ch11 - 11-1 11-2

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Unformatted text preview: 11-1 11-2 36.1ftpsi7.1ftmin.1psimin/ft2.)4(23===mpvcOLKKKKK223min62.636.1min)3min)(3(==ττ=τOLIK2 ζ3 τ3= min21.5336.136.21=×=τ+IOLOLKK121.21)121.2()1.3(13)()(+=++++=′′sssssHsHspFor sssHsp1)23()(=-=′21..2/1)(teth--=′[ ])(1ln21.2tht′--=ft5.2)(=thft5.)(=′thmin53.1=tft.3)(=thft.1)(=′th∞→tTherefore, ft5.2min)53.1(==thft.3)(=∞→th11-3 11.3 ma/ma5)(==ccKsGAssume τm= 0, τv = 0, and K1= 1, in Fig 11.7. a)Offset = FFFTTsp86.14.45)()(=-=∞′-∞′b)+τ++τ=′′111)()(22sKKKKKsKKKKKsTsTvIPcmvIPcmspUsing the standard current range of 4-20 ma, Fma/32.50ma4ma20=-=FKm2.1=vK, psi/ma75.=IPK, τ=5 min , ssTsp5)(=′)440.115(20.7)(22KssKsT++=′)440.11(20.7)(lim)(22KKsTsTs+=′=∞′→F14.4)(=∞′Tpsi/F34.32=Kc)From Fig. 11-7, since =′iT)()(2∞′=∞′TKKPvt, psi03.1)(=∞′tPand TKTKKPivt=+′12, psi74.3=tPpsi77.4)()(=∞′-=∞tttPPP11-4 11.4 a) b) smmmeKsGθ-=)(assuming τm= 0 ssmeesG2323sol/ftlbma67.2ftsollb)39(ma)420()(--=--=τ+=sKsGIcc11)(psi/ma3.)(==IPIPKsGpsiUSGPM67.1)612(USGPM)2010()(-=--==psiKsGvvOverall material balance for the tank, hCqqdtdhAv-+=213ftUSgallons481.7(1) Component balance for the solute, 322113)()(481.7chCcqcqdthCdAv-+=(2) Linearizing (1) and (2) gives 11-5 hhCqdthdAv′-′=′2481.72(3) ( 29332222332481.7chChhCccqqcdtcdhdthdcAvv′-′-′+′=′+′Subtracting (3) times 3cfrom the above equation gives )(481.7323ccdtcdhA-=′( 293222chCcqqv′-′+′Taking Laplace transform and rearranging gives )(1)(1)(22213sCsKsQsKsC′+τ+′+τ=′where USGPMsol/ftlb08.3321=-=hCccKv6.22==hCqKvmin15481.7==τvChAsince 22ft6.124/=π=DA, and ftCqqCqhvv422123=+==Therefore, 11508.)(+=ssGp1156.)(+=ssGd11-6 c) The closed-loop responses for disturbance changes and for setpoint changes can be obtained using block diagram algebra for the block diagram in part (a). Therefore, these responses will change only if any of the transfer functions in the blocks of the diagram change. i.2cchanges. Then block transfer function )(sGpchanges due to K1. Hence Gc(s) does need to be changed, and retuning is required. ii. Kmchanges. Block transfer functions do change. Hence Gc(s) needs to be adjusted to compensate for changes in block transfer functions. The PI controller should be retuned. iii. Kmremains unchanged. No block transfer function changes. The controller does not need to be retuned. 11.5 a) One example of a negative gain process that we have seen is the liquid level process with the outlet stream flow rate chosen as the manipulated variable With an "air-to-open" valve, wincreases if pincreases. However, hdecreases as wincreases. Thus Kp<0 since ∆h/∆wis negative....
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This homework help 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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ch11 - 11-1 11-2

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