Reference no: EM132152714
Electricity Industry Operation and Control Assignment -
Question 1 - Calculate the optimal unit commitment program using backward dynamic programming for the small power system described below. The power system has three generating units. The third unit (n = 3) has no start-up or no-load running costs, or minimum operating level, and hence can effectively be kept committed at all times without penalty. The other two units are available for commitment and de-commitment and will be run under economic dispatch when committed. Ignore generator start up times and ramp rate limits. The total period of unit commitment is 4 hours split into four one hour intervals. Just before the beginning of the first time period, Unit 1 and Unit 3 are on and Unit 2 is off.
The operational characteristics of each unit (n = 1, 2, 3) can be described by the following parameters and equations:
start up cost: Cn($/start)
variable cost curve: Vn(Pn) = Vn0 + vn(Pn - PnMin) ($/hr)
limits on power output: PnMin ≤ Pn ≤ PnMax (MW)
where Vn0 = no-load running cost ($/hr)
vn = incremental variable cost ($/MWh)
Pn = power output (MW)
Data for these units are:
Unit n
|
vn
|
Vn0
|
PnMin
|
PnMax
|
Cn
|
Initial state
|
1
|
50
|
56,000
|
700
|
1200
|
280,000
|
On
|
2
|
80
|
30,000
|
300
|
800
|
120,000
|
Off
|
3
|
150
|
0
|
0
|
1200
|
0
|
On
|
Load data are as follows:
Hour period
|
1
|
2
|
3
|
4
|
Load (MW)
|
1100
|
1900
|
1500
|
900
|
Note that the cost of not meeting load is $14700 per MWh of unmet load. An answer template is provided to assist in solving the question.
Question 2 - There is a spreadsheet for this assignment available on the moodle which includes 30 minute pricing for the South Australian region for calendar year 2017-2018
Use this to estimate the value of a $300 cap for each financial quarter of that year ($/MW)
- Summer (January to March)
- Autumn (April to June)
- Winter (July to September)
- Spring (October to December)
Question 3 - A small power system has two thermal generators with capacity, cost and forced outage rate parameters as shown in the table below. It also has a 1200MW wind farm. The output of this windfarm has the following probability distribution - 1200MW for 25% of the time and 300MW for 75% of the time. These generators serve a load with the inverted load duration curve shown below. Note that the value of lost load (VoLL) is $14700/MWh. Assume that the wind farm output and load are not correlated.
Generator
|
Capacity (MW)
|
Incremental variable cost ($/MWH)
|
Forced outage rate
|
1
|
2500
|
40
|
0.05
|
2
|
1000
|
70
|
0.025
|
Wind farm
|
1200
|
0
|
0
|
The simulation period is one year ahead. Using a table to present your results, enumerate all possible scenarios of generator availability, wind farm output and load demand in terms of their probabilities, hence hours per year, ability to meet load, and associated power system production costs ($/hr) to calculate over a typical year:
(a) Loss of load probability (total hours/year).
(b) Expected unserved energy (USE) (% of load).
(c) Expected production cost over the year ($000) noting that any USE over the year costs $14700/MWh.
Question 4 - A power system has two generators, GenA and GenB, supplying a constant load of 1900 MW. GenA has an incremental variable cost of $50/MWh and a capacity of 2000 MW. It is an ageing coal plant and suffers from random failures and then associated repairs. GenB is a large gas peaker plant and has an incremental variable cost of $150/MWh and a capacity of 2000 MW. It is very carefully maintained and can be considered to be completely reliable (i.e. always available for service if and when required).
The time between failures for GenA (i.e. the time for which it is continuously available for service) has an approximate normal distribution with a mean of 400 hours and standard deviation of 150 hours (but noting of course that there can be no negative times between failures).
The time to repair GenA (i.e. the time for which it is out of service) also has an approximate normal distribution with a mean of 30 hours and standard deviation of 15 hours (again, there can be no negative repair times but there may well be a number of '0' hour repairs where the unit can be bought back into service almost immediately.
At the beginning of the study period, GenA has just been repaired and become available. Calculate one sample of performance of the system using Monte-Carlo simulation of the operation of this system over a study period of one year (that is, 8760 hours), using the following technique.
(1) Using the random numbers available in the spreadsheet made available for this assignment, determine consecutive samples of times to repair and times to failure for GenA. (Use the random numbers in the order provided to calculate a time to failure, then the next number to calculate a time to repair, and so on)
(2) Calculate the production cost rate in $/hour for each of the systems states (i.e. for GenA available and for GenA not available).
(3) Hence calculate the total production cost for this sample outcome.
(4) (optional task). Use the random number generator available in Excel to undertake power system simulations using 10 runs, 100 runs, 1000 runs and say 10,000 Monte-Carlo runs for the above problem, calculating the total average production cost over one year for this power system for each simulation. Comment on the likely number of runs required to get a reasonable (say within 1%) production cost estimate for this problem.
Attachment:- Assignment Files.rar