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| | Specific energy calculation example |
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Let's say that we have to select between two pumps in order to find the most suitable one for a given pipe.
The inflow to the station where we will accommodate the pump is 5 l / s. The performance curves for the two pumps and the system curve for the pipe system is shown in the graph below:
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The overall efficiency of the large pump in the duty point would be 70 % (red curve) and the overall efficiency of the small pump (blue pump) in its duty point would be 50 %.
With the formula:
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Where:
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The consumed power can be calculated for a pump. Looking at the inflow to the station and the flow that the pumps deliver in their duty points we can see that the small pump will run half the time and the large one, one fourth of the time.
The consumed energy can be calculated with the formula:
Where t is the time that the pump is running.
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If we use these two formulas to calculate the energy consumption during one year for each pump alternative we see that the energy becomes the following:
Pump 1 (small one ), E = 8 476 kWh
Pump 2 (big one), E = 16 952 kWh
So what is the reason? In this example we see that the efficiency of the large pump is better and that it delivers twice as high flow compared to the smaller pump. As a consequence the large pump runs half the time of the small one. What we also can see is that the head of the large pump becomes greater than twice the head of the small one. This depends on the fact that friction head losses are higher, they are proportional to the square of the flow through the pipe system. Even since the efficiency of the large pump is better most of the energy becomes heat because of the increased friction losses in the system!
We therefore see that efficiency, as a key number is not enough in order to determine the best pump alternative in an energetic point of view.
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Engineering
Using specific energy to determine energy consumption for a pump.