Calculation of the Approximate Costs of Pumping Sewage
(A rule of thumb)
As alternatives to draining sewage from a neighborhood, frequently there are other choices, such as a canalization that makes a detour, a canalization deep underground or a pumping well. So, you have to choose between investment and operation costs. Many engineers hesitate to calculate pumping costs, because they believe it's laborious.
The following considerations will lead to a surprisingly simple solution to the problem of calculating these costs easily, and with pretty good accuracy, by the way.
The pumping performance is calculated by the equation:
| E = |
γ•Q•h
|
• |
1
|
(kW) |
|
|
|
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|
ηtot•t
|
102 (m•kg/s)
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E = Energy (kW) Q = flow (l/s) h = manometric hight (m) γ = densiy of the water/sludge (kg/l) η = efficiency of pump plus motor t = time (s) 102 = factor (1 kW = 102 m.kg/s) |
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| Where: | |||
From the above equation we can be conclude that at a given efficiency of a pump, the current demand is solely a question of the flow and height the water is pumped! On this basis a simplified computation of the current demand and the ensueing costs is possible:
| E = | 1•8'760 h/yr |
= 172 kWh/yr
|
|
0.5•102 |
| Where: |
Manometric hight Pump-η
Motor-η & totally Density γ |
= 1.0 meter = 0.6 = 0.8 = 0.5 = 1 kg/l (for sewage or diluted sludge) |
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In Switzerland the current costs for large consumers is SFr. 0.15/kWh. Dear reader, you certainly are familiar with the current costs in your country. In my country this will lead to the following conclusion:
| A constant flow of 1 l/s, pumped 1.0 meters high will cost SFr. 25 per year |
Example: There there is a project where you have to pump an average of 7.5 l/s sewage 12 meters high, then you have to expect anual energy costs of
7.5 l/s • 12 m • Fr. 25.-/yr = ca. SFr. 2'250.- per year