# Angle difference of source/load for real power flow

simulate this circuit – Schematic created using CircuitLab

'Because Vs has an angle of 0, θ must be negative in order for real power (P) to flow'

Why is the above statement true? Does real power only flow from large to small angled voltage phasors?

Thanks

• pf defines the real power which depends on matching the phase angle received after inductive source. The capacitor attempts to balance the inductive load to match the conjugate impedance of the supply which may result in either polarity of Vload phase. Both L and C store energy so balancing the net phase of V/I results in max real/total VAR = pf. Conjugate Impedance means ZL2//ZC = -ZL1 is matched for max pf. From source. Jun 11, 2021 at 14:06
• Hi @TonyStewartEE75, does that relate to his question? Jun 11, 2021 at 14:50
• @relayman357. Phase angle must be negative for real power in load? ? Yes that’s what I answered (No) however supply power is always negative. You can also control power by changing source frequency to lead or lag phase Jun 11, 2021 at 14:58

Real power flow (Watts) is dominantly controlled by angular differences, flowing from leading angles toward lagging as you surmised. This assumes a system where all bus voltages are reasonably near 1 p.u. and relatively small angles. Likewise, reactive power (var) is dominantly controlled by voltage magnitudes, flowing from higher voltage toward lower.

Referring to the circuit above (with $$\X=X_G+X_S\$$) the power flow equations are,
$$P = \frac{EVsin\theta}{X}$$
$$Q = \frac{E^2-EVcos\theta}{X}$$
So, looking at $$\P\$$ we can see that in a system where the bus voltages are near 1 p.u. varying the angle $$\\theta\$$ will control real power flow.
Looking at $$\Q\$$ we can see that for normal angular differences (few degrees) between buses the reactive power flow is controlled by varying the bus voltages.