Answer:
Step-by-step explanation:
Let m(t) represent mass of the salt at any time (t)
Given that the rate of flow is 99 L/mins
Then dV/dt=99 L/min
Given that the input concentration of brine is 0.04 kg/L
dM/dV = 0.04Kg/L
Then, the rate of the input mass into the tank.
dM/dt= dM/dV ×dV/dt
dM/dt= 0.04×99
dM/dt= 3.96Kg/min
Now, output rate.
Concentrate on of the salt in the tank at any time (t) is given as
Since it holds initially holds 100L of brine then the mass rate is m(t)/100
dM/dt= dV/dt × M(t)/100
dM/dt= 99×M(t)/100
dM/dt= 0.99M
Given the initial value condition
M(0) = 0.2Kg
Then, the rate of mass rate is given as the input mass rate minus output mass rate
dM/dt= 3.96-0.99M
Using variable separation
1/(3.96-0.99M) dM= dt
Integrate both side
∫1/(3.96-0.99M) dM= ∫dt
-0.99In(3.96-0.99M)=t+C
Divide through by -0.99
In(3.96-099M)=-1.01t-1.01C
-1.01C is still a constant, let say D
In(3.96-099M)=-1.01t+D
Take exponential of both side
3.96-0.99M=exp(-1.01t+D)
3.96-0.99M=exp(-1.01t)exp(D)
exp(D) is a constant, let say A
3.96-0.99M=Aexp(-1.01t)
-0.99M=-3.96+Aexp(-1.01t)
Divide through by -0.99
-0.99M=-3.96-1.01Aexp(-1.01t)
-1.01A is still a constant, let say B
M(t)= 4+Bexp(-1.01t)
Using the initial condition M(0)=0.2kg
0.2=4+Bexp(0)
0.2-4=B
B=-3.8
Then, the equation becomes
M(t)= 4-3.8exp(-1.01t)
2. When will salt concentration be 0.03kg/L
The mass is salt concentration × volume of liquid.
Mass=0.03kg/L × 100L
The mass is 3kg
M(t)= 4-3.8exp(-1.01t)
3=4-3.8exp(-1.01t)
3-4=-3.8exp(-1.01t)
-1=-3.8exp(-1.01t)
-1/-3.8=exp(-1.01t)
0.2632=exp(-1.01t)
Take In of both side
In(0.2632)=-1.01t
t=In(0.2632)÷-1.01
t=1.32seconds
