Hi everyone,
"A 200L tank initially contains 100L of water with a salt concentration of 0.1grams/L. Water with a salt conc. of 0.4g/L flows INTO the tank at 20L/minute. Assuming there is instantaneous mixing of the two salty solutions it then leaves the tank at a rate of 10L/min. Let C(t) denote concentration and V(t) denote the volume (both at a given time t).
Now the last part of the question says "What is the concentration of the salt in the tank when the tank first overflows" which I can do (by way of help from an online source).
However, I'm having trouble with the first two parts of the Q:
Set up the differential eqns governing V(t) and C(t)
Solve the governing equations to find the particular solutions for V(t) and C(t).
Because I can do the final concentration question I have probably answered the above 2, but I just don't know what the hell they are asking for!!
All help greatly appreciated,
sorry for being long winded,
thanks!
"A 200L tank initially contains 100L of water with a salt concentration of 0.1grams/L. Water with a salt conc. of 0.4g/L flows INTO the tank at 20L/minute. Assuming there is instantaneous mixing of the two salty solutions it then leaves the tank at a rate of 10L/min. Let C(t) denote concentration and V(t) denote the volume (both at a given time t).
Now the last part of the question says "What is the concentration of the salt in the tank when the tank first overflows" which I can do (by way of help from an online source).
However, I'm having trouble with the first two parts of the Q:
Set up the differential eqns governing V(t) and C(t)
Solve the governing equations to find the particular solutions for V(t) and C(t).
Because I can do the final concentration question I have probably answered the above 2, but I just don't know what the hell they are asking for!!
All help greatly appreciated,
sorry for being long winded,
thanks!
-
initial mass of salt = 0.1 * 100 = 10 grams
V = (20 - 10)t + 100 = 100 + 10t
dC/dt = (0.4)(20) - C/(100 + 10t) * 10
dC/dt + 10C/(10t + 100) = 8
integrating factor,
k = e^(∫ 10/(10t + 100) dt)
k = e^(ln (t + 10))
k = (t + 10)
general solution,
C k = ∫ 8k dt
C (t + 10) = ∫ 8(t + 10) dt
C (t + 10) = 4t² + 80t + constant
initial condition, at t = 0 ⇒ C = 10 grams
10 (0 + 10) = 4 * 0² + 80 * 0 + constant
constant = 100
C (t + 10) = 4t² + 80t + 100
when the tank will overflow at t = T
V = 100 + 10t
200 = 100 + 10T
T = 10 minutes
mass of salt ant T = 10 min
C (10 + 10) = 4(10)² + 80(10) + 100
C = 65 grams
concentration of solution is
G = 65/200
G = 0.325 gram/L
EDIT:
as your question : "But, I can't see where you've factored in the 10L/min exiting the tank?"
rate of mass of salt = rate of input mass - rate of output mass
rate of mass of salt = rate of input mass - (mass of salt)/(volume of solution) * (volume change of solution which exit from the tank)
dC/dt = (0.4)(20) - C/(100 + 10t) * 10
V = (20 - 10)t + 100 = 100 + 10t
dC/dt = (0.4)(20) - C/(100 + 10t) * 10
dC/dt + 10C/(10t + 100) = 8
integrating factor,
k = e^(∫ 10/(10t + 100) dt)
k = e^(ln (t + 10))
k = (t + 10)
general solution,
C k = ∫ 8k dt
C (t + 10) = ∫ 8(t + 10) dt
C (t + 10) = 4t² + 80t + constant
initial condition, at t = 0 ⇒ C = 10 grams
10 (0 + 10) = 4 * 0² + 80 * 0 + constant
constant = 100
C (t + 10) = 4t² + 80t + 100
when the tank will overflow at t = T
V = 100 + 10t
200 = 100 + 10T
T = 10 minutes
mass of salt ant T = 10 min
C (10 + 10) = 4(10)² + 80(10) + 100
C = 65 grams
concentration of solution is
G = 65/200
G = 0.325 gram/L
EDIT:
as your question : "But, I can't see where you've factored in the 10L/min exiting the tank?"
rate of mass of salt = rate of input mass - rate of output mass
rate of mass of salt = rate of input mass - (mass of salt)/(volume of solution) * (volume change of solution which exit from the tank)
dC/dt = (0.4)(20) - C/(100 + 10t) * 10