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).

Answers

Answer 1

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]


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Answers

Answer:

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Explanation:

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attached below is a detailed solution of the problem

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Answers

Answer:

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Explanation:

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attached below is the detailed solution

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(c) f(t) = 3e^(-t) + (2t+1)e^(-t) - 2e^(-2t) - 5te^(-2t)

What is fraction?
A fraction is a type of mathematical expression that represents a part of a whole. It is written as a ratio of two numbers, with a line separating them. For example, the fraction 1/2 represents one part of a whole that has been divided into two equal parts. Fractions are used to express amounts that are less than one, such as a half, a quarter, or a third. They can also be used to express amounts greater than one, such as two thirds, three quarters, or four fifths. Fractions are used in many different areas of mathematics, such as addition, subtraction, multiplication and division.

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The statements below refer to the design and operation of the power sub-system on a spacecraft. Select all those statements that are correct.

Answers

the battery and ODC input voltages, as well as the main parameters for the power source, fireworks, relays, anti-static resistances, etc. The statements below refer to the design and operation of the power sub-system on a spacecraft.

How do spaceship electric power systems work?

Solar panels on these spacecraft convert solar energy into the electricity they require for propulsion. The electricity generated by the solar panels is used to recharge a battery within the spacecraft. These batteries can power the spaceship even as it is traveling away from the sun.

What make up a spacecraft's subsystems?

The engine, temperature management, power and power distribution, attitude control, telemetry command and control, transmitters/antenna, computers/on-board processing/software, and structural components are only a few of the many subsystems that make up a spaceship bus.

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Where is the airhorn of a 2014 kenworth t700?

Answers

In the front I think

A flow i decribed by velocity field V=ayibxj where a = 2 (m/) and b = 1 (1/) , coordinate are meaured in meter. A) Obtain the equation for the treamline paing through the point (2,5). B) At t=2 , what are the coordinate of the particle that paed through point (0,4) at t=0. C) At t=3 , what are the coordinate of the particle that paed through point (1,4. 25) 2 econd earlier

Answers

The equation for the streamline passing through point (2,5) is y-5=b/a(x-2).

What is streamline?

Streamlining is a process of simplifying or improving a process or system to increase efficiency, reduce complexity, and reduce costs. It can involve reorganizing, restructuring, rearranging, or redefining a system, procedure, or process to make it more efficient and productive.

B) At t=2, the coordinate of the particle that passed through point (0,4) at t=0 is (6,7). This is because since the velocity field is V = ayibxj, the particle will experience a displacement of 2a in the x-direction and 2b in the y-direction in the span of 2 seconds. Thus, the displacement of the particle is (2a,2b) = (4,2). Adding this to the initial position (0,4), the new position of the particle is (4,6).
C) At t=3, the coordinate of the particle that passed through point (1,4.25) 2 seconds earlier is (7.25,6.25). This is because since the velocity field is V = ayibxj, the particle will experience a displacement of 3a in the x-direction and 3b in the y-direction in the span of 3 seconds. Thus, the displacement of the particle is (3a, 3b) = (6,3). Adding this to the initial position (1,4.25), the new position of the particle is (7,7.25).

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What are the two major grains found on ALL lumber?

Answers

The two major grains found on all lumber are the longitudinal grain and the radial grain.

The longitudinal grain runs parallel to the length of the board, while the radial grain runs perpendicular to the length of the board.

What is the Longitudinal Grain?

The longitudinal grain, also known as the "edge grain," runs parallel to the length of the board.

This grain is created by the growth rings of the tree, which are formed by the tree's annual growth.

When a tree is cut down and milled into lumber, the edge grain is exposed on the edges of the board. This grain is typically considered to be the strongest and most stable of the two grains.

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