Introduction to Electric Circuits

120 Three-phase a.c. c#'cuits

Example 5.6

A 440 V three-phase motor has a useful output of 50 kW and operates at a

power factor of 0.85 lagging with an efficiency of 89 per cent. Calculate the

readings on two wattmeters connected to measure the input power.

Solution

Since efficiency is 77- Po/Pi, the input power Pi- Po/rl- 50/0.89 - 56.2 kW.

Let the readings on the two wattmeters be P~ and P2. Then P~ + P2 = 56.2 kW.

Now ~b = cos -1 0.85 = 31.79 ~ so

tan 4, = 0.6197

From Equation (5.14) tan 4)- V3(P~ - Pz)/(P1 + P2), so

P~- P2 = tan 4) • (P1 + P2)/V/3 - (0.6197 • 56.2)//V/3 = 20.1 kW

We now have

P1 -% P2 - 56.2 kW

P1 - P2 = 20.1 kW

Adding gives

2P1 = 76.3 kW

P1 = 38.15 kW

It follows that

P2 = 18.05 kW

5.6 SELF-ASSESSMENT TEST

1 Give an advantage of three-phase systems over single-phase systems for

the purpose of power transmission.

2 Give an advantage of three-phase systems over single-phase systems for

the purpose of power generation.

3 Explain how three-phase emfs are generated.

4 Explain the difference between a star-connected system and a delta-

connected system.

5 Explain why a star connection is more usual for power distribution

purposes.

6 State the meaning of the term 'phase sequence' when applied to three-

phase systems.