On The Definition of Sustainable Energy

Let’s finally cover what I meant by truly sustainable energy production.

Listing electrical energy generation methods portrayed as sustainable, many would consider solutions such as solar power, wind turbines, hydroelectric dams, geothermal plants, sun farms, wave generators, nuclear power and nuclear fusion to fit the image of futuristic green power. But while all of the above may be green, not all of them are sustainable. To me at least, it seems that green energy is measured from the present as a bench point: anything that produces less greenhouse emissions or causes less ecological destruction than methods currently employed is considered "green." However, sustainable energy production is a more definite construct that can be described through thermodynamics.

From thermodynamics we know that every work-energy conversion in our universe is irreversible, meaning that energy is lost as heat every time something is done. Taking the planet earth as a giant engine, which is accurate in the sense that most life processes convert available energy into work and expend heat, we can conclude that the earth needs constant energy input in order to continue running. Aside from ambient space radiation produced by other sources, the sun is the only source of energy earth knows consistently from day to day. This is something many of us should experientially understand, contrasting the death of winter when days are short to the blossoming of spring when long sunlight returns.

Fig. 1: Boston sun angle during the seasons, affecting solar energy influx (Science Blogs)

Therefore, in order to be sustainable within the lifetime of our sun, an extremely high life expectancy for planet earth, our energy generation tactics should be geared towards the sun’s life-giving energy. Does this mean that solar is the only way to go? No, absolutely not. The sun’s energy hitting the earth enacts a grand cascade of events that grow wind, help generate tides and keep the water cycle flowing. This extends the definition of sustainable energy generation to include wind power, wave power, and hydroelectric generation.

Another side category of what I would consider sustainable energy sources includes those whose failure would correspond with the end of life on earth. One example is geothermal energy. Sure, eventually the earth’s core may cool down, but there would be bigger problems associated with this scenario than humans running out of electricity, such as the dissipation of the earth’s magnetic field letting a bombardment of solar radiation char the planet. Wind power may also be considered part of this category, as wind is in part generated by the Coriolis Effect that depends on the earth continuing to spin. If the earth stopped spinning, bad things would ensue.

So what is not covered under the definition of sustainable electricity generation that is commonly perceived as such? The top two that really get to me are nuclear energy and fusion energy. The selling point of nuclear energy is that it is clean in terms of greenhouse gas generation. However, clean does not directly translate into sustainable. The input into nuclear reactors is uranium, and lots of it. Uranium is not an uncommon element on earth, but a large 1000MWe nuclear power plant requires about 200 tons of refined uranium to operate for one year [1]. Refining uranium is itself a wasteful process, and mining large amounts of ore has an ecological impact as well. Nuclear fusion is often sold as the sustainable energy solution of the future, but in actuality nuclear fusion requires the same input as the sun does: hydrogen. With so much hydrogen on earth, why would this be a problem? Call me paranoid, but any time someone wants to convert something necessary for life into a luxury item (yes, electricity is a luxury item) and some useless helium, then I begin to worry. And relative to the sun’s supply of hydrogen, earth’s hydrogen bank is chump change. The hydrogen fusion reaction that generates the most energy is that of deuterium (₁H²) with tritium (₁H³) as described by this equation [2]:

                                                        1.  ₁H² + ₁H³ = ₂He⁴ + neutron

If we calculate the mass difference

       2.  m(₂He⁴) + m(neutron) – m(₁H²) – m(₁H³) = 4.002602 + 1.008665 – 2.01412 – 3.016050  =                   5.011267 – 5.030152= -0.018885 amu

then insert this mass difference into Einstein’s mass-energy equation

                  3.  ΔE=Δmc²

                  4.  Δm=0.018885 amu(1.66053892 x 10^-27 kg/amu)

                  5.  E=(3.135928 x 10^-29 kg)(3 x 10⁸ m/s)² = 2.82 x 10^-12 J/He nucleus formed

and convert the result into a meaningful number

       6.  2.82 x 10^-12 J/He(2.778 x 10^-7 kWh/J)(1 He nucleus formed/2H nuclei consumed)=
            3.92 x 10^-19 kWh/H nucleus consumed(6.022 x 10²³ H nuclei/mol H)(1 mol H/((2.014102 +                   3.0160492)/2) g) = 9.39 x 10⁴ kWh/g H

we get that 9.39 x 10⁴ kWh of power can be generated from one gram of hydrogen gas (assumption made that hydrogen composition is half deuterium, half tritium). Considering that the average American consumes 10,908 kWh annually [3] and that the population of America is about 321 million [4], approximately 37,300 kg of hydrogen gas would be consumed annually to support America alone by fusion energy. This is not zero input.

All in all, what I am trying to say is that science has laid out a very specific definition of what sustainable energy truly is, so it is the job of every person to evaluate whether what is being portrayed as sustainable energy generation is truly backed by the facts.

Does your definition of sustainable energy sourcing differ? Let me know your thoughts in the comments below!

Wind Turbine Efficiency, Part 3: The 40.3% That Got Away

You made it to Part 3! If you haven’t already please go back to read Part 1 and Part 2.

Let's continue. From calculus, we know that to find a maximum of a given single-variable function, we take the derivative, or the slope of the function, with respect to the variable and find where the derivative equals to zero. This is because the slope of a function is zero where the function changes from increasing to decreasing or decreasing to increasing output values (aka peaks and valleys). Equation 24 from Part 2 is rather ugly, however, so let’s set the derivative with respect to k=v2/v1 (the ratio of final and initial velocity) as our plan of attack (reminder: ve, v1 and v2 are all known constants in this expression). Let’s do this now.

            25.     P=1/4rAv1³(1-k²+k-k³) (substitution of k=v2/v1 into equation 24)

            26.     dP/dk=1/4rAv1³(0-2k+1-3k²)=1/4rAv1³(-2k+1-3k²)=0

            27.     -2k+1-3k²=0 ((1/4rAv1³) is a giant multiplicative non-zero constant term, so it can be
             dumped into the zero never to emerge again (except by integration))

            28.     k=1/3, -1 (solutions to equation 27, obtained by factoring the second-degree polynomial and
             letting each set of terms equal to 0 since if ab=0, either a=0 or b=0 makes the statement true)

Obviously, a negative value of k=v2/v1 does not make sense since this requires that the wind hit the turbine and flow backwards, so we will only keep the k=1/3 value. Plugging the value for k into equation 25, we get:

                                           29.     Pmax=1/4rAv1³(1-1/9+1/3-1/27)=16/27(1/2rAv1³)

To interpret this result, all that must be remembered is that the total power in the original cylinder of wind was found to be Pwind=1/2rAv1³ in equation 7. Substituting in this expression reveals the following relationship:

                                                                      30.     Pmax=16/27Pwind

In other words, only 16/27, or 59.3%, of the total power in a column of wind can ever be extracted by wind turbines (this is assuming 100% turbine internal efficiency relative to the 59.3% limitation). This means that there is an inherent limitation in the production efficiency of wind power relative to available energy to be harvested.

Fig. 4: Turbine efficiency cartoon (Wikimedia)

And there you have it, Betz's Law in only 30 steps.

So, knowing this, should we abandon wind power? The answer is a definite no. Wind turbines have an installation cost in both money and materials and also severe limitations as to practical operation, such as a decreasing efficiency in high wind velocities to prevent structural damage, but they are an important member of a small set of what I would consider to be true renewable energy strategies. I won’t talk about this now since I feel guilty about the density of the topic I just covered, but I will make this the topic of my next post.

If you have any questions, leave them in the comments section below and I’ll try to answer them as best I can.

Wind Turbine Efficiency, Part 2: Too Many V Terms

Welcome back! If you haven’t already, please read Wind Turbine Efficiency, Part 1: A Windy Cylinder of Power.

All calculations thus far have dealt with air in front of the wind turbine. We should take a look at what happens behind it as well. From fluid dynamics we know the following equations to be true through logical progression:

                               8.     V1=V2 (V1 is air volume before turbine, V2 is air volume after)

                               9.     dV1/dt=dV2/dt

                               10.     dV/dt=Av (equation 5)

                               11.     A1v1=A2v2

Equations 9-11 simply state that the volume of wind entering our cylinder from Part 1 must change at the same rate as the volume of air exiting the turbine from behind. This is true if we assume no changes in temperature or atmospheric pressure before to after the turbine (beyond those induced by the slowing wind, which we are calculating for now). From energy conservation, we know that if work is performed on the wind turbine, then the kinetic energy of the wind after the turbine, and accordingly the wind velocity, must be less than that before. In other words, v1>v2. The result is that A2>A1, so the wind after the turbine must effect a cone of area expanding away form the turbine blades.

Fig. 3: Air flow through a wind turbine (Danish Wind Industry Association)

Let’s try to piece apart the maximum power obtainable from the wind cylinder calculated above. Wind force is defined as follows:

                                     12.     F=ma=d(mv)/dt=(dm/dt)v+m(dv/dt) (force equation)

For simplicity’s sake, we are assuming that the wind before the turbine has velocity v1 and slows to velocity v2 after performing work on the turbine, both of which are know constant values. When wind blows through our wind cylinder, the mass of air, and as a result the effective length of the cylinder since base A is constant, changes with time. Therefore, we are not concerned with how force changes with velocity, so the m(dv/dt) term can go to zero and we can evaluate the change in wind force over the turbine as follows:

                                  13.     ΔF=(dm/dt)Δv=(dm/dt)(v1-v2)

                                  14.     ΔF=r(dV/dt)(v1-v2)=rAve(v1-v2) (from equations 4 and 5)

A new variable has been introduced in equation 14, namely the effective velocity ve, which corresponds to the wind velocity that the turbine experiences as part of work production (in other words, if the wind were to lose all kinetic energy after contacting the turbine, a wind velocity of ve would generate a quantity of power the same as from wind with velocity v1 slowing to v2). Concerning the power obtained from the change in force, we know that since the change in wind force is equal to the force exerted on the turbine, power can be calculated as follows concerning the wind contacting the turbine:

                         15.     W=FL (instantaneous work equation, L is analogous displacement here)

                         16.     dW/dL=F, therefore dW=FdL

                         17.     P=dW/dt=d/dt(FdL)=F(dL/dt)=Fve (power equation)

                         18.     P=rAve²(v1-v2) (from equation 14)

From expanding equation 4 to accept that the wind velocity after the turbine is not zero, we obtain the equation:

               19.    P=1/2(dm/dt)(v1²-v2²) (this is just the difference in kinetic energy with changing mass)

               20.    P=1/2rAve(v1²-v2²) (from equations 5 and 6, parallel to equation 7)

Now we have two different equations for power, one from the work equation and the other from the kinetic energy equation! Setting these equal to each other, we obtain the expression:

                   21.     P=1/2rAve(v1²-v2²)=rAve²(v1-v2)

                   22.   v=1/2(v1+v2) (simplification of equation 21 using (v1²-v2²)=(v1+v2)(v1-v2) identity)

This equation tells us what the relationship between the effective velocity ve and the initial and final wind velocities v1 and v2 are. Specifically, the relationship is that the effective velocity is the average of the initial and final wind velocities. Using this relationship, let’s redefine the kinetic energy-derived power equation in terms of v1 and v2 in preparation for calculating maximum power.

                            23.    P=1/2rAv(v1²-v2²)=1/2rA(1/2(v1+v2))(v1²-v2²) (from equation 20)

                            24.    P=1/4rAv1³(1-(v2/v1)²+(v2/v1)-(v2/v1)³) (simplification of equation 23)

Things are starting to get exciting! Read on to Part 3 to finally start calculating the external efficiency limitation of wind turbines, and let me know any questions in the comments below.

Wind Turbine Efficiency, Part 1: A Windy Cylinder of Power

When people think of renewable energy, probably what comes to mind is a mixture of solar panels, wave motion generators, geothermal plants, hydroelectric dams and wind turbines. Of these, the two most associated with the image of a sustainable society are probably solar panels and wind turbines. That’s why when I caught a glimpse of this wind farm while at the Jersey shore, I felt like I had been transported into some sort of futuristic society apart from our own. 

Fig. 1: Wind farm near Atlantic City (personal photo, hence the quality)

Assumedly, we all know that every energy-harvesting method has efficiency limitations, even though some people like to pretend they don’t. With solar panels, calculating efficiency can be complicated as it comes down to a matter of quantum efficiency, but the efficiency limitations of wind turbines, also called Betz’s Law, can be readily calculated with some physics. Let’s explore this.

(Note: though not particularly difficult conceptually, this derivation is math-heavy. I will do my best to explain the underlying logic as we go along. Also, this explanation will be separated into three parts so that your eyes don’t bleed from the endless stretch of math. Backbone calculations for Betz’s Law come from the Wikipedia article and this MIT presentation, but I hope you will find that my explanation, though lengthy, is more intuitive than these sources. Enjoy!)

Let’s start with calculating how much power there exists in a cylinder of wind with base area A corresponding to the arm span of the wind turbine.

Fig. 2: Cylinder with base A and wind influx at velocity v (MIT)

                          1.  KE=1/2 mv²  (kinetic energy equation)
                   2.   KE=W (work-energy equation)
                   3.   P=d/dt(W)=1/2(dm/dt)v²+mv(dv/dt)  (power equation, product rule)

For equation two, let’s assume all work is converted to electrical energy, not friction or blade deformation or other stuff. Now in equation three, what normally happens is that the dm/dt term, or the change in mass over time, is assumed to be zero, producing the familiar mv(dv/dt) or (d(mv)/dt)v or Fv term associated with power. Instead, let’s assume that the air cylinder has a constant known speed v1. Therefore, the term mv(dv/dt) goes to zero, and we are left with equation 4:

                                                                   4.  P=1/2(dm/dt)v1²

From fluid dynamics, we know that

                                                         5.  dm/dt=(m/V)(dV/dt)=r(dV/dt)

or that the mass per volume, density, multiplied by the change in volume over time is equal to the change in mass over time. Makes sense, right? This is essentially just dimensional analysis. With constant wind cylinder base A and changing length L, change in volume over time is defined as

                                                         6.  dV/dt=d/dt(AL)=A(dL/dt)=Av1

Combining equations 4, 5 and 6, we get the general power equation for a cylinder of wind to be

                                                                       7.  P=1/2rAv1³

where r is density, A is cylinder base area (also area of turbine span) and v is a constant wind velocity. 

I’ll stop the first post here since it’s a good spot to take a break. Read on to the next post to learn how to calculate the effective wind speed acting on a wind turbine, and let me know in the comments below if you have any questions.