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 ( ) − − + = R C C s C 1 2 2 ( ) 3 7 4 4 ( )1 4 (2 )1 + = = + + + + R s C R C s s 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 ) 2 25 1 2 − + = X s s Y
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. ( 1)(2 )1 5 1 ( 1)(2 )1 5 + + + + + = s s K s s R C 2 3 9 8 2 + + = R s s C s R 1.0 = b) .0 0889 9 8.0 2 3 9 8.0 ( ) 2 0 = = + + ∞ = → s s C Lt S offset = 0.0111 c) 2 2 1 3 1 2; 3 2 ; 9 8.0 K = τ = ξτ = ⇒ξ =
overshoot = .0 305 1 exp 2 = − − ξ πξ = Maximum vslue of C = 1.0305*0.0889=0.116 Maximum value of C = 0.116 − − + − = − − − ξ ξ τ ξ ξ τ ξ 2 2 1 2 1 sin 1 tan 1 1 1 9 8.0 .0 116 t e t 6.1 1 tan 1 2 1 2 = − − = − ξ ξ ξ τ t Time at which Cmax occurs = 1.6 (c ) Period of ociullation is .3 166 1 2 2 = − = ξ πτ T T =3.166 Decay ratio = (overshoot)2 = 0.093
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, D τ , I τ and 1 τ . (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. + + + + + + + = s s s k s k s s R C a I D I D τ τ τ τ τ τ 1 1 1 1 1 1 1 1 ) 1 1 ( ) + + + + + = s k s s s U C I D τ τ τ τ 1 1 1 1 1 1 1 1 1 = 1 2 1 + + + + s k k s k k s I I I I D I τ τ τ τ τ τ + + + + + + = s s k s s k s R C I D I D τ τ τ τ τ 1 1 1 1 1 1
( ) k s k s k k s s R C I D I I D I I + + + + + + = τ τ τ τ τ τ τ τ ( ) ( )1 1 2 1 2 k k k k I D I I τ τξ τ τ τ τ ( )1 2; ( ) 2 + = + = k k k k I D I I τ ξ τ ( τ τ ) ( )1 2 + = + = × 2 ( ) ( )1 1 τ τ τ ξ + + = = D I k k k 2 ( ) 4 ( ) ( )1 ( ) 2 1 2 1 2 1 1 2 τ τ τ τ τ τ τ τ π ξ τπ + + − + + × = − = = D D D I D k k k k k k k T 1 2 1 4 ( )1 4 ( ) − + + + = k k k k T I I D D τ τ τ τ π τ τ B) D τ = I τ =1; 1 τ .=2 For k = 0.5 ; ξ =0.75 .0 671 )5.2(5.0 1 = For k = 2 ; ξ =1.5 .0 530 2 3 1 = × C) + + = + + = = k k k k k I D I D I τ τ τ τ τ τ ξ 2 1 2 1 1 2 1 ( ) ( )1 2 1
As .0 3535 2 1 , 1 → ∞ = = τ τ ξ I k 1 2 1 4 ( )1 4 ( ) − + + + = k k k k T I I D D τ τ τ τ π τ τ k k I D I τ τ τ τ τ + 1 = k I I D 1 τ τ τ = τ τ + K I D τ = τ τ →∞ =1 .7 2552 4 1 4 , = − → ∞ = I D D As k T τ τ πτ (d) + + + + + = s s s k s U C I D τ τ τ τ 1 1 1 1 1 1 1 1 k s k s k s U C I D I I + + + + = 2 1 ( )1 τ (τ τ )τ τ s U 1 = k s k s k C I D I I + + + + = 2 1 ( )1 τ (τ τ )τ τ 0 0 ( ) 0 = = → k Lt sf s S
K=2 For a unit step change in U C(∞) = 0 Offset = 0 (e) k = 0.5, ξ =0.671 & τ =2.236 18.95 1 2 2 = − = ξ πτ T If k = 0.5 5 3 1 2 2 + + = s s s U C ; 5 3 1 2 + + = s s s C If k = 2 2 1.5 1 5.0 2 + + = s s s U C ; 2 1.5 1 5.0 2 + + = s s C In general τ ξ τ ξ τ ξ t C t e t 2 2 sin 1 1 1 1 ( ) − − = − The maximum occurs at ξ ξ ξ τ 2 1 2 1 tan 1 − − = − t If k = 0.5 tmax = 2.52 Cmax=0.42 If k = 2 tmax = 1.69 Cmax=0.19
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. ; 0 1 1 = U2 = s U C s s S R C U = + + − + 2 1 1 2 1 2 ( ( ) ) 1
R = 0 4 4 11 2 2( )1 1 2 5 2( )1 2 2 2 2 1 + + = + + × + = s s s s U C 4 4 11 2 2 1 + + = U s s C 11 1 11 4 2; 11 2 ; 11 2 K = τ = ξτ = ⇒ξ = .0 2516 2 10 2 1 1 2 1 1 2 = = − = = = τ π ξ T π Frequency f C(∞) = 2/11 Offset = 2/11 =0.182 U1=0;U2=1/s C s R C U s = + − + + ⇒ × 2 1 1 ( ) 2 1 2 5 2 R=0
1 2 1 1 2 1 10 2 1 1 2 + + × + + ⇒ = s s s U C 4 4 11 2 1 2 2 + + + = s s s U C C(∞) = 1/11 Offset = 1/11=0.091 a) if ; 1 1 s U = frequency = 0.2516 if ; 1 2 s U = frequency = 0.2516 b)if ; 1 1 s U = frequency = 0.182 if ; 1 2 s U = frequency = 0.091
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. ( 1)( )1 1( ) 1 ( )1 1( ) ) 1 1 + + + + + + = s s K s s K s R C a m C D C D τ τ τ τ τ
( ) ( )1 1( )(1 ) 1 2 1 + + + + + + + = m m C D C C D m s K s K K s s R C τ τ τ τ τ τ τ = 1 2 1 + = C m K τ τ τ 1 1 + = C m K τ τ τ 2 ( )1 1 1 1 + + + = m C m C D k k τ τ τ τ τ ξ b) 7.0 1min; 10 ; τ 1 = τ m = s ξ = 1) 600( )1 70 2 1 7.0 0 + ⇒ = × = C D k τ kc=3.167 2) .5 255 600( )1 70 3 2 1 7.0 3 = + + ⇒ = × = C C C D k k k τ s c)for 1 ; ( ) 1 + = ∞ = C C k k c s R
for s period s period for period period for s C offset For C offset D D D D = = − = = = = − × = = = ∞ = = = ∞ = = 86.17 1 ( )7.0 .6 255 600 2 3 ; 105.57 1 .4 167 600 2 ;0 3 ; ( ) .0 84; .0 16 ;0 ( ) .0 76; .0 24 2 2 π τ ξ π τ τ τ 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 ) +WT2C WT2C = ρCV3 (T3 −T2 ) +WT3C ( ) ( ) 0 1 T1 W C CV1 T WC+ ρCV + q = + ρ 1 1 0 WC CV q T T + ρ = + T1= T2 = T3 ⇒ + = + 1 3 0 WC CV q T T ρ ( ) WC CV s q s T s 1 3 ( ) ( ) + ρ = 13.6 (b) ( 1)( .1 25 1)( 5.1 )1 2 1 1( ) ( 1)( .1 25 1)( 5.1 )1 2 1( ) ( ) ( ) ' 3 + + + + + + + + + = s s s k s s s s k s R s T s C D C D τ τ = ( 1)( .1 875 .2 75 )1 2 1( ) 2 1( ) ( ) ( ) 2 ' 3 s s s k s k s R s T s C D C D τ τ + + + + + + = .1 875 .4 625 .3( 75 2 ) 2 1 2 1( ) ( ) ( ) 3 2 ' 3 + + + + + + = C D c C D s s k s k k s R s T s τ τ
c) kC=3; s offset s D 1 ,5.0 ?, ( ) ' τ = = τ 0 = .1 875 .4 625 .3( 75 2 ) 2 1 1 ( ) ( ) 3 2 0 ' ' 3 i s→ s + s + + kC D s + kC + Lt T s T s τ = .0 143 7 1 2 1 1 = = kC + 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. C s R C U s s KC = − + + + × 1 ( ) .0 25 1 1 = .0 25 1 1 1 . 1 + + + = s s s K s U C C
.0 25 ( )1 .0 25 1 2 + + + + = s s K s s U C C C KC s K s s U C (4 )1 4 4 2 + + + + = b) ξ=2.3 C C C K K K 1 2; 4 1 + τ = ξτ = = C C C K K K 1 2 3.2 4 1 + × = 3.2 1 = + = C C K K KC=2.952 C) KC=4,U = 1/s = 20 16 1 4 2 + + + = × s s s s C 4 1 C(∞)= offset = 0.25. 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.
(a) ( )1 4 1 ( )1 4 + + + = s s s s R C 4 4 2 + + = R s s C b) C(∞) =2*1=2 C(∞) =2 C) offset = 0 d) 4 1 4 1 2; 2 1 τ = ξτ = ⇒ξ = + − = − − − tan 15 4 15 sin 4 1 1 1 1 2 ( ) 2 1 2 τ t e C t t = = − + − − tan 15 4 15 sin 15 4 )5.0( 2 1 4 1 1 C e 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). 2 1` 1 1 1 1 1 1 + + = + + + = s s R C s s R C 2 2 1 ( ) 1 2 1 1 1 2( )1 1 1 t C t e s s s s s C s R − = − + − = + + + = = 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 : ( + ( +1)) = C C s s K K C +ve feed back 1 1 1 1 1 + − + × = s K s K R C C C ( 1( )) C C s s K K C + − = For KC = 0.5 , response of -ve feed back is 2 3 3 2 3 1 2( )3 1 + − = + + = s s s s C 1( ) 3 1 3 1 3 1 ( ) 2 3 2 3t t C t e e − − = − = − response of +v feed back is
2 ( ) 1 2 1 1 2 2( )1 1 t C t e s s s s C − = − + − = + + = For KC = 1, response of -ve feed back is t C t e s s s s C 2 2 1 2 1 ( ) 2 2 1 2 1 ( )2 1 − = − + − = + + = response of +ve feed back is C t t s C = = ( ) 1 2 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 11 6( 6 ) 0 ( 6 11 6( 6 ) 0 ( 1)( 2)( )3 6 0 0 ( 1)( 2)( )3 6 1 3 2 2 + + + + = + + + + = + + + + = = + + + + s s s Kc s s s Kc or s s s Kc s s s Kc Routh array 1(6 ) 6 6( 6 ) 1 11 2 3 s Kc s Kc s + + 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 0 2 4 10 10 2 3 1 2 1 2 = + + + + s s s
2 25 60 0 2 4 50 120 0 2 4 10 40 120 0 2( 4 10) (2 3). .2 10 0 3 2 3 2 3 2 2 + + + = + + + = + + + + = + + + + = s s s s s s s s s s s s s s Routh Array 1 25 2 60 -10/2 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 : .................................. 0 1 0 + 1 + + = − n n n a x a x a ( / ....................... / 0 ) 0 1 0 + 1 0 + = − a x a a x an a n n , ............................. 0 ( )( )....................( ) 0 1 2 0 1 2 < − − − = n n We have a x x x α α α α α α As we know the second co-efficient a1/a0 is sum of all the roots ( )1 2/ 1 1 2 0 1 = − ∑ ∑= = n j i j n a i a α α Therefore sum of all possible products of two roots will happen twice as α1α 2 dividing the total by 2. And / 0 0 0 ( 0 )0 ∴ 2 0 > ⇒ 2 > > < < a a a α iα j α i α j Similarly
(0 1,....... ) / 0 ( )1 ( )1 0 0 ( )1 1 0 / 0 ( )1 ( ) 0 0 0 0 so a for j n in both case a a is again a a if j odd is and the sum is so if j even is and the sum is so a a sum of aoll possible products of j roots a a j j j j j j j j > = > = − − < > = − > > = − 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 0 2 3 3 2 1 0 1 3 1 2 2 3 s s s s s s s − + + + 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 ................................. 0 1 0 + 1 + + = − n n n a x a x a 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 ( ) ( ) ................................. 0 1 0 − + 1 − + + = − n n n a X σ a X σ a Hence if we apply Routh criteria, We will actually be looking for roots with real part >σ rather than >0 ................................. 0 2 2 1 0 + 1 + + = − − n n n n a x a x a x a Routh criterion detects if any root α j is greater than zero. Is there any , ,..............., ,.......... 0 )1( x = α1 α 2 α j α n > − − − − − Now we want to detect any root )1( 0 From j j > > − α α σ
α σ α σ σ σ α σ σ α σ σ α σ σ α σ σ α α α α α α α α > − + > = + + = + > + = + > + = + > + = + > = > = > = > = > = > j j n j n j j n or and apply Routh criteria to ect any so Let X x x x x x is there any add on both sides x x x x implies is there any x det 0 , 0 . . . . 0 . . . 0 0 0 . . . 0 . . . 0 0 , ,............................ ,.......................... , 0 2 1 2 1 1 2 14.7 Show that any complex no S1 satisfying S < ,1 yields a value of s s Z − + = 1 1 that satisfies Re(Z)>0, Let S=x+iy, 1 2 2 x + y <
ss Z −+ = 11
( ) − + = + > < + + − < = = = − = + + − − + > + < − < < − < < + < + < − + + > > − + > − + + − + = − + + − + + = + − + − + = + − − − + − + − − + + − 5.0( )5.0 1 5.0( )5.0 : Re( ) 0 0 1( ( ) 2 ) 4 1 & 0 0 1& 0 4 1 ( ) 2 1 ( ) 0 1 1 1 int 1 1 1 1 1 2 ( ) 0 Re( ) 0 1 ( ) 0 1 2 ( ) 1 ( ) Re( ) 1 2 ( ) 1 ( ) 2 1 2 1( 1( 1 ) ) 1( ) 1( ) 1( ) 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 s i L if s i the system isunstable due to the real part example z x y x if x y then it is if x y then it is x y x Now x y is true therefore x y y Po s in the unit circle Ranges are x x y we have x y x x y if z then x y and x x y x y Z x x y x y iy x x y x x x iy y x iy x iy x iy x iy
( 5.0 5.0 ) 5.0( )5.0 1 1 5.0 e Cos t Sin t s i L t = + − + − 14.8 For the output C to be stable, we analyze the characteristic equation of the system ( )1 0 ( 1)( )1 1 1 1 3 1 2 × + = + + + s s s s I τ τ τ τ ( ) ( ) 1 0 ( )1 1 0 3 2 1 2 3 1 2 2 1 2 3 2 1 + + + + + = + + + + + = s s s s s s s s I I I I τ τ τ τ τ τ τ τ τ τ τ τ τ τ Routh Array 1 0 ( ) 1 0 1 2 2 1 2 3 3 s s s s I I I α τ τ τ τ τ τ τ τ + + ( ) ( )( ) 1 2 1 2 3 1 2 τ τ τ τ τ τ τ τ τ τ τ α + + + − = I I I I Now )1( 0 τ I τ 2 τ 1 > Since 1 & 2 τ τ are process time constant they are definitely +ve ;0 0 τ 1 > τ 2 > (2) ( ) 0 τ I τ 1 +τ 2 > (3) 1 2 3 1 2 α 0 τ (τ τ )(τ τ ) τ τ τ > ⇒ I + I + > I
0 ( ) ( ) 0 3 1 2 1 2 1 2 1 2 3 1 2 1 1 3 2 2 3 1 2 > − + > + > − + + + + − > I I I I I also τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ τ 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 6 11 6( 6 ) 0 ( 3 2)( )3 6 0 ( 1)( 2)( )3 6 0 0 ( 1)( 2)( )3 6 1 3 2 2 + + + + = + + + + = + + + + = = + + + + s s s Kc s s s Kc or s s s Kc s s s Kc Routh array + − + ) 3 1 3 3 1 1 3 2 3 Kc s s Kc s
For verge of instability 8 ) 3 1 3 = + = Kc Kc Characteristics equation 0 ( 3)1 10 1( ) 1 = + + + s kcs 3 3( 10 ) 11 0 3 2 s + s + s + Kcs + = Routh Array /2 30 30 2 3(3 10 ) 11 3 11 1 3 10 2 3 > > + > + D S s D for vege s s s τ τ τ τ 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
2 3 1( ) 2 0 ( )(2 )1 ( )2 0 0 ( )1 1 2 1 1 3 2 1 3 2 2 + + + + = + + + + = = + = + + s s kc s kc s s s kc s s s s kc 3 3 1( ) 0 3 2 s + s + s + + kc = Kc=4Routh array 3 ;0 3 0 3 1(3 ) 4 3/1 3 8 2 5 2 3 − = = = + − − Kc Kc kc kc not stable s s s For verge of instability 14.11 for the control shown, the characteristics equation is 4 6 4 1( ) 0 4 3 2 s + s + s + s + + k = (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. 0 4 6 4 1( ) 4 3 2 s + s + s + s + + k =
k s k s k s s k + − + + + 1 1 1( ) 5 4 4 5 1 4 4 1 1(6 ) 2 3 4 For the system to be unstable 1 1 1 0 4 5 1 1 0 5 1 4 1 > − < − + < > + < < + − k k k k k k The system is stable at -1<k<4 (b) For two imaginary roots 1( ); 4 5 4 4 = + k k = Value of complex roots s i s = ± 5 + 5 = 0 2 1 4 6 4 5 4 5 2 4 3 2 2 s + s + s + s + s + s + s +
4 2 s + 0 + s 0 5 5 5 5 4 0 4 4 5 2 2 3 3 2 + + + + + s s s s s s SOLUTION: i i s = − ± − ± = − ± − = 2 2 4 2 2 4 16 20 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
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