powerthree phase
When calculating the active power in the 3 phase Star, and Delta formations, there are two different currents used depending on the layout. I have shown my working below, with what I have been taught.
However, I am wondering why when calculating the Active power for the star connected formation we use \$I_{phase}\$ for the calculation. Whereas in the Delta connected formation we use \$I_{line}\$. What is the reason for this?
No guarantees as to correctness, but this shows the general method.
By convention, phase rotation is anti-clockwise. An inductive load will make the current lag behind the voltage. (Remember, inductors resist changes in voltage dv/dt.)
Diagrams below are not to scale.
I don't see any way to do this. You need the power factor of the individual phases and that's not available. The equation is incorrect.
\$P_{actL1} = P_{appL1}\cdot\cos\varphi_1\$
\$P_{act,tot}/P_{app,tot}= \$
\$= (U\cdot I_{l1}\cdot \cos\varphi_1 + U\cdot I_{l2}\cdot \cos\varphi_2 + U\cdot I_{l3}\cdot \cos\varphi_3)/(I_{l1}\cdot U + I_{l2}\cdot U + I_{l3}\cdot U)\$
\$= ( I_{l1}\cdot \cos\varphi_1 + I_{l2}\cdot \cos\varphi_2 + I_{l3}\cdot \cos\varphi_3)/(I_{l1} + I_{l2} + I_{l3})\$
and that is not equal to \$\cos\varphi_1\$.
Best Answer
You are using \$I_{LINE}\$ in both.
There is no dot, so the current that flows in phase must flow in transmission line. So in a star or wye, \$I_{LINE}\ =\ I_{PHASE}\$ (and \$V_{LINE}\ =\ √ (3)\ V_{PHASE}\$). You show this in your answer.
In a delta, \$I_{LINE}\ =\ √ (3)\ I_{PHASE}\$ (and \$V_{LINE}\ =\ V_{PHASE}\$). A component of two phase currents make up the line current. There is a dot.
$$P_T = √ (3)\ V_{LINE}\ I_{LINE}\ cos\ θ $$ $$P_T = 3\ V_{PHASE}\ I_{PHASE}\ cos\ θ $$
So in your first answer, \$V_{L-L}\$ which is line voltage \$V_{LINE}\ =\ 415V\$, which means a \$V_{PHASE}\ =\ 415V / √ (3)\ =\ 239.6V\$. $$ I_{PHASE}\ =\ V_{PHASE} / Z\ = 239.6V / 10Ω = 24.0A$$ $$ P\ = I_{PHASE}^2\ R = (24.0A)^2\ x\ 8Ω\ =\ 4.59kW$$ $$ P_T\ = 3\ P = 3\ x\ 4.59kW\ =\ 13.8kW$$