Question

A 140 gallon tank initially contains 5 lbs salt dissolved in 40 gallons of water. Brine containing 1 lb salt per gallon begins to flow into the tank at the rate of 3 gal/min and the well-mixed solution is drawn off at the rate of 1 gal/min. How much salt is in the tank when it is about to overflow? (Round your answer to the nearest integer.)

Answer 1

Answer:

96.46

Step-by-step explanation:

Volume of the Gallon =140 gallons

At time t=0, V(0)=40 gallons

$\frac{dV}{dt}=Rate \:In - Rate \:Out \\\frac{dV}{dt}=3\frac{gal}{min} - 1\frac{gal}{min} =2\frac{gal}{min}$

V(t)=2t+k, Where K is a constant.

Since V(0)=40

V(0)=2(0)+k=40

k=40

V(t)=40+2t

The Change in Salt Concentration

$\frac{dC}{dt}=Rate \:In - Rate \:Out \\\frac{dC}{dt}=1\frac{lb}{gallon}*3\frac{gallon}{minute}-1\frac{lb}{gallon}*\frac{C}{40+2t} \frac{gallon}{minute}\\\frac{dC}{dt}=3-\frac{C}{40+2t}$

Rearranging

$\frac{dC}{dt}+\frac{C}{40+2t}=3$

Using Integrating Factor: $e^{\int\frac{1}{40+2t} dt} =e^{ln|40+2t|}=40+2t$

$\frac{dC(40+2t)}{dt}=3(40+2t)$

$\int d[C(40+2t)]=\int(120+6t)dt\\C(40+2t)=120t+3t^2+k_0\\C(t)=\frac{120t+3t^2}{40+2t}+\frac{k_0}{40+2t}$

At t=0, C(0)=0.125, k_0=5

$C(t)=\frac{120t+3t^2}{40+2t}+\frac{5}{40+2t}\\C(t)=\frac{120t+3t^2+5}{40+2t}$

At V(t)=140

V(t)=40+2t=140

t=50

$C(50)=96.46$

Just  the tank  overflows, C(t)=96.46