BOB_5 Calculations/Questions (15%) -- Suppose you are to measure the BOD removal

Question

BOB_5 Calculations/Questions (15%) -- Suppose you are to measure the BOD removal rate for the DC wastewater treatment plant. For comparison purpose, you take two samples the same time of raw sewage on its way into the plant and two samples of the effluent leaving the plant. The measured data of BOD_5 tests, without seeding, are below: (a) Find BOD_5 for samples #1, #3 and the percentage removal. (b) Find DO_final in Sample #2 and dilution ratio of sample 14. (c) What does the BOD of wastewater indicate? Is a high BOD of water good or bad? Why? (d) Why is the BODs test conducted in the darkroom? Why it is tested for only 5 days? (e) Why is it often necessary to dilute the wastewater? And why is it often to seed the test sample?

Explanation / Answer

Solution:

a) To determine the value of the BOD5 in mg/L for sample-1, we can use the following formula:

BOD (mg/L) = [(Initial DO Final DO) x 300]/mL sample --------> (1)

Given here, 30 mL dilution per mL of sample

Initial DO = 9.5 mg/L

Final DO = 2.2 mg/L

Therefore,

BOD (mg/L) = [(9.5 2.2) x 30]/1 = (7.3 x 30)/1 = 219 mg/L

Also, percent removal = (9.5 – 2.2)x100/9.5 = 730/9.5 = 76.8%

Similarly, for sample-3 with a dilution ratio of 20:1 we have:

BOD (mg/L) = [(9.0 2.0) x 20]/1 = (7.0 x 20)/1 = 140 mg/L

Percent removal = (9.0 – 2.0)x100/9.0 = 700/9.0 = 77.77%

b) DOfinal of sample-2 can be calculated using equation (1) as follows:

[BOD (mg/mL) x sample volume (mL)]/dilution (mL) = Initial DO Final DO

Or, Final DO = Initial DO - [BOD (mg/mL) x sample volume (mL)]/dilution (mL)

Or, Final DO = 9.5 - [219 x 1]/25

Or, Final DO = 9.5 – 8.76

Or, Final DO = 0.74 mg/L

Also dilution ratio of sample-4 can be calculated as follows:

Sample volume (mL)/dilution (mL) = (Initial DO Final DO)/ BOD (mg/mL)

Or, dilution (mL)/Sample volume (mL) = BOD (mg/mL)/(Initial DO Final DO)

                                                               = 140/(9 - x);            where, x>0

Or, Dilution ratio = 140/(9-x)

c) The BOD test measures the strength of the wastewater by measuring the amount of oxygen used by the bacteria as they stabilize the organic matter under controlled conditions of time and temperature.

High BOD of water is not good, because it means high demand of dissolved oxygen in water for the microorganisms to act on the organic matter found in wastewater (or water).

d) The BOD test is conducted in absence of sunlight to avoid the side reactions like other oxidative reactions etc. which may give erroneous test result.

A large numbers of nitrifying organisms are developed during BOD treatment process. These organisms can exert an oxygen demand as they convert nitrogenous compounds like ammonia and organic nitrogen to more stable forms of nitrites and nitrates. At least part of this oxygen demand it is normally measured in a five day BOD test protocol.

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