A water tank

[BMAD]

A tank contains 10 kg of salt dissolved in 3000 L of water. Brine containing 0.05 kg of salt per liter of water enters the tank at a rate of 10 L/min. The solution is kept thoroughly mixed and drains from the tank at a rate of 5 L/min. How much salt is in the tank after 1 hour?

 

[wolfgang]

after one hour there will be 38.90722kg salt in 3300 L water.

[antel0pe]

after one hour there will be 38.90722kg salt in 3300 L water.

 

Can I ask how did you come about the answer?

[dgreening]

The elegant solution would include integrals, but...

 

I chose to solve it as a summation in a spread sheet [over 3600 seconds

 

We know the  starting conditions

• 10 kg of salt and 3000 L of water

​We know that every minute "0.05 kg of salt per liter of water enters the tank at a rate of 10 L/min."

• 0.50 kg salt + 10 L of water/ min; or
• 0.00833 kg of salt + 0.1667 L of water/ sec

 

We also know that liquid leaves at 5L/ min [or 0.0833 L/ sec].

 

The tricky part is calculating the amount of salt that leaves. Since the concentration changes constantly, the amount of salt leaving also changes.

 

I calculated the concentration [kg of salt/ L] every second and then calculated how much salt left the tank every second.

.

after 3600 Sec I calculated that there is 37.73 kg of salt and 3300 L of water.

 

a solution using integrals might be a little more precise, but I suspect the difference is very small.

 

I will be glad to share the spread sheet if anyone want to look it over.

[plasmid]

Initial mass of salt = S(0) = 10 kg

Initial volume = V(0) = 3000 L
Change in volume over time = dV/dt = 5 L/min (net from water coming in and going out)
Volume at any specific time = V(0) + dV/dt * t = 3000 + 5t
Change in mass of salt over time = dS/dt =
= 0.5 kg/min (that's 0.05 kg/L * 10 L/min coming in) - S(t)/V(t) * 5 L/min (going out)
= 0.5 - 5S(t) / (3000 + 5t)

dS/dt + 5S(t) / (3000 + 5t) = 0.5
This is of the form dy/dx + y*f(x) = g(x), which can in general be solved as
y = (Integral [ g(x) e^{Integral [f(x) dx]} dx] + C) / e^{Integral[f(x) dx]}
S(t) = (Integral [0.5 * e^{Integral [(5/(3000+5t) dt]} dt] + C) / e^{Integral [5/(3000+5t) dt]}

Use the following substitution
e^{Integral [(5/(3000+5t) dt]} = e^ln(3000+5t) = 3000+5t

S(t) = (Integral [ 0.5 * (3000+5t) dt] + C) / (3000+5t)
S(t) = (Integral [(1500+2.5t) dt] + C) / (3000+5t)
S(t) = (1500t + 1.25t² + C) / (3000+5t)

Since S(0) = 10 kg, C must be 300000
S(t) = (1500t + 1.25t² + 30000) / (3000+5t), with time in units of minutes and salt in units of kg

Check that by calculating the derivative
S(t) = (1500t + 1.25t² + 30000) / (3000+5t)
By the multiplication rule
dS/dt = (2.5t + 1500) / (5t + 3000) - 5 (1500t + 1.25t² + 30000) / (3000+5t)²
= 0.5 - 5 (1500t + 1.25t² + 30000) / (3000+5t)²
= 0.5 - 5 S(t) / (3000 + 5t)
Which is exactly what it should be.

[dgreening]

Nicely done plasmid!