{"id":2784,"date":"2024-03-07T14:45:07","date_gmt":"2024-03-07T13:45:07","guid":{"rendered":"https:\/\/www.bf-hydraulik.com\/encyclopedia\/calculate-pump-power\/"},"modified":"2026-04-15T11:52:51","modified_gmt":"2026-04-15T09:52:51","slug":"calculate-pump-power","status":"publish","type":"encyclopedia","link":"https:\/\/www.bf-hydraulik.com\/en\/calculate-pump-power\/","title":{"rendered":"Calculate pump power"},"content":{"rendered":"\t\t<div data-elementor-type=\"wp-post\" data-elementor-id=\"2784\" class=\"elementor elementor-2784 elementor-901\">\n\t\t\t\t\t\t<section class=\"elementor-section elementor-top-section elementor-element elementor-element-735b55f8 elementor-section-boxed elementor-section-height-default elementor-section-height-default\" data-id=\"735b55f8\" data-element_type=\"section\">\n\t\t\t\t\t\t<div class=\"elementor-container elementor-column-gap-default\">\n\t\t\t\t\t<div class=\"elementor-column elementor-col-50 elementor-top-column elementor-element elementor-element-2820cb8b\" data-id=\"2820cb8b\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-widget-wrap elementor-element-populated\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-9fb9997 elementor-widget elementor-widget-text-editor\" data-id=\"9fb9997\" data-element_type=\"widget\" data-widget_type=\"text-editor.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t\t\t\t\t<h1>Calculate pump power<\/h1><p>Power is work per unit of time, and work is force times distance. <br>Force, in turn, is mass times acceleration.<\/p><p>The basic parameters of mass, acceleration, time, and distance can be found within a pump. A fluid to be circulated naturally has a defined mass that is moved over a certain distance. This requires force. If you then add the factor of time, you arrive back at power.    <\/p><h2>What is the purpose of calculating pump power?<\/h2><p>A pump performs two tasks in a <a href=\"https:\/\/www.bf-hydraulik.com\/en\/hydraulic-system\/\" target=\"_self\" title=\"Hydraulic system: design &amp; applications explained simplyA hydraulic system is used to apply large forces in a targeted manner over a defined distance with minimal effort. The underlying hydraulic principle&hellip;\" class=\"encyclopedia\">hydraulic system<\/a>:<\/p><ul><li><strong>Generating operating pressure<\/strong><br>The higher the operating pressure, the greater the forces that can be produced. Operating pressure is independent of time. However, it defines the capacity of a <a href=\"https:\/\/www.bf-hydraulik.com\/en\/hydraulic-system\/\" target=\"_self\" title=\"Hydraulic system: design &amp; applications explained simplyA hydraulic system is used to apply large forces in a targeted manner over a defined distance with minimal effort. The underlying hydraulic principle&hellip;\" class=\"encyclopedia\">hydraulic system<\/a> in that it determines the work to be performed (force times distance).  <\/li><li><strong>Generating a volume flow<\/strong><br>The volume flow ultimately indicates the speed of a <a href=\"https:\/\/www.bf-hydraulik.com\/en\/hydraulic-system\/\" target=\"_self\" title=\"Hydraulic system: design &amp; applications explained simplyA hydraulic system is used to apply large forces in a targeted manner over a defined distance with minimal effort. The underlying hydraulic principle&hellip;\" class=\"encyclopedia\">hydraulic system<\/a>. The faster the fluid flows in the system, the faster the connected actuators move. <\/li><\/ul><h3>Factors of pump performance<\/h3><p>The performance of a pump depends on three factors:<\/p><ol><li><a href=\"https:\/\/www.bf-hydraulik.com\/en\/displacement\/\" target=\"_self\" title=\"Displacement as a Basis for Flow CalculationDisplacement is the amount of hydraulic fluid that a hydraulic motor requires for one working stroke. This volume is used for the design of&hellip;\" class=\"encyclopedia\">Displacement<\/a> volume of the pump unit<\/li><li>Rotational speed of the drive<\/li><li>Power of the drive<\/li><\/ol><p>A drive can be designed to be as large as possible, but if the pump can only handle a small volume, then higher force or speed will not change anything. Conversely, a powerful pump can never reach its full potential if it is equipped with a drive that is too weak. Motor and pump must therefore be matched to one another in order to work as efficiently as possible.  <\/p><h3>Resistance to pumping work<\/h3><p>In a <a href=\"https:\/\/www.bf-hydraulik.com\/en\/hydraulic-system\/\" target=\"_self\" title=\"Hydraulic system: design &amp; applications explained simplyA hydraulic system is used to apply large forces in a targeted manner over a defined distance with minimal effort. The underlying hydraulic principle&hellip;\" class=\"encyclopedia\">hydraulic system<\/a>, a pump always works against a certain resistance. Typical factors influencing this resistance are the viscosity of the hydraulic medium and the system&rsquo;s own backpressure. Channels that are too narrow, incorrectly selected valves, or too many branches can significantly reduce the performance of a pump.  <\/p><p>The <a href=\"https:\/\/www.bf-hydraulik.com\/en\/hydraulic-system\/\" target=\"_self\" title=\"Hydraulic system: design &amp; applications explained simplyA hydraulic system is used to apply large forces in a targeted manner over a defined distance with minimal effort. The underlying hydraulic principle&hellip;\" class=\"encyclopedia\">hydraulic system<\/a> and the performance of the pump with its drive must also be matched. In pump calculation, it is therefore preferable to work with the efficiency factor 510. <\/p><p><em>510 = 600 (conversion factor) * 0.85 (total efficiency of electric motor\/pump)<\/em><\/p><h3>Formulas for calculating a pump<\/h3><p>The design of a pump unit for a <a href=\"https:\/\/www.bf-hydraulik.com\/en\/hydraulic-system\/\" target=\"_self\" title=\"Hydraulic system: design &amp; applications explained simplyA hydraulic system is used to apply large forces in a targeted manner over a defined distance with minimal effort. The underlying hydraulic principle&hellip;\" class=\"encyclopedia\">hydraulic system<\/a> takes place in three steps:<\/p><p><strong>#1 Determination of the flow rate Q <\/strong><strong><br><\/strong>The flow rate &ldquo;Q&rdquo; is the volume that the pump must move and is calculated using the pump <a href=\"https:\/\/www.bf-hydraulik.com\/en\/displacement\/\" target=\"_self\" title=\"Displacement as a Basis for Flow CalculationDisplacement is the amount of hydraulic fluid that a hydraulic motor requires for one working stroke. This volume is used for the design of&hellip;\" class=\"encyclopedia\">displacement<\/a> &ldquo;Vg&rdquo; and the number of revolutions &ldquo;n&rdquo;.<br><br>The formula is: <br>Flow rate &ldquo;Q&rdquo; equals <a href=\"https:\/\/www.bf-hydraulik.com\/en\/displacement\/\" target=\"_self\" title=\"Displacement as a Basis for Flow CalculationDisplacement is the amount of hydraulic fluid that a hydraulic motor requires for one working stroke. This volume is used for the design of&hellip;\" class=\"encyclopedia\">displacement<\/a> &ldquo;Vg&rdquo; times revolutions &ldquo;n&rdquo; divided by 1,000<br><br><em>Q = (Vg x n) \/ 1,000<\/em><\/p><p><strong>#2 Determination of the drive power P<br><\/strong>The power &ldquo;P&rdquo; of the connected motor is calculated using the pump pressure &ldquo;p&rdquo; and the efficiency &ldquo;510&rdquo; (also known as the pump constant). <br><br>The drive power &ldquo;P&rdquo; is equal to pump pressure &ldquo;p&rdquo; times flow rate &ldquo;Q&rdquo; divided by the efficiency &ldquo;510&rdquo;: <br><br><em>P = (p x Q) \/ 510<\/em><br><br><\/p><p>With these factors and these two formulas, every pump and its drive can be designed for the respective situation.<\/p><p><strong>#3 Calculating pump power &ndash; using a practical example<\/strong><\/p><p>A lifting platform requires a pump unit. <br>It requires a pressure of 100 bar and a flow rate of 10 l\/min.<\/p><p>Calculation of motor power &ldquo;P&rdquo;:<br><em>P = (100 bar X 10 l\/min) \/ 510 = 1.96 &#8793; 2 kW<\/em><\/p><p>An electric motor with 2 kW of power and a speed of 1,000 rpm is sufficient in this case.<\/p><p>Now the pump is designed. <br>To do this, the formula for determining the flow rate &ldquo;Q = (Vg x n) \/ 1,000&rdquo; is rearranged for <a href=\"https:\/\/www.bf-hydraulik.com\/en\/displacement\/\" target=\"_self\" title=\"Displacement as a Basis for Flow CalculationDisplacement is the amount of hydraulic fluid that a hydraulic motor requires for one working stroke. This volume is used for the design of&hellip;\" class=\"encyclopedia\">displacement<\/a> &ldquo;Vg&rdquo;.<\/p><p><em>Vg = Q * 1,000 \/ n.<\/em> <br><br>This results in: <br><em>Vg = 10 l\/min x 1,000 \/ 1,000 rpm <\/em><br><em>Vg = <\/em><strong><em>10 cm&sup3;\/rev<\/em><br><\/strong><br>A pump with a capacity of 10 cm&sup3; \/ revolution is sufficient for this application.<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t<div class=\"elementor-column elementor-col-50 elementor-top-column elementor-element elementor-element-17578f6\" data-id=\"17578f6\" data-element_type=\"column\">\n\t\t\t<div class=\"elementor-widget-wrap elementor-element-populated\">\n\t\t\t\t\t\t<div class=\"elementor-element elementor-element-18d5d30 elementor-widget elementor-widget-sidebar\" data-id=\"18d5d30\" data-element_type=\"widget\" data-widget_type=\"sidebar.default\">\n\t\t\t\t<div class=\"elementor-widget-container\">\n\t\t\t\t\t<div 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