Answer:
love
Explanation: you
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please solve part two
Answer:
Wat part 2
Explanation:
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:
Explanation:
From the given information:
The BOD of the wastewater can be determined by using the formula:
[tex]BOD _5 = \dfrac{DO_i-DO_f}{\dfrac{V_s}{V_b}}[/tex]
where;
[tex]BOD _5 =[/tex]biochemical oxygen demand = ???
[tex]DO_i=[/tex] initial dissolved oxygen = 9 mg/L
[tex]DO_f=[/tex] final dissolved oxygen = 4 mg/L
[tex]V_b =[/tex] sample of the bottle, which is normally 300 mL
[tex]V_s[/tex] = sample volume = 30 mL
[tex]BOD _5 = \dfrac{9-4}{\dfrac{30}{300}}[/tex]
[tex]BOD _5 = \dfrac{5}{\dfrac{1}{10}}[/tex]
[tex]BOD _5 = 5\times{\dfrac{10}{1}}[/tex]
[tex]\mathbf{BOD _5 = 50 \ mg/L}[/tex]
The ultimate Carbonaceous BOD is estimated from the formula:
[tex]y_t = L_o-L_t \\ \\ y_t = L_o-L_oe^{-kt} \\ \\ y_t = L_o (1-e^{-kt})[/tex]
Making [tex]L_o[/tex] the subject, we have:
[tex]L_o = \dfrac{y_t}{(1-e^{-kt}} \\ \\ L_o = \dfrac{50}{1 - e^{-0.25*5}}[/tex]
[tex]L_o[/tex] = 70 mg/L
c) The BOD left over after five days = [tex]L_oe^{-kt}[/tex]
= [tex]70 \times e^{-0.25 *5}[/tex]
= 20 mg/L
d) The reaction constant rate is estimated as follows:
Recall that:
[tex]\mathbf{BOD _5 = 50 \ mg/L}[/tex]
Since DO measure 2 mg/L;
After 30 days, [tex]BOD_{30} = (9-2) \times 10 = 70 \ mg/L[/tex]
Therefore, the reaction rate constant is:
[tex]\dfrac{50 \ mg/L}{70 \ mg/L} =\dfrac{1-e^{-k*5}}{1-e^{-k*30}} \\ \\ 50 (1-e^{-k*30}) = 70 (1-e^{-k*5}) \\ \\ K = 0.25 /day[/tex]
5 Systems Modeling
es / SPE(2201 / General / Business System Modelling CAT
2. Business Process Modelling is important to a business due to the following advantages except:
(2 marks)
O a. None of the above
h
O b. Enhances Customization of Business Processes
O c Enhances Competitive advantage
O d. Enhances Process Communication
age
Next pag
Answer: None of the above
Explanation:
Business process modeling refers to the graphical representation of the business processes of a company, which is vital in the identification of potential improvements.
Business pticess modelling can be done through graphing methods, like data-flow diagram, flowchart etc. It is vital as business managers can effectively and quickly communicate their ideas.
It also enhances the customization of business processes, enhances the competitive advantage and enhances the process communication as well.
Therefore, the answer to the question will be "None of the above".