Three-phase AC Power.pptx

Three-phase AC Power

• _{A three-phase a.c. supply is carried by three conductors, called ‘lines’}

Difference between line and phase

quantities

• _{Line voltage is nothing but the output voltage measured at}

transformer externally i.e. Voltage available between any 2 wires out of 3 wires. We call it as V_L.

• _{But Phase voltage is the output voltage available internally i.e.}

Voltage available between single phase wire and neutral. We call it as V_ph.

• _{And the same definition holds good for current available which shall}

• _{Our basic rule is to measure power available at secondary which is}

nothing but the product of Phase Voltage x Phase Current and summation of all three phases,

• _{P = 3 x VphIph ….(1)}

• _{But as you see it is very difficult to measure the phase voltages from a}

transformer which may involve complex connections and may be impossible. Rather it’s easy to measure from the lines present

outside.

• _{As we know in a star connected transformer, the line voltage is}

• _{Here we take the case of star connected transformer,}

• Since the line voltage is combination of two phase voltages we cannot just add together

to make V_L = 2 x V_ph which doesn’t make sense since there is a phase shift of 120°.

•

• _{So, VL = 2 x Vph x sin(120)}

• _{VL = 2 x Vph x √3/2} • _{VL = √3 x Vph}

• _{Vph = VL /√3 …. (2)} •

• _{We get P = 3 x VL /√3 x IL since Iph = IL}

P = √3 .V_L .I_L

• _{This holds good for transformers and for other loads like motors,}

generators etc. Another factor called power factor gets introduced in the equation and converted as,

ECE 441 8

• _{4 wires}

• _{3 “active” phases, A, B, C} • _{1 “ground”, or “neutral”}

• _{Color Code}

• _{Phase A Red}

• _{Phase B Black American system} • _{Phase C Blue}

ECE 441 9

• _{4 wires}

• _{3 “active” phases, A, B, C} • _{1 “ground”, or “neutral”}

• _{Color Code}

• _{Phase A Red}

• _{Phase B Yellow British system} • _{Phase C Blue}

ECE 441 10

ECE 441 11

What is Three-Phase Power?

• _{Three sinusoidal voltages of equal amplitude and frequency out of}

phase with each other by 120°. Known as “balanced”.

• _{Phases are labeled A, B, and C.}

ECE 441 12

• _{Typically, transmission lines consist of four wires, rather than two as}

you might have guessed. One of these wires is the ground; the

remaining three are used to transmit three-phase ac power which is a superposition of three ac voltages 120⁰ out of phase with each other:

• _{V₁ = V₀ sinωt}

Why is three-phase power used?

• _{Single-phase power—just the voltage V₁ by itself—delivers voltage to a load}

in pulses. A much smoother flow of power can be delivered if we use three-phase power.

• _{Suppose that each of the three voltages making up the three-phase source}

is hooked up to a resistor R. Then the power delivered is:

• _{P = (V₁² + V₂² + V₃²)}

• _{It can be shown that this power is a constant equal to 3V₀²/2R, which is}

three times the rms power delivered by a single-phase source (V₀/√2 is the rms voltage). This smooth flow of power makes electrical equipment run

Reasons why three-phase power is

superior to single phase power

1. The horsepower rating of three-phase motors and the KVA (kilo-volt-amp)

rating of three-phase transformers is about 150% greater than for single-phase motors or transformers with a similar frame size.

2. The power delivered by a single-phase system pulsates, The power falls to zero

three times during each cycle. The power delivered by a three-phase circuit

pulsates also, but it never falls to zero. In a three-phase system, the power delivered to the load is the same at any instant. This produces superior

operating characteristics for three-phase motors.

ECE 441 16

Phasor (Vector) Form for abc

V_a=V_m/0°

V_b=V_m/-120°

Single-phase vs three-phase

With the wave form of single-phase power, when the wave passes through zero, the power supplied at that moment is zero. In the U.S., the wave cycles 60 times per second.

THREE-PHASE POWER

• _{Students sometimes become confused when computing power in}

three phase circuits. One reason for this confusion is that there are actually two formulas that can be used. If line values of voltage and current are known, the power (watts) of a pure resistive load can be computed using the formula:

• _{VA = √ 3 X ELine X ILine}

• _{Likewise, if the phase values of voltage and current are known, then}

power (watts) can be computed using the formula:

• _{Notice that in the first formula, the line values of voltage and current}

are multiplied by the square root of 3. In the second formula, the phase values of voltage and current are multiplied by 3. The first

THREE PHASE CIRCUITS ) )( 240 cos( ) ( ) )( 120 cos( ) ( ) )( cos( ) ( V t V t v V t V t v V t V t v m c m bn m an           Voltages Phase ous Instantane

2

120



V

V _{c n} V _{b n}

V _{a n}

0 1 2 0 2 4 0 3 p h a s e

_

_ _

+ +

+

n

a

b c

c

b

a

V  0

V  - 1 2 0 V  - 2 4 0

Delta Source

V_ab = | V_ab |  0 V_bc = V_ab  -120 V_ca = V_ab  -240

a

b c

a

b

c

+ +

+

_

_

_ _{D e l t a}

Wye – Wye System

Z _l

Z _l

Z _l

Z _L

Z _L Z _L a

n

b c

A

B C

Delta – Wye System

Z _l

Z _l

Z _l

a

b c

A

B C

+

+ +

_

_

_

Z _L

Z _L

The voltage given for a three-phase system is always the line voltage unless it is stated otherwise.

Wye – Delta System

_ _ _ + + + n a b c c b a

V  0

V  - 1 2 0

V  - 2 4 0

A

B C

Z _Z

Z

I _{a A I}

C A

• _{Problem 1. Three loads, each of resistance 30 Ω , are connected in star to a 415 V,}

3-phase supply. Determine (a) the system phase voltage, (b) the phase current and (c) the line current.

• _{A ‘415 V, 3-phase supply’ means that 415 V is the line voltage, VL}

• _{(a) For a star connection, VL = √3 Vp}

• _{Hence phase voltage, Vp = = = 239.6 V or 240 V}

correct to 3 significant figures

• _{(b) Phase current, Ip = = = 8 A}

• _{(c) For a star connection, Ip = IL}