Q2. The level of liquid in a vessel is controlled by pure proportional control. The controller acts on avalve on the outlet stream by means of an analogue current signal in the range 4-20 mA. The valvestem position responds linearly to the current over the range 0-100% and for this installed valve it isknown that the % of max flow is roughly equal to the % stem position (c.f. course notes p34, Fig 23).The maximum flowrate through the valve is 100 m³ hr:¹.At an initial steady-state condition, the level in the tank is at its set-point and the flow rates into andout of the tank are both 50 m³ hr¹.After a step-change disturbance in the inlet flowrate of 5 m³ hr¹, the process response results in asteady-state offset of 50 cm in the tank level.a) Calculate the gain of the controller (in mA/m)b) How should the gain be altered to limit level changes to ±1 m when the inlet flowrate mayvary between 20 and 80 m³hr¹. Q1. Advantages and disadvantages of using control actions in these combinations were outlined inthe lecture and are covered in detail in the lecture notes.There was a lot of discussion of the concept of steady-state error for pure proportional control.Remember that this can be understood by considering the control algorithm when only theproportional part is present: c.a. = Kce + BiasConsider a step disturbance to the flow in to a tank. When steady-state is reached, the flow out ofthe tank should be the same as the flow in. This is largely achieved by having the exit valve open abit wider. To open the valve wider, we need a different control signal compared to the initial one(since the bias is constant by definition). In pure proportional control this can only exist when thereis a finite error at steady state.Q2. It was useful to start by writing down the control algorithm for pure proportional control.c.a. Kçe + BiasFollowing the comments on Q1 above, the change in the control signal required to accommodatethe additional flow corresponds to a steady-state error.We can relate the change in flowrate to a change in the control signal (in mA) given the informationprovided about the installed control valve.Since we're told the corresponding steady-state error (in m), we're able to calculate the gain (inmA/m). (note that this gain is strictly the product of the controller gain and the instrument gain).For an increase of 5 m³ hr¹ in outlet flowrate, the valve must move by 5% of its full rangecorresponding to a change in the control signal of (20-4)*0.05 = 0.8 mA.Numerical answers: a) 1.6 mA m³¹; b) 4.8 mA m²¹.

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