Question 10.7: Quasi-Autothermal Operation of a Semibatch Reactor Using Add...

Quasi-Autothermal Operation of a Semibatch Reactor Using Addition of Cold Feed

At the time Kladko’s article (5) was prepared, market conditions dictated that it would not be financially remunerative to develop a continuous process for producing B by the reaction discussed in Illustration 10.1 and other illustrations in this chapter. Consequently, they considered the possibility of using a semibatch reactor with continuous addition of cold feed to maintain the temperature of the reactor contents within prescribed limits. The basic problem was considered in Illustration 8.11. However, in addition to the material balance aspects of the design, we now wish to consider the heat transfer requirements for operation in accordance with the filling schedule outlined earlier. For the conditions indicated in Illustration 8.11 (isothermal, 163°C); determine the direction and magnitude of the heat transfer requirements.

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The material balance equations have been solved earlier with the result that

m_{ A }=m_{ A i} e^{-k\left(t-t_i\right)}+\frac{\phi_{ A 0}}{k}\left[1-e^{-k\left(t-t_i\right)}\right]

where m_{A} is the mass of A remaining in the reactor at time t, m_{Ai} is the mass of A in the reactor at time t_{i}, and \phi_{A0} is the mass feed rate of A appropriate to the time interval between t and t_{i}.

An energy balance on the reactor can be derived from equation (10.1.1) by omission of appropriate terms:

\begin{aligned}\frac{d E_{\text {sys }}}{d t}= & \dot{Q}-\dot{W}_s+\sum_{\substack{\text { input } \\\text { streams }}}\left(h_{\text {in }}+\frac{v_{\text {in }}^2}{2 g _c}+\frac{ g }{ g _c} Z_{\text {in }}\right) \dot{m}_{\text {in }} \\& -\sum_{\substack{\text { output } \\ \text { streams }}}\left(h_{\text {out }}+\frac{v_{\text {out }}^2}{2 g _c}+\frac{ g }{ g _c} Z_{\text {out }}\right) \dot{m}_{\text {out }}\end{aligned}              (10.1 .1)

\frac{d E_{ sys }}{d t}=\dot{Q}+h_{ in } \dot{m}_{ in }

Multiplying by dt and integrating between time zero and time t gives

E_{sys} − E_{sys , 0} = Q + h_{in}m_{in}

If the temperature of the inlet stream is taken as the datum temperature for enthalpy and internal energy calculations,

h_{in} = 0

and

E_{ sys }-E_{ sys , 0}=Q

However, for condensed phases, the difference between internal energy and enthalpy is usually negligible:

Q \approx H_{ sys }-H_{ sys , 0}            (A)

The enthalpy difference is given by

\begin{aligned}H_{ sys }-H_{ sys , 0}=\int_{20^{\circ} C }^{163^{\circ} C }& \left(m_A+m_B-m_{B 0}\right) \hat{C} d T \\ +\Delta H_{R \text { at } 20^{\circ} C} & \left(m_{ B }-m_{ B , 0}\right)\end{aligned}                      (B)

where Ĉ is the average heat capacity per unit mass and where we have assumed that the initial material sump (m_{B0}) is at 163°C.

Equations (A) and (B) can be combined with appropriate numerical values to obtain

\begin{aligned}Q & =\left(m_{ A }+m_{ B }-m_{ B 0}\right)(0.5)(163-20)-83\left(m_{ B }-m_{ B 0}\right) \\& =\left(m_{ A }+m_{ B }-m_{ B 0}\right) 71.5-83\left(m_{ B }-m_{ B , 0}\right) \\& =71.5 m_{ A }-11.5\left(m_{ B }-m_{ B 0}\right)\end{aligned}

where Q is expressed in calories and m is expressed in grams. If m is expressed in pounds and Q in Btu, then

Q=128.7 m_{ A }-20.7\left(m_{ B }-m_{ B 0}\right)

The numerical values characterizing m_{A} \text{  and  } m_{B} at various times are obtained as indicated in Illustration 8.11. They may be used to determine the net heat required between time zero and time t. The pertinent results are presented in Table I10.7-1. It is evident that heat must be supplied to the reactor during the first several hours, but the reactor then reaches a point beyond which cooling must be supplied.

Table I10.7-1 Material and Energy Balance Analysis
Time (h) m_{A} (lb) Tota mass (lb) Q_{total} \times 10^{-3}  (Btu) \begin{aligned}&\dot{Q}\times10^{-3} \\& ( Btu / h)^a\end{aligned}
0 0 1500 0 22.52
1 120  1675 14.31 8.18
2 175 1850 18.90 1.61
3 199 2025 18.86 5.18
4 244 2250 20.93 – 0.20
5 265 2475 19.41 – 2.71
6 274 2700 16.10 2.65
7 312 2975 16.08 4.54
8 364 3300 17.12 7.98
9 439 3700 20.05 – 0.98
10 473 4100 16.85 – 5.04
11 488 4500 10.81 – 16.49
12 443 4825 -2.64 – 17.55
13 388 5100 -16.55 – 17.41
14 329 5325 -30.02 – 16.79
15 268 5500 -42.76 – 19.16
16 189 5600 -56.63 – 16.15
17 119 5650 -68.13 – 14.22
18 54 5650 -77.84 – 6.45
19  24 5650 -82.32 – 2.87
20  11 5650 -84.26 – 1.31
^{a} Instantaneous heat transfer rates are calculated after changes are made in the mass flow rates at the times indicated.

Instantaneous heat transfer requirements may be determined by recognizing that the thermal energy supplied must be used to effect either a sensible heat change or the reaction.
If \dot{Q} is the heat transfer rate,

\dot{Q}=\phi_{ A 0} \int_{T_0}^T \hat{C} d T+r V_R \Delta H_R

where V^{\prime}_{R} is the liquid-phase volume at time t. Because
C_{A} = m_{A}/V^{\prime}_{R} and the reaction obeys first-order kinetics,

\dot{Q}=\phi_{ A 0} \int_{T_0}^T \hat{C} d T+k m_{ A } \Delta H_R

Substitution of numerical values gives

\begin{aligned}\dot{Q} & =\phi_{ A 0}(0.5)(163-20)+0.8 m_{ A }(-83) \\& =71.5 \phi_{ A 0}-66.4 m_{ A }\end{aligned}

where \dot{Q} is expressed in cal/h, ϕ_{A0} is expressed in g/h, and m_{A} is expressed in grams.

In more conventional engineering units,

\dot{Q}=128.7 \phi_{ A 0}-119.5  m_{ A }

where \dot{Q} is now expressed in Btu /h, ϕ_{A0} is expressed in lb/h, and m_{A} is expressed in pounds. Calculated values of \dot{Q} are also presented in Table I10.7-1.

If one were to operate this semibatch reactor under a filling schedule, which for the first 11 h is identical to that considered previously, and then proceed to feed A at the maximum rate of 400 lb/h for an additional 2.875 h, the same total amount of A would have been introduced to the reactor. However, in this case, the heat transfer
requirements would change drastically. There would be a strong exotherm beginning at the moment the cold feed is stopped. The results for this case are presented in Table I10.7-2.

Table I10.7-2 Results of Analysis for Abrupt
Termination of Feed at 13.875 h
Time

(h)

m_{A} (lb) mass (lb) Q_{\text {total }} \times 10^{-3} (Btu) \dot{Q} \times10^{-3} (Btu/h)^{a}
11 488 4500 10.81 −6.84
12 494 4900 3.42 – 7.55
13 498 5300 – 4.26 – 8.03
13.875 499 5650 – 11.35 – 59.63
14.00 452 5650 – 18.38 – 54.01
14.50 303 5650 – 40.64 – 36.21
15.0 203 5650 – 55.58 – 24.26
15.50 136 5650 – 65.59 – 16.25
16.00 91 5650 – 72.31 – 10.87
17 41 5650 – 79.78 – 4.90
18 18 5650 – 83.22 – 2.15
19 8 5650 – 84.71 – 0.96
^{a} Instantaneous heat transfer rates are calculated after changes are made in the mass flow rates at the times indicated.

Consequently, if one desires to minimize the possibility that this reaction will run away (be out of control) as a result of exceeding the capacity of the cooling network, it would be advisable to operate in a mode such that the feed rate is gradually diminished instead of being abruptly terminated at a high feed flow srate.

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