{"id":40623,"date":"2019-09-24T21:28:13","date_gmt":"2019-09-24T20:28:13","guid":{"rendered":"https:\/\/www.thermal-engineering.org\/que-es-el-ejemplo-del-ciclo-de-brayton-problema-con-la-solucion-definicion\/"},"modified":"2020-01-10T07:50:44","modified_gmt":"2020-01-10T06:50:44","slug":"que-es-el-ejemplo-del-ciclo-de-brayton-problema-con-la-solucion-definicion","status":"publish","type":"post","link":"https:\/\/www.thermal-engineering.org\/es\/que-es-el-ejemplo-del-ciclo-de-brayton-problema-con-la-solucion-definicion\/","title":{"rendered":"El ejemplo del ciclo de Brayton &#8211; Problema con la soluci\u00f3n. Definici\u00f3n"},"content":{"rendered":"<div class=\"su-quote su-quote-style-default\">\n<div class=\"su-quote-inner su-clearfix\">Ejemplo de ciclo de Brayton: problema con la soluci\u00f3n.\u00a0Calcule los par\u00e1metros termodin\u00e1micos clave del ciclo de Brayton cerrado, como temperaturas, presiones y transferencias de calor.\u00a0Ingenieria termal<\/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>Ciclo de Brayton: problema con la soluci\u00f3n<\/h2>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Brayton-cycle-example-calculation.png\"><img loading=\"lazy\" class=\"alignright size-medium wp-image-17724 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Brayton-cycle-example-calculation-300x196.png\" alt=\"Ciclo de Brayton - ejemplo - c\u00e1lculo\" width=\"300\" height=\"196\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Brayton-cycle-example-calculation-300x196.png\" \/><\/a>Supongamos el\u00a0<strong>ciclo<\/strong>\u00a0cerrado de\u00a0<strong>Brayton<\/strong>\u00a0, que es uno de los\u00a0<strong>ciclos termodin\u00e1micos<\/strong>\u00a0m\u00e1s comunes\u00a0que se pueden encontrar en los motores modernos de turbina de gas.\u00a0En este caso, suponga una\u00a0<strong>turbina de gas helio<\/strong>\u00a0con compresor \u00fanico y disposici\u00f3n de turbina \u00fanica.\u00a0Uno de los par\u00e1metros clave de tales motores es la temperatura m\u00e1xima de entrada de la turbina y la relaci\u00f3n de presi\u00f3n del compresor (PR = p\u00a0<sub>2<\/sub>\u00a0\/ p\u00a0<sub>1<\/sub>\u00a0) que determina la eficiencia t\u00e9rmica de dicho motor.<\/p>\n<p>En esta turbina, la etapa de alta presi\u00f3n recibe gas (punto 3 en la figura) de un intercambiador de calor:<\/p>\n<ul>\n<li>p\u00a0<sub>3<\/sub>\u00a0= 6,7 MPa;<\/li>\n<li>T\u00a0<sub>3<\/sub>\u00a0= 1190 K (917 \u00b0 C))<\/li>\n<li>la eficiencia de la turbina isentr\u00f3pica es\u00a0<em>\u03b7\u00a0<\/em><em><sub>T<\/sub><\/em>\u00a0= 0.91 (91%)<\/li>\n<\/ul>\n<p>y expulsarlo a otro intercambiador de calor, donde la presi\u00f3n de salida es (punto 4):<\/p>\n<ul>\n<li>p\u00a0<sub>4<\/sub>\u00a0= 2.78 MPa<\/li>\n<li>T\u00a0<sub>4, es<\/sub>\u00a0=?<\/li>\n<\/ul>\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<p><span>Por lo tanto, la relaci\u00f3n de presi\u00f3n del compresor es igual a PR = 2.41.\u00a0Adem\u00e1s, sabemos que el compresor recibe gas (punto 1) en la figura:<\/span><\/p>\n<ul>\n<li><span>p\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0= 2,78 MPa;<\/span><\/li>\n<li><span>T\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0= 299 K (26 \u00b0 C)<\/span><\/li>\n<li><span>La eficiencia del compresor isentr\u00f3pico \u03b7\u00a0<\/span><sub><span>K<\/span><\/sub><span>\u00a0= 0.87 (87%).<\/span><\/li>\n<\/ul>\n<p><span>La relaci\u00f3n de capacidad calor\u00edfica, para helio es igual a\u00a0<\/span><strong><span>= c\u00a0<\/span><sub><span>p<\/span><\/sub><span>\u00a0\/ c\u00a0<\/span><sub><span>v<\/span><\/sub><span>\u00a0= 1.66<\/span><\/strong><\/p>\n<p><strong><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Brayton-cycle-problem-with-solution.png\"><img loading=\"lazy\" class=\"alignright size-medium wp-image-17725 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Brayton-cycle-problem-with-solution-300x271.png\" alt=\"Ciclo de Brayton: problema con la soluci\u00f3n\" width=\"300\" height=\"271\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/Brayton-cycle-problem-with-solution-300x271.png\" \/><\/a><span>Calcular:<\/span><\/strong><\/p>\n<ol>\n<li><span>El calor agregado por el intercambiador de calor (entre 2 \u2192 3)<\/span><\/li>\n<li><span>la temperatura de salida del compresor del gas (T\u00a0<\/span><sub><span>2, es<\/span><\/sub><span>\u00a0)<\/span><\/li>\n<li><span>El verdadero trabajo realizado en este compresor, cuando la eficiencia del compresor isentr\u00f3pico es \u03b7\u00a0<\/span><sub><span>K<\/span><\/sub><span>\u00a0= 0.87 (87%)<\/span><\/li>\n<li><span>la temperatura de salida de la turbina del gas (T\u00a0<\/span><sub><span>4, es<\/span><\/sub><span>\u00a0)<\/span><\/li>\n<li><span>El trabajo real realizado por esta turbina, cuando la eficiencia de la turbina isentr\u00f3pica es \u03b7\u00a0<\/span><sub><span>T<\/span><\/sub><span>\u00a0= 0.91 (91%)<\/span><\/li>\n<li><span>la eficiencia t\u00e9rmica de este ciclo<\/span><\/li>\n<\/ol>\n<p><strong><span>Soluci\u00f3n:<\/span><\/strong><\/p>\n<p><span>1) + 2)<\/span><\/p>\n<p><span>A partir de la\u00a0<\/span><a title=\"Primera ley de la termodin\u00e1mica\" href=\"https:\/\/www.thermal-engineering.org\/es\/que-es-la-primera-ley-de-la-termodinamica-definicion\/\"><span>primera ley de la termodin\u00e1mica<\/span><\/a><span>\u00a0, el calor neto agregado est\u00e1 dado por\u00a0<\/span><strong><span>Q\u00a0<\/span><sub><span>add, ex<\/span><\/sub><span>\u00a0= H\u00a0<\/span><sub><span>3<\/span><\/sub><span>\u00a0&#8211; H\u00a0<\/span><sub><span>2<\/span><\/sub><\/strong><span>\u00a0\u00a0[kJ] o Q\u00a0<\/span><sub><span>add<\/span><\/sub><span>\u00a0= C\u00a0<\/span><sub><span>p<\/span><\/sub><span>\u00a0. (T\u00a0<\/span><sub><span>3<\/span><\/sub><span>\u00a0-T\u00a0<\/span><sub><span>2s<\/span><\/sub><span>\u00a0)\u00a0<\/span><strong><span>,<\/span><\/strong><span>\u00a0pero en este caso no sabemos La temperatura (T\u00a0<\/span><sub><span>2s<\/span><\/sub><span>\u00a0) en la salida del compresor.\u00a0Resolveremos este problema en\u00a0<\/span><a title=\"Propiedades extensivas e intensivas\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/thermodynamics\/extensive-and-intensive-properties\/\"><span>variables intensivas<\/span><\/a><span>\u00a0.\u00a0Tenemos que reescribir la ecuaci\u00f3n anterior (para incluir\u00a0<\/span><strong><span>\u03b7\u00a0<\/span><\/strong><strong><sub><span>K<\/span><\/sub><\/strong><span>\u00a0) usando el t\u00e9rmino (\u00a0<\/span><strong><span>+ h\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0&#8211; h\u00a0<\/span><sub><span>1<\/span><\/sub><\/strong><span>\u00a0) para:<\/span><\/p>\n<p><span>Q\u00a0<\/span><sub><span>add<\/span><\/sub><span>\u00a0= h\u00a0<\/span><sub><span>3<\/span><\/sub><span>\u00a0&#8211; h\u00a0<\/span><sub><span>2<\/span><\/sub><span>\u00a0= h\u00a0<\/span><sub><span>3<\/span><\/sub><span>\u00a0&#8211; h\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0&#8211; (h\u00a0<\/span><sub><span>2s<\/span><\/sub><span>\u00a0&#8211; h\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0) \/ \u03b7\u00a0<\/span><sub><span>K<\/span><\/sub>\u00a0<strong>\u00a0<\/strong><span>\u00a0[kJ \/ kg]<\/span><\/p>\n<p><span>Q\u00a0<\/span><sub><span>add\u00a0<\/span><\/sub><strong><span>=<\/span><\/strong><span>\u00a0c\u00a0<\/span><sub><span>p<\/span><\/sub><span>\u00a0(T\u00a0<\/span><sub><span>3<\/span><\/sub><span>\u00a0-T\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0) &#8211; (c\u00a0<\/span><sub><span>p<\/span><\/sub><span>\u00a0(T\u00a0<\/span><sub><span>2s<\/span><\/sub><span>\u00a0-T\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0)\u00a0<\/span><strong><span>\/<\/span><\/strong><span>\u00a0\u03b7\u00a0<\/span><sub><span>K<\/span><\/sub><span>\u00a0)<\/span><\/p>\n<p><span>Luego calcularemos la temperatura,\u00a0<\/span><strong><span>T\u00a0<\/span><sub><span>2s<\/span><\/sub><\/strong><span>\u00a0, utilizando la relaci\u00f3n p, V, T (de la\u00a0<\/span><a title=\"Ley del gas ideal\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/thermodynamics\/ideal-gas-law\/\"><span>Ley de gases ideales<\/span><\/a><span>\u00a0) para el\u00a0<\/span><a title=\"Proceso adiab\u00e1tico\" href=\"https:\/\/www.thermal-engineering.org\/es\/que-es-el-proceso-adiabatico-definicion\/\"><span>proceso adiab\u00e1tico<\/span><\/a><span>\u00a0entre (1 \u2192 2).<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/pVT-relation-isentropic-process.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-17281 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/pVT-relation-isentropic-process.png\" alt=\"Relaci\u00f3n p, V, T - proceso isentr\u00f3pico\" width=\"209\" height=\"72\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/pVT-relation-isentropic-process.png\" \/><\/a><\/p>\n<p><span>En esta ecuaci\u00f3n, el factor para helio es igual a = c\u00a0<\/span><sub><span>p<\/span><\/sub><span>\u00a0\/ c\u00a0<\/span><sub><span>v<\/span><\/sub><span>\u00a0= 1.66.\u00a0De la ecuaci\u00f3n anterior se deduce que la temperatura de salida del compresor, T\u00a0<\/span><sub><span>2s<\/span><\/sub><span>\u00a0, es:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/isobaric-process-example.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-17429 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/isobaric-process-example.png\" alt=\"proceso isob\u00e1rico - ejemplo\" width=\"492\" height=\"84\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/isobaric-process-example.png\" \/><\/a><\/p>\n<\/div>\n<\/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-accordion\">\n<div class=\"su-spoiler su-spoiler-style-default su-spoiler-icon-plus su-spoiler-closed\">\n<div class=\"su-spoiler-content su-clearfix\">Usando esta temperatura y la <a title=\"Eficiencia isentr\u00f3pica - Turbina \/ Compresor \/ Boquilla\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/thermodynamics\/thermodynamic-processes\/isentropic-process\/isentropic-efficiency-turbinecompressornozzle\/\">eficiencia del compresor isentr\u00f3pico<\/a>\u00a0, podemos calcular el calor agregado por el intercambiador de calor:<\/div>\n<\/div>\n<\/div>\n<p><strong><span>Q\u00a0<\/span><sub><span>add<\/span><\/sub><span>\u00a0=<\/span><\/strong><span>\u00a0c\u00a0<\/span><sub><span>p<\/span><\/sub><span>\u00a0(T\u00a0<\/span><sub><span>3<\/span><\/sub><span>\u00a0-T\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0) &#8211; (c\u00a0<\/span><sub><span>p<\/span><\/sub><span>\u00a0(T\u00a0<\/span><sub><span>2s<\/span><\/sub><span>\u00a0-T\u00a0<\/span><sub><span>1<\/span><\/sub><span>\u00a0)\u00a0<\/span><strong><span>\/<\/span><\/strong><span>\u00a0\u03b7\u00a0<\/span><sub><span>K<\/span><\/sub><span>\u00a0) = 5200. (1190 &#8211; 299) &#8211; 5200. (424-299) \/0.87 = 4.633 MJ \/ kg &#8211; 0.747 MJ \/ kg =\u00a0<\/span><strong><span>3.886 MJ \/ kg<\/span><\/strong><\/p>\n<p><span>3)<\/span><\/p>\n<p><span>El trabajo realizado en el gas por el compresor en el proceso de compresi\u00f3n isentr\u00f3pica es:<\/span><\/p>\n<p><strong><em><span>W\u00a0<\/span><\/em><\/strong><strong><em><sub><span>C, s<\/span><\/sub><\/em><\/strong><strong><em><span>\u00a0= c\u00a0<\/span><\/em><\/strong><strong><em><sub><span>p<\/span><\/sub><\/em><\/strong><strong><em><span>\u00a0(T\u00a0<\/span><\/em><\/strong><strong><em><sub><span>2s<\/span><\/sub><\/em><\/strong><strong><em><span>\u00a0&#8211; T\u00a0<\/span><\/em><\/strong><strong><em><sub><span>1<\/span><\/sub><\/em><\/strong><strong><em><span>\u00a0) = 5200 x (424 &#8211; 299) = 0.650 MJ \/ kg<\/span><\/em><\/strong><\/p>\n<p><span>El verdadero trabajo realizado en el gas por el compresor en la compresi\u00f3n adiab\u00e1tica es entonces:<\/span><\/p>\n<p><strong><em><span>W\u00a0<\/span><\/em><\/strong><strong><em><sub><span>C, real<\/span><\/sub><\/em><\/strong><strong><em><span>\u00a0= c\u00a0<\/span><\/em><\/strong><strong><em><sub><span>p<\/span><\/sub><\/em><\/strong><strong><em><span>\u00a0(T\u00a0<\/span><\/em><\/strong><strong><em><sub><span>2s<\/span><\/sub><\/em><\/strong><strong><em><span>\u00a0&#8211; T\u00a0<\/span><\/em><\/strong><strong><em><sub><span>1<\/span><\/sub><\/em><\/strong><strong><em><span>\u00a0).\u00a0\u03b7\u00a0<\/span><\/em><\/strong><strong><em><sub><span>C<\/span><\/sub><\/em><\/strong>\u00a0<strong><em><span>= 5200 x (424 &#8211; 299) \/ 0.87 = 0.747 MJ \/ kg<\/span><\/em><\/strong><\/p>\n<p><span>4)<\/span><\/p>\n<p><span>La temperatura de salida de la turbina del gas,\u00a0<\/span><strong><span>T\u00a0<\/span><sub><span>4, se<\/span><\/sub><\/strong><span>\u00a0puede calcular utilizando la misma relaci\u00f3n\u00a0\u00a0<\/span><strong><span>p, V, T<\/span><\/strong><span>\u00a0\u00a0que en 2) pero entre los estados 3 y 4:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/pVT-relation-isentropic-process.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-17281 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/pVT-relation-isentropic-process.png\" alt=\"Relaci\u00f3n p, V, T - proceso isentr\u00f3pico\" width=\"209\" height=\"72\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/pVT-relation-isentropic-process.png\" \/><\/a><\/p>\n<p><span>De la ecuaci\u00f3n anterior se deduce que la temperatura de salida del gas, T\u00a0<\/span><sub><span>4<\/span><\/sub><span>\u00a0, es:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/isentropic-process-example.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-17283 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/isentropic-process-example.png\" alt=\"proceso isentr\u00f3pico - ejemplo\" width=\"503\" height=\"84\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/isentropic-process-example.png\" \/><\/a><\/p>\n<p><span>5)<\/span><\/p>\n<p><span>El trabajo realizado por la turbina de gas en la expansi\u00f3n isentr\u00f3pica es entonces:<\/span><\/p>\n<p><strong><em><span>W\u00a0<\/span><\/em><\/strong><strong><em><sub><span>T, s<\/span><\/sub><\/em><\/strong><em><span>\u00a0= c\u00a0<\/span><\/em><em><sub><span>p<\/span><\/sub><\/em><em><span>\u00a0(T\u00a0<\/span><\/em><em><sub><span>3<\/span><\/sub><\/em><em><span>\u00a0&#8211; T\u00a0<\/span><\/em><em><sub><span>4s<\/span><\/sub><\/em><em><span>\u00a0) = 5200 x (1190 &#8211; 839) =\u00a0<\/span><\/em><strong><em><span>1.825 MJ \/ kg<\/span><\/em><\/strong><\/p>\n<p><span>El verdadero trabajo realizado por la turbina de gas en la expansi\u00f3n adiab\u00e1tica es entonces:<\/span><\/p>\n<p><strong><em><span>W\u00a0<\/span><\/em><\/strong><strong><em><sub><span>T, real<\/span><\/sub><\/em><\/strong><em><span>\u00a0= c\u00a0<\/span><\/em><em><sub><span>p<\/span><\/sub><\/em><em><span>\u00a0(T\u00a0<\/span><\/em><em><sub><span>3<\/span><\/sub><\/em><em><span>\u00a0&#8211; T\u00a0<\/span><\/em><em><sub><span>4s<\/span><\/sub><\/em><em><span>\u00a0).\u00a0\u03b7\u00a0<\/span><\/em><em><sub><span>T<\/span><\/sub><\/em>\u00a0<em><span>= 5200 x (1190 &#8211; 839) x 0.91 =\u00a0<\/span><\/em><strong><em><span>1.661 MJ \/ kg<\/span><\/em><\/strong><\/p>\n<p><span>6)<\/span><\/p>\n<p><span>Como se dedujo en la secci\u00f3n anterior, la\u00a0<\/span><strong><span>eficiencia t\u00e9rmica<\/span><\/strong><span>\u00a0de un ciclo Brayton ideal es una funci\u00f3n de la\u00a0<\/span><strong><span>relaci\u00f3n<\/span><\/strong><span>\u00a0de\u00a0<strong>presi\u00f3n<\/strong>\u00a0y\u00a0<\/span><strong><span>\u03ba<\/span><\/strong><span>\u00a0:<\/span><\/p>\n<p><a href=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/thermal-efficiency-brayton-cycle-pressure-ratio-equation.png\"><img loading=\"lazy\" class=\"aligncenter size-full wp-image-17696 lazy-loaded\" src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/thermal-efficiency-brayton-cycle-pressure-ratio-equation.png\" alt=\"eficiencia t\u00e9rmica - ciclo de brayton - relaci\u00f3n de presi\u00f3n - ecuaci\u00f3n\" width=\"187\" height=\"66\" data-lazy-type=\"image\" data-src=\"https:\/\/thermal-engineering.org\/wp-content\/uploads\/2019\/05\/thermal-efficiency-brayton-cycle-pressure-ratio-equation.png\" \/><\/a><\/p>\n<p><span>por lo tanto<\/span><\/p>\n<p><strong><em><span>\u03b7\u00a0<\/span><\/em><em><sub><span>th<\/span><\/sub><\/em><\/strong><em><span>\u00a0=<\/span><\/em><span>\u00a00.295 =<\/span><strong><span>\u00a029.5%<\/span><\/strong><\/p>\n<p><span>La eficiencia t\u00e9rmica tambi\u00e9n se puede calcular utilizando el trabajo y el calor (sin \u03b7\u00a0<\/span><sub><span>K<\/span><\/sub><span>\u00a0):<\/span><\/p>\n<p><strong><em><span>\u03b7\u00a0<\/span><\/em><\/strong><em><sub><strong><span>th, s<\/span><\/strong>\u00a0<\/sub><\/em><em><span>= (\u00a0<\/span><\/em><em><span>W\u00a0<\/span><\/em><em><sub><span>T, s<\/span><\/sub><\/em><em><span>\u00a0&#8211; W\u00a0<\/span><\/em><em><sub><span>C, s<\/span><\/sub><\/em><em><span>\u00a0)\u00a0<\/span><\/em><strong><em><span>\/<\/span><\/em><\/strong><span>\u00a0Q<\/span><sub><span>\u00a0add, s<\/span><\/sub><span>\u00a0= (1.825 &#8211; 0.650) \/ 3.983<\/span><strong><em><span>\u00a0\u00a0=<\/span><\/em><\/strong><span>\u00a00.295\u00a0=<\/span><strong><span>\u00a029.5%<\/span><\/strong><\/p>\n<p><span>Finalmente, la eficiencia t\u00e9rmica que incluye la eficiencia de la turbina isentr\u00f3pica \/ compresor es:<\/span><\/p>\n<p><strong><em><span>\u03b7\u00a0<\/span><\/em><em><sub><span>th, real\u00a0<\/span><\/sub><\/em><\/strong><em><span>\u00a0= (\u00a0<\/span><\/em><em><span>W\u00a0<\/span><\/em><em><sub><span>T, real<\/span><\/sub><\/em><em><span>\u00a0\u00a0&#8211; W\u00a0<\/span><\/em><em><sub><span>C, real<\/span><\/sub><\/em><em><span>\u00a0)\u00a0<\/span><\/em><strong><em><span>\/<\/span><\/em><\/strong><span>\u00a0Q<\/span><sub><span>\u00a0add<\/span><\/sub><span>\u00a0\u00a0= (1.661 &#8211; 0.747) \/ 3.886<\/span><strong><em><span>\u00a0\u00a0=<\/span><\/em><\/strong><span>\u00a00.235\u00a0\u00a0=<\/span><strong><span>\u00a023.5%<\/span><\/strong><\/p>\n<\/div>\n<\/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>Ejemplo de ciclo de Brayton: problema con la soluci\u00f3n.\u00a0Calcule los par\u00e1metros termodin\u00e1micos clave del ciclo de Brayton cerrado, como temperaturas, presiones y transferencias de calor.\u00a0Ingenieria termal Ciclo de Brayton: problema con la soluci\u00f3n Supongamos el\u00a0ciclo\u00a0cerrado de\u00a0Brayton\u00a0, que es uno de los\u00a0ciclos termodin\u00e1micos\u00a0m\u00e1s comunes\u00a0que se pueden encontrar en los motores modernos de turbina de gas.\u00a0En este &#8230; <a title=\"El ejemplo del ciclo de Brayton &#8211; Problema con la soluci\u00f3n. Definici\u00f3n\" class=\"read-more\" href=\"https:\/\/www.thermal-engineering.org\/es\/que-es-el-ejemplo-del-ciclo-de-brayton-problema-con-la-solucion-definicion\/\" aria-label=\"M\u00e1s en El ejemplo del ciclo de Brayton &#8211; Problema con la soluci\u00f3n. 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>El ejemplo del ciclo de Brayton - Problema con la soluci\u00f3n. Definici\u00f3n<\/title>\n<meta name=\"description\" content=\"Ejemplo de ciclo de Brayton: problema con la soluci\u00f3n. Calcule los par\u00e1metros termodin\u00e1micos clave del ciclo de Brayton cerrado, como temperaturas, presiones y transferencias de calor. 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