{"id":214388,"date":"2022-12-27T13:48:26","date_gmt":"2022-12-27T12:48:26","guid":{"rendered":"https:\/\/educelec.com\/?page_id=214388"},"modified":"2026-01-09T11:06:35","modified_gmt":"2026-01-09T10:06:35","slug":"mot-synch","status":"publish","type":"page","link":"https:\/\/educelec.com\/en\/mot-synch\/","title":{"rendered":"moteurs synchrone"},"content":{"rendered":"<p>[et_pb_section fb_built=\u00a0\u00bb1&Prime; admin_label=\u00a0\u00bbsection\u00a0\u00bb _builder_version=\u00a0\u00bb4.19.1&Prime; global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_row _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb custom_margin=\u00a0\u00bb-50px||||false|false\u00a0\u00bb locked=\u00a0\u00bboff\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_column type=\u00a0\u00bb4_4&Prime; _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_text _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb custom_margin=\u00a0\u00bb|0px||-30px|false|false\u00a0\u00bb custom_padding=\u00a0\u00bb|||0px|false|false\u00a0\u00bb hover_enabled=\u00a0\u00bb0&Prime; locked=\u00a0\u00bboff\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb sticky_enabled=\u00a0\u00bb0&Prime;]<\/p>\n<p><a href=\"https:\/\/educelec.com\/en\/\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone wp-image-216559\" src=\"https:\/\/educelec.com\/wp-content\/uploads\/2025\/12\/chap-300x209.png\" alt=\"\" width=\"17\" height=\"12\" \/><\/a><a href=\"https:\/\/educelec.com\/en\/\">Accueil<\/a><a href=\"https:\/\/educelec.com\/en\/\">\u00a0&gt;<\/a> <a href=\"https:\/\/educelec.com\/en\/electrotech\/\">Electrotech <\/a><a href=\"https:\/\/educelec.com\/en\/electrotech\/\">&gt; <\/a><strong>Moteurs synchrones<\/strong><\/p>\n<p>[\/et_pb_text][et_pb_text _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb text_font=\u00a0\u00bbAcme|700|||||||\u00a0\u00bb background_color=\u00a0\u00bb#88C4DD\u00a0\u00bb text_text_shadow_style=\u00a0\u00bbpreset1&Prime; text_text_shadow_blur_strength=\u00a0\u00bb2em\u00a0\u00bb text_text_shadow_color=\u00a0\u00bb#0047E0&Prime; locked=\u00a0\u00bboff\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb text_font_size__hover_enabled=\u00a0\u00bbon|desktop\u00a0\u00bb]<\/p>\n<blockquote>\n<h1 style=\"text-align: center;\"><strong>Exercices corrig\u00e9s<\/strong><\/h1>\n<h3 style=\"text-align: center;\"><strong>Etude des moteurs synchrones<\/strong><\/h3>\n<\/blockquote>\n<p>[\/et_pb_text][\/et_pb_column][\/et_pb_row][et_pb_row _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb custom_margin=\u00a0\u00bb-50px||||false|false\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_column type=\u00a0\u00bb4_4&Prime; _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_text _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb background_color=\u00a0\u00bb#69D0FC\u00a0\u00bb min_height=\u00a0\u00bb40px\u00a0\u00bb height=\u00a0\u00bb40px\u00a0\u00bb custom_padding=\u00a0\u00bb7px|||18px|false|false\u00a0\u00bb border_radii=\u00a0\u00bbon|15px|15px|15px|15px\u00a0\u00bb border_width_all=\u00a0\u00bb2px\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb]<\/p>\n<ul>\n<li>\n<h3><strong>Exercice N\u00b0 1<\/strong><\/h3>\n<\/li>\n<\/ul>\n<p>[\/et_pb_text][\/et_pb_column][\/et_pb_row][et_pb_row _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb custom_margin=\u00a0\u00bb-25px||||false|false\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_column type=\u00a0\u00bb4_4&Prime; _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_text _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb]<\/p>\n<h4>Un alternateur triphas\u00e9 \u00e9toile \u00e0 une tension (entre phases) U = 660 V et d\u00e9bite un courant de 500 A sous un cos j = 0,8 (inductif) \u00e0 la fr\u00e9quence f = 50 Hz.<\/h4>\n<h4>1) Calculer la puissance apparente, la puissance active et la puissance r\u00e9active.<\/h4>\n<h4>2) Sachant que l&rsquo;induit comporte 372 conducteurs et que le flux sous un p\u00f4le est de 0,0027 Wb. Calculer le coefficient de Kapp en admettant que E est \u00e9gal \u00e0 la tension sur une phase \u00e0 la sortie de l&rsquo;alternateur.<\/h4>\n<p>[\/et_pb_text][\/et_pb_column][\/et_pb_row][et_pb_row _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb custom_margin=\u00a0\u00bb||12px||false|false\u00a0\u00bb custom_padding=\u00a0\u00bb0px||||false|false\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_column type=\u00a0\u00bb4_4&Prime; _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_toggle title=\u00a0\u00bbToucher pour afficher la solution\u00a0\u00bb _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb title_font_size=\u00a0\u00bb13px\u00a0\u00bb custom_margin=\u00a0\u00bb-30px||-25px||false|false\u00a0\u00bb locked=\u00a0\u00bboff\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb]<\/p>\n<h4>1) <u>Puissance apparente, active et r\u00e9active<\/u><\/h4>\n<h4><u>Puissance active<\/u><\/h4>\n<h4 style=\"text-align: center; margin: 0cm 0cm .0001pt 7.1pt;\" align=\"center\">P=<span style=\"font-size: 10.5pt; font-family: 'Arial',sans-serif; color: #4d5156; background: white;\">\u221a 3U.I,cos<\/span><span style=\"font-family: Symbol;\">f = 457,26 K<\/span> W<\/h4>\n<h4><u>Puissance r\u00e9active<\/u><\/h4>\n<h4 style=\"text-align: center; margin: 0cm 0cm .0001pt 7.1pt;\" align=\"center\">Q=<span style=\"font-size: 10.5pt; font-family: 'Arial',sans-serif; color: #4d5156; background: white;\">\u221a 3U.I,sin<\/span><span style=\"font-family: Symbol;\">f = 342,95 K<\/span> VAR<\/h4>\n<h4><u>Puissance apparente<\/u><\/h4>\n<h4 style=\"text-align: center; margin: 0cm 0cm .0001pt 7.1pt;\" align=\"center\">S=<span style=\"font-size: 10.5pt; font-family: 'Arial',sans-serif; color: #4d5156; background: white;\">\u221a 3U.I<\/span><span style=\"font-family: Symbol;\"> = 571,58 K<\/span> W<\/h4>\n<h4>2) <u>Coefficient de Kapp<\/u><\/h4>\n<h4>Pour un alternateur triphas\u00e9, la fem par phase s&rsquo;exprime par la relation:\u00a0\u00a0\u00a0 E = Knff<sub>m<\/sub><\/h4>\n<h4>O\u00f9 K est le coefficient de Kapp<br \/>n = 372 \/ 3 = 124\u00a0\u00a0 le nombre de conducteurs par phase<br \/>f = 50Hz\u00a0\u00a0 la fr\u00e9quence<br \/>f<sub>m<\/sub> = 0,0027Wb\u00a0 le flux sous un p\u00f4le.<\/h4>\n<h4>En confondant la tension de sortie et la fem, c&rsquo;est \u00e0 dire en n\u00e9gligeant les chutes de tension dans l&rsquo;alternateur, E = 381V et <strong>K= 22,7 SI<\/strong><\/h4>\n<p>[\/et_pb_toggle][\/et_pb_column][\/et_pb_row][\/et_pb_section][et_pb_section fb_built=\u00a0\u00bb1&Prime; _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb custom_margin=\u00a0\u00bb||-100px||false|false\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_row _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb locked=\u00a0\u00bboff\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_column type=\u00a0\u00bb4_4&Prime; _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_text _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb background_color=\u00a0\u00bb#69D0FC\u00a0\u00bb min_height=\u00a0\u00bb40px\u00a0\u00bb height=\u00a0\u00bb40px\u00a0\u00bb custom_padding=\u00a0\u00bb7px|||18px|false|false\u00a0\u00bb border_radii=\u00a0\u00bbon|15px|15px|15px|15px\u00a0\u00bb border_width_all=\u00a0\u00bb2px\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb]<\/p>\n<ul>\n<li>\n<h3><strong>Exercice N\u00b0 2<\/strong><\/h3>\n<\/li>\n<\/ul>\n<p>[\/et_pb_text][\/et_pb_column][\/et_pb_row][et_pb_row _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb custom_margin=\u00a0\u00bb-25px||||false|false\u00a0\u00bb locked=\u00a0\u00bboff\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_column type=\u00a0\u00bb4_4&Prime; _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_text _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb]<\/p>\n<h4 style=\"margin: 0cm 0cm .0001pt 7.1pt;\">Un alternateur monophas\u00e9 fournit un courant de 50 A sous une tension de 240 V et avec un facteur de puissance de 0,8 (charge inductive). Le rotor consomme 8 A sous une tension de 35 V, les pertes constantes sont de 450 W et la r\u00e9sistance de l&rsquo;enroulement du stator est R=0,2 <span style=\"font-family: Symbol;\">W<\/span>.<\/h4>\n<p>&nbsp;<\/p>\n<h4 style=\"margin: 0cm 0cm .0001pt 7.1pt;\">1) Calculer la puissance utile de l&rsquo;alternateur et son rendement.<\/h4>\n<h4 style=\"margin: 0cm 0cm .0001pt 7.1pt;\">2) Pour la m\u00eame excitation on a relev\u00e9: E<sub>v<\/sub> = 280 V et I<sub>cc<\/sub>= 40 A. Calculer l&rsquo;imp\u00e9dance et la r\u00e9actance interne de l&rsquo;alternateur et d\u00e9terminer la f.e.m. (E<sub>v<\/sub>) par le graphique de Behn-Eschenburg.<\/h4>\n<p>[\/et_pb_text][\/et_pb_column][\/et_pb_row][et_pb_row _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb locked=\u00a0\u00bboff\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_column type=\u00a0\u00bb4_4&Prime; _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_toggle title=\u00a0\u00bbToucher pour afficher la solution\u00a0\u00bb _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb title_font_size=\u00a0\u00bb13px\u00a0\u00bb custom_margin=\u00a0\u00bb-30px||-80px||false|false\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb]<\/p>\n<h4><a href=\"http:\/\/assocampus.ifrance.com\/pages\/altcor3.htm\">Correction<\/a><\/h4>\n<h4>1) <u>Puissance utile et rendement<\/u><\/h4>\n<h4>La puissance utile est donn\u00e9e par la relation:<\/h4>\n<h4>P<sub>u<\/sub> = UI cos f = 9,6 kW<\/h4>\n<h4>La puissance absorb\u00e9e \u00e9tant donn\u00e9e par:<\/h4>\n<h4>P<sub>abs<\/sub> = P<sub>u<\/sub> + P<sub>js<\/sub> + P<sub>jr<\/sub> + P<sub>c<\/sub> = 10,83 kW<\/h4>\n<h4>Avec<\/h4>\n<h4>P<sub>jr<\/sub> = ui = 280W<br \/>P<sub>js<\/sub> = RI<sup>2<\/sup> = 500W<br \/>P<sub>c<\/sub> = 450W<\/h4>\n<h4>Le rendement est<\/h4>\n<h4>h = P<sub>u<\/sub> \/ P<sub>abs<\/sub> = 0,89<\/h4>\n<h4>2) <u>Imp\u00e9dance et r\u00e9actance internes, f\u00e9m.<\/u><\/h4>\n<h4>L&rsquo;imp\u00e9dance interne est obtenue \u00e0 partir de la mesure de la fem et du courant de court-circuit. Elle est peu diff\u00e9rente de la r\u00e9actance interne.<\/h4>\n<p><a href=\"https:\/\/educelec.com\/wp-content\/uploads\/2022\/12\/sexo1-2.png\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-214415 alignnone size-full\" src=\"https:\/\/educelec.com\/wp-content\/uploads\/2022\/12\/sexo1-2.png\" alt=\"\" width=\"224\" height=\"48\" \/><\/a><\/p>\n<h4>Pour obtenir la fem on trace le diagramme de Behn-Eschenburg<\/h4>\n<p><a href=\"https:\/\/educelec.com\/wp-content\/uploads\/2022\/12\/Sexo-2-2.png\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-214416 alignnone size-full\" src=\"https:\/\/educelec.com\/wp-content\/uploads\/2022\/12\/Sexo-2-2.png\" alt=\"\" width=\"280\" height=\"156\" \/><\/a><\/p>\n<h4>En projetant sur un axe horizontal Ox et un axe vertical Oy, on obtient<\/h4>\n<h4>E<sub>vx<\/sub> = U + RI cosf + LwI sin f = 458 V<br \/>E<sub>vy<\/sub> = &#8211; RI sin f + LwI cos f = 274 V<\/h4>\n<h4>E<sub>v<\/sub><sup>2<\/sup> = E<sub>vx<\/sub><sup>2<\/sup> + E<sub>vx<\/sub><sup>2<\/sup> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 E<sub>v<\/sub> = 534 Vbit de 400 A sous 420 V de tension compos\u00e9e.<\/h4>\n<p>[\/et_pb_toggle][\/et_pb_column][\/et_pb_row][\/et_pb_section][et_pb_section fb_built=\u00a0\u00bb1&Prime; _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb locked=\u00a0\u00bboff\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_row _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb locked=\u00a0\u00bboff\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_column type=\u00a0\u00bb4_4&Prime; _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_text _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb background_color=\u00a0\u00bb#69D0FC\u00a0\u00bb min_height=\u00a0\u00bb40px\u00a0\u00bb height=\u00a0\u00bb40px\u00a0\u00bb custom_padding=\u00a0\u00bb7px|||18px|false|false\u00a0\u00bb border_radii=\u00a0\u00bbon|15px|15px|15px|15px\u00a0\u00bb border_width_all=\u00a0\u00bb2px\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb]<\/p>\n<ul>\n<li>\n<h3><strong>Exercice N\u00b0 3<\/strong><\/h3>\n<\/li>\n<\/ul>\n<p>[\/et_pb_text][\/et_pb_column][\/et_pb_row][et_pb_row _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb custom_margin=\u00a0\u00bb-25px||||false|false\u00a0\u00bb locked=\u00a0\u00bboff\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_column type=\u00a0\u00bb4_4&Prime; _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_text _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb]<\/p>\n<h4 style=\"margin: 0cm 0cm .0001pt 7.1pt;\">Un alternateur triphas\u00e9 \u00e9toile fournit un courant de 400 A sous une tension compos\u00e9e de 420 V et avec un facteur de puissance de 0,9 (charge inductive). La r\u00e9sistance mesur\u00e9e entre phases du stator est R = 0,03 <span style=\"font-family: Symbol;\">W<\/span> et l&rsquo;ensemble des pertes constantes et par effet Joule au rotor est P = 6 kW.<\/h4>\n<p>&nbsp;<\/p>\n<h4 style=\"margin: 0cm 0cm .0001pt 7.1pt;\">1) Calculer la puissance utile de l&rsquo;alternateur et son rendement<\/h4>\n<h4 style=\"margin: 0cm 0cm .0001pt 7.1pt;\">2) Pour la m\u00eame excitation on a relev\u00e9 : E<sub>ve<\/sub> = 510 V (entre phases) et I<sub>cc<\/sub> = 300 A. Calculer la r\u00e9actance interne (R est ici n\u00e9glig\u00e9e) et d\u00e9terminer la f.\u00e9.m. (E<sub>ve<\/sub>) entre phases qui correspond \u00e0 un d\u00e9bit de 400 A sous 420 V de tension compos\u00e9e.<\/h4>\n<p>[\/et_pb_text][\/et_pb_column][\/et_pb_row][et_pb_row _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb custom_margin=\u00a0\u00bb-20px||-100px||false|false\u00a0\u00bb locked=\u00a0\u00bboff\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_column type=\u00a0\u00bb4_4&Prime; _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_toggle title=\u00a0\u00bbToucher pour afficher la solution\u00a0\u00bb _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb title_font_size=\u00a0\u00bb13px\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb]<\/p>\n<h4><a href=\"http:\/\/assocampus.ifrance.com\/pages\/altcor4.htm\">Correction<\/a><\/h4>\n<h4>1) <u>Puissance utile et rendement<\/u><\/h4>\n<h4>La puissance utile est donn\u00e9e par<\/h4>\n<h4><strong>Pu=<span style=\"font-size: 10.5pt; font-family: 'Arial',sans-serif; color: #4d5156; background: white;\">\u221a 3.U.I,cos<\/span><span style=\"font-family: Symbol;\">f = 261,89 K<\/span> W<\/strong><\/h4>\n<h4>La puissance absorb\u00e9e \u00e9tant quant \u00e0 elle<\/h4>\n<h4>P<sub>abs<\/sub> = P<sub>u<\/sub> + P<sub>js<\/sub> + P<sub>jr<\/sub> + P<sub>fs<\/sub> + P<sub>m<\/sub> = P<sub>u<\/sub> + P<sub>js<\/sub> + P<sub>jr<\/sub> + P<sub>c<\/sub><\/h4>\n<h4>P<sub>js<\/sub> = ( 3\/2 )RI<sup>2<\/sup> = 7938 W repr\u00e9sente les pertes Joule au stator<br \/>P<sub>jr<\/sub>, les pertes Joule au rotor<br \/>P<sub>fs<\/sub>, les pertes fer au stator<br \/>P<sub>m<\/sub> les pertes m\u00e9caniques<br \/>P<sub>c<\/sub> = P<sub>fr<\/sub> + P<sub>m<\/sub>, les pertes constantes<\/h4>\n<h4>D\u2019o\u00f9 la puissance absorb\u00e9e<\/h4>\n<h4>P<sub>abs<\/sub> = 275,82 kW<\/h4>\n<h4>Et le rendement<\/h4>\n<h4>h = P<sub>u<\/sub> \/ P<sub>abs<\/sub> = 0,95<\/h4>\n<h4>2) <u>R\u00e9actance interne et fem<\/u><\/h4>\n<h4>La r\u00e9sistance interne \u00e9tant n\u00e9glig\u00e9e on a la r\u00e9actance interne<\/h4>\n<h4>On d\u00e9termine alors la fem sur une phase \u00e0 partir du diagramme de Behn Eschenburg<\/h4>\n<p><a href=\"https:\/\/educelec.com\/wp-content\/uploads\/2022\/12\/Sexo-3-3.png\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-214417 alignnone size-full\" src=\"https:\/\/educelec.com\/wp-content\/uploads\/2022\/12\/Sexo-3-3.png\" alt=\"\" width=\"280\" height=\"156\" \/><\/a><\/p>\n<h4>En projetant sur des axes, horizontal (Ox) et Vertical (Oy)<\/h4>\n<h4>E<sub>vx<\/sub> = V + RI cos f + LwI sin f = <br \/>E<sub>vy<\/sub> = &#8211; RI sin f + LwI cos f =<\/h4>\n<h4>\u00a0<\/h4>\n<h4>\u00a0<\/h4>\n<h4>La fem entre phases est alors<\/h4>\n<h4>\u00a0<\/h4>\n<p>[\/et_pb_toggle][\/et_pb_column][\/et_pb_row][\/et_pb_section][et_pb_section fb_built=\u00a0\u00bb1&Prime; admin_label=\u00a0\u00bbsection\u00a0\u00bb _builder_version=\u00a0\u00bb4.19.1&Prime; custom_margin=\u00a0\u00bb||-120px||false|false\u00a0\u00bb locked=\u00a0\u00bboff\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_row _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_column type=\u00a0\u00bb4_4&Prime; _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_text _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb background_color=\u00a0\u00bb#69D0FC\u00a0\u00bb min_height=\u00a0\u00bb40px\u00a0\u00bb height=\u00a0\u00bb40px\u00a0\u00bb custom_padding=\u00a0\u00bb7px|||18px|false|false\u00a0\u00bb border_radii=\u00a0\u00bbon|15px|15px|15px|15px\u00a0\u00bb border_width_all=\u00a0\u00bb2px\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb]<\/p>\n<ul>\n<li>\n<h3><strong>Exercice N\u00b0 4<\/strong><\/h3>\n<\/li>\n<\/ul>\n<p>[\/et_pb_text][\/et_pb_column][\/et_pb_row][et_pb_row _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb custom_margin=\u00a0\u00bb-25px||||false|false\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_column type=\u00a0\u00bb4_4&Prime; _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_text _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb]<\/p>\n<h4>On consid\u00e8re un alternateur monophas\u00e9 (circuit magn\u00e9tique non satur\u00e9), ayant les caract\u00e9ristiques suivantes:<\/h4>\n<h4>&#8211; Tension d&rsquo;induit U = 380 V;<br \/>&#8211; Fr\u00e9quence f = 60 Hz;<br \/>&#8211; Vitesse de rotation N = 900 tr\/min;<br \/>&#8211; R\u00e9sistance d&rsquo;induit r = 0,02 W.<\/h4>\n<h4>\u00a0Lorsque le courant d&rsquo;excitation vaut 9 A, la tension \u00e0 vide est \u00e9gale \u00e0 420 V. De plus, pour un courant d&rsquo;excitation de 5 A, l&rsquo;alternateur d\u00e9bite un courant de court-circuit de 307 A.<\/h4>\n<h4>1) D\u00e9terminer le nombre de p\u00f4les de l&rsquo;alternateur.<\/h4>\n<h4>2) D\u00e9terminer la r\u00e9actance synchrone.<\/h4>\n<h4>3) Le facteur de puissance de l&rsquo;installation \u00e9tant de 0,9 (inductif), trouver la fem \u00e0 avoir pour U = 380 V et\u00a0\u00a0\u00a0 I = 120 A en utilisant le diagramme de Behn-Eshenburg.<\/h4>\n<h4>4) En d\u00e9duire le courant d&rsquo;excitation correspondant (on consid\u00e8re que la courbe E(i) est lin\u00e9aire entre 380 et 450 V).<\/h4>\n<h4>Le rotor consomme un courant de i = 5 A sous une tension de 17 V, et les pertes constantes sont \u00e9gales \u00e0 700 W.<\/h4>\n<h4>5) Calculer pour les conditions des questions 3\/ et 4\/, la puissance utile ainsi que son rendement<\/h4>\n<p>[\/et_pb_text][\/et_pb_column][\/et_pb_row][et_pb_row _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb custom_margin=\u00a0\u00bb||12px||false|false\u00a0\u00bb custom_padding=\u00a0\u00bb0px||||false|false\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_column type=\u00a0\u00bb4_4&Prime; _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_toggle title=\u00a0\u00bbToucher pour afficher la solution\u00a0\u00bb _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb title_font_size=\u00a0\u00bb13px\u00a0\u00bb custom_margin=\u00a0\u00bb-30px||-25px||false|false\u00a0\u00bb locked=\u00a0\u00bboff\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb]<\/p>\n<h4>\u00a0<\/h4>\n<h4>1) Nombre de p\u00f4les de l&rsquo;alternateur<\/h4>\n<h4>Le nombre de paires de p\u00f4les de l&rsquo;alternateur est donn\u00e9 par la relation:<\/h4>\n<h4>p = f \/ N\u00a0\u00a0\u00a0 (f en Hz et N en tr\/s) \u00a0 p = 4 soit 8 p\u00f4les<\/h4>\n<h4>2) R\u00e9actance synchrone<\/h4>\n<h4>En supposant que la courbe Icc(i) du courant de court-circuit \u00e0 l&rsquo;induit en fonction du courant d&rsquo;excitation est lin\u00e9aire on obtient<\/h4>\n<h4>Icc (i = 9A) = (9\/5) Icc (i = 5A) = 553 A. La r\u00e9actance synchrone est alors donn\u00e9e par la relation:<\/h4>\n<h4><a href=\"https:\/\/educelec.com\/wp-content\/uploads\/2022\/12\/sexo1.png\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-214401 alignnone size-full\" src=\"https:\/\/educelec.com\/wp-content\/uploads\/2022\/12\/sexo1.png\" alt=\"\" width=\"224\" height=\"48\" \/><\/a><\/h4>\n<h4>3) fem pour U = 380 et I = 120 A<\/h4>\n<h4>En se pla\u00e7ant dans l&rsquo;hypoth\u00e8se de behn-Eshenburg<\/h4>\n<h4><a href=\"https:\/\/educelec.com\/wp-content\/uploads\/2022\/12\/sexo1-1.png\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-214402 alignnone size-full\" src=\"https:\/\/educelec.com\/wp-content\/uploads\/2022\/12\/sexo1-1.png\" alt=\"\" width=\"280\" height=\"156\" \/><\/a><\/h4>\n<h4>En projetant sur un axe horizontal Ox et un axe vertical Oy, on obtient<\/h4>\n<h4>E<sub>vx<\/sub> = U + RI cos f + LwI sin f = 421 V<br \/>E<sub>vy<\/sub> = &#8211; RI sin f + LwI cos f = 81 V<\/h4>\n<h4>E<sub>v<\/sub><sup>2<\/sup> = E<sub>vx<\/sub><sup>2<\/sup> + E<sub>vx<\/sub><sup>2<\/sup> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 E<sub>v<\/sub> = 429 V<\/h4>\n<h4>4) <u>Courant d&rsquo;excitation<\/u><\/h4>\n<h4>La caract\u00e9ristique interne \u00e9tant consid\u00e9r\u00e9e comme lin\u00e9aire on en d\u00e9duit le courant d&rsquo;excitation:<\/h4>\n<h4>i ( E<sub>v<\/sub> = 429 V) = i ( E<sub>v<\/sub> = 420 V) (429 \/ 420) = 9,2 A<\/h4>\n<h4>3) <u>Puissance utile et rendement<\/u><\/h4>\n<h4>La puissance utile est la puissance active fournie \u00e0 l&rsquo;induit par l&rsquo;alternateur:<\/h4>\n<h4>P<sub>u<\/sub> =UI cos f = 41,04 kW<\/h4>\n<h4>Pour d\u00e9terminer le rendement nous devons \u00e9valuer les diff\u00e9rentes pertes:<\/h4>\n<h4>Pertes Joule \u00e0 l&rsquo;induit: P<sub>js<\/sub> = rI<sup>2<\/sup> = 288 W<br \/>pertes Joule \u00e0 l&rsquo;inducteur: P<sub>jr<\/sub> = Ri<sup>2<\/sup> = (17 \/ 5) i<sup>2<\/sup> = 287,8 W<br \/>pertes constantes: P<sub>c<\/sub> = 700 W<\/h4>\n<h4>La puissance absorb\u00e9e est donc<\/h4>\n<h4>P<sub>abs<\/sub> = P<sub>u<\/sub> + P<sub>js<\/sub> + P<sub>jr<\/sub> + P<sub>c<\/sub> = 42,31 kW<\/h4>\n<h4>Et le rendement<\/h4>\n<h4>h = P<sub>u<\/sub> \/ P<sub>abs<\/sub> = 0,97<\/h4>\n<h4>\u00a0<\/h4>\n<p>[\/et_pb_toggle][\/et_pb_column][\/et_pb_row][et_pb_row _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb custom_margin=\u00a0\u00bb-15px||||false|false\u00a0\u00bb locked=\u00a0\u00bboff\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_column type=\u00a0\u00bb4_4&Prime; _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][et_pb_button button_url=\u00a0\u00bbhttps:\/\/educelec.com\/electrotech\/\u00a0\u00bb button_text=\u00a0\u00bbRetour a la page pr\u00e9c\u00e9dente\u00a0\u00bb _builder_version=\u00a0\u00bb4.19.1&Prime; _module_preset=\u00a0\u00bbdefault\u00a0\u00bb global_colors_info=\u00a0\u00bb{}\u00a0\u00bb][\/et_pb_button][\/et_pb_column][\/et_pb_row][\/et_pb_section]<\/p>","protected":false},"excerpt":{"rendered":"<p>Accueil\u00a0&gt; Electrotech &gt; Moteurs synchronesExercices corrig\u00e9s Etude des moteurs synchrones Exercice N\u00b0 1 Un alternateur triphas\u00e9 \u00e9toile \u00e0 une tension (entre phases) U = 660 V et d\u00e9bite un courant de 500 A sous un cos j = 0,8 (inductif) \u00e0 la fr\u00e9quence f = 50 Hz. 1) Calculer la puissance apparente, la puissance active [&hellip;]<\/p>","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_et_pb_use_builder":"on","_et_pb_old_content":"","_et_gb_content_width":"","om_disable_all_campaigns":false,"_uf_show_specific_survey":0,"_uf_disable_surveys":false,"footnotes":""},"class_list":["post-214388","page","type-page","status-publish","hentry"],"aioseo_notices":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.6 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>moteurs synchrone - Exercices et Solutions en Electricit\u00e9, Electronique, Electrotechnique et Automatisme industrielle<\/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:\/\/educelec.com\/en\/mot-synch\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"moteurs synchrone - Exercices et Solutions en Electricit\u00e9, Electronique, Electrotechnique et Automatisme industrielle\" \/>\n<meta property=\"og:description\" content=\"Accueil\u00a0&gt; Electrotech &gt; Moteurs synchronesExercices corrig\u00e9s Etude des moteurs synchrones Exercice N\u00b0 1 Un alternateur triphas\u00e9 \u00e9toile \u00e0 une tension (entre phases) U = 660 V et d\u00e9bite un courant de 500 A sous un cos j = 0,8 (inductif) \u00e0 la fr\u00e9quence f = 50 Hz. 1) Calculer la puissance apparente, la puissance active [&hellip;]\" \/>\n<meta property=\"og:url\" content=\"https:\/\/educelec.com\/en\/mot-synch\/\" \/>\n<meta property=\"og:site_name\" content=\"Exercices et Solutions en Electricit\u00e9, Electronique, Electrotechnique et Automatisme industrielle\" \/>\n<meta property=\"article:modified_time\" content=\"2026-01-09T10:06:35+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/educelec.com\/wp-content\/uploads\/2025\/12\/chap-e1766263997563.png\" \/>\n\t<meta property=\"og:image:width\" content=\"288\" \/>\n\t<meta property=\"og:image:height\" content=\"200\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/png\" \/>\n<meta name=\"twitter:label1\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data1\" content=\"9 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"WebPage\",\"@id\":\"https:\/\/educelec.com\/mot-synch\/\",\"url\":\"https:\/\/educelec.com\/mot-synch\/\",\"name\":\"moteurs synchrone - Exercices et Solutions en Electricit\u00e9, Electronique, Electrotechnique et Automatisme industrielle\",\"isPartOf\":{\"@id\":\"https:\/\/educelec.com\/#website\"},\"primaryImageOfPage\":{\"@id\":\"https:\/\/educelec.com\/mot-synch\/#primaryimage\"},\"image\":{\"@id\":\"https:\/\/educelec.com\/mot-synch\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/educelec.com\/wp-content\/uploads\/2025\/12\/chap-300x209.png\",\"datePublished\":\"2022-12-27T12:48:26+00:00\",\"dateModified\":\"2026-01-09T10:06:35+00:00\",\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/educelec.com\/mot-synch\/\"]}]},{\"@type\":\"ImageObject\",\"inLanguage\":\"en-US\",\"@id\":\"https:\/\/educelec.com\/mot-synch\/#primaryimage\",\"url\":\"https:\/\/educelec.com\/wp-content\/uploads\/2025\/12\/chap-300x209.png\",\"contentUrl\":\"https:\/\/educelec.com\/wp-content\/uploads\/2025\/12\/chap-300x209.png\"},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/educelec.com\/#website\",\"url\":\"https:\/\/educelec.com\/\",\"name\":\"Exercices et Solutions en Electricit\u00e9, Electronique, Electrotechnique et Automatisme industrielle\",\"description\":\"WordPress\",\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/educelec.com\/?s={search_term_string}\"},\"query-input\":{\"@type\":\"PropertyValueSpecification\",\"valueRequired\":true,\"valueName\":\"search_term_string\"}}],\"inLanguage\":\"en-US\"}]}<\/script>\n<!-- \/ Yoast SEO plugin. -->","yoast_head_json":{"title":"moteurs synchrone - Exercices et Solutions en Electricit\u00e9, Electronique, Electrotechnique et Automatisme industrielle","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/educelec.com\/en\/mot-synch\/","og_locale":"en_US","og_type":"article","og_title":"moteurs synchrone - Exercices et Solutions en Electricit\u00e9, Electronique, Electrotechnique et Automatisme industrielle","og_description":"Accueil\u00a0&gt; Electrotech &gt; Moteurs synchronesExercices corrig\u00e9s Etude des moteurs synchrones Exercice N\u00b0 1 Un alternateur triphas\u00e9 \u00e9toile \u00e0 une tension (entre phases) U = 660 V et d\u00e9bite un courant de 500 A sous un cos j = 0,8 (inductif) \u00e0 la fr\u00e9quence f = 50 Hz. 1) Calculer la puissance apparente, la puissance active [&hellip;]","og_url":"https:\/\/educelec.com\/en\/mot-synch\/","og_site_name":"Exercices et Solutions en Electricit\u00e9, Electronique, Electrotechnique et Automatisme industrielle","article_modified_time":"2026-01-09T10:06:35+00:00","og_image":[{"width":288,"height":200,"url":"https:\/\/educelec.com\/wp-content\/uploads\/2025\/12\/chap-e1766263997563.png","type":"image\/png"}],"twitter_misc":{"Est. reading time":"9 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"WebPage","@id":"https:\/\/educelec.com\/mot-synch\/","url":"https:\/\/educelec.com\/mot-synch\/","name":"moteurs synchrone - Exercices et Solutions en Electricit\u00e9, Electronique, Electrotechnique et Automatisme industrielle","isPartOf":{"@id":"https:\/\/educelec.com\/#website"},"primaryImageOfPage":{"@id":"https:\/\/educelec.com\/mot-synch\/#primaryimage"},"image":{"@id":"https:\/\/educelec.com\/mot-synch\/#primaryimage"},"thumbnailUrl":"https:\/\/educelec.com\/wp-content\/uploads\/2025\/12\/chap-300x209.png","datePublished":"2022-12-27T12:48:26+00:00","dateModified":"2026-01-09T10:06:35+00:00","inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/educelec.com\/mot-synch\/"]}]},{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/educelec.com\/mot-synch\/#primaryimage","url":"https:\/\/educelec.com\/wp-content\/uploads\/2025\/12\/chap-300x209.png","contentUrl":"https:\/\/educelec.com\/wp-content\/uploads\/2025\/12\/chap-300x209.png"},{"@type":"WebSite","@id":"https:\/\/educelec.com\/#website","url":"https:\/\/educelec.com\/","name":"Exercices et Solutions en Electricit\u00e9, Electronique, Electrotechnique et Automatisme industrielle","description":"WordPress","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/educelec.com\/?s={search_term_string}"},"query-input":{"@type":"PropertyValueSpecification","valueRequired":true,"valueName":"search_term_string"}}],"inLanguage":"en-US"}]}},"_links":{"self":[{"href":"https:\/\/educelec.com\/en\/wp-json\/wp\/v2\/pages\/214388","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/educelec.com\/en\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/educelec.com\/en\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/educelec.com\/en\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/educelec.com\/en\/wp-json\/wp\/v2\/comments?post=214388"}],"version-history":[{"count":28,"href":"https:\/\/educelec.com\/en\/wp-json\/wp\/v2\/pages\/214388\/revisions"}],"predecessor-version":[{"id":217470,"href":"https:\/\/educelec.com\/en\/wp-json\/wp\/v2\/pages\/214388\/revisions\/217470"}],"wp:attachment":[{"href":"https:\/\/educelec.com\/en\/wp-json\/wp\/v2\/media?parent=214388"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}