{"id":4499,"date":"2026-04-01T06:56:01","date_gmt":"2026-04-01T06:56:01","guid":{"rendered":"https:\/\/hydrathpower.com\/?p=4499"},"modified":"2026-04-01T06:56:01","modified_gmt":"2026-04-01T06:56:01","slug":"berechnung-der-hydraulikformel","status":"publish","type":"post","link":"https:\/\/hydrathpower.com\/de\/hydraulic-formulas-and-calculations-power-flow-force-and-speed-explained\/","title":{"rendered":"Hydraulische Formeln und Berechnungen: Leistung, Durchfluss, Kraft und Geschwindigkeit erkl\u00e4rt"},"content":{"rendered":"<h3>Was ist diese Seite?<\/h3>\n<p>Auf dieser Seite sind die Hydraulikformeln zusammengestellt, die Ingenieure t\u00e4glich verwenden. Jede Gleichung wird mit Einheiten dargestellt, die der industriellen Praxis entsprechen, gefolgt von kleinen Beispielen, die Sie an Ihr System anpassen k\u00f6nnen.<\/p>\n<h3>Wichtige Formeln<\/h3>\n<ul>\n<li>Druck-Kraft-Fl\u00e4che (imperial): p(psi) = F(lbf) \/ A(in\u00b2); F = p \u00d7 A; A = F \/ p<\/li>\n<li>Hydraulische Leistung: hp = p(psi) \u00d7 Q(gpm) \/ 1714; kW = Q(l\/min) \u00d7 p(bar) \/ 600<\/li>\n<li>Durchfluss der Pumpe: Q(gpm) = n(rpm) \u00d7 D(in\u00b3\/rev) \/ 231; Q(l\/min) = D(cm\u00b3\/rev) \u00d7 n(rpm) \/ 1000<\/li>\n<li>Motordrehmoment: T(in-lb) = p(psi) \u00d7 D(in\u00b3\/rev) \/ 6,28; T(Nm) = D(cm\u00b3\/rev) \u00d7 p(bar) \/ (20\u03c0)<\/li>\n<li>Fl\u00e4che des Zylinders: A = \u03c0d\u00b2\/4; Stangenseitenfl\u00e4che = A<sub>Kolben<\/sub> - A<sub>Stab<\/sub><\/li>\n<li>Kraft des Zylinders: F = p \u00d7 A<\/li>\n<li>Zylindergeschwindigkeit: v(ft\/min) = 19,25 \u00d7 Q(gpm) \/ A(in\u00b2); oder v = Q\/A in einheitlichen Einheiten<\/li>\n<li>Richtwert f\u00fcr die Leitungsgeschwindigkeit: v(ft\/s) = Q(gpm) \u00d7 0,3208 \/ A(in\u00b2)<\/li>\n<li>W\u00e4rme\u00e4quivalenz: 1 PS = 2545 BTU\/Std. \u2248 746 W<\/li>\n<\/ul>\n<h3>Praktische Beispiele<\/h3>\n<ol>\n<li>Hydraulische Leistung bei 150 bar und 20 L\/minPh(kW) = Q \u00d7 p \/ 600 = 20 \u00d7 150 \/ 600 = 5 kW (\u2248 6,7 PS)<\/li>\n<li><a href=\"\/hydraulic-cylinders\/\">Zylinder<\/a> Schubkraft mit 50 mm Bohrung bei 140 barFl\u00e4che = \u03c0 \u00d7 (0,05 m)\u00b2 \/ 4 = 0,001963 m\u00b2Kraft = 14 MPa \u00d7 0,001963 m\u00b2 \u2248 27,5 kN<\/li>\n<li>Motordrehmoment, D = 25 cm\u00b3\/U, p = 140 barT = D \u00d7 p \/ (20\u03c0) = 25 \u00d7 140 \/ 62,832 \u2248 55,7 Nm<\/li>\n<li>Pr\u00fcfung der Schlauchgeschwindigkeit, Q = 30 L\/min (\u2248 7,93 gpm), 1\/2 Zoll IDA = \u03c0 \u00d7 (0,5 Zoll)\u00b2 \/ 4 = 0,196 Zoll\u00b2v = 7,93 \u00d7 0,3208 \/ 0,196 \u2248 13,0 ft\/s<\/li>\n<\/ol>\n<h3>Wie man eine HPU dimensioniert (f\u00fcnf Schritte)<\/h3>\n<ol>\n<li>Definieren Sie den Betrieb: Zykluszeit, Spitzendruck, Umgebungstemperatur.<\/li>\n<li>Berechnen Sie die Antriebsfl\u00e4che und -kraft aus der erforderlichen Last; legen Sie den Auslegungsdruck mit einem Spielraum von 10-20% fest.<\/li>\n<li>Ableitung des Durchflusses aus Zielgeschwindigkeit und -fl\u00e4che; Umrechnung in Pumpenf\u00f6rdermenge mit erwarteter Drehzahl.<\/li>\n<li>Sch\u00e4tzen Sie die hydraulische Leistung anhand von p und Q; wenden Sie den Gesamtwirkungsgrad an, um den Antriebsmotor auszuw\u00e4hlen.<\/li>\n<li>\u00dcberpr\u00fcfen Sie die Leitungsgeschwindigkeit und die W\u00e4rmebelastung; dimensionieren Sie den Speicher und die K\u00fchlung entsprechend.<\/li>\n<\/ol>\n<h3>H\u00e4ufig gestellte Fragen<\/h3>\n<h4>Wie lautet die Formel f\u00fcr hydraulische Pferdest\u00e4rken?<\/h4>\n<p>hp = p(psi) \u00d7 Q(gpm) \/ 1714. F\u00fcr metrische Systeme: kW = Q(l\/min) \u00d7 p(bar) \/ 600.<\/p>\n<h4>Wie berechne ich die Zylindergeschwindigkeit?<\/h4>\n<p>Teilen Sie den Durchfluss durch die Kolbenfl\u00e4che in einheitlichen Einheiten. Imperiale Einheiten: v(ft\/min) = 19,25 \u00d7 Q(gpm) \/ A(in\u00b2).<\/p>\n<h4>Wie ermittle ich das Motordrehmoment aus Druck und Hubraum?<\/h4>\n<p>Verwenden Sie T(in-lb) = p(psi) \u00d7 D(in\u00b3\/rev) \/ 6,28 oder T(Nm) = D(cm\u00b3\/rev) \u00d7 p(bar) \/ (20\u03c0).<\/p>\n<h4>Was sind typische Geschwindigkeitsgrenzen f\u00fcr Schl\u00e4uche?<\/h4>\n<p>Druckleitungen werden in der Regel bei 10-20 ft\/s gehalten; Saugleitungen sind viel niedriger. Pr\u00fcfen Sie immer anhand der Spezifikationen Ihrer Komponenten.<\/p>\n<h4>Wie sch\u00e4tze ich die W\u00e4rmebelastung anhand eines Druckabfalls?<\/h4>\n<p>Verlustleistung (hp) \u2248 \u0394p(psi) \u00d7 Q(gpm) \/ 1714, dann in BTU\/hr oder Watt umrechnen.<\/p>","protected":false},"excerpt":{"rendered":"<p>What is This Page? This page collects the hydraulic formulas engineers use every day. Each equation is shown with units [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":4505,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"set","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center 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