Monday, January 25, 2010

Transformers in Parallel

Objective:

· To learn how to connect transformers in parallel.

· To determine the efficiency of parallel-connected transformers.


Discussion:


Transformers may be connected in parallel to furnish load currents greater than the rated current of each transformer. There are two precautions to be observed when connecting transformers in parallel.


1. The windings to be paralleled must have identical output voltage ratings.

2. The windings to be paralleled must have identical polarities.


Very large short-circuit currents can be developed if these rules are not followed. In fact, transformers, circuit breakers and associated circuitry can be severely damaged, or may even explode, if these short-circuit currents are large enough.


The efficiency of any machine or electrical device is given by the ratio of output power to input power. (Apparent power and reactive power are not used in calculating transformer efficiency). The equation for percent efficiency is:





Equipment Required:

  • A Single Phase Transformer
  • Resistive Load
  • Power Supply
  • AC Ammeter
  • AC Voltmeter
  • Wires


Procedure:

CAUTION!!!

High voltages are Present In the Experiment! Do not make any connections with the power on! The power should be turned off after completing each individual measurement!!!

1. Using your Single-Phase Transformer, Power Supply, Single-Phase Wattmeter, Resistive Load, AC Ammeter and AC Voltmeter, connect the circuit shown in Figure – 1. Note that the two transformers are connected in parallel. The primary windings (1 and 2) are connected together to the 220 V ac power source. The wattmeter will indicate the input power. Each secondary winding (3 to 4) is connected in parallel to the load RL. Ammeters are inserted to measure load current IL and transformer secondary currents I1 and I2.

Connect two resistance sections of the Variable resistance module in series to implement the resistive load RL shown in Figure – 1. This is required to dissipate the large amount of power involved in this exercise.

2. Place all the resistance switches in their open positions for zero load current. Note that the windings are connected for voltage step-up operation (220 V primary to 380 V secondary).

3. Please your circuit wiring approved by the instructor before proceeding.

4. a. Turn on the power supply and slowly advance the voltage output control knob while noting the transformer secondary current meters I1, if the windings are properly phased, no load or secondary currents should be flowing.

b. Adjust the power supply voltage to 220 V ac as indicated by the voltmeter connected across the wattmeter.

5. a. Gradually increase the load RL until the load current IL equals 250mA ac. Check to see that the input voltage is exactly 220 v ac.

b. Measure and record the load voltage, load current, transformer secondary currents and the input power.

EL = 340 V ac

IL = 0.24 A ac

I1 = 0.12 A ac

I2 = 0.12 A ac

Pin = 95 W

c. Return the voltage to zero and turn off the power supply.

6. a. Calculate the load power,

E1 (340) × IL (0.24) = 81.6 W

b. Calculate the circuit efficiency,




c. Calculate the transformer losses

Pin (95) – Pout (81.6) = 13.4 W

d. Calculate the power delivered by transformer-1.

I1 (0.12) × EL (340) = 40.8 W

e. Calculate the power delivered by transformer – 2.

I2 (0.12) × EL (340) = 40.8 W

7. Is the load reasonably distributed between the two transformers? Yes.


Review Questions:

1. Show how you would parallel connect the transformers to the

source and the load in Figure – 2. Windings 1 to 2 and 3 to 4 are rated for 2.4 KV ac and windings 5 to 6 and 7 to 8 are rated for 400 V ac.











Ans: Figure – 3




In a parallel connection of two transformers, should be properly connected with regard to polarity. The percentage impedance should be equal in magnitude. The regulation must be same.

2.

The efficiency of transformers which supplies a pure capacitive load is zero. Explain.

Ans: We know the efficiency of a transformer,




For capacitive load, Pout = VI cosθ = VI cos 90o = 0

3. Name the losses which cause a transformer to heat up.

Ans: The I2R losses that means copper losses and core losses are cause a transformer to heat up.

4. How does the efficiency of our Single-Phase Transformer compare to the efficiency of DC Motor/Generator? Explain.


Ans: In DC generator or motor; there is a friction loss which plays an important role on efficiency calculation. But in Transformers there is no friction loss, so transformer is more efficiency than a dc motor or generator of same rating.

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