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Q2. The level of liquid in a vessel is controlled by pure proportional control. The controller acts on a valve on the outlet stream by means of an analogue current signal in the range 4-20 mA. The valve stem position responds linearly to the current over the range 0-100% and for this installed valve it is known 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 and out 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 a steady-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 may vary between 20 and 80 m³hr¹¹.

Q2. The level of liquid in a vessel is controlled by pure proportional control. The controller acts on a valve on the outlet stream by means of an analogue current signal in the range 4-20 mA. The valve stem position responds linearly to the current over the range 0-100% and for this installed valve it is known 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 and out 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 a steady-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 may vary between 20 and 80 m³hr¹¹.

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Q2. The level of liquid in a vessel is controlled by pure proportional control. The controller acts on a valve on the outlet stream by means of an analogue current signal in the range 4-20 mA. The valve stem position responds linearly to the current over the range 0-100% and for this installed valve it is known 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 and out 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 a steady-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 may vary between 20 and 80 m³hr¹.
Transcribed Image Text:Q2. The level of liquid in a vessel is controlled by pure proportional control. The controller acts on a valve on the outlet stream by means of an analogue current signal in the range 4-20 mA. The valve stem position responds linearly to the current over the range 0-100% and for this installed valve it is known 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 and out 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 a steady-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 may vary between 20 and 80 m³hr¹.
Q1. Advantages and disadvantages of using control actions in these combinations were outlined in the 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 the proportional part is present: c.a. = Kce + Bias Consider a step disturbance to the flow in to a tank. When steady-state is reached, the flow out of the tank should be the same as the flow in. This is largely achieved by having the exit valve open a bit 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 there is 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 + Bias Following the comments on Q1 above, the change in the control signal required to accommodate the 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 information provided about the installed control valve. Since we're told the corresponding steady-state error (in m), we're able to calculate the gain (in mA/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 range corresponding 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.
Transcribed Image Text:Q1. Advantages and disadvantages of using control actions in these combinations were outlined in the 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 the proportional part is present: c.a. = Kce + Bias Consider a step disturbance to the flow in to a tank. When steady-state is reached, the flow out of the tank should be the same as the flow in. This is largely achieved by having the exit valve open a bit 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 there is 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 + Bias Following the comments on Q1 above, the change in the control signal required to accommodate the 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 information provided about the installed control valve. Since we're told the corresponding steady-state error (in m), we're able to calculate the gain (in mA/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 range corresponding 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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