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3-phase calculation

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

Reply to
pietgeeroms
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P= ?I sin (wt) * U sin (wt)

Integral calculation: P = ? U sin(? t) I sin(? t) dt Substitution: x = ? t dx = ? dt ? dx / ? = dt P = U * I ? sin²( ? t) dt

P= (U * I)/ ? ? sin² x dx sin² x is equal to (1-cos2x)/2 so: P=(U * I)/ ? ? ((1-cos2x)/2)dx => P=(U * I)/ (2*?) (? 1 dx - ?

cos2x dx) solution: P=(U * I)/(2*?) (x-(sin2x)/2) + C so P=(U * I)/(2*?) (? t -(sin2 ? t)/2) + C if x = ? t is included

Now, the exact formula is: P=UI?3

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° phase switch? ...

all help is welcome

You are trying to do the calculation of average power for a single phase. Every text handles this. Hint: the power you are wanting is the average over a cycle. Use symmetry and integrate over a half cycle (0 to pi). What is the average of cos(2wt) over this period- by inspection).

Actually, you needn't go through all the integration above but it is apparently something you missed earlier so it is a good exercise.

Once you get the result. you have the power for one phase (ignoring power factor) based on phase voltage and phase current. How many phases? Your "exact" result is based on line values of voltage and current (how are these related to phase values?) and assumes a resistive load.

Reply to
Don Kelly

ur integration is correct.but since we are interested in POWER and not ENERGY,you have to calculate the AVERAGE POWER.you have calculated the energy and not power. since sinusoidal signals are periodic,avg. power over 1 cycle and 10 or

100 or any no: of cycles is same. hence replace the above by limits from 0 to 2PI,and divide by the same.you will get the answer(u have to!!).
Reply to
gautham

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