Differential Equation word problem

Question

A 100-gallon tank holds a (well) mixed brine solution with a total of 10
pounds of salt. Fresh water comes in at a rate of 2 gallons per minute, and a brine
solution leaves the tank at a rate of 2 gallons per minute. How long will it take before
the amount of salt in the tank is less than 5 pounds?

Answer 1

V = 100 gal brine solution

C0 = initial salt mass/V = 10 lb salt/100 gal solution = 0.10 lb salt/gal sol'n

 

RateIN = 2 gal fresh water/min

RateOut = 2 gal brine sol'n/min   (well-mixed)

Since RateIN = RateOut, there is no (or insignificant) volume change in brine tank; therefore, V = constant = 100 gal

 

t? when the mass salt in tank is at 5 lb salt or C = 5 lb salt/100 gal = 0.05 lb salt/gal sol'n

 

Need to provide a salt mass balance around the control volume or brine tank

 

IN: 0      (no salt sol'n in - only fresh water)

 

OUT:  C(gal salt/gal sol'n)*RateOut(gal sol'n/min)

 

ACCUM:  V*dC/dt       (note: V does not change so C*dV/dt = 0 from chain rule) 

 

 

Therefore, ACCUM = IN - OUT

 

V*dC/dt = 0 - C*RateOut

 

dC/C = -(RateOut/V)*dt

 

Integrating both sides,

 

ln(C/C0) = -(RateOut/V)*t

 

Solving for t:

 

t = (V/RateOut)*ln(C0/C) = (100 gal/2 gal/min)*ln[(0.10 lb salt/gal sol'n)/(0.05 lb salt/gal sol'n)]

 

t = (50 min)*ln(2) = 34.66 min

 

Therefore when t < 34.66 min (or 34 min 40 sec) the salt in the tank is less than 5 lbs