The Delta Connected Generator

Facebook
Whatsapp
Twitter
LinkedIn
If we rearrange the coils of the generator in Fig. 1(a) as shown in Fig. 1(b), the system is referred to as a three-phase, three-wire, $\Delta$-connected ac generator.
Fig. 1: $\Delta$-connected generator.
In this system, the phase and line voltages are equivalent and equal to the voltage induced across each coil of the generator; that is,
$$\left.\begin{array}{lll}\mathbf{E}_{A B}=\mathbf{E}_{A N} & \text { and } & e_{A N}=\sqrt{2} E_{A N} \sin \omega t \\\mathbf{E}_{B C}=\mathbf{E}_{B N} & \text { and } & e_{B N}=\sqrt{2} E_{B N} \sin \left(\omega t-120^{\circ}\right) \\\mathbf{E}_{C A}=\mathbf{E}_{C N} & \text { and } & e_{C N}=\sqrt{2} E_{C N} \sin \left(\omega t+120^{\circ}\right)\end{array}\right\} \begin{array}{l}\text { Phase } \\\text { sequence } \\A B C\end{array}$$
or $$ \mathbf{E}_{L}=\mathbf{E}_{\phi g} $$ Note that only one voltage (magnitude) is available instead of the two available in the Y-connected system. Unlike the line current for the Y-connected generator, the line current for the $ \Delta $-connected system is not equal to the phase current. The relationship between the two can be found by applying Kirchhoff's current law at one of the nodes and solving for the line current in terms of the phase currents; that is, at node $ A $,
$$\begin{array}{l}\mathbf{I}_{B A}=\mathbf{I}_{A a}+\mathbf{I}_{A C} \\\mathbf{I}_{A a}=\mathbf{I}_{B A}-\mathbf{I}_{A C}=\mathbf{I}_{B A}+\mathbf{I}_{C A}\end{array}$$
The phasor diagram is shown in Fig. 2 for a balanced load.
Fig. 2: Determining a line current from the phase currents of a $\Delta$-connected, three phase generator.
Using the same procedure to find the line current as was used to find the line voltage of a $ \mathrm{Y} $-connected generator produces the following:
$$\begin{array}{l}I_{A a}=\sqrt{3} I_{B A} \angle-30^{\circ} \\I_{B b}=\sqrt{3} I_{C B} \angle-150^{\circ} \\I_{C c}=\sqrt{3} I_{A C} \angle 90^{\circ}\end{array}$$
In general:
$$I_{L}=\sqrt{3} I_{\phi g}$$
with the phase angle between a line current and the nearest phase current at $30^{\circ}$. The phasor diagram of the currents is shown in Fig. 3.
Fig. 3: The phasor diagram of the currents of a three-phase, $\Delta$-connected generator.
It can be shown in the same manner employed for the voltages of a Y-connected generator that the phasor sum of the line currents or phase currents for $\Delta$-connected systems with balanced loads is zero.

Do you want to say or ask something?

Only 250 characters are allowed. Remaining: 250
Please login to enter your comments. Login or Signup .
Be the first to comment here!
Terms and Condition
Copyright © 2011 - 2024 realnfo.com
Privacy Policy