13.6 Return to the Motivating Example

Consider Example 13-1, where 40% ethanol (stream 1) is being blended with pure water (stream 2) to produce a product of 30% ethanol. Here we wish to use the RGA to help us develop the control structure (pair inputs and outputs).

We first develop the steady-state model, then linearize to find the process gain matrix, and finally calculate the RGA to decide on variable pairing. For simplicity we neglect density differences between the streams and components.

Total Material Balance

Neglecting density differences between the streams and components, the total blend product volumetric flow rate is equal the sum of the volumetric flow rates of the two feed streams,

Component Material Balance on Ethanol

Assume a binary system with z₁ = volume fraction of ethanol in stream 1. Let z = volume fraction of ethanol in the blend stream. Since there is no ethanol in the water stream, we find

Solving these equations for the volume fraction of ethanol in the blend stream,

Let the volume fraction of ethanol be output 1 and total flow rate output 2. Also, inputs 1 and 2 are the flow rates of the whiskey and water feedstreams, respectively. The following steady-state values were used in Example 13.1: F₁ = 3 gpm, F₂ = 1 gpm, z = 0.3 mole fraction ethanol, and F = 4 gpm.

The steady-state input-output gains are

which yields the process gain matrix

The relative gain relating input 1 to output 1 is

This value of 0.25 indicates that the effective gain relating input 1 to output 1 with loop 2 closed is four times the value of the gain with loop 2 open. This explains the result shown in Figure 13-11.

The RGA is then

which indicates that the blend total flow (y₂) should be paired with the whiskey flow rate (u₁) and the blend ethanol concentration (y₁) should be paired with the water flow (u₂). Notice that the largest component stream flow rate then controls the total flow.