Sewage Parameters 3: Population Equivalent (PE) Part 2

May 23, 2018 Tommy Butler

Introduction

Following up on our previous post on Population Equivalent (PE), this installment provides additional examples and calculations to illustrate how PE is used to size sewage treatment systems for various types of developments.

Example A: Housing Development

Scenario:

10 No. 2-Bedroom Old Persons Dwellings
25 No. 3-Bedroom Houses
10 No. 4-Bedroom Houses

Relevant Figures:

1 PE = 200 l of flow/day & 60 g of BOD/day.
Occupants per house type:
- 2-bedroom: 2 persons
- 3-bedroom: 4 persons
- 4-bedroom: 5 persons

Calculation:
10×2+25×4+10×5=170 PE10 \times 2 + 25 \times 4 + 10 \times 5 = 170 \, \text{PE}10×2+25×4+10×5=170PE

Flow and BOD:

Flow:
170 PE×200 l=34,000 l/day170 \, \text{PE} \times 200 \, \text{l} = 34,000 \, \text{l/day}170PE×200l=34,000l/day
BOD:
170 PE×60 g=10,200 g/day170 \, \text{PE} \times 60 \, \text{g} = 10,200 \, \text{g/day}170PE×60g=10,200g/day

Solution:
A BMS BL2000 Blivet Package Sewage Treatment Plant is ideal for this development and provides room for expansion.

Example B: Hotel

Scenario:

50 Bedrooms (2 persons per room)
100 extra meals/day
300 function room guests
50 bar-only guests
30 non-residential staff

Relevant EPA Figures:

CategoryFlow (l/person)BOD (g/person)Bedroom (2 persons)25075Meals2525Function room guests1010Bar-only guests1010Non-residential staff6030

Calculation:

Bedrooms:
- Flow:
  50×2×250=25,000 l/day50 \times 2 \times 250 = 25,000 \, \text{l/day}50×2×250=25,000l/day
- BOD:
  50×2×75=7,500 g/day50 \times 2 \times 75 = 7,500 \, \text{g/day}50×2×75=7,500g/day
Extra Meals:
- Flow:
  100×25=2,500 l/day100 \times 25 = 2,500 \, \text{l/day}100×25=2,500l/day
- BOD:
  100×25=2,500 g/day100 \times 25 = 2,500 \, \text{g/day}100×25=2,500g/day
Function Room Guests:
- Flow:
  300×10=3,000 l/day300 \times 10 = 3,000 \, \text{l/day}300×10=3,000l/day
- BOD:
  300×10=3,000 g/day300 \times 10 = 3,000 \, \text{g/day}300×10=3,000g/day
Bar-Only Guests:
- Flow:
  50×10=500 l/day50 \times 10 = 500 \, \text{l/day}50×10=500l/day

Sewage treatment plants should always be sized on the basis of BOD as it is the biological load that has to be treated. Flows can be higher or lower that the BOD and this can vary from country to country. A BMS BL3000 Blivet with a balance tank to deal with peak early morning and evening flows typically produced by a hotel would be ideal for the above project. A Balance Tank can also help screen out non-soluble solids and also be used to lower Total Nitrogen by Denitrification.

Below you can see a BMS BL3000 Blivet Package Sewage Treatment System and Balance Tank being installed at a new hotel

Next Topic

In the next post, we’ll dive into Phosphorus.

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