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solucionario de control de procesos de ingeniería química tercera edición Carlos A. Smith, Armando B. Corripio
Tipo: Ejercicios
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The controller is On/Off and the control i feedback with respect to temperature.,
M, D, A: temperature element/controller TE/C. Usually a bi-metallic strip that pushes the latch on the mechanism holding the slice of bread. The bread is released and the heating element de-energized when the temperature reaches the value set by the set point. (^) SP TE/C
The controller is On/Off and the control is feedback on the temperature variable.
M: Temperature element TE , usually a gas-filled bulb
D: Temperature controller TC.
A: Solenoid S that operates the heating element in the oven
TE
TC
S
SP
The controller is On/Off and the control is feedback.
M: Temperature element TE in thermostat TE/C
D: Mercury switch in thermostat TE/C
A: Solenoid S that turns unit (AC/H) on and off. TE/C^
S
AC /H
SP
Identificaton of the M-D-A components, controller type, instrumentation diagram, and type of control.
Smith & Corripio, 3rd edition
Smith & Corripio, 3rd edition
TC TE
Hot water
Cold water
S
SP
M: temperature sensor TE , a gas-filled bulb
D: temperature controller TC , mechanilly integrated to the sensor, but with a signa output
A: solenoid operated control valve on the hot water line.
The cold water valve is operated manually.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
F s( ) s s^2 +ω^2
s −i ⋅ω
s +i ⋅ω
= s^ −^ i^ ⋅ω+^ s+i^ ⋅ω
= 2 s⋅
= s s^2 +ω^2
s − i ⋅ω
e −^ (^ s^ −i^ ⋅ω)t ∞ 0
s + i ⋅ω
e −^ (^ s^ +i^ ⋅ω)t ∞ 0
∞ e −(^ s^ −i^ ⋅ω)t t
d 0
∞ e −(^ s^ +i⋅^ ω)t t
F s( ) 0
∞ cos ⋅ ωt⋅e −st t
= d 0
∞ e (^) t i ⋅ ωt−e − i⋅ωt 2
e −st
F s( ) 1 s +a
F s( ) 0
∞ e −^ ate −st t
= d 0
∞ e −(^ s^ +a)t t
= d −^1 s + a
e −^ (^ s^ +a)t ∞ 0
s +a
F s( ) 1 s^2
F s( ) −t s
e −^ st ∞ 0
s (^0)
∞ e −st t
= + ⋅ d 0 − 0 1 s^2
e −^ st ∞ 0
s^2
v −^1 s
du = dt = e −st
F s( ) By parts: u = t dv = e −^ stdt 0
∞ t e⋅ −st t
F s( ) 0
∞ f t( ) e −st t
= d
s
s + 2
s + 1
s
s + 2
s + 1
F s( ) 1 s
s + 2
s + 1 Used the linearity property. = −
s
s + 1
(s + 1 ) 2
F s( ) 1 s
s + 1
( s + 1 ) 2
Used the linearity property. =^ +
F s( ) = e −^2 s^ G s( ) e −^2 s^1 s
(s + 2 ) 2 + 1
Used the real translation theorem and linearity. (^) F s( ) e −^2 s^1 s
(s + 2 ) 2 + 1
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
s
s^2
s^3
= + ⋅ F s( ) 1 s
s^2
s^3
Used the linearity property.
s + 2
s
s^2
s^3
s + 2
(s + 2 ) 2
(s + 2 ) 3
F s( ) 1 s + 2
(s + 2 ) 2
(s + 2 ) 3
Used the complex translation theorem. =^ +
Must apply L'Hopital's rule:
s ∞
2 s( + 2 )
3 s( + 2 ) 2
lim = 1 Final value: →
t ∞
→ s 0
s 1 s + 2
(s + 2 ) 2
(s + 3 ) 2
lim = 0 L'Hopital's rule:^ →
t ∞
2e 2t
2e 2t
lim = 0 → Check!
s
s + 2
s + 1
Initial value:
t 0
→ s ∞
s 1 s
s + 2
s + 1
lim = → L'Hopital's rule:
s ∞
lim = 0 → Final value:
t ∞
→ s 0
s 1 s
s + 2
s + 1
lim = 1 + 0 + 0 = 1 →
s
s^2
s^3
Initial value:
t 0
→ s ∞
s 1 s
s^2
s^3
s ∞
s
s^2
lim = 1 →
lim = → Final value:
t ∞
→ s 0
s
s^2
lim = ∞ →
Check!
s + 2
(s + 2 ) 2
(s + 2 ) 3
Initial value:
t 0
→ s ∞
s 1 s + 2
(s + 2 ) 2
(s + 2 ) 3
lim = →
= 1 1( + 0 + 0 ) = 1
f t( ) = H u t⋅ ( )−H u t⋅ ( −T)
F s( ) H^1 s
⋅ H e⋅ −sT^1 s
= − ⋅ H^1 e^ − −sT s
= ⋅ F s( ) H s
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
(^00 2 )
2 fd ( )t
t
(^00 2 )
2 f t( )
t
f t( ) e
t 0 τ (^) e
−t := ⋅ τ fd ( )t^ u t(^ −t^0 )^ e
−( t t− 0 )
:= ⋅ τ
u t( ) 0 if t < 0 1 ift ≥ 0
Sketch the functions: (^) t 0 := 1 τ := 1 := F s( )^
τ e
− t 0 ⋅s ⋅ τ ⋅s + 1
The result to part (b) agrees with the real translation theorem.
e
− t 0 ⋅ s − 1
s 1 τ
⋅ e
s 1 τ
− ⋅λ ⋅ ∞ 0
= ⋅ e
− t 0 ⋅s
s 1 τ
= τ^ e
− t 0 ⋅s ⋅ τ ⋅s + 1
F s( ) −t 0
∞
u (^ λ)^ e λ
− λ τ (^) e −s^ (^ λ^ +t 0 )
= (^) ⌡ d e
− t 0 ⋅s
0
∞
e λ
s 1 τ
− λ
= ⋅⌡ d Let λ = t −t 0
F s( ) 0
∞
u t( −t 0 ) e t
−( t t− 0 ) τ (^) e −st
f t( ) u t( −t 0 ) e = d
−( t t− 0 )
= τ
F s( ) τ^ e
t 0
⋅^ τ τ ⋅ s+ 1
F s( ) e =
t 0 τ 1 s 1 τ
= τ^ e
t 0
⋅^ τ τ ⋅ s+ 1
f t( ) e =
t 0 τ (^) e
−t = τ^ (from Table 2-1.1)
f t( ) e
−( t t− 0 )
= τ
9 d^ Y 0( ) = 0
(^2) ⋅Y t( )
dt^2
⋅ 12 d Y t⋅ ( ) dt
Subtract initial steady state: + ⋅ + 4 Y t( ) = 8 X t( )
9 d^
(^2) ⋅y t( )
dt^2
⋅ 12 d y t⋅ ( ) dt
Invert usingTable 2-1.1: Y t( ) = ( − 1 +1.134i)e ( − 0.5+0.441i)t (^) + (− 1 −1.134i)e ( − 0.5−0.441i)t+2 u t( )
Y s( ) −^1 +1.134i s + 0.5−0.441i
− 1 −1.134i s + 0.5+0.441i
s
s 0
9s^2 + 9s+ 4
lim = 2 →
A 2 = − 1 −1.134i =
9 2 0.441i( ⋅ ) ( −0.5 +0.441i)
A 1 =− 1 +1.134i s − 0.5+0.441i
9 s( + 0.5+0.441i) s
lim →
s + 0.5−0.441i
s + 0.5+0.441i
s
Y s( ) 8 9 s( + 0.5−0.441i) s( + 0.5+0.441)s
Solve for Y(s), expand: =
s −1.
9 s( +0.255)s
lim = = 0. →
s 0
9 s( +0.255) s( +1.745)
lim = = 2. →
Y s( ) −2. s +0.
s +1.
s
Invert with Table 2-1.1: (^) Y t( ) = −2.342 e − 0.255t (^) + 0.342e − 1.745t+2 u t( )
(c) 9 d^
(^2) ⋅y t( )
dt^2
⋅ 9 d y t⋅ ( ) dt
Subtract initial steady state: 9 d 2 ⋅Y t( ) dt^2
⋅ 9 d Y t⋅ ( ) dt
s
r 1 −^9
:= r 2 −^9
Find roots: :=^ r^1 =−0.5^ +0.441i r 2 =−0.5 −0.441i
(^3) A 2 = 0.027 −0.022i 2 2 2.598i( ⋅ ) ( − 1 +2.598i) (− 1.5 +2.598i)
=0.027 +0.022i
s − 1.5+2.598i
2 s( + 1.5+2.598i) s( +0.5)s
lim = 0.027 +0.022i →
s + 1.5−2.598i
s + 1.5+2.598i
s +0.
s
Y s( ) 3 2 s( + 1.5−2.598i) s( + 1.5+2.598i) s( +0.5)s
Solve for Y(s) and expand: =
polyroots
− 1.5 −2.598i − 1.5 +2.598i − 0.
Find roots: =
s
Laplace transform: = ⋅
2 d^
(^3) ⋅Y t( )
dt^3
⋅ 7 d^
(^2) ⋅ (^) Y t( )
dt^2
Subtract initial steady state: + ⋅ + 9 Y t( ) = 3 X t( )
2 d^
(^3) ⋅y t( )
dt^3
⋅ 7 d^
(^2) ⋅y t( )
dt^2
Y t( ) −^4 3
Invert using Table 2-1.1: = e −^ 0.667t^ +2 u t( )
s 0
9 s( +0.667) 2
lim = 2 →
s −0.
d ds
9s
9s^2
lim = − 2 →
lim = →
s −0.
9s
lim = →
Y s( ) 8 9 s( +0.667) 2 s
( s +0.667) 2
s +0.
s
Solve for Y(s) and expand: = +
r 2 =−0.
r 2 −^12 12 r 1 =−0.
r 1 −^12 12 :=
Find roots: :=
s
Laplace transform: = ⋅
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Y t( ) u t( − 1 ) −^8 3
Apply the real translation theorem in reverse to this solution:
Y s( ) −^8 3
( s +0.667) 2
s +0.
s +0.
= e −s
The partial fraction expansion of the undelayed signal is the same:
(Real translation theorem)
X s( ) e^
−s
s 1 3
X t( ) u t( − 1 ) e =
−( t 1−)
Y t( ) −^8 3
Invert using Table 2-1.1: = e −^ 0.667t^ +8e −^ 0.333t
Y s( ) −^8 3
(s +0.667) 2
s +0.
s +0.
s −0.
d ds
9 s( +0.333)
9 s( +0.333) 2
lim = − 8 →
lim = →
s −0.
9 s( +0.667) 2
lim = 8 →
s −0.
9 s( +0.333)
lim = →
9 s( +0.667) 2 ( s +0.333)
(s +0.667) 2
s +0.
s +0.
Y s( ) 8
X s( ) 1 s 1 3
X t( ) e From Table 2-1.1: =
−t
9 d^ Y 0( ) = 0
(^2) ⋅Y t( )
dt^2
⋅ 12 d Y t⋅ ( ) dt
(Final value theorem)
dt^2
⋅ 18 d y t⋅ ( ) dt
Subtract initial steady state: 9 d^
(^2) ⋅Y t( )
dt^2
⋅ 18 d Y t⋅ ( ) dt
Laplace transform and solve for Y(s): Y s( ) 8 9s^2 + 18s+ 4
= X s( )
Find roots: r 1 −^18
2 9⋅ min
:= r 2 −^18
2 9⋅ min
:= r 1 =−0.255 min −^1
r 2 =−1.745 min −^1
Invert using Table 2-1.1: Y t( ) = A 1 ⋅ e −^ 0.255t+A 2 ⋅e −^ 1.745t
The response is stable and monotonic. The domnant root is: r 1 =−0.255 min −^1
Time for the response to decay to 0.67% of its initial value: −^5 r 1
=19.6 min
Final steady-state value for unit step input: s 0
s 8 9s^2 + 18s+ 4
s
lim →
(Final value theorem)
Smith & Corripio, 3rd edition
dt
Initial steady state: 2 y 0( ) = 5 x 0( ) + 3
Subtract: d Y t⋅ ( ) dt
Laplace transform: s Y s⋅ ( )+ 2 Y s( ) = 5 X s( ) Y 0( ) = y 0( ) −y 0( ) = 0
Solve for Y(s): Y s( ) 5 s + 2
= X s( )
s + 2
= + terms of X(s)
Invert using Table 2-1.1: Y t( ) = A 1 ⋅e −^2 t + terms of X(t)
The response is stable and monotonic.The dominant and only root is r :=− 2 min −^1
Time for response to decay to within 0.67% of its initial value: −^5 r
=2.5 min
Final steady-state value for unit step input: s 0
s 5 s + 2
s
lim →
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.
(Final value theorem) s^0
s 3 2s^3 + 7s^2 + 21s+ 9
s
lim →
Final steady state value for a unit step input: →
r 2
e −^ 1.5min^ Time for response to die out: =10 min
− (^1) T Decay ratio: =0.
T 2 π T =2.42 min 2.598min −^1
The period of the oscillations is: :=
The response is stable and oscillatory. The dominant root is r 2 =−0.5 min −^1
r
−1.5 −2.598i − 1.5 +2.598i − 0.
r polyroots = min −^1
:= min −^1
Find roots:
Y s( ) 3 2s^3 + 7s^2 + 21s+ 9
Laplace transform and solve for Y(s): = X s( )
2 d^
(^3) ⋅Y t( )
dt^3
⋅ 7 d^
(^2) ⋅ (^) Y t( )
dt^2
Subtract initial steady state: + ⋅ + 9 Y t( ) = 3 X t( )
2 d^
(^3) ⋅y t( )
dt^3
⋅ 7 d^
(^2) ⋅y t( )
dt^2
(Final value theorem) s^0
s 8 9s^2 + 12s+ 4
s
lim →
Final steady state value for a unit step input: → 2
r 1
Time required for the response to decay within 0.67% of its initial value: =7.5 min
The response is stable and monotonic. The dominant root is r 1 =−0.667 min −^1
Value of k: (^) k −M^ ⋅g y 0
:= k 1.816 N m
Laplace transform: (^) M s ⋅ 2 Y s( ) + k Y s⋅ ( ) = F s( )
Solve for Y(s): (^) Y s( ) 1 M s ⋅ 2 +k
= F s( )
s i k M
s i k M
:= ⋅ + terms of f(t)
The mobile will oscillate forever with a period of (^) T 2 π M k
:= ⋅ T =1.043 s
Smith & Corripio, 3rd edition
f ( t )
y ( t )
-ky ( t )
y = 0
Problem data: (^) M := 50gm y 0 :=− 27 cm
Force balance:
M d v t⋅ ( ) dt
⋅ = −M ⋅ g− k y t⋅ ( )+f t( )
Velocity: d y t⋅ ( ) dt
= v t( )
Initial steady state: (^0) = −M ⋅g −k y⋅ 0
Subtract and substitute:
M d^
(^2) ⋅Y t( )
dt^2
⋅ = −k ⋅Y t ( )+f t( )
Y 0( ) = 0