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2 years ago

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A lake with constant volume 1.1 x 10^6 m^3 is fed by a stream with a non-conservative pollutant of 2.3 mg/L and flow rate 35 m^3

/s. A factory dumps 4.3 m^3 /s of waste with 100 mg/L of the same non-conservative pollutant into the lake. The pollutant has a first order decay coefficient K of 0.10/day. Assuming the lake is well mixed, find the steady-state concentration of pollutant in the lake.

1 answer:

Jet001 [13]2 years ago

4 0

Answer:

12.84 mg/L

Explanation:

We are given;

Volume of lake; V = 1.1 x 10^(6) m³

decay coefficient; K = 0.10/day = 0.1/(24 × 60 × 60) /s = 0.00000115741 /s

Factory rate: Q_f = 4.3 m³/s

Factory concentration: C_f = 100 mg/L

Stream rate: Q_s = 34 m³/s

Stream Concentration: C_s = 2.3 mg/L

Now, to find the steady state concentration of pollutant in the lake, we will use the formula;

(Q_s•C_s) + (Q_f•C_f) = (Q_f + Q_s)C_L + (KV•C_L)

Where C_L is the steady state concentration of pollutant in the lake.

Thus, making C_L the subject, we have;

C_L = [(Q_s•C_s) + (Q_f•C_f)]/(Q_f + Q_s + K•V)

Plugging in the relevant values gives;

C_L = ((34 × 2.3) + (4.3 × 100))/(4.3 + 34 + (0.00000115741 × 1.1 × 10^(6)))

C_L = 12.84 mg/L

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b. The time it takes for the tank to empty the remaining half is presented as follows;

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The time, 'dt', it takes for the water to drop by a level, dh, is given as follows;

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$t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }$

The time it takes for the tank to empty the remaining half, t₂, is presented as follows;

$t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} }$

The total time, t, to empty the tank is given as follows;

$t = t_1 + t_2 = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \left (\sqrt{2} -1 \right) + t_2 = { \dfrac{ D_0^2 }{D} \cdot\sqrt{\dfrac{H}{g} } = \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} } \cdot \sqrt{2}$

$t = \sqrt{2} \cdot \dfrac{D_0^2 }{D^2 } \cdot \sqrt{ \dfrac{H}{g} }$

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