Design of a Tubular Reactor

Ethyl acetate is to be produced at a rate of 1250 kg/hr in a tubular reactor operating at 100ºC. If the feed is the same as that used in the previous examples, how large should the reactor be to produce the desired amount of ester by achieving the same conversion as the batch reactor? What are the heat transfer specifications for this reactor?

Question

Design of a Tubular Reactor

Ethyl acetate is to be produced at a rate of 1250 kg/hr in a tubular reactor operating at 100ºC. If the feed is the same as that used in the previous examples, how large should the reactor be to produce the desired amount of ester by achieving the same conversion as the batch reactor? What are the heat transfer specifications for this reactor?

Accepted Answer

The steady-state mass and energy balances for the tubular reactor are

[latex]\frac{q}{A} \frac{d C_{ i }}{d z}=r_{ i }[/latex] (a)

and

[latex]\frac{q}{A} \frac{d}{d z} \sum_{ i } C_{ i } \bar{H}_{ i }=\frac{q}{A}\left[\sum_{ i } \frac{d C_{ i }}{d z} \bar{H}_{ i }+\sum_{ i } C_{ i } \frac{d \bar{H}_{ i }}{d z} ight]=\frac{\dot{ L }}{A}[/latex] (b)

Since we are neglecting solution nonidealities, and the temperature is constant, [latex]d \bar{H}_{ i } / d z=0[/latex], so that Eq. b can be rewritten as

[latex]\frac{q}{A} \sum_{ i } \bar{H}_{ i } \frac{d C_{ i }}{d z}=\frac{\dot{ Q }}{A}[/latex] (c)

Setting τ = Az/q, we obtain

[latex]\frac{d C_{ EA }}{d au}=r_{ EA }=k C_{ A } C_{ E }-k^{\prime} C_{ EA } C_{ W }[/latex] (d)

and using Eq. a in Eq. c yields

[latex]\sum \frac{d C_{ i }}{d au} \bar{H}_{ i }=\sum r_{ i } \bar{H}_{ i }=\frac{d \hat{X}}{d au} \sum u_{ i } \bar{H}_{ i }=\Delta_{ rxn } H \frac{d \hat{X}}{d au}=\frac{\dot{ Q }}{A}[/latex] (e)

If we equate the time variable t with the distance variable τ , Eqs. d and e become identical to Eqs. a and c of the preceding batch reactor illustration. That is, the composition in a batch reactor t minutes after start-up will be the same as that at a distance z feet down a tubular reactor, where z is the distance traversed by the fluid in the time t. That is, z = qt/A or t = Az/q.

Thus, to convert 37.2 percent of the acid to ester, the reaction time τ in the tubular reactor must be such that τ = AL/q = V/q = 119.3 min, where L is the total reactor length and V = AL is its volume. Therefore, V = (119.3/139.3)Vbatch reactor = 18.22 [latex]m ^{3}[/latex], since we do not have to allow time for discharging, cleaning, and charging of the tubular reactor, as we did with the batch reactor. Also

[latex]\frac{\dot{ Q }}{A}=\Delta_{ rxn } H \frac{d \hat{X}}{d au} \quad ext { or } \quad \dot{ Q }=A \Delta_{ rxn } H \frac{d \hat{X}}{d au}[/latex]

Since [latex]\hat{X}[/latex] is known as a function of t from the previous illustration, we can use the same figure to obtain [latex]\hat{X}[/latex] , and hence d[latex]\hat{X}[/latex]/dτ , as a function of τ (or distance down the reactor). Thus, the heat flux at each point along the tubular reactor needed to maintain isothermal conditions can be computed from the heat program as a function of time in a batch reactor.

Question 14.2.1: Design of a Tubular Reactor Ethyl acetate is to be produced ...

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