for the pwoer calculation of a 3-phase user, you need following formula:
P=3D =E2=88=ABI sin (wt) * U sin (wt)
Integral calculation: P =3D =E2=88=AB U sin(=CF=89 t) I sin(=CF=89 t) dt Substitution: x =3D =CF=89 t dx =3D =CF=89 dt =EF=83=A0 dx / =CF=89 =3D dt P =3D U * I =E2=88=AB sin=C2=B2( =CF=89 t) dt
P=3D (U * I)/ =CF=89 =E2=88=AB sin=C2=B2 x dx sin=C2=B2 x is equal to (1-cos2x)/2 so: P=3D(U * I)/ =CF=89 =E2=88=AB ((1-cos2x)/2)dx =3D> P=3D(U * I)/ (2*=CF=89)= (=E2=88=AB 1 dx - =E2=88=AB
cos2x dx) solution: P=3D(U * I)/(2*=CF=89) (x-(sin2x)/2) + C so P=3D(U * I)/(2*=CF=89) (=CF=89 t -(sin2 =CF=89 t)/2) + C if x =3D =CF=89 t= is included
Now, the exact formula is: P=3DUI=E2=88=9A3
Can anybody help my working further? I don't know anymore.
What are the integral limits? Any error in mixiung phase powers and line power? Is the formula to use more difficult?(taking into account the 120=C2=B0 phase switch? =2E..=20
all help is welcome