Water Quality L11, 12, 13

Question 1

Describe the BOD Reaction model in words.

Answer 1

The curve going up is the organic matter oxidized and the curve going down is the organic matter remaining.

Question 2

Describe the BOD Reaction model in terms of variables.

Answer 2

The curve going up is the BODt, the oxygen demand exerted. The point at which this curve plateaus to is the L0. The curve going down is Lt, the oxygen demand remaining. The difference between the Y values of these curves at certain x values is L0 - Lt. And BODt + Lt = Lo.

Question 3

What is BOD_t?

Answer 3

DO consumed in undiluted sample after t days of incubation

Question 4

What is L0?

Answer 4

Oxygen demand remaining in sample at time 0. The ultimate BOD.

Question 5

What is Lt?

Answer 5

Oxygen demand remaining in a sample at time t

Question 6

At any time, how are BODt and Lt related?

Answer 6

L0 = BODt + Lt. The oxygen demand exerted plus oxygen demand remaining to be exerted.

Question 7

As temperature increases,

Answer 7

Microbial metabolism and thus the rate of DO utilization increase.
Equation: Kt = K20(theta)^(t-20), for typical domestic wastewater

Question 8

In natural waters, we are often worried about the dissolved oxygen concentration getting too low/high

Answer 8

Low

Question 9

Nitrogen and phosphorus are nutrients/toxic metals that are used to synthesize new cells.

Answer 9

Nutrients

Question 10

The nature of waste (e.g., sugar vs. cellulose) affects how quickly the waste will be biodegraded. (True/False)

Answer 10

True

Question 11

Theoretical oxygen demand is always equal to the biochemical oxygen demand. (True/False)

Answer 11

False. It is always less than

Question 12

An increase in temperature usually causes biological processes to

Answer 12

Speed up

Question 13

The purpose of the BOD test is to measure

Answer 13

The amount of oxygen that is consumed by the oxidation of the waste

Question 14

Oxygen demand remaining (dL/dt = -kL) is modeled as a ___ order reaction

Answer 14

First

Question 15

One source of oxygen in a river is

Answer 15

re-aeration from the atmosphere

Question 16

For deep, slow-moving streams

Answer 16

Assume K_d=K_(BOD)

Question 17

For shallow, turbulent, rapidly moving streams

Answer 17

Assume K_d=K_(BOD) + (v/H)n, where
v = average velocity in stream (m/s)
H = average depth in stream (m)
n = bed activity coefficient
k_BOD = BOD rate constant (day^-1)

*Bed activity coefficient ranges from 0.1 for stagnant or deep water to 0.6 for rapidly flowing water