A tank contains 30 lb of salt dissolved in 200 gallons of water. A brine solutio

Question

A tank contains 30 lb of salt dissolved in 200 gallons of water. A brine solution is pumped into the tank at a rate of 2 gal/min; it mixes with the solution there. , and then the mixture is pumped out at a rate of 2 gal/min. Determine A(t), the amount of salt in the tank at time t, if the concentration of salt in the inflow is variable and given by cin(t) = 2 + sin (t/4) lb/gal. A(t) = lb

Explanation / Answer

just for help: Initially, 50 lbs of salt is dissolved in a tank containing 300 gallons of water. A salt solution with 2 lb/gal concentration is poured into the tank at 3 gal/min. The mixture, after stirring, flows from the tank at the same rate the brine is entering the tank. In all of the problems below, we measure t in minutes. (accumulation, lb/min) = (rate in, lb/min) - (rate out, lb/min) dQ/dt = 2*3 - (Q/300)*3 = 6 - Q/100 dQ/(6-Q/100) = dt -100*ln(6-Q/100) = t + D ln(6-Q/100) = -t/100 + C 6-Q/100 = B*exp(-t/100) 600-Q=A*exp(-t/100) Q = 600-A*exp(-t/100) initial condition ==> t=0, Q=50 ==> A=550 Q = 600-550*exp(-t/100) (lb) B) C = Q/300 C = (600-550*exp(-t/100))/300 C = 2-11/6*exp(-t/100) (lb/gal) C) steady-state ==> dQ/dt=0 dQ/dt = 6 - Q/100 0 = 6 - Q/100 Q = 600 lb D) steady-state ==> Q=600 (from part C) C = Q/300 C = 600/300 C = 2 lb/gal

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.