# Chapter 7 - CHAPTER 7 3-7—1 131 substituting the given...

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Unformatted text preview: CHAPTER 7 3-7—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—7-2. We shall solve this problem by using two different approaches: cmhaudantheaxaatmttwdandtMotherbaaedmtheuscofmmz-age 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. SOUL-lava 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 .- d-H _ 2-5 - 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 11-05) = 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 vol-Lave t1 = 0.4904 x 258 = 126.5 s This solution has ham obtained by use of an average resistance. 3-7-3. The equations for the liquid-level 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 Hencethetransferhmctimofthasystanwhen-qistheinmtandqz is theontputisgivmby 02(8) _ 1 ——.— 9(5) - (Rlcls + 13(R2czs + 1) 3-7-4. 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 Ems-44m ('1‘ m nul- (1) (2) (3) (4) CRsH(a)+l-I() '——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 Wain-glamr 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 can-th. Haar. c=r27¢=(-§—)2?c z I 7:: an -.- —o.oosﬁdt or ' 111-5 an - —o.oos ﬁz- dt MmﬂnttremamvesddmfrunH=thoxfortJn60mcmdperiom Then at . . 50 H1'5 dH 3. -0.005 db — 2 0 7...: §-— (x2-5 — 22-5) = —0.01432(60 - 0) which can be reurittm as x2-5 - (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 electrical-liquid-level analogy given below, equations for an mlogous electrical system: can be obtained. ' Electrical systems Liquid-level 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—liquid-level 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—n-ﬂ—“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 ' wan-w— 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 3-7—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: 3-0) > 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 ql-Kl 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 p1-92=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 3-7—11. For the toggle joint shown in Figure 7—35. I!!! have Liz. ’5‘” 11 I F=2—[-‘lR 4‘2 3-7-12. 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 10-5 = 0.0053342 11:2 Thus the minimum area of the piston is 58.84 0112. “whnﬂ—“-——”mm———Mm_mﬂ - 3-7-13. 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 21-8 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- __ (H-H)2+ no Her-I Neglecting the higher—order terms. a linearized equation for the system can be written as Q-ﬂm-am-E where ftﬁ) = 5(4) - 0.2 _£. 1 3-m wml =mms H=§=4_ 2f? Thus, a linearized equation beauties Q - 0.2 = 0.025“! — 4} —_—"me———- 3-7-16. 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:- ‘ 33-2n2yf ~22+1o=32 .3; 3? =2x+10y -22+so=72 x=11,y=5 Thus, the linearized equation is z—356=32(x-11)+72(y-5) or 32x+72y-z=356 :7-19. 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 ‘Ta-I-b 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 -1mn4-nu- 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) Clad-Ge (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—ﬂHM-ﬂmmﬂ m—nm ...
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Chapter 7 - CHAPTER 7 3-7—1 131 substituting the given...

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