The issue with the pump’s head???????
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For example, if the pump’s head is 10 meters and it is used to spray water from a platform at a height of 8 meters (the distance from the pump’s outlet to the platform), then is the flow rate at this point the one corresponding to 8 meters on the curve? Right? If it’s a closed loop, then the energy consumed due to resistance in the entire loop is equivalent to the head of the pump, right? The flow rate at this point is also the flow rate corresponding to the head. I’m not sure if what I said is correct. . . . . . . Seeking advice
If the inlet is at atmospheric pressure, then it’s basically correct; an additional slight frictional pressure drop is needed for 8m, but the change isn’t significant. Also, don’t make the font size so large next time.
:'(If not increased, it has no impact.) . . . . . . . . . Haha, thank you
1) “The pump’s head is 10 meters. It is currently used to spray water on a platform at a height of 8 meters (the distance from the pump’s outlet to the platform). Is the flow rate at this time the one corresponding to 8 meters on the curve?””； Firstly, the actual flow rate in the pipeline is determined not only by the pump’s performance curve but also by the characteristics of the pipeline itself. The degree to which the valve at the pump’s outlet is open also affects the flow rate; it’s not a coincidence that when the valve is fully open, it happens to correspond to that specific point on the pump’s performance curve ； 2) Closed-loop systems are divided into two types: in one type, the pump’s outlet pipe goes around once before connecting directly to the pump’s inlet; in the other type, the pump’s outlet pipe goes around once before connecting to a container at atmospheric pressure, and the pump then draws in the fluid from this container ； In any operating condition, the pipe friction loss is not necessarily equal to the head (unless the piping technician is truly a genius who can manage to offset the head with the pipe friction loss) ； Because the pressure loss in typical industrial pipelines is on the order of a few thousand pascals (excluding losses due to equipment), while the outlet pressure of pumps is at least several tens of thousands of pascals
There is no problem with the concept mentioned by the original poster.
Isn’t the head corresponding to the pipe resistance? The head is also used for generating heat; so if the pipe resistance is low, does that mean there’s extra head available? In a closed loop, wouldn’t the inlet pressure of that pump keep increasing, causing it to get damaged? Please provide an explanation
Regarding question one, the original poster has confused the concepts. Strictly speaking, the head of a centrifugal pump and its lift height are two different things. One can look up the theoretical calculation formulas for head in the principles of chemical engineering. The head of a centrifugal pump refers to the effective energy that the pump imparts to each unit weight of liquid; with this energy, the liquid can be lifted to a certain height ΔZ. This energy is also used to increase the static pressure head by ΔP/ρg, the dynamic pressure head by Δu²/2g, and to overcome the pressure losses in the delivery pipes. The lifting height, on the other hand, is the vertical distance over which the liquid is raised from a lower level to a higher one. It is clear, then, that the lifting height is only a part of the total head, and when the pump is operating, its head is greater than the lifting height. So the so-called 8 meters you mentioned and a head of 8 meters are not the same concept; naturally, based on the relationship curve between flow rate and head, the flow rate you determine will also be inaccurate. Additionally, you need to consider the issue of the increase in static pressure energy for these two points.
The head of a centrifugal pump is different from its lift height; the lift height of a pump refers to the vertical distance over which the pump can transport liquid from a lower level to a higher one. Only when both the inlet and outlet vessels of the pump are at 0.1 MP, the inlet and outlet pipe diameters are the same, and the pipeline resistance can be ignored, will the pump’s head be equal to the lifting height. Right? Is question 2 described correctly?
If one simply wants to know the maximum height that can be pumped, then in fact, the pump outlet pressure multiplied by 10 gives the height that can be reached. For example, if the outlet pressure is 5 KG/CM2, then the corresponding pumping height is 50 meters. As for pressure losses, the calculation formulas are detailed; however, from a chemical engineering perspective, the key is to control valves, equipment, and height... As for pressure losses in the pipelines, they can basically be ignored... as their proportion is very small. The above information is provided for reference only