solution hw1

solution hw1 - Problem 1 Overall Material Balance: dn d(V)...

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Problem 1 Overall Material Balance : ii dn d( V) dV d F F V F F dt dt dt dt ρρ == ρ ρ ρ + = ρ ρ (S1.1) Assuming constant density and reactor volume, equation (S1.1) yields: () F F 0 F F 0 ρ−= = (S1.2) Therefore, the input and output flow rates are equal at each point in time. Component A Material Balance : 2 A iA f A 1A 3A d(Vc ) Fc k c V k c V dt =− (S1.3) Taking into account equation (S1.1) and the fact that the reaction volume is constant, equation (S1.3) can be rewritten as: 2 A Af A 1 A 3 A dc F cc k c k c dt V (S1.4) Component B Material Balance : B B1 A 2 B d(Vc ) Fc k c V k c V dt + (S1.5) Since the reaction volume is constant, equation (S1.5) becomes: B A2 B dc F ck c dt V + (S1.6) Component C Material Balance : C C2 B d(Vc ) Fc k c V dt + (S1.7) As the reaction volume is constant, equation (S1.7) is rewritten as follows:
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C C2 B dc F ck c dt V =− + (S1.8) Component D Material Balance : 2 D D3 A d(Vc ) 1 Fc k c V dt 2 + (S1.9) As the reaction volume is constant, equation (S1.9) yields: 2 D A dc F 1 c dt V 2 + (S1.10) In equations (S1.4) and (S1.10), the reaction rate is defined with respect to the number of moles of A that are consumed due to reaction (1.3). Therefore, the factor 1/2 appearing in equation (S1.10) accounts for the fact that for each mole of A consumed, half mole of D is produced. Thus, a) The model is given by equations (S1.4), (S1.6), (S1.8) and (S1.10) b) State variables: c A , c B , c C , c D Input: F i , c Ao Output: F, c A , c B , c C , c D Parameters: ρ , V, k 1 , k 2 , k
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solution hw1 - Problem 1 Overall Material Balance: dn d(V)...

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