Question

8(10pts) A tank with a capacity of 500 gal originally contains 200 gal of water with 100 lb of salt in the solution. Water containing 1 lb of salt per gallon in entering at a rate of 3 gal/min, and the mixture is allowed to flow out of the tank at a rate of 2 gal/min. Find the amount of salt in the tank at any time prior to the instant when the solution begins to overflow.

Answer 1

Answed. Givendatar- capacity of the tank -500 gal 089;nally contains = 200gal of water with 100lb of salt in solution Now Let X(t) be the amount of salt in the tank at time it". V(t) is the volume of the water in tank, 50 V(t) = = 2007 X(t) = 3 - 2x (6) V (1), 3- 2x4) 200+ Also given that the initial condition 460)=100 . The integrating factor ,M(t)= J dt 20076 e 2109 (2007) C = (200+)² : The general solution is can be written as 34)- (200 H² (JE 3 (200++)*dt +c) Where, "G" is integrating constant, 200+ [-]3+ (2007)

70-100 100 (200) + c (20002 -4406 :: 200 - goixh- & Groot TC+007) 2 The amount of salt in the tank at any time it is X(t) = (2007.)3-4X106 (200+13) -*