A gas-phase reaction between butadiene (A) and ethene (B) is conducted in a PFR, producing cyclohexene (C). The feed contains equimolar amounts of each reactant at 525°C (T,) and a total pressure of 101 kPa. The enthalpy of reaction is - 115 kJ (mol A)-‘, and the reaction is first-order with

Question

A gas-phase reaction between butadiene (A) and ethene (B) is conducted in a PFR, producing cyclohexene (C). The feed contains equimolar amounts of each reactant at 525°C (T,) and a total pressure of 101 kPa. The enthalpy of reaction is - 115 kJ (mol A)-‘, and the reaction is first-order with respect to each reactant, with [latex] k_A = 32,000 e^{-13,850/T} m^3 mol^{-1} s^{-1} [/latex] . Assuming the process is adiabatic and isobaric, determine the space time required for 25% conversion of butadiene.
Data: [latex] C_{PA} = 150 J mol^{-1} K^{-1} ; C_{PB} = 80 J mol^{-1} K^{-1} ; C_{PC} = 250 J mol^{-1} K^{-1} [/latex]

Accepted Answer

The reaction is [latex] C_4H_6(A)+C_2H_4(B) ightarrow C_6H_{10}(C)[/latex] . Since the molar ratio of A to B in the feed is 1: 1, and the ratio of the stoichiometric coefficients is also 1: 1 , [latex]c_A = c_B[/latex] throughout the reaction. Combining the material-balance equation (15.2-2) with the rate law, we obtain

[latex]V = F_{Ao}\int{{df_A}/{(-r_A)}}[/latex] (15.2-2)

[latex]V = F_{Ao}\int{\frac{df_A}{(-r_A)}} = F_{Ao}\int{\frac{df_A}{k_Ac_Ac_B}} = F_{Ao}\int{\frac{df_A}{k_Ac_A^2} } \ = F_{Ao}\int{\frac{df_A}{k_A(F_A/q)^2}} = frac{1}{F_{Ao}}\int_{0}^{f_A}{\frac{q^2df_A}{k_A(1 - f_A)^2}} [/latex] (A)

Since [latex]k_A[/latex] depends on T, it remains inside the integral, and we must relate T to [latex]f_A[/latex]. Since the density (and hence q) changes during the reaction (because of changes in temperature and total moles), we relate q to [latex]f_A[/latex] and T with the aid of a stoichiometric table and the ideal-gas equation of state.

Species	Initial moles	Change	Final moles
A	[latex] F_{Ao}[/latex]	[latex]-f_AF_{Ao}[/latex]	[latex]F_{Ao}(1 - f_A)[/latex]
B	[latex] F_{Ao}[/latex]	[latex]- f_AF_{Ao} [/latex]	[latex]F_{Ao}(1 - f_A)[/latex]
C	0	[latex]f_AF_{Ao}[/latex]	[latex]F_{Ao} f_A[/latex]
total	[latex]2 F_{Ao}[/latex]		[latex]F_t = F_{Ao}(2 - f_A)[/latex]

Since at any point in the reactor, [latex]q = F_tRT/P [/latex], and the process is isobaric, q is related to the inlet flow rate [latex]q_o[/latex] by

[latex]\frac{q}{q_o} = \frac{F_{Ao}(2 - f_A)T}{2F_{Ao}T_o} [/latex]

That is ,

[latex]q = q_o(1 - 0.5f_A)T/T_o[/latex] (B)

Substitution of equation (B) into (A) to eliminate q results in

[latex] au =\frac{V}{q_o} = \frac{1}{c_{Ao}T_o^2}\int_{0}^{f_A}{\frac{(1 - 0.5f_A)^2T^2}{k_A(1 - f_A)^2}df_A} [/latex] (C)

To relate [latex]f_A[/latex] and T, we require the energy balance (15.2-9). From this equation, for adiabatic operation (the heat transfer term on right side vanishes),

[latex]\dot{m}c_PdT = \left[(-\Delta H_{RA}) + \frac{4U(T_S - T)}{D(-r_A)}\underset{_I}{}F_{Ao}df_A ight. [/latex] (15.2-9)

[latex]\dot{m}c_pdT \equiv F_tC_pdT = (-\Delta H_{RA})F_{Ao}df_A[/latex] (D)

where [latex]C_P[/latex] is the molar heat capacity of the flowing steam, and

[latex]F_tC_p = F_AC_{PA} + F_BC_{PB} + F_CC_{PC} \ = { F_{Ao}}(1 - f_A)C_{PA} + F_{Ao}(1 - f_A)C_{PB} + F_{Ao}f_AC_{PC} \ = F_{AO}[(C_{PA} + C_{PB}) + (C_{PC} - C_{PA} - C_{PB})f_A][/latex] (E)

Substituting equation (E) in (D), and integrating on the assumption that [latex]( -\Delta H_{RA}) [/latex] is constant, we obtain

[latex]T = T_o + (-\Delta H_{RA})\int_{0}^{f_A}{\frac{df_A}{(C_{PA} + C_{PB}) + (C_{PC} - C_{PA} - C_{PB})f_A}} [/latex] (F)

With numerical values inserted, this becomes

[latex]T = 798 + 115,000 \int_{0}^{f_A}{\frac{df_A}{230 + 20f_A}} \ = 798 + \frac{115,000}{20}[\ln(230 + 20f_A) \ln230] \ = 798 + 5750 [ \ln (230 + 20 f_A) - 5.4381 [/latex] (G)

The value of τ in equation (C) may now be determined by numerically (or graphically) evaluating the right side, with T given by equation (G), and [latex]k_A[/latex] obtained from the given Arrhenius equation at this T. For use in equation (C),

[latex]c_{Ao} = F_{Ao}/q_o = p_{Ao}/RT_o = 0.5(101,000)/8.314(798) = 7.6 mol m^{-3}[/latex]

For numerical evaluation, we use the simple trapezoidal rule, and a stepwise procedure similar to that in Example 12-5, which can be readily implemented by a spreadsheet program. For convenience, in equation (C), we let

[latex]G = \frac{1}{c_{Ao}T^2_o}\left\{\frac{(1 - 0.5f_A)^2[T(f_A)]^2}{k_A(T)(1 - f_A)^2} ight\} [/latex] (H)

Table 15.1 Results for Example 15-7

[latex] f_A [/latex]	T/K	[latex] k_A/ m^3 mol^{-1} s^{-1}[/latex]	G/s	G*/S	[latex] G^*\Delta f_A/s [/latex]
0.000	798.0	0.928	141.78
0.025	810.5	1.213	114.82	128.30	3.21
0.050	822.9	1.571	93.85	104.33	2.61
0.075	835.4	2.018	77.38	85.61	2.14
0.100	847.8	2.572	64.34	70.86	1.77
0.125	860.2	3.253	53.95	59.14	1.48
0.150	872.5	4.086	45.59	49.77	1.24
0.175	884.8	5.097	38.83	42.21	1.06
0.200	897.1	6.317	33.32	36.07	0.90
0.225	909.4	7.781	28.80	31.06	0.78
0.250	921.7	9.526	25.08	26.94	0.67
					[latex] au = \sum G^\ast \Delta f_A = 15.86 s[/latex]

so that

[latex] au = \underset{I}{} Gdf_A \simeq \sum G^\ast\Delta f_A [/latex] (I)

where G* is the average of two consecutive values of G, and [latex]\Delta f_A[/latex] is the increment in [latex]f_A[/latex] over that interval, chosen in the calculation to be 0.025. The following algorithm is used:

(1) Choose a value of [latex]f_A = 0[/latex] .
(2) Calculate T from equation (G).
(3) Calculate [latex]k_A[/latex] , from T and the Arrhenius equation given.
(4) Calculate G using equation (H).
(5) Repeat steps (1) to (4) for [latex]f_A = 0.025 , 0.050 , ... , 0.25 [/latex] (that is, in steps of 0.025).
(6) Calculate τ from equation (I).

The results are given in Table 15.1. Thus, the space time is 15.9 s for 25% conversion of butadiene. The integrals in equations (C) and (F) can also be evaluated numerically using the E-Z Solve software (file exl5-7.msp). The calculated value of τ is 15.8 s, slightly less than that obtained using the trapezoidal approximation.If simplifying conditions such as adiabatic behavior and temperature-independence of [latex](-\Delta H_{RA}) [/latex] and [latex]C_P[/latex] are not valid, the material and energy balances may have to be integrated numerically as a system of two coupled ordinary differential equations .

Question 15.7: A gas-phase reaction between butadiene (A) and ethene (B) is...

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