Trabalho Sisca parte 1 PDF

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Baseado no livro do autor Seborg, este é um trabalho sobre Sistemas de controle e automação...

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Trabalho de Sistemas de Automa¸c˜ao e Controle Parte 1 Vin´ıcius Augusto Hendges de Souza 2 de outubro de 2018

1

1

Exerc´ıcio de Exemplo 2.1

A stirred-tank blending process with a constant liquid holdup of 2m3 is used to blend two streams whose density are both approximately 900kg/m3 . The density does not change during mixing.

a) Assume that the process has been operating for a long period of time with flow rates of w1 = 500kg/min and w2 = 200kg/min, and feed compositions (mass fractions) of x1 = 0.4 and x2 = 0.75. What is the steady-state value of x? b)Suppose that w1 changes suddenly from 500 to 400kg/min and remains at the new value. Determine an expression for x(t) and plot it. c)Repeat part (b) for the case where w2 (instead of w1 ) changes suddenly from 200 to 100kg/min and remains there. d)Repeat part (c) for the case where x1 suddenly changes from 0.4 to 0.6. Esquema de resolu¸ca˜o do Scilab e gr´aficos de resposta:

1

Figura 1: Esquema feito no Scilab.

2

Figura 2: Gr´aficos de respostas dos ´ıtens a, b, c e d.

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2 2.1

Quiz Socrative Feedback and feedforward control both require a measured variable.

R.: True.

2.2

The process variable to be controlled is measured in feedback control.

R.: True.

2.3

Feedforward control can be perfect in the theoretical sense that the controller can take action via the manipulated variable even while the controlled variable remains equal to its desired value.

R.: True.

2.4

Feedforward control can provide perfect control; that is, the output can be kept at its desired value, even with an imperfect process model.

R.: False.

2.5

Feedback control will always take action regardless of the accuracy of any process model that was used to design it and the source of a disturbance.

R.: True.

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2.6

Does a typical microwave oven utilize feedback control to set cooking temperature or to determine if the food is ”cooked”? If not, what mechanism is used? Can you think of any disadvantages to this approach, for example, in thawing and cooking foods?

R.: A temperaruture measure at the center of the food.

2.7

Indique o tipo dos sistemas, se a distˆ ancia entre o transmissor e a v´ alvula ´ e pequena

Figura 3: Sistemas. R.: Sistema A ´e feedback e Sistema B e´ feedforward.

2.8 R.:

Escreva a equa¸c˜ ao de Bernoulli entre os pontos a e b de uma tubula¸c˜ ao. pa ρ

+

Va2 2

+ gha =

pb ρ

+

Vb2 2

+ ghb

5

2.9

Dado um tanque onde a diferen¸ca entre o n´ıvel do l´ıquido e a tubula¸ c˜ ao de sa´ıda ´ e h, a vaz˜ a o na sa´ıda ´ e q e a resistˆ encia a vaz˜ ao na sa´ıda ´ e R, e a sua analogia com a Lei de Ohm. Qual a vari´ avel relacionada com h?

R.: Cv∗ .

2.10

Dado um tanque onde a diferen¸ca entre o n´ıvel do l´ıquido e a tubula¸ c˜ ao de sa´ıda ´ e h, a vaz˜ ao na sa´ıda ´ e q e a resistˆ encia a vaz˜ ao na sa´ıda ´ e R, e a sua analogia com a Lei de Ohm. Qual a vari´ avel relacionada com q?

R.: Corrente i.

2.11

R.:

3 3.1

11. Dado um tanque onde a diferen¸ca entre o n´ıvel do l´ıquido e a tubula¸c˜ ao de sa´ıda ´ e h, a vaz˜ a o na sa´ıda ´ e q e a resistˆ encia a vaz˜ a o na sa´ıda ´ e R, e a sua analogia com a Lei de Ohm. Qual a vari´ avel relacionada com Rv?

h q

Exerc´ıcios Exerc´ıcio 1.1

a) True; b) True; c) True; d) False; e) True.

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3.2

Exerc´ıcio 1.2

Disturbios: perda de calor para o meio, vazamento de gas, perda de pressao, casa muito fria ou muito quente (deve-se colocar um termometro dentro da casa para desligar a fornalha).

3.3

Exerc´ıcio 1.3

N´ıvel da caixa d’´agua por uma boia.

3.4

Exerc´ıcio 1.4

No. It just analyses the opperating time. If you want to cook any longcook-time-food, you need to set the time. And, if the food is toasting, the microwave oven can be stopped any second.

3.5

Exerc´ıcio 1.5

a) The sensor is your eyes, the actuator is the steering wheel. There is a ”set point”, that’s the rood lines. If you are too close to them, your brain tells you to adjust the course, and, same stuff if you are too far from them (because you are closer the right lines). The disturbances would be water on the road, any change at the road’s thickness...

3.6

Exerc´ıcio 1.6

a) Feedback Control: Measured variable: y; Manipulated variable: D,R, or B(schematic shows D). b) Feedforward Control: Measured variable: F; Manipulated variable: D (shown), R or B.

3.7

Exerc´ıcio 1.7

Both flow control loops are feedback control systems. In both cases, the controlled variable (flow) is measured and the controller responds to that measurement.

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3.8

Exerc´ıcio 1.8

a) Outputs: Tp , L Inputs : Q(t), Fw Disturbances: Tw , Ta. b) Either Tw orTa or both can be measured in order to add feedforward control. c) The magnitudes of the terms U A(Tp − Ta) and Fw ρC(Tp − Tw ) relative to the magnitude of Q(t) will determine whether Ta or Tw (or both/neither) is the important disturbance variable. d) Determine which disturbance variable is important as suggested in part c) and investigate the economic feasibility of using its measurement for feedforward control.

3.9

Exerc´ıcio 2.1

a) Constant volume, so w3 = w1 + w2 , and the mass conservation law is = w1 + w2 − w3 . ρ dV dt ′ = w1 cT1′ + w2 cT2′ − w3 cT3′. Energy conservation law is dmcT dt The transient model is T ′ (ρ)

dT ′ dV + (ρ)Vss = w1 T1′ + w2 T2′ − w3 T 3′ . dt dt

b) V = Vss , we can say ′ = w1 T1′ + w2 T 2′ − (w1 + w2 )T3′ or, if (ρ)Vss dT dt ′ dV (ρ)T dt = T ′ (w1 + w2 ) − T ′ (w1 + w2 ) dV = 0. dt

3.10

Exerc´ıcio 2.2

a) If the flow rate varies and the volume remains constant, so Always this is true wi = wo is true, from the mass conservation law C is also constant. From energy conservation law, a in ) At the liquid dQ ) − ( dQdtcout + dQ = ( dQdtcoil + dQ dt dt dt Vc

d(ρ)T = hcAc(Tc − T ) + wi ci (Ti − T ) − (haAa(T − Ta)). dt 8

b) If Ti and]or wi increase, the energy ratiowill increase, the system will get hotter.

3.11

Exerc´ıcio 2.3

a) Mass conservation law: w1 = w2 +w3 , so

dma dt

= w1 −w2 −w3 , and

d((π)(r 2 )aha) = w1 − w2 − w3 dt dh d(ha) = w1 − (π)(r 2 )b b − w3 . (π)(r 2 )a dt dt

b) Nf = Nv − Ne , so Nf = 3 − 2 Nf = 1. This system has infinite solutions. 1 input and 4 outputs.

3.12

Exerc´ıcio 2.4 d(ρAh + mg ) = ρ(qi − q) dt dh A = (qi − q) dt mg RT M ρgh dh + = qi − Cv ( A − pa). dt A(H − h) gc

3.13

Exerc´ıcio 2.5

a) We know that wb = (p1R−bp2 ), so wa = For mass conservation law: ρ

(pdt −p1 ) Ra

and wc =

dV1 (pdt − p1 ) (p1 − p2 ) = − Rb dt Ra 9

(p2 −pf ) . Rc

mb dt

= w2 .

and ρ

(p1 − p2 ) (p2 − pf ) dV2 . = − Rc dt Rb

So, for both of them V1 M dP1 V2 M dP2 (pdt − p1 ) (p2 − p1 ) . + = − RT dt RT dt Rc Ra b) 8 parameters, 5 variables and 2 equations. Nf = Nv − Ne Nf = 3.

3.14

Exerc´ıcio 2.6

a) Tank 1: q1 = q. V1 ρc

dT1 = dt

Tank 2: q = q1 + q2 Enegy conservation equation: V2 ρc

dT2 = ρqc(T1 − T2 ) + U A(T1 − T2 ) − UcAc(T2 − T1 ) dt

b) Nf = Nv + Ne Nf = 3 c) 3 new variables would appear: C1 , C2 and Ci . 2 new equations and 2 new variables: C1 and C2 . Ci must be known.

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3.15

Exerc´ıcio 2.8

If V is constant, then, through mass conservation law, dm = qf − q dt qf = q With energy conservation law, for the tank d(mcT ) = (qf ρcTf ) − (Kq0.8 A(T − Ti )) dt dT = (qf ρcTf ) − (Kq0.8 A(T − Ti )) dt Energy conservation law for the jacket, ρV c

ρj Vj cj

3.16

dTj = (qj ρj cj Tj ) − (Kqj0.8A(T − Tj )). dt

Exerc´ıcio 2.10

Energy conservation law, when the reactor, when there is no C: V

dca −E = qi cai − qca − V K1 e RT ca dt

−E −E dcb = qi cbi − qcb − V (K1 e RT ca − K2 e RT cb ) dt dcc −E V = −qcc + V K2 e RT cb . dt And, for the jacket,

V

ρj cj Vj

3.17

dTc = ρj cj qci (Tci − Tc) + U A(T − Tc). dt

Exerc´ıcio 2.13

a) Mass conservation law: At the first moment: dm = w1 + w2 − w3 − q4 dt 11

√ dh = w1 + w2 − w3 − 0.025 h − 1. dt But, to reach the corosion point, Aρ

dh = w1 + w2 − w3 dt dh 3.14 ∗ 800 = 120 + 100 − 200 dt dh m . = 7.96 ∗ 10−3 min dt Aρ

So, to reach 1 meter, t1m = 125.6min. b) If w3 + q4 (h = 2.5m) > w1 + w2 is true, it won’t overflow. And, if the h = 1.75m point does not change w1 and w2 , we can assume that w3 + q4 (h = 2.5m) > w1 + w2 √ 200 + 0.025 2.5 − 1 > 120 + 100 0.3 > 20. That is not true! So, it will overflow sometime, because the input is higher then the output.

3.18

Exerc´ıcio 3.1

a) The equation is

ω . ω 2 + (s + b)2

b) ω2

3.19

s+b . + (s + b)2

Exerc´ıcio 3.4

a) f (t) = 5S (t)–4S (t − 2)–S(t − 6) 1 F (s) = (5 − 4e−2s − e−6s ). s b) X(s) =

a (1 − e−2ts + e−3ts ). s2 12

3.20

Exerc´ıcio 3.17

With V

dC + qC = qCi dt

we can reach to the equation Vt

c(t) = ci (1 − e− q ).

3.21

Exerc´ıcio 4.1

a) III; b) III; c) V; d) V;

3.22

Exerc´ıcio 4.12

a) 1 H ′ (s) = ′ Q (s) sA +

C√v 2 h

.

b) 1 Q′ (s) . = Qi (s) τ s + 1 c)Thus level in the square root outflow TF is twice as sensitive to changes in q i and reacts only 21 as fast (two times more slowly) since τ = 2τ ′ .

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4 4.1

Exerc´ıcios em Scilab - XCos Exemplo 3.7

Figura 4: Esquema feito no Scilab.

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