The volumes of wastewater
Example 1.1
What is the spread on the maximum hourly flow to the Ejby M01le treatment plant on a
dry-weather day?
What is the percentage of the dry-weather days by which the maximum hourly flow
rate is less than 3,650 m 3 /h?
s = f (84%)- f (50%) = 3,525- 3,175 = 350 m 3 !h.
The spread, s, can be determined as the difference between the curve values for 84
per cent probability and 50 per cent probability, see Table 1.1.
By means of the curve it can be established that 90 per cent of the days the maximum
hourly flow rate will be less than or equal to 3,650 m 3 !h.
1.1.3 Estimates
If adequate measurements of the volumes of wastewater are not available, estimates
and calculations must be made. For that purpose the wastewater is divided into
parts which typically consist of
- domestic wastewater
- industrial and public institutions wastewater
- infiltration
In respect of domestic wastewater, the calculation can be made as shown in Fig 1.6.
The basis is a number of persons and their annual wastewater production, Qyr,per9
Table 1.2 gives an idea of the quantity of Qyr,pern However, rough average figures are
stated. Based on the annual volume of wastewater, the other calculations I estimates
can be carried out as shown in Fig 1.6.
16
Example 1.2
The town of Heraklion is situated on the northern coast of the Greek island of Crete. It
has a fine museum with objects from Knossos which is situated in the vicinity and certainly is worth a visit.
Calculate the maximum hourly flow for domestic wastewater exclusive of infiltration
and exfiltration for Heraklion on the island of Crete. The population counts 120,000 persons.
From Table 1.2 the annual volume of wastewater per capita for Greece is estimated at
60m 3 .
3
Oyr,pers = 60 m /(yr · pers)
N = 120,000 persons
3
Oyr = Oyr,pers · N = (60 m /(yr · pers)) · (120,000 pers)
6 3
Oyr=7.2· 10 m /yr
Example 1.1
What is the spread on the maximum hourly flow to the Ejby M01le treatment plant on a
dry-weather day?
What is the percentage of the dry-weather days by which the maximum hourly flow
rate is less than 3,650 m 3 /h?
s = f (84%)- f (50%) = 3,525- 3,175 = 350 m 3 !h.
The spread, s, can be determined as the difference between the curve values for 84
per cent probability and 50 per cent probability, see Table 1.1.
By means of the curve it can be established that 90 per cent of the days the maximum
hourly flow rate will be less than or equal to 3,650 m 3 !h.
1.1.3 Estimates
If adequate measurements of the volumes of wastewater are not available, estimates
and calculations must be made. For that purpose the wastewater is divided into
parts which typically consist of
- domestic wastewater
- industrial and public institutions wastewater
- infiltration
In respect of domestic wastewater, the calculation can be made as shown in Fig 1.6.
The basis is a number of persons and their annual wastewater production, Qyr,per9
Table 1.2 gives an idea of the quantity of Qyr,pern However, rough average figures are
stated. Based on the annual volume of wastewater, the other calculations I estimates
can be carried out as shown in Fig 1.6.
16
Example 1.2
The town of Heraklion is situated on the northern coast of the Greek island of Crete. It
has a fine museum with objects from Knossos which is situated in the vicinity and certainly is worth a visit.
Calculate the maximum hourly flow for domestic wastewater exclusive of infiltration
and exfiltration for Heraklion on the island of Crete. The population counts 120,000 persons.
From Table 1.2 the annual volume of wastewater per capita for Greece is estimated at
60m 3 .
3
Oyr,pers = 60 m /(yr · pers)
N = 120,000 persons
3
Oyr = Oyr,pers · N = (60 m /(yr · pers)) · (120,000 pers)
6 3
Oyr=7.2· 10 m /yr