How to calculate the actual power consumption of a pump per hour?

The power consumption of the original two pumps can be calculated using the actual operating current, voltage, and power factor. Now, the power consumption of a new pump can be calculated by using the design parameters of the pump such as flow rate, head, medium density, and efficiency to determine the shaft power, which can then be roughly converted into electricity consumption. By comparing the two, an approximate estimate of the energy saved can be obtained.

Power and KWH can both indicate the approximate technical power consumption

Thank you. The new pump has a head of 70, a flow rate of 800, a medium density of 722, and an efficiency of 0.8. Could you please explain how these values are calculated?

This post was last edited by *nht1 on 2015-7-7 at 21:53. Normally, it is calculated in this way: w=pt. Measure the current during normal operation, and then use this formula to calculate: P=√3×U×I×cosΦ, where cosΦ is the power factor, which can be found on the motor’s nameplate – go and check it. There is also an abnormal situation in which, when using a clamp meter to measure the three phases, the currents turn out to be unequal (to be precise, not close to each other). This indicates phase imbalance. In such cases, it is necessary to measure the current of each phase separately, calculate them individually, and then add them together: P=∑(√3×U×Ii×cosΦ). I hope this helps; give it a try in practice.

Pump shaft power = liquid density x flow rate x head / (102 x pump efficiency), where: Pump shaft power – in kilowatts; Liquid density – in kg/m3; Flow rate – in m3/second; Head – in meters. Since I don’t know the units of the parameters you provide, you can calculate it using the formula above

If the efficiency of the old pump and the new pump is similar, replacing it with a new one won’t save much electricity.

For the existing two pumps, the actual power consumption can be known from the electrical readings; for the new pump, it can be determined by calculating based on flow rate and head, or by referring to the pump’s performance curve, allowing for comparison

Just compare the shaft power. Add up the shaft power of the two original pumps and compare it with the shaft power of the large pump to find out.

This post was last edited by 0432108985 on 2015-7-14 at 16:35. Since there is no change in your pipeline and it is assumed that the flow rate remains constant, then the power required for the same flow rate (hydraulic power) will also be the same. Motor active power consumption * Motor efficiency * Pump head efficiency (as per the performance curve) = Power required to move water. Therefore, the amount of energy used to do work on the water remains unchanged; the energy-saving effect comes primarily from the efficiencies of both the pump and the motor. -- In addition to the calculation formula forfld8882005 on the 6th floor, there is another algorithm: Water power = flow rate (m^3/h) * head (m) * density * 9.81/3600. Pump shaft input power = Water power / Pump efficiency. Motor output power = Motor input power * Motor efficiency. [Correction] Pump shaft input power = Motor output power – Based on calculations using 140A and 148A, the load capacity of a 110kW motor is approximately 73%, resulting in an operating output power of around 80kW ； The 185kW motor meets the power requirements, and its operating efficiency is close to 90%. By using a 185kW motor, the load rate increases, and the power factor also improves ； The 185kW-4P motor has a slightly higher efficiency than the 110kW-2P motor; therefore, overall it can be expected to result in energy savings. Current data do not allow for an exact calculation of this effect.

You can install an electricity meter to check it: lol