Question

Problem #3: Biochemical Oxygen Demand (BOD) and Stream Quality

A wastewater
treatment plant designed for a community of 45,000 people has a
flow of 100 gal/person/day and a BOD_{5} loading of 0.2
lb/person/day. The upstream characteristics of the receiving waters
are:

stream flow rate =
20 cfs and BOD_{5} = 2 mg/L

The water quality
standards require the in-stream BOD_{5} to be less than 7
mg/L in this stretch of the river.

The minimum BOD removal efficiency that the treatment plant should achieve is most nearly:

(a) 96 (b) 87 (c) 84 (d) 91

Problem #4: Wastewater Treatment

A wastewater plant receives a flow of 6 MGD with an influent BOD of 200 mg/L. The plant consists of aerated grit chamber, primary clarifier, activated sludge tank, final clarifier and disinfection through UV.

The primary
clarifier removes 60% solids and 35% BOD. The parameters for the
activated sludge process are volume is 175,000 ft^{3},
mixed liquor suspended solids concentration is 2500 mg/L,
biological yield is 0.6 and decay coefficient is 0.06 /day. Final
plant effluent needs to have a BOD of 10 mg/L. the peaking factor
is assumed to be 3.

- What are the dimensions (ft) of the grit chamber if the detention time is 1 min at peak flow and the l:w:d is 1.5:1.5:1 ?

(a) 10.5 x 10.5x7 (b) 13.5 x 13.5 x 9 (c) 12x12x8 (d) 15x15x10

- What is the depth (closest in ft) of the circular primary
clarifier if the overflow rate is 1000 gpd/ft
^{2}and the detention time is 2 hours for the average flow ?

(a) 11 (b) 15 (c) 9 (d) 13

Answer #1

You have been retained to design a wastewater treatment plant using
a conventional activated sludge process for a town of 40,000
people. The industrial flow contribution to the sewer system is
300,000 gallons per day. Effluent discharge standards require
effluent BOD5 and total suspended solids concentration of <10
mg/l.
Assume the following design criteria are applicable:
Domestic sewage flow = 100 gal/cap/day (including I/I)
BOD loading = 0.18 lbs/cap/day
TSS loading = 0.20 lbs/cap/day
BOD removal in primary = 33%...

The volumetric flow to the ACME wastewater treatment plant is 25
MGD. The influent soluble BOD5 is 200 mg/L and the effluent soluble
BOD5 is 8 mg/L. The plant is designed to operate with a mean cell
(biomass) residence time of 10 days. The yield coefficient is 0.6
mg biomass/mg substrate, and the biomass decay rate is 0.09/day.
The reactor is to operate at a biomass concentration of 2500 mg/L.
The thickened underflow biomass concentration from the secondary
clarifier is...

A wastewater plant with the activated sludge process received 5
× 106 L/day of wastewater with a BOD5 of 500
mg/L. The primary clarifier removes 30% of the BOD and primary
clarify effluent has no biomass (Xi=0 mg/L). Recycle
(Qr) and waste sludge (Qw) flows are 25% and
2% of Qin, respectively. Effluent BOD5 from secondary
clarifier and waste sludge is 20 mg/L. Microorganism concentration
in the waste sludge (Xw) is 6,000 mg/L. Soluble
BOD5 is biologically converted into CO2...

A wastewater treatment plant with a zero dissolved oxygen,
discharges 10,000 m3/d to a small stream. The stream
average flow is 0.4 m3/s with a dissolved oxygen level
of 8 mg/l and a BOD5 of 2 mg/l. The minimum oxygen demand that must
be maintained in the stream is 4 mg/l. Determine the maximum
BOD5 that can be discharged if the initial ultimate BOD
after the mixing is 8.1 mg/l. The BOD reaction rate is 0.23
d-1.

A wastewater treatment plant receives an average of 3.0 MGD of
wastewater flow. Two rectangular primary clarifiers operating in
parallel will treat the flow. If the length of each clarifier is
four times its width, the width is 20 ft, and the depth is 10 ft,
determine:
a) the overflow rate in gpd/ft2.
b) the detention time of each clarifier (in days).

A 14 mgd wastewater treatment plant has a primary clarifier that
treats an influent with 800 mg/L of solids and a removal efficiency
of 55%. Sludge at the bottom of the tank is pumped at a rate of 0.1
mgd.
a. What is the effluent solids concentration from the clarifier
(in mg/L)?
b. What is the solids concentration in the sludge at the bottom
of the tank (in mg/Land %)?
c. The primary sludge is further processed in a digester...

Wastewater treatment
The Ampang Jaya Municipal Council with a population of 350,000
and several industries discharged 11 MGD of raw sewage into an
activated sludge system treatment plant. The average concentration
of BOD5 is 240 mg/L and the suspended solids is 260
mg/L.
If the average domestic per capita sewage flow is 225 L/c.day,
what is the average flow of the industrial wastewater (in
MGD)?
marks)
If the average BOD5 load from the domestic sewage is
55gcapita-1day-1, determine the average...

A wastewater treatment plant that treats wastewater to secondary
treatment standards uses a primary sedimentation process followed
by an activated sludge process composed of aeration tanks and a
secondary clarifiers.
This problem focuses on the activated sludge process.
The activated sludge process receives a flow of 15 million
gallons per day (MGD).
The primary sedimentation process effluent has a BOD5
concentration of 180 mgBOD5/L (this concentration is
often referred as S). Before flowing into the activated sludge
process, the primary...

Effluent from a wastewater-treatment plant is discharged to a
surface stream. The characteristics of the effluent and the stream
are given in the table below. Determine the stream characteristics
after mixing has occurred.
Parameter
Effluent
Stream
Flow,
8500 m3/day
1.5 m3/sec
BOD5, mg/L
25.0
2.0
Ammonia, mg/L
10.0
0.0
Nitrate, mg/L
7.0
3.0
Chloride, mg/l
15
5.0

. A wastewater treatment plant is to be designed for the 5-d
sustained peak flowrate given an average daily flow of 13 MGD. Two
aerated grit chambers in parallel will provide a total detention
time of 4 minutes at peak flow. Determine: a) The volume of each
chamber, in m3 . b) Chamber dimensions, in m, (length and width) if
the design depth is 3.5 m and W:D is 1.2:1. Is the calculated L:W
within the recommended range? c) Using...

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