SOLUTIONS MANUAL FOR SELECTED
PROBLEMS IN
PROCESS SYSTEMS ANALYSIS AND
CONTROL
DONALD R. COUGHANOWR
COMPILED BY
M.N. GOPINATH BTech.,(Chem)
CATCH ME AT gopinathchemical@gmail.com
Disclaimer: This work is just a compilation from various sources believed to be
reliable and I am not responsible for any errors.
CONTENTS
PART 1: SOLUTIONS FOR SELECTED PROBLEMS
PART2:
LIST OF USEFUL BOOKS
PART3:
USEFUL WEBSITES
PART 1
1.1 Draw a block diagram for the control system generated when a human
being steers an automobile.
1.2 From the given figure specify the devices
Solution:
Inversion by partial fractions:
3.1(a)
dx 2 dx
+
+ x = 1 x ( 0) = x ' ( 0) = 0
2
dt
dt
 dx 2 
L  2  = s 2 X ( s ) − sx(0) − x ' (0)
 dt 
 dx 
L   = s X ( s ) − x ( 0)
 dt 
L(x) = X(s)
L{1} = 1/s
s 2 X ( s ) − sx(0) − x ' (0) + s X ( s ) − x (0) + X ( s ) =
= ( s 2 + s + 1) X ( s ) =
X ( s) =
1
s
1
s
1
s( s + s + 1)
2
Now, applying partial fractions splitting, we get
X ( s) =
X ( s) =
1
s +1
− 2
s ( s + s + 1)
3
2
s +1
1
 1  2 
−
−  

2
2
2
2
s 
 2  3  
1   3
1  3

s +  +  
 s +  + 
2  2 
2   2 


L−1 ( X ( s )) = 1 − e
X (t ) = 1 − e
b)
1
− t
2
1
− t
2
1
Cos
3
1 −2t
3
t−
e sin
t
2
2
3

 3 
3
1
 Cos
t +


Sin


 2 t
2
3




dx 2
dx
+ 2 + x = 1 x ( 0) = x ' ( 0) = 0
2
dt
dt
when the initial conditions are zero, the transformed equation is
( s 2 + s + 1) X ( s ) =
1
s
X ( s) =
1
s( s + s + 1)
1
A
Bs + C
= + 2
s ( s + s + 1) s s + 2 s + 1
2
1 = A(s2 + 2s +1) + Bs2 + Cs
0 = A + B(by equating the co − effecient of s 2 )
0 = 2 A + C (by equating the co − effecients of s )
1 = A(by equating the co − effecients of const)
A+ B = 0
B = −1
C = −2 A
A = 1, B = −1, C = −2
1
s+2
− 2
s s + 2s + 1
 1 (s + 1) + 1 
L−1{ X ( s )} = L−1  −
(s + 1)2 
s
X ( s) =
 1
1 
{ X (t )} = 1 − L−1 
+
2 
 s + 1 (s + 1) 
{ X (t )} = 1 − e − t (1 + t )
dx 2
dx
3.1 C
+ 3 + x = 1 x ( 0) = x ' ( 0) = 0
2
dt
dt
by Applying laplace transforms, we get
= ( s 2 + 3s + 1) X ( s ) =
X ( s) =
1
s( s + 3s + 1)
2
1
s
2
X ( s) =
A
Bs + C
+ 2
s s + 3s + 1
1 = A( s 2 + 3s + 1) + Bs 2 + Cs
0 = A + B(by equating the co − effecient of s 2 )
0 = 3 A + C (by equating the co − effecients of s)
1 = A(by equating the co − effecients of const)
A+ B = 0
B = −1
C = −3 A = −3
A = 1, B = −1, C = −3
s+3 
1
L−1{ X ( s )} = L−1  − 2

 s s + 3s + 1 


−1
−1  1
L { X ( s )} = L  −
s 
s +





1
L−1{ X ( s )} = L−1  −
s 
s +



X (t ) = 1 − e
−
3t
2
(Cos
s+3
2
3 
 −
2  



2 
5 

2  
3 2 
3

 .
s+
2 5

2
−
2
2
2
3 
3  5 



+
−
s


 −
2  
2   2 

5t
3
5
+
t
sinh
2
2
5
3.2(a)
dx 4 d 3 x
+ 3 = Cos t; x (0) = x ' (0) = x ''' (0) = 0
4
dt
dt
x11 (0) = 1



2 
5 

2  
5
2
Applying Laplace transforms, we get
s 4 X ( s ) − s 3 x (0) − s 2 x1 (0) − sx '' (0) − x ''' (0) + s 3 X ( s ) − s 2 x (0) − sx ' (0) − x '' (0) =
X ( s ) ( s 4 + s 3 ) − ( s + 1) =
s
s +1
2
s
s +1
2
 s

+ 1) + ( s + 1)  s 4 + s 3
X ( s ) =  ( 2
 s +1

3
s + s + s + s 2 + 1 s 3 + s 2 + 2s + 1
= 3 2
= 3 2
s ( s + 1)( s + 1)
s ( s + 1)( s + 1)
s 3 + s 2 + 2s + 1 A B C
D
Es + F
= + 2 + 3+
+ 2
3
2
s ( s + 1)( s + 1) s s
s
s +1 s +1
s 3 + s 2 + 2 s + 1 = As 2 ( s + 1)( s 2 + 1) + Bs( s + 1)( s 2 + 1) + c( s + 1)( s 2 + 1) + Ds 3 ( s 2 + 1) + ( Es + F ) s 3 ( s + 1)
A+B+E=0 equating the co-efficient of s5.
A+B+E+F=0 equating the co-efficient of s4.
A+B+C+D+F=0 equating the co-efficient of s3.
A+B+C=0 equating the co-efficient of s2.
B+C=2 equating the co-efficient of s.
A+B+E=0 equating the co-efficient of s2.
C=1equating the co-efficient constant.
C=1
-B=-C+2=1
A=1-B-C=-1
D+F=0
E+F=0D+E=1
D-E=0
2D=1
A=-1; B=1; C=1
D=1/2; E=1/2; F =-1/2
 − 1 1 1 1 / 2 1 / 2( s − 1) 
+
L−1{( s)} = L−1  + 2 + 3 +

s
s +1
s2 + 1 
s s
1 1 / 2 1 / 2( s − 1) 
−1 1
L−1 {X ( s )} = L−1  + 2 + 3 +
+

s
s
s +1
s2 + 1 
 s
2
{X (t )} = −1 + t + t + 1 e −t + 1 Cos t − 1 S int
2 2
2
2
d 2 q dq
+
= t 2 + 2t q(0) = 4; q1 (0) = −2
dt 2 dt
applying laplace transforms,we get
s 2Q ( s ) − sq(0) − q ' (0) + sQ (( s ) − q(0) =
Q ( s )( s 2 + s ) − 4 s + 2 − 4 =
2 1 
 + 1
s2  s

2( s + 1)
+ ( 4 s + 2)
3
s
Q( s) =
( s 2 + s)
=
2 s + 2 + 4 s 4 + 2s 3
s 4 ( s + 1)
2
2*3
 1 
+ 4
Q ( s ) = 4
+
 s + 1  s( s + 1) s ( s + 1)
1
L−1 (Q ( s )) = q(t ) = 4e −t + 2(1 − e −t ) + t 3
3
therefore q(t ) = 2 +
t3
+ 2 e −t
3
2
2
+ 2
3
s
s
3s
3s  1
1 
=  2
− 2
2
( s + 1)( s + 4) 3  s + 1 s + 4 
3.3 a)
2
1 
 1
= 2 2 − 2
s + 2 2 
s +1
1 
 1
L−1  2 2 − 2
= Cost − Cos 2t
s + 2 2 
s +1
b)
1
1
A
B+C
=
= + 2
2
2
s ( s − 2 s + 5) s ( s − 1) + 2
s s − 2s + 5
2
[
]
A+B=0
-2A+C=0
5A=1
A=1/5 ;B=-1/5;C=2/5
We get
X ( s) =
1 1
2−s 
+ 2

5  s s − 2 s + 5 
Inverting,we get
=
1
1 t

t
1
2
2
+
e
Sin
t
−
e
Cos
t

5 
2
=
1
1

1 + e t  Sin 2t − Cos 2t 

5
2

3s 2 − s 2 − 3s + 2 A B
C
D
= + 2+
+
c)
2
2
s ( s − 1)
s s
s − 1 ( s − 1) 2
As( s − 1) 2 + B( s − 1) 2 + Cs 2 ( s − 1) + Ds 2 = 3s 3 − s 2 − 3s + 2
A( s 3 − 2 s + s ) + B ( s 2 − 2 s + 1) + C ( s 3 − s 2 ) + Ds 2 = 3s 3 − s 2 − 3s + 2
A+C=3
-2A+B-C+D=-1
A-2B=-3
B=2;
A=2(2)-3=1
C=3-1=2
D=2(1)-2+2-1=1
We get X ( s ) =
1 2
2
1
+ 2+
+
s s
s + 1 ( s − 1) 2
By inverse L.T
L−1 [X (t )] = 1 + 2t + 2e t + te t
L−1 [X (t )] = 1 + 2t + e t ( 2 + t )
3.4 Expand the following function by partial fraction expansion. Do not evaluate
co-efficient or invert expressions
X ( s) =
2
( s + 1)( s + 1) 2 ( s + 3)
X ( s) =
A
Bs + C
Ds + E
F
+ 2
+ 2
+
2
s + 1 s + 1 ( s + 1)
s+3
2
= A( s 2 + 1) 2 ( s + 3) + ( Bs + C )( s + 1)( s + 3)( s 2 + 1) + ( Ds + E )( s + 1)( s + 3) + F ( s + 1)( s 2 + 1) 2
= A( s 4 + 2 s 2 + 1)( s + 3) + ( Bs + C )( s 2 + 4 s + 3)( s 2 + 1) + ( Ds + E )( s 2 + 4 s + 3) + F ( s + 1)( s 2 + 4 s + 1)
= s 5 ( A + B + F ) + s 4 (3 A + C + 4 B + F ) + s 3 ( 2 A + B + 4C + 3B ) + s 2 (6 A + C + 4 B + 3C ) + s( A + 4C + 3B
+ 4 E + F ) + 3 A + 3 AC + 3E + F = 2
A+B+F=0
-3A+C+4B+F=0
2A+B+4C+3B=0
6A+C+4B+3C=0
A+4C+3B+3D+4E+F=0
3A+3C+3E+F=2
by solving above 6 equations, we can get the values of A,B,C,D,E and
1
.
X ( s) = 3
s ( s + 1)( s + 1) ( s + 3) 3
X ( s) =
A B C
D
E
F
G
H
+ 2 + 3+
+
+
+
+
2
s s
s
s +1
s + 2 s + 3 ( s + 3)
( s + 3) 3
by comparing powers of s we can evaluate A,B,C,D,E,F,G and H.
1
c) X ( s ) =
s ( s + 2)( s + 3) ( s + 4)
A
B
C
D
+
+
+
s +1 s + 2
s+3
s+4
by comparing powers of s we can evaluate A,B,C,D
X ( s) =
3.5 a) X ( s ) =
Let
1
s ( s + 1)(0.5s + 1)
1
A
B
C
= +
+
s ( s + 1)(0.5s + 1) s s + 1 (0.5s + 1)
 s 2 3s

 s2

= A + + 1 + B  + s  + C ( s 2 + s ) = 1
2
2

2

A=1
A B
B
1
+ + C = 0= +C = −
2 2
2
2
3A
3
+ B + C = 0= B + C = −
2
2
B/2=1/2 *-3/2=-1;
B=-2;
C= -3/2+2=1/2
X ( s) =
1
2
1 1 
−
+ 

s s + 1 2  0.5s + 1 
= L − 1 ( X ( s )) = x ( t ) = 1 − 2 e − t + e − 2 t
b)
dx
+ 2 x = 2; x (0) = 0
dt
Applying laplace trafsorms
sX ( s ) − x (0) + 2 X ( s ) = 2 / s
L−1 ( X ( s )) =
2
s ( s + 2)
 2 
L−1 ( X ( s )) = 2 L−1 

 s ( s + 2) 
1 / 2 1 / 2 
= L−1 ( X ( s )) = 2 L−1 
−
s + 2 
 s
−2 t
1
−
e
=
3.6 a) Y ( s ) =
s +1
s + 2s + 5
2
= Y ( s) =
=
s +1
s + 2s + 5
2
s +1
( s + 1) 2 + 4


s +1
= L−1 (Y ( s )) = L−1 
2

 ( s + 1) + 4 
using the table,we get
Y (t ) = e − t Cos2t
b) Y ( s ) =
Y ( s) =
s 2 + 2s
s4
1
2
+ 3
2
s
s
Y(t)= L−1 (Y ( s )) = t + t 2
c) Y ( s ) =
=
=
2s
( s − 1) 3
2s − 2 + 2
( s − 1) 3
2
2
+
2
( s − 1)
( s − 1) 3
 2

 2 
Y (t ) = L−1 
+ L−1 

2 
3 
 ( s − 1) 
 ( s − 1) 
t2 t
2(te + e = e t (t 2 + 2t )
2
t
=
3.7a)
Y ( s) =
As + B Cs + D
1
=
+
( s 2 + 1)
( s 2 + 1) ( s 2 + 1)
thus ( As + B ) + (Cs + D )( s 2 + 1) = 1
= Cs 3 + Ds 2 + ( A + C ) s + ( B + D ) = 1
C=0,D=0
Also A=0;B=1
1
1
A
B
C
D
Y ( s) = 2
=
=
+
+
+
2
2
2
2
( s + 1)
(s + i) (s − i)
(s + i) (s + i)
(s − i) (s − i)2
A( s + i )( s − i ) 2 + B( s − i ) 2 + C ( s − i )( s + i ) 2 + D( s + i ) 2 = 1
( A + C ) s 3 + ( − Ai + B + Ci + D ) s 2 + ( A − 2 Bi + C + 2 Di ) + ( − Ai − B + Ci − D ) = 1
Thus,A+C=0
-Ai+B+Ci+2Di=0 ; B=D
A-2Bi+C+2Di=0
-Ai-B+Ci-D=1
Also D=-Ci;B=-Ci,
A=-C,C=-i/4
A=i/4 ; B=-1/4; D=-1/4
Y ( s) =
i/4
− 1/ 4
−i/4
− 1/ 4
+
+
+
2
(s + i) (s + i)
(s − i) (s − i)2
Y (t ) =
Y (t ) =
− 1/ 4
−i/4
−1/ 4
i/4
+
+
+
2
(s + i) (s + i)
( s − i) ( s − i) 2
−1/ 4
−i/4
−1/ 4
i/4
+
+
+
2
(s + i) (s + i)
(s − i) (s − i)2
Y (t ) = i / 4e −it − 1 / 4e −it −1 / 4e it − 1 / 4te it
Y (t ) = 1 / 4(ie − it − te −it − ie it − te it )
Y (t ) = 1 / 4(i (Cost − iSin t ) − t (Cost − i Sin t ) − i(Cos t + iSin t ) − t (Cos t + i Sin t ) )
Y (t ) = 1 / 4( 2 Sin t − 2t Cos t )
Y ( t ) = 1 / 2 ( Sin
3.8
f ( s) =
= f ( s) =
t − t Cos
t)
1
s ( s + 1)
2
A B
C
+ +
2
s
s s +1
= A( s + 1) + Bs( s + 1) + Cs 2 = 1
Let s=0 ; A=1
s=1; 2A+B+C=1
s=-1: C=1
B=-1
1 1
1
f ( s) = 2 + +
s
s s +1
f (t ) = (t − 1) + e − t
PROPERTIES OF TRANSFORMS
4.1 If a forcing function f(t) has the laplace transforms
f ( s) =
=
1 e − s − e −2 s e −3s
+
−
s
s2
s
1 − e −3s
e − s − e −2 s
+
s
s2
f (t ) = L−1{ f ( s )} = [u(t ) − u(t − 3)] + [(t − 1)u (t − 1) − (t − 2)u (t − 2)]
= u(t ) + (t − 1) u(t − 1) − (t − 2)u(t − 2) − u (t − 3)
graph the function f(t)
4.2 Solve the following equation for y(t):
t
∫ y (τ ) dτ
0
=
dy (t )
y ( 0) = 1
dt
Taking Laplace transforms on both sides
t
 dy (t ) 
L{∫ y (τ ) dt} = L 

 dt 
0
1
. y ( s ) = s. y ( s ) − y (0)
s
1
. y ( s ) = s. y ( s ) − 1
s
y ( s) =
s
s −1
2
 s 
y (t ) = L−1{ y ( s )} = L−1  2
 cosh(t )
 s − 1
4.3 Express the function given in figure given below the
s-
t – domain and the
domain
This graph can be expressed as
= {u(t − 1) − u(t − 5)} + { (t − 2)u(t − 2) − (t − 3)u (t − 3)} + {u(t − 5) − (t − 5)u (t − 5) + (t − 6)u (t − 6)}
f (t ) = u (t − 1) + (t − 2)u(t − 2) − (t − 2)u(t − 3) − (t − 5)u(t − 5) + (t − 6)u(t − 6)
f ( s ) = L{ f (t )} =
e − s e −2 s e −3s e −3s
e −5 s e −6 s
+ 2 − 2 −
− 2 + 2
s
s
s
s
s
s
=
e − s − e −3s e −2 s + e −6 s − e −3s − e −5 s
+
s
s2
4.4 Sketch the following functions:
f (t ) = u (t ) − 2u(t − 1) + u(t − 3)
f (t ) = 3tu(t ) − 3u(t − 1) + u(t − 2)
4.5 The function f(t) has the Laplace transform
f ( S ) = (1 − 2e − s + e −2 s ) / s 2
obtain the function f(t) and graph f(t)
1 − 2 e − s + e −2 s
f ( s) =
s2
=
1 − e − s e − s − e −2 s
−
s2
s2
f (t ) = L−1{ f ( s )} = − (t − 1)u(t − 1) + tu(t ) − {(t − 1)u (t − 1) − (t − 2)u(t − 2)]
= tu(t ) − 2(t − 1)u (t − 1) + (t − 2)u(t − 2)
4.6 Determine f(t) at t = 1.5 and at t = 3 for following function:
f (t ) = 0.5u(t ) − 0.5u(t − 1) + (t − 3)u (t − 2)
At t = 1.5
f (t ) = 0.5u(t ) − 0.5u (t − 1) + (t − 3)u (t − 2)
f (1.5) = 0.5u(t ) − 0.5u(t − 1)
f (1.5) = 0.5 − 0.5 = 0
At t = 3
f (3) = 0.5 − 0.5 + (3 − 3) = 0
RESPONSE OF A FIRST ORDER SYSTEMS
5.1 A thermometer having a time constant of 0.2 min is placed in a
temperature bath and after the thermometer comes to equilibrium with
the bath, the temperature of the bath is increased linearly with time at the
rate of I deg C / min what is the difference between
the indicated
temperature and bath temperature
(a) 0.1 min
(b) 10. min
after the change in temperature begins.
© what is the maximum deviation between the indicated temperaturew
and bath temperature and when does it occurs.
(d) plot the forcing function and the response on the same graph. After the
long enough time buy how many minutes does the response lag the input.
Consider thermometer to be in equilibrium with
temperature Xs
X (t ) = X S + (1° / m )t , t > 0
as it is given that the temperture varies linearly
X(t)-Xs = t
Let X(t) = X(t) - Xs = t
temperature
bath at
Y(s) = G(s).X(s)
Y ( s) =
1 1
A
B C
=
+ + 2
2
1 + τs s 1 + τs s s
A = τ 2 B = − τ C =1
τ2
τ 1
Y ( s) =
− + 2
1 + τs s s
Y ( t ) = τe − t / τ − τ + t
(a) the difference between the indicated temperature and bath temperature
at t = 0.1 min = X(0.1)_ Y(0.1)
= 0.1 - (0.2e-0.1/0.2 - 0.2+0.1) since T = 0.2 given
= 0.0787 deg C
(b) t = 1.0 min
X(1) - Y(1) = 1- (0.2e-1/0.2 - 0.2 +1) = 0.1986
(c) Deviation D = -Y(t) +X(t)
= -τe-t/T+T =τ (-e-t/T+1)
For maximum value dD/dT = τ (-e-t/T+(_-1/T) = 0
-e-t/ = 0
as t tend to infinitive
D = τ (-e-t/T+(_-1/T) = τ =0.2 deg C
5.2 A mercury thermometer bulb in ½ in . long by 1/8 in diameter. The
glass envelope is very thin. Calculate the time constant in water flowing
at 10 ft / sec at a temperature of 100 deg F. In your solution , give a
summary which includes
(a) Assumptions used.
(b) Source of data
(c) Results
T = mCp/hA =
( ρAL)C p
h ( A + πDL)
Calculation of
NU d =
Re d =
Pr =
hD
= CRem (Pr) n
K
Dvρ
µ
Cpµ
K
=
(1 / 8 * 2.54 * 10 −2 )(10 * 0.3048)103
= 9677.4
10 −3
= 4.2 KJ / KgK
Source data: Recently, Z hukauskas has given c,m ,ξ,n values.
For Re = 967704
C = 0.26 & m = 0.6
NuD = hD/K = 0.193 (9677.4)*(6.774X10-3) = 130
.h = 25380
5.3 Given a system with the transfer function Y(s)/X(s) = (T1s+1)/(T2s+1).
Find Y(t) if X(t) is a unit step function. If T1/T2 = s. Sktech Y(t) Versus
t/T2. Show the numerical values of minimum, maximum and ultimate values
that may occur during the transient. Check these using the initial value
and final value theorems of chapter 4.
Y ( s) =
T1s + 1
T2 s + 1
X(s) =unit step function = 1 X(s) = 1/s
Y ( s) =
T1 s + 1
A
B
= +
s (T2 s + 1)
s iT2 s
A = 1 B = T1 - T2
Y ( s) =
1 T1 − T2
+
s 1 + T2 s
Y (t ) = 1 +
T1 − T2 −t / T2
e
T2
If T1/T2 = s then
Y (t ) = 1 + 4e − t / T2
Let t/T2 = x then Y (t ) = 1 + 4e
−x
Using the initial value theorem and final value theorem
Lim Y (T ) = Lim sY ( s )
S →∞
T →0
1
T s +1
s = T1 = 5
Lim 1
= Lim
S →∞ T s + 1
S →∞
1
T2
2
T2 +
s
T1 +
=
Lim Y (T ) = Lim sY ( s ) = Lim
T →0
Figure:
S →∞
S →0
T1 s + 1
=1
T2 s + 1
5.4 A thermometer having first order dynamics with a time constant of 1
min is placed in a temperature bath at 100 deg F. After the thermometer
reaches steady state, it is suddenly placed in bath at 100 deg F at t = 0 and
left there for 1 min after which it is immediately returned to the bath at
100 deg F.
(a) draw a sketch showing the variation of the thermometer reading with
time.
(b) calculate the thermometer reading at t = 0.5 min and at t = 2.0 min
1
Y ( s)
(τ = 1 min)
=
X ( s) s + 1
1 e−s 
( s ) = 10  −
s 
s
 1 − e−s 
Y ( s ) = 10

 s 

− e−s 
1
Y ( s ) = 10

 s ( s + 1) s ( s + 1) 
Y (t ) = 10(1 − e − t ) t < 1
Y (t ) = 10( (1 − e − t ) − (1 − e − ( t −1) ) ) t ≥ 1
At t = 0.5 T = 103.93
At = 2 T =102.325
5.5 Repeat problem 5.4 if the thermometer is in 110 deg F for only 10 sec.
If thermometer is in 110 deg F bath for only 10 sec
T = 110 − 10e − t / 60
0 < t < 10 sec & T = 60 sec
T (t = 10 sec) = 101.535
T = 100 + 1.535e − ( t −10 ) / 60 t > 10 sec
T(t=30sec) = 101.099 deg F
T(t=120sec) = 100.245 deg F
5.6 A mercury thermometer which has been on a table for some time,is
registering the room temperature ,758 deg F. Suddenly, it is placed in a 400
deg F oil bath. The following data are obtained for response of the
thermometer
Time (sec)
Temperature, Deg F
0
75
1
107
2.5
140
5
205
8
244
10
282
15
328
30
385
Give two independent estimates of the thermometer time constant.
τ =
t
 325 
ln

 400 − T 
From the data , average of 9.647,11.2,9.788,10.9,9.87,9.95, and 9.75 is 10.16 sec.
5.7 Rewrite the sinusoidal response of first order system (eq 5.24) in
terms of a cosine wave. Re express the forcing function equation (eq 5.19)
as a cosine wave and compute the phase difference between input and
output cosine waves.
1
 
1
Aω  τ 
( s) = 2
Y ( s) =
τs + 1
s +ω2 s + 1
τ
splitting into partial fractions then converting to laplace transforms
Y (t ) =
Aαωτ −t /τ
e
+
τ 2ω 2 + 1
A
τ 2ω 2 + 1
sin(ωt + φ )
where φ = tan-1 (ωτ)
As t →∝
Y (t ) s =
Y (t ) =
Y (t ) =
A
τ 2ω 2 + 1
sin(ωt + φ ) =
π

cos(ωt −  − φ 
2

τ 2ω 2 + 1
A
π

A sin(ωt + φ ) = A cos − ωt 
2

π

A cos ωt − 
2

The phase difference = φ −
π
 π
− −  = φ
2  2
)
5.8 The mercury thermometer of problem 5.6 is allowed to come to
equilibrium in the room temp at 75 deg F.Then it is immersed in a oil
bath for a length of time less than 1 sec and quickly removed from the
bath and re exposed to 75 deg F ambient condition. It may be estimated
that the heat transfer coefficient to the thermometer in air is 1/ 5th that in
oil bath.If 10 sec after the thermometer is removed from the bath it reads
98 Deg F. Estimate the length of time that the thermometer was in the bath.
t < 1 sec
T1 = 400 − 325e − t1 / τ
Next it is removed and kept in 75 Deg F atmosphere
Heat transfer co-efficient in air = 1/5 heat transfer co-efficient in oil
hair = 1/5 hoil
τ =
mC
τ oil = 10 sec
hA
τ air = 50 sec
TF = 75 + (T1 − 75)e − t2 / 50
TF = Final Temp = 98 deg C
98 = 75 + (325 − 325e − t1 / 10 )e −10 / 50
e −t / 10 = 0.91356
t 1 = 0.904 sec.
5.9 A thermometer having a time constant of 1 min is initially at 50 deg C.
it is immersed in a bath maintained at 100 deg C at t = 0 . Determine the
temperature reading at 1.2 min.
τ = 1 min for a thermometer initially at 50 deg C.
Next it is immersed in bath maintained at 100 deg C at t = 0
At t = 1.2
Y (t ) = A(1 − e − t / τ )
Y (1.2) = 50(1 − e −1.2 / 1 ) + 50
Y(1.2) = 84.9 deg C
5.10 In Problem No 5.9 if at, t = 1.5 min thermometer having a time
constant of 1 minute is initially at 50 deg C.It is immersed in a bath
maintained at 100 deg C at t = 0.Determine the temperature reading at t =
1.2 min.
At t = 1.5
Y (1.5) = 88.843°C
Max temperature indicated = 88.843 deg C
AT t = 20 min
T = 88.843 − 13.843(1 − e −18.8 / 1 )
T = 75 Deg C.
5.11 A process of unknown transfer function is subjected to a unit impulse
input. The output of the process is measured accurately and is found to be
represented by the function Y(t) = t e-t. Determine the unit step response in
this process.
X(s) = 1 Y(t) = te-t
Y (s) =
G( s) =
1
( s + 1) 2
1
Y ( s)
=
X ( s ) ( s + 1) 2
For determining unit step response
Y ( s) =
1
( s + 1) 2
Y ( s) =
1
A
B
C
= +
+
2
( s + 1)
s s + 1 ( s + 1) 2
A = 1 B = -1 C = -1
Y ( s) =
1
1
1
−
−
s s + 1 ( s + 1) 2
Y (t ) = 1 − e − t − te − t
Response of first order system in series
7.1 Determine the transfer function H(s)/Q(s) for the liquid level shown in
figure P7-7. Resistance R1 and R2 are linear. The flow rate from tank 3 is
maintained constant at b by means of a pump ; the flow rate from tank
3 is independent of head h. The tanks are non interacting.
Solution :
A balance on tank 1 gives
q − q1 = A1
dh1
dt
where h1 = height of the liquid level in tank 1
similarly balance on the tank 2 gives
q1 − q2 = A2
dh2
dt
and balance on tank 3 gives
q2 − q0 = A3
here q1 =
dh
dt
h1
h
q2 = 2
R1
R2
q0 = b
So we get
q−
dh
h1
= A1 1
dt
R1
dh2
h1 h2
−
= A2
dt
R1 R2
dh
h2
− b = A3
R2
dt
writing the steady state equation
qS −
dh
h1s
= A1 1S = 0
dt
R1
dh2 S
h1S h2 S
−
= A2
dt
R1 R2
h2 S
−b = 0
R2
Subtracting and writing in terms of deviation
Q −
dH 1
H
= A1
dt
R1
dH 2
H1 H 2
−
= A1
dt
R1
R2
dH
H2
= A3
dt
R2
where Q = q –qS
H1= h1-h1S
H1= h2-h2S
H = h - hS
Taking Laplace transforms
Q( s) −
H 1 ( s)
= A1 s H 1 ( s ) ---------(1)
R1
H 1 (s) H 2 (s)
−
= A2 s H 2 ( s ) --------(2)
R1
R2
H 2 (s)
= A3 s H ( s ) ----------(3)
R2
We have three equations and 4 unknowns(Q(s),H(s),H1(s) and H2(s). So we
can express one in terms of other.
From (3)
H 2 (s) =
H 2 (s)
-------------(4)
R1 A3 s
R2 H 1 ( s )
where τ2 =R2 A2------------(5)
R1 (τ 2 s + 1)
H 2 (s) =
From (1)
H 1 ( s) =
R1Q( s )
,
(τ 1 s + 1)
τ 1 = R1 A1 ---------(6)
Combining equation 4,5,6
H ( s) =
Q( s)
( A3 s )(τ 1 s + 1)(τ 2 s + 1)
H ( s)
1
=
Q( s)
( A3 s )(τ 1 s + 1)(τ 2 s + 1)
Above equation can be written as
i.e, if non interacting first order system are there in series then there overall
transfer function is equal to the product of the individual transfer function in
series.
7.2 The mercury thermometer in chapter 5 was considered to have all its
resistance in the convective film surrounding the bulb and all its
capacitance in the mercury. A more detailed analysis would consider both
the convective resistance surrounding the bulb and that between the bulb
and mercury. In addition , the capacitance of the glass bulb would be
included.
Let
Ai = inside area of bulb for heat transfer to mercury.
Ao = outside area of bulb, for heat transfer from surrounding fluid.
.m = mass of the mercury in bulb.
mb = mass of glass bulb.
C = heat capacitance of mercury.
Cb = heat capacity of glass bulb.
.hi = convective co-efficient between the bulb and the surrounding fluid.
.ho = convective co-efficient between bulb and surrounding fluid.
T = temperature of mercury.
Tb = temperature of glass bulb.
Tf = temperature of surrounding fluid.
Determine the transfer function
resistance and capacitance on
inclusion of the bulb results in
overall transfer function different
between Tf and T. what is the effect of bulb
the thermometer response? Note that the
a pair of interacting systems, which give an
from that of Eq (7.24)
Writing the energy balance for change in term of a bulb and mercury
respectively
Input - output = accumulation
h0 A0 (T f −Tb )−hi Ai (Tb − T ) = mb C b
hi Ai (Tb −T )−0 = m C
dTb
dt
dT
dt
Writing the steady state equation
h0 A0 (T fs −Tbs )−hi Ai (Tbs − Ts ) = mb C b
hi Ai (Tbs −Ts ) = 0
dTbs
=0
dt
Where subscript s denoted values at steady subtracting and writing these
equations in terms of deviation variables.
h0 A0 (T f −Tb )−hi Ai (Tb − Tm ) = mb C b
hi Ai (Tb −Tm )−0 = m C
dTb
dt
dTm
dt
Here TF = Tf - TfS
TB = Tb - TbS
Tm = T - TS
Taking laplace transforms
h0 A0 (TF ( s ) −TB ( s ))−hi Ai (TB − Tm ) = mb C b TB ( s ) ----(1)
And hi Ai (TB ( s ) −Tm ( s )) = mC sTB ( s ) ------(2)
= h0 A0 (TF ( s ) −TB ( s ))−mCSTm ( s ) = mb C b sTB ( s )
From (2) we get

 mC
s + 1 = Tm ( s ) (τ i s + 1)
TB ( s ) = Tm ( s ) 

 hi Ai
Where τ i =
mC
hi Ai
Putting it into (1)

mC 
TF ( s ) − Tm ( s ) (τ i s + 1))(τ 0 s + 1) +
s = 0
h0 A0 


mC 
= TF ( s ) = Tm ( s ) (τ i s + 1))(τ 0 s + 1) +
s
h0 A0 

=
Tm ( s )
=
TF ( s )
=
1
τ iτ 0 s 2 + (τ i + τ 0 +
Tm ( s )
=
TF ( s )
mC
)s + 1
h0 A0
1
τ iτ 0 s 2 + (τ i + τ 0 +
mC
)s + 1
h0 A0
Or we can write
T (s)
=
T f (s)
τi =
1
τ iτ 0 s 2 + (τ i + τ 0 +
mC
)s + 1
h0 A0
mC
mC
and τ 0 = b b
hi Ai
h0 A0
We see that a loading term mC/ hoAo is appearing in the transfer function.
The bulb resistance and capacitance is appear in τ 0 and it increases the
delay i.e Transfer lag and response is slow down.
7.3 There are N storage tank of volume V Arranged so that
when
water is fed into the first tank into the second tank and so on. Each tank
initially contains component A at some concentration Co and is equipped
with a perfect stirrer. A time zero, a stream of zero concentration is
fed into the first
tank at volumetric
rate q. Find the resulting
concentration in each tank as a function of time.
Solution:
. ith tank balance
qC i −1 − qC i = V
dC i
dt
qC ( i −1) s − qC is = 0
C (i −1) − C i =
V dC i
q dt
 V
τ = 
q

Taking lapalce transformation
C (i −1) ( s ) − C i ( s ) = τ sCi ( s )
C (i −1) ( s ) = (1 + τ s )Ci ( s )
Ci ( s)
1
=
C i −1 ( s ) 1 + τ s
Similarly
Ci ( s) C1 ( s) C 2 ( s)
C ( s) Ci ( s)
1
=
×
× − − − − − − − − − i −1
×
=
Co( s) C 0 ( s) C1 ( s)
Ci −2 ( s) Ci ( s) (1 + τ s)i
Or
C N (s)
1
=
Co( s ) (1 + τ s ) N
C N ( s) =
− C0
s (1 + τ s ) N
1
τ
τ
τ 
C N ( s) = − C0  −
−
−−−−−−−

N
N −1
1 + τs 
(1 + τ s )
(1 + τ s )
s
t
t
−
−

t 
N −1
τ
τ
−
e
t
e
t N −2

− N −2 .
−−−−−e τ 
C N (t ) = − C 0 1 − N −1 .
 τ

( N − 1)! τ
( N − 2)!




t
− 
τ 

C N (t ) = − C0 1 − e .




N −1
t
 
τ 
+.
( N − 1)!
N −2

t
 

τ 
− − − − − +1

( N − 2)!


7.4 (a) Find the transfer functions H2/Q and H3/Q for the three tank system
shown in Fig P7-4 where H1,H3 and Q are deviation variables. Tank 1 and
Tank 2 are interacting.
7.4(b) For a unit step change in q (i.e Q = 1/s); determine H3(0) , H3(∞)
and sketch H3(t) vs t.
Solution :
Writing heat balance equation for tank 1 and tank 2
q − q1 = A1
q1 − q 2 = A2
q1 =
dh1
dt
dh2
dt
h1 − h2
h
q2 = 2
R1
R2
Writing the steady state equation
q s − q1s = 0
q1s − q 2 s = 0
Writing the equations in terms of deviation variables
Q − Q1 = A1
dH 1
dt
Q1 − Q2 = A2
dH 2
dt
Q1 =
H1 − H 2
H
Q2 = 2
R1
R2
Taking laplace transforms
Q ( s ) − Q1 ( s ) = A1 sH 1 ( s )
Q1 ( s) − Q2 ( s) = A1 sH 2 ( s)
R1 Q1 ( s ) = H 1 ( s ) − H 2 ( s )
R2 Q2 ( s ) = H 2 ( s )
Solving the above equations we get
H 2 ( s)
R2
=
2
Q(s)
τ 1τ 2 s + (τ 1 + τ 2 + A1 R2 )s + 1
[
Here
]
τ1 = R1 A1
τ 2 = R2 A2
Now writing the balance
q 2 − q3 = A3
dh3
dt
Steady state equation
q 2 S − q3 S = 0
q3 =
h3
R3
Q2 −
dh
H3
= A3 3
dt
R3
for third tank
Taking laplace transforms
H 3 (s)
= A3 sH ( s )
R3
Q2 ( s ) −
H 3 ( s)
(τ 3 s + 1) where
R3
Q2 ( s ) =
τ 3 = R3 A3
From equation 1,2,3,4 and 5 we got
Qs ( s)
Q( s )
=
1
[τ τ
1 2
]
s + (τ 1 + τ 2 + A1 R2 ) s + 1
2
Putting it in equation 6
H 3 ( s)
Q( s)
=
R3
τ 1τ 2 s + (τ 1 + τ 2 + A1 R2 ) s + 1 (τ 3 s + 1)
[
]
2
Putting the numerical values of R1,R2 and R3 and A1,A2,A3
H 3 ( s)
Q( s)
H 2 (s)
Q( s )
4
4 s + 6 s + 1 (2s + 1)
=
[
=
2
4s + 6s + 1
[
Solution (b)
Q (s) =
1
s
2
2
]
]
H 3 (s) =
1
4
2
s 4s + 6s + 1 (2 s + 1)
[
]
From initial value theorem
H 3 (0) = Lim sH 3 ( s)
S →∞
=
Lim
S →∞
4
(2 s + 1)(4s 2 + 6s + 1)
4
Lim
=
S →∞
(2 s + 1) s 3 (4 +
6 1
+ 2)
s
s
H3 (0) = 0
From final value theorem
H 3 (∞) = Lim sH 3 ( s )
S →0
=
H3
Lim
S →0
(∞) = 4
4
(2s + 1)(4 s 2 + 6 s + 1)
7.5 Three identical tanks are operated in series in a non-interacting fashion
as shown in fig P7.5 . For each
tank R=1, τ = 1. If the deviation in
flow rate to the first tank in an impulse function of magnitude 2,
determine
(a) an expression for H(s)
where H is the deviation in level in the third
tank.
(b) sketch the response H(t)
(c) obtain an expression for H(t)
solution :
writing energy balance equation for all tanks
q − q1 = A
dh1
dt
q1 − q2 = A
dh2
dt
q2 − q3 = A
q1 =
h
h1
q2 = 2
R
R
dh
dt
q3 =
h
R
So we get
qS − q1S = 0
q1S − q2 S = 0
q 2 S − q3 S = 0
writing in terms of deviation variables and taking laplace transforms
Q( s) −
H1 ( s)
= AS H 1 ( s )
R
Q1 ( s ) H 2 ( s )
−
= AS H 2 ( s )
R
R
H 2 ( s) H ( s)
−
= AS H ( s )
R
R
solving we get
H ( s)
Q( s)
H ( s) =
=
R
1
=
3
(τ s + 1)
( s + 1) 3
2
Q( s)
=
3
(τ s + 1)
( s + 1) 3
H (t ) = L−1 {H ( s )} = 2
t 2 −t
e
2
H (t ) = t 2 e − t
dH (t )
= 2te −t − te −t = 0
dt
= 2t = t 2
at t = 2 max will occur.
7.6 In the two- tank mixing process shown in fig P7.6 , x varies from 0 lb
salt/ft3 to 1 lb salt/ft3 according to step function. At what time
3
does the
salt concentration in tank 2 reach 0.6 lb/ ft ? The hold up volume of each
tank is 6
ft3.
Solution
Writing heat balance equation for tank 1 and tank 2
qx − q y = V
dy
dt
q y − qc = V
dl
dt
steady state equation
q xs − q ys = 0
q ys − qcs = 0
writing in terms of deviation variables and taking laplace transforms
X ( s) − Y ( s) =
V
s Y ( s)
q
1
1
V
Y ( s)
;τ =
=
=
q
X ( s)  V
 τ s +1
 s + 1
q

C ( s) =
Y ( s)
X ( s)
=
(τ s + 1)
(τ s + 1) 2
C ( s)
1
=
X ( s ) (τ s + 1) 2
X ( s) =
τ=
1
s
V 6
= =2
q 3
C ( s) =
X ( s)
s( 2 s + 1) 2
C ( s) =
(1 / 4)
1
s( s + ) 2
2
1
C ( s) =  
4



1
1

 s ( s + 1 )2
2
1
 
1
1
2
C ( s) = −   2 −
1
s  1 
s + 
+
1
 2  
2
t
C (t ) = 1 −
t
−
1 −2
te − e 2
2
C ( t ) = 0 . 61 lb salt / ft 3
t = 4.04 min
7.7 Starting
from first
and H2(s)/Q(s)
for
the
principles, derive
liquid level
the transfer functions H1(s)/Q(s)
system
shown in
resistance are linear and R1= R2 = 1. Note that two
figure P7.7. The
streams are flowing
from tank 1, one of which flows into tank 2. You are expected to give
numerical values
of the parameters
and
in the transfer functions and to
show clearly how you derived the transfer functions.
Writing heat balance equation for tank 1
q − qa − q1 = A1
q1 =
dh1
dt
h1
h
qa = 1
R1
Ra
=q −
dh
h1
h
− 1 = A1 1
dt
Ra R1
writing the balance equation for tank 2
q1 − q2 = A2
dh2
dt
dh2
h1 h2
−
= A2
dt
R1 R2
writing steady state equations
qs −
hs
hs
− 1 =0
Ra
R1
h1 s h2 s
−
=0
R1
R2
writing the equation in terms of deviation variables
 1
dH 1
1 
Q − H 1 
+  = A1
dt
 Ra R1 
dH 2
H1 H 2
−
= A2
dt
R1 R2
taking laplace transforms
 R + R2 
 = A1S H 1 s -----------(1)
Q ( s ) − H 1 ( s ) 1
 R1 Ra 
and
H1 ( s) H 2 ( s)
−
= A2 s H 2 ( s ) -----------(2)
R1
R2
from (1) we get
H1 ( s)
1
=
Q( s)

R1 + Ra 
 A1 s + R R 
1 a 

 R1 Ra 
R + R 
H1 ( s)
a 
 1
=
Q( s)

 R1 Ra A1
 R + R s + 1
a

 1
 R1 Ra 


RR A
H 1 ( s )  R1 + Ra 
; τ1 = 1 a 1
=
[τ 1s + 1]
R1 + Ra
Q( s)
and from (2 ) we get
 R2 Ra  R2 
 R + R  R 
H1 ( s)
a  1 
τ =R A
=  1
[ τ 1s + 1](τ 2 s + 1) 2 2 2
Q( s)
putting the numerical values of parameters
2
 
H1 ( s)
3
=  
Q( s)
4

 s + 1
3


2
 
H 2 (s)
3
=
Q(s)
4

 s + 1(s + 1)

3
8.1 A step
transfer
change of magnitude 4 is introduced into a system having the
Y ( s)
10
= 2
X ( s ) s + 1.6s + 4
Determine
(a) % overshoot
(b)Rise time
(c)Max value of Y(t)
(d)Ultimate value of Y(t)
(e) Period of Oscillation.
Given X ( s ) =
4
s
Y ( s) =
The transfer function is
40
s ( s + 1.6s + 4)
2
2.5
Y ( s)
10 × 0.25
=
=
2
0.25s + 0.4 s + 1)
X ( s ) 0.2( s 2 ) + ( 1.6 ) s + 1)
4
τ 2 = 0.25 ; τ = 0.5
and 2ξ τ = 0.4
ξ =
0.4
= 0.4 (< 1 = system is underdamped )
2(0.5)
we find ultimate value of Y(t)
40 s
40
=
= 10
S →0 s ( s + 1.6 + 4)
4
Lt Y (t ) = Lt sY ( s ) = Lt
t →∞
S →0
2
thus B= 10
now, from laplace transform tables
ξt


−
1
Y (t ) = 101 −
e τ sin(α + φ )
1−ξ 2


where α =
1−ξ 2
τ
, φ = tan −
1−ξ 2
ξ
(a) Over shoot =
 − πξ
A
= exp
 1−ξ 2
B


 = exp  − π × 0.4  = 0.254
 0.84 




thus % overshoot = 25.4
c)thus, max value of Y(t) = A+B = B(0.254)+B
= 2.54+10 = 12.54
e) Period of oscillation =
2πτ
1−ξ 2
= 3.427
b) For rise time, we need to solve
ξt


−
1
e τ sin(αt + φ ) = 10 for t = tr
101 −
1−ξ 2


= e
= e
−
ξτ r
τ
0.4τ r
−
0.5
sin(α t r + φ ) = 0
sin(1.833 t r + 1.1589) = 0
solving we get tr = 1.082
thus
SOLUTION: % Overshoot = 25.4
Rise time = 1.0842
Max Y(t) = 12.54
U(t) Y(t) = 10
Period of oscillation = 3.427
Comment : we see that the Oscillation period is small and the decay ratio
also small = system is efficiently under damped.
8.2 The tank system operates at steady state. At t = 0, 10 ft3 of wateris
added to tank 1. Determine the maximum deviation in level in both tanks
from the ultimate steady state values, and the time at which each
maximum occurs.
A1 = A2 = 10 ft3
R1 = 0.1ft/cfm R2 = 0.35ft/cfm.
As the
tanks are non interacting the transfer functions are
H ( s)
K1
0.1
=
=
Q ( s ) τ 1 s + 1 ( s + 1)
H 2 ( s)
R2
0.35
=
=
Q( s)
(τ 1 s + 1)(τ 2 s + 1) ( s + 1)(3.5s + 1)
Now, an impulse of ∂(t ) = 10 ft 3 is provided
Q ( s ) = 10 = H 1 ( s ) =
and H 2 ( s ) =
1
= e −t
s +1
3.5
3.5
=
2
( s + 1)( 3.5s + 1) 3.5s + 4.5s + 1
Now τ 2 = 3.5 = τ = 1.871
2ξτ = 4.5 = ξ =
4.5
= 1.202
2τ
thus, this is an ovedamped system
Using fig8.5, for ξ = 1.2 , we see that maximum is attained at
t
= 0.95, t = 1.776 min
τ
And the maximum value is around τ 2 = 0.325 Y2 (t) = 0.174
= H2(t) = 0.174x3.5 = 0.16ft
thus max deviation is H1 will be at t = 0 = H1 = 1 ft
max deviation is H2 will be at t = 1.776 min = H2Max = 0.61 ft.
comment : the first tank gets the impulse and hence it max deviation turns out
to be higher than the deviations for the second tank. The second tank exhibits an
increase response ie the deviation increases, reaches the H2Max falls off to zero.
8.3 The tank liquid level shown operates at steady state when a step
change is made in the flow to tank 1.the transaient response in critically
damped, and it takes 1 min for level in second tank to reach 50 % of total
change. If A1/A2 = 2 ,find R1/R2 . calculate τ for each tank. How long does
it take for level in first tank to reach 90% of total change?
For the first tank, transfer function
For the second tank
H 1 (s)
R1
=
Q (s)
τ s + 11
R
H 2 ( s)
=
Q( s)
(τ 1 s + 1)(τ 2 s + 1)
=
H 2 ( s)
R2
=
2
Q( s)
τ 1τ 2 s + (τ 1 + τ 2 ) s + 1
Q( s) =
R2
1
1
; H 2 ( s) =
2
s
s τ 1τ 2 s + (τ 1 + τ 2 ) s + 1
τ ( parameter) = τ 1τ 2
For τ ( parameter ) = τ 1τ 2
 
t
for ξ = 1, H 2 (t ) = R2 1 − 1 +

τ 1τ 2
 
 −
e


t
τ 1τ 2



given, t = 1 for τ ( parameter) = τ 1τ 2
H 2 (t → ∞) = R2 (1 − (0) ) = R2
 
1
= R2 1 − 1 +
 
τ 1τ 2
 
 −
e


1
τ 1τ 2
 R
 = 2 −I

2

also 2ξτ = τ 1 + τ 2
ξ =1=
τ1 + τ 2
2
= τ = A1 R1 = A2 R2 =
from I
1
 1 −
1 − 0.5 = 1 +  e τ
 τ
R1 A2
=
= 0.5
R2 A1
t
−
R1
8.3) H 1 ( s ) =
; H 1 (t ) = R1 (1 − e τ 1 )
s (τ 1 s + 1)
0.94(t → ∞) = R1 (1 − e
0.9 R1 = R1 (1 − e
e
−
t
0.596
−
t
0.596
−
t
τ1
)
)
= 0.1; t =1.372 min
thus
R1
= 0.5
R2
τ 1 =τ 2 = 0.596 min
t 90% = 1.372 min
Comment :
Small values of τ 1 ,τ 2 indicate the system regains the steady state quickly. Also as R1 > R2 , the sec ond tan
responds more slowly to changes than first tan k .
8.4 Assuming the flow in the manometer to be laminar function between
applied pressure P1 and the manometer reading h. Calculate a) steady sate
gain ,b) τ ,c) ξ . Comment on the parameters and their relation to the
physical nature of this problem.
Assumptions:
Cross-sectional area =a
Length of mercury in column = L
Friction factor = 16/Re (laminar flow)
Mass of mercury = mrg
Writing a force balance on the mercury
Mass X acceleration = pressure force - drag force - gravitational force
( ALρ )
d 2h
ρu 2
=
Ap
−
Af
− A( ρgh)
1
2
dt 2
p
L d 2 h 8µ dh
+
+h= 1
2
ρgD dt
ρg
g dt
At Steady state,
hs =
p1s
ρg
=
p
L d 2 H 8µ dH
+
+H = 1
2
ρgD dt
ρg
g dt
=
p (s)
8µ
L 2
s H ( s) +
sH ( s ) + H ( s ) = 1
g
ρgD
ρg
[
[
]
]
= k1 s 2 + k 2 s + 1 H ( s ) = k 3 p1 ( s )
=
H 1 (s)
k3
=
2
p1 ( s ) (k1 s + k 2 s + 1)
Where
Thus
Where
k1 =
8µ
1
L
; k2 =
; k3 =
;
g
ρgD
ρg
H 1 (s)
R
=
2 2
p1 ( s ) ( τ s + 2ξτ s + 1)
R =
Now b)τ =
1
8µ
L
; τ 2 = ; 2ξτ =
;
g
ρgD
ρg
L
;
g
8µ 1  4 µ  L 


. =
c)ξ =
ρgD 2τ  ρgD  g 
−1
Steady state gain
Lt G ( s ) = R =
S →0
1
;
ρg
Comment : a) τ is the time period of a simple pendulum of Length L.
b) ξ is inversely proportional to τ , smaller the τ ,the system will
tend to move from under damped to over damped characteristics.
8.5 Design a mercury manometer that will measure pressure of upto 2
atm, and give responses that are slightly under damped with ξ = 0.7
Parameter to be decide upon :
.a) Length of column of mercury
.b) diameter of tube.
Considering hmax to be the maximum height difference to be used
p1 = ρghmax = hmax =
2 * 1.01325 * 10 5
;
9.81 *13600
hmax =1.51 m;
Assuming the separation between the tubes to be 30 cm,
We get an additional length of 0.47 m;
Which gives us the total length L= 1.5176.47
L = 2 M
Now, ξ = 0.7 =
 4 µ  g 
 = 0.7



 ρgD  L 
9.81
g
4 * 1.6 *10 −3 *
2 = 1.5 * 10 −7
L=
D=
0.74 * 13600 * 9.81
0.7 ρg
= 0.00015
4µ
As can be seen, the values yielded are not proper, with too small a diameter
and too large a length. A smaller ξ value and lower measuring range of
pressure might be better.
8.6 verify that for a second order system subjected to a step response,
1
Y (t ) = 1 −
1− ξ 2
e
−
ξt
τ
[
sin 1 − ξ 2
]τt + tan
−1
1− ξ 2
ξ
With ξ <1
1
1
2 2
s (τ s + 2ξτs + 1)
Y (s) =
τ 2 s 2 + 2ξτs + 1 = ( s − s1 )(s − s 2 )
where s1 =
s2 =
−ξ
τ
+
−ξ
τ
+
ξ 2 −1
=− a + b
τ
ξ 2 −1
=− a −b
τ
 1 
 2
τ 
Y (s) =
s ( s − s1 )( s − s 2 )
Y (s) =
1 A
B
C 
+
+

2 
τ  s ( s − s1 ) ( s − s 2 ) 
Y (s) =
1 A
B
C 
+
+

2 
τ  s ( s − s1 ) ( s − s 2 ) 
A( s 2 − ( s1 + s s ) s + s1 s 2 ) + Bs( s − s 2 ) + Cs ( s − s1 ) = 1
A+ B +C = 0
− A( s1 + s s ) − Bs 2 − Cs1 = 1
As1 s s = 1; A =
1
1
= B+C = −
s1 s 2
s1 s 2
1
s1
= Bs 2 + Cs 2 = −
= Cs1 + Cs 2 =
s1 + s 2 1 1
− =
s1 s 2
s1 s 2
=C=
1
1
1
1
=B=−
−
=−
s 2 ( s 2 − s1 )
s 2 ( s 2 − s1 ) s1 s 2
s1 ( s 2 − s1 )
Y (s) =
1  1

τ 2   s1 s 2
1
1 
1
1
1
. −
+
.
.

 s s1 (s 2 = s1 ) (s − s1 ) s 2 ( s 2 − s1 ) (s − s 2 ) 

1  1
1
1
−
e S1t +
e s2t 
2 
s2 ( s 2 − s1 )
τ  s1 s2 s1 ( s2 − s1 )

8.6 Y (t ) =
1
= 1
τ s1 s2
2
Y (t ) = 1 −
 1 S1t
1 S2t 
s e − s e 
τ ( s2 − s1 )  1
2

Y (t ) = 1 −
1
2
[
1
s2 e S1t − s1e S2t
( s2 − s1 )
Y (t ) = 1 +
τ
2 ξ −1
2
[s e
2
S1t
]
− s1e S2t
]
Y (t ) = 1 −
τje
−
ξt
τ
2 ξ −1
2
[(−a − jb)(cos bt + j sin bt) − (−a + jb)(cos bt − j sin bt]
Y (t ) = 1 −
τje
−
ξt
τ
2 ξ 2 −1
Y (t ) = 1 −
e
ξ
−
2
[− 2 jb(b cos bt + a sin bt )]
ξt
τ
[ 1−ξ
−1
2
cos(αt ) + ξ sin(αt )
]
[ 1−ξ ]
2
α =
ξ
[
]
 1−ξ 2 

 ξ

φ = tan −1 
verified
8.14 From the figure in your text Y(4) for the system response is
expressed
b) verify that for ξ = 1, and a step input
t

Y (t ) = 1 − 1 +
 τ
t
 −τ
e

Y ( s) =
Y ( s)
1
1
s τ 2 s 2 + τs + 1
A Bs + C
1
= +
2
s (τs + 1)
B (τ + 1) 2
2
A(τ 2 s 2 + 2τs + 1) + Bs Cs = 1
Aτ 2 + B = 0
2 Aτ + C = 0
A=1; B = τ 2 ; C = 2τ
Y ( s) =
1 τ (τs + 1) + τ
−
s
τ (s + 1)2
Y ( s) =
τ
τ
1
−
−
s (τs + 1) (τs + 1) 2
Y (t ) = 1 − e
−
t
τ
t
1 −
− te τ
τ
t
Y (t ) = 1 − (1 + )te
τ
−
t
τ
proved
c) for ξ > 1, prove that the step response is
Y (t ) = 1 − e
−
ξt
τ
[cosh(αt ) + β sinh(αt )]
ξ 2 −1
ξ
β = 2
α =
τ
ξ −1
Now Y ( s ) =
1/τ 2
s ( s − B1 )( s − s2 )
Where
s1 = −
ξ 2 −1
ξ
+
τ
τ
ξ 2 −1
ξ
s2 = − −
τ
τ
from 8.6(a)
Y (t ) =
1  1 1
1
1
1
1 
−
+

2 
τ  s1 s2 s s1 ( s2 − s1 ) ( s − s1 )
s2 ( s2 − s1 ) ( s − s2 ) 
Y (t ) = 1 −
[
1
s2 e S1t − s1e S2t
( s2 − s1 )

 − ξ − 1 − ξ 2
τ
Y (t ) = 1 +

τ
2 ξ 2 − 1 







e
Y (t ) = 1 +
− ξ e 
2 ξ 2 −1 

−
ξt
τ
 ξ 2 −1 


t
τ 

 −ξt 
 e τ e

1
ξ 2 −1 
τ
t


]
+ ξe
−
ξt
τ
 −ξ + 1−ξ 2
−

τ

− ξ 2 − 1e
−
ξ 2 −1 t
τ
 − ξt −
e τ e


− ξ 2 − 1e e
−
ξ 2 −1
τ
ξ 2 −1
t
τ


t







Y (t ) = 1 + e
−

ξ  eαt − e −αt

−
2
 ξ 2 − 1 
ξt
τ
Y (t ) = 1 − e
−
ξt
τ
  eαt + e −αt
 − 
2
 



[cosh(αt ) + β sinh(αt )]
8.7 Verify that for a unit step-input
 − πξ 

(1) overshoot = exp
 1−ξ 2 


 − 2πξ
(2) Decay ratio = exp
 1−ξ 2





For a unit step input the response (ξ<1):
 ξt 
− 
τ 

 1 − ξ 2 
t

Sin  1 − ξ 2 + tan −1 
2



τ
ξ
1−ξ



(1) we have to find time t where the maxima occurs
e
Y (t ) = 1 −
= dY/dt = 0
 ξt 
− 
τ 

 1−ξ 2
dY ) ξe
t
Sin  1 − ξ 2 + tan −1 
=

 ξ
dt τ 1 − ξ 2
τ


−
e
 ξt 
− 
τ 
τ

= tan 



 1 − ξ 2 
t
 = 0
Cos  1 − ξ 2 + tan −1 


τ
ξ  



2 

1−ξ 2
2 t
−1  1 − ξ  
=
+ tan
1−ξ

 ξ
τ
ξ


1−ξ 2t
ξ
= nπ
for maxima
=
1−ξ 2t
ξ
= 2 nπ




πt
= t=
1−ξ 2
8.8 Verify that for X(t) =A sin ωt, for a second order system,
A
Y (t ) =
(1 − (ωt ) ) + (2ξτ )
2 2
φ = − tan −1
sin(ωt + φ )
2
2ξωτ
1 − (ωτ ) 2
Y ( s) =
A
1
2
2 2
( s + ω ) (τ s + 2ξτs + 1)
Y ( s) =
Aω  A1
B1
C1
D1 
+
+
+
τ 2  ( s − jω ) s + jω ( s − s1 ) ( s − s2 ) 
2
Now as t → ∞, Y (t ) = A11 cos ωt + B11 sin ωt
Where A11 = A1 + B1
B11 = j( A1 − B1 )
to determine A1 , B1 put s = jω ,− jω in the order
A1 =
− j
j
B1 =
2ω ( jω − s1 )( jω − s2 )
2ω ( jω + s1 )( jω + s2 )
A11 =

j 
1
1
−

2ω  ( s1 + jω )( s2 + jω ) ( jω − s1 )( jω − s2 ) 
A11 =
j  ( −ω 2 − js1ω − js2ω + s1 s2 ) − ( −ω 2 + js1ω + js2ω + s1 s2 ) 


2
2ω 
( s1 + ω 2 )( s22 + ω 2 )





( s1 + s2 )
( s1 s2 − ω 2 )
11
similarly
A11 =  2
B
=


2
2
2
2
2
2
2 
 ( s1 + ω )( s2 + ω ) 
 ω ( s1 + ω )( s2 + ω ) 
using s1 + s2 =
2
2
= s1 + s2 =
− 2ξ
4ξ 2
τ2
τ
−
2
τ2
s1 s2 =
1
τ2
2(2ξ 2 − 1)
=
τ2


2ξ
−


A
ω
τ
A11 = 2 

2
τ  1 2ω
2
4
+ 2 ( 2ξ − 1) + ω
τ
 τ 4

− 2 Aωξ
=
τ3
2
 2 1   2ξω 
ω − 2  + 

τ   τ 



Aϖ 
11
and B =
τ   1
ω  
   τ 2

=
=
2
− 2 Aωξτ
(1 − (ωτ ) 2 ) 2 + ( 2ξωτ ) 2



2
2 
  2ξω   
−ω2  + 

  τ   
A(1 − (ωτ ) 2 )
(1 − (ωτ ) 2 ) 2 + ( 2ξωτ ) 2
 1
2
 2 −ω 
τ

Thus tan φ =
And, A New =
Thus, Y (t ) =
− 2ωτξ
A11
=
11
1 − (ωτ ) 2
B
A
((1 − (ωτ ) 2 ) 2 + ( 2ξωυ ) 2
A
((1 − (ωτ ) 2 ) 2 + ( 2ξωυ ) 2
(using
2
2
A11 + B11 = ANew
Sin(ωt + φ )
proved
8.9) If a second- order system is over damped, it is more difficult to
determine the parameters ξ & τ experimentally. One method for determining
the parameters from a step response has been suggested by R.c Olderboung
and H.Sartarius (The dynamics of Automatic controls,ASME,P7.8,1948),as
described below.
(a) Show that the unit step response for the over damped case may
be written in the form.
s (t ) = 1 −
r1e r2t − r2 e r1t
r1 r2
Where r1 and r2 are the roots of
τ 2 s 2 + 2ξτs + 1 = 0
(b) Show that s(t) has an inflection point at
ti =
ln(r2 / r1 )
(r2 − r1 )
© Show that the slope of the step response at the inflection point
d ( s)
dt
t − ti
= s 1 (t i )
Where, s 1 (t i ) = − r1e r1t = −r2 e r2ti
r1 ( r1 − r2 )
r 
= − r1  2 
 r1 
(d) Show that the value of step response at the inflection point is
rr
s (t )
1 1
s 1 (t i ) =1 + 1 2 s 1 (t i ) and that hence 1 − 1 i = − −
r1 r2
r1 r2
s (t i )
(e) on a typical sketch of a unit
step response show distances equal to
s (t i )
1
& 1
1
s (t i ) s (t i )
(f) Relate ξ & τ to r1 & r2
1−
1
(a) G ( s ) =
1
τ
=
τ s + 2ξτs + 1)  2  2ξ 
1
 s +   s + 2
τ  τ

1
2
2
2



=
τ2
( s − r1 )( s − r2 )
1
= Y (s) =
τ2
s ( s − r1 )( s − r2 )
1
A
B
C
= +
+
s ( s − r1 )( s − r2 ) s s − r1 s − r2
1 = A( s − r1 )( s − r2 ) + Bs ( s − r2 ) + cs( s − r1 )
Put s= 0 = Ar1r2 =1 ; A = τ 2
1
r1 (r1 − r2 )
1
s = r2 = Cr2 (r2 − r1 ) = 1; C =
r2 (r2 − r1 )
Put s = r1 = Br1 (r1 − r2 ) = 1; B =
Y (s) =

1 τ 2
1
1
 +

+
2 
τ  s r1 (r1 − r2 )( s − r1 ) r2 (r2 − r1 )( s − r2 ) 
Y (t ) =

e r1t
e r2t
1  2


τ
+
+
2 
r1 (r1 − r2 ) r2 (r2 − r1 ) 
τ 
 1
Y (t ) =1 − 
r1e r2t − r2 e r1t
−
r
r
 1 2
[
Y (t ) = 1 −
]

φr1e r t − r2 e r t
2
1
(r1 − r2 )
(b) For inflection point ,
d 2s
d 3s
=
0
&
=0
dt 2
dt 2
r r (e r2t − e r2t )
ds
=− 1 2
r1 − r2
dt
r1 r2 (r2 e r2t − r1e r2t )
d 2s
=
−
=0
r1 − r2
dt 2
= u 2 e r2t = r1e r1t =
r2
= e ( r1 −r2 ) ti
r1
r 
ln 2 
r
= ti =  1 
r1 − r2
(c )
ds (t )
dt
t = ti

r1 r2  r2

=−
r1 − r2  r1

= s ' (t i ) ]
r



r1 − r2
r
−  2
 r1



r1
r2 − r r1





r1 r2  r2

=−
r1 − r2  r1

r




r1 (r1 − r2 )  r2

= −
r1 − r2  r1

ds (t )
dt
Also
ds (t )
dt
ds (t )
dt
t = ti
t = ti
r
=−
t = ti




r1

 r2 − r r1 




r1 r2 (e r2t − e r1t )
=−
( r1 − r2 )
r1 r2 − e r1t
(r1 − r2 )
 r1

 − 1
 r2

= −r1e r1t = − r2 e r2t
(d) s (t i ) = 1 −
r1e
− r2 e
r1 − r2
r2t i
rt i1
r r 
s 1 (t i )  1 − 2 
 r2 r1 
=1+
r1 − r2
r r 
s 1 (t i )  1 − 2 
 r2 r1 
= s (t i ) = = 1 +
r1 − r2
Now



r1
r2 − r r1
r1

 r2 − r r1 





r
= −r1  2
 r1

ds (t )
dt
t = ti
 r2

−
r1− r2  r1
2
1
r1 − r2
1 1 
s (t i ) − 1 = = s 1 (t i )  + 
 r1 r2 
 1 1
1 − s (t i )
= = − − 
'
s (t )
 r1 r2 
r1 + r2 =
1
τ
2
− 2ξ
r1 + r2 =
τ
1
r1 r2 = ;τ =
=
τ
1
r1 r2
=; r1 + r2 = −2 r1r2ξ
r 
1  r1
+ 2 
2  r2
r1 


ξ =− 
proved.
8.10 Y(0),Y(0.6),Y( ∞ ) if
Y ( s) =
1 25( s + 1)
s ( s 2 + 2 s + 1)
1
1

Y ( s) =  1 +  2
2s
s s

( +
+ 1)
25 25
Y(s) impulse response + step response of G(s)
Where G ( s ) =
1
s
2s
( +
+ 1)
25 25
2
Y (t ) =
1
1−ξ 2 τ
e
−
ξt
τ
sin 1 − ξ 2
t
τ
+ tan −1
1−ξ 2
ξ
Y(t) = 1+5.0.3e-t sin (4.899t)-1.02e-t sin(4.899t+1.369)
Y(0)= 1-1=0
Y(0.6) = 1+0.561+0.515
Y( ∞ ) =1
Comment : as we can see ,the system exhibits an inverse response by
increasing from zero to more than 1 and as t tend to ∞ ,will reach the
steady state value of 1.
8.11 In the system shown the dev in flow to tank 1 is an impulse of
magnitude 5 . A1 = 1 ft2, A2 = A3 = 2 ft2 , R1 = 1 ft/cfm R2 = 1.5 ft/cfm .
(a) Determine H1(s), H2(s),
H3(s)
Transfer function for tank 1
H1 ( s)
1
=
Q( s)
(τ 1s + 1)
H 1 ( s) =
5
( s + 1)
from tank 2,
H 2 ( s)
R2
1.5
=
=
Q( s)
(τ 1 s + 1)(τ 2 s + 1) ( s + 1)(3s + 1)
for tank 3, q2 − qc = A3
dh3
dt
q3 = qc (const ) = Q2 = A3
dh3
dt
dh
H2
= A3 3
dt
R2
H 2 ( s ) H 3 ( s) 1
=
=
R2
H 2 ( s ) 3s
thus, A3 SH 3 ( s ) =
A3 SH 3 ( s ) =
H 2 ( s) H 3 ( s) 1
=
=
R2
H 2 ( s ) 3s
H 3 ( s)
0.5
=
Q ( s ) s( s + 1)(3s + 1)
8.11© H 1 ( s ) =
5
( s + 1)
H 1 (t ) = 5e − t
H 1 (3.46) = 0.155 A
H 2 ( s)
1.5
= 2
Q( s)
3s + 4 s + 1
Q (1) = 5 = H 2 ( s ) =
7.5
3s + 4 s + 1
2
τ= 3
2ξτ = 4
ξ=
4
4
=
= 1.155
2τ
2 3
from fig 8.5
ξ = 1.155 and
t
τ
= ξ=
t
τ
=
3.46
=2
3
τH 2 (t ) = 0.265 X 7.5
H 2 (t ) =
0.265 X 7.5
τ
= 1.147
H 3 ( s)
0.5
=
2
Q( s)
s(3s + 4 s + 1)
Q( s) = 5 = H 3 ( s) =
τ= 3ξ=
2
3
from fig 8.2 at
t
τ
= 2, ξ = 1.155
Y(t) =0.54
H3(t) =0.54*2.5 = 1.35
2.5
s (3s + 4 s + 1)
2
8.12 sketch the response Y(t) if
Y ( s) =
e −2 S
( s 2 + 1.2 s + 1)
Determine Y(t) for t = 0,1,5, ∞
e −2 S
e −2 S
1
e −2 S (0.8)
=
=
( s 2 + 1.2 s + 1) ( s + 0.6) 2 + (0.8) 2
0.8 ( s + 0.6) 2 + (0.8) 2
Y (t ) = 1.25e −.6( t −2 ) sin(0.8(t − 2)) t ≥ 2
Y ( s) =
for t = 0 Y (0) = 0
t = 1, Y (1) = 0
t = 5, Y (5) = 0.14
t = ∞, Y ( ∞ ) = 0
Problem 8.13 The system shown is at steady state at t = 0, with q = 10
cfm
A1 = 1ft2,A2=1.25ft2, R1= 1 ft/cfm, R2= 0.8 ft/cfm.
a) If flow changes fro 10 to 11 cfm, find H2(s).
b) Determine H2(1),H2(4),H2( ∞ )
c) Determine the initial levels h1(0),h2(0) in the tanks.
d) obtain an expression for H1(s) for unit step change.
Writing mass balances,
q−
(h1 − h2 ) = A
1
R1
dh1
( for tan k 1)
dt
At steady state q 2 −
(h1s − h2 s )
R1
= h1S − h2 S
Also for tank 2
(h1 − h2 ) − h2 = A dh2
2
R1
R2
dt
At steady state
(h1s − h2 s )
1
=
h2 S
= h2 S = 0.8 * 10 = 8
0.8
h1S = 18
C) h1S = 18 ft h2 (0) = 8 ft
The equations in terms of deviation variables
Q − Q1 = A1
H − H2
dH 1
where Q1 = 1
dt
R1
H
dH 2
Q2 = 2
dt
R2
H 2 ( s)
R2
0.8
=
= 2
2
Q( s ) τ 1τ 2 s + (τ 1 + τ 2 + A1 R2 ) s + 1 s + 2.8s + 1
Q1 − Q2 = A2
H 2 ( s) =
0.8
( Ans 8.31(a ))
s ( s + 2.8s + 1)
2
Step response of a second order system
τ 2 =1=τ =1
2ξτ = 2.8; ξ =
a)t = 1 =
b)t = 4 =
t
τ
t
τ
2.8
= 1.4
2
= 1; H 2 (t ) = 0.8(0.22) = 0.176 ft ( from fig )
= 4; H 2 (t ) = 0.8(0.78) = 0.624 ft ( from fig )
c)t → ∞ = H 2 (t → ∞) = 0.8 ft
Thus
H 2 (1) = 0.176 ft
H 2 (4) = 0.624 ft
H 2 (∞) = 0.8 ft
8.13(d) we have
Q( s )Q1 ( s ) = A1 ( s ) H 1 ( s )
Q1 ( s ) − Q2 ( s ) = A2 ( s ) H 2 ( s )
Q( s ) − Q2 ( s ) = A1 ( s ) H 1 ( s ) + A2 ( s ) H 2 ( s )
 1

+ A2 ( s ) H 2 ( s ) 
Q( s ) − A1 ( s ) H 1 ( s ) = 
 R2

1 + τ 2s 

 H 2 ( s ) 
=  

  R2 

 R (Q( s ) − A1 sH 1 ( s ) 

H 2 ( s ) =  2
Hτ 2 s


We have

H 2 ( s) 
R2
R
 = 2
= 
2
Q( s )  τ 1τ 2 s + (τ 1 + τ 2 + A1 R2 ) s + 1  Deg
R2 H 2 ( s )  R2 (Q( s ) − A1 sH 1 ( s ) 

= 
(1 + τ 2 s )
Deg


H (s) 
1 +τ 2s 
= 1 − A1 s − 1 
Deg
Q( s) 

H ( s)
Q( s)
H ( s)
Q( s )
H ( s)
Q( s)
=
1  Deg − 1 − τ 2 s 
1  Hτ 2 ( s ) 
 =
1 −

sA1 
Deg  A1 s 
Deg

=
1  τ 1τ 2 s 2 + (τ 1 + τ 2 + A1 R2 ) s + 1 − 1 − τ 2 s 


sA1 
Deg

=
1  τ 1τ 2 s + τ 1 + A1 R2 ) s 
 =

sA1 
Deg



R2 + R1 (1 + τ 2 s )

= 
2
Q( s )  τ 1τ 2 s + (τ 1 + τ 2 + A1 R2 ) s + 1 
H ( s)
8.14
Y ( s) =
(2 s + 4)
2
2
s ( 4 s + 0.8s + 1)
Y ( s) =
(s + 2)
4
2
s ( 4 s + 0.8s + 1)
2
1

Y ( s ) = 4 1 + 
2
s  ( 4 s + 0.8s + 1)

Y ( s) =
8
4
+
2
s ( 4 s + 0.8s + 1) ( 4 s + 0.8s + 1)
2
= (step response) + (impulse response)
Now,τ = 4 = 2 ; 2ξτ = 0.8
ξ = 0.2
also,
t
τ
=
4
=2
2
impulse response τY(t) = 4*0.63 = 2.52 (from figure)
step response = 8*1.15 = 9.2 (from figure)
Y(4) = 1.26+9.2
Y(4) =10.46
Q 9.1. Two tank heating process shown in fig. consist of two identical,
well stirred tank in series. A flow of heat can enter tank2. At t = 0 , the flow rate of heat
to tank2 suddenly increased according to a step function to 1000 Btu/min. and the temp
of inlet Ti drops from 60oF to 52oF according to a step function. These changes in heat
flow and inlet temp occurs simultaneously.
(a)
Develop a block diagram that relates the outlet temp of tank2 to inlet
temp of tank1 and flow rate to tank2.
(b)
Obtain an expression for T2’(s)
(c)
Determine T2(2) and T2(∞)
(d)
Sketch the response T2’(t) Vs t.
Initially Ti = T1 = T2 = 60oF and q=0
W = 250 lb/min
Hold up volume of each tank = 5 ft3
Density of the fluid = 50 lb/ft3
Heat Capacity = 1 Btu/lb (oF)
q
w
Ti
Solution:
(a) For tank 1
T1
w
T2
Input – output = accumulation
WC(Ti – To) - WC(T1 – To) = ρ C V
dT1
-------------------------- (1)
dt
At steady state
WC(Tis – To) - WC(T1s – To) = 0 ------------------------------------(2)
(1) – (2) gives
WC(Ti – Tis) - WC(T1 – T1s) = ρ C V
dT '1
dt
dT '1
WTi - WT1 = ρ V
dt
’
’
Taking Laplace transform
WTi(s) = WT1(s) + ρ V s T1(s)
T1 ( s )
1
, where τ = ρ V / W.
=
Ti ( s ) 1 + τs
From tank 2
q + WC(T1 – To) - WC(T2 – To) = ρ C V
dT2
-------------------------- (3)
dt
At steady state
qs + WC(T1s – To) - WC(T2s – To) = 0 ------------------------------------(4)
(3) – (4) gives
Q ‘ + WC(T1 – T1s) - WC(T2 – T2s) = ρ C V
Q ‘ + WCT1’ - WCT2’ = ρ C V
Taking Laplace transform
dT ' 2
dt
dT ' 2
dt
Q (s) + WC(T1(s) - T2(s)) = ρ C V s T2(s)
T2 ( s ) =
(b)
1  Q( s)

+ T1 ( s ) , where τ = ρ V / W.

1 + τs  WC

τ = 50*5/250 = 1 min
WC = 250*1 = 250
Ti(s) = -8/s and Q(s) = 1000/s
Now by using above two equations we relate T2 and Ti as below and after taking laplace
transform we will get T2(t)
T2 ( s ) =
1 Q( s)
1
+
Ti ( s )
1 + τs 250 (1 + τs )2
T2 ( s ) =
4
8
−
(1 + s ) (1 + s )2
1
1
1 
1  1

 − 8 −
−
T2 ( s ) = 4 −
2 
 s (1 + s )   s s + 1 (1 + s ) 
T2 (t ) = (4 + 8t )e −t − 4
(c)
T2’(2) = -1.29
T2(2) = T2’(2) + T2s = 60 – 1.29 = 58.71 oF
T2’(∞) = -4
T2(∞) = T2’(∞) + T2s = 60 – 4 = 56 oF
0.85
T2’(t)
0
0.5
t
-4
Q – 9.2. The two tank heating process shown in fig. consist of two identical , well stirred
tanks in series. At steady state Ta = Tb = 60oF. At t = 0 , temp of each stream changes
according to a step function
Ta’(t) = 10 u(t)
Tb’(t) = 20 u(t)
(a) Develop a block diagram that relates T2’ , the deviation in the temp of tank2,
to Ta’ and Tb’.
(b) Obtain an expression for T2’(s)
(c) Determine T2(2)
W1 = W2 = 250 lb/min
V1 = V2 = 10 ft3
ρ1 = ρ2 = 50 lb/ft3
C = 1 Btu/lb (oF)
W1
T1
Ta
W1
Tb
W2
W3=W1+W2
T2
Solution:
(a)
For tank1
T1 ( s)
1
, where τ1= ρ V / W1.
=
Ta( s) 1 + τ 1 s
For tank2
W1C(T1 – To) +W2C(Tb – To) – (W1+W2)C(T2 – To)= ρ C V
dT2
------ (1)
dt
At steady state
W1C(T1s – To) +W2C(Tbs – To) – (W1+W2)C(T2s – To)= 0 -----------------(2)
(1) – (2)
W1T1’ + W2Tb’- W3T2’ = ρ V
dT ' 2
dt
Taking L.T
W1T1(s) + W2Tb(s)- W3T2(s) = ρVs T2(s)
T2 ( s ) =
(b)
[
]
1 W1
T (S ) + W 2
T ( S ) where τ= ρ V / W3.
W3 1
W3 b
1 + τs
τ1 = 50*10/250 = 2 min
τ = 50*5/250 = 1 min
W1/W3 = 1/2 = W2/W3
Ta(s) = 10/s and Tb(s) = 0/s
Now by using above two equations we relate T1 and Ta as below and after taking laplace
transform we will get T2(t)
1 T (s) 1 T (s)
b
1
T2 ( s ) = 2
− 2
(1 + s )
(1 + s )
1 Tb ( s )
1 Ta ( s )
2
T2 ( s ) =
− 2
(1 + s )(1 + 2s ) (1 + s )
5
10
T2 ( s ) =
−
s (1 + s )(1 + 2s ) s (1 + s )
T2 ( s ) =
15 + 20 s
s (1 + s )(1 + 2s )
 15
5
20 

T2 ( s ) =  −
−
 s s + 1 (1 + 2s ) 
T2 (t ) = 15 − 5e −t − 10e
(c)
−t
2
T2’(2) = 10.64 oF
T2(2) = T2’(2) + T2s = 60 + 10.64 = 70.64 oF
Q – 9.3. Heat transfer equipment shown in fig. consist of tow tanks, one nested inside the
other. Heat is transferred by convection through the wall of inner tank.
1. Hold up volume of each tank is 1 ft3
2. The cross sectional area for heat transfer is 1 ft2
3. The over all heat transfer coefficient for the flow of heat between the tanks is 10
Btu/(hr)(ft2)(oF)
4. Heat capacity of fluid in each tank is 2 Btu/(lb)(oF)
5. Density of each fluid is 50 lb/ft3
Initially the temp of feed stream to the outer tank and the contents of the outer tank are
equal to 100 oF. Contents of inner tank are initially at 100 oF. the flow of heat to the inner
tank (Q) changed according to a step change from 0 to 500 Btu/hr.
(a) Obtain an expression for the laplace transform of the temperature of inner
tank T(s).
(b) Invert T(s) and obtain T for t= 0,5,10, ∞
10 lb/hr
Q
T1
T2
Solution:
(a)
For outer tank
WC(Ti – To) + hA (T1 – T2)- WC(T2 – To) = ρ C V2
dT2
-------------------------- (1)
dt
At steady state
WC(Tis – To) + hA (T1s – T2s)- WC(T2s – To) = 0 ------------------------------------ (2)
(1) – (2) gives
WCTi’ + hA (T1’ – T2’)- WCT2’ = ρ C V2
dT2 '
dt
Substituting numerical values
10 Ti’ + 10 ( T1’ – T2’) – 10 T2’ = 50
dT2 '
dt
Taking L.T.
Ti(s) + T1(s) – 2T2(s) = 5 s T2(s)
Now Ti(s) = 0, since there is no change in temp of feed stream to outer tank. Which gives
T2 ( s)
1
=
T1 ( s ) 2 + 5s
For inner tank
dT1
--------------------- (3)
dt
Q - hA (T1 – T2) = ρ C V1
Qs - hA (T1s – T2s) = 0 ------------------------------- (4)
(3) – (4) gives
Q’ - hA (T1’– T2’ ) = ρ C V1
dT1 '
dt
Taking L.T and putting numerical values
Q(s) – 10 T1(s) + 10 T2(s) = 50 s T1(s)
Q(s) = 500/s and
T2(s) = T1(s) / (2+ 5s)
10T1 ( s )
500
− 10T1 ( s ) +
= 50sT1 ( s )
s
2 + 5s
50
1


= T1 ( s ) 5s −
+ 1
s
2 + 5s 

T1 ( s ) =
50(2 + 5s )
s (25s 2 + 15s + 1)
T1 ( s ) =
2( 2 + 5 s )
s s + 3.82
s + 26.18
50
50
T1 ( s ) =
100
94.71
5.29
−
−
3
.
82
26
s
s+
s + .18
50
50
(
)(
(
'
T1 (t) = 100 - 94.71 e
)
) (
− 3.82 t
50
- 5.29 e
50
- 5.29 e
− 26.18t
50
and
T1 (t) = 200 - 94.71 e
For t=0,5,10 and ∞
T(0) = 100 oF
T(5) = 134.975 oF
T(10) = 155.856 oF
T(∞) = 200 oF
− 3.82 t
− 26.18 t
50
)
Q – 10.1. A pneumatic PI controller has an output pressure of 10 psi,
when the set point and pen point are together. The set point and pen point are suddenly
changed by 0.5 in (i.e. a step change in error is introduced) and the following data are
obtained.
Time,sec
Psig
0-
10
0+
8
20
7
60
5
90
3.5
Determine the actual gain (psig per inch displacement) and the integral time.
Soln:
e(s) = -0.5/s
for a PI controller
Y(s)/e(s) = Kc ( 1 + τI-1/s)
Y(s) = -0.5Kc ( 1/s + τI-1/s2)
Y(t) = -0.5Kc ( 1 + τI-1 t )
At t = 0+ y(t) = 8 Y(t) = 8 – 10 = -2
2=0.5Kc
Kc = 4 psig/in
At t=20 y(t) = 7 Y(t) = 7-10 = -3
3 = 2 ( 1 + τI-1 20 )
τI = 40 sec
Q-10.2. a unit-step change in error is introduced into a PID controller. If KC = 10 , τI = 1
and τD = 0.5. plot the response of the controller P(t)
Soln:
P(s)/e(s) = KC ( 1 + τD s+ 1/ τIs)
For a step change in error
P(s) = (10/s)(1 + 0.5 s + 1/s )
P(s) = 10/s + 5 + 10/s2
P(t) = 10 + 5 δ(t) + 10 t
10(1+t)
P(t)
15
10
t
Q – 10.3. An ideal PD controller has the transfer function
P/e = KC ( τD s + 1)
An actual PD controller has the transfer function
P/e = KC ( τD s + 1) / (( τD/β) s + 1)
Where β is a large constant in an industrial controller
If a unit-step change in error is introduced into a controller having the second
transfer function, show that
P(t) = KC ( 1 + A exp(-βt/ τD))
Where A is a function of β which you are to determine. For β = 5 and KC =0.5,
plot P(t) Vs t/ τD. As show that β ∞, show that the unit step response approaches that
for the ideal controller.
Soln:
P/e = KC ( τD s + 1) / (( τD/β) s + 1)
For a step change, e(s) = 1/s
P(s) = KC s( τD s + 1) / (( τD/β) s + 1)



 1 τ D 1 − 1 β  

= KC  + 
s
τ
s


1+ D


β
P(t)




 τ D 1 − 1 β  − βt 
 e τD 
= K C 1 + 
τ


D


β
− βt


= K C 1 + ( β − 1)e τ D 


So, A = β – 1
P(t) = 0.5 ( 1 + 4 exp(-5t/ τD))
2.5
P(t)
0.5
t/τD
As β ∞ then τD/β 0 and
P/e = KC ( τD s + 1) / (( τD/β) s + 1) becomes
P/e = KC ( τD s + 1) that of ideal PD controller
Q – 10.4. a PID controller is at steady state with an output pressure of a psig. The set
point and pen point are initially together. At time t=0, the set point is moved away from
the pen point at a rate of 0.5 in/min. the motion of the set point is in the direction of lower
readings. If the knob settings are
KC = 2 psig/in of pen travel
τI = 1.25 min
τD = 0.4 min
plot output pressure Vs time
Soln:
Given de/dt = -0.5 in/min
s e(s) = -0.5
Y(s)/e(s) = KC ( 1 + τD s+ 1/ τIs)
Y(s) = -( 1/s + 1/ τIs2 + τD )
Y(t) = -( 1 + t/1.25 + 0.4 δ(t) )
Y(t) = y(t) – 9 = - ( 1 + t/1.25 + 0.4 δ(t) )
y(t) = 8 – 0.8 t – 0.4 δ(t)
9
8
7.6
y(t)
t
10
Q – 10.5. The input (e) to a PI controller is shown in the fig. Plot the output of the
controller if KC = 2 and τI = 0.5 min
0.5
2
e
0
-0.5
1
3
4
t, min
e(t) = 0.5 ( u(t) - u(t-1) - u(t-2) + u(t-3) )
e(s) = (0.5/s) ( 1 – e-s - e-2s + e-3s )
P(s)/e(s) = KC ( 1 + (1 / τI s) ) = 2 ( 1+ 2/s )
P(s) = ( 1/s + 2/s2 ) (1 – e-s - e-2s + e-3s )
P(t)
= 1 + 2t
0≤t<1
=2
1≤t<2
= 5 – 2t
2≤t<3
=0
3≤t<∞
Q – 12.1. Determine the transfer function Y(s)/X(s) for the block diagrams shown.
Wxpress the results in terms of Ga, Gb and Gc
Soln.
(a)
Balances at each node
(1) = GaX
(2) = (1) – Y = GaX – Y
(3) = Gb(2) = Gb(GaX – Y)
(4) = (3) + X = Gb(GaX – Y) + X
Y = Gc(4) = Gc (Gb(GaX – Y) + X)
= GaGbGcX – GbGcY + GcX
Y Gc(GaGb + 1)
=
X
1 + GbGc
(b)
Balances at each node
(1) = X – (4)
(2) = Gb(1) = Gb( X – (4))
(5) = GcX/Ga
(3) = Gc(2) = GbGc( X – (4))
(4) = (3) + (5)
--------------------------- 5
= GbGc( X – (4)) + GcX/Ga
Y = Ga(4)
From the fifth equation
(4) = GbGcX – GbGc(4) + GcX/Ga ----------- 6
( 4) =
(GaGbGc + Gc) X
(1 + GbGc)Ga
From the sixth equation
Y (GaGb + 1)Gc
=
X
(1 + GbGc)
Q – 12.2
Find the transfer function y(s)/X(s) of the system shown
Soln:
Balance at each node
(1) = X – Y
---------(a)
(2) = (1) + (3)
----------(b)
(3) = G1(2) where G1 = 1/(τ1s + 1)
----------(c)
(4) Y = G2(3) where G2 = 0.5/(τ1s/2 + 1)
----------(d)
From (d) and (c)
Y = (2)G1G2
= G1G2 (X – Y + (3) )
----------(e)
Also from (b) and (c)
(3) = G1((1) + (3))
(3)(1 – 1/(τ1s + 1)) = 1/(τ1s + 1)
(3) τ1s = 1
(3) = 1/(τ1s ) = (X – Y) / (τ1s)
Substitute this in (e)
Y=
0.5
τ 1s
(τ 1 s + 1)(
2

1 
1 +
( X − Y )
τ
s


1
+ 1)
1
Y
= 2 2
X τ 1 s + 2τ 1 s + 1
Q – 12.3. For the control system shown determine the transfer function C(s)/R(s)
Soln.
Balances at each node
(1) = R – C
------------------(a)
(2) = 2 (1) = 2(R – C)
------------------(b)
(3) = (2) – (4) = 2(R – C) – (4)
-------------------(c)
(4) = (3)/s = (2(R – C) – (4))/s
-------------------(d)
(5) = (4) – C
-------------------(e)
C = 2(5)
-------------------(f)
Solving for (4) using (d)
s (4) = 2(R – C) – (4)
(4) = 2(R – C) / (s +1)
Using (e)
(6) = 2(R – C) / (s +1) – C
 2
(R − C ) − C 
C = 2

s +1
4 R = C (( s + 1) + 4 + 2( s + 1) )
C
4
=
R 3s + 7
Q – 12.4. Derive the transfer function Y/X for the control system shown
Soln.
Balance at each node
(1) = (5) + X
-----------------(a)
(2) = (1) – (4)
-----------------(b)
(3) = (2)/s
------------------(c)
Y = (3)/s
------------------(d)
(5) = 2 (3)
------------------(e)
(4) = 25Y
------------------(f)
From (b)
(4) = (1) – (2)
= (1) – s (3)
from (c)
= (1) – s2 Y
from (d)
= (5) + X - s2 Y
from (a)
= 2 (3) + X - s2 Y from (e)
= 2 s Y + X - s2 Y
From (f)
Y = (2 s Y + X - s2 Y)/25
X = Y( 25 – 2s + s2 )
Y
1
= 2
X s − 2 s + 25
13.1 The set point of the control system in fig P13.1 given a step change
of 0.1 unit. Determine
(a) The maximum value of C and the time at which it occurs.
(b) the offset
(c) the period of oscillation.
Draw a sketch of C(t) as a function of time.
5
C
( s + 1)( 2 s + 1)
=
5
R
1+ K
( s + 1)( 2 s + 1)
C
8
= 2
R 2 s + 3s + 9
R=
0.1
s
0.8
0.8
=
= 0.0889
S →0 2 s + 3s + 9
9
b) C ( ∞) = Lt
2
offset = 0.0111
c) K =
1
1
2
0.8
;2ξτ = ⇒ ξ =
;τ =
3
3
9
2 2

πξ
overshoot = exp −

1−ξ 2


 = 0.305


= Maximum vslue of C = 1.0305*0.0889=0.116
Maximum value of C = 0.116
0.116 =
t=
ξt

 1−ξ 2
−
1
0.8 
t
1−
e τ sin  1 − ξ 2 + tan −1 
 ξ
τ
9 

1−ξ 2


τ
1−ξ 2
tan
−1
1−ξ 2
ξ
= 1.6
Time at which Cmax occurs = 1.6
(c ) Period of ociullation is T =
2πτ
1−ξ 2
T =3.166
Decay ratio = (overshoot)2 = 0.093
= 3.166
 

 
 
13.2 The control system shown in fig P 13.2 contains three-mode controller.
(a) For the closed loop, develop formulas for the natural period of oscillation
τ and the damping factor ξ in terms of the parameters K, τ D , τ I and τ 1 .
(b) Calculate ξ when K is 0.5 and when K is 2.
(c ) Do ξ & τ approach limiting values as K increases, and if so, what are
these values?
(d ) Determine the offset for a unit step change in load if K is 2.
(e ) Sktech the response curve (C vs t) for a unit-step change IN LOAD
WHEN k is 0.5 and when K is 2.
(f) In both cases of part (e) determine the max value of C and the time at
which it occurs.
a)
C
=
R

1
1 

k 1 + τ D s +
τ 1s + 1 
τ I s 
1+
k 
1 
1 + τ D s +

τ 1s + 1 
τ I s 
1
(τ 1 s + 1)
C
=
U
1  
1  

1+
k 1 + τ D s +

τ 1s + 1  
τ I s  
τIs
=
k
τ Iτ I  2  k + 1 

s + 
τ I τ D +
τ I s + 1
k 
 k 


1 

k 1 + τ D s +
τ I s 
C

=
R

1 

τ 1 s + 1 + k 1 + τ D s +
τ I s 

(
)
k τ Dτ I s 2 + τ I s + 1
C
=
R (kτ I τ D + τ 1τ I ) s 2 + (k + 1)τ I s + k
τ2 =
τ I (kτ D + τ I )
=2×
=ξ =
=T =
k
;2τξ =
τ I (kτ D + τ I )
k
ξ=
(k + 1)τ I
k
(k + 1)τ I
k
τI
(k + 1)
2
k (kτ D + τ 1 )
2π ×
2τπ
1− ξ
2
=
τ I (kτ D + τ 1 )
k
4k (kτ D + τ 1 ) − (k + 1) 2 τ D
2 k (kτ D + τ 1 )
T=
4π (kτ D + τ 1 )
 τ
4k  k  D
 τI
  τ1
 + 
 τ I

  − (k + 1) 2


B) τ D = τ I =1; τ 1 .=2
For k = 0.5 ; ξ =0.75
For k = 2 ; ξ =1.5
1
= 0.671
0.5(2.5)
1
= 0.530
2×3
2
 1
1 +  τ I
2
1 (k + 1) τ I
1  k
C) = ξ =
=
τ 
2 k (kτ D + τ 1 ) 2 
τ D + I 
k 

As k → ∞, ξ =
4π (kτ D + τ 1 )
T=
 τ
4k  k  D
 τI
  τ1
 + 
 τ I

  − (k + 1) 2


kτ I τ D + τ 1τ I
k
τ=
τ = τ Iτ D +
τ
1 τI
= 0.3535
2 τ1
K →∞
τ Iτ 1
k
= τ I τ D =1
As k → ∞, T =
4πτ D
τ
4 D −1
τI
= 7.2552
1
τ 1s + 1
C
(d)
=
U
1 
k 
1 + τ D s +

1+
τ 1s + 1 
τ I s 
τIs
C
=
U (k + 1)τ I s + (τ 1 + kτ D )τ I s 2 + k
U=
C=
1
s
τI
(k + 1)τ I s + (τ 1 + kτ D )τ I s 2 + k
Lt sf ( s ) =
S →0
0
=0
k
K=2
For a unit step change in U
C (∞) = 0
Offset = 0
(e) k = 0.5, ξ =0.671 & τ =2.236
T=
2πτ
1− ξ 2
= 18.95
If k = 0.5
C
2s
s
;C= 2
= 2
U 5s + 3s + 1
5s + 3s + 1
If k = 2
C
0.5s
0.5
;C = 2
= 2
U 2 s + 1. 5 s + 1
2s + 1.5s + 1
In general C (t ) =
1
1
τ 1− ξ
2
e
The maximum occurs at t =
−
ξt
τ
sin 1 − ξ 2
τ
1− ξ 2
If k = 0.5 tmax = 2.52 Cmax=0.42
If k = 2 tmax = 1.69 Cmax=0.19
tan −1
t
τ
1− ξ 2
ξ
13.3 The location of a load change in a control loop may affect the system
response. In the block diagram shown in fig P 13.3, a unit step chsange
in load enters at either location 1 and location 2.
(a) What is the frequency of the transient response when the load enters at
location Z?
(b) What is the offset when the load enters at 1 & when it enters at 2?
(c) Sketch the transient response to a step change in U1 and to a step
change in U2.
1
U 1 = ;U 2 = 0
s
 2  1 
( S ( R − C ) + U 1 )
=C

 2 s + 1  2 s + 1 
R=0
2
C
2
(2 s + 1) 2
=
= 2
2
U1
4 s + 4 s + 11
5×
2
( 2 s + 1) + 1
2
C
= 2
U 1 4 s + 4 s + 11
K=
4
1
2
2
;2ξτ = ⇒ ξ =
;τ =
11
11
11
11
1
1
Frequency = f = =
T 2π
1−ξ 2
τ
=
1
2π
C(∞) = 2/11
Offset = 2/11 =0.182
U1=0;U2=1/s
2

 1 
⇒5×
( R − C ) + U 2 
=C
 2s + 1
 2 s + 1 
R=0
10
= 0.2516
2
1
C
2s + 1
⇒
=
10
1
U2
×
+1
2s + 1 2 s + 1
C
2s + 1
= 2
U 2 4 s + 4 s + 11
C(∞) = 1/11
Offset = 1/11=0.091
1
a) if U 1 = ; frequency = 0.2516
s
1
if U 2 = ; frequency = 0.2516
s
1
b)if U 1 = ; frequency = 0.182
s
1
if U 2 = ; frequency = 0.091
s
13.5A PD controller is used in a control system having first order process and a
measurement lag as shown in Fig P13.5.
(a) Find the expressions for ξ and τ for the closed –loop response.
(b) If τ1 = 1 min, τm = 10 sec, find KC so that
ξ = 0.7 for the two cases: (1)
τD =0,(2) τD =3 sec,
(c) Compare the offset and period realized for both cases, and comment on the
advantage of adding derivative mode.
K C (1 + τ D s )
(τ 1 s + 1)
C
a) =
K
R
C (1 + τ D s )
1+
(τ 1 s + 1)(τ m s + 1)
K C (1 + τ D s )(1 + τ m s )
C
=
2
R τ 1τ m s + (τ 1 + τ m + K Cτ D ) s + ( K C + 1)
= τ2 =
τ 1τ m
KC + 1
τ 1τ m
τ=
KC + 1
ξ=
1 τ 1 + τ m + k Cτ D
2 τ 1τ m (k C + 1)
b) τ 1 = 1 min; τ m = 10s;ξ = 0.7
τD = 0
1)
1
70
⇒ 0.7 = ×
2
600(k C + 1)
kc=3.167
τ D = 3s
2) ⇒ 0.7 =
70 + 3k C
1
×
2
600(k C + 1)
k C = 5.255
k
1
c)for R = ; c(∞) = C
s
kC + 1
For τ D = 0; C (∞) = 0.76; offset = 0.24
for τ D = 3s; C (∞) = 0.84; offset = 0.16
600
4.167 = 105.57 = period
1− ξ 2
2π ×
for τ D = 0; period =
600
6.255 = 86.17 s = period
for τ D = 3s; period =
1 − ( 0.7) 2
2π
Comments:
Advantage of adding derivative mode is lesser offset lesser period
13.6The thermal system shown in fig P 13.6 is controlled by PD controller.
Data ; w = 250 lb/min; ρ = 62.5 lb/ft3;
V1 = 4 ft3,V2=5 ft3; V3=6ft3;
C = 1 Btu/(lb)(°F)
Change of 1 psi from the controller changes the flow rate of heat of by 500
Btu/min. the temperature of the inlet stream may vary. There is no lag in the
measuring element.
(a) Draw a block diagram of the control system with the appropriate transfer
function in each block.Each transfer function should contain a numerical values of
the parameters.
(b) From the block diagram, determine the overall transfer function relating the
temperature in tank 3 to a change in set point.
(c ) Find the offset for a unit steo change in inlet temperature if the controller
gain KC is 3psi/°F of temperature error and the derivative time is 0.5 min.
WT0C + q = ρCV1 (T1 − T0 ) + WT1C
WT1C = ρCV2 (T2 − T1 ) + WT2 C
WT2 C = ρCV3 (T3 − T2 ) + WT3C
T0 (WC + ρCV1 ) + q = T1 (W C + ρCV1 )
T1 = T0 +
q
WC + ρCV1
T1= T2 = T3
T3 = T0 +
q
q( s )
⇒ T3 ( s ) =
(WC + ρCV1 )s
WC + ρCV1
13.6 (b)
2
T ( s)
( s + 1)(1.25s + 1)(1.5s + 1)
=
2
R( s )
1 + k C (1 + τ D s)
( s + 1)(1.25s + 1)(1.5s + 1)
'
3
=
k C (1 + τ D s)
T3' ( s)
2k C (1 + τ D s)
=
2
R( s) ( s + 1)(1.875s + 2.75s + 1) + 2k C (1 + τ D s)
T3' ( s)
2k C (1 + τ D s)
=
3
R( s) 1.875s + 4.625s 2 + (3.75 + 2k Cτ D ) s + 2k c + 1
c) kC=3; τ D = 0.5, offset = ?,τ 0' ( s) =
1
s
T3' ( s)
1
Lt
'
3
2
s
→
0
Ti ( s) 1.875s + 4.625s + (3.75 + 2k Cτ D ) s + 2k C + 1
=
1
2k C + 1
=
1
= 0.143
7
Offset =0.143
13.7 (a) For the control system shown in fig P 13.7, obtain the closed loop
transfer function C/U.
(b) Find the value of KC for whgich thre closed loop response has a ξ of
2.3.
(c) find the offset for a unit-step change in U if KC = 4.
s +1

1
(R − C) + U  = C
 KC ×
0.25s + 1

s
C
=
=
U
1
s
1+
KC
s +1
.
s 0.25s + 1
C
0.25s + 1
=
2
U 0.25s + s + K C ( s + 1)
C
s+4
= 2
U s + 4( K C + 1) s + 4 K C
b) ξ=2.3
τ=
=
=
K +1
1
;2ξτ = C
KC
4K C
K +1
1
2 × 2. 3 = C
KC
4KC
KC + 1
= 2.3
KC
KC=2.952
C) KC=4,U = 1/s
1
s+4
=C= × 2
s s + 20s + 16
1
4
offset = 0.25.
C (∞ ) =
13.8 For control system shown in Fig 13.8
(a) C(s)/R(s)
(b) C(∞)
(c) Offset
(d) C(0.5)
(e) Whether the closed loop response is oscillatory.
4
C
s( s + 1)
(a) =
4
R
1+
s( s + 1)
C
4
= 2
R s +s+4
b) C(∞) =2*1=2
C(∞) =2
C) offset = 0
1
1
1
d) τ = ;2ξτ = ⇒ ξ =
2
4
4
C (t )
= 1−
2
1
1
1−  
4
2
e
−
t
2
 15 t

+ tan −1 15 
sin 
 4 τ

1

−
 15

4
+ tan −1 15  
e 4 sin 
= C (0.5) = 2 1 −
15

 4
 
C(0.5)=0.786
.e) ξ<1, the response is oscillatory.
13.9 For the control system shown in fig P13.9,determine an expression for
C(t)
if a unit step change occurs in R. Sketch the response C(t) and compute
C(2).
C
=
R
1+
1
s
 1
1 + 1 + 
 s
C
s +1
=
R 2 s + 1`
R=
1
s
C=
1
s +1
−1
= +
s (2 s + 1) s 2 s + 1
t
1 −
C (t ) = 1 − e 2
2
C(2) = 0.816
13.10 Compare the responses to a unit-step change in a set point for the
system shown in fig P13.10 for both negative feedback and positive feedback.Do
this for KC of 0.5 and 1.0. compare the responses by sketching C(t).
-ve feed back :
C=
KC
s( s + ( K C + 1))
+ve feed back
C
=
R
1
s +1
1
1 − KC
s +1
C=
KC
s ( s + (1 − K C ))
KC ×
For KC = 0.5 , response of -ve feed back is
1
2
−
1
3
C=
= 3+
s ( 2 s + 3) s 2 s + 3
3t
3t
−
1 1 −
1
C (t ) = − e 2 = (1 − e 2 )
3 3
3
response of +v feed back is
C=
1
1
−2
= +
s( 2 s + 1) s 2 s + 1
C (t ) = 1 − e
−
t
2
For KC = 1, response of -ve feed back is
1
1
−
1
C=
= 2+ 2
s ( s + 2) s s + 2
1 1
C (t ) = − e −2 t
2 2
response of +ve feed back is
C=
1
s2
C (t ) = t
14.1 Write the characteristics equation and construct Routh array for
the control system shown . it is stable for (1) Kc= 9.5,(ii) KC =11; (iii) Kc=
12
Characteristics equation
6 Kc
=0
( s + 1)( s + 2)( s + 3)
or ( s + 1)(s + 2)(s + 3) + 6 Kc = 0
1+
( s 2 + 6 s + 11s + (6 + 6 Kc) = 0
s 3 + 6 s 2 + 11s + (6 + 6 Kc) = 0
Routh array
s3
s
s
2
1
11
6
(6 + 6 Kc)
6(1 + Kc)
For Kc=9.5
= 10-(Kc)= 10-9.5=0.5>0 therefore stable.
For Kc=11
= 10-(Kc)= 10-11=-1<0 therefore unstable
For Kc=121
= 10-(Kc)= 10-12=--2<0 therefore unstable
14.2 By means of the routh test, determine the stability of the system shown
when KC = 2.
Characteristic equation
10

 3 
1 + 21 + 2 2
=0
 s   2 s + 4s + 10 
(2 s 2 + 4 s + 10) s + 2( s + 3).2.10 = 0
2s 3 + 4 s 2 + 10s + 40 s + 120 = 0
2s 3 + 4 s 2 + 50s + 120 = 0
s 3 + 2s 2 + 25s + 60 = 0
Routh Array
1
2
-10/2
25
60
The system is unstable at Kc = 2.
14.4 Prove that if one or more of the co-efficient (a0,a1,….an) of the
characteristic equation are negative or zero, then there is necessarily an
unstable root
Characteristic equation :
a 0 x n + a1 x n −1 + .................................. + a n = 0
a 0 ( x n + a1 / a 0 x n −1 + .......................a n / a 0 ) = 0
a 0 ( x − α 1 )( x − α 2 )....................( x − α n ) = 0
We have α 1 , α 2 .............................α n < 0
As we know the second co-efficient a1/a0 is sum of all the roots
 n n

a1
= (−1) 2 ∑ ∑ α iα j  / 2
a0
 i =1 j =1

Therefore sum of all possible products of two roots will happen twice as
α 1α 2 dividing the total by 2.
α iα j > 0 (α i < 0 α j < 0)
And
∴ a 2 / a0 > 0 ⇒ a 2 > 0
Similarly
aj
a0
= (−1) j ( sum of aoll possible products of j roots)
if j = even (−1) j is 1 and the sum is > 0 so a j / a 0 > 0
if j = odd (−1) j is (−1) and the sum is < 0 so
aj
a0
is again > 0
in both case a j / a 0 > 0
so a j > 0( for j = 1,.......n)
14.5 Prove that the converse statement of the problem 14.4 that an
unstable root implies that one or more co-efficient will be negative or
zero is untrue for all co-efficient ,n>2.
Let the converse be true, always .Never if we give a counter example we
can contradict.
Routh array
s 3 + s 2 + 2s + 3
s3 1 2
s2 1 3
s −1 0
s0
System is unstable even when all the coefficient are greater than 0; hence a
contradiction,
14.6 Deduce an expression for Routh criterion that will detect the
Presence of roots with real parts greater than σ for any rectified σ >0
Characteristic equation
a 0 x n + a1 x n −1 + ................................. + a n = 0
Routh criteria determines if for any root, real part > 0.
Now if we replace x by X such that
.x + σ =X
Characteristic equation becomes
a 0 ( X − σ ) n + a1 ( X − σ ) n −1 + ................................. + a n = 0
Hence if we apply Routh criteria,
We will actually be looking for roots with real part > σ rather than >0
a 0 x n + a1 x n −1 + a 2 x n −2 ................................. + a n = 0
Routh criterion detects if any root α j is greater than zero.
Is there any x = α 1 , α 2 ,...............,α j ,..........α n > 0 − − − − − (1)
Now we want to detect any root
α j > −σ
α j> 0
From(1)
x = α 1 , α 2 ,............................α j ,..........................α n , > 0
implies is there any
x = α1 > 0
x = α2 > 0
.
.
.
x =αj >0
.
.
.
x = αn > 0
add σ on both sides
is there any
x + σ = α1 + σ > 0
x +σ = α2 +σ > 0
.
.
.
.x + σ = α j + σ > 0
.
.
.
x +σ = αn +σ > 0
so, Let X = x + σ
and apply Routh criteria to det ect any α j + σ > 0
or α j > −σ
14.7 Show that any complex no S1 satisfying S < 1, yields a value of
Z=
1+ s
that satisfies Re(Z)>0,
1− s
Let S=x+iy,
x2 + y2 < 1
Z=
1+ s
1− s
(1 + x ) + iy (1 − x) + iy
(1 − x) − iy (1 − x) + iy
(1 − x 2 + (1 = x + 1 − x)iy − y 2 )
1 + x 2 − 2x + y 2
=
1 − ( x 2 + y 2 ) + 2iy
1 − 2x + (x 2 + y 2 )
Re(Z ) =
1 − (x2 + y 2 )
1 − 2x + (x 2 + y 2 )
if Re( z ) > 0 then 1 − ( x 2 + y 2 ) > 0 and
1 − 2x + (x 2 + y 2 ) > 0
we have x 2 + y 2
<1
x2 + y2 < 1
Ranges are − 1 < x < 1
−1 < y < 1
Po int s in the unit circle
1 − ( x 2 + y 2 ) > 0 is true therefore x 2 + y 2 < 1
Now
1 + (x 2 + y 2 ) − 2x
if x = −1& y = 0 then it is 4
if x = 1 & y = 0 then it is 0
0 < (1 + ( x 2 + y 2 ) − 2 x) < 4
Re( z ) > 0
example:
if s = (0.5 + i0.5) the system is unstable due to the real part


1
L−1 

 s − (0.5 + i 0.5) 


1
L−1 
= e 0.5t (Cos 0.5t + Sin 0.5t )

 s − (0.5 + i0.5) 
14.8 For the output C to be stable, we analyze the characteristic
equation of the system
1+
1
1
× (τ 3 s + 1) = 0
τ I s (τ 1 s + 1)(τ 2 s + 1)
τ I s(τ 1 s 2τ 2 + τ 1 s + τ 2 s + 1) + τ 3 s + 1 = 0
τ I τ 1τ 2 s 3 + τ I (τ 1 + τ 2 ) s 2 + (τ I + τ 3 ) s + 1 = 0
Routh Array
s3
τ I τ 1τ 2
s2
τ I (τ 1 + τ 2 )
α
s
s0
α =
τ I + τ3
1
0
1
τ I (τ 1 + τ 2 )(τ I + τ 3 ) − τ I τ 1τ 2
τ I (τ 1 + τ 2 )
Now
(1) τ I τ 2τ 1 > 0
Since τ 1 & τ 2
τ 1 > 0;τ 2 > 0
are process time constant they are definitely +ve
(2)
τ I (τ 1 + τ 2 ) > 0
(3) α > 0 ⇒ τ I (τ 1 + τ 2 )(τ I + τ 3 ) > τ I τ 1τ 2
τ 1τ I + τ 1τ 3 + τ 2τ I + τ 2τ 3 − τ 1τ 2 > 0
τ I (τ 1 + τ 2 ) > τ 1τ 2 − τ 3 (τ 1 + τ 2 )
τI >
τ 1τ 2
−τ 3
τ1 +τ 2
also τ I > 0
14.9 In the control system shown in fig find the value of Kc for which
the system is on the verge of the instability. The controller is replaced by a
PD controller, for which the transfer function is Kc(1+s). if Kc = 10,
determine the range for which the system is stable.
Characteristics equation
1+
6 Kc
=0
( s + 1)( s + 2)( s + 3)
or ( s + 1)(s + 2)(s + 3) + 6 Kc = 0
( s 2 + 3s + 2)(s + 3) + 6 Kc = 0
s 3 + 6 s 2 + 11s + (6 + 6 Kc) = 0
Routh array
s3
1 3
s2
3 1 + Kc
s
 1 + Kc 
3−
)
 3 
For verge of instability
 1 + Kc 
)
3=
 3 
Kc = 8
Characteristics equation
1 +
10(1 + kcs)
=0
( s + 1)3
s 3 + 3s 2 + s (3 + 10 Kcs) + 11 = 0
Routh Array
s3
1
3 + 10τ D
s2 3
11
s
3(3 + 10τ s ) > 11 for vege
30τ S > 2
τ D > 2 / 30
14.10 (a) Write the characteristics equation for the central system shown
(b) Use the routh criteria to determine if the system is stable for Kc=4
© Determine the ultimate value of Kc for which the system is unstable
(a) characteristics equation
 s + 2  1  1
1 + kc
=0


 3  2s = 1  ( s + 1)
( s 2 + s )(2 s + 1) + kc( s + 2) = 0
2s 3 + 3s 2 + (1 + kc) s + 2kc = 0
s 3 + 3s 2 + 3s + (1 + kc) = 0
Kc=4Routh array
s3 2
s
s
2
5
3 8
− 1/ 3
not stable
3(1 + kc) − 4kc
=0
3
3 − Kc = 0; Kc = 3
For verge of instability
14.11
for the control shown, the characteristics equation is
4
3
s + 4 s + 6 s 2 + 4 s + (1 + k ) = 0
(a) determine value of k above which the system is unstable.
(b) Determine the value of k for which the two of the roots are on
the imaginary axis, and determine the values of these imaginary roots and
remaining roots are real.
s 4 + 4 s 3 + 6 s 2 + 4s + (1 + k ) = 0
s4 1
6(1 + k )
s3 4
4
s
2
s
1
1+ k
4
4 − (1 + k )
5
1+ k
5
For the system to be unstable
 1 + k 
41 − 
  < 0
  5 
1<
1+ k
5
k>4
1+ k < 0
k < −1
k > −1
The system is stable at -1<k<4
(b) For two imaginary roots
4=
4
(1 + k ); k = 4
5
Value of complex roots
5s 2 + 5 = 0
s = ±i
s 2 + 1 s 4 + 4 s 3 + 6s 2 + 4s + 5
s 2 + 4s + 5
s4 + 0 + s2
4s 3 + 5s 2
4s 3 + 0 + 4 s
5s 2 + 5
5s 2 + 5
0
SOLUTION:
s=
− 4 ± 16 − 20
− 4 ± 2i
=
= −2 ± i
2
2
PART 2
LIST OF USEFUL BOOKS FOR PROCESS
CONTROL
1.
PROCESS CONTROL BY R.P VYAS, CENTRAL TECHNO
PUBLICATIONS, INDIA ( WIDE VARIETY OF SOLVED PROBLEMS ARE
AVAILABLE IN THIS BOOK)
2. ADVANCED CONTROL ENGINEERING BY RONALD.S .BURNS ,
BUTTERWORTH AND HIENEMANN.
3. PROCESS MODELLING SIMULATION AND CONTROL FOR
CHEMICAL ENGINEERS, WILLIAM.L.LUYBEN, MCGRAW HILL.
4.
A MATHEMATICAL INTRODUCTION TO CONTROL THEORY BY
SCHOLOMO ENGELBERG, IMPERIAL COLLEGE PRESS
LIST OF USEFUL WEBSITES
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als,Discussion Group,Video Lectures
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***********************