Calculation of shaft power for centrifugal pumps
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Seeking help: Regarding the calculation of the shaft power of centrifugal pumps. 1) P = Q × H × 9.81 × specific gravity of the medium ÷ 3600 ÷ pump efficiency. 2) P = Q × H × specific gravity of the medium ÷ 102 ÷ pump efficiency. Both formulas are used to calculate shaft power, but I’m confused. I hope someone experienced can give me some guidance.
P=Q×H×specific gravity of the medium÷367.2÷pump efficiency
Three centrifugal pumps were purchased for the project, with the following specifications: the medium is 31% hydrochloric acid containing 2% fine solid particles, with a density of around 1200; the flow rate is 20 m3/h and the head is 58 meters. The manufacturer’s model for these pumps is KJF80-50-400W, and the power of the motor used with them is 15 KW. After installation, the system was tested with water; the current was around 12A when there was no load, but it rose to 30A–32A when a load was applied. The pump tripped when the flow rate exceeded 10 M3/H. I’d like to ask, what exactly is the reason? ! ! Thank you. 1) Shaft power: Assuming an efficiency of 0.6, Pa = HQρ/102η = (58×20×1200)/(3600×102×0.6) = 6.32 kW; therefore, a motor capacity of 15 kW is more than sufficient. 2) Shaft power: With an efficiency of 0.6, P = QH×9.81×density of the medium÷3600÷0.6 = 58×20×9.81×1200÷3600÷0.6 = 6315 W = 6.32 kW. These two calculation methods differ, but the results are similar. How should this be understood? In the first formula, what does 3600 mean? Is it a coefficient or something else? Seeking advice from experts!
3600 means that one hour is equal to 3600 seconds; just remember the formula – there’s no need to focus too much on the meaning of each number. By idle operation, do you mean when the pump runs alone without any fluid flowing through it, or when fluid is flowing but the outlet valve is closed and the flow rate is zero? And how is your data usage measured? Is there a flow meter installed?
What is the outlet pressure at 10 cubic hours?
Axial power: Assuming an efficiency of 0.6, Pa = HQρ/102η = (58x20x1200)/(3600x102x0.6) = 6.32 kW. What I’m asking is: where does 3600 come from? Seeking advice from experts
1) P = Q × H × 9.81 × density of the medium ÷ 3600 ÷ pump efficiency. The correct formula is therefore: P = Q × H × 9.81 × density of the medium ÷ 3600 ÷ pump efficiency. All units must be converted to international units; the flow rate unit needs to be converted to m^3/s, which is why 3600 is used in the calculation. The unit of measurement resulting from this calculation is watts. 2) P = Q × H × density of the medium ÷ 102 ÷ pump efficiency. When applying this formula, you divide by 3600 as well; the result is more or less the same, only the unit is in kW. Here, 1/102 is equivalent to ×9.8*10^-3. In other words, this formula is just a modified version of the previous one, but it’s incorrect. The correct formula should be: P = Q × H × density of the medium / (102 × pump efficiency × 3600). Since an extra 10^-3 is included, the resulting unit is kilowatt. Check it and see if it’s like this.
It is recommended to go back and study the derivation of the formulas; then you will understand
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The second formula is incorrect; it lacks a /3.6 factor. All formulas are derived from the original one, and 9.81 is used in the conversion for head pressure; in the denominator, this becomes /102. 3600 represents the result of converting the flow rate unit to cubic meters per second