Question

5. A series of BOD (ATU) determinations was made in a sample to enable calculation of the ultimate BOD and rate constant. Incubation was carried out on a 5 per cent dilution of the sample at 20°C when the initial saturation DO for samples and blanks was 9.10 mg/. Day Final Do in sample (mg/l) Final Do in dilution water (mg/l) 1 2 3 4 5 6 7 7.10 6.10 5.10 4.20 3.90 3.50 3.00 9.00 9.00 8.90 8.90 8.80 8.70 8.60 Use the Thomas method to calculate the L and k values for the sample. (126 mg/, 0.145/day)

Answer 1

BODt = (DOsat - DOst) * DF (DF = dilution factor = total vol after dilution/sample volume = 100/5 = 20(since 5% dilution given)

since in this case d=water used for dilution is also showing DO depletion that need to be corrected. so BOD wiill be

BOD = ((DOsat-DOst)-(DOsat - Dow)*0.95)*20

0.95 factor is used for dilution water since 5% dilution so fraction of dilution water is 0.95

DOsat = saturation DO, DOst = DO of sample at time t, DOw = DO of dilution water

t(days)	Do in sample (Dos)	DO in dilution water (DOw)	BOD = ((DOsat-DOs)-(DOsat - Dow)*0.95)*20		t(days)	(t/BOD)^(1/3)
1	7.1	9	38.1		1	0.297
2	6.1	9	58.1		2	0.325
3	5.1	8.9	76.2		3	0.340
4	4.2	8.9	94.2		4	0.349
5	3.9	8.8	98.3		5	0.371
6	3.5	8.7	104.4		6	0.386
7	3	8.6	112.5		7	0.396
	DOsat	9.1

tvs (t/BOD)^(1/3) 0.410 y = 0.016x + 0.287 R? = 0.982 0.390 0.370 0.350 (t/BOD)^(1/3) 0.330 0.310 (t/BOD)^(1/3) -Linear ((t/BOD)^(1/3)) 0.290 0.270 0.250 0 2 4 6 00 t(days)

according to thomas (t/BOD)^(1/3)  = [(1/(2.3*K*L₀)^(1/3)) + ((2.3*K)^(2/3) / 6*L₀^(1/3))*t ]

so from line equation obtained in graph intercept 0.287 =1/(2.3*K*L₀)^(1/3) & slope 0.016 = ((2.3*K)^(2/3) / 6*L₀^(1/3))

on solving this

K = 0.145 /day

Lo = 126.46 mg/l

hope you like it. thank you

tvs (t/BOD)^(1/3) 0.410 y = 0.016x + 0.287 R? = 0.982 0.390 0.370 0.350 (t/BOD)^(1/3) 0.330 0.310 (t/BOD)^(1/3) -Linear ((t/BOD)^(1/3)) 0.290 0.270 0.250 0 2 4 6 00 t(days)

tvs (t/BOD)^(1/3) 0.410 y = 0.016x + 0.287 R? = 0.982 0.390 0.370 0.350 (t/BOD)^(1/3) 0.330 0.310 (t/BOD)^(1/3) -Linear ((t/BOD)^(1/3)) 0.290 0.270 0.250 0 2 4 6 00 t(days)