Power station

power station

1. Course: Power Station

Faculty Name: Taskin Jamal

AIUB

2. Energy Requirements
•For Power plant design the first step is to define the
Energy Requirements that the plant must meet.

•The conditions are:
1.Maximum demand
2.Total energy requirements and
3.Distribution of energy demand

3. Maximum Demand
•The capacity of a plant depends upon the maximum
power demand made by the consuming devices.

•Power consuming devices are of different types
including the machines, lights, heaters, processors
energy converters etc.

•Loads are connected to the distribution side of the
power grid.

4. Single Line Diagram
•Single Line Diagram of Power System

5. •Elec. Energy is generated at high potential varying
from 4 to 18 kV in most cases.

•Transmission over the feeders may take place at
voltage level as high as 400 kV.

•The longer the distance the higher the line voltage
employed

•At the distribution buses the voltage is transferred to
the voltage required by the consuming devices.

6. Demand Factor
•Demand Factor =

•If all the devices run to their fullest extent at the same
time, the maximum demand of the consumer on the
system would equal his connected load.

•Experience demonstrates that generally Maximum
Demand < Connected Load.

7. Group Diversity Factor (GDF)
•Max demands of individual customers don’t occur simultaneously
but are spread out over a period of time.

•The time distribution of maximum demands for similar types of
consumers is measured by the diversity factor.

•Group Diversity Factor =

•This holds for consumers whose activities and energy requirements
are very similar.

•This value is always greater than the unity. For example, factors for
residential consumers are usually highest at about 5.0, whereas for
industrial consumers may have values as low as 1.3.

8. Peak Diversity Factor (PDF)
•The peak demand of a system is made up of the individual
demands of the devices that happen to be functioning at the
time of the peak.

•At the time of the system peak demand, the demand of a
particular group of similar consumers is seldom at the
maximum value that it may reach at some other time of the
year.

•Peak Diversity Factor =

9. Maximum Demand Determination
•The method of determining the max demand to be
expected on a system when the foregoing data are
available is:

•Let,
–C
1, C
1
’
, C
1
”
… C
1
n
= individual connected loads of group 1
–C
2, C
2
’
, C
2
”
… C
2
n
= individual connected loads of group 2
–d
1 = demand factor of group 1
–d
2 = demand factor of group 2
–D
1 = GDF among the consumers of group 1
–D
2 = GDF among the consumers of group 2

10. –M
1 = max demand of group 1
–M
2 = max demand of group 2
–r
1 = PDF for group 1
–r
2 = PDF for group 2
–L
m = system max demand
–L
1, L
2, L
3, … L
n = demands of each type of consumers at
the time of system max demand.

Then, GDF =

So, D
1 = (C
1d
1 + C
1’d
1’ + … + C
1
n
d
1
n
) / M
1
or, M
1 = C
1d
1 / D
1 and similarly, M
2 = C
2d
2 / D
2

11. –PDF =
– r
1 = M
1 / L
1
–So, L
1 = M
1 / r
1
–Similarly, L
2 = M
2 / r
2
–
and L
M = L
1 + L
2 + L
3 + … L
n

12. Example 30.1
•A new housing development is to be added to the
lines of a public utility. There are 1000 apartments
each having a connected load of 4 kW, also stores and
services are included of the characteristics shown in
the following tabulation.

13. •The demand factor of the apartments is 45%. The
GDF of residential load is for the system is 3.5, the
PDF is 1.4. The commercial load group GDF is 1.5
and the peak diversity factor is 1.1.

•Find the increase in the peak demand on the total
system delivery from the station bus from addition of
this development of the distribution system. Assume
line losses at 5% of delivered energy.

14. Solution of Ex: 30.1
•Max demand per apartment = 4 * 0.45 = 1.8 kW

•Actual Max demand of 1000 apartments =
(1.8 * 1000 / 3.5 ) = 514 kW

•Demand of 1000 apartments at time of system peak =
514 / 1.4 = 367 kW

15. •Commercial loads as in the following tabulation:

16. •Actual Max demand of the commercial group =
140.1 / 1.5 = 94 kW

•Commercial demand at time of system peak =
94 / 1.1 = 86 kW

•So, demand at point of consumption at time of system
maximum demand = 367 + 86 = 453 kW

•So, Total increase in max demand at station bus =
{453 + (453 * 0.5)} = 476 kW.

17. Load Curves
•The chronological variation in demand for energy on
the source of supply is plotted graphically for study
and easy comprehension. Such graphs for electrical
energy demands are generally termed as load curves.

18. Load Curves

19. Load Curves

20. Load Curve Variations
•Weekdays and weekend variation
•Seasonal variation
•Weather Effect
•Variation due to special occasions
•Variation in urban and rural areas
•Variation in types of services ( e.g. inter-city trains
and intra-city trains)

21. Load Curve Analysis
•The area under the curve of a daily chronological load
curves measures the total energy consumed by the
load during the day. This energy is evaluated by
, the unit being the kilowatt-hour (kWh).

22. Load Duration Curve

23. Energy-Load Curve
•Energy-load curve plots the cumulative integrations
of the are under the load curve starting at zero load vs
a particular load.

•It is the plot of vs kW.

24. Load Factor
Load Factor:

The degree of variation of load over a period of time is measured by the load factor
which may be defined as



Where L
max = Peak load for period
L
avg = Average load for period
E
= Total Energy in load curve for period
h
= total number of hours in period

•Measures the variation of the load
•Doesn’t provide the exact shape of the load duration curve
•Always less than 1 because the average load is smaller than the maximum load
•Higher load factor means lesser maximum demand indicating reduced cost per
unit generated maxmax
avg
L
h/E
L
L
factor Load 

25. Load Factor












•if a load factor approaches to 0; the curve approaches to L shape with very low
or no load during the major portion of time
•if a load factor approaches to unity; the curve approaches to a rectangular shape
indication high sustained load

26. Capacity Factor
Capacity Factor:

The extent of use of the generating plant is measured by the capacity factor also
called plant factor or use factor

• If during a given period a plant is kept fully loaded it is evident that it is used to the
maximum extent or operated at 100% capacity factor
• If no energy was produce the capacity factor would be 0%








factorLoad
Cap
L
Cap
L
FactorCapacity
avg
 )(
max

27. Utilization Factor
Utilization Factor:

The utilization factor measures the use made of the total installed capacity of the
plant


• A low utilization factor may mean that the plant is used only for standby purpose or
that the capacity has been installed well in advance of need
• A high utilization factor indicates the plant is probably most efficient in the system
• In a isolated system a high utilization factor refers a good design with some reserve
capacity allowance
• It might be higher than unity which pointing towards that the loads have been
carried in excess of the rated capacity of the equipment



Cap
L
FactornUtilizatio
max
 factorLoadFactornUtilizatioFactorCapacity 

28. Mathematical Problem

Exercise: 30.4
The yearly duration curve can be considered as a straight line from 20,000 to
3,000kW. To meet this load 3turbine generator units 2 rated at 10,00kW each and 1
at 5,000kW are installed. Determined:

i.Installed Capacity
ii.Plant Factor
iii.Maximum demand
iv.Load factor
v.Utilization Factor

29. Mathematical Problem
Solution:
The time has to be converted in to hour format










i. Installed Capacity = (10,000 x 2) + 5,000 = 25,000kW
ii. Plant Factor:
Total load should be area under the above curve

3000
0
5,000
10,000
15,000
20,000
25,000
0 8760
Load (kW)

Load Curve

30. Mathematical Problem
Total load = (1/2 x 8760 x 17,000) + (8760 x 3000) = 100740 kW
L
avg = (100740 / 8760) = 11,500kW
So the plant factor = (11,500/25,000) = 0.46 = 46%

iii. Maximum Demand:

From the graph the we can observe that the maximum load is 20,000kW (L
max)

iv. Load Factor:

Load factor = L
avg /L
max = 11,500/20,000 = 0.575

v. Utilization Factor

Utilization Factor = L
max /C
ap = 20,000/25,000 = 0.8

31. Mathematical Problem
30.5 Determine the maximum demand for the group of energy consumers shown in
the following table:





Class of Service Total Connected Load (kW) Demand Factor (%) Group Diversity Factor Peak Diversity Factor
Public Building 100 35 1.6 1
Apartment 1000 55 4 1.2
Hospital 200 45 1.5 1.05
Theater 150 60 1.6 1
Laundries 50 70 1.8 1.05
Residence 3000 40 4.5 1.2
Stores 500 65 1.6 1.05
Offices 100 70 1.8 1.05
Lighting (Street) 600 100 1 1
Foundry 3500 80 1.1 1.05
Boiler Factory 4000 90 1.1 1.05
Hotel 700 25 1.8 1.2
Motor Factory 5000 75 1.1 1.05

32. Mathematical Problem





Class of
Service
Total Connected
Load (kW)
Demand
Factor (%)
Group Diversity
Factor
Peak Diversity
Factor
Maximum
Demand( kW)
Max Demand
Group (kW)
Public
Building
100 35 1.6 1 35 21.88
Apartment 1000 55 4 1.2 550 137.50
Hospital 200 45 1.5 1.05 90 60.00
Theater 150 60 1.6 1 90 56.25
Laundries 50 70 1.8 1.05 35 19.44
Residence 3000 40 4.5 1.2 1200 266.67
Stores 500 65 1.6 1.05 325 203.13
Offices 100 70 1.8 1.05 70 38.89
Lighting
(Street)
600 100 1 1 600 600.00
Foundry 3500 80 1.1 1.05 2800 2545.45
Boiler
Factory
4000 90 1.1 1.05 3600 3272.73
Hotel 700 25 1.8 1.2 175 97.22
Motor
Factory
5000 75 1.1 1.05 3750 3409.09
13320 10728.24
Total Connected Load
(kW)

X

Demand Factor

33. Mathematical Problem





Class of
Service
Total Connected
Load (kW)
Demand
Factor (%)
Group Diversity
Factor
Peak Diversity
Factor
Maximum
Demand
Maximum
Demand Group
Public
Building
100 35 1.6 1 35 21.88
Apartment 1000 55 4 1.2 550 137.50
Hospital 200 45 1.5 1.05 90 60.00
Theater 150 60 1.6 1 90 56.25
Laundries 50 70 1.8 1.05 35 19.44
Residence 3000 40 4.5 1.2 1200 266.67
Stores 500 65 1.6 1.05 325 203.13
Offices 100 70 1.8 1.05 70 38.89
Lighting
(Street)
600 100 1 1 600 600.00
Foundry 3500 80 1.1 1.05 2800 2545.45
Boiler
Factory
4000 90 1.1 1.05 3600 3272.73
Hotel 700 25 1.8 1.2 175 97.22
Motor
Factory
5000 75 1.1 1.05 3750 3409.09
13320 10728.24
Maximum Load (kW)
/

Group Diversity Factor

34. Mathematical Problem
Exercise 30.2: An electrical railway system has such severely fluctuating loads that instantaneous
wattmeter reading do not define its load curve adequately. The readings of the following table
has taken from the station totalizing watt-hour meter at the times indicated. The station meter
constant is 10,000 to convert to kilowatt-hours.









a.Plot the average hourly chronological and load duration curve
b.Plot the load-energy curve for the average hourly loads
c.Find the load factor based on the average hourly peak
d.If the instantaneous peak is 85MW, what is the load factor?
e.What is the utilization factor and capacity factor of the plant serving this load if its capacity
is 100MW?

Time Meter Reading Time Meter Reading
12pm 5595 2 pm 5639
1am 5597 3 5643
6 5602 4 5648
7 5605 5 5654
8 5611 6 5661
9 5618 7 5667
10 5624 8 5672
11 5629 9 5676
12 pm 5633 10 5678
1 pm 5636 12 am 5682

35. Mathematical Problem
Time Meter Reading Duration (Hours) kWhr MW
12pm 5595 0 0 0
1am 5597 1 20000 20
6 5602 5 50000 10
7 5605 1 30000 30
8 5611 1 60000 60
9 5618 1 70000 70
10 5624 1 60000 60
11 5629 1 50000 50
12m 5633 1 40000 40
1pm 5636 1 30000 30
2pm 5639 1 30000 30
3 5643 1 40000 40
4 5648 1 50000 50
5 5654 1 60000 60
6 5661 1 70000 70
7 5667 1 60000 60
8 5672 1 50000 50
9 5676 1 40000 40
10 5678 1 20000 20
12 5682 2 40000 20
kWhr = ( Present Reading – Previous Reading ) x Meter Constant
MW = kWhr / (1000 x hour duration)

36. Mathematical Problem
Time MW
12pm 0
1am 20
6 10
7 30
8 60
9 70
10 60
11 50
12m 40
1pm 30
2pm 30
3 40
4 50
5 60
6 70
7 60
8 50
9 40
10 20
12 20

37. Mathematical Problem

38. Mathematical Problem
10 => 10 X 24 = 240

20 => 19 X10 = 190
240 + 190 = 430

30=> 15 X 10 = 150
430 + 150 = 580

40=> 10 X 12 = 120
580 + 120 = 700

50=> 10 X 9 = 90
700 + 90 = 790

60=> 10 X 6 = 60
790 + 60 = 850

70=> 10 X 2 = 20
850 + 20 = 870

39. Mathematical Problem
10 => 10 X 24 = 240

20 => 19 X10 = 190
240 + 190 = 430

30=> 15 X 10 = 150
430 + 150 = 580

40=> 10 X 12 = 120
580 + 120 = 700

50=> 10 X 9 = 90
700 + 90 = 790

60=> 10 X 6 = 60
790 + 60 = 850

70=> 10 X 2 = 20
850 + 20 = 870

40. Mathematical Problem
c. Load Factor = L
avg/ L
max

L
avg = 870/24 = 36.25
MW and L
max = 70MW

Load Factor = 36.25 / 70 = 0.5178 = 51.78%



d. If L
max = 85MW

Load Factor = 36.25/85 = 0.4267 = 42.67%



e. Utilization Factor = L
max / Cap

Utilization Factor = 85/100 = 85%

Capacity Factor = L
avg / Cap
= 36.25/100 = 36.25%