Waterwaste Technology

Question 1

Name the two most important nutrients in wastewater and specify two forms of each nutrient.

Answer 1

Nitrogen: Nitrate and Ammonia
Phosphorus: Organic phosphorus and Phosphate

Question 2

Which parameters have an influence on the amount of infiltration water entering the sewer system?

Answer 2

+ Absorption of water by dry soil (Type of soil)
+ Saturation of water in soil
+ Wrong connections or leaks in sewer systems.
+ Hydraulic conductivity

Question 3

What is the rain intensity (in l/(sha)) of a rain event with the depth hn = 8 mm and
the duration T = 120 min?

Answer 3

i = intensity = depth/duration = 8/120 = 0.0666 mm/min
r = rainfall intensity = 166.67 x i = 11.11 l/s.ha

Question 4

The runoff coefficient  describes the percentage of precipitation that will be
translated into runoff. (6)
Which parameters determine the value of the runoff coefficient?

Answer 4

% of slope of drained area
% impervious area
Rainfall intensity and duration
Type of pavement
Type of soil and vegetation

Question 5

Calculate the runoff coefficient  for a rain event with 15 mm precipitation depth,
if 1 mm are retained in hollows, 2 mm are lost due to evapotranspiration and 40%
of the total precipitation is infiltrated.

Answer 5

Precipitation = 15 mm
Ψ = flow rate / intensity = 15 – 1 – 2 – 40%(15) / 15 = 0.4

Question 6

Give three advantages and three disadvantages of a modified sewer system in
comparison to a combined sewer system (CSS)!

Answer 6

Modified sewer system:
\+Advantages:
- Pipe diameters can be reduced
- Hydraulic load in pipes and WWTP is
low
- Volumes of storm water treatment
tanks can be reduced
\+Disadvantages:
- Expenses can be higher than those for
a conventional combined sewer system
- Planning and construction requirements are higher
- Wrong connections are possible

Combined sewer system
\+Advantages:
- Costs are lower because no separation of sewage and storm water is
necessary
- Wrong connections are not possible
- Less planning requirements construction requirements
\+Disadvantages:
Pipe diameters must to be large
- High load of pollutants in pipe and
WWTP´s
- Storm water treatment facilities are
always necessary

Question 7

Which functions have stormwater tanks retaining the first flush of stormwater
(STRs)? What’s the difference between on-line and off-line STRs? Sketch the
scheme of an STR in an off-line arrangement!

Answer 7

STR´s store the first flush of combined wastewater flow and retain the pollution load washed
out with the first flush. Try to avoid that high polluted water reach the water bodies, and only
the less polluted combined wastewater flow is discharged directly into the receiving water.
STR´s off-line: the storage tank is fed via a separation weir; after that, the water is translated
directly to the WWTP with pumps. After tank is filled, overflow is discharged directly to the
receiving water (without treatment) via an overflow weir.
STR´s on-line: the flow transported to the WWTP passes trough the storage tank. When the
storage tank is filled, the water goes directly to the receiving water without treatment.

Check sketch at the exam

Question 8

During dimensioning of the sewer system for a development area, the planners
find out that for rainfall events, wastewater cannot be discharged into the combined sewer system in the old part of the town, as the sewer system would be
overloaded. Find two solutions for the problem and shortly give the advantages
and disadvantages of each solution!

Answer 8

Assuming that the old part of the town does not have stormwater polluted with harmful substances can be treated trough infiltration. Two ways to make the infiltration is trough Permeable Pavement and Infiltration Basins.
Permeable Pavement:
Advantages:
- Easy maintenance
- Low technical requirements
- Cheap
- Very good treatment efficiency for grassy areas
Disadvantages:
- Low storage capacity
- Large areas necessary
Infiltration Basins:
Advantages:
- Good treatment efficiency
- Good retention capacity
- Low requirements of water quality
- Can be integrated into the landscape
- Easy maintenance
Disadvantages:
- Risk for children
- Improper maintenance can seal the basin bottom

Question 9

Sketch the scheme of a WWTP with all necessary sludge flows, which contains
the following elements:

Screens (a)
 Digester (b)
 Secondary sedimentation tank (c)
 Grit removal (d)
 Prethickener (e)
 Primary sedimentation tank (f)
 Sludge drying (g)
 Pump for return sludge flow (h)
 Aeration tank (i)
 Mechanical sludge dewatering (j)

Answer 9

Check sketch at the exam

Question 10

The inflow of a WWTP consists of a pollutant loading of 20,000 inhabitants and
10,000 population equivalents. The daily inflow Qd amounts to 5,000 m3
/d. (8)
With the help of the appendix, calculate the average concentrations of BOD5, TSS
and Ntotal (in mg/l) in the inflow to the WWTP and in the inflow to the biological
treatment step, assuming that the primary sedimentation tank has a volume of
420 m³.
Which parameter is mainly eliminated in the primary sedimentation step?

Answer 10

PT = P + PE = 20,000 + 10,000 = 30,000 inhabitants.
Qd = 5,000 m3
/d (to WWTP)
Loads (table 1 Apendix):
BOD5 = 60 g/C.d * 30,000 = 1,800 kg/d
TSS = 70 g/C.d * 30,000 = 2,100 kg/d
NTotal = 11 g/C.d * 30,000 = 330 kg/d
a) Average concentrations in the inflow of WWTP
CBOD5 = 1800 kg/d / 5000 m3
/d = 0.36 kg/m3 = 360 mg/l
CTSS = 2100 kg/d / 5000 m3
/d = 0.42 kg/m3 = 420 mg/l
CNTOTAL = 330 kg/d / 5000 m3
/d = 0.066 kg/m3 = 66 mg/l
b) Average concentrations at inflow of biological treatment step (after Primary Sedimentation Tank)
a d f i c
Return Sludge h
Primary Sludge
Inflow
Secondary Sludge or Exceso Sludge
e
Return flow
b
Raw Sludge
j
g Stabilized dried
Sludge
Outflow
Institute for Sanitary Engineering, Water Quality and Solid Waste Management 5
Time t = Vtank / Q = 420 m3
/ 5000 m3
/d = 0.084 d = 2 h
Loads (table 3 Apendix)
BOD5 = 40 g/C.d * 30,000 = 1,200 kg/d
TSS = 25 g/C.d * 30,000 = 750 kg/d
NTotal = 10 g/C.d * 30,000 = 300 kg/d
CBOD5 = 1200 kg/d / 5000 m3
/d = 0.24 kg/m3 = 240 mg/l
CTSS = 750 kg/d / 5000 m3
/d = 0.15 kg/m3 = 150 mg/l
CNTOTAL = 300 kg/d / 5000 m3
/d = 0.060 kg/m3 = 60 mg/l

Which parameter is mainly eliminated in the primary sedimentation step?
The amount of suspended solids (between 50 – 70%)

Question 11

What is the difference between heterotrophic and autotrophic microorganisms?
Which consequence does that have for dimensioning of the WWTP?

Answer 11

Heterotrophic microorganisms use organic substances in water as carbon source and can
reduce BOD5 (remove of carbon).
Autotrophic microorganisms do not need organic substances as carbon source, they use CO2
or HCO3- as carbon source, and reduce nitrogen compounds.
Both, heterotrophic and autotrophic microorganisms form the activated sludge, used for the dimensioning of the WWTP (Sludge volume index SVI and total suspended solids XTSS). Also, the degradation process is faster by heterotrophic microorganisms than by autotrophic
ones.