{"id":4144,"date":"2026-04-22T11:27:03","date_gmt":"2026-04-22T09:27:03","guid":{"rendered":"https:\/\/sti2tech.com\/?page_id=4144"},"modified":"2026-04-22T11:32:44","modified_gmt":"2026-04-22T09:32:44","slug":"bts-die-synthese-n1-2","status":"publish","type":"page","link":"https:\/\/sti2tech.com\/?page_id=4144","title":{"rendered":"BTS – DIE – Synth\u00e8se n\u00b02"},"content":{"rendered":"\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t\t
\n\t\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t
\n\t\t\t\t\t\t\t\t\t

Synth\u00e8se n\u00b02 : Bilan de puissance<\/h2>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t\t
\n\t\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t
\n\t\t\t\t\t\t\t\t\t

Objectifs :\u00a0<\/h3>
  • Etablir<\/strong> un bilan de puissance<\/li>
  • Compenser<\/strong>\u00a0l’\u00e9nergie r\u00e9active<\/li>
  • D\u00e9terminer <\/strong>la puissance d\u00e9formante<\/li>
  • R\u00e9aliser<\/strong> des calculs avec rigueur<\/li><\/ul>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t\t
    \n\t\t\t
    \n\t\t\t\t\t\t
    \n\t\t\t\t
    \n\t\t\t\t\t\t\t\t\t

    Partie 1 – Les machines \u00e9lectriques<\/strong><\/h4>
    Calcul des Puissances en monophas\u00e9<\/strong><\/h5>
    • Un moteur monophas\u00e9 est aliment\u00e9 sous une tension , U<\/span>=<\/span><\/span>230<\/span>V<\/span><\/span><\/span><\/span><\/span> et absorbe un courant <\/span>I<\/span>=<\/span><\/span>5<\/span>A\u00a0<\/span><\/span><\/span><\/span><\/span>cos<\/span>(<\/span>\u03d5<\/span>)<\/span>=<\/span><\/span>0.8<\/span><\/span><\/span><\/span>.<\/span>
      <\/mi><\/mrow><\/semantics><\/span><\/span><\/span><\/span>
      1. Calculer la puissance active P.<\/li>
      2. Calculer la puissance r\u00e9active Q.<\/li>
      3. Calculer la puissance apparente S.<\/li><\/ol><\/li><\/ul>
        Calcul des Puissances en Triphas\u00e9<\/strong><\/h5>
        • Un atelier est aliment\u00e9 en triphas\u00e9 400<\/span>V<\/span><\/span><\/span><\/span><\/span> et absorbe un courant \u00a0I<\/span>=<\/span><\/span>10<\/span>A <\/span><\/span><\/span><\/span><\/span>cos<\/span>(<\/span>\u03d5<\/span>)<\/span>=<\/span><\/span>0.75<\/span><\/span><\/span><\/span>.<\/span><\/mn><\/mrow><\/semantics><\/span><\/span><\/span><\/span>
          1. Calculer la puissance active P<\/span><\/span><\/span><\/span>.<\/li>
          2. Calculer la puissance r\u00e9active Q<\/span><\/span><\/span><\/span>.<\/li>
          3. Calculer la puissance apparente S<\/span><\/span><\/span><\/span>.<\/li>
          4. Proposer une solution pour am\u00e9liorer le facteur de puissance.<\/li><\/ol><\/li><\/ul>
            Calcul de la puissance d\u00e9formante<\/strong><\/h5>

            • Pour le m\u00eame syst\u00e8me, les puissances mesur\u00e9es sont :\u00a0Puissance active\u00a0<\/span>P<\/span>=<\/span><\/span>5<\/span>kW,\u00a0<\/span><\/span><\/span>Puissance r\u00e9active <\/span>Q<\/span>=<\/span><\/span>3<\/span>kvar &\u00a0<\/span><\/span><\/span><\/span><\/span>Puissance apparente\u00a0<\/span>S<\/span>=<\/span><\/span>6.5<\/span>kVA<\/span><\/span><\/span><\/p>

              1. Calculer la <\/span>puissance d\u00e9formante D<\/span><\/span><\/span><\/span><\/strong>.<\/span><\/p><\/li>

              2. Expliquer l’origine de cette puissance d\u00e9formante et son impact sur l’installation.<\/li><\/ol><\/li><\/ul>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t
                \n\t\t\t
                \n\t\t\t\t\t\t
                \n\t\t\t\t
                \n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"\"\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
                \n\t\t\t\t\t\t
                \n\t\t\t\t\t
                \n\t\t\t
                \n\t\t\t\t\t\t
                \n\t\t\t\t
                \n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"\"\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t
                \n\t\t\t
                \n\t\t\t\t\t\t
                \n\t\t\t\t
                \n\t\t\t\t\t\t\t\t\t

                Partie II – La compensation d’\u00e9nergie r\u00e9active<\/h4>
                • Une installation triphas\u00e9e absorbe une puissance active\u00a0P<\/span>=<\/span><\/span>20<\/span>kW<\/span><\/span><\/span><\/span><\/span>\u00a0et une puissance r\u00e9active\u00a0Q<\/span>=<\/span><\/span>15<\/span>kvar<\/span><\/span><\/span><\/span><\/span>.<\/span>
                  1. Calculer le facteur de puissance initial.<\/li>
                  2. D\u00e9terminer la puissance r\u00e9active Q<\/span>c<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> \u00e0 compenser pour obtenir un facteur de puissance de <\/span>0.95<\/span><\/span><\/span><\/span>.<\/span><\/li>
                  3. D\u00e9terminer la valeur des condensateurs<\/li><\/ol><\/li><\/ul>

                    Partie III – Calcul du THD<\/h4>
                    • Un syst\u00e8me triphas\u00e9 est aliment\u00e9 par une tension fondamentale\u00a0U<\/span>1<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>400<\/span>V<\/span><\/span><\/span><\/span> et un courant fondamental\u00a0<\/span><\/span><\/span><\/span>I<\/span>1<\/span><\/span><\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>10<\/span>A<\/span><\/span><\/span><\/span>. <\/span><\/li>
                    • Les mesures suivantes ont \u00e9t\u00e9 relev\u00e9es pour les harmoniques :<\/span><\/li><\/ul>
                      Rang h<\/th>Tension Uh (V)<\/th>Courant Ih (A)<\/th><\/tr><\/thead>
                      3<\/td>15<\/td>2<\/td><\/tr>
                      5<\/td>8<\/td>1<\/td><\/tr>
                      7<\/td>5<\/td>0.5<\/td><\/tr><\/tbody><\/table>
                        1. Calculer le THD en tension<\/strong>.<\/li>
                        2. Calculer le THD en courant<\/strong>.<\/li>
                        3. Analyser les r\u00e9sultats : le syst\u00e8me respecte-t-il les normes :
                          • THD < 8% pour le courant<\/li>
                          • THD < 5% pour la tension<\/li><\/ul><\/li><\/ol><\/li><\/ol>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<\/div>\n\t\t","protected":false},"excerpt":{"rendered":"

                            Synth\u00e8se n\u00b02 : Bilan de puissance Objectifs :\u00a0 Etablir un bilan de puissance Compenser\u00a0l’\u00e9nergie r\u00e9active D\u00e9terminer la puissance d\u00e9formante R\u00e9aliser des calculs avec rigueur Partie 1 – Les machines \u00e9lectriques Calcul des Puissances en monophas\u00e9 Un moteur monophas\u00e9 est aliment\u00e9 sous une tension , U=230V et absorbe un courant I=5A\u00a0cos(\u03d5)=0.8. Calculer la puissance active P….<\/p>\n","protected":false},"author":4,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"_kad_post_transparent":"","_kad_post_title":"hide","_kad_post_layout":"fullwidth","_kad_post_sidebar_id":"","_kad_post_content_style":"unboxed","_kad_post_vertical_padding":"hide","_kad_post_feature":"hide","_kad_post_feature_position":"","_kad_post_header":false,"_kad_post_footer":false,"_kad_post_classname":"","footnotes":""},"folder":[22],"class_list":["post-4144","page","type-page","status-publish","hentry"],"_links":{"self":[{"href":"https:\/\/sti2tech.com\/index.php?rest_route=\/wp\/v2\/pages\/4144","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/sti2tech.com\/index.php?rest_route=\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/sti2tech.com\/index.php?rest_route=\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/sti2tech.com\/index.php?rest_route=\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/sti2tech.com\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=4144"}],"version-history":[{"count":5,"href":"https:\/\/sti2tech.com\/index.php?rest_route=\/wp\/v2\/pages\/4144\/revisions"}],"predecessor-version":[{"id":4277,"href":"https:\/\/sti2tech.com\/index.php?rest_route=\/wp\/v2\/pages\/4144\/revisions\/4277"}],"wp:attachment":[{"href":"https:\/\/sti2tech.com\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=4144"}],"wp:term":[{"taxonomy":"folder","embeddable":true,"href":"https:\/\/sti2tech.com\/index.php?rest_route=%2Fwp%2Fv2%2Ffolder&post=4144"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}