3. Result and Discussion
3.1. Question 01
For determining the mean and standard deviation, by month filter was applied and then the average and standard deviation of every month was taken. For the yearly mean and standard deviation, 8760 data were analyzed to get the values for 2014, 2015, and 2016.
Mean speed for 2015 = 4.875 m/s
Mean speed for 2014 = 5.07 m/s
Mean speed for 2016 = 5.15 m/s
Standard Deviation for 2015 = 2.6585
Standard Deviation for 2014 = 2.73
Standard Deviation for 2016 = 2.79
For the yearly mean and standard deviation for the year 2014, the rand () function in Excel was used to create a random sample and then 1000 data points were collected from that group. From the 1000 random data points, the month wise mean and standard deviation for 2014 was calculated and they are as follows:
Month |
Mean (M/S) |
Standard Deviation |
January |
6.09 |
2.93 |
February |
5.93 |
3.27 |
March |
5.29 |
2.86 |
April |
5.36 |
2.84 |
May |
4.41 |
2.75 |
June |
4.56 |
2.2 |
July |
4.18 |
2.09 |
August |
4.27 |
2.09 |
September |
4.10 |
2.09 |
October |
5.22 |
2.34 |
November |
4.49 |
2.52 |
December |
5.49 |
1.54 |
3.2. Question 02
Figure 2.
Probability plot of wind speed with normal distribution.
Figure 2.
Probability plot of wind speed with normal distribution.
Figure 3.
Probability plot of wind speed with exponential distribution.
Figure 3.
Probability plot of wind speed with exponential distribution.
Figure 4.
Probability plot of wind speed with lognormal distribution.
Figure 4.
Probability plot of wind speed with lognormal distribution.
Using Minitab to fit a random sample of 1000, we concluded that the Weibull distribution is the best fit but still has a low p-value of <0.01. The second-best distributions were normal distribution with a p-value <0.005.
Figure 5.
Probability plot of wind speed with Weibull distribution.
Figure 5.
Probability plot of wind speed with Weibull distribution.
3.3. Question 03
- ✓
Hypothesis testing on windspeed means, Null Hypothesis: Ho: uo = u1
- ✓
Alternative hypothesis: H1: uo≠u1
- ✓
The mean of the 1000 random value of wind speed of 2015 is = 4.85 m/s
- ✓
The variance of the 1000 random value of wind of 2015 speed is = 7.05
- ✓
Standard deviation = 2.64
- ✓
The mean of the 1000 random value of wind speed of 2016 is = 5.05m/s
- ✓
The variance of the 1000 random value of wind of 2016 speed is = 7.78 m/s
- ✓
Standard deviation = 2.78 m/s
- ✓
Sp= 2.72
- ✓
to = -0.51
- ✓
to falls in between the range. Hence, we fail to reject the null hypothesis. So, the mean wind speed is not significantly different between two adjacent years.
Here, F value is less than the F critical value, so we fail to reject the null hypothesis using 95% confidence level.
Figure 6.
F-test of two samples for variances.
Figure 6.
F-test of two samples for variances.
3.4. Question 04
We have taken 1000 sample data from the 8760 data to estimate the yearly wind speed. We have used Vh = Vg (80/8)3 to find the wind speed at turbine height. We have also removed the data which are unable to operate wind turbine. So, we have eliminated the wind speed less than 2m/s and greater than 25m/s. This gives us 330 data where wind turbine can operate between wind speed 12 m/s to 25 m/s.
The output will be = 2 MW/h*330 hrs. = 650 MWh. Between 0 to 2m/s numbers of samples = 32 Between 2 to 4m/s numbers of samples = 106 Between 4 to 6m/s numbers of samples = 134 Between 6 to 8m/s numbers of samples = 196 Between 8 to 10m/s numbers of samples = 150 Between 10 to 12m/s numbers of samples = 52
Wind Speed (m/s) |
12-25 |
10-12 |
8-10 |
6-8 |
4-6 |
2-4 |
Power (MW) |
2 |
1.75 |
1.5 |
1.25 |
0.75 |
0.25 |
From 1000 random data we have calculated total power =
(330Hrs x 2MWh) + (52Hrs x 1.75MWh) + (150Hrs x 1.5MWh) + (196Hrs x 1.25MWh) +
(134Hrs x .75MWh) + (106Hrs x .25MWh) = 1348 MWh total
For total 8760 data it will be = (1348/1000) *8760 = 11,808.48 MWh total
So, the number of homes wind turbine could power = (11,808.48*1000)/7300 = 1618 homes
3.5. Question 05
Figure 7.
Sample calculation for question 5.
Figure 7.
Sample calculation for question 5.
For PV output all-night data have been deleted and set the PV angle to the same as the latitude of NYC to maximize output. After calculating the sunrise and sunset time, excel was used to find solar irradiance at every hour. To find the power output I mapped the given data condition range of 1-9 to the required 1-0. Then the power output at every hour could be established for the panels. After power output at every hour was found the sum has been calculated and found the yearly power output.
In total, the solar panels generated 867.37 MWh.
3.6. Question 06
Figure 8.
Hourly wind speed forecasting for January 02, 2014.
Figure 8.
Hourly wind speed forecasting for January 02, 2014.
Using the second order of polynomial line, the equation for forecasting is:
For the linear fit model, equation is, y = 0.2608x + 14.732.
Figure 9.
Monthly wind speed forecasting for January 2014.
Figure 9.
Monthly wind speed forecasting for January 2014.
For monthly wind speed estimation, we averaged the daily windspeed for January 2014, using this data then plotted the daily windspeed average vs the day of the month. For a second-order polynomial trend line, the equation obtained is, y = 6E-06x2 - 0.0041x + 13.268
A linear trend line gives the equation: y = 0.0004x + 12.705.
Figure 10.
Yearly wind speed forecasting for January 2014.
Figure 10.
Yearly wind speed forecasting for January 2014.
For yearly wind speed, the average wind speed for every month for the year 2014 was calculated and plotted vs the number of the month. The formula for the second-degree polynomial is y = 0.0473x2 - 0.7101x + 7.0016
Using a Linear line of fit we got, y = -0.0949x + 5.5662.
3.7. Question 07
Factors that are significant and/or correlated:
Factor 1. The Size of PV system: The size of the PV is one of the important factors. The energy generation will increase with the PV size.
Factor 2. PV Capacity Cost: The cost is the most important factor that can be calculated by the multiplication of the number of PV panels and the installation cost. The equation is 106x. Here, x
= PV size in MW.
Factor 3. PV Efficiency: The PV efficiency has been selected as maximum as 25%. If we don’t care much about cost, the efficiency can be increased in a limited size.
Factor 4. Weather Condition: In perfect weather conditions, the factory needs 100,000 m2 panels to handle roughly 27 MW assuming each panel generates 270 W/m2. The angle of the sun and weather conditions have a great effect on the fluctuation of power, on average in NYC is 130 W/m2.
In January 2014 average weather condition in New York is 0.412 representing the solar cells generate about 41% of their maximum output during the day in winter. Whereas in summer the value is 0.453 or 45% of maximum generation
Factor 5 and 6. Ambient Temperature and Wind Speed: Ambient temperature and wind speed have also an effect on panel efficiency. Both are helping to cool down the PV systems, making them run more efficiently.
Factor 7. Hour of a Day: The power can be generated in daytime, not at night and the maximum power can be found at noon. Based on the formula, in January (winter) it is 1274.76 W/m2 and in summer it is 1468.6 W/m2.
Factor 8 & 9. Utility Pricing Scheme and Government Incentives: Utility scheme could offer an income possibility if the PV system is generating more power than it is consuming, and Government incentives could drastically lower the upfront price of panel installation.
Factor 10. The batter capacity: battery capacity can reduce the size of the PV system by storing energy when power generation is over the factory power usage and discharging it during bad weather or at night.
In winter, the factory has a peak energy draw of 17.4 MW during hours 13 and 16. So the expected power generation in January 1274.76 W × 25% × 1m2 × 0.412 × 1.0309 (from temperature) = 135.4.
In summer, 1468.6 W × 25% × 1m2 × 0.453 × 0.8621 = 143.4.
In winter, factory draws an average of 14.78 MW with a peak of 17.4 MW. Minimum we need
17.4 × 106/ 135.4 = 128,509 panels at 1 m2 each. In summer, minimum we need 22.6 × 106/143.4
= 157,602 panels at 1 m2 each.
If the panels make an average of 17.4 MW during the daytime (from hour 7 to 17), we can create
17.4 × 10 = 174 MWh per day. The factory needs 14.78 × 24 = 354.8 MWh. So, the system would need an average of 35.48 MW/hour for 10 hours to generate enough energy for daytime power draw peaks and night usage.
If we use a 60 MW system, simulation gives us average excess energy storage of 62.3 MW/day.
If a 510 MW battery is installed, it would take 510/62.3 = 8.2 day to fully charge and can be capable of providing a little over one day of full power outage.
3.8. Question 08
Life cycle cost analysis
Cost of Electricity by source in US
Power Plant Type |
Cost (LCOE) |
|
$/kW-hr |
Coal with CCS |
$0.12-0.13 |
CC Natural Gas |
$0.04 |
CC with CCS |
$0.08 |
Nuclear |
$0.09 |
Wind onshore |
$0.04 |
Wind offshore |
$0.11 |
Solar PV |
$0.04 |
Solar Thermal |
$0.17 |
Geothermal |
$0.04 |
Biomass |
$0.09 |
Hydro |
$0.03 |
(Levelized cost of electricity, LCOE) Source: Adapted from US DOE