{"id":395,"date":"2025-11-08T06:47:45","date_gmt":"2025-11-08T06:47:45","guid":{"rendered":"https:\/\/diagramademollier.com\/?p=395"},"modified":"2025-11-08T07:03:25","modified_gmt":"2025-11-08T07:03:25","slug":"potencia-frigorifica","status":"publish","type":"post","link":"https:\/\/diagramademollier.com\/?p=395","title":{"rendered":"C\u00e1lculo potencia frigor\u00edfica"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Con del diagrama de Mollier <strong>(p\u2013h o h\u2013s seg\u00fan convenci\u00f3n)<\/strong> se puede hacer el  c\u00e1lculo de la <strong>potencia frigor\u00edfica<\/strong> que se necesita para un circuito frigor\u00edfico de <strong>R-134a<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">1) Definici\u00f3n de estados y convenciones<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Usaremos la convenci\u00f3n habitual del ciclo por compresi\u00f3n de vapor con 4 estados:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>1 \u2014 salida del <a href=\"https:\/\/frigoristatorpe.com\/el-evaporador\" target=\"_blank\" rel=\"noreferrer noopener\">evaporador<\/a> \/ entrada al compresor<\/strong>: vapor (saturado o algo sobrecalentado). Entalp\u00eda = <strong>h\u2081<\/strong>.<\/li>\n\n\n\n<li><strong>2 \u2014 salida del compresor<\/strong>: vapor comprimido. Entalp\u00eda = <strong>h\u2082<\/strong>.<\/li>\n\n\n\n<li><strong>3 \u2014 salida del condensador<\/strong>: l\u00edquido (saturado o subenfriado). Entalp\u00eda = <strong>h\u2083<\/strong>.<\/li>\n\n\n\n<li><strong>4 \u2014 salida de la v\u00e1lvula de expansi\u00f3n \/ entrada al evaporador<\/strong>: mezcla l\u00edquido\u2013vapor (isent\u00e1lpico en la v\u00e1lvula), entalp\u00eda = <strong>h\u2084<\/strong> \u2248 <strong>h\u2083<\/strong> (la expansi\u00f3n isentr\u00f3pica real no es, pero la expansi\u00f3n por estrangulamiento es isenthalpica).<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">En la pr\u00e1ctica, la potencia frigor\u00edfica del evaporador (capacidad frigor\u00edfica) es: Q\u02d9L=m\u02d9\u2009(h1\u2212h4)\\dot{Q}_L = \\dot{m}\\,(h_1 &#8211; h_4)Q\u02d9\u200bL\u200b=m\u02d9(h1\u200b\u2212h4\u200b)<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">donde Q\u02d9L\\dot{Q}_LQ\u02d9\u200bL\u200b en W, m\u02d9\\dot{m}m\u02d9 es el caudal m\u00e1sico en kg\/s, y hhh en kJ\/kg (convertir a J\/kg para consistencia en unidades).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">2) Procedimiento pr\u00e1ctico usando el diagrama (p\u2013h \/ Mollier)<\/h2>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Fijar temperaturas de dise\u00f1o<\/strong>: temperatura de evaporaci\u00f3n TevapT_{evap}Tevap\u200b y temperatura de condensaci\u00f3n TcondT_{cond}Tcond\u200b (o las presiones de trabajo).<\/li>\n\n\n\n<li><strong>En el<a href=\"https:\/\/diagramademollier.com\/?wpdmpro=diagrama-de-mollier-r-134a\"> diagrama p\u2013h (Mollier para R-134a)<\/a><\/strong>:\n<ul class=\"wp-block-list\">\n<li>Localiza la <strong>isoterma<\/strong> o la <strong>presi\u00f3n de evaporaci\u00f3n<\/strong> correspondiente a TevapT_{evap}Tevap\u200b. Sobre la l\u00ednea de saturaci\u00f3n, lee la <strong>entalp\u00eda del vapor saturado<\/strong> hgh_ghg\u200b. Si el refrigerante sale del evaporador con <strong>sobrecalentamiento<\/strong> \u0394T_s, despl\u00e1zate por la isobara hasta la temperatura correspondiente y lee <strong>h\u2081<\/strong> (entalp\u00eda de vapor sobrecalentado).<\/li>\n\n\n\n<li>Localiza la <strong>presi\u00f3n\/temperatura de condensaci\u00f3n<\/strong> TcondT_{cond}Tcond\u200b. En la l\u00ednea de saturaci\u00f3n lee la <strong>entalp\u00eda del l\u00edquido saturado<\/strong> hfh_fhf\u200b. Si existe <strong>subenfriamiento<\/strong> \u0394T_sub, despl\u00e1zate por la isobara hacia la izquierda y toma <strong>h\u2083<\/strong> (entalp\u00eda del l\u00edquido subenfriado).<\/li>\n\n\n\n<li>Como la expansi\u00f3n por v\u00e1lvula es isentr\u00f3pica? No \u2014 es <strong>isenthalpica<\/strong>: por tanto <strong>h\u2084 = h\u2083<\/strong>.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Calcula la diferencia de entalp\u00eda en el evaporador<\/strong>: \u0394hevap=h1\u2212h4\\Delta h_{evap} = h_1 &#8211; h_4\u0394hevap\u200b=h1\u200b\u2212h4\u200b.<\/li>\n\n\n\n<li><strong>Si conoces la carga t\u00e9rmica requerida<\/strong> Q\u02d9req\\dot{Q}_{req}Q\u02d9\u200breq\u200b (W), obt\u00e9n el caudal m\u00e1sico:<\/li>\n<\/ol>\n\n\n\n<p class=\"wp-block-paragraph\">m\u02d9=Q\u02d9reqh1\u2212h4\\dot{m} = \\dfrac{\\dot{Q}_{req}}{h_1 &#8211; h_4}m\u02d9=h1\u200b\u2212h4\u200bQ\u02d9\u200breq\u200b\u200b<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">(aseg\u00farate de usar hhh en J\/kg si Q\u02d9\\dot{Q}Q\u02d9\u200b est\u00e1 en W).<br>5. <strong>Potencia del compresor (te\u00f3rica)<\/strong>: W\u02d9comp=m\u02d9\u2009(h2\u2212h1)\\dot{W}_{comp} = \\dot{m}\\,(h_2 &#8211; h_1)W\u02d9comp\u200b=m\u02d9(h2\u200b\u2212h1\u200b). Para h2h_2h2\u200b obt\u00e9n el punto 2 en el diagrama: partir de h\u2081 y seguir la compresi\u00f3n real (usando eficiencia adiab\u00e1tica del compresor para estimar h2h_2h2\u200b).<br>6. <strong>COP frigor\u00edfico<\/strong>: COP=Q\u02d9LW\u02d9comp\\mathrm{COP} = \\dfrac{\\dot{Q}_L}{\\dot{W}_{comp}}COP=W\u02d9comp\u200bQ\u02d9\u200bL\u200b\u200b.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">3) Consideraciones pr\u00e1cticas<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Para hacer un correcto c\u00e1lculo de potencia frigor\u00edfica, debes de considerar:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Usar entalp\u00edas reales<\/strong> del diagrama: incluye sobrecalentamiento y subenfriamiento reales.<\/li>\n\n\n\n<li><strong>P\u00e9rdidas de presi\u00f3n<\/strong> en evaporador\/condensador afectan las presiones y, por tanto, las entalp\u00edas.<\/li>\n\n\n\n<li><strong>Eficiencia del compresor<\/strong>: calcular h2h_2h2\u200b real a partir del h2sh_{2s}h2s\u200b isentr\u00f3pico y la eficiencia adiab\u00e1tica \u03b7c\\eta_c\u03b7c\u200b:<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">h2=h1+h2s\u2212h1\u03b7ch_2 = h_1 + \\dfrac{h_{2s} &#8211; h_1}{\\eta_c}h2\u200b=h1\u200b+\u03b7c\u200bh2s\u200b\u2212h1\u200b\u200b<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Unidades<\/strong>: si hhh en kJ\/kg y Q\u02d9\\dot{Q}Q\u02d9\u200b en kW, usar m\u02d9=Q\u02d9\u2009[kW]\/(h1\u2212h4)\u2009[kJ\/kg]\\dot{m} = \\dot{Q}\\,[\\text{kW}] \/ (h_1 &#8211; h_4)\\,[\\text{kJ\/kg}]m\u02d9=Q\u02d9\u200b[kW]\/(h1\u200b\u2212h4\u200b)[kJ\/kg] \u2192 resultado en kg\/s (1 kW = 1 kJ\/s).<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">4) Ejemplo num\u00e9rico <strong>ilustrativo<\/strong> (valores ficticios \u2014 no extra\u00eddos del Mollier real)<\/h2>\n\n\n\n<p class=\"wp-block-paragraph\">Supongamos: Q\u02d9req=50&nbsp;kW\\dot{Q}_{req} = 50\\ \\text{kW}Q\u02d9\u200breq\u200b=50&nbsp;kW. Del diagrama tomamos:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>h1=390\u00a0kJ\/kgh_1 = 390\\ \\text{kJ\/kg}h1\u200b=390\u00a0kJ\/kg (vapor con algo de sobrecalentamiento),<\/li>\n\n\n\n<li>h4=h3=250\u00a0kJ\/kgh_4 = h_3 = 250\\ \\text{kJ\/kg}h4\u200b=h3\u200b=250\u00a0kJ\/kg (l\u00edquido subenfriado).<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">C\u00e1lculo del caudal m\u00e1sico:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>\u0394h=h1\u2212h4=390\u2212250=140\u00a0kJ\/kg\\Delta h = h_1 &#8211; h_4 = 390 &#8211; 250 = 140\\ \\text{kJ\/kg}\u0394h=h1\u200b\u2212h4\u200b=390\u2212250=140\u00a0kJ\/kg.<\/li>\n\n\n\n<li>m\u02d9=50\u00a0kW140\u00a0kJ\/kg=50\u00a0kJ\/s140\u00a0kJ\/kg=0,357142857\u00a0kg\/s\\dot{m} = \\dfrac{50\\ \\text{kW}}{140\\ \\text{kJ\/kg}} = \\dfrac{50\\ \\text{kJ\/s}}{140\\ \\text{kJ\/kg}} = 0{,}357142857\\ \\text{kg\/s}m\u02d9=140\u00a0kJ\/kg50\u00a0kW\u200b=140\u00a0kJ\/kg50\u00a0kJ\/s\u200b=0,357142857\u00a0kg\/s.<br>(operaci\u00f3n: 50 \u00f7 140 = 0.357142857&#8230;)<\/li>\n<\/ol>\n\n\n\n<p class=\"wp-block-paragraph\">Si la compresi\u00f3n implica h2\u2212h1=120&nbsp;kJ\/kgh_2 &#8211; h_1 = 120\\ \\text{kJ\/kg}h2\u200b\u2212h1\u200b=120&nbsp;kJ\/kg, potencia compresor: W\u02d9comp=0,357142857&nbsp;kg\/s\u00d7120&nbsp;kJ\/kg=42,85714284&nbsp;kW&nbsp;(\u224842,86&nbsp;kW).\\dot{W}_{comp} = 0{,}357142857\\ \\text{kg\/s} \\times 120\\ \\text{kJ\/kg} = 42{,}85714284\\ \\text{kW}\\ (\\approx 42{,}86\\ \\text{kW}).W\u02d9comp\u200b=0,357142857&nbsp;kg\/s\u00d7120&nbsp;kJ\/kg=42,85714284&nbsp;kW&nbsp;(\u224842,86&nbsp;kW).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">COP \u2248 50\/42,857=1,166750 \/ 42{,}857 = 1{,}166750\/42,857=1,1667.<\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p class=\"wp-block-paragraph\">Nota: el ejemplo es meramente ilustrativo; para dise\u00f1o real debes extraer hhh exactas del diagrama Mollier de R-134a para las temperaturas\/presiones y considerar eficiencias y p\u00e9rdidas.<\/p>\n<\/blockquote>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1536\" height=\"1024\" src=\"https:\/\/diagramademollier.com\/wp-content\/uploads\/2025\/11\/diagrama-r-134a.png\" alt=\"Calculo potencia frigor\u00edfica en diagrama de Mollier R-134a\" class=\"wp-image-398\" srcset=\"https:\/\/diagramademollier.com\/wp-content\/uploads\/2025\/11\/diagrama-r-134a.png 1536w, https:\/\/diagramademollier.com\/wp-content\/uploads\/2025\/11\/diagrama-r-134a-300x200.png 300w\" sizes=\"auto, (max-width: 1536px) 100vw, 1536px\" \/><figcaption class=\"wp-element-caption\">Diagrama de Mollier R-134a<\/figcaption><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h2 class=\"wp-block-heading\">5) Checklist para hacer el c\u00e1lculo real<\/h2>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Temperatura de evaporaci\u00f3n y condensaci\u00f3n (o presiones).<\/li>\n\n\n\n<li>Sobrecalentamiento real y subenfriamiento real.<\/li>\n\n\n\n<li>Entalp\u00edas h1,h3h_1, h_3h1\u200b,h3\u200b le\u00eddas en el diagrama p\u2013h (Mollier).<\/li>\n\n\n\n<li>Eficiencia adiab\u00e1tica del compresor (para h2h_2h2\u200b).<\/li>\n\n\n\n<li>P\u00e9rdidas de presi\u00f3n en tuber\u00edas y accesorios.<\/li>\n\n\n\n<li>C\u00e1lculo del caudal m\u00e1sico con la f\u00f3rmula m\u02d9=Q\u02d9\/(h1\u2212h4)\\dot{m}=\\dot{Q}\/(h_1-h_4)m\u02d9=Q\u02d9\u200b\/(h1\u200b\u2212h4\u200b).<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n","protected":false},"excerpt":{"rendered":"<p>Con del diagrama de Mollier (p\u2013h o h\u2013s seg\u00fan convenci\u00f3n) se puede hacer el c\u00e1lculo de la potencia frigor\u00edfica que se necesita para un circuito frigor\u00edfico de R-134a. 1) Definici\u00f3n de estados y convenciones Usaremos la convenci\u00f3n habitual del ciclo por compresi\u00f3n de vapor con 4 estados: En la pr\u00e1ctica, la potencia frigor\u00edfica del evaporador [&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":[103,101,102],"class_list":["post-395","post","type-post","status-publish","format-standard","hentry","category-entender-el-diagrama","tag-calculo-potencia-frigorifica","tag-diagrama-de-mollier-r-134a","tag-potencia-frigorifica"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.3 - 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