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).

[tex]\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:[/tex]

[tex]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})[/tex]

Omm2 Omm2 · Answer 2 · 2022-03-30T11:26:34+02:00

A) So, the required value is [tex]1.01-\frac{11S}{1000}[/tex].

B) The required value is [tex]\frac{1010}{11} (1-e^{-11/1000t} )[/tex]

Differential Equation:

A differential equation is an equation that contains one or more terms and the derivatives of one variable (i.e., dependent variable) with respect to the other variable (i.e., independent variable).

[tex]\frac{dy}{dx} =f(x)[/tex]

(A) Let [tex]S(t)[/tex] denotes the number of kg of the salt in the tank at the time [tex]t[/tex] minutes.

Then,

[tex]\frac{ds}{dt}=[/tex](rate of salt in to tank)-(rate of salt of tank)

[tex]\frac{ds}{dt}=\left ( 0.08\times 0.09\times 5 \right )-\left ( \frac{s\left (t\right )}{1000}\times 11 \right ) \\ \therefore \frac{ds}{dt}=1.01-\frac{11S}{1000}[/tex]

(B)

[tex]\frac{ds}{dt}=\frac{101}{100}-\frac{11S}{1000} \\ \Rightarrow \frac{ds}{dt}=\frac{1010-11S}{1000} \\ \Rightarrow \frac{ds}{1010-11S}=\frac{dt}{1000}[/tex]

Integrating we get,

[tex]-\frac{1}{11}I_n|1010-11S|=\frac{t}{1000}+In \ C_1 \\ \Rightarrow 1010-11S=C_1e^{-11/1000t} \\ S(t)=\frac{1010-C_1e^{-11/1000t}}{11}[/tex]

Initially, [tex]S(0)=0\\\Rightarrow C_1=1010[/tex]

Then the required solution is,

[tex]\therefore S(t)=\frac{1010}{11} (1-e^{-11/1000t} )[/tex]

Learn more about the topic of Differential Equation: https://brainly.com/question/24546081