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he function of a power station is to de- liver power to a large number of consum ers. However, the power demands of dif- ferent consumers vary in accordance with their activities. The result of this variation in demand is that load on a power station is never constant, rather it varies from time to time. Most of the complexities of modern power plant operation arise from the inherent variability of the load de- manded by the users. Unfortunately, electrical power cannot be stored and, therefore, the power station must produce power as and when de- manded to meet the requirements of the consum- ers. On one hand, the power engineer would like that the alternators in the power station should run at their rated capacity for maximum efficiency and on the other hand, the demands of the con- sumers have wide variations. This makes the design of a power station highly complex. In this chapter, we shall focus our attention on the prob- lems of variable load on power stations.
3.13.13.13.13.1 StructurStructurStructurStructurStructure of Electric Powere of Electric Powere of Electric Powere of Electric Powere of Electric Power
SystemSystemSystemSystemSystem
The function of an electric power system is to connect the power station to the consumers’ loads
C H A P T E RC H A P T E RC H A P T E R C H A P T E RC H A P T E R
Variable Load on Power Stations
3.1 Structure of Electric Power System 3.2 Variable Load on Power Station 3.3 Load Curves 3.4 Important Terms and Factors 3.5 Units Generated per Annum 3.6 Load Duration Curve 3.7 Types of Loads 3.8 Typical Demand and Diversity Fac- tors 3.9 Load Curves and Selection of Gener- ating Units 3.10 Important Points in the Selection of Units 3.11 Base Load and Peak Load on Power Station 3.12 Method of Meeting the Load 3.13 Interconnected Grid System
4141414141
IntrIntrIntrIntrIntroductionoductionoductionoductionoduction
T
he function of a power station is to de- liver power to a large number of consum ers. However, the power demands of dif- ferent consumers vary in accordance with their activities. The result of this variation in demand is that load on a power station is never constant, rather it varies from time to time. Most of the complexities of modern power plant operation arise from the inherent variability of the load de- manded by the users. Unfortunately, electrical power cannot be stored and, therefore, the power station must produce power as and when de- manded to meet the requirements of the consum- ers. On one hand, the power engineer would like that the alternators in the power station should run at their rated capacity for maximum efficiency and on the other hand, the demands of the con- sumers have wide variations. This makes the design of a power station highly complex. In this chapter, we shall focus our attention on the prob- lems of variable load on power stations.
3.13.13.13.13.1 StructurStructurStructurStructurStructure of Electric Powere of Electric Powere of Electric Powere of Electric Powere of Electric Power
SystemSystemSystemSystemSystem
The function of an electric power system is to connect the power station to the consumers’ loads
C H A P T E RC H A P T E RC H A P T E R C H A P T E RC H A P T E R
Variable Load on Power Stations
3.1 Structure of Electric Power System 3.2 Variable Load on Power Station 3.3 Load Curves 3.4 Important Terms and Factors 3.5 Units Generated per Annum 3.6 Load Duration Curve 3.7 Types of Loads 3.8 Typical Demand and Diversity Fac- tors 3.9 Load Curves and Selection of Gener- ating Units 3.10 Important Points in the Selection of Units 3.11 Base Load and Peak Load on Power Station 3.12 Method of Meeting the Load 3.13 Interconnected Grid System
4444444444 Principles of Power System
3.33.33.33.33.3 Load CurvesLoad CurvesLoad CurvesLoad CurvesLoad Curves
The curve showing the variation of load on the power station with respect to ( w.r.t ) time is known as a load curve****. The load on a power station is never constant; it varies from time to time. These load variations during the whole day ( i.e., 24 hours) are recorded half-hourly or hourly and are plotted against time on the graph. The curve thus obtained is known as daily load curve as it shows the variations of load w.r.t. time during the day. Fig. 3.2. shows a typical daily load curve of a power station. It is clear that load on the power station is varying, being maximum at 6 P.M. in this case. It may be seen that load curve indicates at a glance the general character of the load that is being imposed on the plant. Such a clear representation cannot be obtained from tabulated figures. The monthly load curve can be obtained from the daily load curves of that month. For this purpose, average* values of power over a month at different times of the day are calculated and then plotted on the graph. The monthly load curve is generally used to fix the rates of energy. The yearly load curve is obtained by considering the monthly load curves of that particular year. The yearly load curve is generally used to determine the annual load factor.
Importance. The daily load curves have attained a great importance in generation as they sup- ply the following information readily : ( i ) The daily load curve shows the variations of load on the power station during different hours of the day. ( ii ) The area under the daily load curve gives the number of units generated in the day. Units generated/day = Area (in kWh) under daily load curve. ( iii ) The highest point on the daily load curve represents the maximum demand on the station on that day. ( iv ) The area under the daily load curve divided by the total number of hours gives the average load on the station in the day.
Average load =
Area (in kWh) under daily load curve 24 hours ( v ) The ratio of the area under the load curve to the total area of rectangle in which it is con- tained gives the load factor.
Load factor =
Average load Max. demand
Average load 24 Max. demand 24
×
×
Area (in kWh) under daily load curve Total area of rectangle in which the load curve is contained
- For instance, if we consider the load on power station at mid-night during the various days of the month, it may vary slightly. Then the average will give the load at mid-night on the monthly curve.
Variable Load on Power Stations 4545454545
- It will be shown in Art. 3.9 that number and size of the gener- ating units are selected to fit the load curve. This helps in operating the generating units at or near the point of maximum efficiency. ** It is the sequence and time for which the various generating units ( i.e., alternators) in the plant will be put in operation.
( vi ) The load curve helps in selecting* the size and number of generating units. ( vii ) The load curve helps in preparing the operation schedule** of the station.
3.43.4 3.43.43.4 Important TImportant TImportant TImportant TImportant Terererererms and Factorsms and Factorsms and Factorsms and Factorsms and Factors
The variable load problem has introduced the following terms and factors in power plant engineering: ( i ) Connected load. It is the sum of continuous ratings of all the equipments connected to supply system. A power station supplies load to thousands of consumers. Each consumer has certain equipment installed in his premises. The sum of the continuous ratings of all the equipments in the consumer’s premises is the “connected load” of the consumer. For instance, if a consumer has connections of five 100-watt lamps and a power point of 500 watts, then connected load of the consumer is 5 × 100 + 500 = 1000 watts. The sum of the connected loads of all the consumers is the connected load to the power station. ( ii ) Maximum demand : It is the greatest demand of load on the power station during a given period. The load on the power station varies from time to time. The maximum of all the demands that have occurred during a given period ( say a day) is the maximum demand. Thus referring back to the load curve of Fig. 3.2, the maximum demand on the power station during the day is 6 MW and it occurs at 6 P.M. Maximum demand is generally less than the connected load because all the con- sumers do not switch on their connected load to the system at a time. The knowledge of maxi- mum demand is very important as it helps in de- termining the installed capacity of the station. The station must be capable of meeting the maximum demand. ( iii ) Demand factor. It is the ratio of maximum demand on the power station to its connected load i.e.,
Demand factor =
Maximum demand Connected load The value of demand factor is usually less than 1. It is expected because maximum demand on the power station is generally less than the connected load. If the maximum de- mand on the power station is 80 MW and the connected load is 100 MW, then demand factor = 80/100 = 0·8. The knowl- edge of demand factor is vital in determining the capacity of the plant equipment. ( iv ) Average load. The average of loads occurring on the power station in a given period ( day or month or year ) is known as average load or average demand.
Maximum demand meter
Energy meter
Variable Load on Power Stations 47
Thus if the considered period is one year,
Annual plant capacity factor = Annual kWh output Plant capacity × 8760 The plant capacity factor is an indication of the reserve capacity of the plant. A power station is so designed that it has some reserve capacity for meeting the increased load demand in future. Therefore, the installed capacity of the plant is always somewhat greater than the maximum demand on the plant. Reserve capacity = Plant capacity − Max. demand It is interesting to note that difference between load factor and plant capacity factor is an indica- tion of reserve capacity. If the maximum demand on the plant is equal to the plant capacity, then load factor and plant capacity factor will have the same value. In such a case, the plant will have no reserve capacity. ( viii ) Plant use factor. It is ratio of kWh generated to the product of plant capacity and the number of hours for which the plant was in operation i.e.
Plant use factor = Station output in kWh Plant capacity ×Hours of use Suppose a plant having installed capacity of 20 MW produces annual output of 7·35 × 10 6 kWh and remains in operation for 2190 hours in a year. Then,
Plant use factor =
⋅ ×
× ×
e j
3.53.5 3.53.53.5 Units Generated per AnnumUnits Generated per AnnumUnits Generated per AnnumUnits Generated per AnnumUnits Generated per Annum
It is often required to find the kWh generated per annum from maximum demand and load factor. The procedure is as follows :
Load factor =
Average load Max. demand ∴ Average load = Max. demand × L.F. Units generated/annum = Average load (in kW) × Hours in a year = Max. demand (in kW) × L.F. × 8760
3.63.6 3.63.63.6 Load Duration CurveLoad Duration CurveLoad Duration CurveLoad Duration CurveLoad Duration Curve
When the load elements of a load curve are arranged in the order of descending magnitudes, the curve thus obtained is called a load duration curve.
4848484848 Principles of Power System
The load duration curve is obtained from the same data as the load curve but the ordinates are arranged in the order of descending magnitudes. In other words, the maximum load is represented to the left and decreasing loads are represented to the right in the descending order. Hence the area under the load duration curve and the area under the load curve are equal. Fig. 3.3 ( i ) shows the daily load curve. The daily load duration curve can be readily obtained from it. It is clear from daily load curve [See Fig. 3.3. ( i )], that load elements in order of descending magnitude are : 20 MW for 8 hours; 15 MW for 4 hours and 5 MW for 12 hours. Plotting these loads in order of descending magnitude, we get the daily load duration curve as shown in Fig. 3.3 ( ii ). The following points may be noted about load duration curve : ( i ) The load duration curve gives the data in a more presentable form. In other words, it readily shows the number of hours during which the given load has prevailed. ( ii ) The area under the load duration curve is equal to that of the corresponding load curve. Obviously, area under daily load duration curve (in kWh) will give the units generated on that day. ( iii ) The load duration curve can be extended to include any period of time. By laying out the abscissa from 0 hour to 8760 hours, the variation and distribution of demand for an entire year can be summarised in one curve. The curve thus obtained is called the annual load duration curve.
3.73.7 3.73.73.7 TTTTTypes of Loadsypes of Loadsypes of Loadsypes of Loadsypes of Loads
A device which taps electrical energy from the electric power system is called a load on the system. The load may be resistive ( e.g., electric lamp), inductive ( e.g., induction motor), capacitive or some combination of them. The various types of loads on the power system are : ( i ) Domestic load. Domestic load consists of lights, fans, refrigerators, heaters, television, small motors for pumping water etc. Most of the residential load occurs only for some hours during the day ( i.e., 24 hours) e.g., lighting load occurs during night time and domestic appliance load occurs for only a few hours. For this reason, the load factor is low (10% to 12%). ( ii ) Commercial load. Commercial load consists of lighting for shops, fans and electric appli- ances used in restaurants etc. This class of load occurs for more hours during the day as compared to the domestic load. The commercial load has seasonal variations due to the extensive use of air- conditioners and space heaters. ( iii ) Industrial load. Industrial load consists of load demand by industries. The magnitude of industrial load depends upon the type of industry. Thus small scale industry requires load upto 25 kW, medium scale industry between 25kW and 100 kW and large-scale industry requires load above 500 kW. Industrial loads are generally not weather dependent. ( iv ) Municipal load. Municipal load consists of street lighting, power required for water sup- ply and drainage purposes. Street lighting load is practically constant throughout the hours of the night. For water supply, water is pumped to overhead tanks by pumps driven by electric motors. Pumping is carried out during the off-peak period, usually occurring during the night. This helps to improve the load factor of the power system. ( v ) Irrigation load. This type of load is the electric power needed for pumps driven by motors to supply water to fields. Generally this type of load is supplied for 12 hours during night. ( vi ) Traction load. This type of load includes tram cars, trolley buses, railways etc. This class of load has wide variation. During the morning hour, it reaches peak value because people have to go to their work place. After morning hours, the load starts decreasing and again rises during evening since the people start coming to their homes.
3.83.8 3.83.83.8 TTTTTypical Demand and Diversity Factorsypical Demand and Diversity Factorsypical Demand and Diversity Factorsypical Demand and Diversity Factorsypical Demand and Diversity Factors
The demand factor and diversity factor depend on the type of load and its magnitude.
50 Principles of Power System
farther from the ultimate consumer in making measurements, one should expect decreasing numeri- cal values of diversity factor as the power plant end of the system is approached. This is clear from the above table showing diversity factors between different elements of the power system. Example 3.1. The maximum demand on a power station is 100 MW. If the annual load factor is 40% , calculate the total energy generated in a year. Solution. Energy generated/year = Max. demand × L.F. × Hours in a year = (100 × 103 ) × (0·4) × (24 × 365) kWh = 3504 ××××× 10 5 kWh Example 3.2. A generating station has a connected load of 43MW and a maximum demand of 20 MW; the units generated being 61·5 × 106 per annum. Calculate ( i ) the demand factor and ( ii ) load factor. Solution.
( i ) Demand factor = Max. demand Connected load
( ii ) Average demand = Units generated / annum Hours in a year
61 5⋅ × 10
6 = 7020 kW
∴ Load factor = Average demand Max. demand
×
= 0·351 or 35·1%
Example 3.3. A 100 MW power station delivers 100 MW for 2 hours, 50 MW for 6 hours and is shut down for the rest of each day. It is also shut down for maintenance for 45 days each year. Calculate its annual load factor. Solution. Energy supplied for each working day = (100 × 2) + (50 × 6) = 500 MWh Station operates for = 365 − 45 = 320 days in a year ∴ Energy supplied/year = 500 × 320 = 160,000 MWh
Annual load factor = MWh supplied per annum Max. demand in MW × Working hours
× 100
a 100 f b × 320 × (^24) g
× 100 = 20·8%
Example 3.4. A generating station has a maximum demand of 25MW, a load factor of 60%, a plant capacity factor of 50% and a plant use factor of 72%. Find ( i ) the reserve capacity of the plant ( ii ) the daily energy produced and (iii) maximum energy that could be produced daily if the plant while running as per schedule, were fully loaded.
Solution.
( i ) Load factor =
Average demand Maximum demand
or 0·60 =
Average demand 25 ∴ Average demand = 25 × 0·60 = 15 MW
Plant capacity factor = Average demand Plant capacity
∴ Plant capacity = Average demand Plant capacity factor
= 30 MW
Variable Load on Power Stations 5151515151
∴ Reserve capacity of plant = Plant capacity − maximum demand = 30 − 25 = 5 MW ( ii ) Daily energy produced = Average demand × 24 = 15 × 24 = 360 MWh ( iii ) Maximum energy that could be produced
=
Actual energy produced in a day Plant use factor = 360 0 72⋅
= 500 MWh/day
Example 3.5. A diesel station supplies the following loads to various consumers : Industrial consumer = 1500 kW ; Commercial establishment = 750 kW Domestic power = 100 kW; Domestic light = 450 kW If the maximum demand on the station is 2500 kW and the number of kWh generated per year is 45 × 105 , determine ( i ) the diversity factor and ( ii ) annual load factor. Solution.
( i ) Diversity factor =
( ii ) Average demand = kWh generated / annum Hours in a year
= 45 × 105 /8760 = 513·7 kW
∴ Load factor =
Average load Max. demand
Example 3.6. A power station has a maximum demand of 15000 kW. The annual load factor is 50% and plant capacity factor is 40%. Determine the reserve capacity of the plant. Solution.
Energy generated / annum = Max. demand × L.F. × Hours in a year
= (15000) × (0·5) × (8760) kWh = 65·7 × 106 kWh
Plant capacity factor = Units generated / annum Plant capacity ×Hours in a year
∴ Plant capacity = 65 7^10
⋅ ×^6
0 4⋅ × 8760
= 18,750 kW
Reserve capacity = Plant capacity − Max. demand = 18,750 − 15000 = 3750 kW Example 3.7. A power supply is having the following loads : Type of load Max. demand ( k W ) Diversity of group Demand factor Domestic 1500 1·2 0· Commercial 2000 1·1 0· Industrial 10,000 1·25 1 If the overall system diversity factor is 1·35, determine ( i ) the maximum demand and ( ii ) con- nected load of each type.
Solution. ( i ) The sum of maximum demands of three types of loads is = 1500 + 2000 + 10,000 = 13, kW. As the system diversity factor is 1·35,
∴ Max. demand on supply system = 13,500/ 1·35 = 10,000 kW
Variable Load on Power Stations 5353535353
- Since the tubewells operate together, the diversity factor is 1.
Solution. Sum of max. demands of houses = (1·5 × 0·4) × 1000 = 600 kW Max. demand for domestic load = 600A2·5 = 240 kW Max. demand for factories = 90 kW Max. demand for tubewells = 7* × 7 = 49 kW The sum of maximum demands of three types of loads is = 240 + 90 + 49 = 379 kW. As the diversity factor among the three types of loads is 1·2,
∴ Max. demand on station = 379A1·2 = 316 kW ∴Minimum capacity of station requried = 316 kW Example 3.10. A generating station has the following daily load cycle : Time ( Hours ) 0 —6 6 —10 10 — 12 12 — 16 16 — 20 20 — Load ( M W ) 40 50 60 50 70 40 Draw the load curve and find ( i ) maximum demand ( ii ) units generated per day ( iii ) average load and ( iv ) load factor.
Solution. Daily curve is drawn by taking the load along Y -axis and time along X -axis. For the given load cycle, the load curve is shown in Fig. 3.6.
( i ) It is clear from the load curve that maximum demand on the power station is 70 MW and occurs during the period 16 — 20 hours.
∴ Maximum demand = 70 MW
( ii ) Units generated/ day = Area (in kWh) under the load curve
= 10^3 [40 × 6 + 50 × 4 + 60 × 2 + 50 × 4 + 70 × 4 + 40 × 4]
= 10^3 [240 + 200 + 120 + 200 + 280 + 160] kWh = 12 ××××× 105 kWh
( iii ) Average load = Units generated / day 24 hours
5 = ×^ = 50,000 kW
( iv ) Load factor = Average load Max. demand 70 103
×
5454545454 Principles of Power System
Example 3.11. A power station has to meet the following demand : Group A : 200 kW between 8 A.M. and 6 P.M. Group B : 100 kW between 6 A.M. and 10 A.M. Group C : 50 kW between 6 A.M. and 10 A.M. Group D : 100 kW between 10 A.M. and 6 P.M. and then between 6 P.M. and 6 A.M. Plot the daily load curve and determine ( i ) diversity factor ( ii ) units generated per day ( iii ) load factor. Solution. The given load cycle can be tabulated as under :
Time (Hours) 0 — 6 6 —8 8 — 10 10 — 18 18 — 24 Group A — — 200 kW 200 kW — Group B — 100 kW 100 kW — — Group C — 50 kW 50 kW — — Group D 100 kW — — 100 kW 100 kW
Total load on power station 100 kW 150 kW 350 kW 300 kW 100 kW
From this table, it is clear that total load on power station is 100 kW for 0 — 6 hours, 150 kW for 6— 8 hours, 350 kW for 8 — 10 hours, 300 kW for 10 — 18 hours and 100 kW for 18 — 24 hours. Plotting the load on power station versus time, we get the daily load curve as shown in Fig. 3.7. It is clear from the curve that maximum demand on the station is 350 kW and occurs from 8 A.M. to 10 A. M. i.e., Maximum demand = 350 kW Sum of individual maximum demands of groups = 200 + 100 + 50 + 100 = 450 kW
( i ) Diversity factor = Sum of individual max. demands Max. demand on station
= 450A350 = 1·
( ii ) Units generated/day = Area (in kWh) under load curve = 100 × 6 + 150 × 2 + 350 × 2 + 300 × 8 + 100 × 6 = 4600 kWh
( iii ) Average load = 4600 / 24 = 191·7 kW
∴ Load factor = 191 7 350
⋅ × 100 = 54·8%
Example 3.12. The daily demands of three consumers are given below : Time Consumer 1 Consumer 2 Consumer 3 12 midnight to 8 A.M. No load 200 W No load 8 A.M. to 2 P.M. 600 W No load 200 W 2 P.M. to 4 P.M. 200 W 1000 W 1200 W 4 P.M. to 10 P.M. 800 W No load No load 10 P.M. to midnight No load 200 W 200 W
5656565656 Principles of Power System
∴ Load factor =
× 100 = 33·3%
Example 3.14. A power station has a daily load cycle as under : 260 MW for 6 hours ; 200 MW for 8 hours : 160 MW for 4 hours, 100 MW for 6 hours. If the power station is equipped with 4 sets of 75 MW each, calculate ( i ) daily load factor ( ii ) plant capacity factor and ( iii ) daily requirement if the calorific value of oil used were 10,000 kcal/kg and the average heat rate of station were 2860 kcal/kWh.
Solution. Max. demand on the station is 260 × 103 kW. Units supplied/day = 10^3 [260 × 6 + 200 × 8 + 160 × 4 + 100 × 6] = 4400 × 103 kWh
( i ) Daily load factor = 4400 10 260 10
3 3
×
× × 24
× 100 = 70·5%
( ii ) Average demand / day = 4400 × 103 /24 = 1,83,333 kW
Station capacity = (75 × 103 ) × 4 = 300 × 103 kW
∴ Plant capacity factor = 183 333,^ , 300 × 10 3
× 100 = 61·1 %
( iii ) Heat required/day = Plant heat rate × units per day = (2860) × (4400 × 10 3 ) kcal
Fuel required/ day = 2860 4400 10
× × 3
= 1258·4 × 103 kg = 1258·4 tons
Example 3.15. A power station has the following daily load cycle : Time in Hours 6 —8 8 —12 12 —16 16 — 20 20 — 24 24 — Load in MW 20 40 60 20 50 20 Plot the load curve and load duratoin curve. Also calculate the energy generated per day. Solution. Fig. 3.9 ( i ) shows the daily load curve, whereas Fig. 3.9 ( ii ) shows the daily load duraton curve. It can be readily seen that area under the two load curves is the same. Note that load duration curve is drawn by arranging the loads in the order of descending magnitudes.
Units generated/day = Area (in kWh) under daily load curve = 10^3 [20 × 8 + 40 × 4 + 60 × 4 + 20 × 4 + 50 × 4] = 840 ××××× 103 kWh
Fig. 3.
Variable Load on Power Stations 5757575757
Alternatively : Units generated/day = Area (in kWh) under daily load duration curve = 10^3 [60 × 4 + 50 × 4 + 40 × 4 + 20 × 12] = 840 ××××× 103 kWh which is the same as above.
Example 3.16. The annual load duration curve of a certain power station can be considered as a straight line from 20 MW to 4 MW. To meet this load, three turbine-generator units, two rated at 10 MW each and one rated at 5 MW are installed. Determine ( i ) installed capacity ( ii ) plant factor ( iii ) units generated per annum ( iv ) load factor and ( v ) utilisation factor.
Solution. Fig. 3.10 shows the annual load duration curve of the power station. ( i ) Installed capacity = 10 + 10 + 5 = 25 MW ( ii ) Referring to the load duration curve,
Average demand =
[20 + 4] = 12 MW
∴ Plant factor = Average demand Plant capacity
( iii ) Units generated/annum = Area (in kWh) under load duration curve
=
[4000 + 20,000] × 8760 kWh = 105·12 ××××× 106 kWh
( iv ) Load factor = 12, 20,
× 100 = 60%
( v ) Utilisation factor = Max.demand Plant capacity
Example 3.17. At the end of a power distribution system, a certain feeder supplies three distri- bution transformers, each one supplying a group of customers whose connected load are listed as follows :
Transformer 1 Transformer 2 Transformer 3 General power Residence lighting Store lighting and power service and lighting a : 10 H.P., 5kW e : 5 kW j : 10 kW, 5 H.P. b : 7·5 H.P., 4kW f : 4 kW k : 8 kW, 25 H.P. c : 15 H.P. g : 8 kW l : 4 kW d : 5 H.P., 2 kW h : 15 kW i : 20 kW Use the factors given in Art. 3.8 and predict the maximum demand on the feeder. The H.P. load is motor load and assume an efficiency of 72%.
Variable Load on Power Stations 5959595959
( iii ) the maximum energy that could be produced daily if the plant was running all the time ( iv ) the maximum energy that could be produced daily if the plant was running fully loaded and oper- ating as per schedule. [( i ) 288 ××××× 103 kWh ( ii ) 0 ( iii ) 4·80 ××××× 103 kWh ( iv ) 600 ××××× 10 3 kWh]
6. A generating station has the following daily load cycle : Time (hours) 0—6 6—10 10—12 12—16 16—20 20— Load (MW) 20 25 30 25 35 20 Draw the load curve and find ( i ) maximum demand, ( ii ) units generated per day, ( iii ) average load, ( iv ) load factor, [( i ) 35 MW ( ii ) 560 ××××× 10 3 kWh ( iii ) 23333 kW ( iv ) 66·67%] 7. A power station has to meet the following load demand : Load A 50 kW between 10 A.M. and 6 P.M. Load B 30 kW between 6 P.M. and 10 P.M. Load C 20 kW between 4 P.M. and 10 A.M. Plot the daily load curve and determine ( i ) diversity factor ( ii ) units generated per day ( iii ) load factor. [( i ) 1·43 ( ii ) 880 kWh ( iii ) 52·38%] 8. A substation supplies power by four feeders to its consumers. Feeder no. 1 supplies six consumers whose individual daily maximum demands are 70 kW, 90 kW, 20 kW, 50 kW, 10 kW and 20 kW while the maximum demand on the feeder is 200 kW. Feeder no. 2 supplies four consumers whose daily maximum demands are 60 kW, 40 kW, 70 kW and 30 kW, while the maximum demand on the feeder is 160 kW. Feeder nos. 3 and 4 have a daily maximum demand of 150 kW and 200 kW respectively while the maximum demand on the station is 600 kW. Determine the diversity factors for feeder no. 1. feeder no. 2 and for the four feeders. [1·3, 1·25, 1·183] 9. A central station is supplying energy to a community through two substations. Each substation feeds four feeders. The maximum daily recorded demands are : POWER STATION........ 12,000 KW Substation A ...... 6000 kW Sub-station B .... 9000 kW Feeder 1 ............ 1700 kW Feeder 1 ............ 2820 kW Feeder 2 ............ 1800 kW Feeder 2 ............ 1500 kW Feeder 3 ............ 2800 kW Feeder 3 ............ 4000 kW Feeder 4 ............ 600 kW Feeder 4 ............ 2900 kW Calculate the diversity factor between ( i ) substations ( ii ) feeders on substation A and ( iii ) feeders on sub- station B. [( i ) 1·25 ( ii ) 1·15 ( iii ) 1·24] 10. The yearly load duration curve of a certain power station can be approximated as a straight line ; the maximum and minimum loads being 80 MW and 40 MW respectively. To meet this load, three turbine- generator units, two rated at 20 MW each and one at 10 MW are installed. Determine ( i ) installed capacity ( ii ) plant factor ( iii ) kWh output per year ( iv ) load factor. [( i ) 50MW ( ii ) 48% ( iii ) 210 ××××× 10 6 ( iv ) 60%]
3.93.9 3.93.93.9 Load Curves and Selection of Generating UnitsLoad Curves and Selection of Generating UnitsLoad Curves and Selection of Generating UnitsLoad Curves and Selection of Generating UnitsLoad Curves and Selection of Generating Units
The load on a power station is seldom constant; it varies from time to time. Obviously, a single generating unit ( i.e., alternator) will not be an economical proposition to meet this varying load. It is because a single unit will have very poor* efficiency during the periods of light loads on the power station. Therefore, in actual practice, a number of generating units of different sizes are installed in a power station. The selection of the number and sizes of the units is decided from the annual load curve of the station. The number and size of the units are selected in such a way that they correctly
- The efficiency of a machine (alternator in this case) is maximum at nearly 75% of its rated capacity.
6060606060 Principles of Power System
fit the station load curve. Once this underlying principle is adhered to, it becomes possible to operate the generating units at or near the point of maximum efficiency. Illustration. The principle of selection of number and sizes of generating units with the help of load curve is illustrated in Fig. 3.11. In Fig. 3.11 ( i ), the annual load curve of the station is shown. It is clear form the curve that load on the station has wide variations ; the minimum load being some- what near 50 kW and maximum load reaching the value of 500 kW. It hardly needs any mention that use of a single unit to meet this varying load will be highly uneconomical.
As discussed earlier, the total plant capacity is divided into several generating units of different sizes to fit the load curve. This is illustrated in Fig. 3.11( ii ) where the plant capacity is divided into three* units numbered as 1, 2 and 3. The cyan colour outline shows the units capacity being used. The three units employed have different capacities and are used according to the demand on the station. In this case, the operating schedule can be as under :
Time Units in operation From 12 midnight to 7 A.M. Only unit no.1 is put in operation. From 7 A.M. to 12.00 noon Unit no. 2 is also started so that both units 1 and 2 are in operation. From 12.00 noon to 2 P.M. Unit no. 2 is stopped and only unit 1operates. From 2 P.M. to 5 P.M. Unit no. 2 is again started. Now units 1 and 2 are in operation. From 5 P.M. to 10.30 P.M. Units 1, 2 and 3 are put in operation. From 10. 30 P.M. to 12.00 midnight Units 1 and 2 are put in operation. Thus by selecting the proper number and sizes of units, the generating units can be made to operate near maximum efficiency. This results in the overall reduction in the cost of production of electrical energy.
3.103.10 3.103.103.10 Important Points in the Selection of UnitsImportant Points in the Selection of UnitsImportant Points in the Selection of UnitsImportant Points in the Selection of UnitsImportant Points in the Selection of Units
While making the selection of number and sizes of the generating units, the following points should be kept in view : ( i ) The number and sizes of the units should be so selected that they approximately fit the annual load curve of the station.
- It may be seen that the generating units can fit the load curve more closely if more units of smaller sizes are employed. However, using greater number of units increases the investment cost per kW of the capacity.
