Transformer, power factor of two transformers on load “EX11-V2/2”.

Determine the upstream power factor of two single-phase transformers on load, without calculating active power (P), reactive power (Q) or apparent power (S).
Method:
The power factor is FP=COS(phi°)=COS(phiV° - phiI°) , where phi° = (phiV° - phiI°) and phiV° is the phase angle of the voltage and phiI° that of the current. phi° therefore represents the phase shift of the current with respect to the voltage. When we speak of a lagging or leading FP power factor, it's always the current that is referred to in relation to the voltage.
Even if the readings taken in situ are only RMS values, without maximum amplitudes or phase angles ( Vs1=123[KVeff] , Is1=433[Aeff] and Vs2=18[KVeff] , Is2=810[Aeff] ) , we can deduce the phase angles of currents and voltages via power factors. Only the source voltage is given in phase-free amplitude Vp=321.026 [KV].
In this case, the load Z1 on the secondary side of the first transformer generates a power factor FP1=0.8 lagging, so the current Is1 lags behind the voltage Vs1, while Z2 on the secondary side of the second transformer generates a power factor FP2=0.1 leading, so the current Is2 leads the voltage Vs2.
1) Currents and voltages are expressed in modulus and phase, or phasors. RMS values are therefore multiplied by the root of 2 .
2) The phase angle of the source's Vp is unknown, but there's nothing to stop us from setting the source's phase angle to any value, such as phase(0°) for example, and then calculating the other phase angles of the currents and voltages in relation to Vp=321.026.phase(0°) [KV].
3) This gives Vp=312.026.phase(0°) [KV].
Is1=433.2^(1/2).phase(phiIs1°) with FP1=0.8 late, so Is1 current is late.
Is2=810.2^(1/2).phase(phiIs2°) with FP2=0.1 early, so the Is2 current is early.
4) The aim is to find the phase shift between the total current and the total voltage upstream of the two transformers, and then deduce the total power factor.
5) Although the transformation ratios n1 and n2 of the transformers are unknown, n1 and n2 are used as intermediate constants to calculate the total current Ip, given that Ip=Ip1+Ip2.
4) Calculation of phiIs1° and phiIs2° :
phiIs1° = ARCcos(0.8) = 36.87° .
Now COS(-36.87°)=COS(36.87°)=0.8 because the cosine function is even. So we don't know whether FP1=0.8 comes from (-36.87°) or (36.87°), but the fact that FP1 is lagging means that the angle is POSITIVE, as in an inductive circuit where the current lags behind the voltage. The solution for phiIs1° is: phiIs1°= +36.87°.
The same reasoning applies to phiIs2°,
FP2=ARCcos(0.1)=84.261° and COS(84.261°)=COS(-84.261°)=0.1 or FP2 is leading, so it's the opposite of the lagging case, i.e. the angle is negative, so the solution is: phiIs2°= -84.261° .
5) Ip = Ip1 + Ip2 = 433.2^(1/2).phase(36.87°) + 810.2^(1/2).phase(-84.261°). Finally, we obtain the total upstream current: Ip=295,265.phase(-158,398°).
6) Ip and Vp are therefore out of phase by -158.398° . The total power factor is therefore FP=COS(-158.398°) = -0.93 .
7) Note: when a SYNCHRONOUS motor operates at a certain speed where the so-called critical rotor excitation current is reached, it generates reactive power (Q) to the network, so the power factor is much lower than unity. In this case, the load Z2 may be a synchronous motor or other. FP= -0.93 , negative, which implies that reactive power is generated/produced by the load located downstream of the transformers. Synchronous motors are large motors.