conversion_11_10_23.py

Created by famille-bvc

Created on October 10, 2023

3.13 KB


Expressions temporelles et 
complexes des tensions simples

v1(t)=Vsqrt(2)sin(wt)
v2(t)=Vsqrt(2)sin(wt- 2pi/3)
v3(t)=Vsqrt(2)sin(wt+ 2pi/3)

V1 = V
--
V2 = a^2 V = Ve^-j120 deg
--
V3 = aV = Ve^j120 deg
--

///////////////////////////
représentation vectorielle 
des tensions simples et 
composées (U vert V noir)
(V plus petit que U)

          V3    _>U12
  U31<_   ^   _/
       \_ \  /
          \\ ____> V1
           /|
          / |
      V2 v  |
            vU23
            
///////////////////////////
expressions complexes et 
temporelles des tensions 
composées

u12(t)=Usqrt(2)sin(wt- pi/6)
u23(t)=Usqrt(2)sin(wt- pi/2)
u31(t)=Usqrt(2)sin(wt+ 5pi/6)

U12 = Ue^j30 deg
---
U23 = Ue^-j90 deg
---
U31 = Ue^j150 deg
---

///////////////////////////
schéma équivalent monophasé
de linstallation

     IT
     --
_____>_____________________
^         |     |     |
|V        R ZM1 R ZM2 R ZRL
|-        | --- | --- | ---
---------------------------

/\/\/\/\/\/\/\/\/\/\/\/\/\/\
\/\/\/\/\/\/\/\/\/\/\/\/\/\/

Reseau Tri 230/400V - 50Hz
alimente :
  1 moteur 3kw fact de 0.7
  1 moteur 2kw fact de 0.6
  1 charge 1kw fact de 0.5

cos(phi1)=0.7 
=> phi1=45.57 deg
=> tan(phi1) = 1.02
... a calculer

///////
puissances active, réactive
et apparente totales

PT = P1 + P2 + P3
   = 3k+2k+1k = 6000W

QT = Q1 + Q2 + Q3    
   = P1 tan phi1 + ...
   = 3060.61 + 2666.67
   + 1732.05 = 7459.33 VAR
   
ST = sqrt(PT^2 + QT^2)
   = sqrt(6000^2 + 7459,3^2)
   = 9572.96 VA
   
///////
facteur de puissance total

cos(PhiT) = PT/ST
= 6000/9572.96 = 0.63
=> PhiT = 51.19 deg
=> tan(PhiT) = 1.24

///////
courant total

IT = ST/3V = 9572.96/3*230
   = 13.87 A
   
=> IT = ITe^-jphiT 
   --
 = 13.87e^-j51.19 deg [A] 

///////
représentation vectorielle
des courants totaux de 
ligne => meme schéma qu'av
+au centre  > I3T
           /
          /
I2T <-----
          \
           \
            > I1T
            
angle PhiT entre V1 et I1T

///////
On utilise la méthode des
2 wattmètres pour mesurer
les puissances active et 
réactive totales

PT = P1 + P2
QT/sqrt(3) = P1 - P2

=> 

P1 = PT/2 + QT/2sqrt(3)
   = 5153.32W
P2 = PT/2 - QT/2sqrt(3)
   = 846.68W
   
///////
On désire relever le 
facteur de puissance à 
0.93 à l'aide de trois 
condos

=> cos(Phi'T)= 0.93
phi'T= 21.57 deg
tan(Phi'T)=0.4

puissance reactive fournie
par les 3 condos

QC=PT(tan(PhiT)-tan(Phi'T))
=6000(1.24-0.4)=5040VAR

///////
deduire valeur des condos
en etoile U -> V

C = Qc/3wU^2
=5040/ 3 x 314.16 x 400^2
=33.4 microF (uF)

///////
nouvelle valeur du courant
total

I'T = PT / 3Vcos(Phi'T)
= 6000 / 3x230x0.93
= 9.35 A

I'T = I'T e^-jPhi'T
---
    = 9.35e^-j21.57 deg [A]

I'T < IT => Diminution des
pertes en ligne

/\/\/\/\/\/\/\/\/\/\/\/\/\/\
\/\/\/\/\/\/\/\/\/\/\/\/\/\/

Fact de puissance moyen

Q/P = Sin(Phi)/Cos(Phi)
    = Tan(Phi)

avec P = Wa/T et Q = Wr/t

intensité du courant dans 
chaque fil

I = P/ sqrt(3)Ucos(Phi)

si on donne U0 et cos(phi0)
P0 = sqrt(3)U0I0Cos(Phi0)
Q0 = sqrt(3)U0I0Sin(Phi0)

Puissance active perdue
PL = P0 - P [W]
en % => 100xPL/P0
Puissance reactive perdue
QL = Q0 - Q [VAR]
Resistance r
r = PL/ 3I^2 [ohms]
Réactance Lw
Lw = QL/ 3I^2 [ohms]

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