Question 14.3.1: Comparison of the Calculated Heat Loads with the Overall and...

Comparison of the Calculated Heat Loads with the Overall and Specific Energy Balances

Ethyl acetate is to be produced from the acetic acid and ethyl alcohol feed used in the illustrations of Secs. 14.1 and 14.2. If the feed and the product streams are both at 100ºC, and 37.2 percent of the acid is converted to the ester, determine the total heat input into any reactor to produce 2500 kg of the ester. Show that this heat input is equivalent to that obtained in the illustrations in the previous sections.

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From Eqs. 14.3-3 and 14.3-4 we can immediately write

 

\left(N_{ i }\right)_{\text {out }}=\left(N_{ i }\right)_{\text {in }}+\sum_{ j =1}^{ M } \nu_{ ij } X_{ j } (14.3-3)

 

Q+W=\sum_{ i =1}^{ C }\left(N_{ i } \bar{H}_{ i }\right)_{\text {out }}-\sum_{ i =1}^{ C }\left(N_{ i } \bar{H}_{ i }\right)_{\text {in }} (14.3-4)

 

\left(N_{ i }\right)_{\text {out }}=\left(N_{ i }\right)_{ in }+\nu_{ i } X

 

and

 

\begin{aligned}Q &=\sum\left(N_{ i } \underline{H}_{ i }\right)_{\text {out }}-\sum\left(N_{ i } \underline{H}_{ i }\right)_{ in } \\&=\sum_{ i }\left(N_{ i }\right)_{ in }\left[\underline{H}_{ i }\left(T_{ out }\right)-\underline{H}_{ i }\left(T_{ in }\right)\right]+X \sum \nu_{ i } \underline{H}_{ i }\left(T_{\text {out }}\right) \\&=X \Delta_{ rxn } H\left(T_{\text {out }}\right)=13.746 X kJ\end{aligned}

 

In writing this last equation we have neglected solution nonidealities; that is, we have set \bar{H}_{ i }= \underline{H}_{ i } and noted from Illustration 14.1-1 that \Delta_{ rxn } H\left(T=100^{\circ} C \right)=13.746 kJ / mol.

The total molar extent of reaction can be computed by writing the species mass balance, Eq. 14.3-3, for ethyl acetate to obtain

 

\begin{aligned}X &=\frac{\left(N_{ i }\right)_{\text {out }}-\left(N_{ i }\right)_{\text {in }}}{\nu_{ i }}=\frac{2500 kg \times 1000 g / kg \times(1 mol ) /(88 g )-0}{+1} \\&=2.841 \times 10^{4} mol\end{aligned}

 

\text { Thus } Q=2.841 \times 10^{4} mol \times 13.7 kJ / mol =3.892 \times 10^{5} kJ \text {. }

 

From Illustration 14.1-1 we have that for the continuous-flow stirred-tank reactor \dot{Q} = 3254 kJ/min and that 120.0 min are required to produce 2500 kg of ethyl acetate. Therefore, the total heat load is

 

Q=3.243 \times 10^{3} kJ / min \times 1.20 \times 10^{2} min =3.892 \times 10^{5} kJ

 

To obtain the total input to either the batch or tubular reactor, we must integrate the area under the \dot{Q} versus t curve given in Illustration 14.1-2. The result is

 

Q=3.892 \times 10^{5} kJ

 

Thus, the energy required to produce a given amount of product from specified amounts of reactants all at the same temperature is independent of the type of reactor used to accomplish the transformation. This is still another demonstration that the change in thermodynamic state properties of a system, here ΔH, between any two states is independent of the path between the states.

Comment

This example illustrates both the advantages and disadvantages of the black-box style of thermodynamic analysis. If we are interested in merely computing the total heat requirement for the reaction, the black-box or overall analysis is clearly the most expeditious and does not require any kinetic data. However, black-box thermodynamics gives us no information about the reactor size or the details of the heat program (that is, the heat flow as a function of time in the batch reactor or as a function of distance in the tubular reactor). The decision of whether to use the black-box or more detailed thermodynamic analysis will largely depend on the amount of kinetic information available and the degree of detail desired in the final solution.

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