Transcription

SOLUTIONS MANUAL FOR SELECTEDPROBLEMS INPROCESS SYSTEMS ANALYSIS ANDCONTROLDONALD R. COUGHANOWRCOMPILED BYM.N. GOPINATH BTech.,(Chem)CATCH ME AT [email protected]: This work is just a compilation from various sources believed to bereliable and I am not responsible for any errors.

CONTENTSPART 1: SOLUTIONS FOR SELECTED PROBLEMSPART2:LIST OF USEFUL BOOKSPART3:USEFUL WEBSITES

PART 11.1 Draw a block diagram for the control system generated when a humanbeing 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) 02dtdt 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/ss 2 X ( s ) sx(0) x ' (0) s X ( s ) x (0) X ( s ) ( s 2 s 1) X ( s ) X ( s) 1s1s1s( s s 1)2Now, applying partial fractions splitting, we getX ( s) X ( s) 1s 1 2s ( s s 1)32s 11 1 2 2222s 2 3 1 3 1 3 s s 2 2 2 2 L 1 ( X ( s )) 1 eX (t ) 1 eb)1 t21 t21Cos31 2t3t e sint223 3 3 1 Cos t Sin 2 t 23 dx 2dx 2 x 1 x ( 0) x ' ( 0) 02dtdtwhen the initial conditions are zero, the transformed equation is( s 2 s 1) X ( s ) 1s

X ( s) 1s( s s 1)1ABs C 2s ( s s 1) s s 2 s 121 A(s2 2s 1) Bs2 Cs0 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 0B 1C 2 AA 1, B 1, C 21s 2 2s s 2s 1 1 (s 1) 1 L 1{ X ( s )} L 1 (s 1)2 sX ( s) 11 { X (t )} 1 L 1 2 s 1 (s 1) { X (t )} 1 e t (1 t )dx 2dx3.1 C 3 x 1 x ( 0) x ' ( 0) 02dtdtby Applying laplace transforms, we get ( s 2 3s 1) X ( s ) X ( s) 1s( s 3s 1)21s2

X ( s) ABs C 2s s 3s 11 A( s 2 3s 1) Bs 2 Cs0 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 0B 1C 3 A 3A 1, B 1, C 3s 3 1L 1{ X ( s )} L 1 2 s s 3s 1 1 1 1L { X ( s )} L s s 1L 1{ X ( s )} L 1 s s X (t ) 1 e 3t2(Coss 323 2 2 5 2 3 2 3 .s 2 5 2 2223 3 5 s 2 2 2 5t35 tsinh2253.2(a)dx 4 d 3 x 3 Cos t; x (0) x ' (0) x ''' (0) 04dtdtx11 (0) 1 2 5 2 52

Applying Laplace transforms, we gets 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) ss 12ss 12 s 1) ( s 1) s 4 s 3X ( s ) ( 2 s 1 3s s s s 2 1 s 3 s 2 2s 1 3 2 3 2s ( s 1)( s 1)s ( s 1)( s 1)s 3 s 2 2s 1 A B CDEs F 2 3 232s ( s 1)( s 1) s sss 1 s 1s 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 1A 1-B-C -1D F 0E F 0D E 1D-E 02D 1A -1; B 1; C 1D 1/2; E 1/2; F -1/2

1 1 1 1 / 2 1 / 2( s 1) L 1{( s)} L 1 2 3 ss 1s2 1 s s1 1 / 2 1 / 2( s 1) 1 1L 1 {X ( s )} L 1 2 3 sss 1s2 1 s2{X (t )} 1 t t 1 e t 1 Cos t 1 S int2 222d 2 q dq t 2 2t q(0) 4; q1 (0) 2dt 2 dtapplying laplace transforms,we gets 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)3sQ( s) ( s 2 s) 2 s 2 4 s 4 2s 3s 4 ( s 1)22*3 1 4Q ( s ) 4 s 1 s( s 1) s ( s 1)1L 1 (Q ( s )) q(t ) 4e t 2(1 e t ) t 33therefore q(t ) 2 t3 2 e t322 23ss

3s3s 11 2 22( s 1)( s 4) 3 s 1 s 4 3.3 a)21 1 2 2 2s 2 2 s 11 1L 1 2 2 2 Cost Cos 2ts 2 2 s 1b)11AB C 222s ( s 2 s 5) s ( s 1) 2s s 2s 52[]A B 0-2A C 05A 1A 1/5 ;B -1/5;C 2/5We getX ( s) 1 12 s 2 5 s s 2 s 5 Inverting,we get 1 1 t t122 eSint eCost 5 2 1 1 1 e t Sin 2t Cos 2t 5 2 3s 2 s 2 3s 2 A BCD 2 c)22s ( s 1)s ss 1 ( s 1) 2

As( s 1) 2 B( s 1) 2 Cs 2 ( s 1) Ds 2 3s 3 s 2 3s 2A( s 3 2 s s ) B ( s 2 2 s 1) C ( s 3 s 2 ) Ds 2 3s 3 s 2 3s 2A C 3-2A B-C D -1A-2B -3B 2;A 2(2)-3 1C 3-1 2D 2(1)-2 2-1 1We get X ( s ) 1 221 2 s ss 1 ( s 1) 2By inverse L.TL 1 [X (t )] 1 2t 2e t te tL 1 [X (t )] 1 2t e t ( 2 t )3.4 Expand the following function by partial fraction expansion. Do not evaluateco-efficient or invert expressionsX ( s) 2( s 1)( s 1) 2 ( s 3)X ( s) ABs CDs EF 2 2 2s 1 s 1 ( s 1)s 32 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 2A B F 0-3A C 4B F 02A B 4C 3B 06A C 4B 3C 0A 4C 3B 3D 4E F 03A 3C 3E F 2by solving above 6 equations, we can get the values of A,B,C,D,E and1.X ( s) 3s ( s 1)( s 1) ( s 3) 3X ( s) A B CDEFGH 2 3 2s sss 1s 2 s 3 ( s 3)( s 3) 3by comparing powers of s we can evaluate A,B,C,D,E,F,G and H.1c) X ( s ) s ( s 2)( s 3) ( s 4)ABCD s 1 s 2s 3s 4by comparing powers of s we can evaluate A,B,C,DX ( s) 3.5 a) X ( s ) Let1s ( s 1)(0.5s 1)1ABC s ( s 1)(0.5s 1) s s 1 (0.5s 1) s 2 3s s2 A 1 B s C ( s 2 s ) 12 2 2 A 1A BB1 C 0 C 2 2223A3 B C 0 B C 22

B/2 1/2 *-3/2 -1;B -2;C -3/2 2 1/2X ( s) 121 1 s s 1 2 0.5s 1 L 1 ( X ( s )) x ( t ) 1 2 e t e 2 tb)dx 2 x 2; x (0) 0dtApplying laplace trafsormssX ( s ) x (0) 2 X ( s ) 2 / sL 1 ( X ( s )) 2s ( 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 t1 e 3.6 a) Y ( s ) s 1s 2s 52

Y ( s) s 1s 2s 52s 1( s 1) 2 4 s 1 L 1 (Y ( s )) L 1 2 ( s 1) 4 using the table,we getY (t ) e t Cos2tb) Y ( s ) Y ( s) s 2 2ss412 32ssY(t) L 1 (Y ( s )) t t 2c) Y ( s ) 2s( s 1) 32s 2 2( s 1) 322 2( s 1)( s 1) 3 2 2 Y (t ) L 1 L 1 2 3 ( s 1) ( s 1) t2 t2(te e e t (t 2 2t )2t

3.7a)Y ( s) As B Cs D1 ( 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 ) 1C 0,D 0Also A 0;B 111ABCDY ( s) 2 2222( s 1)(s i) (s i)(s i) (s i)(s i) (s i)2A( 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 ) 1Thus,A C 0-Ai B Ci 2Di 0 ; B DA-2Bi C 2Di 0-Ai-B Ci-D 1Also D -Ci;B -Ci,A -C,C -i/4A i/4 ; B -1/4; D -1/4Y ( s) i/4 1/ 4 i/4 1/ 4 2(s i) (s i)(s i) (s i)2Y (t ) Y (t ) 1/ 4 i/4 1/ 4i/4 2(s i) (s i)( s i) ( s i) 2 1/ 4 i/4 1/ 4i/4 2(s i) (s i)(s i) (s i)2

Y (t ) i / 4e it 1 / 4e it 1 / 4e it 1 / 4te itY (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 ( Sin3.8f ( s) f ( s) t t Cost)1s ( s 1)2A BC 2ss s 1 A( s 1) Bs( s 1) Cs 2 1Let s 0 ; A 1s 1; 2A B C 1s -1: C 1B -11 11f ( s) 2 ss s 1f (t ) (t 1) e tPROPERTIES OF TRANSFORMS4.1 If a forcing function f(t) has the laplace transformsf ( s) 1 e s e 2 s e 3s ss2s1 e 3se s e 2 s ss2

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 ) dτ0 dy (t )y ( 0) 1dtTaking Laplace transforms on both sidest dy (t ) L{ y (τ ) dt} L dt 01. y ( s ) s. y ( s ) y (0)s

1. y ( s ) s. y ( s ) 1sy ( s) ss 12 s y (t ) L 1{ y ( s )} L 1 2 cosh(t ) s 1 4.3 Express the function given in figure given below thes-t – domain and thedomainThis 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 3se 5 s e 6 s 2 2 2 2ssssss e s e 3s e 2 s e 6 s e 3s e 5 s ss2

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 transformf ( S ) (1 2e s e 2 s ) / s 2obtain the function f(t) and graph f(t)1 2 e s e 2 sf ( s) s2 1 e s e s e 2 s s2s2f (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 1.5f (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 0At t 3f (3) 0.5 0.5 (3 3) 0RESPONSE OF A FIRST ORDER SYSTEMS5.1 A thermometer having a time constant of 0.2 min is placed in atemperature bath and after the thermometer comes to equilibrium withthe bath, the temperature of the bath is increased linearly with time at therate of I deg C / min what is the difference betweenthe indicatedtemperature and bath temperature(a) 0.1 min(b) 10. minafter the change in temperature begins. what is the maximum deviation between the indicated temperaturewand bath temperature and when does it occurs.(d) plot the forcing function and the response on the same graph. After thelong enough time buy how many minutes does the response lag the input.Consider thermometer to be in equilibrium withtemperature XsX (t ) X S (1 / m )t , t 0as it is given that the temperture varies linearlyX(t)-Xs tLet X(t) X(t) - Xs ttemperaturebath at

Y(s) G(s).X(s)Y ( s) 1 1AB C 221 τs s 1 τs s sA τ 2 B τ C 1τ2τ 1Y ( s) 21 τs s sY ( t ) τe t / τ τ t(a) the difference between the indicated temperature and bath temperatureat 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 minX(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/ 0as t tend to infinitiveD τ (-e-t/T ( -1/T) τ 0.2 deg C5.2 A mercury thermometer bulb in ½ in . long by 1/8 in diameter. Theglass envelope is very thin. Calculate the time constant in water flowingat 10 ft / sec at a temperature of 100 deg F. In your solution , give asummary which includes(a) Assumptions used.(b) Source of data(c) Results

T mCp/hA ( ρAL)C ph ( A πDL)Calculation ofNU d Re d Pr hD CRem (Pr) nKDvρµCpµK (1 / 8 * 2.54 * 10 2 )(10 * 0.3048)103 9677.410 3 4.2 KJ / KgKSource data: Recently, Z hukauskas has given c,m ,ξ,n values.For Re 967704C 0.26 & m 0.6NuD hD/K 0.193 (9677.4)*(6.774X10-3) 130.h 253805.3 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) Versust/T2. Show the numerical values of minimum, maximum and ultimate valuesthat may occur during the transient. Check these using the initial valueand final value theorems of chapter 4.Y ( s) T1s 1T2 s 1X(s) unit step function 1 X(s) 1/s

Y ( s) T1 s 1AB s (T2 s 1)s iT2 sA 1 B T1 - T2Y ( s) 1 T1 T2 s 1 T2 sY (t ) 1 T1 T2 t / T2eT2If T1/T2 s thenY (t ) 1 4e t / T2Let t/T2 x then Y (t ) 1 4e xUsing the initial value theorem and final value theoremLim Y (T ) Lim sY ( s )S T 01T s 1s T1 5Lim 1 LimS T s 1S 1T22T2 sT1 Lim Y (T ) Lim sY ( s ) LimT 0Figure:S S 0T1 s 1 1T2 s 1

5.4 A thermometer having first order dynamics with a time constant of 1min is placed in a temperature bath at 100 deg F. After the thermometerreaches steady state, it is suddenly placed in bath at 100 deg F at t 0 andleft there for 1 min after which it is immediately returned to the bath at100 deg F.(a) draw a sketch showing the variation of the thermometer reading withtime.(b) calculate the thermometer reading at t 0.5 min and at t 2.0 min1Y ( s)(τ 1 min) X ( s) s 1 1 e s ( s ) 10 s s

1 e s Y ( s ) 10 s e s 1Y ( s ) 10 s ( s 1) s ( s 1) Y (t ) 10(1 e t ) t 1Y (t ) 10( (1 e t ) (1 e ( t 1) ) ) t 1At t 0.5 T 103.93At 2 T 102.3255.5 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 secT 110 10e t / 60

0 t 10 sec & T 60 secT (t 10 sec) 101.535T 100 1.535e ( t 10 ) / 60 t 10 secT(t 30sec) 101.099 deg FT(t 120sec) 100.245 deg F5.6 A mercury thermometer which has been on a table for some time,isregistering the room temperature ,758 deg F. Suddenly, it is placed in a 400deg F oil bath. The following data are obtained for response of thethermometerTime (sec)Temperature, Deg F07511072.514052058244102821532830385Give two independent estimates of the thermometer time constant.

τ t 325 ln 400 T From