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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)


ηtot•t
102 (m•kg/s)

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)
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)

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