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• Andres Felipe Schrader Florez

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TALLER NO. 3 “CASOS DE ESTUDIO”

TK BELTRAN SANCHEZ LUIS ALBERTO TK SCHRADER FLOREZ ANDRES FELIPE

Profesor: ING. CARLOS GUTIÉRREZ

CONTROL AUTOMÁTICO

FUERZAS MILITARES DE COLOMBIA ARMADA NACIONAL ESCUELA NAVAL DE CADETES ALMIRANTE PADILLA CARTAGENA, BOLÍVAR 20 JUNIO 2019

1. Consider the vacuum filter shown in fig. P7-1. This process is part of a waste treatment plant. The sludge enters the filter at about 5 % solids. In the vacuum filter the sludge is dewatered to about 25 % solids. The filterability of the sludge in the rotating filter depends on the pH of the sludge entering the filter. One way to control the moisture of the sludge to the incinerator is by is by the addition of chemicals (ferric chloride) to the sludge feed to the sludge feed to maintain the necessary pH. Figure P7-1 shows a proposed control scheme. The moisture transmitter has a range 55 to 90 %. The following data have been obtained from a step test on the output of the controller (MIC70) of +25 %CO. Time (min) 0 1 1,5 2,5 3,5 4,5 5,5 6,5 7,5 8,5 9,5

Moisture, Time Moisture, % (min) % 75 10,5 70,9 75 11,5 70,3 75 13,5 69,3 75 15,5 68,6 74,9 17,5 68 74,6 19,5 67,6 74,3 21,5 67,4 73,6 25,5 67,1 73 29,5 67 72,3 33,5 67 71,6 Tabla 1, datos problema 1

𝑉𝑖# = 55% 𝑉𝑓( = 95% 𝑉𝑖* = 75% 𝑉𝑓, = 67% ∆𝑚 = 12,5%

Imagen 1, Vacuum filter for problem 1.

Se procede a pasar el valor inicial del “Moisture” a 𝐶(%𝑇6 )

𝐶𝑉𝑖* =

𝑉𝑖* − 𝑉𝑖# 𝑉𝑓( − 𝑉𝑖#

𝐶𝑉𝑓* =

𝑉𝑓* − 𝑉𝑖# 𝑉𝑓( − 𝑉𝑖#

𝐶𝑉𝑖* =

75% − 55% 95% − 55%

𝐶𝑉𝑓* =

67% − 55% 95% − 55%

𝐶𝑉𝑖* = 0,5

𝐶𝑉𝑓* = 0,3

A continuacion de halla el ∆𝐶 ∆𝐶 = 𝑉𝑓* − 𝑉𝑖*

∆𝐶 = 𝐶𝑉𝑓* − 𝐶𝑉𝑖*

∆𝐶 = 67% − 75%

∆𝐶 = 30% − 50%

∆𝐶 = −8%

∆𝐶 = −20%

Se halla el valor de k

𝐾=

∆𝐶 −20% = = −1,6 ∆𝑚 12,5%

Se hallan los valores de 𝑡> , 𝑡? . 𝑡> = 𝑉𝑖* + ∆𝐶 ∙ 0,283

𝑡? = 𝑉𝑖* + ∆𝐶 ∙ 0,632

𝑡> = 75 + (−8) ∙ 0,283

𝑡? = 75 + (−8) ∙ 0,632

𝑐(𝑡> ) = 72,736

𝐶(𝑡? ) = 69,944

En la tabla 1 se procede a ubicar 𝑡> , 𝑡?

Para 𝒕𝟏

Para 𝒕𝟐

𝑦> = 73

𝑥> = 7,5

𝑦> = 70,3

𝑥> = 11,5

𝑦? = 72,3

𝑥? = 8,5

𝑦? = 69,3

𝑥? = 13,5

𝑦6 = 72,736 𝑦6 −𝑦> 𝑥6 = + 𝑥> 𝑚 𝑦? − 𝑦> 𝑚= 𝑥? − 𝑥> 𝑚=

𝑥6 = ?

72,3 − 73 8,5 − 7,5

𝑚=

𝑚 = −0,7 𝑥6 =

𝑦6 = 69,944 𝑦6 −𝑦> 𝑥6 = + 𝑥> 𝑚 𝑦? − 𝑦> 𝑚= 𝑥? − 𝑥>

𝑥6 = ?

69,3 − 70,3 13,5 − 11,5

𝑚 = −0,5

72,736 − 73 + 7,5 −0,7

𝑥6 = 7,877 𝑠𝑒𝑔

𝑥6 =

69,944 − 70,3 + 11,5 −0,5

𝑥6 = 12,212 𝑠𝑒𝑔

Hallamos el valor de 𝜏 𝜏=

3 3 (𝑡? − 𝑡> ) = (12,212 − 7,877) = 6,502 2 2

Se halla el valor de 𝑡6 . 𝑡6 = 𝑡? − 𝜏 = 12,212 − 6,502 = 5,71

2. Consider the absorber shown in Fig. P7-2 . The gas entering the absorber has a composition of 90 mole % air and 10 mole % ammonia (𝑁𝐻Q ). Before this gas is vented to the atmosphere, it is necessary to remove most of the 𝑁𝐻Q concentration in the exit gas steam cannot be above 200 ppm. The absorber has been designed so that the outlet 𝑁𝐻Q concentration in the vapor is 50 ppm. From dynamic simulations of the absorber, the following data were obtained.

Time (seg)

Water flow (gpm)

Outlet NH3 concetration

0

250

50

0

200

50

20

200

50

30

200

50,12

40

200

50,3

50

200

50,6

60

200

50,77

70

200

50,9

80

200

51,05

90

200

51,2

100

200

51,26

110

200

51,35

120

200

51,48

130

200

51,55

140

200

51,63

160

200

51,76

180

200

51,77

250

200

51,77

Imagen 2, Absorver for the problem 2 (Figure p7-2)

Tabla 2, Response to a step change in water flow to the absorber

𝑉𝑖# = 50%

𝑉𝑓, = 51,77%

𝑉𝑓( = 200%

∆𝑚 = 10%

𝑉𝑖* = 50%

Se procede a pasar el valor inicial del “Moisture” a 𝐶(%𝑇6 )

𝐶𝑉𝑖* =

𝑉𝑖* − 𝑉𝑖# 𝑉𝑓( − 𝑉𝑖#

𝐶𝑉𝑓* =

𝑉𝑓* − 𝑉𝑖# 𝑉𝑓( − 𝑉𝑖#

𝐶𝑉𝑖* =

50% − 50% 200% − 50%

𝐶𝑉𝑓* =

51,77% − 50% 200% − 50%

𝐶𝑉𝑖* = 0

𝐶𝑉𝑓* = 0,019

A continuación de halla el ∆𝐶 ∆𝐶 = 𝑉𝑓* − 𝑉𝑖*

∆𝐶 = 𝐶𝑉𝑓* − 𝐶𝑉𝑖*

∆𝐶 = 51,77% − 50%

∆𝐶 = 1,9% − 0, %

∆𝐶 = 1,77%

∆𝐶 = 1,9%

Se halla el valor de k

𝐾=

∆𝐶 1,9% = = 1,9 ∆𝑚 10%

Se hallan los valores de 𝑡> , 𝑡? . 𝑡> = 𝑉𝑖* + ∆𝐶 ∙ 0,283

𝑡? = 𝑉𝑖* + ∆𝐶 ∙ 0,632

𝑡> = 50 + 1,77 ∙ 0,283

𝑡? = 50 + 1,77 ∙ 0,632

𝐶(𝑡> ) = 50,501

𝐶(𝑡? ) = 51,12

En la tabla 1 se procede a ubicar 𝐶(𝑡> ), 𝐶(𝑡? ).

Para 𝒕𝟏

Para 𝒕𝟐

𝑦> = 50,6

𝑥> = 50

𝑦> = 51,05

𝑥> = 80

𝑦? = 50,77

𝑥? = 60

𝑦? = 51,2

𝑥? = 90

𝑦6 = 50,501 𝑦6 −𝑦> 𝑥6 = + 𝑥> 𝑚 𝑦? − 𝑦> 𝑚= 𝑥? − 𝑥> 𝑚=

𝑥6 = ?

50,6 − 50,3 50 − 40

𝑚=

𝑚 = 0,03 𝑥6 =

𝑦6 = 50,12 𝑦6 −𝑦> 𝑥6 = + 𝑥> 𝑚 𝑦? − 𝑦> 𝑚= 𝑥? − 𝑥>

𝑥6 = ?

50,2 − 50,05 90 − 80

𝑚 = 0,015

50.501, −50,3 + 40 0,03

𝑡> = 46,697 𝑠𝑒𝑔

𝑥6 =

50,12 − 50,05 + 80 0,015

𝑡? = 84,67 𝑠𝑒𝑔

Hallamos el valor de 𝜏 𝜏=

3 3 (𝑡? − 𝑡> ) = (84,67 − 46,697) = 56,96 2 2

Se halla el valor de 𝑡6 . 𝑡6 = 𝑡? − 𝜏 = 84,67 − 56,96 = 27,71

3. Consider the furnace shown in Fig. P7-3, which is used to heart the supply air to a catalyst regenerator. The temperature transmitter is calibrated for 300 − 500 °𝐹. The following response data were obtained for a step change of +5 % in the output of the controller.


, 𝑡? . 𝑡> = 𝑉𝑖* + ∆𝐶 ∙ 0,283

𝑡? = 𝑉𝑓* + ∆𝐶 ∙ 0,632

𝑡> = 425 + 20 ∙ 0,283

𝑡? = 425 + 20 ∙ 0,632

𝑡> = 430,66

𝑡? = 437,64

En la tabla 1 se procede a ubicar 𝑡> , 𝑡?

Para 𝒕𝟏

Para 𝒕𝟐

𝑦> = 430,6

𝑥> = 3,5

𝑦> = 437,6

𝑥> = 6

𝑦? = 432,4

𝑥? = 4

𝑦? = 439,4

𝑥? = 7

𝑦6 = 430,66 𝑦6 −𝑦> 𝑥6 = + 𝑥> 𝑚 𝑦? − 𝑦> 𝑚= 𝑥? − 𝑥> 𝑚=

432,4 − 430,6 4 − 3,5

𝑚 = 3,6

𝑥6 = ?

𝑦6 = 437,64 𝑦6 −𝑦> 𝑥6 = + 𝑥> 𝑚 𝑦? − 𝑦> 𝑚= 𝑥? − 𝑥> 𝑚=

439,4 − 437,6 7−6

𝑚 = 1,8

𝑥6 = ?

𝑥6 =

430,66 − 430,6 + 3,5 3,6

𝑥6 = 3,516 𝑠𝑒𝑔

𝑥6 =

437,64 − 437,6 +6 1,8

𝑥6 = 6,02 𝑠𝑒𝑔

Hallamos el valor de 𝜏 3 3 𝜏 = (𝑡? − 𝑡> ) = (6,02 − 3,516) = 3,756 2 2 Se halla el valor de 𝑡6 . 𝑡6 = 𝑡? − 𝜏 = 6,02 − 3,756 = 2,264