{"id":43660,"date":"2019-10-13T01:46:22","date_gmt":"2019-10-13T00:46:22","guid":{"rendered":"https:\/\/www.thermal-engineering.org\/was-ist-ein-beispiel-fur-brayton-cycle-problem-mit-der-losung-definition\/"},"modified":"2020-03-09T08:32:07","modified_gmt":"2020-03-09T07:32:07","slug":"was-ist-ein-beispiel-fur-brayton-cycle-problem-mit-der-losung-definition","status":"publish","type":"post","link":"https:\/\/www.thermal-engineering.org\/de\/was-ist-ein-beispiel-fur-brayton-cycle-problem-mit-der-losung-definition\/","title":{"rendered":"Das Beispiel f\u00fcr Brayton Cycle &#8211; Problem mit der L\u00f6sung &#8211; Definition"},"content":{"rendered":"<div class=\"su-quote su-quote-style-default\">\n<div class=\"su-quote-inner su-clearfix\">Beispiel f\u00fcr einen Brayton-Zyklus &#8211; Problem mit der L\u00f6sung.\u00a0Berechnen Sie f\u00fcr den geschlossenen Brayton-Zyklus die wichtigsten thermodynamischen Parameter wie Temperaturen, Dr\u00fccke und W\u00e4rme\u00fcbertragungen.\u00a0W\u00e4rmetechnik<\/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>Brayton-Zyklus &#8211; Problem mit L\u00f6sung<\/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=\"Brayton-Zyklus - Beispiel - Berechnung\" 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>Angenommen, der geschlossene\u00a0<strong>Brayton-Zyklus<\/strong>\u00a0ist einer der h\u00e4ufigsten\u00a0<strong>thermodynamischen Kreisprozessen<\/strong>\u00a0, die in modernen Gasturbinentriebwerken zu finden sind.\u00a0In diesem Fall sei eine\u00a0<strong>Heliumgasturbine<\/strong>\u00a0mit Einzelverdichter und Einzelturbinenanordnung angenommen.\u00a0Einer der Schl\u00fcsselparameter solcher Motoren ist die maximale Turbineneinlasstemperatur und das Kompressordruckverh\u00e4ltnis (PR = p\u00a0<sub>2<\/sub>\u00a0\/ p\u00a0<sub>1<\/sub>\u00a0), die den thermischen Wirkungsgrad eines solchen Motors bestimmen.<\/p>\n<p>In dieser Turbine erh\u00e4lt die Hochdruckstufe Gas (Punkt 3 in der Abbildung) von einem W\u00e4rmetauscher:<\/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>der Wirkungsgrad der isentropen Turbine betr\u00e4gt\u00a0<em>\u03b7\u00a0<\/em><em><sub>T<\/sub><\/em>\u00a0= 0,91 (91%)<\/li>\n<\/ul>\n<p>und lasse es zu einem anderen W\u00e4rmetauscher ab, wo der Ausgangsdruck ist (Punkt 4):<\/p>\n<ul>\n<li>p\u00a0<sub>4<\/sub>\u00a0= 2,78 MPa<\/li>\n<li>T\u00a0<sub>4 ist<\/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>Somit ist das Kompressordruckverh\u00e4ltnis gleich PR = 2,41.\u00a0Au\u00dferdem wissen wir, dass der Kompressor in der Abbildung Gas (Punkt 1) erh\u00e4lt:<\/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>der Wirkungsgrad des isentropischen Kompressors \u03b7\u00a0<\/span><sub><span>K<\/span><\/sub><span>\u00a0= 0,87 (87%).<\/span><\/li>\n<\/ul>\n<p><span>Das W\u00e4rmekapazit\u00e4tsverh\u00e4ltnis f\u00fcr Helium ist gleich\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=\"Brayton-Zyklus - Problem mit der L\u00f6sung\" 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>Berechnung:<\/span><\/strong><\/p>\n<ol>\n<li><span>die vom W\u00e4rmetauscher zugef\u00fchrte W\u00e4rme (zwischen 2 \u2192 3)<\/span><\/li>\n<li><span>die Kompressorauslasstemperatur des Gases (T\u00a0<\/span><sub><span>2, ist<\/span><\/sub><span>\u00a0)<\/span><\/li>\n<li><span>die eigentliche Arbeit an diesem Kompressor, wenn der Wirkungsgrad des isentropischen Kompressors \u03b7\u00a0<\/span><sub><span>K<\/span><\/sub><span>\u00a0= 0,87 (87%)\u00a0betr\u00e4gt<\/span><\/li>\n<li><span>die Turbinenaustrittstemperatur des Gases (T\u00a0<\/span><sub><span>4, ist<\/span><\/sub><span>\u00a0)<\/span><\/li>\n<li><span>die eigentliche Arbeit dieser Turbine, wenn der Wirkungsgrad der isentropischen Turbine \u03b7\u00a0<\/span><sub><span>T<\/span><\/sub><span>\u00a0= 0,91 (91%)\u00a0betr\u00e4gt<\/span><\/li>\n<li><span>der thermische Wirkungsgrad dieses Zyklus<\/span><\/li>\n<\/ol>\n<p><strong><span>L\u00f6sung:<\/span><\/strong><\/p>\n<p><span>1) + 2)<\/span><\/p>\n<p><span>Nach dem\u00a0<\/span><a title=\"Erster Hauptsatz der Thermodynamik\" href=\"https:\/\/www.thermal-engineering.org\/de\/was-ist-der-erste-hauptsatz-der-thermodynamik-definition\/\"><span>ersten Hauptsatz der Thermodynamik<\/span><\/a><span>\u00a0ist die hinzugef\u00fcgte Nettow\u00e4rme gegeben durch\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] oder 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>\u00a0aber in diesem Fall wissen wir es nicht die Temperatur (T\u00a0<\/span><sub><span>2s<\/span><\/sub><span>\u00a0) am Auslass des Kompressors.\u00a0Wir werden dieses Problem in\u00a0<\/span><a title=\"Umfangreiche und intensive Eigenschaften\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/thermodynamics\/extensive-and-intensive-properties\/\"><span>intensiven Variablen<\/span><\/a><span>\u00a0l\u00f6sen\u00a0.\u00a0Wir m\u00fcssen die vorherige Gleichung (um\u00a0<\/span><strong><span>\u03b7\u00a0<\/span><\/strong><strong><sub><span>K<\/span><\/sub><\/strong><span><strong>\u00a0einzuschlie\u00dfen<\/strong>\u00a0) unter Verwendung des Terms (\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)\u00a0umschreiben,\u00a0um:<\/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>addiere\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>Dann berechnen wir die Temperatur\u00a0<\/span><strong><span>T\u00a0<\/span><sub><span>2s<\/span><\/sub><\/strong><span>\u00a0unter Verwendung der p, V, T-Beziehung (aus dem\u00a0<\/span><a title=\"Ideales Gasgesetz\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/thermodynamics\/ideal-gas-law\/\"><span>idealen Gasgesetz<\/span><\/a><span>\u00a0) f\u00fcr den\u00a0<\/span><a title=\"Adiabatischer Prozess\" href=\"https:\/\/www.thermal-engineering.org\/de\/was-ist-ein-adiabatischer-prozess-definition\/\"><span>adiabatischen Prozess<\/span><\/a><span>\u00a0zwischen (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=\"p, V, T-Beziehung - isentropischer Prozess\" 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>In dieser Gleichung ist der Faktor f\u00fcr Helium gleich = c\u00a0<\/span><sub><span>p<\/span><\/sub><span>\u00a0\/ c\u00a0<\/span><sub><span>v<\/span><\/sub><span>\u00a0= 1,66.\u00a0Aus der vorherigen Gleichung folgt, dass die Kompressorauslasstemperatur T\u00a0<\/span><sub><span>2s<\/span><\/sub><span>\u00a0ist:<\/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=\"isobarer Prozess - Beispiel\" 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-title\" tabindex=\"0\" role=\"button\"><\/div>\n<\/div>\n<\/div>\n<div class=\"su-divider su-divider-style-dotted\"><\/div>\n<p><span>Mit dieser Temperatur und dem\u00a0<\/span><a title=\"Isentropische Effizienz - Turbine \/ Kompressor \/ D\u00fcse\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/thermodynamics\/thermodynamic-processes\/isentropic-process\/isentropic-efficiency-turbinecompressornozzle\/\"><span>Wirkungsgrad<\/span><\/a><span>\u00a0des\u00a0<a title=\"Isentropische Effizienz - Turbine \/ Kompressor \/ D\u00fcse\" href=\"https:\/\/www.nuclear-power.com\/nuclear-engineering\/thermodynamics\/thermodynamic-processes\/isentropic-process\/isentropic-efficiency-turbinecompressornozzle\/\">isentropischen Kompressors<\/a>\u00a0k\u00f6nnen wir die vom W\u00e4rmetauscher zugef\u00fchrte W\u00e4rme berechnen:<\/span><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>Die Arbeit des Kompressors am Gas im isentropischen Kompressionsprozess ist:<\/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 \u00d7 (424 &#8211; 299) = 0,650 MJ \/ kg<\/span><\/em><\/strong><\/p>\n<p><span>Die eigentliche Arbeit, die der Kompressor bei der adiabatischen Kompression am Gas leistet, ist dann:<\/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>Die Turbinenauslasstemperatur des Gases\u00a0<\/span><strong><span>T\u00a0<\/span><sub><span>4<\/span><\/sub><\/strong><span>\u00a0kann unter Verwendung der gleichen\u00a0<\/span><strong><span>p, V, T-Beziehung<\/span><\/strong><span>\u00a0\u00a0wie in 2)\u00a0berechnet werden\u00a0\u00a0<strong>,<\/strong>\u00a0jedoch zwischen den Zust\u00e4nden 3 und 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=\"p, V, T-Beziehung - isentropischer Prozess\" 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>Aus der vorhergehenden Gleichung folgt, dass die Austrittstemperatur des Gases T\u00a0<\/span><sub><span>4<\/span><\/sub><span>\u00a0ist:<\/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=\"isentropischer Prozess - Beispiel\" 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>Die Arbeit der Gasturbine bei der isentropischen Expansion ist dann:<\/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>4\u00a0<\/span><\/sub><\/em><span><strong><em><sub>s<\/sub><\/em><\/strong><\/span><em><span>\u00a0) = 5200 \u00d7 (1190\u2013839) =\u00a0<\/span><\/em><strong><em><span>1,825 MJ \/ kg<\/span><\/em><\/strong><\/p>\n<p><span>Die eigentliche Arbeit der Gasturbine bei der adiabatischen Expansion ist dann:<\/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>4\u00a0<\/span><\/sub><\/em><span><em><sub>s<\/sub><\/em><\/span><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>Wie im vorherigen Abschnitt abgeleitet, ist der\u00a0<\/span><strong><span>thermische Wirkungsgrad<\/span><\/strong><span>\u00a0eines idealen Brayton-Zyklus eine Funktion des\u00a0<\/span><strong><span>Druckverh\u00e4ltnisses<\/span><\/strong><span>\u00a0und von\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=\"thermischer Wirkungsgrad - Bradyton-Zyklus - Druckverh\u00e4ltnis - Gleichung\" 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>deshalb<\/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>Der thermische Wirkungsgrad kann auch anhand der Arbeit und der W\u00e4rme (ohne \u03b7\u00a0<\/span><sub><span>K<\/span><\/sub><span>\u00a0)\u00a0berechnet werden\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>\u00a0\u00a00,295 =<\/span><strong><span>\u00a029,5%<\/span><\/strong><\/p>\n<p><span>Schlie\u00dflich betr\u00e4gt der thermische Wirkungsgrad einschlie\u00dflich des Wirkungsgrads der isentropischen Turbine \/ des Kompressors:<\/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\u00a0<\/span><strong><em><span>\u00a0\u00a0=<\/span><\/em><\/strong><span>\u00a0\u00a00,235 =<\/span><strong><span>\u00a023,5%<\/span><\/strong><\/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\"><\/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>Dieser Artikel basiert auf der maschinellen \u00dcbersetzung des englischen Originalartikels. Weitere Informationen finden Sie im Artikel auf Englisch. Sie k\u00f6nnen uns helfen. Wenn Sie die \u00dcbersetzung korrigieren m\u00f6chten, senden Sie diese bitte an: translations@nuclear-power.com oder f\u00fcllen Sie das Online-\u00dcbersetzungsformular aus. Wir bedanken uns f\u00fcr Ihre Hilfe und werden die \u00dcbersetzung so schnell wie m\u00f6glich aktualisieren. Danke.<\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Beispiel f\u00fcr einen Brayton-Zyklus &#8211; Problem mit der L\u00f6sung.\u00a0Berechnen Sie f\u00fcr den geschlossenen Brayton-Zyklus die wichtigsten thermodynamischen Parameter wie Temperaturen, Dr\u00fccke und W\u00e4rme\u00fcbertragungen.\u00a0W\u00e4rmetechnik Brayton-Zyklus &#8211; Problem mit L\u00f6sung Angenommen, der geschlossene\u00a0Brayton-Zyklus\u00a0ist einer der h\u00e4ufigsten\u00a0thermodynamischen Kreisprozessen\u00a0, die in modernen Gasturbinentriebwerken zu finden sind.\u00a0In diesem Fall sei eine\u00a0Heliumgasturbine\u00a0mit Einzelverdichter und Einzelturbinenanordnung angenommen.\u00a0Einer der Schl\u00fcsselparameter solcher Motoren ist &#8230; <a title=\"Das Beispiel f\u00fcr Brayton Cycle &#8211; Problem mit der L\u00f6sung &#8211; Definition\" class=\"read-more\" href=\"https:\/\/www.thermal-engineering.org\/de\/was-ist-ein-beispiel-fur-brayton-cycle-problem-mit-der-losung-definition\/\" aria-label=\"Mehr dazu unter Das Beispiel f\u00fcr Brayton Cycle &#8211; Problem mit der L\u00f6sung &#8211; Definition\">Weiterlesen<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":[],"categories":[9],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v15.4 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Das Beispiel f\u00fcr Brayton Cycle - Problem mit der L\u00f6sung - Definition<\/title>\n<meta name=\"description\" content=\"Beispiel f\u00fcr einen Brayton-Zyklus - Problem mit der L\u00f6sung. 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