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Unformatted text preview: CHAPTER 7 37—1. 131' substituting the given nunerical values into' R = (1111:12821.‘
I '35 . 3
gm
weobbain .1“
12011100411104“
Rx:———.———————m=3.2_594x104 E/illz
9.31x3.14x(4x10 )  . 3—72. We shall solve this problem by using two different approaches:
cmhaudantheaxaatmttwdandtMotherbaaedmtheuscofmmzage moisture. . (1) Solutzm 92 the exact method. For the iiquiddml system 10 have By substituting c a 2.112. pi = 0.05 33/3. and 0  0.02 Jﬁ into this last
equation. we obtain _ 2 as. = (0.05  0.0213) dt or I
at = 0.05  0.02 71; d" LetH=x, MstzanddIl='2xdx. SOULlava aunszw uu.m(_:5+_21_+_5_)ax 1 f, 5
tl=5adt3 I 200(1+5_2x)dx
=—.200le:—:+1mo dx 1 mutatattat unlmlreedaesz.5n. men, {:1 is obtained as 5—2::
JET!" f. =  200(./2.s  1) + 1000 (+) 1n(5 ' 2x} = — 200 x 0.531 — mm 1.033 — 1n 3)
=  200 x 0.531  500(O.608?  1.0986)  120.5 a (2) Solution 91 use of an average resistance. Since Q = 0.02 E. the
average resistance R is obtained Eran . dH _ 25  1.0
R — —I—  —I—I—I—I— 2
d9 0.03162  0.02 129 Defineh=H—1. 11mg =Q'—0.02and = . 
system, I i i go h/R For the liquid leval th=(q.q)dt
1 O
A: _ = _.JL
cdt qi go ‘31 R
Hhichcanberewrittenas dh
CR qi Substituting 0 s: 2 m2, R = 129 5/513. and qll = 0.05  0.02 = 0.03 m3/a into
this last equation yield; 253 g3‘» + h = 3.07. 11(0) z 0
dt Taking Laplace transforms of both sides of this last equatim, 1m obtain
‘ 253[sH(s)  h(0)] + Hts) :2 52—37 (2533 + 1.)}!(3)  3% Solving this equation for me). H‘s) 3(2535 + 1)
_ .1...  __._253
' 3'87( a 2585 + 1 The inverse Laplace transform of this last equation gives '  1
1105) = 3.8m — g‘z—ﬂ't) Assume that at t = 1:1, h(t1) —' 1.5. we value of t1 can be determined fran I Rewriting r or So volLave t1 = 0.4904 x 258 = 126.5 s This solution has ham obtained by use of an average resistance. 373. The equations for the liquidlevel system are Cldhl = (Q + q — Q  q1)dt
c2th = (Q + q1 — 0 ~ q2)dt Since R:l = hl/ql and R2 = hZ/qZ, the system equations can be rewritten as cTh‘I ' hl
C1 Et— = q  a . E dh h h
2 1 2
C _" = q  q = "— ' "'— (2)
2 dt 1 2 R1 R2
Fran Equation: (1) and (2) we obtain
dh dh h
1 2 2
c —— + c ——= ———
1 dt 2 at q 22 (3)
By differentiating Equation (2) with respect to t. we get
c1211 an an By eliminating dhl/dt from Equations (3) and (4); we obtain dzh dh 
2 +(R1C +RC)—2 +h =R2q RCRC 1 22c": 2 1l22 C“:2
Substitution of 112 = qu2 into this last equation yields 2
‘3 q2 dqz _
R1C1R2CZ dtz + (“101 + R2C2J EYE" + q2 ' q Hencethetransferhmctimofthasystanwhenqistheinmtandqz is
theontputisgivmby 02(8) _ 1 ——.— 9(5)  (Rlcls + 13(R2czs + 1) 374. The equatioﬁs for the system are Cldhl = qldt czd"2 = “31 .' 91 ‘ ‘12)“: mus. uehava Fran Equation (1) we obtain Claws! = 111: tags) — 31(3)] 31(3) ll
3: N
A m
“.0 Rlcla + 1 M Ems44m ('1‘ m nul (1) (2) (3) (4) CRsH(a)+lI() '——R3H() “LB
3 = 5 
3 3 3 3 R2 2 R2 Hafﬁ) or
H( ) R2 R3
2 s  R3 (R3035 + 1 + R2 mats) (5)
By adding Equations (1): (2). and (3), and taking the Laplace transform
of the resulting Wainglamr we obtain ' C15H1(s) + C28H2(S) + C3BH3(6) = 01(8)  113(5) By substituting Equations (4) and (5) into Equation (6), we 92!: C13
Rlclﬁ + 1 Since 113(5) : R39°(s). this last equation can be written an (c1 + 02:5 + thlczsz
Rlcla 4' 1 Rs 1 32
R3 )(R303B+ 1 +E) + tags +' R 3 )]H3(B) = 01(8) (R3R2C3s 4 R2 4 R3) + (R3C33 "' 1)] 90(3) ' 01“) ﬁrm: which we obtain QOLB) Rlcls + 1 91(8) [(cl + 02)a + 12101023 ](R3R2C35 4 R2 + R3) + (R3c3a + 1)(R1cls + 1)_ ' This is the trmafer function muting Qofs)and 01(8) B~7—5 . For this system canth. Haar. c=r27¢=(§—)2?c z
I 7:: an . —o.oosﬁdt
or '
1115 an  —o.oos ﬁz dt MmﬂnttremamvesddmfrunH=thoxfortJn60mcmdperiom
Then at . . 50
H1'5 dH 3. 0.005 db — 2 0 7...: §— (x25 — 225) = —0.01432(60  0) which can be reurittm as x25  (1.4142315 = — 2.1430
01' I x“? = 5.5559  2.1430 = 3.5039
Ta!ng logaritlm of both sides or this last. equation, we obtain 2.5 =1 3. ‘9
m5'10" °g1o 503 x = 1.652 m 3—7.6. Fm Figure 7.30 we obtain c {ml
lat CI QI
c maz q
zdt ‘11 2
h
$14"
1.
q2:_EL
R2 Using the electricalliquidlevel analogy given below, equations for
an mlogous electrical system: can be obtained. ' Electrical systems Liquidlevel system 9 (voltage) q (flow rate)
I: (head) dh/dt R (resistance)
c (capacitance) Analogous equations for the electrical system are R111 = e ‘ "31 (1) R212 = 91 ‘92. (2) e2 (4) Based on Equations (1) through (4). we obtain the analogous electrical_ system Show} below. Using the table of electrical—liquidlevel analogy 91m in the solution
of Problem 3—7—5, we can obtain an analogous electrical system. The analogous electrical equations are Rlil = a  El R212 = e1  32 (2)
_ 1  1 dt
81 = 1 2) (3)
C1
3 i2 db
Q = (4) CZ Based on Equations (1) through (4} , we obtain the analogous electrical
system sham on next page. _ —————— “ﬂamm—nﬂ—“w—mF mo PV = mRaim'1' In this problem .
p = 7 x 105 + 1.0133 x 105 = 3.0133 x 105 N/m2 abs
IT=273+20=293K The mass m of the air in the taan is W a 3.0133 x 105 x 10
RairT 237 x 293 If the temperature of ompressed air is raised to 40°C, then T =
273 + 40 a 313K and thepressure pbecanes m: = 95.29 kg "’Rair'II .
P: = 95.29x287x313 =3usaox105N/m2m V 10
= 7.547 x 105 Wm? gage = 7.695 )Igf/anz gage
= 109.4 nag/ml gage ' wanw— B—7—9. Note that
C dpo = 4:; at where q is the flow rate through the valve and is given by pl “pa 61“ R dpo _pi '90
at R from which we obtain P003)
.Pits) R05 + 1 .. .1 For the bellows and spring} we have following equation:
APO = '0! The transfer Emotion 3((s)/l='i (s) is then given by X‘s) _ xts) 130(5)
picsf PO{S)P1(S) .4.
ROB+1 :4.
k _..n 37—10. Note that for t s 0. p1  0.5 x 105 N/m2 gage ' 1.5133 x 105 14/13 abs
[:2 = o m2 gage = 1.0133 n”: 105 N/m2 abs
For t) 0, p2 increases. Since p1 stays constant for all t, we have p20: 30) > 0.528 131 . Note that for p2 > 0.528 p1. the speed of air flow is subsonic. So the flour throughout the system is subsonic. The flow rate through
the inlet valve is qlKl p1"”‘2 The flow rate through the outlet valve is ‘12 ='{2‘1132 ‘93 Since both valves have identical flow characteristics. we have K1 = K2 = K.
The equation for the system is dpz GET'KJ‘H '92 ‘KJpz'Pa At steady state. we have dpz/dt = 0 and this last equation becomes
KJPI  P2 = KJPZ  133 p192=p2~93 or p1 + Pa = 1.5133 x 105 + 1.0133 x 105
2 2 2 p = '—' 1.2633 x 105 "/3112 abs = 0.25 x 105 N/I'n2 gage 37—11. For the toggle joint shown in Figure 7—35. I!!! have Liz. ’5‘” 11 I F=2—[‘lR
4‘2 3712. Assume that the minimum area of the piston is A 1112. Then
the minimum force needed to move the load mass is ' F % Afpr P2) =/¢mg Hence A ﬂing _ 0.3 x 1000 x 9.807 p1 — pg 5 x 105
= 588.42 x 105 = 0.0053342 11:2 Thus the minimum area of the piston is 58.84 0112. “whnﬂ—“——”mm———Mm_mﬂ  3713. If we assume that the mass of the power piston is negligible
ounpared with mass m, then the equation for the system is A(p1 p2) mgsino(/(mgcmo( =m'i
The load can be pushed Upward if
Mp; — 1:2) mg sinoc —/&mg 9080‘30
Noting that6=tan‘/4, vehavetan 9:)“. Hence
M131 * 92) ) mg(sin0( +/ucoso() _= mg(SinoC + tan 9 cosoc) cos 9 Sinai + sin 9 (203%
0039 _ mg sinﬁe +°Q zmg mg sin (9 +00
(91  92) cos B A; 8—7—14. Deﬁne the radius and angle of rotatim of the pinicn as r
and 8, respectively. Then, relative displacement between rack C and
pinion B is :9. : ‘ Irol+
‘0
p a0 Are 1 Relative displacement hem rack A and rad: C is 218 and this mist
equal displacmzt 3:. Mature, we have _ 2r9=x
Sincex=r9+y,weobtain
x
=—
y 2 —m———————HWW— o=o¢f§=ﬂm = :0?) +314? dH — 1d% _
H=ﬁ(ﬁ_H) +3352 __ (HH)2+ no
HerI Neglecting the higher—order terms. a linearized equation for the system
can be written as QﬂmamE where
ftﬁ) = 5(4)  0.2
_£. 1
3m wml =mms H=§=4_ 2f? Thus, a linearized equation beauties Q  0.2 = 0.025“! — 4} —_—"me——— 3716. 7. at 5x2 = £fo =f(i)+%(x—i)_+—%?g—2£{x—i)2+' M:2
A linearized equation for the system 13
z  E = ab:  i) uherei=2.i=20.ahd a=% =10}: I =20 Thus, a linaarizad equation becclma
I z  20 = 200:  2) 3‘7"].7. I z 3 x2 + m + = t‘x' 3.1: ‘ 332n2yf ~22+1o=32
.3; 3? =2x+10y 22+so=72
x=11,y=5
Thus, the linearized equation is
z—356=32(x11)+72(y5) or 32x+72yz=356 :719. Define displacenents e, x, and y as shown in the figure
low. Fran the figure, we can construct a block diagram as shown below. Y0“) From the block diagram we obtain the transfer function Y(s)/9(s) as follows: h K
1 us) a+b§ L; n. a+b _! b
9(8) K a ‘TaIb a _' T We see that the piston displacement y is proportional to the deflection
angle 9 of the control lever. MaoIr from the system diagram we see that for each value of y, there is a corresponding value of angle 95. 'I'here—
fore. for each angle 9 of the control lever.r there is a corresponding 1...__1. _.L—Ln 1mn4nu 5""1ﬁ at 3—7—20. The heat balance equations for the system are 011161 = (u  q1)dt (1) czar;2 = (ql  q2)dt. a (2)
Notng that
ql = Gee I q2 = Equations (1) and t2) can be modified to C1__]_‘.=u_gcﬂl (:2 ——Z = Gee1  6:62 I (4)
from which we get I
018316) = Uta) — Gceﬂs)
6230205) I: 0001(3) — Gaozts) By eliminating 01(3) Eran the preoeﬁing two equations: we obtain {cza + Galozts) = ——G—°— Ufa) CladGe (619 + Gc)(czs + 9.2592(3) = GcU(s) Thus the transfer fmctim 02(5))!!(5) can be given by 92‘" = Ge
[1(a) (C13 1 Gc)(czs + 6c) wmmﬁ—ﬂHMﬂmmﬂ m—nm ...
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 Spring '08
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 Fran Equation, linearized equation

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