Answer:
a) ![x(t) = 13*e^(^-^\frac{t}{100}^)](https://tex.z-dn.net/?f=x%28t%29%20%3D%2013%2Ae%5E%28%5E-%5E%5Cfrac%7Bt%7D%7B100%7D%5E%29)
b) 10.643 kg
Step-by-step explanation:
Solution:-
- We will first denote the amount of salt in the solution as x ( t ) at any time t.
- We are given that the Pure water enters the tank ( contains zero salt ).
- The volumetric rate of flow in and out of tank is V(flow) = 50 L / min
- The rate of change of salt in the tank at time ( t ) can be expressed as a ODE considering the ( inflow ) and ( outflow ) of salt from the tank.
- The ODE is mathematically expressed as:
( salt flow in ) - ( salt flow out )
- Since the fresh water ( with zero salt ) flows in then ( salt flow in ) = 0
- The concentration of salt within the tank changes with time ( t ). The amount of salt in the tank at time ( t ) is denoted by x ( t ).
- The volume of water in the tank remains constant ( steady state conditions ). I.e 10 L volume leaves and 10 L is added at every second; hence, the total volume of solution in tank remains 5,000 L.
- So any time ( t ) the concentration of salt in the 5,000 L is:
![conc = \frac{x(t)}{1000}\frac{kg}{L}](https://tex.z-dn.net/?f=conc%20%3D%20%5Cfrac%7Bx%28t%29%7D%7B1000%7D%5Cfrac%7Bkg%7D%7BL%7D)
- The amount of salt leaving the tank per unit time can be determined from:
salt flow-out = conc * V( flow-out )
salt flow-out = ![\frac{x(t)}{5000}\frac{kg}{L}*\frac{50 L}{min}\\](https://tex.z-dn.net/?f=%5Cfrac%7Bx%28t%29%7D%7B5000%7D%5Cfrac%7Bkg%7D%7BL%7D%2A%5Cfrac%7B50%20L%7D%7Bmin%7D%5C%5C)
salt flow-out = ![\frac{x(t)}{100}\frac{kg}{min}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%28t%29%7D%7B100%7D%5Cfrac%7Bkg%7D%7Bmin%7D)
- The ODE becomes:
- Separate the variables and integrate both sides:
![\int {\frac{1}{x} } \, dx = -\int\limits^t_0 {\frac{1}{100} } \, dt + c\\\\Ln( x ) = -\frac{t}{100} + c\\\\x = C*e^(^-^\frac{t}{100}^)](https://tex.z-dn.net/?f=%5Cint%20%7B%5Cfrac%7B1%7D%7Bx%7D%20%7D%20%5C%2C%20dx%20%3D%20-%5Cint%5Climits%5Et_0%20%7B%5Cfrac%7B1%7D%7B100%7D%20%7D%20%5C%2C%20dt%20%20%2B%20c%5C%5C%5C%5CLn%28%20x%20%29%20%3D%20-%5Cfrac%7Bt%7D%7B100%7D%20%2B%20c%5C%5C%5C%5Cx%20%3D%20C%2Ae%5E%28%5E-%5E%5Cfrac%7Bt%7D%7B100%7D%5E%29)
- We were given the initial conditions for the amount of salt in tank at time t = 0 as x ( 0 ) = 13 kg. Use the initial conditions to evaluate the constant of integration:
![13 = C*e^0 = C](https://tex.z-dn.net/?f=13%20%3D%20C%2Ae%5E0%20%3D%20C)
- The solution to the ODE becomes:
![x(t) = 13*e^(^-^\frac{t}{100}^)](https://tex.z-dn.net/?f=x%28t%29%20%3D%2013%2Ae%5E%28%5E-%5E%5Cfrac%7Bt%7D%7B100%7D%5E%29)
- We will use the derived solution of the ODE to determine the amount amount of salt in the tank after t = 20 mins:
![x(20) = 13*e^(^-^\frac{20}{100}^)\\\\x(20) = 13*e^(^-^\frac{1}{5}^)\\\\x(20) = 10.643 kg](https://tex.z-dn.net/?f=x%2820%29%20%3D%2013%2Ae%5E%28%5E-%5E%5Cfrac%7B20%7D%7B100%7D%5E%29%5C%5C%5C%5Cx%2820%29%20%3D%2013%2Ae%5E%28%5E-%5E%5Cfrac%7B1%7D%7B5%7D%5E%29%5C%5C%5C%5Cx%2820%29%20%3D%2010.643%20kg)
- The amount of salt left in the tank after t = 20 mins is x = 10.643 kg