Formulas for calculating power and speed important for hydraulic pump maintenance

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As various industries continue to develop, a wide range of equipment, including hydraulic pumps, has seen significant progress and wider application. To ensure that these devices operate more efficiently, Shenzhen Autosys Hydraulics provides here the formulas for calculating power and speed that are useful when repairing such pumps, in the hope that it will be helpful to everyone. 1. Hydraulic pump/motor displacement: 1.3 l/r | Rated pressure: 20 MPa | Peak pressure: 31.5 MPa | Speed range: 2–320 r/min | Rated output torque: 3833 N.M. 2. Maximum flow rate of the hydraulic pump: 320 r/min x 1.3 l/r = 416 liters per minute. 3. Power of the hydraulic pump: 416 l/min x 20 MPa / 60 ((coefficient) = 138 kilowatts. 4. It can be equipped with a 132-kilowatt motor; however, since it is designed for 20 megapascals, a constant-power valve must be used for the hydraulic pump’s maintenance. In this case, both the maximum flow rate and the maximum pressure can be achieved, but not simultaneously. In other words, when the system pressure exceeds 20 MPa, the pump’s flow rate decreases automatically, and the speed of the motor slows down as a result. Special note: If a variable hydraulic pump is used during maintenance, it is recommended to use the A4VG250 model. If that is too expensive, two 125 models can be used instead. The motor has a power of 132 kilowatts and a rotation speed of 1500 revolutions per minute; a regular Y2 series motor will suffice. The tank for the closed-system must be installed in accordance with the B35 standard, and its capacity of around 600 liters is sufficient; it is advisable to equip it with a cooler as well.