Maximum demand / connected load = demand factor. It is the ratio of max demand on power system to its connected load.
A generating station has a maximum demand of 20MW load factor of 60% a plant capacity factor of 50% and a plant us e factor of 72% the maximum energy that could be produced daily is
Load factor = avg. load / max. demand = Avg. demand = 0.60 X 25 = 15MW
Daily energy produced = Avg. demand X 24 = 15 X 24 = 360 MWh
Max. energy that could be produced = Actual energy produced in a day / plant use factor = 360 / 0.72 = 500 MWh/day
A generating station has a maximum demand of 20MW load factor of 60% a plant capacity factor of 50% and a plant us e factor of 72% what is the reserve capacity of the plant if the plant while running as per schedule were fully load?
A generating station has a maximum demand of 50MW a load factor of 60%, a plant capacity factor of 45% and if the plant while running as per schedule were fully loaded. The daily energy produced will be
Load factor = avg. load / max. demand
=> 0.60 = Avg. demand / 50
Avg. demand = 30 MW
Daily energy produced = Avg. demand X 24 = 30 X 24 = 720 MW
The load duration curve for a system is 8 hours 30MW, 16 hours 25 MW and 24 hours 20MW. The load factor will be
Actual energy consumed = (30 X 8 + 25 X 8 + 20 X 8) = 600 MWh
Load factor = avg. load / max. demand = 600 / 24 X 30 = 0.83
A power station supplies the peak load of 50 MW , 40 MW and 70 MW to three localities. The annual load factor is 0.50 p.u. and the diversity factor of the load at the station is 1.55 the maximum demand on the station and average load respectively will be
Diversity Factor = Sum of individual max. demands / Max. demand on the station => 1.55 = (50+40+70)MW / Max. demand on the station.
Max. demand on the station = 160 / 1.55 = 103.22MW
Avgerage load = load factor X max. Demand = 0.5 X 103.22 = 51.61 MW
What is the utilization factor of a power station which supplies the following loads?
Load A motor load of 200 kW between 10 AM to 7 PM.
Load B lighting load of 100 kW between 7 PM to 11 PM.
Load C pumping load of 110 kW between 3 PM to 10 PM.
Sum of individual max. demands = 200+100+110 = 410kW
10 AM to 3 PM 200kW
3 PM to 7PM 200+110 = 310 kW
7PM to 11 PM 100+110 = 210kW
11PM to 10 AM 110kW
Max. demand of the whole system is 310kW
Utiligation Factor = Sum of individual max. demands / Max. demand of the whole system = 410 / 310 = 1.32
The yearly load duration curve of a power plant is a straight line the maximum load is 750 MW and the minimum load is 600 MW the capacity of the plant is 900 MW what is capacity factor?
A consumer consumes 600 kWh per day at a load factor of 0.40 if the consumer the load factor of 0.70 without increases the maximum demand what is the consumption of energy in kWh?
Load factor = Energy consumed in 24 hrs. / Maximum demand in kW X 24 => 0.40 = 600 / Max. Demand in kW X 24
Max. Demand = 600 / 24 X 0.40 = 62.5 kW
When load factor is 0.70. => Energy consumed in 24 hrs = Load factor X max. demand X 24 = 0.70 X 62.5 X 24 = 1050kWh
What is maximum value of a load which consume 500 KWh per day at a load factor of 0.40 if the consumer increases the load factor of 0.50 without increasing the maximum demand?
Load factor = Energy consumed in 24 hrs. / Maximum demand in kW X 24 => 0.40 = 500 / Max. demand X 24
Max. Demand = 500 / 0.40 X 24 = 52.08 kW
Diversity factor is equal to
Diversity factor= sum of max. demand of consumers / max. demand on system
Demand factor= max. demand /connected load , Load factor= average demand/max. demand.
The maximum demand of a consumer is 80 KW & his daily energy consumption is 1200 units. What is his load factor?
Load factor = (1200) / (24 × 80) × 100% = 62.5%.
What is the coincidence factor?
Coincidence factor is the reciprocal of the diversity factor which is always less than unity.
A consumer has a maximum demand of 60 kW at 0.5 p.f. If tariff is RS 750 per kW of maximum demand plus RS 1.20/kWh, then overall cost per kWh will be
annual energy consumption =60X0.50X8760=262800 KWH
COST of annual energy consumption =1.20X262800=RS 315360
Fixed charge /year=RS 750X60=45000
TOTAL CHARGE =360360 ; ans=360360/262800=1.37
Maximum demand of consumer is 80 kW and his daily energy consumption is 1200 units, load factor will be
load factor = 1200/(24X80)= 0.62=62.5% .
Maximum KVA demand of the consumer is related to p.f. in what way?
KVA inversely proportional to the p.f . Consumers having low pf has to contribute more towards the fixed charges.
An industrial consumer has a connected load of 1000 kW the maximum demand is 80 kW. On an average each machine works for shows a day for 300 works for shows a day for 300 working days in a year. If traiff is Rs.(5000) + Rs. 800 per kW of maximum demand + Rs. 1.15 per kWh of energy consumed. The annual bill of the consumer will be
Energy consumed per year. =100 × 8 × 300= 240000 Kwh, Annual bill of the consumer=Rs(5000+800 × 80+240000 × 1.15)=34500.
A consumer has a maximum demand of 200 kW at 40% load factor. If the traiff is Rs 100 per kW of maximum demand plus 10 Paise per kWh, find the overall cost per kWh?
Unit consumed / year = Max demand X Load factor X Hours in a year = 200 X 0.4 X 8760 = 7,00,800 Kwh. Annual charges = Annual maximum demand charges + Annual energy charges = Rs(100 X 200 + 0.1 X 7,00,800)= Rs 90,080, ∴ Overall cost/Kwh = Rs 90,080 ⁄ 7,00,800 = Rs 0.1285 = 12.85 Paise.
A generating plant has a maximum capacities of 100 kW & cost Rs.1,60,000/-. The annual fixed charges are 12% consisting of 5% interest, 5% depreciation & 25% taxes. Find the fixed charges per kWh if the load factor is 100%.
Maximum demand =100Kw,Annual fixed charges Rs.0.12 × 1,60,000=19,200 ,so units generated ⁄annum=Max demand × Load factor × Hours in a year=100 × 1 × 8760=8760Kwh.Fixed charges ⁄Kwh=Rs19200⁄876000=Rs 0.0219 =2.19 Paise.
The maximum demand on a power station is 100MW. If the annual load factor is 40%. Calculate the energy generated in a year?
Energy generated / year = Max demand X Load factor X Hours in a year = (100 X 103) X 0.4 X (24 X 365) Kwh = 3504 X 105 Kwh.