Calculation of electricity savings from energy-saving upgrades to pumps

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This post was last edited by mjaline on 2011-5-29 09:40. The user wishes to carry out energy-saving modifications to the pump currently in use. To determine the feasibility of such modifications, it is necessary to calculate the actual amount of electricity saved as a result of these changes, so as to figure out how long it will take for the savings to cover the cost of the modifications. However, the user does not know how to calculate the actual amount of electricity that will be saved after the modifications. Experts and fellow users are requested to provide guidance: Option 1: Before the modification, the pump was equipped with a motor with a power rating of 110 KW (at 380V voltage, with an operating current of 180A). After the modification, a 90 KW motor can be used (also at 380V voltage, with an assumed operating current of 160A). Option 1: Before the pump modification, the motor used had a power of 110 KW (at 380V voltage, with an operating current of 180A). After the modification, an 110 KW motor is still used (at 380V voltage, with an assumed operating current of 170A). Please help provide the calculation methods for the actual electricity savings resulting from the transformation of each of the above two solutions. Thank you!