Exercise1 - · 1 Relationships between real, reactive and complex power cos sin 1 Example: A load...

Informatik 7

Rechnernetze und

Kommunikationssysteme

Exercise1

21.10.2015

Dr.-Ing. Abdalkarim Awad

Review of Phasors

Goal of phasor analysis is to simplify the analysis of constant frequency ac systems

v(t) = Vmax cos(wt + qv)

i(t) = Imax cos(wt + qI)

Root Mean Square (RMS) voltage of sinusoid

2 max

0

1( )

2

TV

v t dtT

2Dr.-Ing. Abdalkarim Awad

Phasor Representation

j

( )

Euler's Identity: e cos sin

Phasor notation is developed by rewriting

using Euler's identity

( ) 2 cos( )

( ) 2 Re

(Note: is the RMS voltage)

V

V

j t

j

v t V t

v t V e

V

q

w q

q q

w q


Phasor Representation, cont’d

The RMS, cosine-referenced voltage phasor is:

( ) Re 2

cos sin

cos sin

V

V

jV

jj t

V V

I I

V V e V

v t Ve e

V V j V

I I j I

q

qw

q

q q

q q

(Note: Some texts use “boldface” type for

complex numbers, or “bars on the top”)


Advantages of Phasor Analysis

0

2 2

Resistor ( ) ( )

( )Inductor ( )

1 1Capacitor ( ) (0)

C

Z = Impedance

R = Resistance

X = Reactance

XZ = =arctan( )

t

v t Ri t V RI

di tv t L V j LI

dt

i t dt v V Ij C

R jX Z

R XR

w

w

Device Time Analysis Phasor

(Note: Z is a

complex number but

not a phasor)


Example1

A 50-Hz, single-phase source with volts is applied to a circuit element.

A)Determine the instantaneous source voltage. Also determine the phasor and instantaneous currents entering the positive terminal if the circuit element is

B) a 20-ohm resistor,

C) a 10-mH inductor,

D) a capacitor with 25 ohm reactance.

6

30230


Complex Power

max

max

max max

( ) ( ) ( )

v(t) = cos( )

(t) = cos( )

1cos cos [cos( ) cos( )]

2

1( ) [cos( )

2

cos(2 )]

V

I

V I

V I

p t v t i t

V t

i I t

p t V I

t

w q

w q

q q

w q q

Power


Complex Power, cont’d

max max

0

max max

1( ) [cos( ) cos(2 )]

2

1( )

1cos( )

2

cos( )

= =

V I V I

T

avg

V I

V I

V I

p t V I t

P p t dtT

V I

V I

q q w q q

q q

q q

q q

Power Factor

Average

P

Angle

ower


Complex Power

*

cos( ) sin( )

P = Real Power (W, kW, MW)

Q = Reactive Power (var, kvar, Mvar)

S = Complex power (VA, kVA, MVA)

Power Factor (pf) = cos

If current leads voltage then pf is leading

If current

V I V I

V I

S V I j

P jQ

q q q q

lags voltage then pf is lagging

(Note: S is a complex number but not a phasor)


Complex Power, cont’d

2

1

Relationships between real, reactive and complex power

cos

sin 1

Example: A load draws 100 kW with a leading pf of 0.85.What are (power factor angle), Q and ?

-cos 0.85 31.8

100

0.

P S

Q S S pf

S

kWS

117.6 kVA85

117.6sin( 31.8 ) 62.0 kVarQ


Power Consumption in Devices

2Resistor Resistor

2Inductor Inductor L

2

Capacitor Capacitor C

CapaCapacitor

Resistors only consume real power

P

Inductors only consume reactive power

Q

Capacitors only generate reactive power

1Q

Q

C

I R

I X

I X XC

V

w

2

citorC

C

(Note-some define X negative)X


Example2

A certain single phase load draws 5 MW at 0.7 power factor lagging. Determine the reactive power required from a parallel capacitor to bring the power factor of the parallel combination up to 0.9.


Example 3:

A 8 MW/4 Mvar load is supplied at 13.8 kV through a feeder with an impedance of (1+ j2) . The load is compensated with a capacitor whose output, Qcap, can be varied in 0.5 Mvar steps between 0 and 10.0 Mvars. What value of Qcap minimizes the real power line losses? What value of Qcap minimizes the MVA power into the feeder?


Balanced 3 -- Zero Neutral Current

14

* * * *

(1 0 1 1

3

Note: means voltage at point with respect to point .

n a b c

n

an a bn b cn c an a

xy

I I I I

VI

Z

S V I V I V I V I

V x y


Three Phase - Wye Connection

There are two ways to connect 3 systems:– Wye (Y), and

– Delta ().

15

an

bn

cn

Wye Connection Voltages

V V

V V

V V


Wye Connection Line Voltages

16

Van

Vcn

Vbn

Vab

Vca

Vbc

-Vbn

(1 1 120

3 30

3 90

3 150

ab an bn

bc

ca

V V V V

V

V V

V V

Line to linevoltages arealso balanced.

(α = 0 in this case)


Wye Connection, cont’d

We call the voltage across each element of a wye connected device the “phase” voltage.

We call the current through each element of a wye connected device the “phase” current.

Call the voltage across lines the “line-to-line” or just the “line” voltage.

Call the current through lines the “line” current.

17

6

*3

3 1 30 3

3

j

Line Phase Phase

Line Phase

Phase Phase

V V V e

I I

S V I


Delta Connection

18

IcaIc

IabIbc

Ia

Ib

*3

For Delta connection,

voltages across elements

equals line voltages

For currents

3

3

a ab ca

ab

b bc ab

c ca bc

Phase Phase

I I I

I

I I I

I I I

S V I


Delta-Wye Transformation

19

Y

Linephase

To simplify analysis of balanced 3 systems:

1) Δ-connected loads can be replaced by 1

Y-connected loads with 3

2) Δ-connected sources can be replaced by

Y-connected sources with 3 30

Z Z

VV


Example 4

A three-phase line, which has an impedance of (2 + j4) per phase, feeds a balanced Y-connected three-phase load that has an impedance of 22-4j. The line is energized at the sending end from a 50-Hz, three-phase, balanced voltage source of 230√3 V (rms, line-

to-line). Determine:

The current, real power, and reactive power delivered by the sending-end source.

The line-to-line voltage at the load.


Example5

A industrial company has an average power consumption of 500 kW with average power factor of 0.7 and 4000 working hours. Assume that the company must not pay for reactive power if they maintain the power factor above 0.9.

Calculate

Annual electricity consumption (active power)

The electricity cost for active and reactive power if 1 kWh costs 9 cents and 1 kVArh costs 1.5 cents

Would it be profitable to install a 300 kVArcapacitor bank that costs 8000 euros?


“Ideal” Power Market

Ideal power market is analogous to a lake. Generators supply energy to lake and loads remove energy.

Ideal power market has no transmission constraints

Single marginal cost associated with enforcing constraint that supply = demand– buy from the least cost unit that is not at a limit

– this price is the marginal cost.

This solution is identical to the economic dispatch problem solution.


Two Bus Example

Total Hourly Cost :

Bus A Bus B

300.0 MWMW

199.6 MWMW 400.4 MWMW

300.0 MWMW

8459 $/hr

Area Lambda : 13.02

AGC ON AGC ON


Mathematical Formulation of Costs (C) and Incremental Cost (IC)

Generator cost curves are usually not smooth. However the curves can usually be adequately approximated using piece-wise smooth, functions.

Two representations predominate– quadratic or cubic functions

– piecewise linear functions

We'll assume a quadratic presentation

In order to minimize the total operating cost

IC1(PG1)=IC2(PG2)=…ICN(PGN)

2( ) $/hr (fuel-cost)

( )( ) 2 $/MWh

i Gi i Gi Gi

i Gii Gi Gi

Gi

C P P P

dC PIC P P

dP

=λ


Mathematical Formulation of Costs

For the two bus example

25

Euro/MWh 0.00298PGB 11.83IC(PGB)

Euro/MWh 0.00668PGA11.69IC(PGA)

Euro 0.00149PGB11.83PGB616.9C(PGB)

Euro 0.00334PGA11.69PGA 399.8C(PGA)

2

2


Mathematical Formulation of Costs

For the two generator system

PGA+PGB=600

λ =IC(PGA)=13.02

λ = IC(PGA)=13.02

PGA=200 C(PGA)=2870.6

PGB=400 C(PGB)=5587.3

Total cost 8457.926

Euro/MWh 0.00298PGB 11.83IC(PGB)

Euro/MWh 0.00668PGA11.69IC(PGA)

Euro 0.00149PGB11.83PGB616.9C(PGB)

Euro 0.00334PGA11.69PGA 399.8C(PGA)

2

2


Example 6

The fuel-cost curves for two generators are given as follows:

C1(P1)= 600 + 15 P1 + 0.05 P12

C2(P2) =700 +20P2 + 0.04 P22

Assuming the system is lossless, calculate the optimal dispatch values of P1 and P2 for a total load of 1000 MW, the incremental operating cost, and the total operating cost.


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Exercise1 - · 1 Relationships between real, reactive and complex power cos sin 1 Example: A load...

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