The relationship between the pump’s current, pressure, flow rate, and head
Thread Content
1. Current is proportional to flow rate. The greater the effective opening degree of the outlet valve, the higher the flow rate, the more work the pump has to perform, and the greater the current. 2. The smaller the effective opening degree of the outlet valve, the higher the pressure and the lower the flow rate ； The greater the effective opening degree of the outlet valve, the lower the pressure and the higher the flow rate; at this time, the current becomes larger. So, is there any relationship between pressure and flow rate and current? 3. Head – I haven’t quite understood this concept; it can be thought of as the ability to do work……
The pump shaft work is converted into static pressure energy and kinetic energy.
Continuing with the research means utilizing the kinetic energy of the impeller to convert it into the kinetic energy of the liquid
The static pressure energy is affected by the valve opening degree; pressure divided by the cross-sectional area under stress, plus the kinetic energy of the fluid in the pipeline, equals the shaft power – is that correct?
:'(It’s too complex; I still haven’t fully understood the relationships between these.)**