Programacion

Pruebas | Captador #1 | Captador #2 |
Tal Cual | 21.61∠282° | 20.5∠65⁰ |
Wp115gr∠250⁰ | 25.69∠282⁰ | 26.25∠55⁰ |
Wp215gr∠300⁰ |17.605∠300° | 7.3∠350⁰ |

Procederemos a calcular los pesos de prueba
A=N2-NWp1= 25.69 ∠ 250-21.61∠282° 15∠250 °
A==0.4909-.7634i= .9076∠302.7°A= .9076∠302.7°
∝A= F2- F1Wp1= 26.25∠55°-20.5∠65° 15 ∠250°=
∝A=-.3289+.338i=.4686∠134.6°
∝A=.4686∠134.6°∝=∝AA=.4686∠134.6°.9076∠302.7°=0.5163∠191.9°
B=F3-FWp2=7.3∠ 35°-20.5∠65°15 ∠ 300°=0.7415-0.6347i=0.9760∠319.4°
B=0.9760∠319.4

βB=N3-NWp2=17.605 ∠300°-21.61∠282° 15 ∠300°=βB=4.3095+5.8914i15∠300=0.4866∠113.8°
βB=0.4866∠113.8°

β=βBB= 0.4860 ∠113.8°0.9760 ∠319.4°=0.4986∠154.4°
β=0.4986∠154.4°WC1=βF-NA(1-∝β)=.4986∠154.4°20.5∠65°-(21.61∠282°)(.9076∠302.7°)[1∠0°-0.5153∠191.9°.4986∠154.4°]=
WC1=-28.0639-1.4105i=28.1∠182.9°
WC1=28.1∠182.9°WC2=∝N-FB(1-∝β)=0.5163∠191.9°21.61∠282°-(20.5∠65°)0.9760∠319.4°(1∠0°-[(0.5153∠191.9°)(.4986∠154.4)WC2=11.1572∠113.9°-20.5∠65°0.9760∠319.4°(1∠0°-.2569∠346.3°)=
WC2=15.6206∠-147.5607°.9760∠319.4°(.75285∠4.6321)=
WC2=21.3∠248.3°

WC1=28.1∠182.9°
WC2=21.3∠248.3°

Colocando los nuevos pesos encontradosWC1=28.1∠182.9 y WC2=21.3∠248.3° se tiene como resultado lo siguiente

Xr1= 3.53∠165°
Xr2=6.41∠95°

1. Calcular la eficiencia de balanceo%eficiencia1=N-xrNx100=21.61-3.5321.65x100=83.66%
%eficiencia2=F-xrFx100=20.5-6.4120.5x100=68.73%