EET 305 Chapter 5 the Load in Power System

EET 305

Mohd Rafi, Rev Mei 2011 1

The Load in Power System

EET 305

• The Load in Power System

Define and explain basic load forecasting and load characteristic.

Analyze and calculate power, power factor and power factor correction.

2

Load Characteristic

• Load Curve

• The load on a power station is not constant. It variesfrom time to time.

• The load curve is obtained by plotting the loads againstthe time on a graph paper.

• When it plotted from 24 hours of a day it is called dailyload curve and if the time considered is for one year(8760 hours) then it is called annual load curve.

• The yearly load curve is generally used to determine theload factor.

• The areas under the load curve represent the energygenerated in the period considered.

3Mohd Rafi, rev mei 2011

Load Characteristic

Load Curve

• The area under the curve divided by the total number of

hours gives the average load on the power station.

• The peak curve load indicated by the load curve

represents the maximum demand on the power station.

4Mohd Rafi, rev Mei 2011

Example of Typical

Commercial Load Curve

Load Characteristic

Load Curve

• The greatest problem for a power supply company is

varying load. The generation should be matched with the

load consistently.


• The base load power is

supplied by power plants

running continuously.

• The Intermediated/Average

load prevails for some part of

the day.

• The peak load prevails only

a few hours of the day

Load Characteristic

• Terms and Definitions

a) Demand

The demand of the installation or system is the load

at the receiving terminals averaged over a specified

time.

b) Demand Interval (t)

The period over which the load is averaged.

Selected Δt period may be 15 min, 30 min, 1 hr, or

even longer.


Load Characteristic


c) Maximum Demand

The maximum demand of an installation or system is

the greatest of all demands which have occurred

during the specified period of time.

d) Demand Factor (DF)

The ratio of the maximum demand of a system to a

total connected load of the system.


DemandConnectedTotal

DemandMaximumDF

Load Characteristic


e) Diversified Demand / Coincident Demand

It is the demand of the composite group ( residential,

commercial, industrial, miscellaneous) , as a whole,

of somewhat unrelated loads ( loss in transmission &

distribution) over a specified period of time.

f) Utilization Factor (Fu)

It is the ratio of the maximum demand of a system to

the rated capacity of the system.

8Mohd Rafi rev Mei 2011

capacitysystemRated

demandMaximumFu

Load Characteristic


g) Load Factor (FLD)

The load factor is defined as the ratio of the average

load to the maximum load during a given period.

The load factor is always less than one, because the

average load is less than maximum load.

The greater the load factor, the less will be the cost

per unit.

where: T=time, in days, weeks, months or years.


Tx Load Max/Peak

Served Unit

Tx Load Max/Peak

xT LoadAverage FLD

Load Characteristic


h) Plant Capacity Factor

It is defined the ratio of the total actual energy produced

or served over a designated period of time to the energy

that would have been produced or served if the plant or

unit had operated continously at maximum rating.

where:

E=Energy produced, kWh

C=Plant capacity/rating, kW

t=Total number of hours


txC

EfactorcapacityPlant

Load Characteristic


i) Plant Use Factor

It is defined as the ratio of the energy produced in giventime to the maximum possible energy that could havebeen produced during the actual numbers of hours thatwas in operation

where:

E=Energy produced, kWh

C=Plant capacity, kW

t1=The actual number of hours the plant was inoperation

11Mohd Rafi rev mei 2011

1txC

EfactorusePlant

Load Characteristic


j) Diversity Factor , FD

It is defined as the ratio of the sum of individual

maximum demands to maximum demand on power

station.

TCD is Total Connected Demand, DF is Demand Factor

Dg = coincident max demand of group of n loads12

Mohd Rafi rev Mei 2011

g

n

i

ii

g

n

i

i

g

n

D

D

DFxTCD

D

D

D

D.......DD

demandmax Coincident

demandmaxividual Sum of indF

1121

Load Characteristic


k) Loss Factor , FLS

It is defined as the ratio of the average power loss to

the peak load power loss during a specified period of

time.

Coincidence factor, Fc =1/FD

l) Load Diversity, LD

It is the difference between the sum of the peaks of

two or more individual loads and the peak of the

combined load.


loadpeakatlossPower

losspowerAverageFLS

g

n

i

i DDLD

1

Load Characteristic

• Example 1

A generating station had a maximum load of 20,000 kW

and the units generated being 61,500,000 kWh for the

year.

Calculate load factor.

• Solution Example 1


%35

1002436520000

000,500,61

)()(

)(

xhoursxdaysx

kWh

hTxkWloadMax

kWhServedUnitFLD

Load Characteristic

• Example 2

A power station has two 60MW units each running for

7500 hours in a year and one 30MW unit running for

4500 hours in a year. The energy produced per year is

750x106 kWh.

Determine:

(a) Plant load factor

(b) Plant capacity factor

(c) Plant use factor.

Assume maximum demand to be 80% of the plant

capacity.


Load Characteristic


a) Plant load factor

b) Plant capacity factor


71080101508760

10750

8760

10750

6

6

6

..xWxxh

kWhx

demandMaximum

loadAverageF

h

kWhxloadAverage

150MW30 2x60 plant powerthe ofCapacity

LD

570876010150

10750

8760 3

6

.xx

x

xplantofCapacity

generatedEnergyfactorcapacityPlant

Load Characteristic


c) Plant use factor

Energy that can be generated by two 60MW units and

one 30MW unit

=2 units x 60MW x 7500h + 30MW x 4500h

=103.5x104MWh=1035x106kWh


720101035

107506

6

.kWhx

kWhxfactorusePlant

Load Characteristic

• Example 3

Assume the annual peak load of a primary feeder is

2000kW, at which the power loss i.e., total copper, or

I2R, loss is 80kW per three phase. Assuming an annual

loss factor of 0.15.

Determine:

(a) The average annual power loss

(b) The total annual energy loss due to the copper losses

of the feeder circuits.


Load Characteristic


a) Average annual power loss

Average power loss = Power loss at peak load x FLS

= 80 kW x 0.15

=12 kW

b) Total annual energy loss

TAELCU = Average power loss x 8760 h

=12 kW x 8760 h

=105,120 kWh


Example of Transformer Sizing

• An application has been made to the utility to connect a commercial complex to the nearest distribution transformer. The distribution transformer is rated 310 kVA and the current maximum demand on this transformer is 250 kVA. The transformer is 20 years old and considering hot summer, it is suggested not to load it beyond 90% capacity. The load of the commercial complex are:

A gas station consisting of three unleaded petrol pumps, a diesel pump, and a LNG pump. Each pump is rated at 5 hp and has an efficiency of 0.9. A general purpose shop with installed load of 5 kW. A Fish and Chip shop with a connected load of 4 kW. Six street lights each rated at 100 W. The diversity factor of the pumps is 1.8. The demand factor of the general purpose shop is estimated to be 0.7. The Fish and Chip shop has a demand factor of 0.8. The diversity factor for the whole load is estimated to be 1.2. Calculate the the coincident maximum demand of the whole complex. Is it possible for this load to be connected to the existing distribution transformer?

20

21

Load Forecasting

Introduction

• Load forecasting plays an important role in power system

planning, operation and control

• Forecasting means estimating active load at various load

buses ahead of actual load occurrence.

• Planning and operational applications of load forecasting

requires a certain ‘lead time’ also called forecasting

intervals.

• A good forecast reflecting current and future trends,

tempered with good judgment, is the key to all planning,

indeed to financial success.


Load Forecasting

• Factors Affecting Load Forecast


Load Forecasting

• Nature of Forecast


Nature of Forecast Lead time Application

Very Short Term A few seconds to

several minutes

Generation, distribution

schedules, contingency analysis

for system security

Short Term Half an hour to a

few hours

Allocation of spinning reserve;

operational planning and unit

commitment; maintenance

scheduling

Medium Term A few days to a few

weeks

Planning for seasonal peak-

winter, summer

Long Term A few months to a

few years

Planning generation growth

Load Forecasting

• Forecasting Methodology

• Forecasting techniques may be divided into three broad

classes. Techniques may be based on extrapolation or on

correlation or on a combination of both.

Extrapolation

– Extrapolation techniques involve fitting trend curves to

basic historical data adjusted to reflect the growth trend

itself.

Correlation

– Correlation techniques of forecasting relate system loads to

various demographic and economic factor.


Load Growth

Introduction

• In planning to accommodate future electric energy needs,

it is necessary that an estimate of the rate at which those

needs will grow.


• Examples of a typical

energy requirement

projection for a certain

country

Load Growth

Growth Rates

• Suppose a certain quantity M grows at a rate that is

proportional to the amount of M that is present.

• It gives

• Where a is the constant of proportionality, known as the

per-unit growth rate. The above equation may be written

as

• Where M0 is the value of M at t=0.


aMdt

dM

ateMM 0

Load Growth

Growth Rates

• At any two values of time, t1 and t2, the inverse ratio of

the corresponding quantities M1 and M2 is

• Based on above equation, the doubling time td may be

obtain such that M2 = 2M1 and t2 - t1= td


)tt(ae

M

M12

2

1

a

.

a

lntd

69302

Load Growth

Growth Rates

• Power system planners also need to know how much

power will be demanded.


• The approximation equation

in this curve

• Where P0 is the peak power at

t =0, b is the per unit growth

btePP 0

Examples of the peak power demand

for a certain country over several years.

Load Growth

Growth Rates

• The area under this curve over a given period is a

measure of the energy Q consumed during that period.


• If the per unit growth rate has not

changed, then the energy

consumed in one doubling period

equals the energy consumed for

the entire time prior to that

doubling.

Load Growth

• Growth Rates

• Evaluating the energy Q1 consumed up to t1 and the

energy Q2 consumed during the doubling time td = t2 - t1

• Td = (ln 2/b), therefore Q2 becomes


100

21112 Qe

b

Pe)(

b

PQ

btbt

1

2

1

1

1

1002

001

btbtbt

t

t

btbt

t

e)e(b

PdtePQ

eb

PdtePQ

d

Load Growth

• Example 1

• Suppose that the consumption of the energy in a certain

country has a growth rate of 4 percent per year.

In how many years will the energy consumption be

tripled?



years..

lnt

t.lnorethenQ

Q t.

4727040

3

040333 040

1

2

Load Growth

• Example 2

• In certain country the energy consumption is expected to

be double in 10 years.

Calculate the growth rate in percent.



%.x.

a

t

lna

d

93610010

6930

2

Load Growth

• Example 3

• Assume that one of the distribution transformer of the

Riverside substation supplies three primary feeders. The

30-min annual max demands per feeder as listed as

following table, together with the power factor (PF) at the

time of annual peak load. Assume a diversity factor of

1.15 among the three feeder for both real power (P) and

reactive power (Q).


Feeder Demand (kW) PF

1 1800 0.95

2 2000 0.85

3 2200 0.90

Load Growth

• Example 3

a) Calculate the 30-min annual max demand on the substation

transformer in kW and in kVar

b) Find the load diversity in KW

c) Select a suitable substation transformer size if zero load growth is

expected and if the company policy permits as much as 25 % short

time overload on the distribution substation transformer. Use the

suitable standard 3-phase as follow

2500/3125 KVA self-cooled/forced air cooled




35EET 305 Power System Fundamental - Mohd Rafi (SoESE)

Load Growth

• Example 3

d) Now assume that the substation load will increase at aconstant percentage rate per year and will double in 10years. If the 7500/9375 KVA rated transformer isinstalled, in how many years will be loaded to its fans-on rating. Assume the load growth equation is

Pn=P0(1+g)n

where: Pn=Load at the end of the nth year

P0=Initial Load

g=Annual growth rate

n=Numbers of years

36EET 305 Power System Fundamental - Mohd Rafi (SoESE)

Load Growth


a) Given the diversity factor FD = 1.15

Therefore, Annual maximum demand Dg,kW


15122002000180021 .

D

kWkWkW

D

D.......DD demandmax Coincident

demandmaxividual Sum of indF

gg

n

D

kW..

kW

.

)kWkWkW(D kW,g 395217

151

6000

151

220020001800

Load Growth


a) Dg =5217.39kW ( in KW), in KVA, then find the power

factor angle

The total Reactive Power (Q)

38

kVar.).)(tan(

).)(tan().)(tan(tanxPQ i

i

i

79289684252200

7931200021818003

1

PF Angle

0.95 18.2

0.85 31.79

0.90 25.84

kVar..

kVar.D kVar,g 952518

151

792896

Load Growth


a) Therefore Dg

Dg =(P2+Q2)1/2=S

= ( 52172+2518.822)1/2 = 5793.60 KVA

b) The load diversity (LD) is

39

kWkWkWDDLD g

n

i

i 783521760001

Load Growth


c) The transformer size capacity if permits of 25% short

time overload

The maximum demand is 5793.60 KVA

The most suitable distribution transformer is 3750/4687

KVA self-cooled/ forced air cooled

40

Tx Size 25% overload Remarks

2500/3125 KVA 3125x1.25=3906.25 Under size

3750/4687 KVA 4687x1.25=5858.75 Nearest

5000/6250 KVA 6250x1.25=7812.5 Over size

7500/9375 KVA 9375x1.25=11718.75 Over size

Load Growth


d) The term fans-on means the forced air cooled rating.

The increase annual growth rate (g) per year,

given Pn=P0(1+g)n

hence (1+g)10=2 , 1+ g =1.07177

g = 0.07177%/year

Therefore, (1.07177)n x 5793.60 KVA = 9375 KVA

Or

If the 7500/9375 KVA rated transformer is installed, it

will be loaded to its fan-on rating in about 7 years.41

Yearsor..ln

.lnn 79446

071771

61821

Costumer Billing

• Introduction

• Customer billing is done by taking the difference in

readings of the meter at two successive times, usually at

an interval of 1 month.

• The difference in readings indicates the amount of

electricity, in kilowatt hours, consumed by the customer

in that period.

• The amount is multiplied by the appropriate rate or the

series of rates and the adjustment factor, and the bill is

sent to the customer.

42

Costumer Billing

43

• Sample of

Electricity Bill

Costumer Billing

• Tariff

• ‘Electricity Tariff’ can be define as a list of fixed rateelectricity prices which has been approved by a government.

• In Malaysia the tariff for electricity are divided into 7categories as follow:

1) Domestic

2) Commercial

3) Industrial

4) Mining

5) Street Lighting

6) Specific Agriculture

7) Top up & Stand By

44

Costumer Billing

• Tariff

• Examples of Electricity Tariff in Malaysia

45

Costumer Billing

• Example 1

• Assume that the customer A, use the following and

typical , domestic tariff rate schedule.

46

Costumer Billing

• Example 1(cont)

a) Assume that an average month is 730h and find the

monthly load factor

b) Find the reasonable size of continuous KVA rating of the

distribution transformer

c) Calculate the monthly bill

d) What size of capacitor (in kVar) would rise the PF of the

customer to 0.9

47

Costumer Billing


a) Customer A , FLD

b) Continuous KVA rating

The continuous suitable rating for the distribution

transformer is 10 KVA

48

20507308

1200.

hxkW

kWh

TxloadPeak

servedUnitFLD

kVA..

kW

Cos

PS A

A 419850

8

Costumer Billing


c) The monthly bill

First 200kWh = 200kWh x 21.8 sen/kWh = RM43.6

Next800kWh = 800 kWh x 28.9 sen/kWh =RM 231.2

Over 1000kWh = 200 kWh x 31.2 sen/kWh = RM 62.4

The total monthly Bill= RM 337.20

d) Size of capacitor

Current PF=0.85, the kVarh value is

49

kVarh.).Cos(xSin.

kWh69743850

850

1200 1

Costumer Billing


d) Size of capacitor (cont.)

At PF=0.9, the kVarh value is

Therefore, the capacitor size required is

50

kVarh.).Cos(xSin.

kWh1958190

90

1200 1

kVar.h

kVarh)..(220

730

1958169743

Power Factor Surcharge

Percent of

surcharge from

the current bill

Condition

1.5%For every 0.01 less than

0.85 power factor

3%For every 0.01 less than

0.75 power factor

51

52

Power Factor

• Definition of Power Factor

• Power factor is the ratio between actual (true) load power

(kW) and the apparent load power (kVA)

• It is a measure of how effectively the current is being

converted into useful work output and more particularly

is a good indicator of the effect of the load current on

the efficiency of the supply.

53

)KVA(powerloadApparent

)kW(powerloadActualpf

Power Factor

• Fundamental of Basic Electricity - The Power Triangle

54

P - kW

Q - kVar

S - kVA

KVA

kWFactorPower

QPS

22

Power Factor

• Equipment Causing Poor Power Factor

• Lightly loaded induction motor. Examples of this type of

equipment and their approximate power factor are:

70% power factor or better: Air conditioners, pumps,

center less grinders, cold header, up setter, fans or

blower

60% to 70% power factor: Induction furnaces,

standard stamping machines and weaving machines

60% power factor and below: Single-stroke presses,

automated machine tools, finish grinders, welders

55

Power Factor

• Reactive Power Problem (Motor)

• Example that a motor is rated at 10,000W at 0.8 power

factor. The resistance is 5ohm. At 415V, the motor will

require the following amount of current:

I=10000/(√3x0.8x415)=17.39A

Losses when pf =0.8 : I2R=(17.39)2(5)=1,512W

The same motor rated at 0.65 power factor will require:

I=10000/(√3x0.65x415)=21.403A

Losses when pf=0.65 : I2R=(21.403)2(5)=2290.4W

56

Power Factor

• Reactive Power Problem (Transformer)

• Example that 11/0.433 kV 1000kVA transformer has

maximum loading of 800KW and power factor of 0.45

What is the % loading of the transformer?

PF=KW/KVA=0.45

KVA(Load) = kW/PF=800/0.45=1777

%Tx Load = kVA (Load) / Tx Capacity

= (1777 / 1000 ) x 100 = 177%

57

Power Factor


• Example that 11/0.433 kV 1000kVA transformer has

maximum loading of 800KW and power factor of 0.9

What is the % loading of the transformer?

PF=KW/KVA=0.9

KVA(Load) = kW/PF=800/0.9=888.88

%Tx Load = kVA (Load) / Tx Capacity

= (888.88 / 1000 ) x 100 = 88.88%

58

Power Factor


• Condition 1

PF=0.45

TX Size=1000kVA

Load KVA=1777 KVA

%TX Load=177%

• Condition 2

PF=0.9

TX Size=1000kVA

Load KVA=888 KVA

%TX Load=88%

59

Power Factor

• Minimum Power Factor

• Customers are advise to maintain power factor at

minimum of 0.85

60

Power Factor

• Power Factor Improvement

61

Example Of Power Flow Diagram Of Industrial Plant

Power Factor


62

Power Triangle To Illustrate Power Factor Correction

Power Factor


• The amount of reactive compensation supplied by the

capacitor bank is

• Apparent power consumed by the load before adding

capacitors

• Apparent power supplied by the source after adding

capacitors

63

11

1Cos

P

PF

PS LOADLOAD

22

2Cos

P

PF

PS LOADLOAD

21 QQQCAP

Power Factor


• Where PF1, and PF2 are the actual load power factor and

desired system power factors, respectively.

64

2

2

1

22

11 LOADLOAD

LOAD PPF

PPSQ

2

2

2

22

22 LOADLOAD

LOAD PPF

PPSQ

1

11

12

2

2

1 PFPFxPQ LOADCAP

Power Factor

• Example 1

• An industrial plant has an active power demand of

500kW at a power factor of 0.76 lagging. Determine the

reactive power rating of the capacitor bank required to

improve the power factor to the following:

a) 0.8 lagging

b) 0.9 lagging

c) Unity

Assume the capacitor steps are available in 50 kVar

increments

65

Power Factor


66

1

11

12

2

2

1 PFPFxPQ LOADCAP

kVarkVar...

xkWQ)a CAP 506521800

11

760

1500

22

kVarkVar...

xkWQ)b CAP 20041851900

11

760

1500

22

kVarkVar..

xkWQ)c CAP 400642711

11

760

1500

22

Power Factor

• Example 2

Assume that a 700 kVA load has a 65% power factor. It is

desired to improve the power factor to 92%.Using the

power factor correction table (Table 1).

Determine the following:

a) The correction factor required

b) The capacitor size required

c) What would be the resulting power factor if the next

higher standard capacitor size is used. Assume the

capacitor steps are available in 50 kVar increments

67

Power Factor


a) From correction factor table (Table 1), the correctionfactor required can be found as 0.743

b) The 700 KVA load at 65% power factor is

PL=SL x cos W

=700k x 0.65 W

=455 kW

The capacitor size necessary to improve the power factorfrom 65 to 92% can be found as

Capacitor size = PL x (Correction Factor)

=(455)(0.743)

=338.065 kVar

68

Power Factor


c) Assume that the next higher standard capacitor size (or

rating) is selected to be 350 kVar. Therefore the resulting

new correction factor can be found from

From the table by using interpolation method, based on

original power factor (65%) and new correction factor

(0.769), refer to Table 1.

The new corrected % PF = (92)(0.1613)+(93)(0.8387)

= 92.8387 % =0.92869

7690455

350.

kW

kVar

P

RatingCapacitorStandardFactorCorrectionNew

L

70

Documents

EET 305 Chapter 5 the Load in Power System