Question 7
A generating station has a maximum demand of 20 MW, a load factor of 60%, a plant capacity factor of 48% and a plant use factor of 80% . Find :
- The daily energy produced
- The reserve capacity of the plant
- The maximum energy that could be produced daily if the plant was running all the time
- The maximum energy that could be produced daily if the plant was running fully loaded and operating as per schedule.
Provide your response in the comment section using your surname and last two digit of your matriculation number as your identity (e.g. Abacus20)
Click here to like and follow the E-Learning Community on Facebook for more interactive practices and discussions with peers and instructor.
Feel free to tag your engineering friends and hashtag other social networks and learning websites.
Given:
ReplyDeleteMaximum demand =20 MW
Load factor = 0.6
Plant capacity factor =0.48
Plant use factor = 0.8
Load factor = Average load/Maximum demand
Average demand = Load factor × Maximum demand
= 0.6 × 20000
= 12,000 kW
Plant Capacity factor = Average load/Installed capacity
Installed Capacity = Average demand/Plant capacity Factor
= 1200/0.48
= 25000 kW
Reserve Capacity of the plant = Installed capacity – Maximum Demand
= 25000-20000
= 5000 kW
Daily energy produced =Average Demand × 24
= 12000 × 24
= 288000 kWhr
Energy/day corresponding to installed capacity = 25000×24
= 600000 kWhr
Maximum energy that could be produced = Actual energy produced/Plant use factor
= 288000/0.8
= 360000 kWhr
👍👍💡
Delete1. Formula for daily energy produced = average load * 24hrs.
ReplyDeleteTo get our avegate load we use, load factor = average load/ max load
Where load factor is 0.6 and max load is 20MW, so
Average load = 20 *0.6
= 12MW.
So daily energy produced is 12MW * 24h = 288MWh also 288,000kwh.
2. Reserve capacity = installed capacity - max load
Whereas installed capacity = average load/plant capacity factor
=12/0.48
=25MW..
So reserve capacity = 25 -20
=5MW (5000kW)
3. Maximum energy produced daily if the plant works all time is
Installed capacity * 24hrs
= 25MW * 24
=600Mwh(6000000kwh)
4. Plant use factor = actual energy produced / plant use factor
= 288/0.8
=360 MWh
1. Formula for daily energy produced = average load * 24hrs.
ReplyDeleteTo get our avegate load we use, load factor = average load/ max load
Where load factor is 0.6 and max load is 20MW, so
Average load = 20 *0.6
= 12MW.
So daily energy produced is 12MW * 24h = 288MWh also 288,000kwh.
2. Reserve capacity = installed capacity - max load
Whereas installed capacity = average load/plant capacity factor
=12/0.48
=25MW..
So reserve capacity = 25 -20
=5MW (5000kW)
3. Maximum energy produced daily if the plant works all time is
Installed capacity * 24hrs
= 25MW * 24
=600Mwh(6000000kwh)
4. Plant use factor = actual energy produced / plant use factor
= 288/0.8
=360 MWh
Given:
ReplyDeleteMaximum demand =20 MW
Load factor = 0.6
Plant capacity factor =0.48
Plant use factor = 0.8
Load factor = Average load/Maximum demand
Average demand = Load factor × Maximum demand
= 0.6 × 20000
= 12,000 kW
Plant Capacity factor = Average load/Installed capacity
Installed Capacity = Average demand/Plant capacity Factor
= 1200/0.48
= 25000 kW
Reserve Capacity of the plant = Installed capacity – Maximum Demand
= 25000-20000
= 5000 kW
Daily energy produced =Average Demand × 24
= 12000 × 24
= 288000 kWhr
Energy/day corresponding to installed capacity = 25000×24
= 600000 kWhr
Maximum energy that could be produced = Actual energy produced/Plant use factor
= 288000/0.8
= 360000 kWhr
Maximum demand =20 MW
ReplyDeleteLoad factor = 0.6
Plant capacity factor =0.48
Plant use factor = 0.8
Load factor = Average load/Maximum demand
Average demand = Load factor × Maximum demand
= 0.6 × 20000
= 12,000 kW
Plant Capacity factor = Average load/Installed capacity
Installed Capacity = Average demand/Plant capacity Factor
= 1200/0.48
= 25000 kW
Reserve Capacity of the plant = Installed capacity – Maximum Demand
= 25000-20000
= 5000 kW
Daily energy produced =Average Demand × 24
= 12000 × 24
= 288000 kWhr
Energy/day corresponding to installed capacity = 25000×24
= 600000 kWhr
Maximum energy that could be produced = Actual energy produced/Plant use factor
= 288000/0.8
= 360000 kWhr
Given that
ReplyDeleteMaximum demand for a generating station = 20MW(20000)
Load factor = 60%(0.6)
Plant capacity factor = 48%(0.48)
Plant use favtor = 80%(0.8)
Question 1
The daily energy produced is
Average demand =load factor ×maximum demand which is 0.6×20000= 120000KW
So therefore 120000KW×24h= 288000KW
Question 2
Reserve capacity of the plant
Plant capacity factor = average load /installed capacity, where installed capacity = 120000/0.48= 25000KW
Therefore Reserve capacity of the plant =25000- 20000= 5000KW
Question 3
The maximum energy produced daily if the plant was running all the time = installed capacity ×24h which is 25000×24= 600000KW/h
Question 4
Maximum energy produced daily if plant was running fully loaded and operating as per schedule is
plant use factor = Actual energy produced/ plant use factor
Therefore 288000/08= 360000KW.
Solution.
DeleteLoad factor=average load÷ Max load
0.6=average load÷20MW
Average load= 12MW
=12000KW
Plant capacity factor=average load÷installed capacity
0.48=12MW÷installed capacity
Installed capacity=25MW
=25KW
1.daily energy prodeuced=average load× 24
=25MW×24h=288MWh
=288000kwh
2. The reserved capacity =installed capacity- Max load
=25MW-20MW
=5000KW
3 The maximum energy daily when the plant running all the time(day 24h).
= Installed capacity× 24
=25MW×24h
= 600MW
=6000000KW
4. Plant use factor=actual energy produce M Kwh÷ plant capacity× no of hour that the plant has been in operation
0.8= average load×24h÷ Max energy that could be produced daily if the plant were running according to it's operation schedule(y)
0.8=288000÷y
y=360MWh
Good attempt
DeleteLoad factor = 0.6
ReplyDeleteMax. demand =20 MW
P.C.F=0.48
Plant use factor = 0.8
Load factor = Average load/Maximum demand
Average demand = Load factor × Maximum demand
= 0.6 × 20000
= 12,000 kW
P.C.F = Average load/Installed capacity
Installed Capacity = Average demand/Plant capacity Factor
= 1200÷0.48
= 25000 kW
Reserve Capacity of the plant = Installed capacity – Maximum Demand
= 25000-20000
= 5000 kW
Daily energy produced =Average Demand × 24
= 12000 × 24
= 288000 kWhr
Energy/day corresponding to installed capacity = 25000×24
= 600000 kWhr
Maximum energy that could be produced = Actual energy produced/Plant use factor
= 288000/0.8
= 360000 kWhr
👍👍💡
DeleteFormula
ReplyDeleteLoad factor=average load÷ Max load
0.6=average load÷20MW
Average load= 12MW
=12000KW
Plant capacity factor=average load÷installed capacity
0.48=12MW÷installed capacity
Installed capacity=25MW
=25KW
1.daily energy prodeuced=average load× 24
=25MW×24h=288MWh
=288000kwh
2. The reserved capacity =installed capacity- Max load
=25MW-20MW
=5000KW
3 The maximum energy daily when the plant running all the time(day 24h).
= Installed capacity× 24
=25MW×24h
= 600MW
=6000000KW
4. Plant use factor=actual energy produce M Kwh÷ plant capacity× no of hour that the plant has been in operation
0.8= average load×24h÷ Max energy that could be produced daily if the plant were running according to it's operation schedule(y)
0.8=288000÷y
y=360MWh
Ugele1059
DeleteFormula
Load factor=average load÷ Max load
0.6=average load÷20MW
Average load= 12MW
=12000KW
Plant capacity factor=average load÷installed capacity
0.48=12MW÷installed capacity
Installed capacity=25MW
=25KW
1.daily energy prodeuced=average load× 24
=25MW×24h=288MWh
=288000kwh
2. The reserved capacity =installed capacity- Max load
=25MW-20MW
=5000KW
3 The maximum energy daily when the plant running all the time(day 24h).
= Installed capacity× 24
=25MW×24h
= 600MW
=6000000KW
4. Plant use factor=actual energy produce M Kwh÷ plant capacity× no of hour that the plant has been in operation
0.8= average load×24h÷ Max energy that could be produced daily if the plant were running according to it's operation schedule(y)
0.8=288000÷y
y=360MWh
1) Maximum demand =20 MW
ReplyDeleteLoad factor = 0.6
Plant capacity factor =0.48
Plant use factor = 0.8
Load factor = Average load/Maximum demand
Average demand = Load factor × Maximum demand
= 0.6 × 20000
= 12,000 kW
Plant Capacity factor = Average load/Installed capacity
Installed Capacity = Average demand/Plant capacity Factor
= 1200/0.48
= 25000 kW
2. The reserved capacity =installed capacity- Max load
=25MW-20MW
=5000KW
3 The maximum energy daily when the plant running all the time(day 24h).
= Installed capacity× 24
=25MW×24h
= 600MW
=6000000K
4 4. Plant use factor=actual energy produce M Kwh÷ plant capacity× no of hour that the plant has been in operation
0.8= average load×24h÷ Max energy that could be produced daily if the plant were running according to it's operation schedule(y)
0.8=288000÷y
y=360MWh
Solution
ReplyDeleteGiven that
Maximum demand for a generating station = 20MW(20000)
Load factor = 60%(0.6)
Plant capacity factor = 48%(0.48)
Plant use favtor = 80%(0.8)
Question 1 Solution
The daily energy produced is
Average demand =load factor ×maximum demand which is 0.6×20000= 120000KW
So therefore 120000KW×24h= 288000KW
Question 2 Solution
Reserve capacity of the plant
Plant capacity factor = average load /installed capacity, where installed capacity = 120000/0.48= 25000KW
Therefore Reserve capacity of the plant =25000- 20000= 5000KW
Question 3 Solution
The maximum energy produced daily if the plant was running all the time = installed capacity ×24h which is 25000×24= 600000KW/h
Question 4 Solution
Maximum energy produced daily if plant was running fully loaded and operating as per schedule is
plant use factor = Actual energy produced/ plant use factor
Therefore 288000/08= 360000KW.
Solution
ReplyDeleteGiven that
Maximum demand for a generating station = 20MW(20000)
Load factor = 60%(0.6)
Plant capacity factor = 48%(0.48)
Plant use favtor = 80%(0.8)
Question 1 Solution
The daily energy produced is
Average demand =load factor ×maximum demand which is 0.6×20000= 120000KW
So therefore 120000KW×24h= 288000KW
Question 2 Solution
Reserve capacity of the plant
Plant capacity factor = average load /installed capacity, where installed capacity = 120000/0.48= 25000KW
Therefore Reserve capacity of the plant =25000- 20000= 5000KW
Question 3 Solution
The maximum energy produced daily if the plant was running all the time = installed capacity ×24h which is 25000×24= 600000KW/h
Question 4 Solution
Maximum energy produced daily if plant was running fully loaded and operating as per schedule is
plant use factor = Actual energy produced/ plant use factor
Therefore 288000/08= 360000KW.
ReplyDeleteGiven parameters:
*Maximum demand for a generating station : 20MW(20000).
*Load factor : 60%(0.6).
*Plant capacity factor : 48%(0.48).
*Plant use factor : 80%(0.8).
>Load factor=average load÷ Max load
0.6=average load÷20MW
>Average load= 12MW
=12000KW
Plant capacity factor=average load÷installed capacity
0.48=12MW÷installed capacity;
Installed capacity=25MW
=25000KW
1. The daily energy produced
° daily energy produced=average load× 24
=25MW×24h=288MWh
=288000kwh
2. The reserve capacity of the plant
° The reserved capacity =installed capacity- Max load
=25MW-20MW
=5MW
=5000KW.
3. ° The maximum energy that could be produced if the plant is running all the time i.e (day 24h).
= Installed capacity× 24
=25MW×24h
= 600MW
=600000KW.
4. The maximum energy that could be produced daily if the plant was running fully loaded and operating as per schedule.
° Plant use factor=actual energy produce M Kwh÷ plant capacity× number of hour's that the plant has been in operation.
0.8= average load×24h÷ Max energy that could be produced daily if the plant were running according to it's operation schedule(y)
0.8=288000÷X
X=360MWh
The maximum energy that could be produced daily if the plant was running fully loaded and operating as per schedule(X) is 360MWh.
Load factor = 0.6
ReplyDeleteMax. demand =20 MW
P.C.F=0.48
Plant use factor = 0.8
Load factor = Average load/Maximum demand
Average demand = Load factor × Maximum demand
= 0.6 × 20000
= 12,000 kW
P.C.F = Average load/Installed capacity
Installed Capacity = Average demand/Plant capacity Factor
= 1200÷0.48
= 25000 kW
Reserve Capacity of the plant = Installed capacity – Maximum Demand
= 25000-20000
= 5000 kW
Daily energy produced =Average Demand × 24
= 12000 × 24
= 288000 kWhr
Energy/day corresponding to installed capacity = 25000×24
= 600000 kWhr
Maximum energy that could be produced = Actual energy produced/Plant use factor
= 288000/0.8
= 360000 kWhr
GIVENS:
DeleteMaximum demand =20MW
Load Factor=60%=0.6
Plant Capacity Factor=48%=0.48
Pant Use Factor=80%=0.8
1 Unit Gen (daily)=Ave. demand x 24hrs
Where>> Ave. demand = L.F x max demand=0.6 x 20MW=12MW
Therefore>> Unit Gen (daily) = 12MW x 24hrs = 288000kWh
2 Reserve Capacity = Plant Capacity – max demand
Where plant capacity= Ave. demand/Plant cap factor = 12MW/0.48 = 25MW
Therefore >> Reserve Capacity = 25MW – 20MW = 5MW
3 Max Energy Produced if running in all time = 25MW x 24hr = 600000kWh
4 Maxi energy produced fully loaded if running per schedule = Actual energy produced/plant use factor = 288000kWh/0.8 = 360,000kWh
RIBE
ReplyDeleteUnit 4: Electrical Load
on October 11, 2019
Question 7
A generating station has a maximum demand of 20 MW, a load factor of 60%, a plant capacity factor of 48% and a plant use factor of 80% . Find :
The daily energy produced
The reserve capacity of the plant
The maximum energy that could be produced daily if the plant was running all the time
The maximum energy that could be produced daily if the plant was running fully loaded and operating as per schedule.
Provide your response in the comment section using your surname and last two digit of your matriculation number as your identity (e.g. Abacus20)
Click here to like and follow the E-Learning Community on Facebook for more interactive practices and discussions with peers and instructor.
Feel free to tag your engineering friends and hashtag other social networks and learning websites.
Comments
BAMIKOLE2212 October 2019 at 01:45
Given:
Maximum demand =20 MW
Load factor = 0.6
Plant capacity factor =0.48
Plant use factor = 0.8
Load factor = Average load/Maximum demand
Average demand = Load factor × Maximum demand
= 0.6 × 20000
= 12,000 kW
Plant Capacity factor = Average load/Installed capacity
Installed Capacity = Average demand/Plant capacity Factor
= 1200/0.48
= 25000 kW
Reserve Capacity of the plant = Installed capacity – Maximum Demand
= 25000-20000
= 5000 kW
Daily energy produced =Average Demand × 24
= 12000 × 24
= 288000 kWhr
Energy/day corresponding to installed capacity = 25000×24
= 600000 kWhr
Maximum energy that could be produced = Actual energy produced/Plant use factor
= 288000/0.8
= 360000 kWhr
REPLY
http://astoldpost.blogspot.com/12 October 2019 at 09:21
������
ADEYEMI0612 October 2019 at 03:58
1. Formula for daily energy produced = average load * 24hrs.
To get our avegate load we use, load factor = average load/ max load
Where load factor is 0.6 and max load is 20MW, so
Average load = 20 *0.6
= 12MW.
So daily energy produced is 12MW * 24h = 288MWh also 288,000kwh.
2. Reserve capacity = installed capacity - max load
Whereas installed capacity = average load/plant capacity factor
=12/0.48
=25MW..
So reserve capacity = 25 -20
=5MW (5000kW)
3. Maximum energy produced daily if the plant works all time is
Installed capacity * 24hrs
= 25MW * 24
=600Mwh(6000000kwh)
4. Plant use factor = actual energy produced / plant use factor
= 288/0.8
=360 MWh
REPLY
Olusanya.4912 October 2019 at 06:12
1. Formula for daily energy produced = average load * 24hrs.
To get our avegate load we use, load factor = average load/ max load
Where load factor is 0.6 and max load is 20MW, so
Average load = 20 *0.6
= 12MW.
So daily energy produced is 12MW * 24h = 288MWh also 288,000kwh.
2. Reserve capacity = installed capacity - max load
Whereas installed capacity = average load/plant capacity factor
=12/0.48
=25MW..
So reserve capacity = 25 -20
=5MW (5000kW)
3. Maximum energy produced daily if the plant works all time is
Installed capacity * 24hrs
= 25MW * 24
=600Mwh(6000000kwh)
4. Plant use factor = actual energy produced / plant use factor
= 288/0.8
=360 MWh
1. Formula for daily energy produced = average load * 24hrs.
ReplyDeleteTo get our avegate load we use, load factor = average load/ max load
Where load factor is 0.6 and max load is 20MW, so
Average load = 20 *0.6
= 12MW.
So daily energy produced is 12MW * 24h = 288MWh also 288,000kwh.
2. Reserve capacity = installed capacity - max load
Whereas installed capacity = average load/plant capacity factor
=12/0.48
=25MW..
So reserve capacity = 25 -20
=5MW (5000kW)
3. Maximum energy produced daily if the plant works all time is
Installed capacity * 24hrs
= 25MW * 24
=600Mwh(6000000kwh)
4. Plant use factor = actual energy produced / plant use factor
= 288/0.8
=360 MWh
Given that
ReplyDeleteMaximum demand for a generating station = 20MW(20000)
Load factor = 60%(0.6)
Plant capacity factor = 48%(0.48)
Plant use factor = 80%(0.8)
1. Formula for daily energy produced = average load * 24hrs.
To get our average load, we use the relation, load factor = average load/ max load
Where load factor is 0.6 and max load is 20MW, so
Average load = 20 *0.6
= 12MW.
So daily energy produced is 12MW * 24h = 288MWh also 288,000kwh.
2. Reserve capacity = installed capacity - max load
Whereas installed capacity = average load/plant capacity factor
=12/0.48
=25MW..
So reserve capacity = 25 -20
=5MW (5000kW)
3. Maximum energy produced daily if the plant works all time is
Installed capacity * 24hrs
= 25MW * 24
=600Mwh(6000000kwh)
4. Plant use factor = actual energy produced / plant use factor
= 288/0.8
=360 MWh
1. The daily energy produced
ReplyDelete° daily energy produced=average load× 24
=25MW×24h=288MWh
=288000kwh
2. The reserve capacity of the plant
° The reserved capacity =installed capacity- Max load
=25MW-20MW
=5MW
=5000KW.
3. ° The maximum energy that could be produced if the plant is running all the time i.e (day 24h).
= Installed capacity× 24
=25MW×24h
= 600MW
=600000KW.
4. The maximum energy that could be produced daily if the plant was running fully loaded and operating as per schedule.
° Plant use factor=actual energy produce M Kwh÷ plant capacity× number of hour's that the plant has been in operation.
0.8= average load×24h÷ Max energy that could be produced daily if the plant were running according to it's operation schedule(y)
0.8=288000÷X
X=360MWh
The maximum energy that could be produced daily if the plant was running fully loaded and operating as per schedule(X) is 360MWh.
Given parameters
ReplyDeleteMaximum demand = 20MW
Load factor =60%(0.6)
Plant capacity factor =48% (0.48)
Plant use factor =80%(0.8)
1: for daily energy produced = average load ×24h
Load factor = average load/max load. Where load factor=0.6, and max load =20MW
Therefore: Average load = 0.6×20=12000kW
Daily energy produced= 12000×24=288000kWh
2: the reserved capacity= installed capacity - max load.
Installed capacity = average load /plant capacity factor. I.e 12/0.48= 25000kW
Therefore: reserved capacity= 25000- 20000= 5000kW
3: The max. energy produced daily when the plant run all the time = installed capacity × 24h
25000×24= 600000kWhr
4: plant use factor = actual energy produced /plant use factor
=288/0.8
= 360MWh
Adegun03
DeleteLoad factor = 0.6
Max. demand =20 MW
P.C.F=0.48
Plant use factor = 0.8
Load factor = Average load/Maximum demand
Average demand = Load factor × Maximum demand
= 0.6 × 20000
= 12,000 kW
P.C.F = Average load/Installed capacity
Installed Capacity = Average demand/Plant capacity Factor
= 1200÷0.48
= 25000 kW
Reserve Capacity of the plant = Installed capacity – Maximum Demand
= 25000-20000
= 5000 kW
Daily energy produced =Average Demand × 24
= 12000 × 24
= 288000 kWhr
Energy/day corresponding to installed capacity = 25000×24
= 600000 kWhr
Maximum energy that could be produced = Actual energy produced/Plant use factor
= 288000/0.8
= 360000 kWhr
Given;
ReplyDeleteMaximum demand of a generating station=20MW
Load factor of the generating station= 60%= 0.6
Plant capacity factor =48%=0.48
Plant use factor of the generating station=80%=0.8
1). For the daily energy produced
Average load * 24hrs=0.6*20=12MW
For the daily energy
12*24hrs=288Mwh
2).Reserve capacity = installed capacity - max load
Although, installed capacity = average load/plant capacity factor
=12/0.48
=25MW..
So reserve capacity = 25 -20
=5MW .
3. The maximum energy that could be produced if the plant is running all the time i.e (day 24h).
= Installed capacity× 24
=25MW×24h
= 600MW.
4 Maxi energy produced fully loaded if running per schedule = Actual energy produced/plant use factor = 288000kWh/0.8 = 360,000kWh
Maximum demand =20 MW
ReplyDeleteLoad factor = 0.6
Plant capacity factor =0.48
Plant use factor = 0.8
Load factor = Average load/Maximum demand
Average demand = Load factor × Maximum demand
= 0.6 × 20000
= 12,000 kW
Plant Capacity factor = Average load/Installed capacity
Installed Capacity = Average demand/Plant capacity Factor
= 1200/0.48
= 25000 kW
Reserve Capacity of the plant = Installed capacity – Maximum Demand
= 25000-20000
= 5000 kW
Daily energy produced =Average Demand × 24
= 12000 × 24
= 288000 kWhr
Energy/day corresponding to installed capacity = 25000×24
= 600000 kWhr
Maximum energy that could be produced = Actual energy produced/Plant use factor
= 288000/0.8
= 360000 kWhr
Sir I think creating the option to attach pictures would be easier because typing this wasn't easy��
Load factor = 0.6
ReplyDeleteMax. demand =20 MW
P.C.F=0.48
Plant use factor = 0.8
Load factor = Average load/Maximum demand
Average demand = Load factor × Maximum demand
= 0.6 × 20000
= 12,000 kW
P.C.F = Average load/Installed capacity
Installed Capacity = Average demand/Plant capacity Factor
= 1200÷0.48
= 25000 kW
Reserve Capacity of the plant = Installed capacity – Maximum Demand
= 25000-20000
= 5000 kW
Daily energy produced =Average Demand × 24
= 12000 × 24
= 288000 kWhr
Energy/day corresponding to installed capacity = 25000×24
= 600000 kWhr
Maximum energy that could be produced = Actual energy produced/Plant use factor
= 288000/0.8
= 360000 kWhr
Alade 12
ReplyDeleteGiven:
Maximum demand =20 MW
Load factor = 0.6
Plant capacity factor =0.48
Plant use factor = 0.8
Load factor = Average load/Maximum demand
Average demand = Load factor × Maximum demand
= 0.6 × 20000
= 12,000 kW
Plant Capacity factor = Average load/Installed capacity
Installed Capacity = Average demand/Plant capacity Factor
= 1200/0.48
= 25000 kW
Reserve Capacity of the plant = Installed capacity – Maximum Demand
= 25000-20000
= 5000 kW
Daily energy produced =Average Demand × 24
= 12000 × 24
= 288000 kWhr
Energy/day corresponding to installed capacity = 25000×24
= 600000 kWhr
Maximum energy that could be produced = Actual energy produced/Plant use factor
= 288000/0.8
= 360000 kWhr
REPLY
http://astoldpost.blogspot.com/12 October 2019 at 09:21
👍👍💡
ADEYEMI0612 October 2019 at 03:58
1. Formula for daily energy produced = average load * 24hrs.
To get our avegate load we use, load factor = average load/ max load
Where load factor is 0.6 and max load is 20MW, so
Average load = 20 *0.6
= 12MW.
So daily energy produced is 12MW * 24h = 288MWh also 288,000kwh.
2. Reserve capacity = installed capacity - max load
Whereas installed capacity = average load/plant capacity factor
=12/0.48
=25MW..
So reserve capacity = 25 -20
=5MW (5000kW)
3. Maximum energy produced daily if the plant works all time is
Installed capacity * 24hrs
= 25MW * 24
=600Mwh(6000000kwh)
4. Plant use factor = actual energy produced / plant use factor
= 288/0.8
=360 MWh
REPLY
Olusanya.4912 October 2019 at 06:12
1. Formula for daily energy produced = average load * 24hrs.
To get our avegate load we use, load factor = average load/ max load
Where load factor is 0.6 and max load is 20MW, so
Average load = 20 *0.6
= 12MW.
So daily energy produced is 12MW * 24h = 288MWh also 288,000kwh.
2. Reserve capacity = installed capacity - max load
Whereas installed capacity = average load/plant capacity factor
=12/0.48
=25MW..
So reserve capacity = 25 -20
=5MW (5000kW)
3. Maximum energy produced daily if the plant works all time is
Installed capacity * 24hrs
= 25MW * 24
=600Mwh(6000000kwh)
4. Plant use factor = actual energy produced / plant use factor
= 288/0.8
=360 MWh
REPLY
Afolabi.8812 October 2019 at 06:16
Given:
Maximum demand =20 MW
Load factor = 0.6
Plant capacity factor =0.48
Plant use factor = 0.8
Load factor = Average load/Maximum demand
Average demand = Load factor × Maximum demand
= 0.6 × 20000
= 12,000 kW
Plant Capacity factor = Average load/Installed capacity
Installed Capacity = Average demand/Plant capacity Factor
= 1200/0.48
= 25000 kW
Reserve Capacity of the plant = Installed capacity – Maximum Demand
= 25000-20000
= 5000 kW
Daily energy produced =Average Demand × 24
= 12000 × 24
= 288000 kWhr
Energy/day corresponding to installed capacity = 25000×24
= 600000 kWhr
Maximum energy that could be produced = Actual energy produced/Plant use factor
= 288000/0.8
= 360000 kWhr
REPLY
Olaleye.1012 October 2019 at 06:39
Maximum demand =20 MW
Load factor = 0.6
Plant capacity factor =0.48
Plant use factor = 0.8
Load factor = Average load/Maximum demand
Average demand = Load factor × Maximum demand
= 0.6 × 20000
= 12,000 kW
Plant Capacity factor = Average load/Installed capacity
Installed Capacity = Average demand/Plant capacity Factor
= 1200/0.48
= 25000 kW
Reserve Capacity of the plant = Installed capacity – Maximum Demand
= 25000-20000
= 5000 kW
Daily energy produced =Average Demand × 24
= 12000 × 24
= 288000 kWhr
Energy/day corresponding to installed capacity = 25000×24
= 600000 kWhr
Maximum energy that could be produced = Actual energy produced/Plant use factor
= 288000/0.8
= 360000 kWhr