Three Phase Circuits Experiment: Balanced Wye-Delta and Delta-Delta Connections, Assignments of Electrical Engineering

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EXPERIMENT 7

Eldrin Jae T. Ibe

Group 5

Experiment 7 Three Phase Circuits Objective

• To know what is a Balanced Three Phase Circuit  To be able to differentiate
• Balanced Wye-Wye Connection
• Balanced Wye-Delta Connection
• Balanced Delta-Wye Connection
• Balanced Delta-Delta Connection
• To be able to solve for phase currents and line currents Introduction An electric power distribution system looks like: Where the power transmission uses “balanced three-phase” configuration. Why Three-Phase? Three-phase generators can be driven by constant force or torque Industrial applications, such as high-power motors, welding equipment, have constant power output if they are three-phase systems. Balanced Three-Phase Voltages Three sinusoidal voltages of the same amplitude, frequency, but differing by 120  phase difference with one another. There are two possible sequences:

1. abc (positive) sequence: vb ( t) lags va ( t) by 120 .
2. acb (negative) sequence: vb ( t) leads va ( t) by 120 .
3. vb (t) lags va (t) by 120 or T/3.

- Y-Y connection (i.e., Y-connected source with a Y-connected load) • Y-∆ connection • ∆-∆ connection • ∆-Y connection Balanced Wye-Wye Connection A balanced Y-Y system is a three-phase system with a balanced Yconnected source and a balanced Y-connected load. Balanced Wye-Delta Connection A balanced Y- system consists of a balanced Y-connected source feeding a balanced -connected load.

Balanced Delta-Delta Connection A balanced system is one in which both the balanced source and balanced load are -connected. Balanced Delta-Wye Connection A balanced -Y system consists of a balanced -connected source feeding a balanced Y-connected load.

Methodology

1. Draw the circuit in your simulator. Ic
2. Edit the voltage Vab by doing right-click and modify according to below
3. Follow the format for Vbc and Vca. For Vbc, the phase angel phi=-120 deg, and Vca=120 deg. Ia Ib Iab Ica Ibc

4. From the menu bar, choose Simulate>>Run
5. Use transient analysis and use 120ms since the sources use 60Hz.
6. You will see this dialog box so there should be very minimal series resistance placed per voltage source. Very small resistance is used so readings will not be changed that much as to when there is no resistance.

9. Run the simulation again.
10. Check the current waveform for Rab/Lab, Rbc/Lbc and Rca/Lca. The three currents should be equal in magnitude but having different phase but equally spaced. To get the phase angle, measure the difference of two waveforms

Let the difference be td as measured, tp is the whole period 120ms. 𝒕𝒅 𝒑𝒉𝒂𝒔𝒆 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒊𝒏 𝒅𝒆𝒈𝒓𝒆𝒆𝒔 𝟑𝟔𝟎° 𝒕𝒑

11. In order to solve for the current, we should know first the impedance Z. Given:

R=25 ohms

L=10mH

Z= R+jX

12. The phase currents are computed as below: 𝑽𝒂𝒃 𝑰𝒂𝒃 = 𝒁

13. For a delta load, the line current always lags the corresponding phase current by 30° and has a magnitude times that of the phase current.

SCHEMATIC DIAGRAM RESISTORS CURRENT

INDUCTOR CURRENT

5. Solve for line currents Ia, Ib and Ic. Observe the magnitude is times the phase current

and lags the current by 30°.

We get:

6. Based from the circuit diagram, what is the connection demonstrated?

a. Balanced Wye-Wye Connection

b. Balanced Wye-Delta Connection

c. Balanced Delta-Wye Connection

d. Balanced Delta-Delta Connection

The connection demonstration is letter D. balanced delta-delta connection

Sample Computations

Impedance Z

Rab = 25 Ω

Lab = 100 mH

f = 60 Hz

Zab = Rab + 2 πff Lab j

Zab = 25 + 2 πf ( 60 ) ( 100 m ) j = 25 +1.2 πfj Ω = 25 +3.77 πfj Ω

Impedance Z in polar form

| Zab |=√^25 2

+1.2 πf

2

θab =tan

− 1 1.2^ πf

Zab =25.28 Ω∠ 8. 58 °

Current I AB

Iab =

V ab

Zab

I ab =

I ab =

∠ 0 −8. 58 ° Ω =13.05 ∠ −8. 58 ° A

Line Current I^ AB

ZAB = 1 3.05 ∠ −8. 58 ° A

I (^) A =| I (^) A |√ 3 ∠ ( θ ¿¿ ab − 30 ° ) ¿ I (^) A =13.05 (^) √ 3 ∠ (−8. 58 − 30 ° )=22.61 ∠ −38. 58 ° A

Interpretation of Results

 By using the nodal analysis, the sum of all line currents is equal to zero

 It is a sinusoidal voltage with the same amplitude, frequency and the simulation produce similar

waveforms for the voltages and currents but vary by 120° phase difference with one another

 The line current are all equal to the product of the phase current magnitude and (^) √ 3.

 The line current is always lagging the corresponding phase current by 30 degrees and by the

multiplication times the square root of 3 for the delta load.

Conclusion

 The experiment was successful because we balanced out the three phase circuit

 The three phase system’s line currents add up to zero or it is in neutral power because neutral

power is zero.

 The experiment involves the principles and relationship between the line voltages and phasor

voltage and currents in the circuit and by performing the experiment we got our resistive load

balance and equal.

 The line current is always lagging the corresponding phase current by 30 degrees and by the

multiplication times the square root of 3 for the delta load.