I NEED THIS AS SOON AS POSSIBLE PLEASE!!! IN THIS SOLUTION, THE RED CIRCLED AROUND EQUATIONS MUST BE ENTERED MATLAB THEN NEED A PLOT GRAPH FROM MATLAB. PLEASE HELP TO PLOT GRAPH FROM CIRCLED EQUATION'S GRAPH FROM MATLAB

Example & z QO CAO ringa Te R ICM 82 CAZ Q3 db CAB E Tankt Tank? Tank3 The concentration of A To the same at each tones initially - At we to water is ted to the first tones and a constant Q maintained 1.50rusy throughout the system as Qo-Q1 = Q2 =Q = 300 L/min. Determine the time when the concentration of A in the first tank decreases to 6/40. How does the concentrations of the solutions in tonk 2 and 3 change with time? How long does it take for the exit stream to have Az = 60/10? Data: : V-V2Vzc constant -50x108L Soln Component Balonce for Around Tonk 1 Q min Q. CA CA mot mo! Mass balance ecpildi Qo. Co-Q1. CAN dt time ted 0 - 3000 CAJ - 50x103 dCAA dt 1 ma) miny 4 u mal ) dCAL ett 0 unter is Sant of 300 dt * 50300 , since it is time, it must been mpiol cariello InCAL -0.006+'+41 booking Now, we must integration constant At t=0 CAN=Co= constant number inG -0.006(0) + a = Thco Incm=-0.0067 +Inco in ) -- 0.035t . same (CAN -0.006 6te Twe arrange this) GM0 In ( Come) - -0.00647 * t = 3.86min Componort Balance for A around Tank 2 Q1. CAN-Q2. (Az = V. dcAZ dt 300 / CAN-CA2) = 50x103 dCAZ at SCAN-CA2 = 167. dCAZ dt Cool 0.0067 dt dCAZ -0.006 167 dy Pa og 96) dy +PLX) y= Q(x) CAL - 1670 m2 1.CAL 167 dt, 2= eSP(x)dx Solution : Roy = 12. Qx) dx + C) Stadt +/167 0.006 CAze ESTONE -0.0064 0.0067 t + (ace 2.0064 CAR . CA2 e 0.0067 - flot Cat 2006 + dt + C Co* 2006tt.C -0.00lt *[6 *G* 0.006et +c] t=0 (A2=66 Co=eo [o+cJ C=0 -0.006t CA2 = 6*10.0064 +1) e dt CAT dt * + At 0.0040. Component balance for A dround Tank 3 Q2. Caz-Q3. CA3 = Vod(Az 300(CAZ-CA2) 50x40 CAS (CA2-CAz = 167 dCA3 dt 0.006 CAZ-CA) dCA3 +0.006CAz = 0.006 CA2) dCA? +0.006ca3 = 0.00606* (1+0.006t) ( dt, dy ax dy cx a=0 SP(x)dx solution i hy = Shu Qlxlax +C a=0 50.006att 2006t Cazue b.dost - 0.0066o (4+0.0064 )4 -0.0064 20.0964 dit + c 0.006 P(x) y 3(x) + Pllye Olx) - C 0.0067 + Aques (Az. enost e- 2.00664% (+48.0064) H +C 1) + C 0.00666 (t+2003t?) + c CAz = -0.006470.006 * G ***(1+0.0034) +c] to find c we need a relation between t and CA3 Initial cond. to CA360 6 =1[0+c] =) C= 66 (Az = Cote -0.0064 [0.006***(1+0.003+)+1] + t3 -2 (Az - G 2-6-005+ + [0.006+(1+0.32+)+] do Seo If we take in of both sides : -0.0065 -0.006 - 1 (10) = [e-2.06t4 [2.006+ (1+0.0234 )41]] Since Intxay) - laxtlay INO)= lhe +In (0.006*(1+0.003t)+1] 2.3 = - 0.00644[0.006+ + (1+ 2, 093t6 0.0067-23 = IN [0.006+* (140.393+)+1] B A-B-0 t ) A 400 500 600 t3 886 min Example & z QO CAO ringa Te R ICM 82 CAZ Q3 db CAB E Tankt Tank? Tank3 The concentration of A To the same at each tones initially - At we to water is ted to the first tones and a constant Q maintained 1.50rusy throughout the system as Qo-Q1 = Q2 =Q = 300 L/min. Determine the time when the concentration of A in the first tank decreases to 6/40. How does the concentrations of the solutions in tonk 2 and 3 change with time? How long does it take for the exit stream to have Az = 60/10? Data: : V-V2Vzc constant -50x108L Soln Component Balonce for Around Tonk 1 Q min Q. CA CA mot mo! Mass balance ecpildi Qo. Co-Q1. CAN dt time ted 0 - 3000 CAJ - 50x103 dCAA dt 1 ma) miny 4 u mal ) dCAL ett 0 unter is Sant of 300 dt * 50300 , since it is time, it must been mpiol cariello InCAL -0.006+'+41 booking Now, we must integration constant At t=0 CAN=Co= constant number inG -0.006(0) + a = Thco Incm=-0.0067 +Inco in ) -- 0.035t . same (CAN -0.006 6te Twe arrange this) GM0 In ( Come) - -0.00647 * t = 3.86min Componort Balance for A around Tank 2 Q1. CAN-Q2. (Az = V. dcAZ dt 300 / CAN-CA2) = 50x103 dCAZ at SCAN-CA2 = 167. dCAZ dt Cool 0.0067 dt dCAZ -0.006 167 dy Pa og 96) dy +PLX) y= Q(x) CAL - 1670 m2 1.CAL 167 dt, 2= eSP(x)dx Solution : Roy = 12. Qx) dx + C) Stadt +/167 0.006 CAze ESTONE -0.0064 0.0067 t + (ace 2.0064 CAR . CA2 e 0.0067 - flot Cat 2006 + dt + C Co* 2006tt.C -0.00lt *[6 *G* 0.006et +c] t=0 (A2=66 Co=eo [o+cJ C=0 -0.006t CA2 = 6*10.0064 +1) e dt CAT dt * + At 0.0040. Component balance for A dround Tank 3 Q2. Caz-Q3. CA3 = Vod(Az 300(CAZ-CA2) 50x40 CAS (CA2-CAz = 167 dCA3 dt 0.006 CAZ-CA) dCA3 +0.006CAz = 0.006 CA2) dCA? +0.006ca3 = 0.00606* (1+0.006t) ( dt, dy ax dy cx a=0 SP(x)dx solution i hy = Shu Qlxlax +C a=0 50.006att 2006t Cazue b.dost - 0.0066o (4+0.0064 )4 -0.0064 20.0964 dit + c 0.006 P(x) y 3(x) + Pllye Olx) - C 0.0067 + Aques (Az. enost e- 2.00664% (+48.0064) H +C 1) + C 0.00666 (t+2003t?) + c CAz = -0.006470.006 * G ***(1+0.0034) +c] to find c we need a relation between t and CA3 Initial cond. to CA360 6 =1[0+c] =) C= 66 (Az = Cote -0.0064 [0.006***(1+0.003+)+1] + t3 -2 (Az - G 2-6-005+ + [0.006+(1+0.32+)+] do Seo If we take in of both sides : -0.0065 -0.006 - 1 (10) = [e-2.06t4 [2.006+ (1+0.0234 )41]] Since Intxay) - laxtlay INO)= lhe +In (0.006*(1+0.003t)+1] 2.3 = - 0.00644[0.006+ + (1+ 2, 093t6 0.0067-23 = IN [0.006+* (140.393+)+1] B A-B-0 t ) A 400 500 600 t3 886 min