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answers a) we know that efficiency A=(" Past ")/(" pin ")xx100%-(1) pout =(7.5h^(5))(446(c)/(np)) (t){:[" Pout "=(7.57^(6))(746(co)/(nP))],[=5595N],[" PIN "=" VTII "],[=(120)Nxx(58)A=6960W],[" From (1) "=>eta=(5595 omega)/(6960 omega)xx100%=80.4%]:}b) Now, if armature urent is 35Athen input power to the manitor will be" Pin "=VI_(1)=120 xx35=4200 omega4 internal generate voltage is{:[epsilon_(A_(2))=V_(T)-I_(A)(R_(A)-R_(S))],[=120-35(0.25+0.16)],[=107.4V]:}internal generated voltage rated cate is{:[epsilon_(A_(2))=V_(T)-I_(A)(R_(A)-R_(S))=120-58(0.2+0.16)],[=99.1V]:}then final speed is{:[(epsilonA_(2))/(EA_(1))=(KP_(2)omega_(2))/(Kphi_(2)omega_(2))=(EA_(0)2n_(2))/(EA_(0)*n_(1))],[=>n_(2)=(EA_(2))/(EA_(1))*((A_(0)))/(EA_(0),2)","n_(1)quad(epsilon AO_(2))/(CO_(0))", is "sameas (P_(2))/(D_(1))],[(107.4)/(99.1)xx(134)/(115)xx105001min],[=$326r//min],[1326" rimin "]:}   Power converted from electrical to mechanical form;s{:[rho" conv "=E_(A)I_(A)],[=107.4 xx35=3759omega]:}as core 1055s in moter are 200wthen mechanical losses Pmech =((n_(2))/(n_(1)))^(3)240{:[=((1326)/(1050))^(3)xx2400],[=483 omega],[" oubutpower pout "=P_("con ")v-P_("mech ")=int cor],[=375 ... See the full answer