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How do the pump head and flow rate change when the pump bypass valve is opened further? As shown in Figure 1, at the initial state of equilibrium, the resistance in pipeline OA is equal to the resistance in pipeline OB, both being F1, and the flow rate is Q1. If the valve in pipeline OB is suddenly opened wider, how will the pump’s flow rate and head change? And why? 2. Initially, at equilibrium, the resistance in pipeline OA is equal to the resistance in pipeline OB, i.e., R OA = R OB = F1, and the flow rate is Q1. If the opening degree of the valve in pipeline OB suddenly decreases, how will the pump’s flow rate and head change? Why?
In the first case, the total flow rate increases, the head decreases, and the flow rate in route OA decreases. In the second case, the total flow rate decreases, the head increases, and the flow rate in path OA increases.
Is this for a centrifugal pump? Are positive displacement pumps also suitable?
Sir, if the diameter of the OA pipeline remains unchanged but the flow rate decreases significantly, will the head loss increase greatly? ? Although the pipe resistance decreases when the flow rate in the OA pipeline drops, at the original flow rate of the OA pipe the head generated by the pump was sufficient to meet the process requirements. However, after the flow rate drops significantly, it is uncertain whether the head provided by the pump in the OA pipe will still be sufficient to meet those requirements (as the decrease in pipe resistance resulting from the reduced flow rate may represent only a small part of the total losses; factors such as liquid level differences and pressure differences also play a role)
1. Increased flow, reduced dust. 2. Decreased flow. Dust particles become larger
It is recommended to take a closer look at the pump’s characteristic curve and understand it thoroughly
Search for “minimum flow line” on the forum; there are numerous relevant posts available for your reference.
I’ve learned it. I only know it’s more or less like this; P