Question

A large tank is filled to capacity with 700 gallons of pure water. Brine containing 4 pounds of salt per gallon is pumped into the tank at a rate of 7 gal/min. The well-mixed solution is pumped out at a rate of 14 gallons/min. Find the number A(t) of pounds of salt in the tank at time t.

Answer 1

Answer:

$A(t) = 1400(1 - e^{\frac{-t}{50}})$

Step-by-step explanation:

The net mass flow rate dA/dt = mass flow rate in - mass flow rate out

mass flow rate in = concentration of brine in × brine flow rate in = 4 lb/gal × 7 gal/min = 28 lb/min

Let A(t) be the amount of salt at time t. The concentration of salt in the tank at time, thus C = amount of salt/volume of tank = A(t)/700 lb/gal.

Since the well mixed solution is pumped out at a rate of 14 gal/min, the mass flow rate out = concentration of salt in tank × volume flow rate out = A(t)/700 lb/gal × 14 gal/min = m(t)/50 lb/min

So, dA/dt = mass flow rate in - mass flow rate out

dA/dt = 28 lb/min - A(t)/50 lb/min

dA/dt = 28 - A(t)/50

dA/dt = [1400 - A(t)]/50

separating the variables, we have

dA/[1400 - A(t)] = dt/50

integrating both sides, we have

∫dA/[1400 - A(t)] = ∫dt/50

-1/-1 × ∫dA/[1400 - A(t)] = ∫dt/50

1/-1 × ∫-dA/[1400 - A(t)] = ∫dt/50

-1 × ㏑[1400 - A(t)] = t/50 + C

-㏑[1400 - A(t)] = t/50 + C

㏑[1400 - A(t)] = -t/50 - C

taking exponents of both sides, we have

$1400 - A(t) = e^{\frac{-t}{50} - C} \\1400 - A(t) = e^{\frac{-t}{50}}e^{-C}\\1400 - A(t) = Be^{\frac{-t}{50}} ( B = e^{-C}) \\A(t) = 1400 - Be^{\frac{-t}{50}}$

when t = 0, A(0) = 0. So,

$A(t) = 1400 - Be^{\frac{-t}{50}}\\A(0) = 1400 - Be^{\frac{-0}{50}}\\0 = 1400 - Be^{0}\\0 = 1400 - B\\B = 1400$

So,

$A(t) = 1400 - 1400e^{\frac{-t}{50}}\\A(t) = 1400(1 - e^{\frac{-t}{50}})$