{"id":453,"date":"2026-02-23T06:27:11","date_gmt":"2026-02-23T06:27:11","guid":{"rendered":"https:\/\/diagramademollier.com\/?p=453"},"modified":"2026-02-24T06:47:58","modified_gmt":"2026-02-24T06:47:58","slug":"potencia-entregada-por-un-evaporador","status":"publish","type":"post","link":"https:\/\/diagramademollier.com\/?p=453","title":{"rendered":"Potencia entregada por un evaporador"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">En un evaporador frigor\u00edfico o de aire acondicionado tenemos la carga sensible y la carga latente. Es por esto que saber la potencia entregada por un evaporador se debe de calcular en base a la diferencia de entalpias entre la entrada y la salida del evaporador.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Supongamos que tenemos un evaporador (aire acondicionado, camara frigor\u00edfica, etc) y queremos saber la potencia frigor\u00edfica que nos est\u00e1 entregando.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Para este c\u00e1lculo necesitaremos:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>El caudal de aire.<\/li>\n\n\n\n<li>Un termometro (Mejor 2)<\/li>\n\n\n\n<li>Un higrometro (Mejor 2)<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Para el caudal de aire lo mejor es sacar el dato del fabricante del equipo. Si no lo tenemos, se puede medir con un Anem\u00f3metro.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Saber el caudal de aire con anem\u00f3metro<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Mide la secci\u00f3n libre de paso de aire (m\u00b2) :       A = Ancho x Alto<\/li>\n\n\n\n<li>Mide la velocidad media del aire en la superf\u00edcie de salida:\n<ul class=\"wp-block-list\">\n<li>Haz una <strong>malla de puntos<\/strong> (m\u00ednimo 6\u201310 puntos)<\/li>\n\n\n\n<li>Calcula la velocidad media (m\/s)<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Calcula el caudal:     Qv\u200b=A  x  Vmedia<\/li>\n<\/ol>\n\n\n\n<p class=\"wp-block-paragraph\">Obtendr\u00e1s m\u00b3\/s.<br>Multiplica por 3600 para pasar a m\u00b3\/h.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Importante<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">No midas solo el centro. El perfil de velocitat es muy irregular en evaporadores.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">C\u00e1lculo potencia evaporador (con entalp\u00eda)<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">La potencia frigor\u00edfica total es:<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\" display=\"block\"><semantics><mrow><mover accent=\"true\"><mi>Q<\/mi><mo>\u02d9<\/mo><\/mover><mo>=<\/mo><msub><mover accent=\"true\"><mi>m<\/mi><mo>\u02d9<\/mo><\/mover><mrow><mi>a<\/mi><mi>i<\/mi><mi>r<\/mi><mi>e<\/mi><\/mrow><\/msub><mo>\u22c5<\/mo><mo stretchy=\"false\">(<\/mo><msub><mi>h<\/mi><mrow><mi>e<\/mi><mi>n<\/mi><mi>t<\/mi><mi>r<\/mi><mi>a<\/mi><mi>d<\/mi><mi>a<\/mi><\/mrow><\/msub><mo>\u2212<\/mo><msub><mi>h<\/mi><mrow><mi>s<\/mi><mi>a<\/mi><mi>l<\/mi><mi>i<\/mi><mi>d<\/mi><mi>a<\/mi><\/mrow><\/msub><mo stretchy=\"false\">)<\/mo><\/mrow><annotation encoding=\"application\/x-tex\">\\dot Q = \\dot m_{aire} \\cdot (h_{entrada} &#8211; h_{salida})<\/annotation><\/semantics><\/math>Q\u02d9\u200b=m\u02d9aire\u200b\u22c5(hentrada\u200b\u2212hsalida\u200b)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Donde:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><mrow><mover accent=\"true\"><mi>Q<\/mi><mo>\u02d9<\/mo><\/mover><\/mrow><annotation encoding=\"application\/x-tex\">\\dot Q<\/annotation><\/semantics><\/math>Q\u02d9\u200b = potencia frigor\u00edfica (kW)<\/li>\n\n\n\n<li><math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><mrow><mover accent=\"true\"><mi>m<\/mi><mo>\u02d9<\/mo><\/mover><\/mrow><annotation encoding=\"application\/x-tex\">\\dot m<\/annotation><\/semantics><\/math>m\u02d9 = caudal m\u00e1sico de aire seco (kg\/s)<\/li>\n\n\n\n<li><math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><mrow><mi>h<\/mi><\/mrow><annotation encoding=\"application\/x-tex\">h<\/annotation><\/semantics><\/math>h = entalp\u00eda del aire h\u00famedo (kJ\/kg de aire seco)<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">1.- Convertir caudal volum\u00e9trico a caudal m\u00e1sico<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Si el caudal est\u00e1 en <strong>m\u00b3\/h<\/strong>:<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\" display=\"block\"><semantics><mrow><mover accent=\"true\"><mi>m<\/mi><mo>\u02d9<\/mo><\/mover><mo>=<\/mo><mfrac><mrow><msub><mi>Q<\/mi><mi>v<\/mi><\/msub><mo>\u22c5<\/mo><mi>\u03c1<\/mi><\/mrow><mn>3600<\/mn><\/mfrac><\/mrow><annotation encoding=\"application\/x-tex\">\\dot m = \\frac{Q_v \\cdot \\rho}{3600}<\/annotation><\/semantics><\/math>Donde:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><mrow><msub><mi>Q<\/mi><mi>v<\/mi><\/msub><\/mrow><annotation encoding=\"application\/x-tex\">Q_v<\/annotation><\/semantics><\/math>Qv\u200b = caudal en m\u00b3\/h<\/li>\n\n\n\n<li><math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><mrow><mi>\u03c1<\/mi><\/mrow><annotation encoding=\"application\/x-tex\">\\rho<\/annotation><\/semantics><\/math>\u03c1 \u2248 1,2 kg\/m\u00b3 (valor aproximado a 20 \u00b0C)<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Si quieres mayor precisi\u00f3n, puedes calcular la densidad real seg\u00fan temperatura y humedad.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">2.- Calcular entalp\u00eda de entrada y salida<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Necesitas:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Temperatura seca (T)<\/li>\n\n\n\n<li>Humedad relativa (HR)<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Con el term\u00f3metro y el higr\u00f3metro mediremos temperatura y humedad en la entrada y en la salida del evaporador.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">La entalp\u00eda del aire h\u00famedo se puede obtener:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Con <a href=\"\/?wpdmpro=diagrama-psicometrico\">diagrama psicrom\u00e9trico<\/a><\/li>\n\n\n\n<li>Con software t\u00e9cnico (CoolPack, RefTools, etc.)<\/li>\n\n\n\n<li>Con f\u00f3rmula aproximada. No recomendado por su complejidad y inexactitud.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">3.- Calcular la potencia frigor\u00edfica<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Una vez obtenidos:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><mrow><msub><mi>h<\/mi><mn>1<\/mn><\/msub><\/mrow><annotation encoding=\"application\/x-tex\">h_1<\/annotation><\/semantics><\/math>\u200b (entalpia de entrada)<\/li>\n\n\n\n<li><math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><mrow><msub><mi>h<\/mi><mn>2<\/mn><\/msub><\/mrow><annotation encoding=\"application\/x-tex\">h_2<\/annotation><\/semantics><\/math>\u200b (entalpia de salida)<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\"><math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\" display=\"block\"><semantics><mrow><mover accent=\"true\"><mi>Q<\/mi><mo>\u02d9<\/mo><\/mover><mo>=<\/mo><mover accent=\"true\"><mi>m<\/mi><mo>\u02d9<\/mo><\/mover><mo>\u22c5<\/mo><mo stretchy=\"false\">(<\/mo><msub><mi>h<\/mi><mn>1<\/mn><\/msub><mo>\u2212<\/mo><msub><mi>h<\/mi><mn>2<\/mn><\/msub><mo stretchy=\"false\">)<\/mo><\/mrow><annotation encoding=\"application\/x-tex\">\\dot Q = \\dot m \\cdot (h_1 &#8211; h_2)<\/annotation><\/semantics><\/math>El resultado ser\u00e1 en <strong>kW<\/strong> si:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><mrow><mover accent=\"true\"><mi>m<\/mi><mo>\u02d9<\/mo><\/mover><\/mrow><annotation encoding=\"application\/x-tex\">\\dot m<\/annotation><\/semantics><\/math>m\u02d9 est\u00e1 en kg\/s<\/li>\n\n\n\n<li><math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><semantics><mrow><mi>h<\/mi><\/mrow><annotation encoding=\"application\/x-tex\">h<\/annotation><\/semantics><\/math>h est\u00e1 en kJ\/kg<\/li>\n<\/ul>\n\n\n\n<h2 class=\"wp-block-heading\">Ejemplo pr\u00e1ctico<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Necesitaremos el <a href=\"\/?wpdmpro=diagrama-psicometrico\">DIAGRAMA PSICOM\u00c9TRICO<\/a><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Supongamos:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Caudal: 2.400 m\u00b3\/h<\/li>\n\n\n\n<li>Aire entrada: 27 \u00b0C \u2013 50 % HR<\/li>\n\n\n\n<li>Aire salida: 14 \u00b0C \u2013 90 % HR<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">1&#xfe0f;&#x20e3; Caudal m\u00e1sico \u2248 0,79 kg\/s<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">\u03c1\u22481,18&nbsp;kg\/m3 <math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\" display=\"block\"><semantics><mrow><mover accent=\"true\"><mi>m<\/mi><mo>\u02d9<\/mo><\/mover><mo>=<\/mo><mfrac><mrow><mn>2400<\/mn><mo>\u22c5<\/mo><mn>1<\/mn><mo separator=\"true\">,<\/mo><mn>18<\/mn><\/mrow><mn>3600<\/mn><\/mfrac><\/mrow><annotation encoding=\"application\/x-tex\">\\dot m = \\frac{2400 \\cdot 1,18}{3600}<\/annotation><\/semantics><\/math>m\u02d9=0,79&nbsp;kg\/s<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><br>2&#xfe0f;&#x20e3; h entrada \u2248 53 kJ\/kg<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Aire entrada: 27 \u00b0C \u2013 50 % HR<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"993\" height=\"702\" src=\"https:\/\/diagramademollier.com\/wp-content\/uploads\/2026\/02\/Entalpia-entrada-1.jpg\" alt=\"\" class=\"wp-image-462\" srcset=\"https:\/\/diagramademollier.com\/wp-content\/uploads\/2026\/02\/Entalpia-entrada-1.jpg 993w, https:\/\/diagramademollier.com\/wp-content\/uploads\/2026\/02\/Entalpia-entrada-1-300x212.jpg 300w, https:\/\/diagramademollier.com\/wp-content\/uploads\/2026\/02\/Entalpia-entrada-1-768x543.jpg 768w\" sizes=\"auto, (max-width: 993px) 100vw, 993px\" \/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Partimos de la temperatura de +27\u00baC i subimos verticalmente hasta encontrar la curva de la humedad relativa del 50%. En el punto de encuentro buscamos la l\u00ednea de la entalpia y bajamos hasta ver el valor que nos da de 53kJ\/kg.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><br>3&#xfe0f;&#x20e3; h salida \u2248 35 kJ\/kg<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Aire salida: 14 \u00b0C \u2013 90 % HR<\/p>\n\n\n\n<figure class=\"wp-block-image size-full is-resized\"><img loading=\"lazy\" decoding=\"async\" width=\"708\" height=\"500\" src=\"https:\/\/diagramademollier.com\/wp-content\/uploads\/2026\/02\/Entalpia-salida.jpg\" alt=\"\" class=\"wp-image-463\" style=\"width:1100px\" srcset=\"https:\/\/diagramademollier.com\/wp-content\/uploads\/2026\/02\/Entalpia-salida.jpg 708w, https:\/\/diagramademollier.com\/wp-content\/uploads\/2026\/02\/Entalpia-salida-300x212.jpg 300w\" sizes=\"auto, (max-width: 708px) 100vw, 708px\" \/><\/figure>\n\n\n\n<p class=\"wp-block-paragraph\">Igual que el anterior, partimos de la temperatura de +14\u00baC i subimos verticalmente hasta encontrar la curva de la humedad relativa del 90%. En el punto de encuentro buscamos la l\u00ednea de la entalpia y bajamos hasta ver el valor que nos da de 35kJ\/kg.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">4&#xfe0f;&#x20e3;CALCULO FINAL<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Q\u02d9\u200b=m\u02d9\u22c5(h1\u200b\u2212h2\u200b) <br>Q\u02d9\u200b=0,79\u22c5(53\u221235) <br>Q\u02d9\u200b=0,79\u22c518 <br><strong>Q\u02d9\u200b\u224814,22\u00a0kW<\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><\/p>\n","protected":false},"excerpt":{"rendered":"<p>En un evaporador frigor\u00edfico o de aire acondicionado tenemos la carga sensible y la carga latente. Es por esto que saber la potencia entregada por un evaporador se debe de calcular en base a la diferencia de entalpias entre la entrada y la salida del evaporador. Supongamos que tenemos un evaporador (aire acondicionado, camara frigor\u00edfica, [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[104],"tags":[165,164,163],"class_list":["post-453","post","type-post","status-publish","format-standard","hentry","category-entender-el-diagrama","tag-capacidad-evaporador","tag-diferencia-entalpias","tag-potencia-evaporador"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.3 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Potencia entregada por un evaporador - DIAGRAMA DE MOLLIER<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/diagramademollier.com\/?p=453\" \/>\n<meta property=\"og:locale\" content=\"es_ES\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Potencia entregada por un evaporador - DIAGRAMA DE MOLLIER\" \/>\n<meta property=\"og:description\" content=\"En un evaporador frigor\u00edfico o de aire acondicionado tenemos la carga sensible y la carga latente. Es por esto que saber la potencia entregada por un evaporador se debe de calcular en base a la diferencia de entalpias entre la entrada y la salida del evaporador. 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