Homework_1_Solutions_Complete - Homework Problem Set #1:...

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Unformatted text preview: Homework Problem Set #1: Solution Key Problem 1: 1.6 a) Feedback Control : Measured variable: y Manipulated variable: D,R, or B(schematic shows D) b) Feedforward Control: Measured variable: F Manipulated variable: D (shown), R or B 1-4 Overall balance (No accumulation of mass): 0 = ρq − ρq1 thus q1 = q (1) Energy balance (No change in volume): V11C dT1 = 1qC 2Ti − T1 3 − UA2T1 − T2 3 dt (2) Compartment 2: Overall balance: 0 = ρq1 − ρq2 thus q2 = q1= q (3) Energy balance: V21C b) dT2 = 1qC 2T1 − T2 3 + UA2T1 − T2 3 − U c Ac 2T2 − Tc 3 dt (4) Eight parameters: ρ, V1, V2, C, U, A, Uc, Ac Five variables: Ti, T1, T2, q, Tc Two equations: (2) and (4) Thus NF = 5 – 2 = 3 2 outputs = T1, T2 3 inputs = Ti, Tc, q (specify as functions of t) c) Three new variables: ci, c1, c2 (concentration of species A). Two new equations: Component material balances on each compartment. c1 and c2 are new outputs. ci must be a known function of time. Problem 2: 2.7 Let the volume of the top tank be γV, and assume that γ is constant. Then, an overall mass balance for either of the two tanks indicates that the flow rate of the stream from the top tank to the bottom tank is equal to q +qR. Because the two tanks are perfectly stirred, cT2 = cT. 2-6 Component balance for chemical tracer over top tank: 4V dcT 1 = qcTi + qR cT − 2 q + qR 3cT 1 dt (1) Component balance on bottom tank: (1 − 43V or dcT 2 = 2q + qR 3cT 1 − qR cT − qcT dt (1 − 43V dcT = 2q + qR 32cT 1 − cT 3 dt (2) Eqs. 1 and 2 constitute the model relating the outflow concentration, cT, to inflow concentration, cTi. Describing the full-scale reactor in the form of two separate tanks has introduced two new parameters into the analysis, qR and γ. Hence, these parameters will have to be obtained from physical experiments. Problem 3: 2.8 Additional assumptions: (i) Density of the liquid, ρ, and density of the coolant, ρJ, are constant. (ii) Specific heat of the liquid, C, and of the coolant, CJ, are constant. Because V is constant, the mass balance for the tank is: ρ dV = q F − q = 0 ; thus q = qF dt Energy balance for tank: ρVC dT 0.8 = q F ρC (TF − T ) − Kq J A(T − TJ ) dt (1) Energy balance for the jacket: ρ J VJ C J dTJ dt = q J ρ J C J (Ti − TJ ) + Kq J 2-7 0.8 A(T − TJ ) (2) where A is the heat transfer area (in ft2) between the process liquid and the coolant. Eqs.1 and 2 comprise the dynamic model for the system. 2.9 Additional assumptions: i. The density ρ and the specific heat C of the process liquid are constant. ii. The temperature of steam Ts is uniform over the entire heat transfer area iii. Ts is a function of Ps , Ts = f(Ps) Mass balance for the tank: dV = qF − q dt Energy balance for the tank: 1C d V (T − Tref ) dt = qF 1C 2TF − Tref 3 − q1C 2T − Tref 3 (1) (2) +UA(Ts − T ) where: Tref is a constant reference temperature A is the heat transfer area Eq. 2 is simplified by substituting for (dV/dt) from Eq. 1, and replacing Ts by f(Ps), to give ρVC dT = q F ρC (TF − T ) + UA[ f ( Ps ) − T ] dt Then, Eqs. 1 and 3 constitute the dynamic model for the system. 2-8 (3) Problem 4: a) mass balance: ρA1 dh1 = w1 − w2 − w3 dt (1) dh2 = w2 dt (2) ρA2 Flow relations: Let P1 be the pressure at the bottom of tank 1. Let P2 be the pressure at the bottom of tank 2. Let Pa be the ambient pressure. w2 = Then P1 − P2 ρg (h1 − h2 ) = R2 g c R2 (3) P1 − Pa ρg h1 = R3 g c R3 (4) w3 = b) Seven parameters: ρ, A1, A2, g, gc, R2, R3 (Ri is the valve resistance) Five variables : h1, h2, w1, w2, w3 Four equations Thus NF = 5 – 4 = 1 1 input = w1 (specified function of time) 4 outputs = h1, h2, w2, w3 2.4 Assume constant liquid density, ρ . The mass balance for the tank is d (ρAh + m g ) dt = ρ( q i − q ) Because ρ, A, and mg are constant, this equation becomes 2-3 Practice Problems ...
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This document was uploaded on 10/27/2011 for the course ECH 157 at UC Davis.

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