**Physical Address**

304 North Cardinal St.

Dorchester Center, MA 02124

The main goal of this course is to get the necessary knowledge on atmospheric and fluid dynamics in order to quantify the wind resource of a local or regional area.

We’ll learn about basic meteorology, the specific dynamics of turbulent boundary layers and some standard techniques to estimate wind resources regardless of the type of turbine used or the level of efficiency achieved. Then, we will see what are the turbines characteristics to consider in order to estimate the electricity production from an isolated turbine or from a turbine farm. The differences and similarity between wind or marine resource assessment will also be discussed.

Finally, you will have the opportunity to get hands-on experience with real in-situ data sets and apply what you have learned on wind resource assessment.

Q1. Which one of these units is a unit of energy ?

- Kilowatts
- Watt-hours
- Joule
- Ton of oil equivalent

Q2. What energy is of the same order of magnitude as 1 Toe ?

- 100 kWh
- 1 000 kWh
- 42 GWh
- 10 000 kWh

Q3. The amount of USEFUL ENERGY to heat a house during one year depends on

- The energy resource
- The efficiency of the heating system
- The nature of the heating system (oil heater, electric heater, etc. )
- The desired temperature in the house

Q4. The amount of FINAL ENERGY to heat a house during one year depends on

- The efficiency of the heating system
- The price of energy
- The nature of the heating system (oil heater, electric heater, etc. )
- The energy resource

Q5. Which of these power plants has a CAPACITY FACTOR mainly impacted by the variability of the resource ?

- Geothermal
- Photovoltaic panels
- Nuclear power plant
- Hydroelectric

Q6. The CAPACITY FACTOR of a wind farm

- could vary from one year to another
- depends on the installed power
- depends on the demand
- depends on the wind resource

Q7. The nameplate capacity of the off shore Horns Rev I wind farm is 160MW while its average annual energy generation is 600GWh. The mean capacity factor is therefore:

- 32%
- 43%
- 23%
- 48%

Q1. One nuclear reactor having a nominal power of 800MW generate an annual amount of 514k toe (1k =1000) of electricity. What is the correct capacity factor of this power unit ?

- 90%
- 71%
- 73%
- 88%
- 78%
- 85%

Q2. The above capacity factor is mainly due to

- the variability of the resource
- the amount of primary energy
- the variability of the demand
- the maintenance of the reactor

Q3. We consider an offshore wind farm of 150 turbines of 6MW of nominal power each. The annual capacity factor of this large wind farm is around 35%. The annual amount of electric energy produced will be around:

- 238k toe
- 190k toe
- 127k toe
- 582k toe
- 625k toe
- 305k toe

Q4. How many offshore wind turbines of 6MW (same capacity factor as Q3) will be needed to produce the same amount of annual energy as a nuclear reactor of 800MW having a capacity factor of 78%

- 297
- 134
- 182
- 203
- 312
- 245

Q5. We consider the above offshore wind farm of 150 turbines of 6MW of nominal power each. The annual capacity factor is around 35%. The typical seasonal variability of the energy production for this wind farm is given by the following graph:

This graph gives, in percentage, a relative value of the energy produced each month. The monthly mean energy (the annual energy /12) correspond to 100.

Find which statements are true.

- The maximal power output of the wind farm is 315MW
- The maximal power output of the wind farm is 630MW
- The maximal power output of the wind farm is 900MW
- The maximal power output of the wind farm is 1340MW
- The maximal monthly production is around 230GWh in december
- The maximal monthly production is around 340 GWh in december
- The maximal monthly production is around 530 GWh in december
- The maximal monthly production is around 650 GWh in december
- The monthly production in december exceed the production of one nuclear reactor of 800MW (78% capacity factor)

Q1. The atmospheric layer absorb

- more energy (per square meters) in the tropics than the poles
- a small fraction of the solar spectrum
- more energy than the ocean surface
- only the infrared part of the solar spectrum
- only the visible part of the solar radiation

Q2. The will be no winds if

- there is no water vapor in the atmosphere
- the centrifugal force is equal to zero
- the Coriolis force is equal to zero
- there is no differential solar heating between tropics and mid-latitudes

Q3. The temperature of the troposphere decays with height mainly because

- the winds increase with height and therefore reduce the temperature
- the pressure decays with height
- the surface of the earth is warmed by the solar heating
- the water vapor absorb a fraction of the solar radiation

Q1. The vertical component of the Coriolis force

- is much weaker than the horizontal component
- because the Rossby number is small
- depends on the latitude
- is neglected because the vertical velocities are weak
- depends on the longitude

Q2. If we consider a small atmospheric structure having a typical horizontal scale L=H=10km and keeps all the other terms unchanged

- The Rossby number remains small
- We could still neglect the vertical acceleration in the momentum equation for the vertical component
- We could neglect the pressure gradient in comparison with the gravity
- We could neglect the gravity in comparison with the pressure gradient
- The hydrostatic balance is still valid

Q3. In the tropics there is on average

- less clouds due to the strong winds
- easterly winds close to the surface
- westerly winds at high latitude
- more clouds due to the deep convection

Q1. Surface winds

- satisfy the geostrophic balance
- are affected by the bottom friction
- are (on average) weaker than in the free troposphere
- are affected by the Coriolis force
- satisfy the hydrostatic balance

Q2. The bottom drag

- is turbulent
- depends on the air viscosity
- depends on the Coriolis parameter
- induces a rotation in the wind direction

Q3. The boundary layer height

- Is about 100m
- Varies between day and night
- Is about 10 000 m
- stays constant due to the bottom turbulence
- Is about 1000m

Q1. The Coriolis force

- Is high close to the equator and low close to the poles
- Is parallel to the wind velocity
- Is orthogonal to the wind velocity
- Deflect the air parcels towards the right in the northern hemisphere

Q2. The geostrophic wind is a theoretical wind that

- Is a local wind
- Comes from the balance between the Coriolis force and the pressure gradient force
- Is closer to the measured wind on the sea than on the land
- Flow along isobars
- Flow from high pressure area to low pressure area
- Result from the assumption that the Coriolis force on the air is negligible

Q3. Close to the ground winds are also subjected to the bottom friction that result in a force opposite to wind direction. Find which sentence is true

- As a result of high friction close to the ground the wind direction changes with altitude
- Actual wind measured close to the ground/sea level points slightly towards high pressure area.
- Actual wind will be closer to the geostrophic wind at high altitude
- 10m winds are strictly parallel to isobaric lines

Q4. On average, the strongest winds are located

- Close to the poles
- At the equator
- At 2km height
- At mid latitudes
- At 10km height

Q5. The Rossby number is

- the ratio between gravity and Coriolis acceleration
- weak for a large anticyclone
- weak when the flow satisfies the hydrostatic balance
- large when the flow satisfies the geostrophic balance

Q6. Today we mainly extract wind power in

- In the lower stratosphere
- The free upper troposphere
- In the surface layer
- Above the boundary layer
- In the residual layer

Q7. In the boundary layer, the winds tend to be

- weaker than the geostrophic wind
- parallel but weaker to the upper winds
- deviated toward lower pressures
- parallel to the temperature gradient

Q1. We consider an atmospheric boundary layer where the temperature (as the velocity field) can be decomposed in two terms, a steady temperature ⟨T⟩ and turbulent fluctuations. We can then write:

where ⟨. ⟩ corresponds to a statistical mean. The statistical average of the diffusion term

- Option 1: is then equal to:

- Option 2: is then equal to

- Option 3: is then equal to

- Option 4: is then equal to:

- Option 5: is then equal to:

Q2. We keep here the same hypothesis and notations as in the previous question.The statistical average of the non-linear term

- Option 1: is equal to

- Option 2: is equal to

- Option 3: is equal to

- Option 4: is equal to

- Option 5: is equal to

Q3. We keep here the same hypothesis and notations as in the previous question.The statistical average of the non-linear term

- is equal to

- is equal to

- is equal to

- is equal to

Q1. We consider the equation of advection-diffusion of heat within an 2D idealized atmospheric boundary layer :

where the temperature (as the velocity field) can be decomposed in two terms, a steady profile ⟨T⟩ and turbulent fluctuations. We consider that this idealized boundary layer is invariant along Oy, in other words ∂y = 0 for all the variables.

If we take into account the volume conservation in the xOz plane, the flow is non-divergent and satisfies

We can then re-write the equation (1)

- in the following form

- in the following form

- in the following form

- in the following form

- in the following form

Q2. If we apply the statistical mean to the modified equation (1) chosen above and if we assume that

- we will get

- we will get

- we will get

- we will get

- we will get

Q1. The vertical profile of the horizontal wind speed is usually modeled by either a logarithmic law or a power law. Specify which of the following statement is true

- The shear exponent
*alpha*is likely to differ between day and night - The log law assumes a turbulent flow induced by the bottom roughness or a convectively unstable atmosphere
- The power law is justified by the RANS equations
- The log law is valid up to 10km

Q2. The wind profile within a stable a strongly stratified boundary layer is better described by:

- A power law with a wind shear exponent
*alpha*> 0.4 - A logarithmic law corrected with Monin-Obukhov functions
- A 1/7 power law
- A Poiseuille profile
- A standard logarithmic law

Q3. The 1/7 power law is more relevant for

- A convectively unstable layer in the late morning
- A stable and strongly stratified layer
- Strong winds in the boundary layer
- A slightly unstable convective layer

Q4. The turbulence intensity of an atmospheric boundary layer:

- Could reaches 60% during the night
- Increases with height
- Increases with the wind intensity
- Is stronger for an unstable convective layer
- Depends on the temporal averaging of the measurements

Q5. The sea breeze winds

- Are generally associated with cumulus clouds on the land
- Starts to blow in the early afternoon
- Are directed toward the open ocean
- Starts to blow in the early morning

Q1. Which instrument will be the most accurate to quantify the turbulence intensity

- Ultra sonic anemometers
- Radio soundings with atmospheric balloons
- A wind lidar
- Cup anemometers
- A wind sodar

Q2. The sodar measurements

- are less accurate for strong winds
- are affected by strong precipitations
- are affected by cumulus clouds
- are affected by fog
- could identify a stable and stratified layer

Q3. The wind lidar measurements

- are affected by cumulus clouds
- are affected by rains
- are less accurate for strong winds
- could identify an unstable convective layer
- matches the cup anemometers measurements

Q4. Remote sensing instruments (lidar or sonar)

- have the same precision as cup anemometer
- can measure wind at turbine height and above
- need an external energy source
- should be always connected to the electric grid
- can hardly measure turbulent winds fluctuations

Q5. Specific hydrometeors or aerosols moving in the atmosphere

- Induce a Doppler shift on the backscattered waves
- Reflect the electromagnetic wave of the lidar
- Are advected by the wind
- Reflect the sound waves of the sodar
- Are induced by the atmospheric turbulence
- Are emitted by the lidar

Q1. According to the Betz law the optimal turbine efficiency is reached when:

- The power coefficient Cp is equal to 16/27
- The tip speed ratio is large
- The downstream velocity is 1/3 of the upstream wind
- The viscous dissipation is neglected

Q2. The underlying assumptions to derive the Betz law are

- The fluid is inviscid (i.e. non viscous dissipation)
- The thrust (pressure) exerted by the fluid on the turbine is uniform on the swept area
- The flow downstream the turbine is turbulent
- The fluid is air and not water
- The upstream flow is steady

Q3. For a real turbine with fixed wind intensity and a fixed diameter

- A three blades turbine could exceed a power coefficient of 0.35
- A optimized Darrius rotor will have a lower rotation than a two blades turbine
- The Savonius turbine will be more efficient than other ones
- We could reach the Betz limit Cp=16/27 if we adjust the turbine rotation

Q1. To optimize the efficiency of real turbines the tip speed of the blades should

- starts after the cut-in speed
- be adjusted to the upstream flow
- stays constant
- exceed the speed of sound
- be limited by a cut-out speed

Q2. There is generally no cut-out speed for marine turbines because

- the turbulence intensity is weaker in water than in the air
- the directions of tidal currents are predictable
- tidal currents speeds are higher than wind speeds
- there are no uncertainties on the maximal tidal velocities
- there is no unstable convection in water

Q3. The turbulence of the upstream flow could modify the turbine efficiency by

- 40-50%
- no impact
- 0-5%
- 20-30%

Q1. The efficiency of a wind farm depends on

- the accuracy of the weather forecast
- the wind direction
- the wind intensity
- the intensity of unstable convection
- the layout optimization

Q2. The efficiency of a single (isolated) wind turbine depends on

- the hub height
- the wind direction
- the wind intensity
- the intensity of unstable convection
- the accuracy of the weather forecast

Q3. The relative downstream velocity deficit of an single (isolated) turbine wake depends

- the wind direction
- the accuracy of the weather forecast
- on the turbine diameter
- on the intensity of the unstable convection
- the upstream wind velocity

Q1. Wind probability distribution function (PDF) will be accurate for an annual resource assessment

- If the wind data are averaged over 10 min
- If we have at least 1000 wind measurements per year
- If the mean wind value exceeds 10 m/s
- If the wind interval is less than 1m/s
- If we have daily averaged wind values

Q2. At a given location, the wind distribution function is generally approximated by a Weibull law:

with k and C the two parameters of the law. Find which sentence is true ?

- The probability to find a wind speed lower than C is about 60%.
- The mean velocity <V> depends only on C
- The probability to have strong winds increases when k decreases if C stays constant
- The value of the mean wind speed <V> only allows to deduce the parameters k and C.

Q3. From your knowledge on meteorology and Weibull wind probability distribution law find which of the following sentences are true

- The parameters k and C obtained from data at a given height change if we consider a different height
- the Weibull parameters k and C are dimensionless
- For a given site the parameters k and C vary with the season.
- higher C always implies higher mean wind speed
- higher k, for the same C, implies a higher variability in the wind speed

Q4. The Weibull law will fit accurately the PDF of the observed winds when

- The Cramer Von-Mises score W
_{n}^{2 }exceed 5 - The wind speed distribution have a single peak
- Two distinct physical processes (sea breeze and synoptic winds for instance) could occurs
- The Power Density method is closer to data than the graphical method
- Winds are weaker in summer than during winter

Q5. We consider two distinct sites for which the following Weibull parameters have been determined:

Site 1: k=1.5, C=4.0 m/s

Site 2: k=2.2, C=4.1 m/s

To answer the following questions take the time to do some quantitative calculations using the following curve for the gamma function.

- The knowledge of the mean wind speed allows a good estimation of the energy production potential of a site.
- The wind kinetic energy potential is significantly larger in site 2 than in site 1
- With a precision of 10 % the two sites have the same mean wind speed
- With a precision of 10% the two sites have same wind power density (proportional to the cube of the wind speed)
- The wind kinetic energy potential is significantly larger in site 1 than in site 2

Q1. In a first step you are going to study the distribution of the shear exponent \alpha*α* throughout the year.

You need the wind data that you have already downloaded in the “in-video quizz” and the short code that you have written to compute a power law fit to the wind vertical profile of 2014/08/01. Write a code (based on a loop over the 4750 wind vertical profiles of the data file Lidar_wind_vertical_profiles.txt) that compute for each vertical profile the value of \alpha*α*.

You should obtain 4750 values of \alpha*α* with a minimum value of 0.003 and a maximum value of 0.839. Plot the histogram of \alpha*α* using intervals of width 0.02 for dividing the range of alpha from 0 to 0.84 (i.e. first interval: [0-0.02] and last interval [0.82-0.84]).

What kind of distribution of \alpha*α* is revealed by the histogram?

- A unimodal distribution
- A weibull distribution
- A bimodal distribution
- A normal distribution

Q2. What is the median of the distribution of *α* ?

`Answer:`

Q3. The distribution of *α* that cover all measurements is bimodal. It should be possible to extract two distributions when restricting the data considered. Physically, we expect a variation of the shear exponent *α* between day and night. To investigate that point, extract from the set *α* the subset that corresponds to the daytime (defined between 8 a.m. and 7 p.m. included) and the subset that corresponds to the night (defined between 8 p.m. and 7 a.m. included). Afterwards, plot the histograms that represent the distributions of *α* for day and night separately.

What is the median of *α* during the daytime ?

`Answer: `

Q4. Recall that during the day, the heating of the sun tend to homogenize the boundary layer while during the night the boundary layer become stable which causes stratification. Using the two distributions of the shear exponents that you obtain for daytime and for the night and your knowledge of the boundary layer physics, select the good propositions.

- From experimental measures, the boundary layer experiences stronger shear during the daytime.
- Based on the theory, the value of
*α*should be on average higher during the night. - The statistical analysis of the distribution of
*α*during night and daytime gives contradictory result with what we expect from atmospheric boundary layer physics. - From experimental measures, the boundary layer experiences stronger shear during the night.

Q5. We want now to investigate the typical variations of *α* throughout the year. For that purpose compute the median of *α* for each month for daytime and night separately. Plot for daytime on the one hand and for night on the other hand the evolution of the median of alpha as a function of the month.

- The statistical analysis of measures suggests that the wind vertical gradient is stronger in winter.
- For both daytime and night, the median value of
*α*is higher in winter than in summer. - The range of annual variations of
*α*is larger at night than during daytime.

Q6. What is the median value of alpha in July for the daytime period ?

`Answer: `

Q1. Download the file attached below that contains wind velocities at 40m and 140m.

Plot the histogram of the wind distributions at these two different altitudes. Using the method of power density method (described in the session “Weibull Rayleigh distribution”) fit the wind distribution to a Weibull distribution. You obtain for the two altitudes of 40m and 140m the shape parameters k*k* and the scale parameters c*c*.

What is the value in m/s of the scale parameter c*c* for the altitude of 40m ?

Note : at 140 you should obtain the following parameters : k=2.49*k*=2.49, C=7.67*C*=7.67m/s.

`Answer:`

Q2. For each altitude, superpose the histogram of the wind distributions and the probability density function of the Weibull distribution fit that you have computed. Which one of the following propositions is true ?

- On average, winds are slower at the altitude of 40m than at the altitude of 140m.
- The width of the wind speed distribution is larger at the altitude of 40m.
- Comparing graphically the wind distributions and their Weibull fit, the Weibull distribution is a better model for the altitude of 140m.
*k*140*m* < k_{40m}*k*40*m*.

Q3. It is interesting to compare the Weibull fit that are obtained with different method. You can use the Maximum likelihood estimator (MLE as described in the session “Weibull Rayleigh distribution”) to compute another fit to the wind distributions. Which one of the following proposition is true ?

Note : The numerical software that you are using may have a built-in algorithm to compute a Weibull fit, see what method it uses to compute the shape and scale parameters.

- The difference of the scale parameter and shape parameter obtained by the MLE method and POD method remains less than 5%.
- In the two cases, the scale parameter estimated with the MLE method is higher than the one estimated by the POD method.

Q4. We now imagine the situation, described in the previous video, where we only have the wind speed measures at the altitude of 40m and we want to extrapolate a Weibull distribution at the altitude of 140m.

With the shape and scale parameters that you have obtained at the altitude of 40m with the power density method, use the formula of Justus and Mikhail presented in the video to extrapolate from these values the parameters k_{140m}*k*140*m* and c_{140m}*c*140*m*. (Consider *α*=0.31 which corresponds to the median value of the distribution of *α* studied in the previous quizz)

Compare this set of parameters obtained by extrapolation to those directly computed from the wind speed measures at 140m. Keep the two sets of scale and shape parameters at the altitude of 140m, as they will be useful for the next quizz.

What is the value of k_{140m}*k*140*m* obtained by extrapolation ?

`Answer: `

Q1. The goal of that question is to compute the mean power of the wind turbine if it was installed on the site of SIRTA using the full set of measures that we have at 140m.

Discretize the wind speed range 0 to 20 m/s in intervals of width 0.1 m/s and count for each interval, the number of times the wind speed at 140 m/s belongs to that interval. Denote by V_1=0.05\, {m/s}*V*1=0.05m/s, V_2=0.15\, {m/s}*V*2=0.15m/s, … , V_{200}=19.95\, {m/s}*V*200=19.95m/s the 200 mean wind speed values of the intervals. The number of times, the measured wind speed V \in (V_i-0.05,V_i+0.05)*V*∈(*V**i*−0.05,*V**i*+0.05) is denoted by N_i*N**i*.

Based on the power curve V P(V)*V*↦*P*(*V*) of the 2 MW wind turbine given in the previous video and the histogram of wind velocities at 140 m compute the mean power of the wind turbine with the formula:

P_{mean}={\sum_i N_i P(V_i)}{_i N_i} *P**m**e**a**n*=∑*i**N**i*∑*i**N**i**P*(*V**i*) .

Give P_{mean}*Pmean* in kW.

`Answer:`

Q2. Give, in percent (%), the capacity factor of that wind turbine if it was installed on the site of SIRTA.

`Answer:`

Q3. We now want to quantify the error that we would make in estimating the annual mean power of the wind turbine by using the Weibull distribution with the parameters directly computed with the power density method at 140m.

Compute the annual mean power of the wind turbine using the probability density function of the Weibull distribution f(V)*f*(*V*) with the parameters k_{140m}=2.49*k*140*m*=2.49 and c_{140m}=7.67 \, {m/s}*c*140*m*=7.67m/s:

P_{mean}=_{V=0}^{\infty} P(V) f(V){d}V*P**m**e**a**n*=∫*V*=0∞*P*(*V*)*f*(*V*)d*V*

Give P_{mean}*Pmean* in kW.

`Answer: `

Q4. By computing the annual mean power of the wind turbine using the Weibull distribution obtained by extrapolation from the measures at the altitude of 40 m (i.e., k_{140m}^{extra}=2.62*k*140*mextra*=2.62 and c_{140m}^{extra}=7.63 \, \mathrm{m/s}*c*140*mextra*=7.63m/s) we obtain: P_{mean}=646 \mathrm{kW}*Pmean*=646kW.

In that context, it means that an industrial that would install a mast at the height of 40 m and extrapolate the Weibull distribution at 140 m in order to asses the annual power production of a wind turbine at this location will:

- Underestimate the production by 1%.
- Underestimate the production by 10%.
- Overestimate the production by 10%.
- Overestimate the production by 1%.

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