differentials?

Question

A salt water tank has 100 KG of salt and 500L of water. The input of salt water going in is 10L/sec, with a concentration of 1K/L. The output is whatever concentration is in the tank at time T, at a rate of 10L/sec. Assume perfect mixing of salt in the tank. What expression in terms of T can represent the amount of salt F(T)? Will integration be required on any part of this?

Answer 1

To employ your notation, define F as the number of kilograms of salt present in the tank at time T. Fo is defined as amount of salt at time To = 0, 100 KG

The concentration of salt in the tank is therefore F kg of salt dissolved in 500L of water (the volume of water in the tank is constant as liquid flows in at 10 L / sec and also flows out at 10 L / sec).

The law of conservation of mass implies that the change in amount of salt in the tank with respect to time must equal the rate that salt flows in minus the rate that salt flows out:

dF / dT = 10 L / sec * 1 KG / L - 10 L / sec * F kg / 500 L.

This is a first order differential equation (an equation that contains a function and one or more of its derivatives). It will require integration to solve for F. First do a bit of algebra to get F and its differentials on one side, and T and its differentials on the other:

Multiply both sides by dT and divide both sides by 10 - 10F / 500:

dF / (10 - 10F / 500) = dT

Now, do the integration:

int(Fo, F) [ dF / (10 - 10F / 500) ] = int(To, T)

Employ a substitution u = 10 - 10F / 500. du = -10 / 500 dF:

-50 int(10 - 10Fo / 500, 10 - 10F / 500) du / u = int(To, T)

Let's sub in our known constants (Fo and To) to make the algebra less messy:

-50 int(8, 10 - 10F / 500) du / u = int(0, T)

Do the integration:

-50 ln [ (10 - 10F / 500) / 8] = T – 0

Do the algebra to solve for F(T):

ln | (10 - 10F / 500) / 8| = -T / 50

| (10 - 10F / 500) / 8| = e^(-T / 50)

| (10 - 10F / 500) | = 8 e^(-T / 50)
The quantity | (10 - 10F / 500) | will always be greater than 0, so the | | is not necessary.

10F / 500 = 10 -8 e^(-T / 50)

F / 50 = 10 -8 e^(-T / 50)

F(T) = 500 -400 e^(-T / 50)

Answer 2

F(T) =F(0)+T-F(T)/500* 10T
F(T) (1+T/50)= T+100
so F(T) = 50(T+100)/(50+T)
T is time and also kg salt entered as there enter 1kg/s
F(T)/500 is concentration at time T and
F(T)/500*10T is the quantity of salt out of the tank