Homework SourseEach tank contains 60 gal of water Starting at time t 0 bri
Each tank contains 60 gal of water. Starting at time t = 0, brine containing 3 lb/gal of salt flows into tank 1 at the rate of 2 gal/min. The mixture then enters and leaves tank 2 at the same rate. The mixtures in both tanks are stirred uniformly. It can be shown that the amount of salt in tank 2 after t min is given by the function below where
A(t)=180(1-e^-t/30)-6te^-1/30
is measured in pounds. (Round your answers to one decimal place.)
What is the average amount of salt in tank 2 over the first 2 hr?
A(t)=180(1-e^-t/30)-6te^-1/30
is measured in pounds. (Round your answers to one decimal place.)
What is the average amount of salt in tank 2 over the first 2 hr?
Solution
sol: suppose that. so, (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) so the 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 answer
