Chapter 1: Load Curves PDF

EGEL3130 / Power Stations & Distribution Systems Chapter 1 Load Curves This chapter deals with analyses of the load curves and to find the diversity factor, demand factor, load factor, capacity factor and plant factor...

EGEL3130 / Power Stations & Distribution Systems Chapter 1 Load Curves This chapter deals with analyses of the load curves and to find the diversity factor, demand factor, load factor, capacity factor and plant factor from it. OBJECTIVES LEARNING OUTCOMES 1 Explain the concept of load curves 1. Analyses the load curves and find the diversity and tariff system. factor, demand factor, load factor, capacity factor and plant factor from it. 1.1 Introduction An electric power system is a network of electrical components used to supply, transmit and distribute electric power. Namely, the main components of an electric power system are generation plants, transmission lines and distribution systems. The generating plants and distribution systems are connected through transmission lines, which also connect substations from one area to another. These components are operated at different voltages, for example, in Oman the voltages level that are used in distribution systems are 11kV, 33kV and 415V line to line voltage, and the high voltages used in transmission networks are 132kV and 220kV. This classification is varying from one country to another. In some country the high voltage used for transmitting the electrical power can be 400KV or even 800KV. This is depending on how big is the network and the length of transmission line. There are currently two widespread world standards for power system frequency: 50 Hz in most of Europe and Asia include Oman, and 60 Hz in North America and in other places strongly influenced by the U.S. However, the range of possible frequencies for power systems is constrained by practical concerns. For example, a century ago many electric railroads operated at a frequency of 25 Hz, but 25 Hz was never adopted for general use in power systems because frequencies at that level cause electric lights to flicker. At the other end of the scale, frequencies well above 60 Hz result in higher impedances, leading to unacceptably high transmission and distribution losses. Sep 2024 Page 5 of 98  EGEL3130 / Power Stations & Distribution Systems Figure 1.1: Basic component of electrical power system 1.2 Structure of Power System in Oman Figure 1.2 shows the structure of electrical power system in Oman. Normally the output voltage of generator ranges between 11KV to 25 KV and then stepped up to be transmitted. In Oman the voltage is stepped up to 132KV or 220KV and then transmitted for a long distance. Then, the voltage will step down in grid station 132KV/33KV for distribution purpose to 33KV and this grid station normally will be outside the city. After that the voltage will further step down to 11KV and 415/240 V to be suitable distributed to the customer. Figure 1.2: Electrical power system structure in Oman Sep 2024 Page 6 of 98  EGEL3130 / Power Stations & Distribution Systems 1.3 Load Curve The load on a power station is never constant; it varies from time to time. These load variations during the whole day are recorded half-hourly or hourly and are plotted against time on the graph called daily load curves. Fig.1 shows a daily load curve of a power station. Figure 3.3: Daily load curve The curve showing the variation of load on the power station with respect to time is known as a load curve. The curve shows the variation of load with respect to time during the day is known as daily load curve as shown in Figure 1.3. The monthly load curve can be obtained from the daily load curves of that month. The yearly load curve can be obtained from the monthly load curves of that particular year. The load on power station is varying, being maximum at around 6 PM from the above figure 1.3. The daily load curves are important in generation as they provide following information: 1. The daily load curve shows the variations of load on the power station during different hours of the day. 2. The area under the daily load curve gives the number of units generated in the day. 3. Highest point on the daily load curve represents the maximum demand on the station. 4. The area under the daily load curve divided by the total number of hours gives the average load on the station in that day. Average load = Area (in KWh) under daily load curve/ 24 hours 5. The ratio of average load to maximum demand gives load factor. Sep 2024 Page 7 of 98  EGEL3130 / Power Stations & Distribution Systems 6. The load curve helps in selecting the size and number of generating units. 7. The load curve helps in preparing the operating schedule of the station. 1.4 Load 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. The load duration curve is obtained from the same data of the load curve but they are arranged in the order of descending magnitudes. The maximum load is represented to the left & decreasing loads are represented to the right in the descending order. Hence area under load duration curve & area under load curve are equal. Fig.1.4a shows the daily load curve and Fig.1.4b shows the daily load duration curve. The daily load duration curve can be obtained from load curve. The following points may be noted for load duration curve: 1. The load duration curve gives the data in a more presentable form. 2. The area under the load duration curve is equal to that of the corresponding load curve. 3. The load duration curve can be extended to include any period of time. If we extend the (x- axis) from 0 to 8760 hours, the annual load duration curve can be obtained. Figure 1.4 (a) : Daily load curve. Figure 1.4(b): Load duration curve. 1. Connected load: it is the sum of continuous ratings of all the equipment’s connected to supply system. 2. 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. 𝐴𝑟𝑒𝑎(𝑖𝑛 𝑘𝑊ℎ)𝑢𝑛𝑑𝑒𝑟 𝑑𝑎𝑖𝑙𝑦 𝑙𝑜𝑎𝑑 𝑐𝑢𝑟𝑣𝑒 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑙𝑜𝑎𝑑 (𝑖𝑛 𝑘𝑊) = 24 ℎ𝑜𝑢𝑟𝑠 Sep 2024 Page 8 of 98  EGEL3130 / Power Stations & Distribution Systems 𝐴𝑟𝑒𝑎(𝑖𝑛 𝑘𝑊ℎ)𝑢𝑛𝑑𝑒𝑟 𝑎𝑛𝑛𝑢𝑎𝑙 𝑙𝑜𝑎𝑑 𝑐𝑢𝑟𝑣𝑒 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑙𝑜𝑎𝑑 (𝑖𝑛 𝑘𝑊) = 365 ∗ 24 ℎ𝑜𝑢𝑟𝑠 3. Maximum Demand: It is the greatest demand of load on the power station during a given period. 4. Demand Factor: The ratio of maximum demand on the power station to its connected load is called demand factor. Its value is usually less than 1. It is used to refer to the fractional amount of some quantity being used relative to the maximum amount that could be used by the same system. 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑑𝑒𝑚𝑎𝑛𝑑 𝐷𝑒𝑚𝑎𝑛𝑑 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝐶𝑜𝑛𝑛𝑒𝑐𝑡𝑒𝑑 𝑙𝑜𝑎𝑑 5. Load factor: The ratio of average load to the maximum demand during a given period is known as Load factor. It is a measure the utilization rate, or efficiency of electrical energy usage; a high load factor indicates that load is using the electric system more efficiently. (LOAD FACTOR VALUE IS ALSO ALWAYS LESS THAN OR EQUAL TO 1) 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑙𝑜𝑎𝑑 𝐿𝑜𝑎𝑑 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑑𝑒𝑚𝑎𝑛𝑑 6. Diversity factor: The ratio of the sum of individual maximum demands to the maximum demand on power station is known as diversity factor. Its value always greater than or equal to1. 𝑠𝑢𝑚 𝑜𝑓 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑑𝑒𝑚𝑎𝑛𝑑𝑠 𝐷𝑖𝑣𝑒𝑟𝑠𝑖𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑑𝑒𝑚𝑎𝑛𝑑 𝑜𝑛 𝑝𝑜𝑤𝑒𝑟 𝑠𝑡𝑎𝑡𝑖𝑜𝑛 7. Plant capacity factor: It is the ratio of actual energy produced to the maximum possible energy that could have been produced during a given period 𝐴𝑐𝑡𝑢𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 𝑃𝑙𝑎𝑛𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑒𝑛𝑒𝑟𝑔𝑦 𝑡ℎ𝑎𝑡 𝑐𝑜𝑢𝑙𝑑 ℎ𝑎𝑣𝑒 𝑏𝑒𝑒𝑛 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑑𝑒𝑚𝑎𝑛𝑑 𝑃𝑙𝑎𝑛𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝑃𝑙𝑎𝑛𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 Thus, if the considered period is one year: Sep 2024 Page 9 of 98  EGEL3130 / Power Stations & Distribution Systems 𝐴𝑛𝑛𝑢𝑎𝑙 𝑘𝑊ℎ 𝑜𝑢𝑡𝑝𝑢𝑡 𝑃𝑙𝑎𝑛𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝑃𝑙𝑎𝑛𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑥 8760 Reserve capacity = Plant capacity − Max. demand 8. Plant use factor: It is the ratio of kWh generated to the product of plant capacity and the number of hours for which the plant was in operation 𝑆𝑡𝑎𝑡𝑖𝑜𝑛 𝑜𝑢𝑡𝑝𝑢𝑡 𝑖𝑛 𝑘𝑊ℎ 𝑃𝑙𝑎𝑛𝑡 𝑢𝑠𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝑃𝑙𝑎𝑛𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑥 ℎ𝑜𝑢𝑟𝑠 𝑜𝑓 𝑢𝑠𝑒 Types of Loads: ► (i) Domestic or Residential load: Domestic load consists of lights, fans, refrigerators, heaters, television, small motors for pumping water etc. ► (ii) Commercial load: Commercial load consists of lighting for shops, fans and electric appliances used in restaurants etc. ► (iii) Industrial (Municipal) load: load demand by industries. Small scale (up to 25kW), medium scale (25kW to 100kW) and large-scale industries above 500kW. ► (iv) Government load: street lighting, power required for water supply and drainage purposes. ► (v) Irrigation/Agriculture and Fisheries 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) Tourism load. Numerical: 1. A generating station has a connected load of 43 MW and a maximum demand of 20MW; the units generated being 61.5 x 106 per annum. Calculate (i) the demand factor (ii) load factor. Sep 2024 Page 10 of 98  EGEL3130 / Power Stations & Distribution Systems 2. 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 (Plant factor) of 72%. Find (i) the reserve capacity of the plant (ii) the daily energy produced and (iii) the maximum energy that could be produced daily if the plant running as per schedule were fully loaded. 3. A diesel station supplies the following loads to various consumers: Industrial consumer = 1500kW; Commercial establishment = 750kW Domestic power = 100kW; Domestic light = 450kW If the maximum demand on the station is 2500kW and the number of kWH generated per year is 45 x 105, determine (i) the diversity factor (ii) annual load factor Sep 2024 Page 11 of 98  EGEL3130 / Power Stations & Distribution Systems 4. A power station has a maximum demand of 15000kW. The annual load factor is 50% and plant capacity factor is 40%. Determine the reserve capacity of the plant. 5. A power station has a daily load cycle as under; 260MW for 6 hours; 200MW for 8 hours; 160MW for 4 hours; 100MW for 6 hours. If the power station is equipped with 4 sets of 75MW each, calculate (i) daily load factor (ii) plant capacity factor. Sep 2024 Page 12 of 98  EGEL3130 / Power Stations & Distribution Systems 6. A generation station has the following daily load cycle: Time (h) 0-6 6-10 10-12 12-16 16-20 20-24 Load (MW) 40 50 60 50 70 40 i. Draw the load curve and load duration curve. ii. find (a) Maximum Demand (b) Energy generated per day (c) Average load (d) Load factor. (e)Units generated between 2PM to 4PM Sep 2024 Page 13 of 98  EGEL3130 / Power Stations & Distribution Systems Extra question 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 (il) units generated per day (iii) loadf actor. Sep 2024 Page 14 of 98  EGEL3130 / Power Stations & Distribution Systems 1.5 Selection of generating units: The load on the generating station is never constant. A single generating unit will not be an economical to meet this varying load. 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 a way that they correctly fit the station load curve. The units should be of different capacities to meet the load requirements. The capacity of the plant should be made 15% to 20% more than the maximum demand to meet the future load requirements. There should be a spare generating unit so that repairs and overloading of the working units can be carried out. Sep 2024 Page 15 of 98  EGEL3130 / Power Stations & Distribution Systems 7. A generating station is to supply four regions of load whose peak loads are 10MW, 5MW, 8MW, and 7MW. The diversity factor at the station is 1.5 and the average annual load factor is 60%. Calculate: (i) the maximum demand on the station. (ii) annual energy supplied by the station. (iii) Suggest the installed capacity and the number of units. Sep 2024 Page 16 of 98  EGEL3130 / Power Stations & Distribution Systems 1.6 Base load and Peak load on power station Base load: The unvarying load which occurs almost the whole day on the station is known as base load. Peak load: The various peak demands of the load over and above the base load of the station is known as peak load. Interconnected Grid system: The connection of several generating stations in parallel is known as interconnected grid system. Advantages: 1. Exchange of peak loads: if the load curve of a power station shows a peak demand that is greater than the rated capacity of the plant, then the additional load can be shared by other stations interconnected with it. 2. Use of older plants: older plants can be used to carry peak loads of shorter duration. 3. Ensure economical operation: sharing of load among the stations is arranged in such way that more efficient stations work continuously thought the year at a load factor and the less efficient plants work for peak load hours only. 4. Increases diversity factor: maximum demand on the system is much reduced as compared to the sum of individual maximum demands on different stations. 5. Reduces plant reserve capacity: when several power stations are connected in parallel, the reserve capacity of the system is much reduced. This increases the efficiency of the system. 6. Increases reliability of supply: If a major breakdown occurs in one station, continuity of supply can be maintained by other healthy stations. Sep 2024 Page 17 of 98  EGEL3130 / Power Stations & Distribution Systems Base load power plant A base load power plant is a type of power generating plant that usually generates and supplies electrical energy continuously throughout the year. The base load power plant generates electricity continuously with minimum power generating requirements. Therefore, a base load power plant is turned off only during service and maintenance, upgradation, overhaul, etc. The demand response, i.e. mechanism of matching generation with load it supplies, is slow for a base load power plant. What are examples of base load plants? Examples of baseload power plants are: Nuclear power plant. Coal power plant. Hydroelectric plant. Geothermal plant. Biogas plant. Biomass plant. Solar thermal with storage. Ocean thermal energy conversion. What is a Peak Load Power Plant? A power plant that runs only during the hours of peak load demand of electricity is called a peak load power plant. The peak load power plant is also known as peaking power plant or Peaker. The examples of power generating plants that are used as peak load plants are gas turbine power plant, solar power plants, wind turbine power plants, diesel engine power plant, and sometimes small-scale hydroelectric power plants Sep 2024 Page 18 of 98  EGEL3130 / Power Stations & Distribution Systems QUESTIONS 1. What do you understand by load curve? What information you can get from this? 2. What is load factor and what is its importance? 3. What are plant capacity factor and diversity factor? 4. Why the load factor is always less than 1. 5. Write the differences between load curve and load duration curve. 6. Explain how grid system is important in the operation of a power system. Sep 2024 Page 19 of 98  EGEL3130 / Power Stations & Distribution Systems 7. Daily load of an industry is 200 kW for first two hours, 250 kW for next six hours, 100 kW for next seven hours and 20 kW for remaining time. Draw the Load duration curve. Sep 2024 Page 20 of 98  EGEL3130 / Power Stations & Distribution Systems 8. The daily demands of three consumers are given below: Time Consumer 1 Consumer 2 Consumer 3 12 AM to 8 AM No load 2000 W No load 8 AM to 2 PM 6000 W No load 2000 W 2 PM to 4 PM 2000 W 10000 W 12000 W 4 PM to 10 PM 8000 W No load No load 10 PM to 12AM No load 2000 W 2000 W Plot the load curve and find (i) maximum demand of individual consumer (ii) load factor of individual consumer (iii) diversity factor and (iv) load factor of the station. Sep 2024 Page 21 of 98  EGEL3130 / Power Stations & Distribution Systems 9. A generation station of 1MW supplied a region which has the following demands: Sep 2024 Page 22 of 98  EGEL3130 / Power Stations & Distribution Systems Extra question. A base load station having a capacity of 18MW and a standby station having a capacity of 20MW share a common load. Find the annual load factors and plant capacity factors of two power stations from the following data: Annual standby station output = 7.35 x 106 kWh Annual base load station output = 101.35 x 106 kWh Peak load on standby station = 12 MW Hours of use by standby station/year = 2190 hours Sep 2024 Page 23 of 98  EGEL3130 / Power Stations & Distribution Systems Extra question. The information for a generating station of is given as follows. Capacity of station = 100MW Connected load = 80MW Load variation from 12 AM to 3 PM: 20 MW to 70MW. Load variation from 3 PM to 10 PM: 70MW to 30MW. ► Draw a load curve ► Estimate the total energy delivered ► Calculate the daily load factor Sep 2024 Page 24 of 98  EGEL3130 / Power Stations & Distribution Systems Extra question. The daily load cycle of a 75MW power station is given below: Timing: 6-8 8-12 12-16 20-24 Load (MW): 20 40 60 50 Q1: How many hours in day , the power station was supplying the load? Q2: Calculate the daily energy supplied to the load. Q3: Calculate the daily average load on the station. Q4. Calculate the Load Factor. Q4. Calculate the Reserve capacity of the power station. Q4: Calculate the Plant use factor. Q5: Calculate the plant capacity factor. Sep 2024 Page 25 of 98  EGEL3130 / Power Stations & Distribution Systems Extra questions: 1. the maximum demand on a power station is 100MW,if the annual load factor is 40%, calculate the energy generated in a year. 2. A 100 MW power station delivers 100MW for 2 hours,50MW for 6 hours and it is shut down for rest of the each day. It is also shut down for maintenance for 45days each year. Calculate its annual energy production. 3. A consumer has the following connected load; 10 lamps of 60W each and 2 heaters of 1000W each. His maximum demand is 1500W. On the average, he uses 8 lamps for 5 hours a day and each heater for 3 hours a day. Find his total load, monthly energy consumption and load factor. Sep 2024 Page 26 of 98  EGEL3130 / Power Stations & Distribution Systems Chapter 2 Tariff In this chapter we will discuss about the types of tariffs. We will also calculate the tariff for different customers. OBJECTIVES LEARNING OUTCOMES 1. Explain the concept of load curves and 1. Explain the relation between load factor and tariff system. loss factor in a distribution system, objectives of load management through distribution automation and various tariff system and their evaluation. The rate at which electrical energy is supplied to a consumer is known as Tariff. Objectives of tariff: Electrical energy is also at such a rate so that it not only returns the cost butalso earns reasonable profit. Therefore, a tariff should include the following items: (i) Recovery of cost of producing electrical energy at the power station. (ii) Recovery of cost on the capital investment in transmission and distribution systems. (iii) Recovery of cost of operation and maintenance of supply of electrical energy e.g., metering equipment, billing. (iv) A suitable profit on the capital investment. 2.1 Types of Tariffs 1. Simple tariff: When there is a fixed rate per unit of energy consumed, it is called a simple tariff or uniform tariff. In this type of tariff, the price charged per unit is constant i.e., it does not vary with increase or decrease in number of units consumed. 2. Flat rate tariff: When different types of consumers are charged at different uniform per unit rates, it is called a flat rate tariff. In this type of tariff, the consumers are grouped into different classes and each class of consumers is charged at a different uniform rate. 3. Block rate tariff: When a given block of energy is charged at a specified rate and the succeeding blocks of energy are charged at progressively reduced or increased rates, it is called a block rate tariff. Sep 2024 Page 27 of 98  EGEL3130 / Power Stations & Distribution Systems 4. Two-part tariff: When the rate of electrical energy is charged on the basis of maximum demand of the consumer and the units consumed, it is called a two-part tariff. In two-part tariff, the total charge to be made from the consumer is split into two components, fixed charges and running charges. The fixed charges depend upon the maximum demand of the consumer while the running charges depend upon the number of units consumed by the consumer. Thus, the consumer is charged at a certain amount per kW of maximum demand plus a certain amount per kWh of energy consumed i.e., Total charges = (b * kW + c * kWh) where, b = charge per kW of maximum demand c = charge per kWh of energy consumed 5. Three-part tariff: When the total charge to be made from the consumer is split into three parts viz., fixed charge, semi-fixed charge and running charge, it is known as a three-part tariff. i.e., Total charge = (a + b * kW + c * kWh) Omani Rial where a = fixed charge made during each billing period. It includes interest and depreciation on the cost of secondary distribution and labor cost of collecting revenues, b = charge per kW of maximum demand, c = charge per kWh of energy consumed. 6. Power factor tariff : The tariff in which power factor of the consumer load is taken into consideration is known as power factor tariff. A low power factor increases the rating of station equipment and line losses. A consumer having low power factor must be penalized. The following are the important types of power factor tariff: (i) kVA maximum demand tariff : It is a modified form of two-part tariff. In this case, the fixed charges are made on the basis of maximum demand in kVA and not in kW. A low power factor consumer has to contribute more towards the fixed charges. Sep 2024 Page 28 of 98  EGEL3130 / Power Stations & Distribution Systems (ii) Sliding scale tariff : This is also known as average power factor tariff. In this case, an average power factor, say 0·8 lagging, is taken as the reference. If the power factor of the consumer falls below this factor, suitable additional charges are made. If the power factor is above the reference, a discount is allowed to the consumer. (iii) kW and kVAR tariff: In this type, both active power (kW) and reactive power (kVAR) supplied are charged separately. A consumer having low power factor will draw more reactive power and hence shall have to pay more charges. Numerical examples 1. A consumer has a maximum demand of 200kW at 40% load factor. If the tariff is OMR 1 per kW of maximum demand plus 10 Baisa per KWh, find the overall cost per kWh. Sep 2024 Page 29 of 98  EGEL3130 / Power Stations & Distribution Systems 2. A consumer has a maximum demand of 350kW at 65% load factor. If the tariff is 1 OMR per kW of maximum demand plus 10 baiza per kWh, find the overall cost per kWh. 3. The maximum demand of a consumer is 25A at 230V and his total energy consumption is 9000kWh. If the energy is charged at the rate of 10 baiza per unit for 500 hours use of the maximum demand per annum plus 15 baiza per unit for additional units, calculate (i) annual bill (ii) equivalent flat rate. Sep 2024 Page 30 of 98  EGEL3130 / Power Stations & Distribution Systems 4. Annual Energy consumption by an Industrial load is 3×105 units with 3.9 MW as its peak load at 0·95 p.f. lagging. The tariff is OMR 0.78 per MVA of peak demand plus 190 baisa per unit. If power factor of industrial load is decreased to 0.29, a. Calculate the difference in annual bill. b. Determine the flat rate of energy consumption at both the power factors. Sep 2024 Page 31 of 98  EGEL3130 / Power Stations & Distribution Systems 2.2 Tariff Structure in Sultanate of Oman Energy Tariff in the Residential sector: tariff structure is made in Block Rate Tariff. Energy Tariff in Commercial sector: tariff structure is made in Simple Rate Tariff. Energy Tariff in Industrial sector: tariff structure is made in Simple Rate Tariff, but in two seasons. Energy Tariff in Government sector: tariff structure is made in Block Rate Tariff. Energy Tariff in agriculture & Fisheries sector: tariff structure is made in Block Rate Tariff. Energy Tariff in Tourism sector: tariff structure is made in Block Rate Tariff. Proposed tariff for residential consumers. Sl. Customer Category Tariff structure N0 1. Residential (Omanis 0 – 2000 kWh 15 Bz/kWh with one or 2 2001 – 4000 kWh 20 Bz/kWh accounts) 4001 kWh & Above 30 Bz/kWh 2. Residential (Omanis 0 – 500 kWh 20 Bz/kWh with more than 2 501 – 1500 kWh 25 Bz/kWh accounts) 1501 kWh & Above 30 Bz/kWh 3 Non-Omanis 0 – 500 kWh 20 Bz/kWh 501 – 1500 kWh 25 Bz/kWh 1501 kWh & Above 30 Bz/kWh Sep 2024 Page 32 of 98  EGEL3130 / Power Stations & Distribution Systems Questions: 1. What do you understand by tariff? Discuss the objectives of tariff. 2. Discuss the energy tariff in different sectors. 3. Discuss the tariff structure in Oman 4. Explain the types of tariff. Sep 2024 Page 33 of 98  EGEL3130 / Power Stations & Distribution Systems 5. Calculate annual bill of a consumer whose maximum demand is 200 kW, p. f. = 0·8 lagging and load factor = 70%. The tariff used is OMR 5 per kVAr of maximum demand plus 20 baiza per kWh consumed. Sep 2024 Page 34 of 98  EGEL3130 / Power Stations & Distribution Systems 6. Daily load of an industry is 200 kW for first two hours,250 kW for next six hours, 100 kW for next seven hours and 20 kW for remaining time. If tariff is OMR 2 per kW of maximum demand per year plus 20 Baisa per kWh. Estimate the monthly bill. Sep 2024 Page 35 of 98  EGEL3130 / Power Stations & Distribution Systems 7. A consumer has the following connected load; 10 lamps of 60W each and 2 heaters of 1000W each. His maximum demand is 1500W. On the average, he uses 8 lamps for 5 hours a day and each heater for 3 hours a day. Find his total load, monthly energy consumption and load factor. Sep 2024 Page 36 of 98  EGEL3130 / Power Stations & Distribution Systems 8. A consumer has a maximum demand of 20kW at 0.8 p.f. lagging and an annual load factor of 60%, there are two alternative tariffs (i) O.R. 2.00 per kVA of maximum demand plus 10 BZ per kWh consumed and (ii) R.O.0.5 per kVA of maximum demand plus 15 BZ per kWh consumed. Determine which of the tariffs will be economical for him. Sep 2024 Page 37 of 98  EGEL3130 / Power Stations & Distribution Systems  Extra questions  An electric supply company having a maximum load of 50MW generates 18x107 units per annum and the supply consumers have an aggregate demand of 75MW. The annual expenses including capital charges are: For fuel = OMR63000 Fixed charges concerning generation = OMR19500 Fixed charges concerning transmission & distribution = OMR22500 Assuming 90% of the fuel cost is essential to running charges and the loss in transmission and distribution as 15% of KWh generated, deduce a two part tariff to find the actual cost of supply to the consumers Sep 2024 Page 38 of 98  EGEL3130 / Power Stations & Distribution Systems  Extra question  The maximum demand of a consumer is 20A at 220V and his total energy consumption is 8760KWh. If the energy is charged at the rate of 10 Baisa per unit for 500 hours use of the maximum demand per annum plus 5 Baisa per unit for additional units, calculate: (i) annual bill (ii) equivalent flat rate. Sep 2024 Page 39 of 98  EGEL3130 / Power Stations & Distribution Systems Chapter 3 Power Generating stations This chapter deals with the different power stations like hydroelectric power stations, thermal power station, nuclear power station. OBJECTIVES LEARNING OUTCOMES 1. Analyze different types of power stations. 1. Analyze different power stations like thermal power station, nuclear power station and hydraulic plant in terms of (operation, types, pros & cons). Bulk electric power is produced by special plants known as generating stations or power plants. A generating station essentially employs a prime mover coupled to an alternator for the production of an electric power. The prime mover (steam turbine, water turbine, gas turbine) converts energy from some other form into mechanical energy. The alternator converts mechanical energy into electrical energy. The electrical energy produced by the generating station is transmitted and distributed with the help of conductors to various consumers. Depending upon the form of energy converted into electrical energy, the generating stations are classified as under: 1. Hydroelectric power stations. 2. Steam power stations. 3. Nuclear power stations. 4. Diesel power stations. 5. Gas power stations. Sep 2024 Page 40 of 98  EGEL3130 / Power Stations & Distribution Systems 3.1 Hydro – electric power station A generating station which utilizes potential energy of water at a high level for the generation of electrical energy is known as hydro-electric power station. Hydroelectric power stations are generally located in hilly areas where dams can be built and large water reservoirs can be obtained. In a hydroelectric power station, water head is created by constructing a dam across river or lake. From the dam, water is led to a water turbine. The water turbine captures the energy in the falling water and changes the hydraulic energy (product of head and flow of water) into mechanical energy at the turbine shaft. The turbine drives the alternator which converts mechanical energy into electrical energy. Choice of site for hydroelectric power stations: 1. Quantity of water available. 2. Storage of water. 3. Head of Water. 4. Distance of power station from load centers. 5. Accessibility of the site. Principle of working of a Hydroelectric plant: Figure 3.1: Hydro power plant Sep 2024 Page 41 of 98  EGEL3130 / Power Stations & Distribution Systems The dam is constructed across a river or lake and water from the catchments area collects at the back of the dam to form a reservoir, A pressure tunnel is taken off from the reservoir and water brought to the valve house at the start of the penstock. The valve house contains main sluice valves and automatic isolating valves. The former controls the water flow to the powerhouse and cuts off supply of water before the penstock explode. From the valve house, water is taken to water turbine through a huge steel pipe known as penstock. The water turbine converts hydraulic energy — into mechanical energy. The turbine drives the alternator which converts mechanical energy into electrical energy. A surge tank is built just before the valve house and protects the penstock from exploding in case the turbine gates suddenly close due to electrical load being thrown off. When the gates close, there is a sudden stopping of water at the lower end of the penstock and consequently the penstock can explode. The surge tank absorbs this pressure swing by increase in its level of water. Calculation of power: P = 9.81 Q W h η -----watts Where Q = discharge, m3/sec, W = Density of water = 1000 kg/m3, h = Head, & η = efficiency of turbine and generator. Classification of Hydroelectric plants: Classification may be based on: 1. Quantity of water available. 2. Available head. Classification according to Quantity of water available: 1. Run-off river plants without pondage. 2. Run-off river plants with pondage. 3. Reservoir plants. Classification according to available head: (i) Low head plants (below 30 meters). (ii) Medium head plants. (iii) High head plants (above 300 meters). Sep 2024 Page 42 of 98  EGEL3130 / Power Stations & Distribution Systems I. Low head plants: 1. In this case a small dam is built across the river to provide the necessary head. 2. The excess water is allowed to flow over the dam itself. 3. This plants Francis, Propeller or Kaplan type turbines are used. 4. No surge tank is required. Figure 3.2 : Low head Hydro power plant II. Medium head plants: 1. The forebay provided at the beginning of penstock serves as water reservoir for such plants. 2. The forebay itself works as surge tank in this case. 3. The common types of prime mover used in these plants are Francis, Propeller and Kaplan. Figure 3.3: Medium head Hydro power plant Sep 2024 Page 43 of 98  EGEL3130 / Power Stations & Distribution Systems III. High head plants: 1. Water is carried from the main reservoir by a tunnel up to the surge tank and then from the surge tank to the powerhouse in penstocks. 2. For heads above 500 meters Pelton wheels are used while for lower heads Francis turbine is used (499m-301m). Figure 3.4: High head Hydro power plant Functions of different components in hydroelectric plants: Reservoir: It is a basic requirement of a hydroelectric power plant. Its purpose is to store water which may be utilized to run the prime mover to produce electric power. A reservoir stores water during the rainy season and supplies the same during the dry season. Dam: The function of a dam is to provide a head of water to be utilized in the water turbine. It also increases the reservoir capacity. Trash rack: The purpose of providing a trash rack is to prevent entry of debris which might damage the wicket gates and turbine runners or mean choking of nozzles of the impulse turbines. A trash rack is made of steel bars and is placed across the intake. Forebay: The forebay serves as a regulating reservoir storing water temporarily when load on the plant is reduced and providing water for initial increase on account of increasing load during which time water in the channel is being accelerated. This may either be a pond behind the diversion dam or an enlarged section of a canal spread out to accommodate the required width intake. Sep 2024 Page 44 of 98  EGEL3130 / Power Stations & Distribution Systems Surge tank: This is an additional storage space near the turbine usually provided in high head and medium head plants. When there is a considerable distance between the water source and turbine which necessitates a long penstock. As the load on the turbine decreases or during load rejection by the turbine the surge tank provides space for holding water. Similarly, when load on the turbine increases it furnishes additional water. Thus, a surge tank controls the pressure variations resulting from the rapid changes in water flow in penstock and there by prevents water hammer effects. Penstock: It is a conduit system for taking water from the intake works and forebay to the turbines. There are two types: low pressure and high pressure. Penstock pipes are generally of steel for high and medium head plants. They may be of concrete in low head plants. When the distance from the forebay to the powerhouse is short separate penstocks are used for each turbine. Spillway: this may be considered a sort of safety valve for a dam. A spillway serves to discharge excess water in the reservoir beyond the full permissible level. Powerhouse: It is generally located at the foot of the dam and near the storage reservoir. If the powerhouse is near to the dam the loss of head due to friction in the penstock would be less. Powerhouse consists of two main parts: a substructure to support the hydraulic and electric equipment and a super structure to house and to protect this equipment. The generating units and exciters are usually located on the ground floor. Prime mover: The purpose of prime mover is to convert kinetic energy of water into mechanical energy. Commonly used prime movers are Pelton wheel, Francis, Kaplan and Propeller turbines. Advantages: 1. No fuel is required. 2. Clean as no smoke or ash is produced. 3. Very small running charges. 4. Simple in construction and requires less maintenance. 5. Short starting time. 6. Longer life. 7. For operation few experienced persons are required. Sep 2024 Page 45 of 98  EGEL3130 / Power Stations & Distribution Systems Disadvantages: 1. Requires high capital cost due to construction of dam. 2. Uncertainty about availability of water. 3. Skilled and experienced hands are required to build the plant. 4. Requires high cost of transmission lines as the plant is located in hilly areas which are quite away from the consumers. Numerical examples: 1. A hydroelectric generating station is supplied from a reservoir of capacity 5x106 cubic meters at a head of 200 meters. Find the total energy available in kWh if the overall efficiency is 75%. Sep 2024 Page 46 of 98  EGEL3130 / Power Stations & Distribution Systems 2. It has been estimated that a minimum run off of approximately 94 m3/sec will be available at a hydraulic project with a head of 39m. Determine a) firm capacity b) yearly gross output. assume the efficiency of the plant to be 80% 3. Calculate the average power in kW that can be generated in a hydroelectric project from the following data Catchment area = 7 x 109 m2; Mean head, H = 28m; Annual rainfall, F = 1.33m; Yield factor, k = 75%; Overall efficiency = 72%. If the load factor is 45%, What is the rating of generators installed? Sep 2024 Page 47 of 98  EGEL3130 / Power Stations & Distribution Systems 4. Calculate the continuous power that will be available from hydroelectric plant having an available head of 300 meters, catchment area of 150 square km, annual rainfall 1.25meter and yield factor of 50%. Assume penstock, turbine and generator efficiencies to be 96%, 86% and 97% respectively. If the load factor is 40% what should be the rating of the generators installed? 5. A hydroelectric station has an average available head of 100 meters and reservoir capacity of 50 million cubic meters. Calculate the total energy in kWh that can be generated, assuming hydraulic efficiency of 85% and electrical efficiency of 90%. Sep 2024 Page 48 of 98