Answer:
s(t) = 160/3 ( 1 - e^(-3t / 200) )
Step-by-step explanation:
volume of pure water in tank = 1000 L
Brine contains 0.04kg of salt/L
Inflow rate of Brine containing 0.04kg of salt/L = 5L/min
Brine containing 0.06 kg of salt/L
Inflow rate of Brine containing 0.06 kg of salt/L = 10L/min
Solution is thoroughly mixed and drains from tank at 15L/min
a) Determine the amount of salt is in the tank after t minutes
rate of salt entering = 0.2 + 0.6 = 0.8 kg/min
rate of salt leaving = s/1000 * 15
amount of salt at time (t) = s(t)
initial condition s( 0 ) = 0
ds/dt = 0.8 - 15s/1000 = 0.8 - 3s/200
200 ds/dt = ( 160 - 3s )
-200/3 In ( 160 - 3s ) = t + c
Given that ; t = 0 , s = 0
c = - 200/3 In ( 160 )
∴ -200/3 In ( 160 - 3s ) = t - 200/3 In ( 160 )
- 200/3 [ In ( 60 - 3s ) - In ( 160 ) ] = t
therefore:
In ( 160 - 3s / 160 ) = -3t/200
= ( 160 - 3s / 160 ) = e ^ (-3t/200 )
Hence amount of salt in tank after t minutes
s(t) = 160/3 ( 1 - e^(-3t / 200) )
