Steady-state balances:

\begin{aligned}&0=\bar{q}_{5}+\bar{q}_{1}-\bar{q}_{3}           (1) \\&0=\bar{q}_{3}+\bar{q}_{2}-\bar{q}_{4}               (2) \\&0=\bar{x}_{5} \bar{q}_{5}+\bar{x}_{1} \bar{q}_{1} ^{\nearrow ^{0}} -\bar{x}_{3} \bar{q}_{3}            (3) \\&0=\bar{x}_{3} \bar{q}_{3}+\bar{x}_{2} \bar{q}_{2}-\bar{x}_{4} \bar{q}_{4}                 (4)\end{aligned}

Solve (4) for \bar{x}_{3} \bar{q}_{3} and substitute into (3),

0=\bar{x}_{5} \bar{q}_{5}+\bar{x}_{2} \bar{q}_{2}-\bar{x}_{4} \bar{q}_{4}            (5)

Rearrange,

\bar{q}_{2}=\frac{\bar{x}_{4} \bar{q}_{4}-\bar{x}_{5} \bar{q}_{5}}{\bar{x}_{2}}               (6)

In order to derive the feedforward control law, let

\bar{x}_{4} \rightarrow x_{4 s p} \quad \bar{x}_{2} \rightarrow x_{2}(t) \quad \bar{x}_{5} \rightarrow x_{5}(t) \quad \text { and } \quad \bar{q}_{2} \rightarrow q_{2}(t)

Thus,

q_{2}(t)=\frac{x_{4 s p} \bar{q}_{4}-x_{5}(t) q_{5}(t)}{\bar{x}_{2}}             (7)

Substitute numerical values:

q_{2}(t)=\frac{(3400) x_{4 s p}-x_{5}(t) q_{5}(t)}{0.990}                  (8)

or

q_{2}(t)=3434 x_{4 s p}-1.01 x_{5}(t) q_{5}(t)           (9)

Note: If the transmitter and control valve gains are available, then an expression relating the feedforward controller output signal, p(t), to the measurements, x_{5 m}(t) and q_{5 m}(t), can be developed.

Dynamic compensation: It will be required because of the extra dynamic lag introduced by the tank on the left hand side. The stream 5 disturbance affects x_{3} while q_{3} does not.