{"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":"obliczanie-wzoru-hydraulicznego","status":"publish","type":"post","link":"https:\/\/hydrathpower.com\/pl\/hydraulic-formulas-and-calculations-power-flow-force-and-speed-explained\/","title":{"rendered":"Wzory i obliczenia hydrauliczne: Obja\u015bnienie mocy, przep\u0142ywu, si\u0142y i pr\u0119dko\u015bci"},"content":{"rendered":"<h3>Czym jest ta strona?<\/h3>\n<p>Na tej stronie zebrano wzory hydrauliczne u\u017cywane na co dzie\u0144 przez in\u017cynier\u00f3w. Ka\u017cde r\u00f3wnanie jest przedstawione z jednostkami, kt\u00f3re odpowiadaj\u0105 praktyce przemys\u0142owej, a nast\u0119pnie z ma\u0142ymi przyk\u0142adami, kt\u00f3re mo\u017cna dostosowa\u0107 do w\u0142asnego systemu.<\/p>\n<h3>Kluczowe formu\u0142y<\/h3>\n<ul>\n<li>Ci\u015bnienie-si\u0142a-powierzchnia (imperialne): p(psi) = F(lbf) \/ A(in\u00b2); F = p \u00d7 A; A = F \/ p<\/li>\n<li>Moc hydrauliczna: hp = p(psi) \u00d7 Q(gpm) \/ 1714; kW = Q(l\/min) \u00d7 p(bar) \/ 600<\/li>\n<li>Przep\u0142yw pompy: Q(gpm) = n(rpm) \u00d7 D(in\u00b3\/rev) \/ 231; Q(l\/min) = D(cm\u00b3\/rev) \u00d7 n(rpm) \/ 1000<\/li>\n<li>Moment obrotowy silnika: T(in-lb) = p(psi) \u00d7 D(in\u00b3\/obr) \/ 6,28; T(Nm) = D(cm\u00b3\/obr) \u00d7 p(bar) \/ (20\u03c0)<\/li>\n<li>Powierzchnia cylindra: A = \u03c0d\u00b2\/4; Powierzchnia boczna pr\u0119ta = A<sub>t\u0142ok<\/sub> - A<sub>pr\u0119t<\/sub><\/li>\n<li>Si\u0142a cylindra: F = p \u00d7 A<\/li>\n<li>Pr\u0119dko\u015b\u0107 cylindra: v(ft\/min) = 19,25 \u00d7 Q(gpm) \/ A(in\u00b2); lub v = Q\/A w sp\u00f3jnych jednostkach.<\/li>\n<li>Wytyczne dotycz\u0105ce pr\u0119dko\u015bci na linii: v(ft\/s) = Q(gpm) \u00d7 0,3208 \/ A(in\u00b2)<\/li>\n<li>R\u00f3wnowa\u017cno\u015b\u0107 ciep\u0142a: 1 KM = 2545 BTU\/godz. \u2248 746 W<\/li>\n<\/ul>\n<h3>Dzia\u0142aj\u0105ce przyk\u0142ady<\/h3>\n<ol>\n<li>Moc hydrauliczna przy 150 barach i 20 l\/minPh(kW) = Q \u00d7 p \/ 600 = 20 \u00d7 150 \/ 600 = 5 kW (\u2248 6,7 KM)<\/li>\n<li><a href=\"\/hydraulic-cylinders\/\">Cylinder<\/a> ci\u0105g z otworem 50 mm przy 140 barachPowierzchnia = \u03c0 \u00d7 (0,05 m)\u00b2 \/ 4 = 0,001963 m\u00b2 Si\u0142a = 14 MPa \u00d7 0,001963 m\u00b2 \u2248 27,5 kN<\/li>\n<li>Moment obrotowy silnika, D = 25 cm\u00b3\/obr, p = 140 barT = D \u00d7 p \/ (20\u03c0) = 25 \u00d7 140 \/ 62,832 \u2248 55,7 Nm<\/li>\n<li>Kontrola pr\u0119dko\u015bci w\u0119\u017ca, Q = 30 l\/min (\u2248 7,93 galon\u00f3w USA\/min), 1\/2 in IDA = \u03c0 \u00d7 (0,5 in)\u00b2 \/ 4 = 0,196 in\u00b2v = 7,93 \u00d7 0,3208 \/ 0,196 \u2248 13,0 ft\/s<\/li>\n<\/ol>\n<h3>Jak dobra\u0107 rozmiar jednostki HPU (pi\u0119\u0107 krok\u00f3w)<\/h3>\n<ol>\n<li>Zdefiniuj obowi\u0105zki: czas cyklu, ci\u015bnienie szczytowe, temperatura otoczenia.<\/li>\n<li>Oblicz powierzchni\u0119 si\u0142ownika i si\u0142\u0119 na podstawie wymaganego obci\u0105\u017cenia; ustaw ci\u015bnienie projektowe z marginesem 10-20%.<\/li>\n<li>Uzyskanie przep\u0142ywu na podstawie docelowej pr\u0119dko\u015bci i powierzchni; przeliczenie na wydajno\u015b\u0107 pompy przy oczekiwanych obrotach na minut\u0119.<\/li>\n<li>Oszacowa\u0107 moc hydrauliczn\u0105 na podstawie p i Q; zastosowa\u0107 og\u00f3ln\u0105 sprawno\u015b\u0107 do wyboru silnika nap\u0119dowego.<\/li>\n<li>Sprawdzi\u0107 pr\u0119dko\u015b\u0107 linii i obci\u0105\u017cenie cieplne; odpowiednio dobra\u0107 zbiornik i ch\u0142odzenie.<\/li>\n<\/ol>\n<h3>Cz\u0119sto zadawane pytania<\/h3>\n<h4>Jaki jest wz\u00f3r na moc hydrauliczn\u0105?<\/h4>\n<p>hp = p(psi) \u00d7 Q(gpm) \/ 1714. W przypadku system\u00f3w metrycznych kW = Q(l\/min) \u00d7 p(bar) \/ 600.<\/p>\n<h4>Jak obliczy\u0107 pr\u0119dko\u015b\u0107 cylindra?<\/h4>\n<p>Podzieli\u0107 przep\u0142yw przez powierzchni\u0119 t\u0142oka w sp\u00f3jnych jednostkach. W jednostkach imperialnych v(ft\/min) = 19,25 \u00d7 Q(gpm) \/ A(in\u00b2).<\/p>\n<h4>Jak obliczy\u0107 moment obrotowy silnika na podstawie ci\u015bnienia i przemieszczenia?<\/h4>\n<p>U\u017cyj T(in-lb) = p(psi) \u00d7 D(in\u00b3\/obr) \/ 6,28 lub T(Nm) = D(cm\u00b3\/obr) \u00d7 p(bar) \/ (20\u03c0).<\/p>\n<h4>Jakie s\u0105 typowe limity pr\u0119dko\u015bci w\u0119\u017ca?<\/h4>\n<p>Linie ci\u015bnieniowe s\u0105 zwykle utrzymywane w pobli\u017cu 10-20 st\u00f3p\/s; linie ss\u0105ce s\u0105 znacznie ni\u017csze. Zawsze nale\u017cy sprawdzi\u0107 specyfikacje komponent\u00f3w.<\/p>\n<h4>Jak oszacowa\u0107 obci\u0105\u017cenie cieplne na podstawie spadku ci\u015bnienia?<\/h4>\n<p>Utrata mocy (hp) \u2248 \u0394p(psi) \u00d7 Q(gpm) \/ 1714, a nast\u0119pnie przelicz na BTU\/godz. lub waty.<\/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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center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[1],"tags":[],"class_list":["post-4499","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-uncategorized"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v28.8 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Hydraulic Formulas and Calculations: Power, Flow, Force and Speed Explained - Hydrath Power<\/title>\n<meta name=\"description\" content=\"earn the core hydraulic formulas for power, flow, force and speed, plus pump, motor and cylinder calculations with clear, worked examples.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" 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