solution hw2

solution hw2 - Problem 1 Overall Material Balance: dn d(V)...

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Problem 1 Overall Material Balance : ii dn d( V) d( Ah) dh d F F A Ah F F dt dt dt dt dt ρρ ρ == = ρ ρ ρ += ρ ρ (S1.1) In absence of chemical reactions and at constant temperature and pressure, the density of the fluid liquid in the tank can be assumed constant. Therefore, (S1.1) yields: 12 dh AF F F 2 0 h dt =−=− (S1.2) subject to the initial condition h(0)=h s . At steady state, equation (S1.2) provides a functional relationship between the level asymptotically attained by the liquid in the tank and the stationary inlet flow rate: ss s dh 0 F 20h h 9ft dt =⇒ = = (S1.3) The process model in (S1.3) is nonlinear. In order to linearize this model, one can expand the function F i -20h 1/2 in (S1.3) about the steady state value F s -20h s 1/2 as follows: () ( ) [] iS SS is s i s s FF hh i s i s F2 0 h 0 h F 20h F 20h F F h h HOT Fh 0 1 F F 10h h h F F 10h h h −− ⎡⎤⎡⎤ ∂− ⎢⎥⎢⎥ −= −+ + + = ∂∂ ⎣⎦⎣⎦ ⎡⎤ + =−− ⎣⎦ (S1.4) Substituting equation (S1.4) into equation (S1.2) and defining the deviation variables H=h-h s and R=F i -F S provides the linearized process model in deviation form: dH 10 AR H dt 3 =− (S1.5) subject to the initial condition H(0)=0. Let 10/3 be equal to B. Then, solution to equation (S1.5) can be obtained by employing the Laplace transform, as follows: RR Q ( s ) A sH(s) H(0) BH(s) H(s) ( A s B ) P ( s ) = = + (S1.6) Notice that P(s) has two simple (multiplicity=1) roots s=0 and s=-B/A. By using partial fraction expansion, (S1.6) can be rewritten as follows: CC RQ ( s ) H(s) s(As B) P(s) s s B A = + ++ (S1.7) where 1 s0 2 sB A A R Cs H ( s ) R / B As B R C (As B)H(s) AR / B s = = = ⎢⎥ + =+ = = (S1.8)
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Therefore, (S1.7) becomes: R1 A H(s) Bs sBA ⎛⎞ =− ⎜⎟ + ⎝⎠ (S1.9) Using table 7.1 from the book and taking the inverse Laplace transform of (S1.9) yields: [] s R H(t) h(t) h 1 Aexp( B At) B (S1.9) or i 3(F 60) 10 h(t) 9 1 exp t 10 3 ⎡⎤ =+
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solution hw2 - Problem 1 Overall Material Balance: dn d(V)...

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