ThreePhaseExp101804

# ThreePhaseExp101804 - L302-7.5 Drexel University Electrical...

This preview shows pages 1–4. Sign up to view the full content.

L302-7.5 Drexel University Electrical and Computer Engineering Dept. Electrical Engineering Laboratory II, ECE L 302 C. O. Nwankpa Measuring Three-Phase AC Electric Power Table of Contents Educational Objective Introduction Theory Procedure Concluding Activities Educational Objective The object of the experiment is to continue our analysis of AC power onto three-phase AC circuits. We will learn how to appreciate three-phase AC circuits by understanding the main components as well as variables that are influenced by them. These variables are AC voltages and currents. Three-phase AC power will be shown to be the result of calculations involving these variables. Introduction Almost all electric power generation and most of the power transmission in the world today is in the form of three-phase AC circuits. A three-phase power system consists of three-phase generators, transmission lines, and loads. AC power systems have a great advantage over DC systems in that their voltage levels can be changed to reduce transmission losses. Three-phase AC power systems have a great advantage over single- phase power systems because it is possible to get more power per pound of metal from a three-phase machine and also because the power delivered to a three-phase load is constant at all times, instead of pulsing as it does in single-phase systems. Theory Generation of Three-Phase Voltages and Currents A three-phase generator consists of three single-phase generators, the voltage of each one equal in magnitude but differing in phase angle from the others by 120 o . This can be seen in Figure 1. Each of these three generators could be connected to one of three identical loads by a pair of wires, and the resulting power system would be as shown in Figure 2. Such a system is really three single-phase circuits which simply happen to differ in phase angle by 120 o . The current flowing to each load can be found from the equation: I = V / Z 7-1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
L302-7.5 () vt A B C ( ) ( ) V t B o =− 2 120 sin ω V ( ) ( ) V t C o 22 4 0 sin V V A o V =∠ 0 V ( ) V t A = 2s i n V V B o V 120 V V C o V 240 V Figure 1 A B C ( ) it A ( ) B ( ) C Z = Z θ Z Z Z = Z Figure 2 Therefore, the currents flowing in the three phases are: I A o V Z I = =∠− 0 I B o V Z I = ∠− 120 120 I C o V Z I = 240 240 7-2
L302-7.5 It turns out that three of the six wires shown in this power system are not necessary for the generators to supply power to the loads. Suppose for the sake of discussion that the negative ends of each generator and each load are joined together. In that case, the three return wires in the system could be replaced by a single wire (called the neutral), and the current could still return from the loads to the generators. This new configuration can be seen in Figure 3.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This document was uploaded on 09/25/2011.

### Page1 / 13

ThreePhaseExp101804 - L302-7.5 Drexel University Electrical...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online