Question

Question 4 (9 points) A storage tank has a diameter of 20 ftand a height off 10 ft. The output volumetric flow from this tank is given by out)2h(twhere h(t) is the height of liquid in the tank. At a particular time, the tank is at steady state with an input flow of 10 ft/min. dia Vin.s (a) Develop a dynamic model of the process. (b) Derive the transfer fanction H(s)/Qie(s) for the tank system. H' and Q are deviation variables 4in (c i the imput flow is ramped up at the rate of 0.1 ftminlapproximately after how long will the tank reach an overflow condition? (d) Sketch the dynamic response. Go= -22 Assumpiions well mixesl canstant 2s+t k 622-32cs) Qin=0- no heating 9.9 Q'in (s) constant TandP. costant Area. Ain-Phout in-4out -dt (ha) H'(t)= 9.9k(- gin A d O bact of e page @ st-st s 2(he) A dt Geas Res) HCS) in-ins) A d(h-hs) Qin-fin-Fin,s 4'h-hs in= A dt Laplace AH's) Page 3 of 4 Pines) LST10)6a8-32 Pe A= ar

Answer 1

Tank system :-(mass capacity systems) - F. As the inlet changes, level of liquid in the tools . Cheonges which in teem Changes the hypo- static pressure applied on the wall which results in charge in the oultet flow teate: . This system con Eegulate efself on thout the help of any external agency, whether it is maneras or by any controller, thats why its exemple of self regulating systema For a l'm d Foachim

E Applying the mass balance : SF; - SF = d (PAH) → Incompeteilte liquidh of will be Sopome) C at E o constant areg. 9 & Fi-Fo = adhi E I F.- ch'm a Adh. - => Feigen content Adhes i substored on-liig at : (Fi- Fise) - cf4llzweden) - Aldhoodhes) dt - ennem) . This step con not infeodered the deviation vorejable, plinearization : conveęsion of non-lined peeportinolity to linear peoportionality. - two methods :- - 1

o greaphical method for lineaeration then an. - Pozchilm iz f" (alta about ***(zza) Fost-- apyrynja mloc-) Į Fo-Foss) = Foches) (h-hes) i po = Fo'chss) ñ : . hss has & Analytical method for linearization an Taylor form $12] = $(69+62-93$'(a)=@exa}{"conta Folhy - Folhos) +htss) Fiches [hanss} to = Po ches) 5 l from en (*) - - Roz Adh . at Fe - Pol Ches) ħ andñ is frust at t han tahss in El ordinearly diff en e . I with constant coffe a S lees He why is op het sowny is he cause its repo the reporesent day e . Jelines frust order Cheshm

so reso = KP (1+Eps) Here Kp = 1 ER . Fol(chss.) Ep = (Area) folches) : AR (C) T (S) KP , given F (5)=0:12 F (S) (tips) . s2 Fi(t) = o.lt I Fils) = 0.1 n e (H+ cps) sz -0 " z

Take inverse laplace- hld) = 0.1169(+- Clinetler)). KP - Faleminde ring =0.5 hke If = Antep (1701) - 628. min. for tank to overflowin ült) = 10ft 10 5+ = (0-1)(0.5)(-+ - 62€ (i-e tea) . - 200 = 7 828= t_628 +628 é +/628 +628 e +/628 I +=364 min Mit) (A+) gorput (At- AEP) output t -