00:01
Hello students, in this question we have to find the differential equation and initial conditions for modeling the flow process.
00:09
So here two tanks are connected and tank one it initially contains 80 liters of water at 100 grams of salt, tank two initially contains 20 liters of water and 50 gram of salt.
00:35
Therefore we can specify the initial conditions as initial conditions at time t is equal to zero this will be equal to q1 of zero equal to 100 grams, q2 of zero equal to 50 grams and volume one at time t is equal to zero is 80 liters and volume two equal to 20 liters.
01:07
So these are the initial conditions.
01:10
Next we have to write the differential equations for the flow differential equations.
01:19
So this is given by the basic equation that is net rate of change of salt rate of change of salt is equal to a rate of inflow of salt inflow of salt minus rate of outflow of salt.
02:00
So for tank one tank one the process will be dq1 by dt which is the differential flow is given by rate in minus rate out into concentration in that is c in minus rate flow into concentration out into q1 v1 plus rate back into concentration out into q2 v1.
02:49
So this is the equation so by substituting the values in this equation we get dq1 equal to 3 into 15 minus 5 q1 by 80 minus 2 into 35 into q2 by 80 gram per minute.
03:23
So simplifying this we get the dq1 by dt equal to 45 minus q1 by 0 .0625 minus 70 by 80.
03:41
So this will give 0 .875 q2 gram per minute...