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# hw6 - Week 6 3.6 Cooling/Mixing 3.9 Numerical Solutions 4.1...

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Week 6: 3.6 Cooling/Mixing 3.9 Numerical Solutions 4.1 Higher Order DE

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2 Problem: A 200L tank is half full of a solution containing 100g of a desolved chemical. A solution containing 0.5 g/L of the same chemical is pumped into the tank at a rate of 6 L/min. The well-stirred mixture is pumped out at a rate of 4 L/min. Determine the concentration of the chemical in the tank just before overflow. Solution: V = V ( t ) and A = A ( t ) will be the volumn of the solution in the tank (in L) and amount of chemical in the tank (in g), at time t (in min.). The “rate in” is, r 1 = 6 and “rate out”, r 2 = 4; and the “concentration in” is, c 1 = 0 . 5 , while we solve for the “concentration out”, c 2 = A V .
First, the initial volumn V 0 = 100 (one half of the tank’s volumn), so V ( t ) = 100 + ( r 1 - r 2 ) t = 100 + 2 t (“rate in” - “rate out”). Next, the tank overflows when V is 200, which occurs when t = 50 , (why?) so we’re looking for c 2 (50) . Now our main DE says that the rate of change of the amount of the chemical, dA dt , is the difference r 1 c 1 - r 2 c 2 (“rate in” - “rate out”), which we re-write as dA dt + 4 100 + 2 t A = 3 , (using c 2 = A V ).

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