SUBJECT: The oversized pump 7-10
Do a survey of any process plant and you'll find that a high percentage of the centrifugal pumps are oversized. There must be a reason why this is such a common problem, so here are a few possibilities
Cost of operating an oversized inch size pump
Obviously this larger pump and motor required a higher investment, but since we're not using the full power are we really paying too much for the daily operation? The easiest way to find the answer to this question is to look at a typical pump curve and make our calculations from the numbers we get.
You can use any of the following formulas to make your calculations:
Here is as typical pump curve. It can be used for both inch and metric examples.
Let's assume that the application requires a pump that moves the liquid at :
300 gpm. to a 156 foot head with an efficiency rating of 60%
156 x 300 / 5308 = 8.8 kilowatts being produced, and 8.8 / 0.60 efficiency = 14.7 Kilowatts required
As shown in the above drawing, we should be using impeller "E" to do this, but we have an oversized pump so we're using the larger impeller "A" with the pump discharge valve throttled back to 300 gpm. giving us an actual head of 250 feet and a 50% efficiency. Now our Kilowatts look like this:
250 x 300 / 5308 = 14.1 kw being produced, and 14.1 / 0.50 efficiency = 28.2 kw
required to do this. If 28.2 kw is being used and only 14.7 kw are required, it means that we are paying for an extra 13.5 KW to pump against the throttled discharge valve.
If this pump runs 24 hours per day that would be 8760 hours this year, and at a power cost of $0.05 cents per kilowatt hour it would cost your company an additional:
8760 hours. x .05 cents per kilowatt hour x 13.5 kilowatts = $5913.00 per year, extra operating cost.
Now we'll work the same problem in the metric system:
Assume that we need to pump 68 m3/hr. to a 47 meter head with a pump that is 60% efficient at that point.
68 x 47360 = 8.9 kilowatts being produce, and 8.9 / 0.60 efficient = 14.8 kilowatts required to do this.
As shown in the drawing, we should be using impeller "E" to do this, but we have an oversized pump so we are using the larger impeller "A" with the pump discharge valve throttled back to 68 cubic meters per hour, giving us an actual head of 76 meters. Now our kilowatts look like this:
68 x 76360 = 14.3 kilowatts being produced by the pump, and 14.3 / 0.50 efficient = 28.6 Kilowatts required to do this.
Subtracting the amount of kilowatts we should have been using gives us:
28.6 - 14.8 = 13.8 extra kilowatts being used to pump against the throttled discharge valve. If the pump runs twenty-four hours a day that would be 8760 hours per year, times 13.8 extra kilowatts equals 120,880 kw. Multiply this number by how much you spend for a kilowatt hour of electricity and you will see that the over sized pump is costing you a lot of money. In this example the extra cost of the electricity could almost equal the cost of purchasing the pump.
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