The following data have been collected for dissolved and total BOD below a point

Question

The following data have been collected for dissolved and total BOD below a point source of untreated sewage into a stream. Use this data to estimate the BOD removal rates (kr,ks, andkd) for the river. The velocity and depth are 6600 m/d and 2 m, respectively.

The following data have been collected for dissolved and total BOD below a point source of untreated sewage into a stream. Use this data to estimate the BOD removal rates (k, k, and ka) for the river. The velocity and depth are 6600 m/d and 2 m, respectively x (km)0 5 1015 20 253035 Dissolved (m/L20 17 14.7 12.9 11.6 10.5 9.68.9 Total (mg/L) 40 29.4 22.5 17.8 14.6 12.3 10.7 9.6 x(km) | 40 | 45 | 50 | 60 | 70 | 80 | 90 | 100 Dissolved (mg/L) 8.3 7.87.3 6.6 6.0 5.5 5.1 4.7 Total (mg/L)8.78.07.56.7 6.05.5 5.14.

Explanation / Answer

Step 1: Given data:

V=6600 m/day

d=2 m

V=6.6km/day

at x=0 km,

Kd=0.3(H/2.4)-0.434

Kd=0.3(2/2.4)-0.434

Kd=0.324 (for all distances)

Kr=Kd

Kr=0.324/day

5.1

step 2: calculate the reoxygenation coefficient for each and every distance from x=5 km to x=100 km

by using the formula:

lnL=lnL0-(Kr/U)x

ln17=ln(29.4)-(Kr/6.6)x5

Kr=0.723/day

similarly calculate for remaining all the distances by using the same formula for Kr

STEP 3:

The value of Ks is negative in most of the cases , which is not actually possible so ignoring it we calculate only Kr and Kd values, the above table is the final table

x in km dissolved total 5 17 29.4 10 14.7 22.5 15 12.9 17.8 20 11.6 14.6 25 10.5 12.3 30 9.6 10.7 35 8.9 9.6 40 8.3 8.7 45 7.8 8.0 50 7.3 7.5 60 6.6 6.7 70 6 6 80 5.5 5.5 90 5.1

5.1

100 4.7 4.7

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