{"id":41737,"date":"2019-09-30T00:01:27","date_gmt":"2019-09-29T23:01:27","guid":{"rendered":"https:\/\/www.thermal-engineering.org\/que-es-la-teoria-de-la-bomba-ecuacion-de-turbomachine-de-euler-definicion\/"},"modified":"2020-01-20T13:12:03","modified_gmt":"2020-01-20T12:12:03","slug":"que-es-la-teoria-de-la-bomba-ecuacion-de-turbomachine-de-euler-definicion","status":"publish","type":"post","link":"https:\/\/www.thermal-engineering.org\/es\/que-es-la-teoria-de-la-bomba-ecuacion-de-turbomachine-de-euler-definicion\/","title":{"rendered":"\u00bfQu\u00e9 es la teor\u00eda de la bomba? &#8211; Ecuaci\u00f3n de Turbomachine de Euler &#8211; Definici\u00f3n"},"content":{"rendered":"<div class=\"su-quote su-quote-style-default\">\n<div class=\"su-quote-inner su-clearfix\">La ecuaci\u00f3n de turbom\u00e1quina de Euler, o a veces llamada ecuaci\u00f3n de bomba de Euler, desempe\u00f1a un papel central en la turbom\u00e1quina, tambi\u00e9n en la teor\u00eda de las bombas centr\u00edfugas. Ingenier\u00eda t\u00e9rmica<\/div>\n<\/div>\n<div class=\"su-divider su-divider-style-dotted\"><\/div>\n<div class=\"lgc-column lgc-grid-parent lgc-grid-100 lgc-tablet-grid-100 lgc-mobile-grid-100 lgc-equal-heights lgc-first lgc-last\">\n<div class=\"inside-grid-column\">\n<div class=\"su-spacer\"><\/div>\n<h2>Teor\u00eda de la bomba &#8211; Ecuaciones de turbina de Euler<\/h2>\n<p><strong>La ecuaci\u00f3n de turbom\u00e1quina de Euler<\/strong>\u00a0, o a veces llamada\u00a0<strong>ecuaci\u00f3n de bomba de Euler<\/strong>\u00a0, desempe\u00f1a un papel central en la\u00a0<strong>turbom\u00e1quina<\/strong>\u00a0ya que conecta el\u00a0<strong>trabajo espec\u00edfico Y<\/strong>\u00a0y la geometr\u00eda y las velocidades en el impulsor.\u00a0La ecuaci\u00f3n se basa en los conceptos de\u00a0<strong>conservaci\u00f3n del momento angular<\/strong>\u00a0y\u00a0<strong>conservaci\u00f3n de la energ\u00eda<\/strong>\u00a0.<\/p>\n<p>Las\u00a0<strong>ecuaciones de turbom\u00e1quinas de Euler<\/strong>\u00a0son:<\/p>\n<p><strong>Par del eje:\u00a0<sub>eje<\/sub>\u00a0T\u00a0\u00a0 \u00a0 = \u03c1Q (r\u00a0<sub>2<\/sub>\u00a0V\u00a0<sub>t2<\/sub>\u00a0&#8211; r\u00a0<sub>1<\/sub>\u00a0V\u00a0<sub>t1<\/sub>\u00a0)<\/strong><\/p>\n<p><strong>Potencia de agua: P\u00a0<sub>w<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = \u03c9.\u00a0<sub>Eje \u00a0 \u00a0 \u00a0\u00a0<\/sub>\u00a0T\u00a0= \u03c1Q (u\u00a0<sub>2<\/sub>\u00a0V\u00a0<sub>t2<\/sub>\u00a0&#8211; u\u00a0<sub>1<\/sub>\u00a0V\u00a0<sub>t1<\/sub>\u00a0)<\/strong><\/p>\n<p><strong>Altura de la\u00a0bomba: H = P\u00a0<sub>w<\/sub>\u00a0\/ \u03c1gQ = (u\u00a0<sub>2<\/sub>\u00a0V\u00a0<sub>t2<\/sub>\u00a0&#8211; u\u00a0<sub>1<\/sub>\u00a0V\u00a0<sub>t1<\/sub>\u00a0) \/ g<\/strong><\/p>\n<p>d\u00f3nde<\/p>\n<ul>\n<li><strong>r\u00a0<sub>1<\/sub><\/strong>y\u00a0<strong>r\u00a0<sub>2<\/sub><\/strong>\u00a0son los\u00a0<strong>di\u00e1metros del\u00a0<a title=\"Impulsor - Tipos de impulsores\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/fluid-dynamics\/centrifugal-pumps\/impeller-types-of-impellers\/\">impulsor<\/a><\/strong>\u00a0en la entrada y la salida, respectivamente.<\/li>\n<li><strong>u\u00a0<sub>1<\/sub><\/strong>\u00a0y\u00a0<strong>u\u00a0<sub>2<\/sub><\/strong>son las\u00a0<strong>velocidades absolutas del impulsor<\/strong>\u00a0(u\u00a0<sub>1<\/sub>\u00a0= r\u00a0<sub>1<\/sub>\u00a0. \u03c9) a la entrada y salida respectivamente.<\/li>\n<li><strong>V\u00a0<sub>t1<\/sub><\/strong>\u00a0y\u00a0<strong>V\u00a0<sub>t2<\/sub><\/strong>\u00a0son las\u00a0<strong>velocidades tangenciales del flujo<\/strong>\u00a0a la entrada y a la salida, respectivamente.<\/li>\n<\/ul>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Euler-turbomachine-equation.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-14957 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Euler-turbomachine-equation.png\" alt=\"Ecuaci\u00f3n de turbom\u00e1quina de Euler\" width=\"535\" height=\"415\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Euler-turbomachine-equation.png\" \/><\/a><\/p>\n<p><strong>Las ecuaciones de turbom\u00e1quina de Euler<\/strong>\u00a0se pueden usar para predecir el impacto de cambiar la geometr\u00eda del impulsor en la\u00a0<a title=\"Cabezal de bomba: curva de rendimiento de la bomba centr\u00edfuga\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/fluid-dynamics\/centrifugal-pumps\/pump-head-performance-curve\/\">cabeza<\/a>\u00a0.\u00a0No importa cuando trabajamos con una bomba o una turbina.\u00a0Si el par y la velocidad angular son\u00a0<strong>del mismo signo<\/strong>\u00a0, se est\u00e1 trabajando en el fluido (una bomba o compresor).\u00a0Si el par y la velocidad angular son\u00a0<strong>de signo opuesto,<\/strong>\u00a0se est\u00e1 extrayendo trabajo del fluido (una turbina).\u00a0Por lo tanto, para el aspecto de dise\u00f1o de turbinas y\u00a0<a title=\"Bombas centr\u00edfugas\" href=\"https:\/\/www.thermal-engineering.org\/es\/que-es-la-bomba-centrifuga-definicion\/\">bombas<\/a>\u00a0, las ecuaciones de Euler son extremadamente \u00fatiles.<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<h2><span>Ejemplo: c\u00e1lculo del rendimiento de la bomba<\/span><\/h2>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Pump-head-calculation.png\"><img loading=\"lazy\" class=\"alignright size-full wp-image-14958 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Pump-head-calculation.png\" alt=\"C\u00e1lculo del cabezal de la bomba\" width=\"264\" height=\"580\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Pump-head-calculation.png\" \/><\/a><span>En este ejemplo, veremos c\u00f3mo predecir<\/span><\/p>\n<ul>\n<li><strong><span>la descarga de dise\u00f1o<\/span><\/strong><\/li>\n<li><strong><span>potencia de agua<\/span><\/strong><\/li>\n<li><strong><span>la cabeza de la bomba<\/span><\/strong><\/li>\n<\/ul>\n<p><span>de una bomba centr\u00edfuga.\u00a0Estos datos de rendimiento se derivar\u00e1n de la\u00a0<\/span><strong><span>ecuaci\u00f3n de turbom\u00e1quina de Euler:<\/span><\/strong><\/p>\n<p><strong><span>Par del eje:\u00a0<\/span><sub><span>eje<\/span><\/sub><span>\u00a0T\u00a0\u00a0 \u00a0 = \u03c1Q (r\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0V\u00a0<\/span><sub><span>t2<\/span><\/sub><span>\u00a0&#8211; r\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0V\u00a0<\/span><sub><span>t1<\/span><\/sub><span>\u00a0)<\/span><\/strong><\/p>\n<p><strong><span>Potencia de agua: P\u00a0<\/span><sub><span>w<\/span><\/sub><span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = \u03c9.\u00a0<\/span><sub><span>Eje \u00a0 \u00a0 \u00a0\u00a0<\/span><\/sub><span>\u00a0T\u00a0= \u03c1Q (u\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0V\u00a0<\/span><sub><span>t2<\/span><\/sub><span>\u00a0&#8211; u\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0V\u00a0<\/span><sub><span>t1<\/span><\/sub><span>\u00a0)<\/span><\/strong><\/p>\n<p><strong><span>Altura de la\u00a0bomba: H = P\u00a0<\/span><sub><span>w<\/span><\/sub><span>\u00a0\/ \u03c1gQ = (u\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0V\u00a0<\/span><sub><span>t2<\/span><\/sub><span>\u00a0&#8211; u\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0V\u00a0<\/span><sub><span>t1<\/span><\/sub><span>\u00a0) \/ g<\/span><\/strong><\/p>\n<p><span>Se proporcionan los siguientes datos para una bomba de agua centr\u00edfuga:<\/span><\/p>\n<ul>\n<li><strong><span>di\u00e1metros del impulsor<\/span><\/strong><span>\u00a0en la entrada y salida<\/span>\n<ul>\n<li><strong><span>r\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0= 10 cm<\/span><\/strong><\/li>\n<li><strong><span>r\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0= 20 cm<\/span><\/strong><\/li>\n<\/ul>\n<\/li>\n<li><strong><span>Velocidad = 1500 rpm<\/span><\/strong><span>\u00a0(revoluciones por minuto)<\/span><\/li>\n<li><span>el \u00e1ngulo de la cuchilla en la entrada\u00a0<\/span><strong><span>\u03b2\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0= 30 \u00b0<\/span><\/strong><\/li>\n<li><span>el \u00e1ngulo de la cuchilla en la salida\u00a0<\/span><strong><span>\u03b2\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0= 20 \u00b0<\/span><\/strong><\/li>\n<li><span>suponga que los anchos de la cuchilla en la entrada y en la salida son:\u00a0<\/span><strong><span>b\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0= b\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0= 4 cm<\/span><\/strong><span>\u00a0.<\/span><\/li>\n<\/ul>\n<p><span>Soluci\u00f3n:<\/span><\/p>\n<p><span>Primero, tenemos que calcular la\u00a0<\/span><strong><span>velocidad radial del flujo<\/span><\/strong><span>\u00a0en la salida.\u00a0Desde el diagrama de velocidad, la velocidad radial es igual a (suponemos que el flujo entra exactamente normal al impulsor, por lo que el componente tangencial de la velocidad es cero):<\/span><\/p>\n<p><strong><span>V\u00a0<\/span><sub><span>r1<\/span><\/sub><\/strong><span>\u00a0= u\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0tan 30 \u00b0 = \u03c9 r\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0tan 30 \u00b0 = 2\u03c0 x (1500\/60) x 0.1 x tan 30 \u00b0 =\u00a0<\/span><strong><span>9.1 m \/ s<\/span><\/strong><\/p>\n<p><span>El componente radial de la velocidad del flujo determina cu\u00e1nto\u00a0<\/span><strong><span>ingresa el caudal en el impulsor<\/span><\/strong><span>\u00a0.\u00a0Entonces, cuando conocemos\u00a0<\/span><strong><span>V\u00a0<\/span><sub><span>r1<\/span><\/sub><\/strong><span>\u00a0en la entrada, podemos determinar\u00a0<\/span><strong><span>la descarga<\/span><\/strong><span>\u00a0de esta bomba de acuerdo con la siguiente ecuaci\u00f3n.\u00a0Aqu\u00ed b\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0significa el ancho de la pala del impulsor en la entrada.<\/span><\/p>\n<p><strong><span>Q<\/span><\/strong><span>\u00a0=\u00a0<\/span><strong><span>2\u03c0.r\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0.b\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0.V\u00a0<\/span><sub><span>r1<\/span><\/sub>\u00a0<\/strong><span>= 2n x 0,1 x 0,04 x 9,1 =\u00a0<\/span><strong><span>0,229 m\u00a0<\/span><sup><span>3<\/span><\/sup><span>\u00a0\/ s<\/span><\/strong><\/p>\n<p><span>Para calcular la\u00a0<\/span><strong><span>potencia de agua (P\u00a0<\/span><sub><span>w<\/span><\/sub><span>\u00a0)<\/span><\/strong><span>\u00a0requerida, tenemos que determinar la\u00a0<\/span><strong><span>velocidad de flujo tangencial de salida V\u00a0<\/span><sub><span>t2<\/span><\/sub><\/strong><span>\u00a0, porque se supone que la velocidad tangencial de entrada V\u00a0<\/span><sub><span>t1<\/span><\/sub><span>\u00a0es igual a cero.<\/span><\/p>\n<p><span>La velocidad de flujo radial de salida se deriva de la\u00a0<\/span><strong><span>conservaci\u00f3n de Q<\/span><\/strong><span>\u00a0:<\/span><\/p>\n<p><strong><span>Q = 2\u03c0.r\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0.b\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0.V\u00a0<\/span><sub><span>r2<\/span><\/sub><\/strong><span>\u00a0\u00a0\u21d2\u00a0<\/span><strong><span>V\u00a0<\/span><\/strong><strong><sub><span>r2<\/span><\/sub><\/strong><span>\u00a0= Q \/ 2\u03c0.r\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0.b\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0= 0.229 \/ (2\u03c0 x 0,2 x 0,04) =\u00a0<\/span><strong><span>\u00a04,56 m \/ s<\/span><\/strong><\/p>\n<p><span>A partir de la figura (\u00a0<\/span><strong><span>tri\u00e1ngulo de velocidad<\/span><\/strong><span>\u00a0), el \u00e1ngulo de salida de la cuchilla, \u03b2\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0, se puede representar f\u00e1cilmente de la siguiente manera.<\/span><\/p>\n<p><strong><span>cot \u03b2\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0= (u\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0&#8211; V\u00a0<\/span><sub><span>t2<\/span><\/sub><span>\u00a0) \/ V\u00a0<\/span><sub><span>r2<\/span><\/sub><\/strong><\/p>\n<p><span>y, por lo tanto, la velocidad de flujo tangencial de salida V\u00a0<\/span><sub><span>t2<\/span><\/sub><span>\u00a0es:<\/span><\/p>\n<p><strong><span>V\u00a0<\/span><sub><span>t2<\/span><\/sub><\/strong><span>\u00a0=\u00a0<\/span><strong><span>u\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0&#8211; V\u00a0<\/span><sub><span>r2<\/span><\/sub><span>\u00a0.\u00a0cuna 20 \u00b0<\/span><\/strong><span>\u00a0= \u03c9 r\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0&#8211; V\u00a0<\/span><sub><span>r2<\/span><\/sub><span>\u00a0.\u00a0cuna 20 \u00b0 = 2\u03c0 x 1500\/60 x 0.2 &#8211; 4.56 x 2.75 = 31.4 &#8211; 12.5 =\u00a0<\/span><strong><span>18.9 m \/ s.<\/span><\/strong><\/p>\n<p><span>La potencia de agua requerida es entonces:<\/span><\/p>\n<p><strong><span>P\u00a0<\/span><sub><span>w<\/span><\/sub><span>\u00a0\u00a0= \u03c1 Q u\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0V\u00a0<\/span><sub><span>t2<\/span><\/sub>\u00a0<\/strong><span>= 1000 [kg \/ m\u00a0<\/span><sup><span>3<\/span><\/sup><span>\u00a0] x 0.229 [m\u00a0<\/span><sup><span>3<\/span><\/sup><span>\u00a0\/ s] x 31.4 [m \/ s] x 18.9 [m \/ s] = 135900 W =\u00a0<\/span><strong><span>135.6 kW<\/span><\/strong><\/p>\n<p><span>y la cabeza de la bomba es:<\/span><\/p>\n<p><strong><span>H \u2248 P\u00a0<\/span><sub><span>w<\/span><\/sub><span>\u00a0\/ (\u03c1 g Q)<\/span><\/strong><span>\u00a0= 135900 \/ (1000 x 9.81 x 0.229) =\u00a0<\/span><strong><span>60.5 m<\/span><\/strong><\/p>\n<div><\/div>\n<div>\n<p>&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;&#8230;.<\/p>\n<p>Este art\u00edculo se basa en la traducci\u00f3n autom\u00e1tica del art\u00edculo original en ingl\u00e9s. Para m\u00e1s informaci\u00f3n vea el art\u00edculo en ingl\u00e9s. Puedes ayudarnos. Si desea corregir la traducci\u00f3n, env\u00edela a: translations@nuclear-power.com o complete el formulario de traducci\u00f3n en l\u00ednea. Agradecemos su ayuda, actualizaremos la traducci\u00f3n lo antes posible. Gracias.<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>La ecuaci\u00f3n de turbom\u00e1quina de Euler, o a veces llamada ecuaci\u00f3n de bomba de Euler, desempe\u00f1a un papel central en la turbom\u00e1quina, tambi\u00e9n en la teor\u00eda de las bombas centr\u00edfugas. Ingenier\u00eda t\u00e9rmica Teor\u00eda de la bomba &#8211; Ecuaciones de turbina de Euler La ecuaci\u00f3n de turbom\u00e1quina de Euler\u00a0, o a veces llamada\u00a0ecuaci\u00f3n de bomba de &#8230; <a title=\"\u00bfQu\u00e9 es la teor\u00eda de la bomba? &#8211; Ecuaci\u00f3n de Turbomachine de Euler &#8211; Definici\u00f3n\" class=\"read-more\" href=\"https:\/\/www.thermal-engineering.org\/es\/que-es-la-teoria-de-la-bomba-ecuacion-de-turbomachine-de-euler-definicion\/\" aria-label=\"M\u00e1s en \u00bfQu\u00e9 es la teor\u00eda de la bomba? &#8211; Ecuaci\u00f3n de Turbomachine de Euler &#8211; Definici\u00f3n\">Leer m\u00e1s<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":[],"categories":[16],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v15.4 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>\u00bfQu\u00e9 es la teor\u00eda de la bomba? - Ecuaci\u00f3n de Turbomachine de Euler - Definici\u00f3n<\/title>\n<meta name=\"description\" content=\"La ecuaci\u00f3n de la turbom\u00e1quina de Euler, o a veces llamada ecuaci\u00f3n de la bomba de Euler, juega un papel central en la turbom\u00e1quina, tambi\u00e9n en la teor\u00eda de las bombas centr\u00edfugas. 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