Consider a large tank holding 1000 L of pure water into which a brine solution of salt begins to flow at constant rate of 6L/min. The solution inside the tank is kept well stirred and is flowing out of the tank at rate of 6L/min. If the concentration of salt in the brine entering the tank is 0.1 kg/L, determine when the concentration of salt in the tank will reach 0.05 kg/L


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Question Number 131470 by bemath last updated on 05/Feb/21

$C o n s i d e r a l a r g e t a n k h o l d i n g$ $1000 L o f p u r e w a t e r i n t o$ $w h i c h a b r i n e s o l u t i o n o f$ $s a l t b e g i n s t o f l o w a t c o n s t a n t$ $r a t e o f 6 L / m i n . T h e s o l u t i o n$ $i n s i d e t h e t a n k i s k e p t w e l l$ $s t i r r e d a n d i s f l o w i n g o u t o f$ $t h e t a n k a t r a t e o f 6 L / m i n .$ $I f t h e c o n c e n t r a t i o n o f s a l t$ $i n t h e b r i n e e n t e r i n g t h e t a n k$ $i s 0.1 k g / L, d e t e r m i n e w h e n$ $t h e c o n c e n t r a t i o n o f s a l t$ $i n t h e t a n k w i l l r e a c h 0.05 k g / L$
Commented by bemath last updated on 05/Feb/21

$thank all master$
	Answered by EDWIN88 last updated on 05/Feb/21

	$we conclude that the input rate of salt into$ $the tank is (6 L / \min) \times (0.1 kg / \min) = 0.6 kg / \min$ $out put rate of salt is (6 L / \min) [\frac{x (t)}{1000} kg / L] = \frac{3 x (t)}{500} kg / \min$ $where x (t) denote the mass of salt in tank at time t$ $\Leftrightarrow \frac{dx}{dt} = 0.6 - \frac{3 x}{500}; x (0) = 0$ $\Leftrightarrow \frac{dx}{100 - x} = \frac{3}{500} dt; \ln ∣ 100 - x ∣= - \frac{3}{500} t + c$ $\Leftrightarrow 100 - x (t) = λ e^{- \frac{3}{500} t}; t = 0; x (0) = 0$ $\Rightarrow 100 = λ; 100 - x (t) = {100 e}^{- \frac{3}{500} t}$ $\Leftrightarrow x (t) = 100 (1 - e^{- \frac{3 t}{500}})$ $\frac{x (t)}{1000} = 0.1 (1 - e^{- \frac{3 t}{500}}) kg / L$ $to determine when the concentration of$ $salt is 0.05 kg / L this gives$ $0.1 (1 - e^{- \frac{3 t}{500}}) = 0.05; e^{- \frac{3 t}{500}} = 0.5$ $t = \frac{500 . \ln 2}{3} \approx 115.52 \min$
	Answered by mr W last updated on 05/Feb/21

	$S = s a l t i n t a n k (k g)$ $s = s a l t c o n c e n t r a t i o n (k g / l) = \frac{S}{1000}$ $\frac{d S}{d t} = 0.1 \times 6 - \frac{S}{1000} \times 6 = \frac{6}{1000} (100 - S)$ $\frac{d S}{100 - S} = \frac{6}{1000} d t$ $\int_{0}^{S} \frac{d S}{100 - S} = \frac{6}{1000} \int_{0}^{t} d t$ $\ln \frac{100}{100 - S} = \frac{6 t}{1000}$ $S = 100 (1 - e^{- \frac{6 t}{1000}})$ $s a l t c o n c e n t r a t i o n :$ $s = \frac{S}{1000} = \frac{1}{10} (1 - e^{- \frac{6 t}{1000}})$ $s = \frac{1}{10} (1 - e^{- \frac{6 t}{1000}}) = 0.05$ $e^{- \frac{6 t}{1000}} = \frac{1}{2}$ $t = \frac{1000}{6} \ln 2 \approx 115.5 m i n$
	Answered by TheSupreme last updated on 05/Feb/21

	$Q_{i n} = 6 l / m i n (x_{i n} = 0.1 k g / L)$ $Q_{o u t} = 6 l / m i n (x_{o u t} = x (t))$ $V_{t o t} = 1000 L$ $Δ m_{s a l t} = Δ x V = m_{i n} - m_{o u t} = \int_{0}^{t} (Q_{i n} x_{i n} - Q_{o u t} x_{o u t}) d t$ $V \frac{{d x}_{o u t}}{d t} = Q_{i n} x_{i n} - Q_{o u t} x_{o u t}$ $\frac{x^{'}}{(x_{1} - x)} = - \frac{Q}{V}$ $l n (\frac{x_{1} - x}{x_{1}}) = - \frac{Q}{V}$ $(x_{1} - x) = x_{1} e^{- \frac{Q}{V} t}$ $x = x_{1} (1 - e^{- \frac{Q}{V} t})$ $x = x_{1} / 2$ $\frac{1}{2} = e^{- \frac{Q}{V} t}$ $\frac{Q}{V} t = l n (2)$ $t = V l n (2) / Q$ $i n t h i s e x a m p l e t = 115, 52 m i n$
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