A tank contain 120L
Volume V = 120L
Solution of γ g/L of sugar
Rate of entry i.e input
dL/dt=2L/min
Let M(t) be the amount of sugar in tank at any time.
But at the beginning there was no sugar in the tank
i.e, M(0)=0, this will be out initial value problem,
The rate of amount of sugar at anytime t is
dM/dt=input amount of sugar - output amount of sugar.
Now,
Then rate of input is
2L/min × γ g/L
Then, input rate= 2γ g/mins
Output rate is
2L/mins × M(t)/120 kg/L
then, output rate = M(t)/60 g/min
So,
dM/dt=input rate -output rate
dM/dt= 2γ - M/60
Cross multiply through by 60
60dM/dt= 120γ - M
Using variable separation
60/(120γ - M) dM= dt
Integrate both sides
∫60/(120γ - M) dM= ∫dt
-60In(120γ - M)=t +C
In(120γ - M)=-t/60+C/60
C/60 is another constant, let say B
In(120γ - M)=-t/60+B
Take exponential of both side
120γ - M=exp(-t/60+B)
120γ - M=exp(-t/60)exp(B)
exp(B) is a constant let say C
-120γ - M=Cexp(-t/60)
- M=Cexp(-t/60) - 120γ
M= 120γ - Cexp(-t/60)
Now, the initial condition
a. At the start the mass of sugar in the water is 0 because it is just pure water at start.
Therefore M(0)=0,
b. Applying this to M(t)
M= 120γ - Cexp(-t/60)
M=0, t=0
0 = 120γ - Cexp(0)
0 = 120γ - C
C= 120γ
Therefore,
M= 120γ - 120γ exp(-t/60)
M =120γ[1 - exp(-t/60)]
Let know the mass rate as t tends to infinity
At infinity
exp(-∞)=1/exp(∞)=1/∞=0
Then,
The exponential aspect tend to 0
Then, M(t)=120γ as t tend to ∞
