**Blondel's Two Reaction Theory (Theory of Salient Pole Machine) : part 3**In the phasor diagram shown in the Fig. 4, the angles Ψ and δ are not known, through V

_{t}, I_{a}and Φ values are known. Hence the location of E_{f }is also unknown. The components of I_{a}, I_{d}and I_{q}can not be determined which are required to sketch the phasor diagram.Fig. 4 |

Let us find out some geometrical relationships between the various quantities which are involved in the phasor diagram. For this, let us draw the phasor diagram including all the components in detail.

We know from the phasor diagram shown in the Fig. 4 that,

I

_{d}= I_{a}sin Ψ ............. (4) I

_{q}= I_{a}cos Ψ ..............(5) cosΨ = I

_{q}/I_{a}...............(6) The drop I

_{a}R_{a}has two components which are, I

_{d}R_{d}= drop due to R_{a}in phase with I_{d} I

_{q}R_{a}= drop due to R_{a}in phase with I_{q} The I

_{d}X_{d}and I_{q}R_{q}can be drawn leading I_{d}and I_{q}by 90^{o }respectively. The detail phasor diagram is shown in the Fig. 5.Fig. 5 Phasor diagram for lagging p.f. |

In the phasor diagram,

OF = E

_{f } OG = V

_{t} GH = I

_{d}R_{a}and H_{A}= I_{q}R_{a} GA = I

_{a}R_{a} AE = I

_{d}X_{d}and EF = I_{q}X_{a} Now DAC is drawn perpendicular to the current phasor I

_{a}and CB is drawn perpendicular to AE. The triangle ABC is right angle triangle,

But from equations (6), cosΨ = I

_{q}/I_{a}_{ }Thus point C can be located. Hence the direction of E

_{f}is also known.

Now triangle ODC is also right angle triangle,

Now OD = OI + ID = V

_{t }cos Φ + I_{a}R_{a} and CD = AC + AD = I

_{a}X_{q}+ V_{t }sinΦ As I

_{a}X_{q}is known, the angle Ψ can be calculated from equation (10). As Φ is known we can write, δ = Ψ - Φ for lagging p.f.

Hence magnitude of E

_{f}can be obtained by using equation (11).Note : In the above relations, Φ is taken positive for lagging p.f. For leading p.f., Φ must be taken negative.

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