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For the analysis of the pump outlet conditions, see the figure

2018-01-31View Original

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When the valve in the diagram is closed, how do the operating conditions of the centrifugal pump (flow rate, head) change? And how do the operating conditions of the DN40 and DN150 pipes (flow rate, pressure) change?
Reply #22018-01-31
The following condition is missing: 1. Pump parameters; 2. Pressure before the outlet valve before it is closed~~~
Reply #32018-02-01
When the DN150 valve is closed, both the pressure and flow rate in the DN150 pipeline are 0. The DN40 pipeline is the one with the lowest flow rate, used to protect the pump. The flow rate and pressure will increase, but generally a flow control orifice plate is installed on this pipeline, so the impact on flow rate and pressure is not significant.
Reply #42018-02-01
So at this time, is the pump operating in a low-flow, high-head mode?
Reply #52018-02-02
Based on the analysis of the pump’s characteristic curve and the pipeline’s characteristic curve, if the outlet valve with DN150 is closed, the flow rate in the pipeline is 0, and there is static pressure on the pipe walls; If the pump continues to operate normally, with its speed remaining unchanged, then its performance curve stays the same ; Compared to DN150, the pipeline resistance of DN40 is greater, resulting in an upward slope in the pipeline characteristic curve. As a result, the flow rate of the fluid passing through DN40 decreases, while the outlet pressure increases.
Reply #62018-02-03
With the outlet closed, flow decreases while head increases
Reply #72018-02-04
It is equivalent to a pressure-relief backflow: the flow rate decreases while the pressure increases
Reply #82018-02-06
This post was last edited by BigDreamerNewbie on February 6, 2018, at 09:27. Calculating the flow rate and pressure when a pump is in operation essentially means determining the pump’s operating point along the pipeline. This point is determined by both the pump’s characteristic curve and the pipeline’s characteristic curve: He = ∆Z + ∆P/ρg + ∆u²/2g + Hf. As pipeline resistance increases, the pipeline’s characteristic curve becomes steeper; consequently, the flow rate decreases while the head increases. Given the pump’s characteristic curve, its efficiency should also decrease.

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