lecture notes on the different elements regarding power instrumentation

it typically many ideals to study about

1. 2.1 Terms and definitions of load on an
Electrical Power Plant
2.1.1 Connected load:
It is the sum of ratings, in kW, of the equipment installed in the
consumer's premises. If a consumer has:
4 lamps of 60 W each,
a cocker of 500 W and
a radio consuming 60 W
then the total connected load will be:
= 4x60 + 500 + 60 = 800 W.

2. 2.1.2 Maximum demand:
It is the maximum load that a consumer uses at any
time.
If all the electrical appliances in a consumer's house
are running to their fullest simultaneously then, the
maximum demand will be equal to the connected load.
Therefore, maximum demand of a power station is the
maximum load on the power station in a given period.

3. 2.1.3 Demand Factor (DF): load connected
demand maximum
factor demand 

4. 2.1.4 Load Curve:
It is a graphical representation of an electrical load,
in kW, plotted against time, in hours.
The load curve of an EPP shows the load variations
on that power station.
When plotted for 24 hours it is referred to as a daily
load curve.
Similarly, when the considered time is one year
(8760 hours) then it is called annual load curve.

5. Table7.1: Average hourly loads (kW) for the example day
h h

6. Figure7.1:AveragehourlyloadsfortheexampledaykW
h
kWh
time
dayainenegyTotal
loadAverage 45.15
24
371


7. Figure7.2:Variationindemandwithdemandinterval

8. 2.1.5 Load Factor: demand maximum
load average
factor Load 

9. Load Factors and Demand Factors are
always less than unity.
The higher the Load Factor the lesser will be
the cost of generation per unit for the same
maximum demand.

10. 2.1.6 Base Load Electrical Power Plant:
These types of power stations run throughout
the year, are of large capacity and run at high
load factors.
These are the EPP's used to supply the load
of the base portion of the load curve.

11. 2.1.7 Peak Load Electrical Power Plant:
These types of power stations are of smaller
capacity and they run for a short period in the
year and work at low load factors.
These type of stations should be capable of
quick starting .
Because Peak load power stations supplies
the load on the top portion of the load curve,
they are also called Peak Load Plants.

12. 2.1.8 Plant Capacity Factor :
It is defined as the ratio of the actual energy
produced, in kWh, to the maximum possible energy
that could have been produced during the same
period.
Plant capacity factor =
where: E is the energy, in kWh, in a given period
C is the capacity of the plant, in kW
t is the total number of hours in the given
period.

13. 2.1.9 Plant use factor:
It is defined as the ratio of energy produced
in given time to the maximum possible
energy that could have been produced during
the actual number of hours the plant was in
operation.
Plant use factor =
where t
1is actual number of hours the plant
has been in operation.

14. 2.1.10 Diversity factor :
It is defined as the ratio of sum of individual
maximum demands to the simultaneous Maximum
demands of a system.
Therefore, Diversity factor is more than unity.
Usually the maximum demand of individuals do not
occur at the same time and thus simultaneous
maximum demand will be less than their total
maximum demand.

15. 2.1.10 Diversity factor :
It has to be noted that, power stations should
be capable of supplying the simultaneous
maximum demand without any problem.

16. 2.2Types of Loads and Load Curves
Regardless of how one classifies their electrical loads,
the loads that can be connected to the supply from a
power station (or grid) may be put under some
categories as shown below.
Types of Loads
(i) Residential loads
(ii) Commercial loads
(iii) Industrial loads
(iv) Municipal load
(v) Traction load, etc.

17. Load Curves
These are the electrical characteristics that give the following
information:
(i) The variation of load during different hours of the day
(ii) The maximum and minimum values of the load
(iii) Annual maximum and minimum values of the load
(iv) Average annual load on the station
(v) Information whether the installation is working efficiently
or not.
Load curves are commonly represented in rectangular curves
as although, in reality the load curves, if traced to the minute,
are obtained to be curvilinear shapes.

18. Figure7.2:Variationindemandwithdemandinterval

19. Example 2.1
A power station has a maximum demand of 80 MW and its
daily load curve is defined as follows:
(a)Determine the load factor of the power station
(b) What is the load factor of the stand-by equipment rated at 30
MW that takes up all loads in excess of 60 MW?
Time
(Hours)
0-6 6-88-1212-1414-1818-2222-24
Load (MW) 30 50 60 50 70 80 30

20. Solution:
(a)Energy generated
=(30x6+50x2+60x4+50x2+70x4+80x4+30x2)kWh
= 1280x 10
3
kWh

21. Therefore,
Average load = 1280x 10
3
kWh = 53,333 kW
24 h
Max. Demand = 80,000 kW
Load Factor = Average load=53,333 kW
Max. demand 80,000 kW
= 0.67 Answer

22. (b) Energy generated by stand-by equipment
= (l0 x 4+ 20x 4) x 10
3
= 120x 10
3
kWh
NOTE:
10 MW = 70 MW-60 MW between 1400 and
1800 hours
and
20 MW = 80 MW -60 MW between 1800 and
2200 hours.

23. The time in which the stand-by equipment is available
is 24 hours. Therefore,
Average load = 120x 10
3
kWh = 5 x 10
3
kW, then
24h
Load factor = 5 x 10
3
kW= 0.167 Answer
30 x 10
3
kW

24. Example 2.2
A central power station has annual factors as follows:
load factor = 60%,
Capacity factor = 40%,
Use factors = 45%.
The power station has a maximum demand of 15,000
kW. Determine:
(a) Annual energy production
(b) Reserve capacity over and above peak load
(c) Hours per year not in service

25. Solution:
(a)Since Load factor = Average load
Max. demand
Then,
Average load = Load factor x Maximum demand
= 0.6 x 15,000 = 9000 kW
Thus, annual energy produced is:
E = 9000 x 8760 = 78.84 x 10
6
kWh

26. b) We know that, Capacity factor =
where C is capacity of the plant,
t is the time in hours in one year = 8760 hrs
Therefore, the capacity of the plant will be obtained as:
C = 78.84 x 10
6
= 22,500 kW
0.4 x 8760
Then, reserved capacity will be
= C -Maximum demand
= 22,5000 -15,000 = 7500 kW

27. (c) From, Use factor =
where t 1 = Actual number of hours of the year for which
the plant remains in operation, then
t 1 = Energy (kWh) = 78.84 X 10
6
= 7786 hours
C x Use factor 22,500 x 0.45
Therefore,
hours per year not in service = 8760-7786 = 974 hrs

28. Example 2.3
Find the diversity factor of a power station which supplies the
following loads:
Load A: Motor load of 100 kW working between
10.00a.m. and 6.00 p.m.
Load B: Lighting load of 60 kW working between 6.00 p.m.
and 10.00 p.m.
Load C: Pumping load of 40 kW working between 4.00 p.m.
and 10.00 a.m.
Solution:
Time Load (kW) Total load (kW)
10 a.m. to 4 p.m. 100 (motor) 100
4 p.m. to 6 p.m. 100 (motor) + 40 (pumping) 140
6 p.m. to 10 p.m. 60 (lighting) + 40 (pumping) 100
10 p.m. to 10 a.m. 40 (pumping) 40

29. From the above values of the total loads we see that
simultaneous maximum demand on the power station is 140 kW
which happens between 4 p.m. and 6 p.m.
The diversity factor will, therefore, be obtained as:
Diversity factor = 200= 1.43 Answer
140

30. Example 2.4
The annual peak load on a 30 MW power station is 25 MW. The
power station supplies loads having maximum demands of 10
MW, 8.5 MW, 5 MW and 4.5 MW. The annual load factor is 45
%, find:
(a) Average load
(b) Energy supplied per year
(c) Diversity factor
(d) Demand factor
Solution:
The capacity of the power station is given = 30 MW
Maximum (peak) demand on power station =25 MW

31. (a) From Load factor = Average load
maximum demand
we obtain that,
Average load = 0.45 x 25 = 11.25 MW Answer
(b) Energy supplied per year = Average load x 8760
= 11.25 x 10
3
x 8760
= 98.55 x 10
6
kWh Answer
(c) Diversity factor is a ratio of the sum of the individual
maximum demands to simultaneous maximum demand.
Thus:
Diversity factor = 10 + 8.5 + 5 + 4.5 = 1.12 Answer
25

32. (d) Demand factor of the power station will be:
Demand factor = Maximum demand
Connected load
=
= 0.89 Answer

33. Example 2.5:
For a power station the yearly duration curve is a straight line
from 30,000 kW to 4,000 kW. To meet the load, three turbo-
generators are installed. The capacity of two generators is
15,000 kW each and the third is rated at 5,000 kW. Determine
the following:
(a) Load factor
(b) Capacity factor or plant factor
(c) Maximum demand
Solution:
Load factor = Average load
Max. demand

34. 30,000
Load
(kW)
4,000
22
Time, 8760 hours
From the given duration curve, generated per year will be equal
to the area under the curve.
Therefore, the energy E generated in a year will be:

35. E = 4000 x 8760 + 2600 x 8760 = 8760(4000 + 13000)
2
= 8760 x 17000 kWh
Average load = Energy generated per year
Hours in a year
= 8760 x 17000 = 17,000 kW
8760
With Maximum given as 30,000 kW then:
(a)Load factor = 17,000 = 0.57
30,000

36. (b) But Capacity of plant = 15,000 + 15,000 + 5,000 =35,000 kW
Therefore: Capacity factor = 8760 x 17,000 = 0.49 Answer
8760 x 35,000
(c) The Maximum demand = 30 MW
Example 2.6:
A power station has two 60 MW units each running for 7,000
hours a year and one 30 MW unit running for 1,500 hours a
year. The energy produced per year is 700xl0
6
kWh.
Calculate the following:
(a) Plant load factor
(b) Plant use factor or utilisation factor

37. (a) Plant load factor = ?
We know that, Load factor = Average load
Max. demand
But average load in a year will be = E= 700 X 10
6
t 8760
Given the power station capacity = 60 + 60 + 30 = 150 MW,
then assuming power station capacity equal to Maximum
demand we obtain plant load factor:
Load factor = Average load = 700 X 10
6
= 0.53
Max. demand 8760 x 150 x 10
6

38. (b) It is given that the actual generated energy in a year is
700'10
6
kWh.
However, during the 7,000 hours the two units of 60 MW each
were running together with a 30 MW unit which ran for 1,500
hours the energy that could be generated is:
= 2 x 60 x 7000 + 30 x 1500 = 885 x 10
6
kWh
Therefore, Plant use factor = 700x 10
6
= 0.79 Answer
885 x 10
6