As discussed in earlier section of a.c. distribution system we have take into account the power factor. This power factor can be either considered with respect to receiving end voltage or with respect to load voltage itself. Let us consider each case separately.

__1.1 Power Factors Referred to Receiving End Voltage__

Consider an A.C. distribution PQ having concentrated loads of I

_{1 }and I_{2 }tapped off at point Q and R respectively. This is shown in the Fig. 1.Fig. 1 |

Let voltage V

_{Q }which is the voltage at the receiving end be taken as reference vector. The power factors at R and Q are cosΦ_{1 }and cosΦ_{2 }with respect to V_{Q}and they are lagging.
Let, R

_{1 }= Resistance of section PR
X

_{1 }= Reactance of section PR
R

_{2 }= Resistance of section RQ
X

_{2 }= Reactance of section RQ
As shown in the Fig. 2. the receiving end voltage V

_{Q}is taken as reference vector. The currents I_{1 }and I_{2 }are lagging from V_{Q }by angles of Φ_{1}and Φ_{2 }respectively. The vector sum of I_{1 }and I_{2 }gives current I_{PR}. The drop is I_{2}R_{2 }in phase with I_{2 }while I_{2}X_{2}is leading by 90^{o}. The vector sum of V_{Q}, I_{2}R_{2}and I_{2}X_{2 }gives V_{R}. The drop I_{PR}R_{1}is in phase with current I_{PR }while I_{PR}X_{1}is leading by 90. The vector sum of V_{R}, I_{PR}R_{1}and I_{PR}X_{1}gives the sending end voltage.__1.2 Power Factors Referred to Respective Load Voltages__

In previous section we have considered the load power factors with respect to receiving end voltage. Here we will consider these power factors with respect to their respective load voltages. Now Φ

_{1}is the phase angle between V_{R}and I_{1 }while the angle Φ_{2}is the phase angle between V_{Q }and I_{2}.
Here again the receiving end voltage V

_{Q}is the reference phasor. The vector sum of I_{1}and I_{2}gives the current I_{PR}. The drop I_{2}R_{2 }is in phase with I_{2 }while I_{2}X_{2}is leading by 90. The vector sum of V_{Q}, I_{2}R_{2 }and I_{2}X_{2 }gives voltage V_{R}. The drop I_{PR}R_{1 }is in phase with current I_{PR }while the drop I_{PR}X_{1}is leading by 90^{o}. The vector sum of V_{R}, I_{PR}R_{1 }and I_{PR}X_{1}gives the sending end voltage V_{p}.
Thank you very much..it is really useful..

ReplyDeleteIt's an easy and simple way to solve the AC distribution problems. I think this guide will be the enough to understand the whole phenomenon of solving the AC distribution.

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