Question

A tank is filled with 1000 liters of pure water. Brine containing 0.08 kg of salt per liter enters the tank at 6 liters per minute. Another brine solution containing 0.09 kg of salt per liter enters the tank at 5 liters per minute. The contents of the tank are kept thoroughly mixed and the drains from the tank at 11 liters per minute.
A. Determine the differential equation which describes this system. Let SG) denote the number of kg of salt in the tank after t minutes. Then ds/dt=_____.
B. Solve the differential equation for S(t).

Answer 1

Answer:

The answer is below

Step-by-step explanation:

Let s(t) represent the amount of salt in the tank at time t. Therefore:

ds / dt = (rate of salt flowing into the tank) - (rate of salt going out of the tank)

$\frac{ds}{dt}=[(0.08\ kg/L*6\ L/min)+(0.09\ kg/L*5\ L/min)]-(\frac{s\ kg}{1000\ L}*11\ L/min ) \\\\\frac{ds}{dt}=(0.48\ kg/min+0.45\ kg/min)-(\frac{11s}{1000}\ kg/min )\\\\\frac{ds}{dt}=0.93\ kg/min-\frac{11s}{1000}\ kg/min \\\\\frac{ds}{dt}=\frac{930-11s}{1000}\ kg/min \\\\\frac{ds}{930-11s}=\frac{1}{1000}dt\\\\Integrating:\\\\\int\limits { \frac{ds}{930-11s}} \, ds=\int\limits {\frac{1}{1000}} \, dt\\\\-\frac{1}{11}ln(930-11s)=\frac{t}{1000}+C\\\\multiply\ through\ by\ -11:$

$ln(930-11s)= - \frac{11t}{1000}+A\\\\Take \ exponential:\\\\930-11s=e^{ - \frac{11t}{1000}+A}\\\\930-11s=C_1e^{ - \frac{11t}{1000}}\\\\11s=930-C_1e^{ - \frac{11t}{1000}}\\\\s=\frac{930-C_1e^{ - \frac{11t}{1000}}}{11} \\\\To \ find\ C_1,s(0)=0.Hence:\\\\0=\frac{930-C_1e^{ - \frac{11(0)}{1000}}}{11} \\\\0=930-C_1\\\\C_1=930\\\\s(t)=\frac{930-930e^{ - \frac{11t}{1000}}}{11} \\\\s(t)=930(\frac{1-e^{ - \frac{11t}{1000}}}{11})$