Power in Balanced System

This requires that the analysis be done in the time domain. For a Y - connected load, the phase voltages are

                  Van = √2 Vp coswt ,                         Vbn = √2 Vp cos(wt - 120),                                 Vcn = √2 Vp cos(wt + 120)  

where the factor √2 is important because Vp has been defined as the rms value of the phase voltage. If Zy = Z∠Ɵ, the phase currents lag behid their corresponding phase voltage by Ɵ.

                
  ia = √2Ip cos(wt - Ɵ),                        

  ib = √2Ip cos(wt - Ɵ - 120)                    
  ib = √2Ip cos(wt - Ɵ + 120)

TOTAL INSTANTANEOUS POWER

p = pa + pb + pc = VANia + VBNib + VCNic

    = 2VpIp [ coswt cos(wt - Ɵ) + cos(wt  - 120) cos(wt - Ɵ - 120) + cos(wt + 120) cos(wt - Ɵ + 120) ]

p = 3VpIp cosƟ

INSTANTANEOUS POWER PER PHASE IS:

p = VpIp cosƟ

REACTIVE POWER PER PHASE IS:

Qp = VpIp sinƟ

APPARENT POWER PER PHASE IS:

Sp = VpIp

COMPLEX POWER PER PHASE IS :

Sp = Pp + jQp = Vp Ip*

Balance Three-Phase Circuit

The voltages in the three-phase power system are produced by a synchronous generator. In a balanced system, each of the three instantaneous voltages have equal amplitudes but separated from the other voltages by a phase angle of 120 . The three voltages (or phases) are typically labeled a, b and c. The common reference point for the three phase voltages is designated as the neutral connection and is labeled as n. We may define either a positive phase sequence (abc) or a negative phase an bn cn sequence (acb) as shown below. The three sources Van , Vbn and Vcn are designated as the line-to-neutral voltages in the three-phase system.

LINE TO LINE VOLTAGES

 First, the wye - connected voltages Van, Vbn, and Vcn are respectively between lines a, b, c, and the neutral line n. These are called phase voltages. The voltages are said to be balanced phase voltages if the voltages sources have the same amplitude and frequency w and are out of phase by 120degrees. 

                                                                Van = Vp∠0

                                                                Vbn = Vp∠-120

                                                                Vcn = Vp∠-240 = Vp∠120

     where Vp is the effective or rms value of the phase voltage. this is known as abc sequence or positive sequence, Van leads Vbn, which turn leads Vcn. This rotates counterclockwise. For negative or acb sequence, Van leads Vcn, ehich turn leads Vbn. this rotates clockwise direction.

AC Power Analysis

Instantaneous Power

The instantaneous power changes with time and it is therefore difficult to measure. The average power is more convenient to measure. For sinusoidal signals or any other periodic signal we compute average power over one period using the following relation:

• Product of time-domain voltage 

      and time-domain current
           p(t) = v(t) i(t)

• Determine maximum value
• Transients
• Steady-state

ƒ

• v(t) = Vmcos(ωt+θ) and i(t) = Imcos(ωt +ϕ)

ƒ

• p(t) = Vm Imcos(ωt+θ) cos(ωt +ϕ) =

            = ½ Vm Imcos(θ-ϕ) + ½ Vm Imcos(2ωt + θ+ ϕ)

where:

p(t) = instantaneousƒ,    two parts,     ƒaverage,     ƒperiodic

 ½ Vm Imcos(θ-ϕ)  =  constant,     ƒindependent of t,     ƒthe "average",     ƒwantedƒ,    active or real

+ ½ Vm Imcos(2ωt + θ+ ϕ) = periodicƒ,     period is ½T,     ƒaverage is zero,     ƒunwanted

EXAMPLE:

1. Voltage across impedance (V= 100∠25V and Z= 50∠55Ω). Determine active power absorbed.

I   =V / Z
    = 2 ∠25-55°
    = 2 ∠-30°
I   = 1.7 - j A

Z= 28.7 + j41.9 Ω

P   = ½ 2228.7
     = ½ (1.72+12) 28.7
P   = 57.4 W

EFFECTIVE OR RMS VALUE

Effective or RMS Value

The RMS Value is a constant itself which depending on the shape of the function i(t).

The effective value of a periodic current is equal to dc current that delivers the same average power to a resistor as the periodic current.

The RMS Value of a sinusoid i(t) = Imcos(wt) is:

                                                              I rms =  Im 

                                                                           √2

                                             

                                                           P = Irms Vrms cos(θv-θi)

• Take note: Supply is always Vrms, unless stated that it is maximum voltage.
  
  

   

   Apparent Power and Power Factor

  
  
  • Apparent Power S, is theproduct of the RMS values of voltage and current.
  • It is measured in volt-amperes or VA to distinguish it from the average or real power which is measured in watts.
  
  
  
  
                                             P = Vrms Irms cos(θv-θi) = S cos(θv-θi)
                                  where:
                                             S - the apparent power
                                             (θv-θi) - the power factor, PF 
            
                                                         PF = P/S = (θv-θi)
• PF is the cosine of the phase difference between the voltage and current. Also, angle of the load impedance. 

                                                                 P = S 0.9

                                                   where: 

                                                             P - active power (W)

                                                             S - apparent power (VA)

                                                            0.9 - reactive power (VAR)

Purely resistive load (R) - θv-θi = 0

                                        - PF = 1

                                        - P = S

Purely reactive load (L or C) - θv-θi = ±90

                                              - PF =0

                                               - P = 0

Resistive and Inductive load (R and L) - θv-θi > 0

                                                                - P = S x PF

                                                                - 0 < PF < 1, lag

Resistive and Capacitive load (R and C) - θv-θi < 0

                                                                  - 0 < PF < 1, lead

                     

                                                 

REVIEW:

• In a purely resistive circuit, all circuit power is dissipated by the resistor(s). Voltage and current are in phase with each other.
• In a purely reactive circuit, no circuit power is dissipated by the load(s). Rather, power is alternately absorbed from and returned to the AC source. Voltage and current are 90o out of phase with each other.
• In a circuit consisting of resistance and reactance mixed, there will be more power dissipated by the load(s) than returned, but some power will definitely be dissipated and some will merely be absorbed and returned. Voltage and current in such a circuit will be out of phase by a value somewhere between 0deg and 90deg.

COMPLEX POWER

S = 1/2 VI* = Vrms Irms ∠ (θv-θi)

         

                  V = Vm∠θv           I = Im∠θpf

S =  Vrms Irms cos(θv-θi) + jVrms Irms sin(θv-θi)

S = P +jQ

P: is the average power in watts delivered to a load and it is the only useful power.

Q: is the reactive power exchange between the source and the reactive part of the load. It is measured in VAR.

• Q = 0, for resistive loads (unity PF)
• Q < 0, for capacitive loads (leading PF)
• Q > 0, for inductive loads (lagging PF)

Maximum and Average Power

The maximum power transfer theorem for DC circuit, we can determine the condition for an AC load to absorb maximum power in an AC circuit. For an AC circuit, both the thevenin impedance and the load can have a reactive component. Although these reactances do not absorb any average power, they will limit the circuit current unless the load reactances cancels the reactance of the thevenin impedance. For the maximum power transfer, the thevenin and load reactances must be equal in magnitude but opposite in sign.

If the load is purely real, then RL = √(Rth)^2 + (Xth)^2 = |Zth|

Zth = Rth + jXth    ;     ZL = RL + jXL

Prove that:

XL = -Xth    &        RL = Rth

SOLUTION:

P = 1/2 (I)^2 (RL)

I = VTH/Zth + ZL

P = 1/2 |Vth/(Zth+ZL)|^2 (RL)

P = 1/2 |Vth/(Rth+jXth)+(RL+jXL)|^2 (RL)

P = 1/2 |Vth^2 (Rth + jXth + RL + jXL)^-2| (RL)

dP = 1/2 |Vth^2 (Rth + jXth + RL + jXL)^-2| d(RL) + 1/2 (RL) |Vth^2 (dRL) (-2) (Rth + jXth + RL + jXL)^-3|

dP/dRL = 1/2 |Vth^2 (Rth + jXth + RL + jXL)^-2| + 1/2 (RL) |Vth^2 (-2) (Rth + jXth + RL + jXL)^-3| = 0

          (1/2 Vth^2)                -    RL                 (Vth^2)                  = 0

(Rth + jXth + RL + JXL)^2                (Rth + jXth + RL + jXL)^3

1   -                RL                  =   0

2      Rth + jXth + RL +jXL

1    =                      RL               

2              Rth + jXth + RL +jXL

Rth + jXth + RL + jXL = 2RL

RL = Rth + j (Xth + XL)

RL - Rth = j (Xth +XL)

(RL - Rth) - j (Xth +XL)  =  0

-j (Xth +XL) = 0

Xth = -XL

XL = -Xth

RL - Rth = 0

RL = Rth

Thevenin's And Norton's Theorem

Any combination of sinusoidal AC sources and impedances with two terminals can be replaced by a single voltage source e and a single series impedance z. The value of e is the open circuit voltage at the terminals, and the value of z is e divided by the current with the terminals short circuited. In this case, that impedance evaluation involves a series-parallel combination.

Using Thévenin's Theorem is especially advantageous when:

· we want to concentrate on a specific portion of a circuit. The rest of the circuit can be replaced by a simple Thévenin equivalent.

· we have to study the circuit with different load values at the terminals. Using the Thévenin equivalent we can avoid having to analyze the complex original circuit each time. 

We can calculate the Thévenin equivalent circuit in two steps:

1. Calculate Z_Th. Set all sources to zero (replace voltage sources by short circuits and current sources by open circuits) and then find the total impedance between the two terminals.

2. Calculate V_Th. Find the open circuit voltage between the terminals. 

Norton's Theorem, already presented for DC circuits, can also be used in AC circuits. Norton's Theorem applied to AC circuits states that the network can be replaced by a current source in parallel with an impedance.

We can calculate the Norton equivalent circuit in two steps:

1. Calculate Z_Th. Set all sources to zero (replace voltage sources by short circuits and current sources by open circuits) and then find the total impedance between the two terminals.

2. Calculate I_Th. Find the short circuit current between the terminals.

Superposition on AC circuit

Since we already studied the Superposition Theorem in DC on the previous chapter, and so on this article will talk about Superposition Theorem on AC circuit.

We have already studied the superposition theorem for DC circuits. In this chapter we will show its application for AC circuits.

The superposition theorem states that in a linear circuit with several sources, the current and voltage for any element in the circuit is the sum of the currents and voltages produced by each source acting independently. The theorem is valid for any linear circuit. The best way to use superposition with AC circuits is to calculate the complex effective or peak value of the contribution of each source applied one at a time, and then to add the complex values. This is much easier than using superposition with time functions, where one has to add the individual time functions.

To calculate the contribution of each source independently, all the other sources must be removed and replaced without affecting the final result.

When removing a voltage source, its voltage must be set to zero, which is equivalent to replacing the voltage source with a short circuit.

When removing a current source, its current must be set to zero, which is equivalent to replacing the current source with an open circuit.