Ch04 - 4-1 4-2 By using Simulink-MATLAB above solutions can be verified (t).050.10.150.20.250.30.350.40.450.5timey(t)Fig S4.2a Output for part c

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Unformatted text preview: 4-1 4-2 By using Simulink-MATLAB, above solutions can be verified: 510152025303540455012345678910timey(t)51015202530354045500.050.10.150.20.250.30.350.40.450.5timey(t)Fig S4.2a. Output for part c) and d) Fig S4.2b. Output for part e)51015202530354045500.511.522.533.544.55timey(t)2468101214161820-0.3-0.2-0.10.10.20.30.40.50.60.7timey(t)Fig S4.2c. Output for part f) Fig S4.2d. Output for part g) 4.3 a)The dynamic model of the system is given by )(1wwdtdVi-ρ=(2-45) CVQTTVwdtdTiiρ+-ρ=)((2-46) Let the right-hand side of Eq. 2-46 be f(wi,V,T), ( 29TTfVVfwwfTVwfdtdTssisii′∂∂+′∂∂+′∂∂==,,(1) 4-3 )(1TTVwfisi-ρ=∂∂1)(22=-=ρ--ρ-=∂∂siisdtdTVCVQTTVwVfρ-=∂∂VwTfis=dtdTiiwTTV′-ρ)(1TVwi′ρ-, dtTddtdT′=Taking Laplace transform and rearranging 1/)()()(+ρ-=′′swVwTTsWsTiiii(2) Laplace transform of Eq. 2-45 gives ssWsViρ′=′)()((3) If sVf∂∂were not zero, then using (3) 11)()()(+ρ∂∂+-=′′swVsVfwVwTTsWsTisiiii(4) Appelpolscher guessed the incorrect form (4) instead of the correct form (2) because he forgot that sVf∂∂would vanish. b)From Eq. 3, ssWsViρ=′′1)()(4-4 4.4 1KY( s )G( s )X ( s )s( s)==τ +G(s)Interpretation U(s)Interpretation of G(s) of u(t) )1(+τssK2ndorder process *1 δ(0) [ Delta function] 1+τsK1storder processs1S(0) [Unit step function] sKIntegrator 1+τsKτ-τ/1te[Exponential input] KSimple gain )1(1+τssτ--/1te(i.e no dynamics)[Step + exponential input] * 2nd order or combination of integrator and 1storder process 4.5 a) 2dtd1y= -2y1– 3y2+ 2u1(1) dtd2y= 4y1 – 6y2+ 2u1+ 4u2(2) Taking Laplace transform of the above equations and rearranging, (2s+2)Y1(s) + 3Y2(s) = 2U1(s) (3) -4 Y1(s) + (s+6)Y2(s)=2U1(s) + 4U2(s) (4) Solving Eqs. 3 and 4 simultaneously for Y1(s) and Y2(s), 4-5 Y1(s) = )4)(3(2)(U12)(U)3(224142)(U12)(U)62(21221++-+=++-+ssssssssssY2(s) = )4)(3(2)(U)1(8)(U)3(424142)(U)88()(U)124(21221+++++=+++-+ssssssssssssTherefore, 41)(U)(Y11+=sss, )4)(3(6)(U)(Y21++-=ssss42)(U)(Y12+=sss, )4)(3()1(4)(U)(Y22+++=sssss4.6The physical model of the CSTR is (Section 2.4.6) AAAiAVkcccqdtdcV--=)((2-66) )()()(TTUAVkcHTTwCdtdTCVcAi-+∆-+-=ρ(2-68) where: k= ko e-E/RT(2-63) These equations can be written as, ),(1TcfdtdcAA=(1) ),,(2cATTcfdtdT=(2) Because both equations are nonlinear, linearization is required. After linearization and introduction of deviation variables, we could get an expression for )(scA′/ )(sT′. 4-6 But it is not possible to get an expression for )(sT′/)(sTc′from (2) due to the presence of cAin (2). Thus the proposed approach is not feasiblebecause the CSTR is an interacting system....
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This note was uploaded on 04/10/2008 for the course CHE 242 taught by Professor Cummings during the Spring '08 term at Vanderbilt.

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Ch04 - 4-1 4-2 By using Simulink-MATLAB above solutions can be verified (t).050.10.150.20.250.30.350.40.450.5timey(t)Fig S4.2a Output for part c

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