What is the relationship between the pump’s outlet pressure and head?
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Question: How is the outlet pressure of a pump generally determined? Is it determined by the head? What are the consequences if the outlet pressure of the pump is below the specified value? Also, if the outlet pressure of the pump doesn’t increase, what are the common causes? How to solve it?
The outlet pressure of the pump is lower than the designed pressure; check the motor current. Usually, the flow rate is too high – reducing the pump’s outlet flow or adjusting the backflow can resolve this issue； Another possibility is that the clearance of the pump’s impeller is too large, resulting in excessive leakage. It’s still pretty dangerous; it can easily burn out the motor.
You can take a look at Chapter 2 of Tianjin University’s Textbook on Chemical Engineering Principles and page P68 of the Handbook on Chemical Process Design; they provide detailed explanations. In front of the pump impeller is a low-pressure area, while behind it is a high-pressure area. This arrangement allows materials to be transported from high pressure to low pressure. The higher the pressure, the greater the head generated by the pump. Of course, it is also necessary to consider the properties of the liquid itself as well as the pressure losses in the pipes. There are many reasons why the pump outlet pressure cannot increase: air leakage in the pump, blockages in the upstream section, cavitation, worn blades, and so on
The outlet pressure is generally proportional to the head, but not absolutely
Closed-mouth pressure ≈ Young’s modulus × density × 10 + inlet pressure
What was said on the 6th floor is mostly correct, but one thing is that the closed-port pressure is 1.2 times that indicated on the nameplate; as for the open-port pressure at that time…