Question

STUDENT NAME STUDENT ID 01. (60 points) In this question, you are asked to model the process system in Floure which is a heat exchanger followed by a non-isothermal stirred tank chemical reactor. Figure 1 also contains modeling assumptions for this system. Your Overall goal is to determine the concentration of the reactant in the reactor (CA) for a change in the wall temperature (1) in the exchanger. a. Answer the following questions for the process system. (10 points 1. What are the input(s) and output(s) of the system? 2. Which input(s) could be considered as disturbances? b. You are to develop the complete model of the system that governs the response the reactant (CA), for a change in the exchanger wall temperature (TW). (50 Points F.T.C . W V.T.C. V,,,, CST Heater (No Reaction) CST Reactor (No Heat Transfer) -E/RT Assumptions and modeling guidance: 1. The stirrer work is not negligible; 2. Negligible heat is transferred through the open surface; 3. The tanks are well-mixed; 4. Physical properties are constant (density (P), Cp): 5. Tanks are constructed of steel with mass m,. Cp, which is the same for both tanks 6. For the reaction 2A - B+C: The reaction rate is: -PA = koe CA 2. The heat transferred from the wall to the liquid has a heat transfer coefficient (U): o not write an energy balance on the wall; we assume that it is held a temperature by a fluid in a jacket around the CSTR (not shown in the diagram). 9. The reaction is endothermic with heat of reaction AH.; 10. TW, is uniform and > TO. 11. This changed by a built in electric heater with duty WA Figure 1. Process addressed in Question 1. d

Answer 1

F. To, CA Vi Ti CA CST Heatos (Ho reaction) 22 CST Reactor (No heat transtor) Answers: (a) Inputs (s) * To CAO, WL, outputs (s) = V2, Tz & CA, 27 Disturbance Variable ; Wi, Wo. Assuming To is constant throughout (b) for tank-18. Taking mass balance over tank-l; Mass accumulated = mass in- mass our a (en - FS-15 dt CE Energy balance! Enthalpy Accumuand - input-output t Supplied d (ev Cp T = Sotto - PCFT, tw, dt pvcp dT - SCPF (To - T, tw, de 5 (To - Tu) to wo dt dt

At steady state; - SUCO Subtracting at from ☺ I ((T. – Tos) – (T - Ts) + (w-W Suce apalace Transform SVG of Taking Laplace Transtorm {si(s) - 9067 = C)- TiC) ?+ 960 STO=1,6} - $ 1,0 t (946) (9 576 + (str) ī (6) - W. (8) Suce T(S) w (s) - (st Ex evce Ti (s) - Ysee YSCOF wi(s) (vstf) - Cyst1 where, K = 3F , T = 2 where, hea

for tank 2 Component mass balance on Mass accumulated & mass in - mass out & dissapperence d (UC) - FCA, - FCA ₂ - Kettner To a Twa - (C) = F (CA, - CA,) - Kohlwax A -X x = Linearisation : as power of A is 2. -E/RTWA k, екіх, яke A + 2 Elets (AA) 2 ke ((A2 A 1-E/RTW2 is (Two-Twas) 8. koe & + 2 k è lawas to te fatass / E k C X Tango i. Ko é E/RTE 2 at BCA to ī where, &, ß t r are constants grouped together 2= koe B 2864, ko e to e Eletos -E/RT, ERTO Y RT?

from Eq" ① dt vd (CA) F (CA. - CA ) - ( t BTA + rTwo) At steady state vd (CARS) = F (CAIS - CARP) - Q -- (1) vd [a - lass] = R Cla, - Caps) - (Ca - Cass) | - 8 - Ý Tư, A] = F / Cr, - CA ] - BC - Y To V d dt Taking L palace Transform v Ls Zag (G) - anos J = F | Engels - CA,(6)] -ß G, (3) - V Tun (5) СА, 0 Весовые CA = "A = CA, 8. VSTA(s) = -FA (S) -BA () -8 To (1) . (Ust F+B) (s) = - Tw(s) CA (S) ar Tw(s) - (ust (F+B))

Two, (s) - (F+B) (4) St1] FAB where, Kerr (F+B) Tp 2 V FAB