Moling with first orr quations (Sct. 2.3. Main xampl: Salt in a watr tank. Th xprimntal vic. Th main quations. Analysis of th mathmatical mol. Prictions for particular situations. Salt in a watr tank. Problm: Dscrib th salt concntration in a tank with watr if salty watr coms in an gos out of th tank. Main ias of th tst: Sinc th mass of salt an watr is consrv, w construct a mathmatical mol for th salt concntration in watr. Th amount of salt in th tank pns on th salt concntration coming in an going out of th tank. Th salt in th tank also pns on th watr rats coming in an going out of th tank. To construct a mol mans to fin th iffrntial quation that taks into account th abov proprtis of th systm. Fining th solution to th iffrntial quation with a particular initial conition mans w can prict th volution of th salt in th tank if w know th tank initial conition.
Moling with first orr quations (Sct. 2.3. Main xampl: Salt in a watr tank. Th xprimntal vic. Th main quations. Analysis of th mathmatical mol. Prictions for particular situations. Th xprimntal vic. watr salt pip r i q (t i V (t tank (t r o q (t o instantanously mix
Th xprimntal vic. Dfinitions: r i (t, (t: Rats in an out of watr ntring an laving th tank at th tim t. q i (t, q o (t: Salt concntration of th watr ntring an laving th tank at th tim t. V (t: Watr volum in th tank at th tim t. (t: Salt mass in th tank at th tim t. Units: [ ri (t = [ (t = Volum Tim, [ qi (t = [ q o (t = Mass Volum. [ V (t = Volum, [ (t = Mass. Moling with first orr quations (Sct. 2.3. Main xampl: Salt in a watr tank. Th xprimntal vic. Th main quations. Analysis of th mathmatical mol. Prictions for particular situations.
Th main quations. Rmark: Th mass consrvation provis th main quations of th mathmatical scription for salt in watr. Main quations: t V (t = r i(t (t, Volum consrvation, (1 t (t = r i(t q i (t (t q o (t, Mass consrvation, (2 q o (t = (t, Instantanously mix, (3 V (t r i, : Constants. (4 Th main quations. Rmarks: [ V t = Volum Tim = [ r i, [ t = Mass Tim = [ r i q i q o, [r i q i q o = Volum Tim Mass Volum = Mass Tim.
Moling with first orr quations (Sct. 2.3. Main xampl: Salt in a watr tank. Th xprimntal vic. Th main quations. Analysis of th mathmatical mol. Prictions for particular situations. Analysis of th mathmatical mol. Eqs. (4 an (1 imply V (t = (r i t + V, (5 whr V ( = V is th initial volum of watr in th tank. Eqs. (3 an (2 imply Eqs. (5 an (6 imply t (t = r i q i (t (t V (t. (6 t (t = r i q i (t (r i t + V (t. (7
Analysis of th mathmatical mol. Rcall: t (t = r i q i (t (r i t + V (t. Notation: a(t = (r i t + V, an b(t = r i q i (t. Th main quation of th scription is givn by (t = a(t (t + b(t. Linar ODE for. Solution: Intgrating factor mtho. (t = 1 [ t + µ(s b(s s µ(t with ( =, whr µ(t = A(t an A(t = t a(s s. Moling with first orr quations (Sct. 2.3. Main xampl: Salt in a watr tank. Th xprimntal vic. Th main quations. Analysis of th mathmatical mol. Prictions for particular situations.
Prictions for particular situations. Assum that r i = = r an q i ar constants. If r, q i, an V ar givn, fin (t. Solution: Always hols (t = a(t (t + b(t. In this cas: a(t = (r i t + V a(t = r V = a, W n to solv th IVP: b(t = r i q i (t b(t = rq i = b. (t = a (t + b, ( =. Prictions for particular situations. Assum that r i = = r an q i ar constants. If r, q i, an V ar givn, fin (t. Solution: Rcall th IVP: (t = a (t + b, ( =. Intgrating factor mtho: A(t = a t, µ(t = a t, (t = 1 µ(t t [ t + µ(s b s. µ(s b s = b a ( a t 1 (t = a t [ + b a ( a t 1. So: (t = ( b a a t + b a. But b V = rq i a r W conclu: (t = ( q i V rt/v + q i V. = q i V.
Prictions for particular situations. Assum that r i = = r an q i ar constants. If r, q i, an V ar givn, fin (t. Solution: Rcall: (t = ( q i V rt/v + q i V. Particular cass: V > q i ; V = q i, so (t = ; V < q i. = q v i t Prictions for particular situations. Assum that r i = = r an q i ar constants. If r = 2 litrs/min, q i =, V = 2 litrs, /V = 1 grams/litr, fin t 1 such that q(t 1 = (t 1 /V (t 1 is 1% th initial valu. Solution: This problm is a particular cas q i = of th prvious. Sinc (t = ( q i V rt/v + q i V, w gt (t = rt/v. Sinc V (t = (r i t + V an r i =, w obtain V (t = V. So q(t = (t/v (t is givn by q(t = V rt/v. Thrfor, 1 1 V = q(t 1 = V rt 1/V rt 1/V = 1 1.
Prictions for particular situations. Assum that r i = = r an q i ar constants. If r = 2 litrs/min, q i =, V = 2 litrs, /V = 1 grams/litr, fin t 1 such that q(t 1 = (t 1 /V (t 1 is 1% th initial valu. Solution: Rcall: rt 1/V = 1 1. Thn, r ( 1 t 1 = ln = ln(1 V 1 r t 1 = ln(1. V W conclu that t 1 = V r In this cas: t 1 = 1 ln(1. ln(1. Prictions for particular situations. Assum that r i = = r ar constants. If r = 5x1 6 gal/yar, q i (t = 2 + sin(2t grams/gal, V = 1 6 gal, =, fin (t. Solution: Rcall: (t = a(t (t + b(t. In this cas: a(t = (r i t + V a(t = r V = a, b(t = r i q i (t b(t = r [ 2 + sin(2t. W n to solv th IVP: (t = a (t + b(t, ( =. (t = 1 µ(t t µ(s b(s s, µ(t = a t, W conclu: (t = r rt/v t rs/v [ 2 + sin(2s s.