SOLUTIONS MANUAL FOR SELECTED SOLUTIONS MANUAL FOR SELECTED PROBLEMS IN PROBLEMS IN PROCESS SYSTEMS ANALYSIS AND CONTROL DONALD R. COUGHANOWR COMPILED BY M.N. GOPINATH BTech.,(Chem) M.N. GOPINATH BTech.,(Chem) CATCH ME AT [email protected] 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) 1 )0( )0( 0 ' 2 2 + + x = x = x = dt dx dt dx ( ) )0( )0( 2 ' 2 2 s X s sx x dt dx L = − − s X (s) x )0( dt dx L = − L(x) = X(s)
L{1} = 1/s ( ) − )0( − )0( + 2 ' s X s sx x s s X s x X s 1 ( ) − )0( + ( ) = s s s X s 1 ( )1 ( ) 2 = + + = ( )1 1 ( ) 2 + + = s s s X s Now, applying partial fractions splitting, we get ( )1 1 1 ( ) 2 + + + = − s s s s X s 2 2 2 2 2 3 2 1 2 3 3 2 2 1 2 3 2 1 1 1 ( ) + + − + + + = − s s s s X s L X s e Cos t e t t t 2 3 sin 3 1 2 3 ( ( )) 1 2 1 2 1 1 − − − = − − + = − − X t e Cos t Sin t t 2 3 3 1 2 3 ( ) 1 2 1 b) 2 1 )0( )0( 0 ' 2 2 + + x = x = x = dt dx dt dx when the initial conditions are zero, the transformed equation is s s s X s 1 ( )1 ( ) 2 + + =
( )1 1 ( ) 2 + + = s s s X s ( )1 2 1 1 2 2 + + + = + + + s s Bs C s A s s s = A s + s + +Bs +Cs 2 2 1 ( 2 )1 0 2 ( ) 0 ( ) 2 A C by equating the co effecients of s A B by equating the co effecient of s = + − = + − ,1 ,1 2 2 1 0 1 ( ) = = − = − = − = − + = = − A B C C A B A B A by equating the co effecients of const 2 1 1 2 ( ) 2 + + + = − s s s s X s ( ) ( ) + + + = − − − 2 1 1 1 1 1 1 { ( )} s s s L X s L ( ) + + + = − − 2 1 1 1 1 1 { ( )} 1 s s X t L {X (t)} 1 e 1( t) t = − + − 3.1 C 3 1 )0( )0( 0 ' 2 2 + + x = x = x = dt dx dt dx by Applying laplace transforms, we get s s s X s 1 ( 3 )1 ( ) 2 = + + = ( 3 )1 1 ( ) 2 + + = s s s X s
3 1 ( ) 2 + + + = + s s Bs C s A X s = A s + s + + Bs + Cs 2 2 1 ( 3 )1 0 3 ( ) 0 ( ) 2 A C by equating the co effecients of s A B by equating the co effecient of s = + − = + − ,1 ,1 3 3 3 1 0 1 ( ) = = − = − = − = − = − + = = − A B C C A B A B A by equating the co effecients of const + + + = − − − 3 1 1 3 { ( )} 2 1 1 s s s s L X s L − + + = − − − 2 2 1 1 2 5 2 3 1 3 { ( )} s s s L X s L − + − − + + = − − − 2 2 2 2 1 1 2 5 2 3 2 5 5 2 . 2 3 2 5 2 3 2 3 1 { ( )} s s s s L X s L t t X t e Cos t 2 5 sinh 5 3 2 5 ( ) 1 ( 2 3 = − + − 3.2(a) )0( 1 ; )0( )0( )0( 0 11 ' ''' 3 3 4 4 = + = = = = x Cost x x x dt d x dt dx
Applying Laplace transforms, we get 1 ( ) )0( )0( )0( )0( ( ) )0( )0( )0( 2 4 3 2 1 '' ''' 3 2 ' '' + − − − − + − − − = s s s X s s x s x sx x s X s s x sx x 1 ( () ) ( )1 2 4 3 + + − + = s s X s s s s 4 3 2 )1 ( )1 1 ( ) ( s s s s s X s + + + + + = = ( 1)( )1 2 1 ( 1)( )1 1 3 2 3 2 3 2 3 2 + + + + + = + + + + + + s s s s s s s s s s s s s ( 1)( )1 1 1 2 1 3 2 2 3 2 3 2 + + + + = + + + + + + + + s Es F s D s C s B s A s s s s s s 2 1 ( 1)( )1 ( 1)( )1 ( 1)( )1 ( )1 ( ) ( )1 3 2 2 2 2 2 3 2 3 s + s + s + = As s + s + + Bs s + s + + c s + s + + Ds s + + Es + F s s + 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 (2/1 )1 1 1 1 1 2/1 ( ) 2 3 2 1 1 s s s s s s L s L { } + − + + + + + − = − − 1 (2/1 )1 1 1 1 1 2/1 ( ) 2 3 2 1 1 s s s s s s L X s L { } int 2 1 2 1 2 1 2 ( ) 1 2 e Cost S t X t t t = − + + + + − − 2 )0( ;4 )0( 2 2 1 2 2 + = t + t q = q = − dt dq dt d q applying laplace transforms,we get 3 2 2 ' 2 2 ( ) )0( )0( (( ) )0( s s s Q s − sq − q + sQ s − q = + + − + − = +1 2 1 ( )( ) 4 2 4 2 2 s s Q s s s s ( ) 4( )2 (2 )1 ( ) 2 3 s s s s s Q s + + + + = = ( )1 2 2 4 2 4 4 3 + + + + s s s s s ( )1 2 * 3 ( )1 2 1 1 ( ) 4 4 + + + + + = s s s s s Q s 1 3 3 1 L (Q(s)) q(t) 4e 1(2 e ) t t t = = + − + − − − therefore t e t q t − = + + 2 3 ( ) 2 3
3.3 a) + − + = + + 4 1 1 1 3 3 ( 1)( )4 3 2 2 2 2 s s s s s s + − + = 2 2 2 2 2 1 1 1 s s Cost Cos t s s L 2 2 1 1 1 2 2 2 2 1 = − + − + − b) [ ] ( )1 2 2 5 1 ( 2 )5 1 2 2 2 2 − + + = + − + = − + s s B C s A s s s s s A+B=0 -2A+C=0 5A=1 A=1/5 ;B=-1/5;C=2/5 We get − + − = + 2 5 1 2 5 1 ( ) 2 s s s s X s Inverting,we get = + e Sin t − e Cos t t t 2 2 2 1 1 5 1 = + e Sin t − Cos t t 2 2 2 1 1 5 1 c) 2 2 2 2 2 2 ( )1 1 ( )1 3 3 2 − + − = + + − − − + s D s C s B s A s s s s s
( )1 ( )1 ( )1 3 3 2 2 2 2 2 3 2 As s − + B s − + Cs s − + Ds = s − s − s + ( 2 ) ( 2 )1 ( ) 3 3 2 3 2 3 2 2 3 2 A s − s + s + B s − s + + C s − s + Ds = s − s − s + 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 2 2 ( )1 1 1 1 2 2 ( ) − + + = + + s s s s X s By inverse L.T [ ] t t L X t = + t + e + te − ( ) 1 2 2 1 [ ( )] 1 2 2( ) 1 L X t t e t t = + + + − 3.4 Expand the following function by partial fraction expansion. Do not evaluate co-efficient or invert expressions ( 1)( )1 ( )3 2 ( ) 2 2 + + + = s s s X s 1 1 ( )1 3 ( ) 2 2 2 + + + + + + + + + = s F s Ds E s Bs C s A X s 2 2 2 2 2 = A(s + )1 (s + )3 + (Bs + C)(s +1)(s + 3)(s + )1 + (Ds + E)(s + 1)(s + )3 + F(s + 1)(s + )1 ( 2 1)( )3 ( )( 4 3)( )1 ( )( 4 )3 ( 1)( 4 )1 4 2 2 2 2 2 = A s + s + s + + Bs +C s + s + s + + Ds + E s + s + + F s + s + s +
4 ) 3 3 3 2 ( ) 3( 4 ) 2( 4 3 ) 6( 4 3 ) ( 4 3 5 4 3 2 + + + + + + = = + + + + + + + + + + + + + + + + + E F A AC E F s A B F s A C B F s A B C B s A C B C s A C B 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 3 3 ( 1)( )1 ( )3 1 ( ) + + + = s s s s X s . 2 3 2 3 1 2 3 ( )3 ( )3 ( ) + + + + + + + + + = + + + s H s G s F s E s D s C s B s A X s by comparing powers of s we can evaluate A,B,C,D,E,F,G and H. c) ( 2)( )3 ( )4 1 ( ) + + + = s s s s X s 1 2 3 4 ( ) + + + + + + + = s D s C s B s A X s by comparing powers of s we can evaluate A,B,C,D 3.5 a) ( 1)( 5.0 )1 1 ( ) + + = s s s X s ( 1)( 5.0 )1 1 5.0( )1 1 + + + = + + + s C s B s A s s s Let ( ) 1 2 1 2 3 2 2 2 2 + + = + + = + + s C s s s B s s A A=1 2 1 2 0 2 2 + + = = + C = − B C A B 2 3 0 2 3 + B + C = = B + C = − A
B/2=1/2 *-3/2=-1; B=-2; C= -3/2+2=1/2 + + + = − 5.0 1 1 2 1 1 1 2 ( ) s s s X s t t L X s x t e e 1 2 ( ( )) ( ) 1 2 − − − = = = − + b) + 2x = ;2 x )0( = 0 dt dx Applying laplace trafsorms sX (s) − x )0( + 2X (s) = /2 s ( )2 2 ( ( )) 1 + = − s s L X s + = − − ( )2 2 ( ( )) 2 1 1 s s L X s L = + = − − − 2 2/1 2/1 ( ( )) 2 1 1 s s L X s L = t e 2 1 − − 3.6 a) 2 5 1 ( ) 2 + + + = s s s Y s
= 2 5 1 ( ) 2 + + + = s s s Y s ( )1 4 1 2 + + + = s s = + + + = − − ( )1 4 1 ( ( )) 2 1 1 s s L Y s L using the table,we get Y t e Cos t t ( ) 2 − = b) 4 2 2 ( ) s s s Y s + = 2 3 1 2 ( ) s s Y s = + Y(t)= 1 2 L (Y (s)) = t + t − c) 3 ( )1 2 ( ) − = s s Y s = 3 ( )1 2 2 2 − − + s s 2 3 ( )1 2 ( )1 2 − + − = s s − + − = − − 3 1 2 1 ( )1 2 ( )1 2 ( ) s L s Y t L = ( 2 ) 2 (2 2 2 e e t t t tet t t + = +
3.7a) ( )1 ( )1 ( )1 1 ( ) 2 2 2 + + + + + = + = s Cs D s As B s Y s ( ) ( )( )1 1 2 thus As + B + Cs + D s + = = 1 ( ) ( ) 3 2 Cs + Ds + A + C s + B + D = C=0,D=0 Also A=0;B=1 2 2 2 2 2 2 ( ) ( ) ( ) ( ) ( ) ( ) 1 ( )1 1 ( ) s i D s i C s i B s i A s s i s i Y s − + − + + + + = + − = + = ( )( ) ( ) ( )( ) ( ) 1 2 2 2 2 A s + i s − i + B s − i + C s − i s + i + D s + i = ( ) ( ) ( 2 2 ) ( ) 1 3 2 A + C s + −Ai + B + Ci + D s + A − Bi + C + Di + −Ai − B + Ci − D = 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 2 2 ( ) 4/1 ( ) 4/ ( ) 4/1 ( ) 4/ ( ) s i s i i s i s i i Y s − − + − − + + − + + = 2 2 ( ) 4/1 ( ) 4/ ( ) 4/1 ( ) 4/ ( ) s i s i i s i s i i Y t − − + − − + + − + + = 2 2 ( ) 4/1 ( ) 4/ ( ) 4/1 ( ) 4/ ( ) s i s i i s i s i i Y t − − + − − + + − + + =
it it it it Y (t) = i 4/ e − 4/1 e − 4/1 e − 4/1 te − − ( ) (4/1 ) it it it it Y t = ie − te − ie − te − − Y (t) = i((4/1 Cost − iSin t) − t(Cost − i Sin t) − i(Cost + iSin t) − t(Cost + i Sin t)) Y (t) = 2(4/1 Sin t − 2tCost) Y ( t ) = 1 / 2 ( Sin t − t Cos t ) 3.8 ( )1 1 ( ) 2 + = s s f s = 1 ( ) 2 + = + + s C s B s A f s ( )1 ( )1 1 2 = A s + + Bs s + + Cs = Let s=0 ; A=1 s=1; 2A+B+C=1 s=-1: C=1 B=-1 1 1 1 1 ( ) 2 + = + + s s s f s t f t t e − ( ) = ( − )1 + PROPERTIES OF TRANSFORMS 4.1 If a forcing function f(t) has the laplace transforms s e s e e s f s s s 3s 2 2 1 ( ) − − − − − = + 2 3 2 1 s e e s e − s −s − s − + − =
( ) { ( )} [ ( ) ( 3)] [( )1 ( )1 ( )2 ( 2)] 1 = = − − + − − − − − − f t L f s u t u t t u t t u t = 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): )0( 1 ( ) ( ) 0 = = ∫ y dt dy t y d t τ τ Taking Laplace transforms on both sides = ∫ dt dy t L y dt L t ( ) { ( ) } 0 τ . ( ) . ( ) )0( 1 y s s y s y s = −
. ( ) . ( ) 1 1 y s = s y s − s 1 ( ) 2 − = s s y s cosh( ) 1 ( ) { ( )} 2 1 1 t s s y t L y s L − = = − − 4.3 Express the function given in figure given below the t – domain and the sdomain 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 2 6 2 3 5 2 3 2 2 ( ) { ( )} s e s e s e s e s e s e f s L f t −s − s − s − s − s − s = = + − − − + 2 3 2 6 3 5 s e e e e s e e −s − s − s − s − s − s + − − + − =
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 2 2 f (S) 1( 2e e /) s −s − s = − + obtain the function f(t) and graph f(t) 2 2 1 2 ( ) s e e f s −s − s − + = 2 2 2 1 s e e s e −s −s − s − − − = ( ) { ( )} ( )1 ( )1 ( ) {( )1 ( )1 ( )2 ( 2)] 1 = = − − − + − − − − − − − f t L f s t u t tu t t u t t u t = 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) = 5.0 u(t) − 5.0 u(t − )1 + (t − )3 u(t − )2
At t = 1.5 f (t) = 5.0 u(t) − 5.0 u(t − )1 + (t − )3 u(t − )2 )1 f )5.1( = 5.0 u(t) − 5.0 u(t − f )5.1( = 5.0 − 5.0 = 0 At t = 3 f )3( = 5.0 − 5.0 + 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 bath at temperature Xs X (t) = X + 1( ° / m ,) tt > 0 S as it is given that the temperture varies linearly X(t)-Xs = t Let X(t) = X(t) - Xs = t
Y(s) = G(s).X(s) 2 2 1 1 1 1 ( ) s C s B s A s s Y s + + + = + = τ τ A = 1 2 τ B = −τ C = 2 2 1 1 ( ) s s s Y s − + + = τ τ τ Y t e t t = − + − τ τ /τ ( ) (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 = ( ) ( ) h A DL AL Cp π ρ + Calculation of m n d CRe K hD NU = = (Pr) 9677 4. 10 8/1( * .2 54 *10 )(10* .0 3048)10 Re 3 2 3 = = = − − µ Dvρ d KJ KgK K Cp Pr = = 2.4 / µ 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. 1 1 ( ) 2 1 + + = T s T s Y s X(s) =unit step function = 1 X(s) = 1/s
iT s B s A s T s T s Y s 2 2 1 ( )1 1 ( ) = + + + = A = 1 B = T1 - T2 T s T T s Y s 2 1 2 1 1 ( ) + − = + 2 / 2 1 2 ( ) 1 t T e T T T Y t − − = + If T1/T2 = s then 2 / ( ) 1 4 t T Y t e − = + Let t/T2 = x then x Y t e − ( ) = 1+ 4 Using the initial value theorem and final value theorem ( ) ( ) 0 LimY T Lim sY s S T → →∞ = = 5 1 1 1 1 2 1 2 1 2 1 = = + + = + + →∞ →∞ T T s T s T Lim T s T s Lim S S ( ) ( ) 0 LimY T Lim sY s S T → →∞ = = 1 1 1 2 1 0 = + + → T s T s Lim S Figure:
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 min) 1 1 ( ) ( ) = + = τ X s s Y s = − − s e s s s 1 ( ) 10
− = − s e Y s s 1 ( ) 10 + − + = − ( )1 ( )1 1 ( ) 10 s s e s s Y s s ( ) = 10 1( − ) < 1 − Y t e t t ( ) 10( 1( ) 1( ) ) 1 ( )1 = − − − ≥ − − − Y t e e t t t 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 /60 110 10 t T e − = −
0 < t < 10 sec & T = 60 sec T (t = 10sec) =101.535 sec 100 .1 535 10 ( 10 /) 60 = + > − − T e t t 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.
− = T t 400 325 ln τ 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. τ τ ω ω τ 1 1 ( ) 1 1 ( ) 2 2 + + = + = s s A s s Y s splitting into partial fractions then converting to laplace transforms sin( ) 1 1 ( ) 2 2 / 2 2 ω φ τ ω τ ω αωτ τ + + + + = − t A e A Y t t where φ = tan-1 (ωτ) As t →∝ ) − − + + = + = φ π ω τ ω ω φ τ ω 2 cos( 1 sin( ) 1 ( ) 2 2 2 2 t A t A Y t s Y t = A t + = A −ωt π ω φ 2 ( ) sin( ) cos = − 2 ( ) cos π Y t A ωt 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 /τ 1 1 400 325 t T e − = − 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 hA mC τ = = 10sec oil τ sec = 50 air τ / 50 1 2 75 ( 75) t F T T e − = + − TF = Final Temp = 98degC /10 10/50 98 75 (325 325 ) − 1 − = + − e e t .0 91356 /10 = −t e 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
( ) 1( ) t /τ Y t A e − = − )2.1( 50 1( ) 50 1/2.1 = − + − Y e 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 )5.1( = 88.843°C Max temperature indicated = 88.843 deg C AT t = 20 min 88.843 13.843 1( ) −18 1/8. T = − − e 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 2 ( )1 1 ( ) + = s Y s 2 ( )1 1 ( ) ( ) ( ) + = = X s s Y s G s For determining unit step response 2 ( )1 1 ( ) + = s Y s
2 2 ( )1 1 ( )1 1 ( ) + + + = + + = s C s B s A s Y s A = 1 B = -1 C = -1 2 ( )1 1 1 1 1 ( ) + − + = − s s s Y s t t Y t e te − − ( ) = 1− − 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 dt dh q q A 1 − 1 = 1 where h1 = height of the liquid level in tank 1 similarly balance on the tank 2 gives dt dh q q A 2 1 − 2 = 2 and balance on tank 3 gives
dt dh q2 − q0 = A3 here 1 1 1 R h q = 2 2 2 R h q = q = b 0 So we get dt dh A R h q 1 1 1 1 − = dt dh A R h R h 2 2 2 2 1 1 − = dt dh b A R h 3 2 2 − = writing the steady state equation 0 1 1 1 1 − = = dt dh A R h q s S S dt dh A R h R h S S 2S 2 2 2 1 1 − = 0 2 2 − b = R h S Subtracting and writing in terms of deviation dt dH A R H Q 1 1 1 − =
dt dH A R H R H 2 1 2 2 1 1 − = dt dH A R H 3 2 2 = where Q = q –qS H1= h1-h1S H1= h2-h2S H = h - hS Taking Laplace transforms ( ) ( ) ( ) 1 1 1 1 A s H s R H s Q s − = ---------(1) ( ) ( ) ( ) 2 2 2 2 1 1 A s H s R H s R H s − = --------(2) ( ) ( ) 3 2 2 A s H s R H s = ----------(3) 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) R A s H s H s 1 3 2 2 ( ) ( ) = -------------(4)
( )1 ( ) ( ) 1 2 2 1 2 + = R s R H s H s τ where 2=R2 A2 τ ------------(5) From (1) ( )1 ( ) ( ) 1 1 1 + = s R Q s H s τ , 1 = R1 A1 τ ---------(6) Combining equation 4,5,6 ( )( 1)( )1 ( ) ( ) 3 1 + 2 + = A s s s Q s H s τ τ ( )( 1)( )1 1 ( ) ( ) 3 1 + 2 + = Q s A s s s H s τ τ 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 between Tf and T. what is the effect of bulb resistance and capacitance on the thermometer response? Note that the inclusion of the bulb results in a pair of interacting systems, which give an overall transfer function different from that of Eq (7.24) Writing the energy balance for change in term of a bulb and mercury respectively Input - output = accumulation dt dT h A T T h A T T m C b f − b − i i b − = b b ( ) ( ) 0 0 dt dT hi Ai (Tb −T )−0 = m C Writing the steady state equation 0 0 ( − )− ( − ) = = 0 dt dT h A T T h A T T m C bs fs bs i i bs s b b ( − ) = 0 i Ai Tbs Ts h
Where subscript s denoted values at steady subtracting and writing these equations in terms of deviation variables. dt dT h A T T h A T T m C b f − b − i i b − m = b b ( ) ( ) 0 0 dt dT h A T T m C m i i ( b − m )−0 = Here TF = Tf - TfS TB = Tb - TbS Tm = T - TS Taking laplace transforms ( ( ) ( )) ( ) ( ) 0 0 h A T s T s h A T T m C T s F − B − i i B − m = b b B ----(1) And ) h A (T (s) T (s)) mC sT (s i i B − m = B ------(2) = ( ( ) ( )) ( ) ( ) 0 0 h A T s T s mCST s m C sT s F − B − m = b b B From (2) we get ( ) ( ) 1 = ( () + )1 = s + T s s h A mC T s T s m i i i B m τ Where i i i h A mC τ = Putting it into (1) ( ) ( ) ( 1))( )1 0 0 0 0 = − + + + s h A mC T s T s s s F m i τ τ
= = + + + s h A mC T s T s s s F m i 0 0 0 ( ) ( ) (τ 1))(τ )1 = ( ) 1 1 ( ) ( ) 0 0 0 2 0 + + + + = s h A mC s T s T s i i F m τ τ τ τ = ( ) 1 1 ( ) ( ) 0 0 0 2 0 + + + + = s h A mC s T s T s i i F m τ τ τ τ Or we can write ( ) 1 1 ( ) ( ) 0 0 0 2 0 + + + + = s h A mC s T s T s i i f τ τ τ τ i i i h A mC τ = and 0 0 0 h A mbCb τ = We see that a loading term mC/ hoAo is appearing in the transfer function. The bulb resistance and capacitance is appear in 0 τ 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 dt dC qC qC V i i−1 − i = 0 qC(i− )1 s − qCis = = − − = q V dt dC q V C C i i i τ ( )1 Taking lapalce transformation ( ) ( ) ( ) ( )1 C s C s sCi s i i − = τ − ( ) 1( ) ( ) ( )1 C s s Ci s i = +τ − C s s C s i i +τ = − 1 1 ( ) ( ) 1 Similarly C s s i C s C s C s C s C s C s C s Co s C s i i i i i 1( ) 1 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 1 1 2 0 1 +τ = × ×− − − − − − − − − × = − −
Or N N Co s s C s 1( ) 1 ( ) ( ) +τ = N N s s C C s 1( ) ( ) 0 +τ − = + − − − − − − − + − + = − − − s s s s C s C N N N τ τ τ τ τ τ 1( ) 1( ) 1 1 ( ) 0 1 − − − − − − − − = − − − − − − − − − τ τ τ τ τ N t N t N N t N e N e t N e t C t C ( 2)! . ( 1)! ( ) 1 . 2 2 1 0 1 − − − − − + − + − =− − − − − 1 ( 2)! . ( 1)! ( ) 1 . 1 2 0 N t N t C t C e N N t N τ τ τ 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
dt dh q q A 1 − 1 = 1 dt dh q q A 2 1 − 2 = 2 1 1 2 1 R h h q − = 2 2 2 R h q = Writing the steady state equation 0 qs − q1s = 0 q1s − q2s = Writing the equations in terms of deviation variables dt dH Q Q A 1 − 1 = 1 dt dH Q Q A 2 1 − 2 = 2
1 1 2 1 R H H Q − = 2 2 2 R H Q = Taking laplace transforms ( ) ( ) ( ) 1 1 1 Q s − Q s = A sH s ( ) ( ) ( ) 1 2 1 2 Q s − Q s = A sH s ( ) ( ) ( ) 1 1 1 2 R Q s = H s − H s ( ) ( ) 2 2 2 R Q s = H s Solving the above equations we get ( ) [ ] ( ) 1 ( ) 1 2 1 2 2 1 2 2 2 + + + + = s A R s R Q s H s τ τ τ τ Here 1 = R1A1 τ 2 = R2A2 τ Now writing the balance for third tank dt dh q q A 3 2 − 3 = 3 Steady state equation 0 q2S − q3S = 3 3 3 R h q = dt dh A R H Q 3 3 3 3 2 − =
Taking laplace transforms ( ) ( ) ( ) 3 3 3 2 A sH s R H s Q s − = ( ) 1 ( ) ( ) 3 3 3 2 = s + R H s Q s τ where 3 = R3A3 τ From equation 1,2,3,4 and 5 we got [ ] ( ) 1 1 ( ) ( ) 1 2 1 2 2 1 2 + + + + = Q s s A R s Q s s τ τ τ τ Putting it in equation 6 ( ) [ ] ( ) 1( ) 1 ( ) 1 2 1 2 3 2 1 2 3 3 + + + + + = s A R s s R Q s H s τ τ τ τ τ Putting the numerical values of R1,R2 and R3 and A1,A2,A3 [ ] 4 6 1( ) 2 1 4 ( ) ( ) 2 3 + + + = Q s s s s H s [ ] 4 6 1 2 ( ) ( ) 2 2 + + = Q s s s H s Solution (b) s Q s 1 ( ) =
[ ] 4 6 1( ) 2 1 1 4 ( ) 3 2 + + + = s s s s H s From initial value theorem )0( ( ) 3 3 H Lim sH s S→∞ = = 2( 1)(4 6 )1 4 2 →∞ s + s + s + Lim S = 1 ) 6 2( )1 4( 4 2 3 s s s s Lim S + + + →∞ H3 (0) = 0 From final value theorem ( ) ( ) 3 0 3 H Lim sH s S→ ∞ = = 2( 1)(4 6 )1 4 2 →0 s + s + s + Lim S H3 (∞) = 4
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 dt dh q q A 1 − 1 = dt dh q q A 2 1 − 2 =
dt dh q2 − q3 = A R h q 1 1 = R h q 2 2 = R h q3 = So we get 0 qS − q1S = 0 q1S − q2S = 0 q2S − q3S = writing in terms of deviation variables and taking laplace transforms ( ) ( ) ( ) 1 1 A H s R H s Q s − = S ( ) ( ) ( ) 2 1 2 A H s R H s R Q s − = S ( ) ( ) ( ) 2 A H s R H s R H s − = S solving we get 3 3 ( )1 1 ( ) ( )1 ( ) + = + = s s R Q s H s τ 3 3 ( )1 2 ( )1 ( ) ( ) + = + = s s Q s H s τ
{ } t e t H t L H s − − = = 2 ( ) ( ) 2 2 1 t H t t e − = 2 ( ) 2 0 ( ) = − = −t −t te te dt dH t 2 = 2t =t 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 does the salt concentration in tank 2 reach 0.6 lb/ ft3 ? The hold up volume of each tank is 6 ft3 . Solution Writing heat balance equation for tank 1 and tank 2 dt dy qx − qy = V
dt dl qy − qc = V steady state equation qxs − qys = 0 − = 0 ys cs q q writing in terms of deviation variables and taking laplace transforms ( ) ( ) sY (s) q V X s − Y s = q V s s q X s V Y s = + = + = τ τ ; 1 1 1 1 ( ) ( ) 2 ( )1 ( ) ( )1 ( ) ( ) + = + = s X s s Y s C s τ τ 2 ( )1 1 ( ) ( ) + = X s s C s τ s X s 1 ( ) = 2 3 6 = = = q V τ 2 2( )1 ( ) ( ) + = s s X s C s
2 ) 2 1 ( )4/1( ( ) + = s s C s + = 2 ) 2 1 ( 1 1 4 1 ( ) s s C s + − + = − 2 1 1 2 1 1 2 1 1 ( ) 2 s s C s 2 2 2 1 ( ) 1 t t C t te e − − = − − 3 C (t) = .0 61 lb salt / ft t = 4.04 min 7.7 Starting from first principles, derive the transfer functions H1(s)/Q(s) and H2(s)/Q(s) for the liquid level system shown in figure P7.7. The resistance are linear and R1= R2 = 1. Note that two 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.