solution hw4

solution hw4 - Problem 1 a) For noninteracting capacities...

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Problem 1 a) For noninteracting capacities with linear resistances subject to a unit-step change in the input of the first tank, the material balance can be written as follows: 1 11 1 1 dy AR y Ru(t) dt += (S1.1) 22 2 1 1 dy R y y dt R (S1.2) subject to the initial conditions s 1 s y(0 ) y Ru == s 2 1 1 s 2 s R Ry = (S1.3) Defining the deviation variables Y 1 =y 1 -y 1s , Y 2 =y 2 -y 2s and Q=u(t)-u s , equations (S1.1)- (S1.2) become: 1 1 1 dY Y RQ dt (S1.4) 2 1 1 dY R Y Y dt R (S1.5) subject to the initial conditions 1 Y(0) 0 = 2 = (S1.6) Hence, the transfer functions for equations (S1.4)-(S1.5) are: 1 Y(s) R G(s ) Q(s) (A R )s 1 + (S1.7) 1 2 12 2 R R G( s ) (A R )s 1 + (S1.8) Since the two noninteracting tanks are placed in series, the overall transfer function is the following: [] [ ] 2 2 R G(s) G (s)G (s) Q ( s ) ( ) s1( AR) s1 = + + (S1.9) Rewriting (S1.9) in the standard form, gives: 2 R G(s) G (s)G (s) Q(s) (A R )(A R )s 2(A R A R ) 1 = ⎡⎤ + ++ ⎣⎦ (S1.10)
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For critically damped systems, ξ =1. Therefore, comparing equation (S1.10) with the standard form yields: 2 12 1 1 2 2 (A R )(A R ) τ=ττ= (S1.11) 1 1 2 2 AR τ=τ +τ = + (S1.12) Equations (S1.12)-(S1.11) are simultaneously solved if: 1 1 2 2 A R A R τ=τ = (S1.13) In other words, the system is critically damped if the roots of the denominator in (S1.9) are in reality one root with multiplicity equal to 2. This immediately implies (S1.13).
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This note was uploaded on 10/04/2009 for the course CHE 470 taught by Professor Smith during the Spring '09 term at Rice.

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solution hw4 - Problem 1 a) For noninteracting capacities...

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