Consider a two-tank brine system shown to the right. Let x1 be the amount of sal
ID: 3286947 • Letter: C
Question
Consider a two-tank brine system shown to the right. Let x1 be the amount of salt in tank 1 and x2 the amount in tank 2. The volume of tank 1 is 100 gal, and that of tank 2 is 50 gal. Initially there are 50 lb of salt in tank 1 and none in tank 2. The input is 10 gal/min of water containing 0.2 lb of salt per minute. Transfer 1 is of 20 gal/min from tank 1 to tank 2, and transfer 2 is 10 gal/min from tank 2 to tank 1. The output Ls 10 gal/min from tank 2. Assuming both tanks are well-mixed, derive the system of equations x'1 = 2 - 1/5 x1 + 1/5 x2 x'2 = 1/5 x1 - 2/5 x2 modeling the amount of salt in the two tanks. Use the method of elimination to reduce this to a single equation and solve that to find x1 and x2.Explanation / Answer
let at any time t amount of salt in tank1 be x1 and in tank 2 be x2.Now,for tank1 amount of salt entering at any instant=0.2*10 by input and 10/50*x2 from tank2.amount of salt exiting tank1=20/100*x1 to tank2.Therefore,rate of change for tank1 x1'=2-(x1/5)+(x2/5) similarly for tank2 rate of change x2'=20/100*x1-(10/50)*x2-(10/50)*x2=(x1/5)-2*(x2/5) now,for elimination we introduce the differential operator D,the equations are now Dx1=2-(x1/5)+(x2/5) Dx2=(x1/5)-2*(x2/5) 1) (D+1/5)x1-(1/5)x2=2 and 2) (1/5)x1+(D+2/5)x2=0 Multiplying 1 by 1/5 and 2 by (D+1/5) and eliminating x1 we get -x2/25-(D+1/5)(D+2/5)x2=2 similarly find an ODE for x1 and solve the ODEs to get the solution.