Question

Assume that the distribution coefficient (Kd) for Hg(II) is 104.9 (log Kd=4.9), the concentration of suspended sediments (Csed) is 10 mg L-1, the settling velocity (vs) of the silt-size particles is 4.17 cm hr-1, the area (A) of the water body is 50 km2 and the mean depth (H) is 6 m. What is the rate constant for removal of Hg(II) by scavenging, in units of per day?

Answer 1

Ans. Here, given the distribution coefficient, K_d for Hg (II) = 104.9 and its log Kd = 4.9

and given the concentration of suspended sediments, C_sed = 10 mg/L,

Given for the slit size particles, the settling velocity, Vs = 4.17 cm/hr,

and the area of water body is given = 50 km2 with the mean depth, H = 6m.

Required to calculate the rate constant for removal of Hg(II) by scavenging ,

Using the area of water body given as = 50 km² with the mean depth, H = 6m, the volume of water body will be calculated as: Volume = A x H;

Area of water body given as = 50 km² = 50 x 10⁶ m².

Since 1 km = 1000 m;

Volume of water body , ,

and we know: 1 m³ = 1000 L,

Thus, Volume of water body ,,

Thus the amount of sediment in the given volume of water = Volume of water body x the concentration of suspended sediments,

amount of sediment in the water (in mg) = mg.

The rate constant in removal of scavenging (in mg.cm/hr) is given as:

Rate constant (mg.cm/hr)= amount of sediment (mg) x log of distribution coefficient (log Kd) x settling velocity (mg. cm /hr)

we know, 1 mg = 10^-3 g

and 1 cm = 10^-2 m

Thus,

In the units of per day, we know 1 day = 24 hrs,

Thus, the rate constant for scavenging is given as: 1471.17 x10⁷ in units meter.gm per day.

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