{Help request} The level in the sump is inversely related to the pump outlet pressure
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This post was last edited by Panython on 2020-4-29 at 20:37. In a closed system, the level of the sump is inversely related to the pressure at the pump outlet; sudden and significant fluctuations occur, with the pump outlet pressures at multiple points within the system changing in the same manner. Why? As the liquid level rises, according to P=ρgh, shouldn’t the pressure at the pump outlet increase as well? Thank you so much for your help; I’ve forgotten all the knowledge from my four years in college. . . . . :'(
It would be best to include a simple flowchart as an illustration; it makes it easier to discuss things by looking at the diagram.
As the liquid level in the sump rises, the outlet pressure of the pump increases. In case 1, if there is a flow control circuit in the pump’s outlet pipeline, as the level of liquid in the reservoir increases, the control valve will close, causing the outlet pressure of the pump to rise. The amount by which the pressure increases is equal to the pressure generated by the rise in the reservoir’s liquid level ； In the second scenario, there is no flow control system on the pump’s outlet pipe. As the level of liquid in the reservoir increases, the flow rate increases as well, and the outlet pressure of the pump also rises. However, the increase in outlet pressure is less than the pressure resulting from the increase in the reservoir level. This is because, for a pump, an increase in flow rate leads to a decrease in head; head is equal to outlet pressure minus inlet pressure. Therefore, outlet pressure = inlet pressure + head. You can try out the spreadsheet I prepared at https://bbs.hcbbs.com/thread-1764877-1-1.html
Thank you, but in my case there is a negative correlation: as the level in the high-level tank increases, the pressure at the pump outlet decreases. . Many pumps are dropping at the same time, and the trend is very clear
Judging from the diagram you provided, similar processes were used in previous designs for chilled water systems, hot water systems, and heat transfer oil systems. Such systems reduce potential energy losses to 0; the outlet pressure of the pump is independent of the liquid level in the high-level tank. Before the pump starts operating, the system must be filled with liquid so that there is a certain level in the sump. What you’re describing sounds a bit strange; I think it happens when the pump is first started. Since the gas in the pipelines has not been completely removed, the pressure inside the pump is high at that moment. Once the gas is expelled and the flow path is cleared, the pressure difference between the inlet and outlet is eliminated, and the pressure at the pump’s outlet drops.
This process requires valves; a temperature control valve might be needed. The issue you mentioned might be related to the control valve. This diagram doesn’t explain anything. In normal production, based on this schematic diagram without control valves, the phenomenon you mentioned will not occur. It is recommended to check the valves in the process to see if they could affect the phenomenon you mentioned, for your reference.
Do you mean that when there is air in the pump, the pressure at the pump outlet is high, and once the air is removed, the pressure at the pump outlet drops? What’s the principle behind it? What about the traffic? What about the current? How will it change?
There are indeed valves for controlling temperature and flow rate. But it’s not intuitive and is quite difficult to analyze.