Question

QUESTION 2 The system shown below is a fermentation tank. In this, water (with dissolved glucose) enters the tank with a flow rate of q: [L/min) and a glucose concentration of c [mol/L). During reaction, glucose is converted at a rate of n = -VKc (where n is the moles of glucose converted per minute [mol/min), V is the volume of fluid in the tank (L'), and K is the reaction rate). The tank has a cross-sectional area of A. [Variable units are show in square brackets) Fermentation tank Vent Assume • Temperature is constant (ie there is a control system which maintains the temperature at a constant value of T). • Perfect mixing • There is no pressure drop over the vent (ie the pressure of the gas inside the tank is equal to atmospheric pressure P. 4.6. Density is constant (a) If the flowrate through the outlet is defined by q = 0.21h, find a model which describes the behaviour of the system. You maybe keep your model as two separate differential equations. (3 marks) (b) Find an equation for the steady-state height and steady-state concentration, h ande, for a nominal input of qi and . (2 marks) (e) Linearize and find the deviation form of the mass balance equation. (3 marks) (d) Find the transfer function 69 (2 marks)

Answer 1

a Apply flowrate (4 min) balon & arross the tank input) - output 2 Accumulation Qi - q a dlv va volume of tome - - 0 9:-9 = f (AxL) Azarea of tonk naheight of water in tenk 2-q=Adh Ach = 9; -2 92025th Apply Cone. I molar flow rate balance bo (input) toutput) & (generation)-(Consumpt, a Accumulation 94 - 9C - (VKO) = (vs) qili- c (2+ v) = a (Ah) Tail - c(q + kv) = A call + Ande AC Anda dh

equalion for steady state height At steady state thao in weget 0.22 2 = 2 Ti = 0.2 hhs 1. = 9,27 T-steedy State Nalue of lapul For steady state cone af det = 0) from ea. We have qic - ac- Ckv = Ahede+ Accha But Adhy - 9-q q; ; - 9E-ck V = Ahde + eq; -9C | Ahde Qig_ck v- cai] & t = cq; + CkAts o = ailio ĝi + kah

② mars balance equation is (mars flow rate in) - (meus flow rate out) 2 – Accumulation of meu lai - 19 = a (fv) - a (1Ah) sa_9 9; -9 - Adult - Auth H P=dewrity of glucose on ho = arrume to be constant Acerola --9-02 Vh -@ At steady state Adhes = ais - 0.2 Why - subtract & by we get Ad(h-he) = (1-96) - 0-2 (5h-ste) en from taylor expansion, we have about etia fru-a) = f(a) + (1-9)f'(a)t... we neglect higher order of (a) to linearize . expand in about steady state value no ha he h = the+ (hehe) The HA

Showing = (h-hs) 2 hs c put this in er 6 we get A ech he) = (iris) - 0.21h-he) hohe H6 =devation variable of hight 25 a-us=Q10 = deviation dariable of inlet vol flow rate Ady = Qi - 0.2H a the Adat - Qi - 0:24 - demnation from of man balase taking laplace of eqz on both side we get As 418) – H60) = 2; (S) - 0.24 (8) 2V # 10/20 becaurse at time to tankic I empty AS HIS) = Gils) - 0 1 4 (8) 2 he rhy HIS) +1) Astoll VE 1199 - her that test 2: 107,4 Q'IS