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

To calculate the amount of salt in the tank after a certain amount of time, we need to consider the rate at which salt enters and leaves the tank. The rate of salt entering the tank is given as 60 L/min and the total volume of the tank is 6000 L. Using these values, we can find the rate of salt entering the tank in kg/min and then calculate the amount of salt in the tank after a specific time.

Step-by-step explanation:

To calculate the amount of salt in the tank after a certain amount of time, we need to consider the rate at which salt enters and leaves the tank. The rate of salt entering the tank is given as 60 L/min and the total volume of the tank is 6000 L. Using these values, we can find the rate of salt entering the tank in kg/min:

Rate of salt entering = (60 L/min) * (16 kg/6000 L) = 0.16 kg/min

Therefore, the amount of salt in the tank after t minutes is given by:

y = 0.16 kg/min * t min = 0.16t kg

Answer 2

a)
`S_(a)(t)=16Kg-0.16Kg*(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(L)/(min)`

Solution velocity
`V_(s) = 60(L)/(min)`

Brine concentration =
`(6,000L)/(16Kg)=375(L)/(Kg)`

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

`S_(a)(t)=16Kg-(60Kg*L)/(375L)*((t))/(min)=[tex]16Kg-0.16Kg*(t)/(min)`

b)
`S_(a)(20)=16Kg-0.16(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