A)}{F_{Ao}} = 0 [/latex] (15.2-4)

[latex] \frac{df_A}{dx} = \frac{\pi D^2k_A(1 - f_A)}{4q_o}[/latex] (B)

Equation (B) gives the required expression for d [latex] f_A /dx [/latex] . To develop an expression for [latex]dT/dx [/latex] , we use the energy balance of equation 15.2-9,

[latex]\dot{m}c_pdT = \left[( -\Delta H_{RA}) + \frac {4U(T_s -T)}{D( -r_A)}\right] F_{Ao}df_A[/latex] (15.2-9)

divided by dx, with [latex] ( -r_A) [/latex] eliminated by equation (A), and d [latex] f_A /dx [/latex] by equation (B), to result in

[latex] \frac {dT}{dx} = \frac { \pi D^2 k_Ac_{A0}(1 - f_A)(- \Delta H_{RA} )}{4\dot{m}c_p} + \frac {\pi UD}{\dot{m}c_p} (T_s - T) [/latex] (C)

where [latex]T_s[/latex] is the wall temperature. Note that dT/dx is implicitly related to [latex]df_A/dx[/latex] .

(b) The E-Z Solve software may be used to integrate numerically the differential equations (B) and (C) from x = 0 to x = 7 m, using the four wall temperatures specified (see file exl5-4.msp). The resulting [latex]f_A(x)[/latex] profiles are illustrated in Figure 15.5. Note that as the wall temperature ([latex]T_s[/latex]) is increased from 350 K to 425 K, the length of reactor required to achieve a particular fractional conversion is reduced. For example, virtually 100% conversion is achieved within 4 m when [latex]T_s[/latex] is 425 K, but requires nearly 5 m when [latex]T_s[/latex] is 400 K

Question 15.4: A liquid-phase reaction A + B → 2C is conducted in a nonisot...

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