What is the impact of the diameter of the pump’s outlet pipeline on the water output of the pump?
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This post was last edited by lee2033 on 2018-7-30 at 20:23. I have seen many cases where the diameters of the inlet and outlet pipes of the water pump are much larger than the pump’s own inlet and outlet dimensions; at first, I thought there was a problem with the design. It was later learned that if the diameter of the water outlet pipe is a few sizes larger than that of the pump’s outlet, the water resistance will be lower, making it more suitable for long-distance transportation (This statement is not easy to understand; it seems that the thicker the outlet pipe, the more head loss there will be.) ) What is the impact of the diameter of the outlet pipe being greater than or smaller than the diameter of the water pump’s outlet on the pump’s flow rate and head? Additionally, regarding the regulation of the flow rate in the outlet pipeline using a gate valve, the smaller the opening degree of the gate valve, the lower the flow rate and the lower the motor current; but is the head pressure constant? Or does the flow rate increase as the gate valve opening increases, resulting in a lower head pressure, while the flow rate decreases, leading to a higher head pressure?
For a normal centrifugal pump, its performance curve shows that as the flow rate increases, the head decreases. The operating point of the pump is determined by the device’s characteristic curve. A larger pipe diameter results in a lower flow velocity, which in turn reduces pipeline losses. However, a larger diameter inevitably increases costs; therefore, a reasonable economic flow velocity is determined, and the pipe diameter is chosen based on this value.
The finer the pipe, the greater the head loss; the thicker it is, the more suitable it is for long-distance transportation. Theoretically, the smaller the opening of the gate valve, the greater the head; as the opening increases, the head gradually decreases.
My understanding is that head is the energy obtained by the fluid as it exits the pump; when the pump’s speed is constant and the flow rate is fixed, the head is also determined, with this head being lost starting from the pump outlet. The expansion at the pump outlet represents a type of local resistance loss, and thus it also consumes head. However, the expansion at the pump outlet is limited to just one or two size increments, and reducer pipes are used (which can reduce the local resistance coefficient). After the expansion, the flow velocity decreases, resulting in a reduction in both frictional and local resistances. Overall, therefore, this expansion helps to minimize resistance losses, enabling the medium to be transported over greater distances ； When a gate valve is used to regulate the outlet flow rate, the parameters of the pump remain unchanged (including power consumption and head). The smaller the opening degree of the gate valve, the greater the local resistance loss and the more head is consumed, resulting in less head remaining.
That was my original thought as well. But after analyzing it now, it is believed that when the gate valve is opened to a small degree, the flow rate is low, and the head pressure is also not high, as the current is low at this time as well. It seems that the current increases as the valve opening increases and the flow rate increases. As for the head, I don’t think it has much to do with the valve opening degree (when the valve is open slightly, the local pressure near the valve disc increases, but the overall pressure in the pipeline doesn’t increase significantly). I’m not sure if this understanding is correct.
The method of throttling at the pump outlet can increase the head; by closing the valve, the operating point changes from 1 to 2 as shown in the diagram. Additionally, gate valves are generally not used for throttling, as this can damage the valve seals and result in poor sealing
Thank you for the reply. “The energy loss in the system caused by valve pressure drops represents the potential for energy savings. My understanding is that by using valves to reduce flow rate, energy can be saved (the power consumption decreases), as evidenced by a reduction in motor current. Since energy is being saved, the head pressure is likely not to increase as the flow rate decreases; otherwise, there would be no energy savings. I’m not sure if it’s correct.
Pump power = constant * head * flow rate. Although the power has decreased, the flow rate is also dropping; the head may increase
“The energy loss in the system caused by valve pressure drops represents the potential for energy savings; in other words, if it is possible to reduce or eliminate such energy losses, energy can be saved. For example, variable frequency control or the use of segmented impellers can be employed instead of reducing the valve opening to lower the flow rate. In other words, the valve is fully open, which reduces the pump’s output power; energy is utilized fully with no losses, thereby achieving energy savings.
Agree with the view from Floor 9; it is recommended to use methods such as variable frequency control and impeller cutting to achieve energy savings. Relying on the opening degree of valves at present is not a good approach in the long term. Let’s take a simple example: if you reduce the valve’s opening degree, you can save 10,000 yuan in energy costs per month. But if the valves remain partially open over time and cannot be fully closed, it will cause problems when the pump fails and needs to be repaired. . . . At the beginning of this year, I had several pumps that weren’t functioning properly because the valves couldn’t be closed properly. As a solution, they purchased several tons of dry ice to freeze the pipes, thereby allowing the pumps to be repaired; the cost of this approach was incredibly high. . . . Of course, I understand that this isn’t what the original poster was asking; I’m just offering a reminder, like the friend in post 9 did, or giving a one-time solution. . .