LEP (MODELO DE PRODUCCIÓN) CON FALTANTE
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| FUENTE: NORIS MEDINA |
Antes de empezar la demostración relacionemos los triángulos
t1+t4=Q/R à Q=R(t1+t4)
IMAX= t1(R-d)
t1=(R-d)/IMAX
t2= Imax/d
t3= S/d
t4= S/R-d
Entonces
t1(R-d)=t2d à Imax
t2(R-d)=t3d à S
Imax + S=Q
Imax + S=(t1+t4)(R-d)
Imax + S= (t1+t4)(r-d)
(t1+t4)(R-d)= (t1+t4)(r-d)
t1 + t4 = (Imax/R-d) + (s/R-d)
t1+t4= (Imax + s)/(r-d)
t1+t4=Q/R
Q/R= (Imax/R-d) +(Imax/d)
t1+t2 = Imax [(1/R-d) + (R/d)] à d + R –d / (R-d)d à R/(R-d)d
t1+t2= Imax (R/(r-d)] Reemplazo Imax
t1+t2=[(R-d)Q – SR]/(R-d)d
t3+t4= (S/d) + (S/R-d) à [(R-D)S + Sd]/ d(R-d)
t3 + t4 = S[R/d(R-d)]
DEMOSTRACION LED PARA HALLAR Q* Y S*
Tenemos que:
C´(Q,S)= Cu + Cop + Cmi (t1+t2)Imam/2 + Cf(t3+t4)S/2
Reemplazando:
C´(Q,S)= Cu + Cop + (1/2) Cmi {[(R-D)Q/R] -S } [R/(R-D)D] {[(R-D)Q/R]-S } + (CFS2 / 2)[R/(D(R-D)]
C´(Q,S)= Cu + Cop + (Cmi/2) {[(R-D)/R]Q – S }2 [R/(R-D)D] + (CFS2 / 2)[R/(D(R-D)]
N [C´(Q,S)]= (D/Q) CuQ + (D/Q)Cop + (D/Q) (Cmi/2) {[(R-D)/R]Q – S }2 [R/(R-D)D] + (D/Q) (CFS2 / 2)[R/(D(R-D)]
N [C´(Q,S)]= CuD + (D/Q)Cop + (Cmi/2Q) {[(R-D)/R]Q – S }2 [R/(R-D)] + (CFS2/2Q)[R/(R-D)]
Resolvemos lo siguiente y reemplazamos en la ecuación (R/R-D) = (R/R)-(R/D) = 1-(R/D)
(R-D)/R = (R/R) – (D/R) = 1 – (D/R)
CTA (Q,S) = CuD + (D/Q)Cop + (Cmi/2Q) {[1-(D/R)]Q – S}2 [1-(R/D)] + (CFS2/2Q) [1-(R/D)]
CTA (Q,S) = CuD + (D/Q)Cop + (Cmi/2Q) [(1-(D/R))2Q2 – 2SQ(1-(D/R)) + S2 ] [1-(R/D)] + (CFS2/2Q) [1-(R/D)]
CTA (Q,S) = CuD + (D/Q)Cop + { (1/2Q)[1-(D/R)]2Q2 – (2SQ/2Q)[1-(D/R)] + (S2/2Q) } Cmi[1-(R/D)] + (CFS2/2Q) [1-(R/D)]
CTA (Q,S) = CuD + (D/Q)Cop + { (Q/2)[1-(D/R)]2 – (S)[1-(D/R)] + (S2/2Q) } Cmi[1-(R/D)] + (CFS2/2Q) [1-(R/D)]
CTA (Q,S) = CuD + (D/Q)Cop + { (Q/2)[1-(D/R)]2 – (S)[1-(D/R)] + (S2/2Q) } Cmi[1/( 1 – (D/R))]+ (CFS2/2Q)[1/( 1 – (D/R))]
CTA (Q,S) = CuD + (D/Q)Cop + { (Q/2)[1-(D/R)]2 – (S)[1-(D/R)] + (S2/2Q) } {Cmi/[ 1 – (D/R) }+ (CFS2/2Q( 1 – (D/R)]
CTA (Q,S) = CuD + (D/Q)Cop + { (Q[1-(D/R)]2Cmi/2[ 1 – (D/R)]} – {SCmi[1-(D/R)]/ [ 1 – (D/R)]} + (S2Cmi/2Q[ 1 – (D/R)] + (CFS2/2Q( 1 – (D/R)]
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