Question

A tank contains 500 gallons of salt-free water. A brine containing 0.25 lb of salt per gallon runs into the tank at the rate of 1 gal min , and the well-stirred mixture runs out at 1 gal min . In pounds per gallon, what is the concentration of salt in the tank at the end of 10 minutes? (Round your answer to four decimal places.) lbs per gal

Answer 1

Answer:

0.0198 lbs per gallon

Step-by-step explanation:

amount of salt free water = 500 gallons

salt rate in = 1 gal/min

salt rate out = 1 gal/min

amount o salt in brine = 0.25 lb per gallon

Let the amount of salt in the tank be A(t) at any time t.

$\frac{dA(t)}{dt} =salt rate in - salt rate out$

salt rate in = 0.25 x 1 = 0.25

salt rate out = $\frac{A(t)\times 1}{500}$

The differential equation is given by

$\frac{dA(t)}{dt} =1 - \frac{A(t)\times 1}{500}$

where, A(0) = 0

So, the equation becomes

$\frac{dA(t)}{dt} + \frac{A(t)}{500} = 1$

Here the integrating factor is $e^{\frac{dt}{500}}=e^{\frac{t}{500}}$

The solution of the above differential equation is given by

$A(t)\times e^{\frac{t}{500}} = \int e^{\frac{t}{500}}dt$

$A(t)\times e^{\frac{t}{500}} = 500\times e^{\frac{t}{500}}+C$

where, C is the integrating constant.

$A(t)=500+Ce^{-\frac{t}{500}}$

Put, A(0) = 0

C = - 500

$A(t)=500\left ( 1-e^{-\frac{t}{500} \right )$

As concentration is defined as

Concentration = Quantity / Volume

$C(t)=\frac{A(t)}{500}$

$C(t)=1-e^{\frac{-t}{500}}$

Now concentration at t = 10 min

Put, t = 10 min

$C(10)=1-e^{\frac{-10}{500}}$

C (10) = 0.0198 lbs per gallon

Thus, teh concentration after 10 min is 0.0198 lbs per gallon.