eeet-241/_Homework/project1.md

1. Draw the circuit of a 3-phase 4160/2400V generator. Note that the generator is in a WYE configured generator. Ignore any generator impedance. Draw this first before moving on. Connect the generator to the primary of a 3-phase transformer. You will only have 3 lines from the generator to the transformer primary. The transformer is rated at 1000kVA, 4160-208/120V, 60Hz. In this step, there is no load on the transformer, the secondary of the transformer is open-circuited.

2. Clearly label and determine of the following:

   1. V{generator_{\phi}}
   2. V{generator_{line}}
   3. V{primary_{\phi}}
   4. V{primary_{line}}
   5. V{secondary_{\phi}}
   6. V{secondary_{line}}

   

3. Connect a 1k\Omega resistor across each phase of the secondary of the transformer to neutral. Redraw the circuit.

1.  Determine  $V_{load}$ across each resistor. Show this voltage in the circuit diagram.
    $$
    V_\phi=120V=V_{a}\rightarrow V_{neutral}=V_{b}\rightarrow V_{neutral}=V_{c}\rightarrow V_{neutral}\\
    \therefore\\
    V_{load}=120V
    $$

2.  Find $I_{load}$ through each resistor.
    $$
    I_{load}={V_{load}\over R}={120V\over1k\Omega}=120mA
    $$

3.  Determine the current for each line from each of the secondary of the transformer to the load $I_{secondary_{line}}$.
    $$
    I_{load}=I_{secondary_{line}}\\
    \therefore\\
    I_{secondary_{line}}=120mA
    $$

4.  Determine $I_{secondary_{\phi}}$.
    $$
    WYE\ config\\\therefore\\I_{secondary_{\phi}}=120mA
    $$

5.  Determine $I_{primary_{\phi}}$.
    $$
    I_{primary_{\phi}}={I_{secondary_{\phi}}\over a};\ a=34.\overline6\\
    I_{primary_{\phi}}={120mA\over 34.\overline6}\approx3.5mA
    $$

6.  Determine $I_{primary_{line}}$.
    $$
    I_{primary_{line}}=I_{primary_{\phi}}*\sqrt3\\
    I_{primary_{line}}=5.996mA\approx6mA
    $$

7.  Determine $I_{generator_{line}}$.
    $$
    I_{generator_{line}}=I_{primary_{line}}\\\therefore\\I_{generator_{line}}\approx6mA
    $$

8.  Determine $I_{generator_{\phi}}$.
    $$
    I_{generator_{\phi}}=I_{primary_{\phi}}\\\therefore\\I_{generator_{\phi}}\approx3.5mA
    $$
    

![](project1.assets/step4.png)

5. From the numbers above, determine:

   1. P_{generator_\phi} P_\phi=I_\phi V_\phi\ P_{generator_{\phi}}=3.5mA*4160V\ P_{generator_{\phi}}=14.4W

   2. P_{generator_{3\phi}} P_{3\phi}=3P_\phi\ P_{generator_{3\phi}}=14.4W*3\ P_{generator_{3\phi}}\approx43W

   3. P_{primary_\phi} P_\phi=I_\phi V_\phi\ P_{primary_{\phi}}=3.5mA*4160V\ P_{primary_{\phi}}=14.4W

   4. P_{primary_{3\phi}} P_{3\phi}=3P_\phi\ P_{primary_{3\phi}}=14.4W*3\ P_{primary_{3\phi}}\approx43W

   5. P_{secondary_\phi} P_\phi=I_\phi V_\phi\ P_{secondary_{\phi}}=120mA*120V\ P_{secondary_{\phi}}=14.4W

   6. P_{secondary_{3\phi}} P_{3\phi}=3P_\phi\ P_{secondary_{3\phi}}=14.4W*3\ P_{secondary_{3\phi}}\approx43W

   7. P_{load_\phi} P_\phi=I_\phi V_\phi\ P_{load_{\phi}}=120mA*120V\ P_{load_{\phi}}=14.4W

   8. P_{load_{3\phi}} P_{3\phi}=3P_\phi\ P_{secondary_{3\phi}}=14.4W*3\ P_{secondary_{3\phi}}\approx43W