As far as I investigated, a pump has a specific head in a determined flow rate (relating to its power and rotating speed). Then considering the formula ($\Delta P=\rho g H$), $\Delta P$ is adjusted for any fluid (with a different density) to obtain the same head. But my question is: how the extra pressure is created for a fluid with higher density, when using a specific pump with a specified power and therefore max head? This makes it confusing. Because it seems more logical to say the head is reduced/increased in such case; not that the pump produces more power to obtain the same head.
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$\begingroup$ sound to me like an engineering question. $\endgroup$– anna vCommented May 6, 2020 at 8:27
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$\begingroup$ Could be, but also a matter of physics. Am I supposed to change the community to engineering? $\endgroup$– OmishCommented May 7, 2020 at 9:39
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$\begingroup$ Well, there might not be many phsyicists who know the subject . This was the first time I have seen "head" used this way, had to look it up. $\endgroup$– anna vCommented May 7, 2020 at 10:18
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$\begingroup$ I can see that " the pump produces more power to obtain the same head" perusing this pumpfundamentals.com/what%20is%20head.htm , from its simple definition. $\endgroup$– anna vCommented May 7, 2020 at 10:27
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$\begingroup$ It was good for begining, but not answering my question. Just poited at the fact that I asked about. $\endgroup$– OmishCommented May 8, 2020 at 12:20
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2 Answers
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There are vanes in the impeller of a centrifugal pump, and the tips of those vanes are moving at a tangential speed that is a function of the impeller diameter and impeller revolutions per minute. In SI units, this tip speed is given by $v=r\omega$ m/s. Individual parcels of liquid come off the impeller vane tips at this speed, and the pump head is equivalent to how high those parcels of liquid would rise if you threw them straight up at this speed.
The centrifugal pump is question is coupled to an electric motor that MUST turn at the synchronous frequency of the A/C power being supplied to it, which is 60 Hz in the U.S. If the motor becomes more loaded, such as when the specific gravity of the pumped fluid is increasing, the motor draws more amps and hence more power in order to maintain its designed speed (e.g., 3600 rpm). Obviously, if you pump a fluid that has a specific gravity much higher than the pump is designed for, the motor will exceed its amperage rating and either trip an electrical breaker or burn up.
answered May 14, 2020 at 2:51
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$\begingroup$ Thanks you for your answer. I didn't know a pump can work at different power states and thought it has a certain power consumption like other electric apparatus. $\endgroup$– OmishCommented May 15, 2020 at 5:15
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I am assuming you are referring to a centrifugal pump. If that is the case, the term head is used in place of pressure. To understand where the pump will operate you must plot the system head curve onto the pump characteristic performance curve. The pump curve starts on the left side of the chart and normally arcs down from left to right. The system head curve is an inverted arc and starts on the left side of the chart and projects in an upward motion to the right side. The flow rate in gallons per minute (GPM) is plotted on the horizontal axis starting at zero flow on the left side of the chart to maximum flow to the right side of the chart. Where the pump curve and system curve intersect, is the operating point of the pump. Thus, if you close down on a valve in the piping, it creates more resistance to flow and the intersection point moves to the left pumping less flow at a higher head. Likewise, reduce the resistance to flow by opening a valve larger, the flow resistance decreases and the system curve intersects the pump curve at a greater flow rate. Calculate the horsepower at the operating conditions will be the horsepower needed for those conditions. The motor driving the pump must be suitable to cover the operating range of flow vs. head.
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$\begingroup$ Thank you. I didn't know know the horse power is varied and adjustable at different loads. $\endgroup$– OmishCommented May 15, 2020 at 5:18