3 a tank has a capacity of 2000 l and originally contains...

3. A tank has a capacity of 2000 L and originally contains 800 L of water with 50 kg of salt in solution. Water containing 0.10 kg of salt per liter is entering at a rate of 10 L/min, the tank is well mixed, and the mixture flows out of the tank at a rate of 8 L/min. A. Find the amount of salt in the tank at any time before the solution begins to overflow the tank. B. Find the concentration (in kg/L) of salt in the tank when it is just beginning to overflow. C. What is the theoretical limiting concentration if the tank had infinite capacity?

Transcript

00:01 Here it is given that a tank with capacity of 500 gallon contains 200 gallon of water and 100 pounds of salt in the solution in flow of the mixture and salt is 3 gallon per minute and outflow of the mixture from the tank is 2 gallon per minute now solving part a let the amount of salt at any time t b y t then d y by d t be the rate of change of amount of salt here the water containing one pound per gallon of salt flows inside the tank at three gallon per minute show the salt going inside b three pounds so the amount of water at any point of time t is two hundred 3 minus 2 into t is equal to 200 plus t now the art flow of solution is 2 gallon per minute so the concentration of salt leaving the tank at time t b 2 into concentration of salt leaving the tank at time t b 2 into y t divided by 200 plus t now initially the tank contained 100 pounds of salt therefore we get the initial condition y 0 equal to 100 so the differential equation becomes d y by d t is equal to 3 minus 2 y t divided by 200 plus t and y 0 equal to 100 on changing it into the form of d y by d t on changing it into the form of d y by d t we get into the form of d y by d t plus p x into y equal to q x we get d .y by dt plus 2y t divided by 200 plus t is equal to 3 here pt is equal to 2 divided by 200 plus t and qt is equal to 3 hence integrating factor is e to the power p t t d t so this is equal to e to the power integration 2 by 200 plus t d t so this is e to the power log 200 plus t whole square so this is equal to 200 plus t whole square so the general solution is obtained as y t into i f integrating factor is equal to integration integrating factor into q dt so this imply yt into y t into 200 plus t whole square equal to integration 200 plus t whole square into three to d t so this is equal to y t into 200 plus t whole square is equal to three integration 200 plus t whole square d t so this is equal to three 200 plus t divided by three plus t plus solving further we will get y t into 200 plus t whole square equal to 200 plus t all is 4 equal to 200 plus t all cubic plus c now for finding c take the initial condition y 0 equal to 100 so this imply y 0 into 200 plus 0 whole square equal to 200 plus 0 power to the power 3 plus c so this imply 100 into 200 square equal to 200 to the power 3 plus c so c is equal to 100 into 100 to the power 3 plus c so c is equal to 100 into 200 to the power square minus 200 power cubic so this is equal to common taking common 200 square will get 200 square will be taking common so this is equal 100 minus 200 so this is equal to 200 square into minus 100 so this is equal to minus 4 into 10 to the power 6...

Question

Please give Ace some feedback