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ISARE	
  


       (J1)1	
                              (J1)2	
                              (J1)3	
  

    λ	
        ⎛ ...
⎛ n1 ⎞
                                   ⎜ ⎟ = 15
                                   ⎝ n 2 ⎠1

                    ...
(J1)1	
  =	
  (Ip1)1	
                                                    ⎛ n 2 ⎞

       (J1)1	
                       ...
(J1)2	
  =	
  (Ip1)2	
  
                                                             ⎛ 2 ⎞ ⎛ n 2 ⎞          ⎛ 2 ⎞ ⎡...
(Ip1)3	
                                          ⎛ n 2 ⎞          ⎡1 ⎛ n ⎞ ⎤                   ⎛ n 2 ⎞
          ...
(Ip3)3	
                                                                                                                  ...
(J1)3	
  =	
  (Ip1)3	
  -­‐	
  (Ip3)3	
  
                                                                                ...
(J1)1	
  	
                                                     x
                                                        ...
5x
(J1)1	
  	
                                                     x
                                                     ...
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[EP] ETXERAKO LANA (2010/02/25) orain arte ikusitakoaren berrikuspena

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[EP] ETXERAKO LANA (2010/02/25) orain arte ikusitakoaren berrikuspena

  1. 1. ISARE   (J1)1   (J1)2   (J1)3   λ   ⎛ n1 ⎞ λ   ⎛ n1 ⎞ Δ   ⎛ n1 ⎞ ⎜ ⎟ = 15 ⎜ ⎟ = 15 ⎜ ⎟ = 30 ⎝ n 2 ⎠1 ⎝ n 2 ⎠ 2 ⎝ n 2 ⎠ 3 λ   λ   λ   € € € (IC)1  =  30A   (IC)2  =  90A   (IC)3  =  60A  
  2. 2. ⎛ n1 ⎞ ⎜ ⎟ = 15 ⎝ n 2 ⎠1 ⎛ n1 ⎞ ⎛ n1 ⎞ ⎛ n1 ⎞ ⎜ ⎟ = 15 ⇒ ⎜ ⎟ = ⎜ ⎟ € ⎝ n 2 ⎠ 2 ⎝ n 2 ⎠ 2 ⎝ n 2 ⎠1 ⎛ n1 ⎞ ⎛ n1 ⎞ ⎛ n1 ⎞ (V1)2  =  (V13)3     ⎜ ⎟ = 30 ⇒ ⎜ ⎟ = 2 ⋅ ⎜ ⎟ € ⎝ n 2 ⎠ 3 ⎝ n 2 ⎠ 3 ⎝ n 2 ⎠1 30o   (V13)€ 1   (IC)1=  30A   (IC)2=  90A    (IC)2  =  3(IC)1     (IC)3=  60A    (IC)3  =  2(IC)1    
  3. 3. (J1)1  =  (Ip1)1   ⎛ n 2 ⎞ (J1)1   ⎜ ⎟ ⋅ ( IC )1 = x ⎝ n1 ⎠1 € λ   ⎛ n ⎞ − ⎜ 2 ⎟ ⋅ ( IC )1 = −x ⎝ n1 ⎠1 ⎛ n1 ⎞ ⎜ ⎟ = 15 (Is1)1   ⎝ n 2 ⎠1 € λ   (IC)1   -­‐(IC)1   € (VL)1   (V32)1   (V12)1   (V13)1   (V23)1   (V21)1   (V31)1   (V32)1   (IC)1  =  30A   30   60   90   120   150   180   210   240   270   300   330   360  
  4. 4. (J1)2  =  (Ip1)2   ⎛ 2 ⎞ ⎛ n 2 ⎞ ⎛ 2 ⎞ ⎡⎛ n ⎞ ⎤ ⎡⎛ n ⎞ ⎤ ⎜ ⎟ ⋅ ⎜ ⎟ ⋅ ( IC ) 2 = ⎜ ⎟ ⋅ ⎢⎜ 2 ⎟ ⎥ ⋅ [ 3⋅ ( IC )1 ] = 2 ⋅ ⎢⎜ 2 ⎟ ⋅ ( IC )1⎥ = 2x (J1)2   ⎝ 3 ⎠ ⎝ n1 ⎠ 2 ⎝ 3 ⎠ ⎣⎝ n1 ⎠1⎦ ⎣⎝ n1 ⎠1 ⎦ € λ   ⎛ 1 ⎞ ⎛ n ⎞ ⎛ 1 ⎞ ⎡⎛ n ⎞ ⎤ ⎛ n ⎞ − ⎜ ⎟ ⋅⎜ 2 ⎟ ⋅ ( IC ) 2 = − ⎜ ⎟ ⋅⎢⎜ 2 ⎟ ⎥ ⋅ [ 3⋅ ( IC )1 ] = −⎜ 2 ⎟ ⋅ ( IC )1 = −x ⎝ 3 ⎠ ⎝ n1 ⎠ 2 ⎝ 3 ⎠ ⎣⎝ n1 ⎠1⎦ ⎝ n1 ⎠1 ⎛ n1 ⎞ ⎜ ⎟ = 15 ⎝ n 2 ⎠ 2 (Is1)2   € λ   (IC)2   € (VL)2   (V1)2   (V2)2   (V3)2   (IC)2  =  90A   30   60   90   120   150   180   210   240   270   300   330   360  
  5. 5. (Ip1)3   ⎛ n 2 ⎞ ⎡1 ⎛ n ⎞ ⎤ ⎛ n 2 ⎞ ⎜ ⎟ ⋅ ( IC ) 3 = ⎢ ⋅ ⎜ ⎟ ⎥ ⋅ [2 ⋅ ( IC )1 ] = ⎜ ⎟ ⋅ ( IC )1 = x 2 (J1)3   ⎝ n1 ⎠ 3 ⎣2 ⎝ n1 ⎠1⎦ ⎝ n1 ⎠1 € Δ   ⎛ n ⎞ − ⎜ 2 ⎟ ⋅ ( IC ) 3 = −x ⎝ n1 ⎠ 3 ⎛ n1 ⎞ (Is1)3   ⎜ ⎟ = 30 (IC)3   € ⎝ n 2 ⎠ 3 λ   -­‐(IC)3   € (VL)3   (V12)3   (V13)3   (V23)3   (V21)3   (V31)3   (V32)3   (IC)3  =  60A   30   60   90   120   150   180   210   240   270   300   330   360  
  6. 6. (Ip3)3   ⎛ n 2 ⎞ ⎜ ⎟ ⋅ ( IC ) 3 = x (J1)3   ⎝ n1 ⎠ 3 € Δ   ⎛ n ⎞ − ⎜ 2 ⎟ ⋅ ( IC ) 3 = −x ⎝ n1 ⎠ 3 ⎛ n1 ⎞ (Is3)3   ⎜ ⎟ = 30 € (IC)3   ⎝ n 2 ⎠ 3 λ   -­‐(IC)3   € (VL)3   (V12)3   (V13)3   (V23)3   (V21)3   (V31)3   (V32)3   (IC)3  =  60A   30   60   90   120   150   180   210   240   270   300   330   360  
  7. 7. (J1)3  =  (Ip1)3  -­‐  (Ip3)3   2x (J1)3   € x € Δ   ⎛ n1 ⎞ −2x −x ⎜ ⎟ = 30 € ⎝ n 2 ⎠ 3 (Ip1)3   λ   x € € € −x (Ip3)3   € x (IC)3  =  60A   € −x 30   60   90   120   150   180   210   240   270   300   330   360   €
  8. 8. (J1)1     x 1 ⎡ 2π 2 ⎤ 2 (J1 )1 = ⋅ ⎢2 ⋅ ⋅ x ⎥ = x ⋅ 2π ⎣ 3 ⎦ 3 € 2 ⎡⎛ n 2 ⎞ ⎤ 2 ⎡ 1 ⎤ 2 (J1 )1 = x ⋅ = ⎢⎜ ⎟ ⋅ ( IC )1⎥ ⋅ = ⎢ ⋅ 30⎥ ⋅ = 1.63A 3 ⎣⎝ n1 ⎠1 ⎦ 3 ⎣15 ⎦ 3 −x € € (J1)2     2x 1 ⎡ 2π 2 4π 2 ⎤ 1 ⎡2 4 ⎤ 1 ⎡12 ⎤ (J1 ) 2 = ⋅ ⎢1⋅ ⋅ (2x ) + 1⋅ ⋅ ( x ) ⎥ = x ⋅ ⋅ ⎢ ⋅ 4 + ⎥ = x ⋅ ⋅ 2π ⎣ 3 3 ⎦ 2 ⎣ 3 3 ⎦ 2 ⎢ 3 ⎥ ⎣ ⎦ € ⎡⎛ n ⎞ ⎤ ⎡ 1 ⎤ −x ( J1) 2 = x ⋅ 2 = ⎢⎜ 2 ⎟ ⋅ ( IC )1⎥ ⋅ 2 = ⎢ ⋅ 30⎥ ⋅ 2 = 2.83A ⎣⎝ n1 ⎠1 ⎦ ⎣15 ⎦ € € (J1)3     2x x € 1 ⎡ π 2 π 2 ⎤ 1 ⎡ 1 1 ⎤ 1 ⎡12 ⎤ (J1 ) 3 = ⋅ ⎢2 ⋅ ⋅ (2x ) + 4 ⋅ ⋅ ( x ) ⎥ = x ⋅ ⋅ ⎢2 ⋅ ⋅ 4 + 4 ⋅ ⎥ = x ⋅ ⋅ 2π ⎣ 3 3 ⎦ 2 ⎣ 3 3⎦ 2 ⎢ 3 ⎥ ⎣ ⎦ € ⎡⎛ n ⎞ ⎤ ⎡ 1 ⎤ (J1 ) 3 = x ⋅ 2 = ⎢⎜ 2 ⎟ ⋅ ( IC )1⎥ ⋅ 2 = ⎢ ⋅ 30⎥ ⋅ 2 = 2.83A ⎣⎝ n1 ⎠1 ⎦ ⎣15 ⎦ −x −2x € € 30   60   90   120   150   180   210   240   270   300   330   360   €
  9. 9. 5x (J1)1     x 4x ISARE     € € € −x € (J1)2     2x −2x −3x € € −4 x −x € € 30   60   90   120   150   180   210   240   270   300   330   360   € (J1)3     ISARE = 1 ⎡ π 2 π 2 π 2 π 2 ⎤ ⋅ ⎢4 ⋅ ⋅ ( 4 x ) + 2 ⋅ ⋅ (5x ) + 2 ⋅ ⋅ (2x ) + 2 ⋅ ⋅ ( 3x ) ⎥ 2x 2π ⎣ 6 6 6 6 ⎦ x 1 ⎡ π π π π ⎤ € = x⋅ ⋅ ⎢4 ⋅ ⋅16 + 2 ⋅ ⋅ 25 + 2 ⋅ ⋅ 4 + 2 ⋅ ⋅ 9⎥ 2π ⎣ 6 6 6 6 ⎦ € 1 ⎡64 50 8 18 ⎤ 1 ⎡140 ⎤ 70 = x⋅ ⋅ ⎢ + + + ⎥ = x ⋅ ⋅ ⎢ ⎥ = x ⋅ 2 ⎣ 6 6 6 6 ⎦ 2 ⎣ 6 ⎦ 6 −x −2x 35 ⎡⎛ n 2 ⎞ ⎤ 35 ⎡ 1 ⎤ 35 € ISARE = x ⋅ = ⎢⎜ ⎟ ⋅ ( IC )1⎥ ⋅ = ⎢ ⋅ 30⎥ ⋅ = 6.83A 3 ⎣⎝ n1 ⎠1 ⎦ 3 ⎣15 ⎦ 3 30   60   90   120   150   180   210   240   270   300   330   360   € €

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