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close this bookIntroduction to Electrical Engineering - Basic Vocational Knowledge
close this folder8. Three-phase Current
View the document8.1. Generation of Three-phase Current
View the document8.2. The Rotating Field
View the document8.3. Interlinking of the Three-phase Current
View the document8.4. Power of Three-phase Current

8.4. Power of Three-phase Current

For calculating the power of three-phase systems, the same relations are applicable as for the calculation of the power of alternating current systems. In accordance with the phase angle involved, a distinction is also made between effective power, reactive power and apparent power.

The star connection of three equal resistors is shown in Fig. 8.13.

For the total power, we have

P = 3 · UStr · IL · cos j

When the power is to be determined, on the basis of the line-to-line voltage UL, the following holds when using equation u.2.





(8.4.)


Fig. 8.13. Star connection of three resistors

When three equal resistors are connected in delta (Fig. 8.14.), the total power is written as

P = 3 · UL · IStr · cos j


Fig. 8.14. Delta connection of three resistors

When the line-to-line current is used, the following holds when using equation 8.3.





(8.5.)

where:

UL

line-to-line voltage

IL

line-to-line current

cos j

power factor

A comparison of the equations 8.4. and 8.5. shows that, independent of the given type of connection, the same equations for calculating the power are given.

When the phase load is unequal, the total power is obtained in the form of the sum of the powers in the three phases to be determined individually.

Example 8.3.

Three resistors of 800 W each have to be interposed in a three-phase network of 380 V one time in star connection and another time in delta connection. Calculate the effective power involved in each case.

Given:

UL = 380 V
R = 800 W
cos j = 1

To be found:

effective power P for star connection and for delta connection

Solution:

star connection of the three resistors


In star connection, only the phase voltage drops at the three resistors. Hence, for the current IL we have


This expression is entered in the initial equation


P = (380 V)2/800 W

P = 180.5 W

delta connection of the three resistors


P = 3 · UL · IStr · cos j

Since the full line-to-line voltage is applied to each resistor, we have for the phase current

IStr = UL/R

This expression is entered in the initial equation

P = 3 · UL2/R · cos j
P = 3 · (380 V)2/800 W
P = 541.5 W

At the three resistors, a total power of 180.5 W is obtained, in star connection and of 541.5 W in delta connection.

In practice frequently advantage is taken of the possibility of obtaining different powers by changing the type of connection of the various loads. For example, for three-phase motors, there are special switching devices which enable the changing over from star connection to delta connection and vice versa.

The power of a three-phase system can be determined from the sum of the individual powers in the three phases in ease of unequal phase loads or from the relation given in equation 8.4. in case of equal loads irrespective of the type of connection.

Questions and problems:

1. Determine the effective power of three resistors in star connection of 200 W if the latter are connected to a three-phase network with a line-to-line voltage of 220 V!

2. Determine the effective power when the three resistors of problem 1. are in delta connection!

3. Three unequal effective resistances (80 W, 200 W, 500 W) have to be connected in star and in delta arrangement to a three-phase network with a line-to-line voltage of 380 V. Determine the total effective power!