Question

12b. A 100-L beaker contains 10 kg of salt. Pure water is added at the constant rate of 5 L/min with complete mixing and drawn off at the same rate. How much salt is in the beaker after 1 hour? [Let y(t) be the amount of salt in the beaker at any timet dV (Hint: .-C . Please show the values of the constants at each step of your solution) di

Answer 1

Q: if y(4) be the amount of salt in the beaker at any time I then for mixing problem dy = Rate in - Rate out 6 at The water entering the tank at 5 l/ mim has mo salt that is pure water so Rate in= 0 • 0. Mined solution leaves the tank at 5L/min. The tank has 100-L of capacity with lokig of salt initially. The concentration of salt at time it is given by yoo kigil Rate out - Concentrate + Rate

OR I min - 400 5 Kg Imin Rob out = 5.kg Imio = the kg limim from 20 dy = - dt I s I dx = lnx Integrate con both sides Sody = 5-dt en y = - + +D lax=y azed gath = 29.xb y = è t/2070 - $/20, 8 = P = C (Const) Y = c 1/20 initially it = 0, y = 10 kg 10 = c e

10=Cl e = 1 *.0=1 so amount of salt at time ! y = lo è t/20 o t = there = Go min. ŷ = lo 5-60/20 Y = 10 8 = lox 0.0497870684 y = 0.49787