CHE407 Midterm 3 Solutions

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CIIE 407 - SPRING 2013 MIDTERM III MAY zoth,2or3Time allowed is 120 minutes, open books only (no notes), calculators are allowed. All workshould be shown clearly in the exam booklets. No questions will be answered during the exam,provide your solutions to the best of your understanding of the problem statements.

PROBLEM L (25 points)

A certain contol scheme uses the following block diagram. Derive transfer functions for all the

possible input/output pairs for this block diagram.

UI'p+DGJ

[Cer-EJrDGdPROBLEM 2 (25 points)

A feedback conttol system has the open loop transfer function

Gel(s) -0.s&

Determine the values of K, for

stability criterion.

(J" (lt G?-Gil = Y:fl

(10s+1)(5s+1)(2s+1)

which the system is stable by the application of the Routh

( t,,rtt) [ ,nsl t?: t \ )

\00 5) + 10rz * to5 + r0 s7* -151\

[,t (t t G"GF -G,Gy) = \59Go t D Q,Gr



PROBLEM 3 (25 points)

The tuning constants for a PI controller for the following process are to be detemrined

6r(s)G,(s)G-(s) = 7.5r-2'3s(8.5s + 1)

100Ga(s) = il1

A fellow process control engineer has determined. several sets of constants for the controller sainand integral time. Determine which of these sets of constants are acceptable by making a staffianalysis of the system under each condition.

Parameter Case A Case B Case C Case DK" t2 t2 0.3 0.3T1 6 I 6 1

Hint - you would need to use a Pade approximation for the delay which means that

^-os -r 1 /z) sv _@)5

PROBLEM 4 (25 points)Consider the control system shown in the figure below

a) Define the inputs and outputs on this system. What should be the value of K*multiplyingthe set-point input? (5 points)

b) Derive the closed loop transfer function(s) relating all possible input-output pairs (10points)

c) For the given values of the various transfer functions in the loop, determine whether thesystem is closed loop stable with respect to set-point changes (10 points)

d) Does your answer in part (c)rchange if you considered distrbance changes? Support youranswer with proper calculations (5 points)

)c -1s7 <-1 t -3.|s.

v.-

(-Zrr+0.4b*q 1)( ,.r) = --Ls3 - l+s'+a.fs'+ 7 7 L*??s+ bCI'+



t,|(Gp, -E.) t DGr

PRoBLE M L',

-

Y.= UEIDGe

lrt = EG.

E - Y'f - tt (et, -Gr)r DGg

U - \rp G. - tL(E, -cpJGc: DG",ca

tL (t + Gp, Gc- g,Q) : XrG.- DGaG.

Y= trGp, +DG;

= YspC', , DGrG-" .Gr, , D&s

l* Q,G. -q"G.'

: Ys, GGo,=Jr€lrtZ-E,

* DG. t D€t€ilC, - DGlQ,Ga

I t Gp ,G"- Gp.G.

G. Gp,

l*Gp,G.-G3"G.

(t - E"G)Gat tGg,G" -E..G.



PRosuEM z.',

R. s{*ti ti{:

co.rol,r?r"^ t

loo

Bo

(pXr+)rl Ct

8o.

I to.S K.

[+Gu,=l+

=2'?Y', r o

=(l Os+r)(5s+ rXZs+r) +0.5K. = o

-(5os'1 lSs+r)(x+r)+osk - c).

-)-offf1 s*+ WilEG+I + o5 K. =o: I oos3 1 BosL + lls *(t * o.SK)

+ [ +0.6K. )o + K.>-Z

t+.

I t 0.5L. I 18'6 o '-* 0,dC'r 5o K. .). O,]' ;lZ6o ),5g K.' ,' ' , t:

K. < 2,s-z .' ' r"" J

s{*L', U{J -Z <K. 1ZS.z .r.



leoBuEtr,t i:#.

Fo. +h; ca/\4

ItGou: l+ K.(l*

^.oFP fvc -2 '3s '

{*

l1 ?.5 l-l,l5s ,\--tzre) &,5sa1 l+t.lSs

:r+*"('t:i'X#)(H#)=o

= (5l,Es+r)(ttl.6sXZrs) 1 ?,5 K.(?rE+ t)(l -l.lsr) = O .

:W rln**7< t u{t)Ccs) t},5K" (g{<-t,rscrs{-1.ds)=o

K- < 1,lz .

G,se A ar^-J 6ase, B

K"e,

q,

L

L

3

ct3

o_

ArO, -O.cLo-.-+-.#

a.z

oo

.]

:- fusg$ + 3M +J4{+ r,5 K" + rE-l4rG-t.tils

C[e) o )Q.)o +

Qr-) o +

+ +r,^/)

ct, )o ?

Qz

?, )o'K. )o9.6s-e.5Zs(-)o +

c*.d*1.a"r a,lrU.^;n..1*

Cct-se C Car\L D

0,3 0"2.

6Llb 9 o.L6L +

arcl-o

cenno+ d;54rhg"ish tvr4

S" rn.rSS n^o.^€" .F*,.t" "t Ue

R".^$.h a.r/reg .

f,,\"f .*d.&.'a,n 4zc(r - o3Cto) O

qL

c&re- C + l3.B ? :lYc^^e D iS

c*t e- 0 9 -2,46 s+..tte .

+?.5K = o\*\-k,

AD,



PRoB LrM +;

(c.) lnputs a,\L Ysp o^^-J

Y"dp.r,t

is

K* sh o,-[d 6e 4l,ro S,S. g ai',. of G,n =,.* =Srt

(".2)*

"x{'6

2/+t/1s +.1-

? k-: zv

2W Zx3to.bo.6: . in{.r,n.*L ("op.

2sr3.6. -,_ (o.z) (t)rt:--

2st ]

TI.rr^ l-i Go. =I A 0,6 -2 2-l? ,' '

-,-

2sr3,6 Z-s s+?

- Gs+ a, e) (z -s) (sI+) - r z( 2st3,6) (z-s )G, r+)

-6Cstil

@

(zsr ?.sXz-:2s r ].0 ) (z-s Xs ti) - rz sr ?.0)(z->) Csr +) - t z(s**)

[Zs*l.o) O-s)Cs++) z

(4snr f(2s.n z. sXz+Xsr +) - I zl

YYr,

Y

D -

4st I

(b) Tt...4 loop cc.ur- be s )tt pUfi..l ink,rna. ttg 4t) t

ItG--



(c) | +Go,- = o = G=u z.O(z-s)Csr+) -12:(ts - zs'+ ?, L4.6)(str) - t z

: (?,2 - o ,4s - zs") Cst r) - tL= QE 4x- ^*+ 5o,4 - zrds 1-{f - I z): = _ 2s? _ 14,*s" + +,4 + 39.4 .

t[.{ a.Lo'.rc- is o- 3rd o.du F"t$no"ra,t.r0 w rth c[. au d a.rvgq.liw .' - +C. .t*o-J [o-p .? u ns{-LL

tdl itarwer^

do.n ,no{ cb,.a.*7c a,rt ho{L I . Ps stnb; U {I .5

$o*rlr,ad tg l+6or-

*

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CHE407 Midterm 3 Solutions