Question

A tank contains 6,000 L of brine with 16 kg of dissolved salt. Pure water enters the tank at a rate of 60 L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate. (a) How much salt is in the tank after t minutes? y = kg (b) How much salt is in the tank after 20 minutes? (Round your answer to one decimal place.) y = kg

Answer 1

Answer:

a) $S_{a}(t)=16Kg-0.16Kg*\frac{t}{min}$

b)  $S_{a}(20)=12.8Kg$

Step-by-step explanation:

It can be seen in the graph that the water velocity and solution velocity is the same, but the salt concentation will be lower

Water velocity $V_{w} = 60\frac{L}{min}$

Solution velocity $V_{s} = 60\frac{L}{min}$

Brine concentration = $\frac{6,000L}{16Kg}=375\frac{L}{Kg}$

a) Amount of salt as a funtion of time Sa(t)

$S_{a}(t)=16Kg-\frac{60Kg*L}{375L}*\frac{(t)}{min}=[tex]16Kg-0.16Kg*\frac{t}{min}$

b) $S_{a}(20)=16Kg-0.16\frac{Kg}{min}*(20min)=16Kg-3.2Kg=12.8Kg$

This value was to be expected since as the time passes the concentration will be lower due to the entrance to the pure water tank