Alkylation example model

Alkylation units react olefins with isobutane to produce high octane alkylate. Units are typically fed with a mixture of olefins and light end gases, including isobutane. Alkanes such as propane and n-butane pass straight through the unit and are yielded as a waste gas stream, while the olefins are fully saturated to produce the alkylate. Different olefins, such as propylene and isobutylene, react slightly differently, consuming different amounts of isobutane, and yielding alkylate with different properties.

The unit is often run with an excess of isobutane to ensure full reaction of the olefins, but this isobutane is largely recycled within the reactor. As such, the main products are alkylate and waste gases (propane and n-butane separated within the unit).

For each tonne of propylene that enters the unit, approximately 2.7 tonnes of alkylate are manufactured, although this varies depending on unit design and catalyst. As such, 1.7 tonnes of isobutane are required to react one tonne of propylene.

Imagine that the unit is fed with a mixed feed of propylene, propane, i-butane and n-butane. Propane passes straight through the reactor, so if 100 kg of propane enter, then 100 kg of propane leave. If this propane were separated purely, there would be a propane product containing 100% propane. Since it is difficult to purely separate this gas, there is a small amount of n-butane in it too. However, the proportion of n-butane in the C3 product vary with respect to the amount of n-butane in the feed. Therefore the concentration of n-butane and propane in the C3 product vary with respect to the composition of the feed stream and the relative yields of each product. As such, we need to use a property x yield model.

In the above model, incoming propane is distributed 90% to the C3 output, and 10% to the C4 output. For every 1% increase in propane in the input stream, 0.9% goes to C3 product and 0.1% to C4 product. To calculate the proportion of propane in the C3 product, we use a property x yield prediction. This divides the delta-multiplied value by the yield of the stream to determine the final predicted value. As the final propane yield is in percentage, the property x yield value must be multiplied by 100. If the feed contains any propane, there must be less propylene, and so the yield of alkylate (and consumption of isobutane) is reduced. Therefore an increase in propane also drives a decrease in alkylate and isobutane input yields to the same factors as the production vectors.

Incoming n-butane is distributed 95% to the C4 product, and 5% to the C3 product, and again these factors (0.95 and 0.05 for every 1% increase in n-butane) must be multiplied by 100 within the property x yield delta.

The excess isobutane in this model is assumed to be pure isobutane. If the feed stream contains some isobutane, this cancels out like-for-like some of the requirement for excess.

Alternative structure

When fed with pure propylene, an alkylation unit might consume 1.7 tonnes of isobutane for every tonne of propylene. Thus for every tonne of feed (propylene) an input yield of 1.7 tonnes of isobutane is required, and 2.7 tonnes of alkylate product is manufactured.

If the feed was not pure propylene, but also contained some propane (which does not take part in the reaction and passes through the unit) then for every tonne of feed there would be less required isobutane and less finished product.

So the amount of isobutane input required will vary with respect to the concentration of propylene in the feed. If the feed were 90% propylene, for every tonne of feed 1.53 tonnes of isobutane would be required (1.7 × 0.9). If the excess isobutane were pure isobutane, this would require 1.53 tonnes of excess isobutane, but what if this were also not pure isobutane? Then the amount of excess isobutane required would increase as the concentration of isobutane in this excess falls. That is, the flow of excess isobutane required would be equal to (1.7 × % C3=)/% iC4, so if the feed were 90% propylene, and the excess isobutane were 80% isobutane, then for every tonne of feed (1.7 × 0.9)/0.8 = 1.9125 tonnes of excess isobutane would be required.

We imagined the feed contained 90% propylene, and 10% propane. What happens to the propane? Within the unit, a light ends separator typically separates any propane and butane which passes through the unit. However, the separation of these light ends is difficult and typically not perfect. Therefore within the products most of the input propane will end up in a C3 product stream, but some will end up in a C4 product stream. We might imagine that 90% ends up in the C3 product, and 10% in the C4 product.

The final concentration of propane in these products will vary with respect to the yields of these products (as the yield of each product depends on the composition of the input feed). Therefore property x yield predictions are required to calculate the final predicted property value.

The same argument may be said for any n-butane which is present in the feed stream. This will pass straight through the unit, and 95% might end up in the C4 product, and 5% in the C3 product. Again, property x yield drivers are required to calculate the final composition of each of these streams.

To calculate the amount of excess isobutane required, we need to know the proportion of isobutane in this excess feed. It is not possible to add a driver directly to a process unit that uses an input feed quality as the driver value. However, it is possible to add an operating parameter driver to the unit, and then force this operating parameter to be equal to the input feed composition (using a constraint that the parameter minus the composition must be equal to zero).

The excess isobutane feed is imagined to be composed of propane, isobutane and n-butane. These must all have equivalent operating parameters added to the process model to use them as drivers. Each of these operating parameters must be constrained to be equal to the equivalent molecule proportion in excess feed.