Maxtmum demand 3 the ratio is <1 1. Demand factor = “Connected load ‘Average load (demand) 5 the ratio is 2. Load factor = Max. demand is asi [Point to be noted ; Average demand = Average load] Sum of individual maxdemand - the ratio ig > 1 3. Diversity factor = “Max demand on power station Actual energy produced 4. Plant capacity factor = Maxenergy that can be produced — Average demand Plant capacity 5. Reserve capacity = Plant capacity — Max.demand 6. Units generated in year = Average demand x Hours ina year = (Load factor x Max demand) x Hours in a year Station output in KWh(unit) A ctor = 7, Use fa Plant capacityxhours of use [Point to be noted: Hours of Use #4 Qt FARA TABS VAR; CU PH UST RENCE OY CTT WG HR AAA FACS BA] Max demand 8, Utilization factor = Plant capacityi erating station has a maximum demand of 25MW, a Joad factor of 60% 4 plant capacity factor of 50% and plant use factor of 72%, Find the capactty of the plant. even (BCPCL-18, BUET) Solution: Reserve Plant Capacity = Plant Capacity - Maximum demand Here, Plant capacity = acne 7 Again, Average Demand = Load Factor x Maximum Demand = 0.6 x25 MW = 15MW + Plant Capacity = ="™ = 30 mw. « Reserve Plant Capacity = 30 — 25 = 5 MW. 2. A power station has a maximum demand of 15000kW. The annual load factor ls 50% & power capacity factor is 40%. Determine the reserve capacity of the plant THEVENIN [BADC-20, BUET, BPDB-18,BUET] Solution: Reserve Plant Capacity = Plant Capacity — Maximum demand. ‘i Average demand Here, Plant Capacity = 5i5n¢ capacity Factor Again, Average Demand = Load Factor x Maximum Demand =0.5 x 15000 KW = 7500 Kw. Plant Capacity = 2° = 18750 KW. + Reserve Plant Capacity = 18750 — 1500 = 3750 KW.— shat need to consider while Whee: What are the factor z is load factor? What are cabu(GEe)-17-Eden Cllege] designing a power plant? REE JOB ‘Solution: maximum Load Factor: The ratio of average load and OT. load in a given period of time is known as load fact Average demand Load Factor = Plant Capacity Factor Load factor is always <1 Factors for considering while designing a 1. Power plant site, 2 Type of load to be supplied, 3. Availability of fuel, 4. Availability of cooling water, 5S. Cost of Power transmission. Power Plant: 4 Define diversity actory, Joad factor & power factor. How to improve pf for induction type load? (BB(AD-Electrical)-19, Mirpur Ideal] 2 = = 5 3 Solution: Group Diversity Factor: The ratio of sum of individual maximum demands in a group to the actual maximum demand of that grou i io ii oerenerthen group. This ratio is equal ctor: The ratio of average demand of a load and maximum demand of a load in a given period of time is known as load fa io i equal or less than 1. ctor. This ratio is5, Agenerating station has 2 maximum demand of 25MW, a load factor of 60%, # plant capacity factor of 50% and plant use factor of 72%. Find 1) the reserve capacity of the plant it) the daily energy produced end it!) maximum energy that could be produced daily if the plent while running as per schedule, were fully loaded. [RPGCL-17, BUET] Solution: {)Reserve Plant Capacity = Plant Capacity — Maximum demand _ Average demand Here, Plant capacity Plant capacity factor Again, Average Demand = Load Factor x Maximum Demand = 0.6 x 25 MW = 15 MW « Plant Capacity = ae = 30 MW. « Reserve Plant Capacity = 30 — 25 = 5 MW. ii) Daily Energy Produced = Average Demand x hours ina day. = 15 x 24 = 360 MWh. iif) Maximum energy that could be produced is same as plant capacity. capacity = Average Demand Plant Capacity = plant Capacity Factor [ : 15 = 20.83 MW. . Maximum energy daily = 20.83 x 24 = 500 MWh.6. A mew load needs to be added in 2 powrer station. There are two load! each of & 70% respectively. The group = Demand factor x Connected Load =07 x10=7MW i Actual Maximum Demand | | | Total load = 5.04 + 1667 MW = 21.71 MW Total power sent from transformer nsidering transmission loss = 21.71 x (1 + 0.04) = 22S8MW Transformer Rating = SZE2 = 752 = 234‘EEE OB GUIDELINE. 7, Max.demand = 20,000kW : Load factor = 40% Boller efficiency =85% + Turbine 9086 Coal consumption =0.9ke/kW : Costof 1 ton of coal= Rs 300 1e ) thermal efficiency & il) coal bill per annum. Solution: i) Thermal efficiency = Boiler efficiency x turbine efficiency = 0.85 x09 = 0.765 ii) Load to be supplied = Maximum Demand x Load factor = 20000 x 0.4 = 8000KW One Year = 8760hours Per annum energy supply = 8000 x 8760 = 70080000 KW suppl Per annum energy production = maw Say = 91607843.1 KWh 0.9 91607843.1 Coal needed = "55, tonnes = 82447.05 tonnes. Coal Bill = 300 x 82447.05 = 24734117.6 Rs [NWPGCL-17, BUET]a certain feeder supplies 3 8 At the end of the power distribution system, a group of customers Whose distribution transformers, each one supplyin8. z connected loads are as under: S| rranstormer Land Demand factor. Diversity ofgroup 2 Transformer no.1 10kW 0.65 15 é Transformer 0.2 12kW 06 35 é Transformer no.3 15kW a7 15 = Ifthe diversity factor among the transformer Is 1.3, find the maximum load on the feeder. [Dhaka WASA-17, BUET] Solution: Transformer 1 Maximum Load = Demand factor x Connected Load = 10 KW x 0.65 = 6.5 KW 4. Actual Maximum Load = $3 = 4.33 KW Transformer 2 Connected Load xDemand Factor Actual Maximum Load = ity rector _ 12x06 =e = 2.056 KW ‘Connected Load xdemand factor Load = Diversity factor 7 = 7KW Total Load = 4.33 + 2.05 + 7 KW = 13.38 KW Total load at the time of system peak on feeder = ae = 10.3 KW9. A gonierating sation te ta supply 4 regions oflosd whose peak loads are SMW, BMW & 7MW, The diversity factor at the station Is 15) average annual load factor 1s 60%, Calculate J) che maximum demand on tho station 1) annual onergy supplied by the station. [NPPCBL-17, BUET] Solution: =15= 10454847 Maximum load of the system = Maximum load of the system = 22 = 20 MI) Load factor = Average Demand Maximum Demand = Average demand = load factor x Maximum demand = Average Demand = 20 x 0.6 = 12 MW One year = 8760hours Per Annum energy generated = 12MW x 8760 = 105.12MW 10.A power station has reserve capactty of 3000kW, load factor 60%, Plant capactty factor 509. What Is the max demand of the plant? [NWPGCL-18, BUET] Solution: Reserve capacity = Plant capacity — maximum demand Average demand _ maximum demand .. «. (i) => Reserve capaclty = plant capactty factor We know,i stor = Ateragedemand (ii) Again, Load factor = Maimuti demand Let’s say, Maximum demand = x From (ii), 0.6 = Average demand x = Average demand = 0.6x From (i) = 2k 3000KW = os 7% 11. The cost curve of a 600MW coal fired power plant is 0.003P2 +1 500 $/hr/MW. What will be the Cost to increase the output of th 1 MW. Ifit is operating at 100MW? If the ‘Plant is dispatched for ti maximum of S00MW & its annual energy output is 3600GWh, det capacity factor & load factor. [CPGCBL- THEVENIN Solution: i) At 100 MW, the cost is = 0.003(100)? + 6 x 100 + 500 $/hr At (100 + 1)MW, the cost is = 0.003(101 )?+6x 101 +50 So, Increase in cost = 0.003 (101? — 1002) + 6 (101 — 199) = 6.603 $/hr Annualoutput _ 3600GWh _ hoursinayear 6760n_ — 9-411GW Average demand = . _ Average demand ii). Plant Capacity Factor = Pont copcity — 0.411GW =somw * 100% = 68.5%12. The annual load duration curve of a certain power station can be considered as a straight line from Z0MW to 4MW. To meet this load.3 turbine generator units two rated at 1OMW each & one rated at SMW are installed. Determine 1) installed capacity il) plant factor Ili) units generated per annum tv) load factor and v) utilization factor. [DPDC-19, BUET] MEV Erin Solution: The load curve looks like this. i) Installed capacity = 2 x 10+5 = 25MW ii) Plant factor = Sver#sedemand Plant capacity Average demand = 3 (20 +4) =12MW +. Plant factor = 22 x 100% = 48% fil) Unit generated annum = Area under the load curve = Area of the trapizium = 1 (20 + 4) x 8760MWh = 105120MWh. Iv) Load factor = ,Aversse cement = 2 x 100% = 60% Maximum demand ¥) Utilisation factor = Sy capacity * 100% = 3 x 100% = 80%THEVENIN Solution: « Load factor = 13. Determine the load factor & peak demand ofa power station In which the annual load duration curve varies Hnearly between 12GW & 4GW. If the total transmission loss Is 3 % and for frequently regulation reserve margin of 20% Is to be maintained then what should be the total generation capacity of the system? [PGCB-16, BUET] Here, Peak demand = 12GW Average demand = ; (maxtmum demand + mintmum demand) =3(12+4) = 98Gw Average demand Load Factor = Foo demand =< x 100% = 66.67% Power to be supplied considering transmission loss = 12(1 + 0.03) = 12.366w Total generation capacity should be = 12.36(1 + 0,2) = 1483W Average demand Arcee Maximum demand x 100% = 3604167 » 190% = 90.1% 400014.4 power station has a maximum demand of 15000kW, load factor ls 70% & Plant capacity factor 54.596. Find: D) Energy produced dally, 4) reserve capactty of the plant? il) the max energy that could have been produced dally if the plant runs all the tme. JHEVENIN: [PGCL-17, BUET] Load Factor = -Average demang Maximum demand = Average demand = Load factor x Maximum demand | = 0.7 x 1500KW = 1050KW i) Energy produced daily = Average demand x hours ina day = 1050 x 24 = 2520KW ii) Reserve capacity of the plant = Plant capacity ~ Maximum demand = Memes demand _ Maximum demand “Plant use factor = 2252 _ 1500 = 426.61 KW OS45 — 2ctual jucti lil) Maximum daily energy production = “2221 ¢al¥y production epedty hea 25200 = Gees = 46.2KW/day15. The annual load duration curve of a certain power station can be considered as a straight line from 20MW to 4MW. To meet this load,3 turbine generator units two rated at 1OMW each & one rated at SMW are installed. Determine load factor: [DPDC-19, BUET| Average demand = 320 +4) =12KW Maximum Demand = 20KW +. Load factor = 22 x 100% = 60% 16.A diesel station supplies 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 & the number of kWh generated per year is #5x105, determiné i) the diversity factor and /)annual load factor. THEVENIN [CPA-17, Port College] Solution: Sum of maximum demand of all load f) roup diversity factor = em demand of the group = 1800+750+100+450 2500 1.120 Od _ ——__—_ fi) Average demand = Stetay generated per annum hours ina year Deo” = 513.7 KW = 45x10" 8760 Average demand, Maximum demand 513.7 = 3500 * 100% = 20.5% Load factor = x 100% 17, Calculate annual bill of a consumer whose maximum demand is 100kW, , = 0.8 lagging and load factor 60%. The tariffused is 75 per kVA of maximum demand plus 15 paise per kWh consumed. [CPA-17, Port Colleg Solution: Average demand = Load factor x Maximum demand = 0.6 x 100 = 60KW Energy consumed per annum = Average demand x hours ina year = 60 x 8760 = 525600KWh + Bill for energy use = 525600 x 0.15 = 78840 RS Maximum demand Again, Maximum demand in KVA = = = 100 = 2 = 125 KVA « Bill for maximum demand = 125 x 75 = 9375 RS + Total bill = 9375 + 78840 = 88215 RS18. (related) A generating station Is to supply 4 regions of load whose pe; Joads are 10MW, SMW, 8MW & 7MW. The diversity factor at the static 1.5 & the average annual load factor Is 60%. Calculate 1) the maximum demand on the station if) annual ‘energy supplied by the station. Iii)su the installed capacity & the number of the untts. [DESCO-19, BUET, NPPCBL-17,E THEVENIN Solution: Sum of loads at maximum. Maximum load of the system =15= 10+5+8+7 Maximum load of the system i) Diversity Factor = => Maximum load of the system = 22 = 20 if = Average Demand ) Load factor = FF imum Demand =» Average demand = load factor x Maximum demand => Average Demand = 20 x 0.6 = 12 MW One year = 8760hours Per Annum energy generated = 12MW x 8760 = 105.12MW (iii) Installed capacity should be 20% higher than maximum demand. So, Installed capacity = 20 x 1.2 = 24 MW Since, average demand is 12 MW, installing one 12 MW unit and two 6 MW units can be suitable.g 19.4 power station of plant capactty 100MW, average load 90MW, and maximum load 95MW. Find load factor, plant capacity factor, plant utilization factor. (ERL-17, BUET] PUNT. Solution: average demand maximum demand ee Load factor = 2 x 100% = 94.73% : Average demand Plant capacity factor = So aaa x 100% 90 =, % 100% = 90% Maximum demand Utilization factor = prot capcity —* 100% 95. 7 =T0* 100% = 95% Zi. The day base Joad in 2 system is 3500MW the load changes to 4000MW at (PU ooly &persists til 11PM. Flad the load factor Solution: oo ‘Total demand ina day Average demand = Tima day = Remnaseesonens sasooens = 3604.167 MW z _ Average demand ahead ener = eet = BOE x 100% = 90.1% “2000