Page 124 - Analysis, Synthesis and Design of Chemical Processes, Third Edition

Therefore, 34 batches of each product can be made in a 500 h period using single-product campaigns.

For the multiproduct campaign, referring to Figure E3.5, the cycle time for the sequence ABC is 19.5 h.
This is found by determining the time between successive completions of product C: 43.5 – 24 = 19.5 h,
and 63 – 43.5 = 19.5 h. Therefore, the number of batches of A, B, and C that can be produced is given by

This multiproduct sequence is clearly less efficient than the single-product campaign approach, but it does
eliminate intermediate storage. It should be noted that different multiproduct sequences give rise to
different results, and the ABCABC sequence may not be the most efficient sequence for the production of
these products.

3.5 Product and Intermediate Storage and Parallel Process Units

In this section, the effect of intermediate and product storage on the scheduling of batch processes and the
use of parallel process units or equipment are investigated. Both of these concepts will, in general,
increase the productivity of batch plants.

3.5.1 Product Storage for Single-Product Campaigns

When using combinations of single-product campaigns in a multiproduct plant, it is necessary to store
product during the campaign. For example, considering the products produced in Example 3.3, the plant
will produce 43 batches each of products A, B, and C in a 500 h period. If the required production rates
for these three products are 10,000, 15,000, and 12,000 kg/month, respectively, then what is the amount of
storage required? In practice, it is the volume, and not the weight, of each product that determines the

required storage capacity. For this example, it is assumed that the densities of each product are the same
3
and equal to 1000 kg/m . Considering product A first and assuming that demand is steady, the demand rate
3
(r ) is equal to 10,000/500 = 20 kg/h = 0.020 m /h. Note that the demand rate is calculated on the basis of
d
plant operating hours, and not on the basis of a 24-hour day. During the campaign, 10,000 kg of A must be
made in 43 batch runs, with each run taking t cycle, A = 2.5 h. Thus, during production, the production rate

3
(r ) of A is equal to 10,000/(43)(2.5) = 93.0 kg/h = 0.0930 m /h. Results for all the products are given in
p
Table 3.1.

Table 3.1 Production and Demand Rates for Products A, B, and C in Example 3.3