2.1a) 0 [()sin(2)]sin(2) 44 st LuttetdtI ππ ∞ +=+= ∫ Integratingbyparts: 2 1 22 2 (4) s I s + = + 2.1 a) MATLAB Solution syms x y z t s k laplace(sin(2*t+pi/4)) ans = (1/2*s*2^(1/2)+2^(1/2))/(s^2+4) pretty (ans) 1/2 1/2 1/2 s 2 + 2 ----------------2 s + 4 00 0 2 2 11 ()cos(2)()cos(2) 2424 11 [0]cos(2) 224 2 111 ()(0) 224222 2 (1)442 stst st I etsetdt s Ietdt ss II ss I ππ π ∞ ∞ ∞ =−+−−−+ =−−−+ =−−− + += ∫ ∫ Process Systems Analysis and Control 3rd Edition Coughanowr Solutions Manual Full Download: http://testbanktip.com/download/process-systems-analysis-and-control-3rd-edition-coughanowr-solutions-manual/ Download all pages and all chapters at: TestBankTip.com
2.1b) (1) 00 [()cos2]cos2cos2ttstts LutetteedttedtI ∞∞ −−−−+ === ∫∫ Integratingbyparts: (1)(1) 00 (1) 0 2 (1) 0 2 11 ()sin(2)(1)()sin(2) 22 (1)sin(2) (1)cos2(1) 44 (1)1(1)44 tsts ts ts I etsetdt Isetdt sts II e ss I ∞ ∞ −+−+ ∞ −+ ∞ −+ =−−+ =+ ++ =− ++ += ∫ ∫ 2 1 (1)4 s I s + = ++ 2.1 b) MATLAB Solution syms x y z t s k laplace(exp(-t)*cos(2*t)) ans = (s+1)/((s+1)^2+4) pretty(ans) s + 1 -----------2 (s + 1) + 4 2.2a) 3 3 s = MATLAB Solution ilaplace(3/s)
ans = 3
2.2b) 32 3 (2) t e s = +
MATLAB Solution ilaplace (3/(s+2))
ans = 3*exp(-2*t)
2.2c)
211 31 33 (2)(2)
te ss + == ++
MATLAB Solution ilaplace(3/(s+2)^2)
ans =3*t*exp(-2*t)
2.2d) 2 321 3323 22 t ss + ==
MATLAB Solution ilaplace(3/s^3)
ans =3/2*t^2
2.2e) 222
1 2131sin3 (9)636 t ss == ++
MATLAB Solution ilaplace(0.5/(s^2+9))
ans = 1/18*9^(1/2)*sin(9^(1/2)*t) simplify(ans)
ans = 1/6*sin(3*t)
2.2f)
et ssss === ++++++
22222 33323sin2 48(2)22(2)22
2
t
2
t
MATLAB Solution ilaplace(3/(s^2+4*s+8))
ans = -3/16*(-16)^(1/2)*(exp((-2+1/2*(-16)^(1/2))*t)-exp((-2-1/2*(16)^(1/2))*t)) simplify(ans)
ans = 3/4*i*(-exp((-2+2*i)*t)+exp((-2-2*i)*t))
pretty(ans)
3/4 i (-exp((-2 + 2 i) t) + exp((-2 - 2 i) t))
Howdoesthiscomparewiththehandsolution?Recallingthefollowingrelations: sin,cos,cossin 22
αααα α α ααα −+ ===+ ,let'sworkwiththe
iiii i eeee ei i
MATLABsolutionandseeifwecanderivethehand-solution.TheMATLABsolution cleanedupabitis:
2(2)2(2)2(2)(2)33 ()()() 44 ttittittiti i eeeeieee −+=−+
Fromtheaboverelations,2(sin)ieeii α α α =− ,ifwelet2t α = ,thentheMATLAB solutionbecomes:2222 363(2sin2)sin2sin2(check)
MATLAB Solution ilaplace((s+4)/(s^2+4*s+8))
ans = exp(-2*t)*cos(2*t)+exp(-2*t)*sin(2*t)
442 ttt ieitietet ⎛⎞⎛⎞ ⎜⎟⎜⎟−== ⎝⎠⎝⎠ 2.2g) 2 2222224(2)2(2)2(cos2sin2) 48(2)2(2)2(2)2 t sss ett sssss ++++ ==+=+ ++++++++
2.2h) 2 211 11 (2)(2) t te ss + == ++
ans
t*exp(-2*t)
MATLAB Solution ilaplace(1/(s+2)^2)
=
2.3)Findx(s)forthefollowingdifferentialequations:
x = 1/3+1/6*exp(-3*t)-1/2*exp(-t)
ezplot(x,[0,2])
2.3a) 2
dxdx x ut dtdt ++= 1
== 2 11 ()(43)(3)(1) ()12331 111 1,2,3 362 xs ssssss ccc xs sss ccc == ++++ =++ ++ ===− 13 ()362 tt ee xt =+−
243()
(0)(0)0xx
MATLAB Solution x=dsolve('D2x+4*Dx+3*x=1','x(0)=0','Dx(0)=0')
2.3b) 2
dxdx x ut dtdt ++= 1 (0)(0)0
== 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 0.2 0.4 x y x = t, y = 1/3+1/6 exp(-3 t)-1/2 exp(-t)
22()
xx
22 22 2 ()1123 (21)(1)1 11,21 1(21)3() 1(13)(13)1 31 ()1(1) t ccc xs ssssss cc ssscss cscs c xtte ==++ ++++ ==− =++−++ =++++ =− =−+ MATLAB Solution x=dsolve('D2x+2*Dx+x=1','x(0)=0','Dx(0)=0') x = -exp(-t)-exp(-t)*t+1 0 1 2 3 4 5 6 7 8 9 10 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 t -exp(-t)-exp(-t) t+1 2.3c) 2 2 22() dxdx x ut dtdt ++= 1 (0)(0)0xx==
22 22 2 22 1 ()(221)(221) 221 (2)(2)1 1,2,2 11111() ()22 1111()()2424 ABsC xs ssssss AsAsABsCs ABsACsA ABC s s xs ss ss + ==+ ++++ ++++= ++++= ==−=− ++ + =−=− ++++ 2 ()1[cos()sin()] 22 t tt xte=−+ MATLAB Solution x=dsolve('D2x+2*Dx+x=1','x(0)=0','Dx(0)=0') x = 1-exp(-t)-exp(-t)*t
2.4Seesolutiontoproblem2.1.TheMATLABsolutionsareinterspersedwiththehand solutions.
2.5Seesolutiontoproblem2.2.TheMATLABsolutionsareinterspersedwiththehand solutions.
0 1 2 3 4 5 6 0 0.5 1 x y x = t, y = 1-exp(-t)-exp(-t) t
2.6Seesolutiontoproblem2.3.TheMATLABsolutionsareinterspersedwiththehand solutions.
2.7 Ca=dsolve('5*DCa+Ca=2','Ca(0)=3')
Ca = 2+exp(-1/5*t)
T=dsolve('5*DT+T=70','T(0)=80')
T = 70+10*exp(-1/5*t)
Graphingthesolutions: For the mass balance, ezplot(Ca,[0,30])
ThisfigurematchesFigure2-3. For the energy balance, ezplot(T,[0,30])
0 5 10 15 20 25 30 2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 t 2+exp(-1/5 t)
0 5 10 15 20 25 30 70 71 72 73 74 75 76 77 78 79 t 70+10 exp(-1/5 t) ThisfigurematchesFigure2-7.
ilaplace (2/s/(5*s+1)+15/(5*s+1))
Same result as before for the mass balance.
ilaplace(70/s/(5*s+1)+400/(5*s+1))
ans = 70+10*exp(-1/5*t)
2.8FromEq.2.10, () 25(3)215 ()515151 a s Cs ssss + ==+ + ++
2+exp(-1/5*t) ezplot(ans,[0,30]) 0 5 10 15 20 25 30 2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 t 2+exp(-1/5 t)
ans =
Ts ssss
FromEquation2.11, () () 705(80)70400 515151 s
+ ==+ + ++ .
2.9Iftheinletflowofstream1goesfrom10liters/minto20liters/minandtherestof theinputparametersstaythesame,thefollowingchangesoccurintheinletoccursat 3PM:
Thenewflowrateof40L/minmeansthatthenewtimeconstantwillbe:
1503.75min 40/min L L τ == ,assumingthattheoutletflowincreasescorrespondinglyto 40L/min(thiswillbethecaseiftheoutletfromthetankisanoverflowexitstream), otherwisethetankwilleventuallyoverflow.
Thenewmassbalance,fromEq.2.1becomes:
3.752.5(0)3
andthenewenergybalancefromEq.2.3becomes:
ezplot(ans,[0,30]) 0 5 10 15 20 25 30 70 71 72 73 74 75 76 77 78 79 t 70+10 exp(-1/5 t) Sameresultforthetemperatureasbefore.
()
a a gg CLgLLgLgLC LL TCCLCLCLT →+= →+=
() 33 33 32.520/min(1/)(20/min)(4/)(40/)() 454020/min(25)(20/min)(55)(40/min)()
a aa dC CC dt
+==
()()() ()16 3.75401.051066.25(0)80 1000401 dT TTC dt +=+×== Solvingthesetwoequationsinasimilarmannertothatinthetextyields: 3.75 3.75 2.50.5 66.2513.75 t a t Ce Te =+ =+ Process Systems Analysis and Control 3rd Edition Coughanowr Solutions Manual Full Download: http://testbanktip.com/download/process-systems-analysis-and-control-3rd-edition-coughanowr-solutions-manual/ Download all pages and all chapters at: TestBankTip.com
