Question

A BOD test is run using 30 mL of wastewater and 270 mL of dilution water. The initial DO of the mixture is 9.0 mg/L. On day 5 the DO in the bottle measured 4 mg/L. After 30 days, the DO in the bottle measured 2 mg/L, and after 50 days, the DO in the bottle still measured 2 mg/L. At the beginning of the test, we added a nitrification inhibitor so we can assume that nitrification is not occurring, so only the carbonaceous BOD is being measured.

a) What is the BODs of the wastewater?
b) What is the ultimate carbonaceous BOD?
c) How much BOD remains after 5 days?
d) Based on the data above, estimate the reaction rate constant k (1/day).

Answer 1

Step-by-step explanation:

From the given information:

The BOD of the wastewater can be determined by using the formula:

`BOD _5 = (DO_i-DO_f)/((V_s)/(V_b))`

where;

`BOD _5 =`biochemical oxygen demand = ???

`DO_i=` initial dissolved oxygen = 9 mg/L

`DO_f=` final dissolved oxygen = 4 mg/L

`V_b =` sample of the bottle, which is normally 300 mL

`V_s` = sample volume = 30 mL

`BOD _5 = (9-4)/((30)/(300))`

`BOD _5 = (5)/((1)/(10))`

`BOD _5 = 5*{(10)/(1)}`

`\mathbf{BOD _5 = 50 \ mg/L}`

The ultimate Carbonaceous BOD is estimated from the formula:

`y_t = L_o-L_t \\ \\ y_t = L_o-L_oe^(-kt) \\ \\ y_t = L_o (1-e^(-kt))`

Making
`L_o` the subject, we have:

`L_o = (y_t)/((1-e^(-kt)) \\ \\ L_o = (50)/(1 - e^(-0.25*5))`

`L_o` = 70 mg/L

c) The BOD left over after five days =
`L_oe^(-kt)`

=
`70 * e^(-0.25 *5)`

= 20 mg/L

d) The reaction constant rate is estimated as follows:

Recall that:

`\mathbf{BOD _5 = 50 \ mg/L}`

Since DO measure 2 mg/L;

After 30 days,
`BOD_(30) = (9-2) * 10 = 70 \ mg/L`

Therefore, the reaction rate constant is:

`(50 \ mg/L)/(70 \ mg/L) =(1-e^(-k*5))/(1-e^(-k*30)) \\ \\ 50 (1-e^(-k*30)) = 70 (1-e^(-k*5)) \\ \\ K = 0.25 /day`