Week 8 Tutorial Solutions
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University of Manchester**We aren't endorsed by this school
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ENGINEERIN 30091
Subject
Mechanical Engineering
Date
Jan 9, 2025
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5
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Week 8 Tutorial Solutions 1) A PI controller is used on the following process: 𝐺𝑝(𝑠) =𝐾𝑝𝜏2𝑠2+ 2𝜁𝜏𝑠 + 1The process parameters are 𝐾𝑝= 1, 𝜏 = 2, 𝜁 = 0.7. The control tuning parameters are 𝐾𝑐= 5, 𝜏𝐼= 0.2. Is the closed loop system stable or unstable? 𝐺𝑝(𝑠) =14𝑠2+ 2.8𝑠 + 1𝐺𝐶= 𝐾𝐶(1 +1𝜏𝐼𝑠) = 5 +50.2𝑠=5 + 𝑠0.2𝑠The transfer function of the closed loop system is given by 𝑦(𝑠) = 𝐺𝑝𝐺𝐶1 + 𝐺𝑝𝐺𝐶𝑦𝑠𝑝(𝑠)𝐺𝑝𝐺𝐶1 + 𝐺𝑝𝐺𝐶= 14𝑠2+ 2.8𝑠 + 1×5 + 𝑠0.2𝑠1 +14𝑠2+ 2.8𝑠 + 1×5 + 𝑠0.2𝑠= 5 + 𝑠0.2𝑠(4𝑠2+ 2.8𝑠 + 1)1 +5 + 𝑠0.2𝑠(4𝑠2+ 2.8𝑠 + 1)=5 + 𝑠0.2𝑠(4𝑠2+2.8𝑠 +1)(5 + 𝑠) + 0.2𝑠(4𝑠2+ 2.8𝑠 + 1)0.2𝑠(4𝑠2+2.8𝑠 +1)= 𝑠 + 50.84𝑠3+ 0.56𝑠2+ 1.2𝑠 + 5We need to assess the denominator of the closed-loop transfer function. All the coefficients are positive, therefore we need to do Routh-Hurwitz stability analysis to determine if the system is stable.
𝐴1is negative, therefore the system is unstable. 2) A process is described by the transfer function 𝐺(𝑠) =1(5𝑠 + 1)(𝑠 + 1)Find the range of controller settings that yield stable closed-loop systems for: a) A proportional-only controller. b) A proportional-integral controller (𝜏𝐼= 0.1, 1.0, 10.0). c) What can you say about the effect of adding larger amounts of the integral action on the stability of the controlled system; that is, does it tend to stabilize or destabilize the system relative to proportional-only control? Justify your answer. a) 𝐺𝐶= 𝐾𝐶𝐺𝑝(𝑠) =1(5𝑠 + 1)(𝑠 + 1)𝐺 = 𝐺𝑝𝐺𝐶1 + 𝐺𝑝𝐺𝐶Considering just the denominator for stability, and set it equal to 0 1 +𝐾𝐶(5𝑠 + 1)(𝑠 + 1)= 05𝑠2+ 6𝑠 + 1 + 𝐾𝐶= 0Solving using the quadratic formula, 𝑠 =−6 ± √36 − 20(1 + 𝐾𝑐)10To have a stable system, both roots of the characteristic equation must have negative real parts. Thus, 20(1 + 𝐾𝑐) > 0, so 𝐾𝑐> −1. b) Following the same steps as for, this time 𝐺𝐶= 𝐾𝐶(1 +1𝜏𝐼𝑠), therefore the denominator of the controlled process transfer function is given by 𝑎0= 0.8𝑎2= 1.2𝑎1= 0.56𝑎3= 5𝐴1=0.56 × 1.2 − 0.8 × 50.56= −5.9
1 +𝐾𝐶(1 +1𝜏𝐼𝑠)(5𝑠 + 1)(𝑠 + 1)= 0𝜏𝐼(5𝑠3+ 6𝑠2+ (1 + 𝐾𝑐)𝑠) + 𝐾𝑐= 0When 𝜏𝐼= 0.1, 0.5𝑠3+ 0.6𝑠2+ 0.1(1 + 𝐾𝑐)𝑠 + 𝐾𝑐= 0Doing Routh-Hurwitz stability analysis: For all coefficients to be positive, 𝐾𝑐must be greater than 0, which gives us our lower bound. To find the upper bound of 𝐾𝑐use a solver to find the value of 𝐾𝑐for which 𝐴1is positive. This gives us a final answer of 0 < 𝐾𝑐< 0.136. Repeat this for 𝜏𝐼= 1and 𝜏𝐼= 10, which will give a stable system for 𝐾𝑐> 0in both cases. c) We can see that adding integral control destabilises the system, as the range of 𝐾𝑐for which the system is stable is greatest for proportional only control. As we add larger amounts of integral control (i.e.,decreasing 𝜏𝐼) the system becomes more destabilised. 3) Consider the following instrumentation diagram for a heat exchanger 𝑎0= 0.5𝑎2= 0.1(1 + 𝐾𝑐)𝑎1= 0.6𝑎3= 𝐾𝑐𝐴1=0.06(1 + 𝐾𝑐)0.5𝐾𝑐𝑎0= 0.5𝑎2= 0.1(1 + 𝐾𝑐)𝑎1= 0.6𝑎3= 𝐾𝑐𝐴1=0.06(1 + 𝐾𝑐)0.6
a) Construct a control block diagram labelling all signals and transfer functions. Include valves and measurement sensor dynamics. Include at least one important disturbance input. b) What kind of controller will you use? a) The block diagram must be specific to the actual process, not just a generic feedback control block diagram. The set point is a temperature set point, Tsp(s),and the disturbance, Gd,could be something like a change in the ambient temperature. Here, TMis a temperature measurement and TCis a temperature controller. b) The process is a single-input, single-output (SISO) system therefore we expect to only have to use something simple, such as P, PI or PID control. In this case, we cannot have any offset as it may be dangerous (temperatures!), therefore we cannot us simple
proportional control. To determine whether PI or PID control is more suitable we need to think about the response time of the process. If it is relatively fast a PI controller would be suitable, however if it is slow, then we would need to introduce derivative control as well.