A tank contains 100 liters of a fluid, in which 35 grams of Chemical X has been

Question

A tank contains 100 liters of a fluid, in which 35 grams of Chemical X has been dissolved. If fluid containing 4 grams of Chemical X per liter enters the tank at a rate of 3 liters per minute and the well-mixed solution exits the tank at 5 liters per minute, find the function A(t): the amount of Chemical X in the tank at a time t. Also, during this fill and drain process, what is the maximum amount of Chemical X in the tank? Please Answer all parts of this question for a rating. And show steps. Thank you. Mainly need to know how to find the maximum amount of chem x in tank.

Explanation / Answer

(dx/dt) + [r??x / (a + (r? - r?)?t) ] = r??s? Since incoming liquid contains no salt, .i.e. s? = 0, this differential equation simplifies to: (dx/dt) + [r?/(a + (r? - r?)?t)]?x = 0 (dx/dt) = - [r?/(a + (r? - r?)?t)]?x That is a separable 1st order differential equation. It can be solved by separating the variables and integration. Separation of variables gives: (1/x) dx = - [r?/(a + (r? - r?)?t)] dt => ? (1/x) dx = ? - [r?/(a + (r? - r?)?t)] dt => ln(x) = - [r?/(r? - r?)]?ln(a + (r? - r?)?t) + C C is the constant of integration. It can be evaluated from the initial condition: x = x? at t = 0 By applying this initial value to the general solution above one finds: ln(x?) = - [r?/(r? - r?)]?ln(a + (r? - r?)?0) + C C = ln(x?) + [r?/(r? - r?)]?ln(a) Hence, ln(x) = - [r?/(r? - r?)]?ln(a + (r? - r?)?t) + ln(x?) + [r?/(r? - r?)]?ln(a) ln(x) - ln(x?) = - [r?/(r? - r?)] ? [ln(a + (r? - r?)?t) - ln(a)] Since ln(a) - ln(b) = ln(a/b) you can rewrite this as: ln(x/x?) = - [r?/(r? - r?)]?ln(1 + [(r? - r?)/a]?t) with c?ln(a) = ln(a^c) ln(x/x?) = ln( (1 + [(r? - r?)/a]?t)^[- r?/(r? - r?)] ) x/x? = (1 + [(r? - r?)/a]?t)^[- r?/(r? - r?)] x = x??(1 + [(r? - r?)/a]?t)^[- r?/(r? - r?)] with x? = 10kg r? = 3 L?min?

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