ps18 l - university of chicago
TRANSCRIPT
GEOS 24705 / ENST 24705 / ENSC 21100 Problem set #18 Due: Tues. May 30th Problem 1: Can solar power the world?
Background We determined at the very beginning of class that the ultimate source of energy for a renewable world had to be the sun, and we computed a requirement of a 5% conversion efficiency of that sunlight (10 W/m2) for any technology to power the world. The conversion efficiency of plants, wind, and hydro were all too low. (Hydro and wind both capture kinetic energy produced by the atmosphere as a heat engine, and you know that heat engines are not efficient unless they involve large temperature gradients.) Natural conversions of solar energy are not good enough. We can however capture enough sunlight with human-‐made technology.
Two potential technologies for conversion of solar radiation energy that could power the world are solar photovoltaics and solar thermal (also called “concentrating solar power” or CSP).
Solar photovoltaic panels are semiconductors that use the photoelectric effect to convert solar radiation directly to electricity. The average efficiency of conversion for commercial panels sold today is 12-‐18%. Furthermore, you can put them places where the solar radiation is higher than the world or U.S. average. The panels have typically had a nonlinear response to being partially shaded – shade from a passing cloud on even a small part of the panel can cut power dramatically – but if you site them in the desert that is not a problem. (And recent changes in how they are wired reduces the partial-‐shading problem). You can boost your capacity factor by an extra 20% or so (multiply it by 1.2) if you install more expensive tracking solar panels that rotate with the sun to maximize the amount of solar energy they catch. These facilities take up a bit more land – the panels have to be spaced a bit further -‐ but are generally considered the most cost-‐effective for utility-‐scale solar.
Solar thermal plants are less high-‐tech and consist of no elements that would be unfamiliar to a 19th century engineer. They involve concentrating sunlight to power a heat engine. Mirrors capture incoming solar radiation (very efficiently, perhaps 90%) and direct it onto a tube of some substance (usually oil, sometimes molten salt) to heat it to very high temperatures (usually ~ 600 C). The hot fluid is circulated to make steam and run a perfectly ordinary steam turbine (or in some cases to power a Stirling cycle engine); the turbine or engine then spins a generator just as in any other fossil-‐fuel-‐powered power plant. The high temperature means that you get the normal efficiency of industrial heat engines (~30%), rather than the low efficiency of the atmospheric heat engine: concentrating mirrors overcome the low areal energy density of renewables. The chief advantage of CSP that it is easy to store heat, unlike electricity, so a CSP plant can generate power at all times, whether or not the sun is shining.
18 MW tracking solar PV installation at Five Points, CA. Image: Blue Oak Energy
Most solar thermal plants no longer use the trough system (above) Instead, they use a “power tower” system where multiple mirrors focus solar radiation on the top of a tower and heat a boiler there. By concentrating more solar power on one small location they obtain higher temperatures and so higher efficiencies. The towers and tracking systems are obviously more expensive than troughs, so there’s a financial tradeoff, but the extra power generated seems to be worth it.
Problems
A. Rescale the solar PV efficiency to account for the fact that panels have to have some physical spacing – you need extra land from which you can’t generate power. What fraction of the sunlight falling on the whole PV facility is turned to electricity? You can look at the image for a PV facility here in the “Background” section. Try to visualize the panels rotated flat to the ground, and then estimate how much extra space the facility needs beyond the areas of the panels.
B. Now, decide you are the Energy Czar of the U.S. with unlimited powers of eminent domain to seize land for energy production. Pick a general region for installing large-‐scale solar plants based on the NREL map of annual mean solar insolation. The NREL site is buggy but there’s an older map here: http://www.c2es.org/docUploads/Solar-‐1.png. You want to pick locations that are sunny, cloud-‐free, flat, and cheap. The map uses irritating areal energy density units of kWh/m2/day rather than W/m2 so you have to convert units. State the (estimated) average W/m2 of solar insolation in the region you are considering appropriating.
Parabolic trench collectors, solar thermal installation near Barstow, CA, built in 1984. Operated by NextEra Energy. Image copyright unknown.
PS20 power tower, Seveille, Spain, 1255 mirrors, power up to 20 MW. Operational 2013. Image: SWNS.com.
C. Given the effective efficiency of your PV facility, what is the Welect/m2 you can generate from your choice of solar technology in that location?
D. (Optional) Repeat C for solar thermal. Estimate the effective efficiency of solar thermal at producing electricity from the information above and from the photographs shown, and then compute the resulting areal energy density (i.e. Welect/m2),How does it compare to solar PV?
E. Compare your answers in C (and D if you did it) to the target areal energy density for powering a future world that we’ve been using in class. Note: remember that the number you computed above is for generating electricity, so is not truly comparable to the metric we wrote down in class. For solar PV there is no waste heat as you’d have in a heat engine; for solar thermal the number you have in D is “downstream” of that energy loss. So a world powered by solar PV or thermal would really need to meet only about ½ your target number.
F. Appropriate as much land as you need to set up an energy system capable of meeting current U.S. electricity needs. Print out the NREL map and block out on it the land you would have to seize to fill all current U.S. electricity needs with your favorite solar technology.
G. Repeat the above but now assume that you will get rid of fossil fuels entirely – you’ll electrify cars etc. Appropriate enough land to meet all of US current primary energy needs, and mark your appropriation on a new printout of the map. Again, remember that if you move to solar power you can reduce the primary energy usage you need because in the case of PV, you avoid heat losses, and for solar thermal you already accounted for them. Compare to your wind maps and discuss.
To actually implement a transition to solar, you’d have to also build infrastructure to move electricity from where it is generated to where it is needed. There is limited transmission into the sunny desert areas you likely appropriated, so you’ll basically need to build from scratch. The slides handed out for the electricity markets discussion list a rule of thumb for transmission line cost of $500-‐700/(MW km), though other estimates are higher (https://iea-‐etsap.org/E-‐TechDS/PDF/E12_el-‐t&d_KV_Apr2014_GSOK.pdf) Sub
H. (Optional) For your replace-‐electricity generation solution of F, draw on your map the primary transmission lines you’d need to include and estimate their total cost. Since you’re not proposing increasing total electricity usage, you can assume that the existing grid is more or less fine; you just need to build high-‐voltage transmission lines to get your electricity closer to the demand locations.
I. (Optional). Repeat the above for the all-‐primary-‐energy solution of G. Now you’re going to need to expand the entire electrical grid since you’re proposing a radical expansion of electricity use – that is, you’re going to also have to build new low-‐voltage transmission and distributions lines.
Problem 2: Fossil fuel extraction & consequences
The modern “pumpjack” looks somewhat like an early Newcomen or Watt engine:
L. Pumpjack, R. Watt’s “Old Bess” engine, both used for pumping liquids
Both actually use similar pumps. But, in the steam engines, the cycle time of the engine was the same as that of the pump – the piston connected directly to the rocker beam. The pumpjack is driven by a high-‐speed diesel engine, so it needs complicated gearing to allow the beam to rock much more slowly than the engine pistons are oscillating.
A. Watch the video here on the “sucker rod pump” used in oil wells. What reminds you of the older lift pump?
http://www.youtube.com/watch?v=SFJFiyXTOa0
B. Watch the video here on the pumpjack’s gearing system and explain how the 1200 rpm rotation of the engine is downconverted to the speed of the pump. (What is that speed?)
http://www.youtube.com/watch?v=FU0dYV3LvAk&feature=related
C. Watch these fracking videos, first on drilling the wells…
http://www.youtube.com/watch?v=fBQCQ6HL2Yw
D. ..and then on the fracking process itself, and comment on something interesting. Both videos are by a drilling company so obviously are positive on safety issues.
http://www.youtube.com/watch?v=7ned5L04o8w
Problem 3: Energy losses in automobiles / fuel economy
Background: Where does the chemical energy of the fuel you burn in your automobile go? When you think about it, it’s not immediately obvious that you’d need to keep the engine on once you get a car up to speed. Of course you need to apply power to the car to accelerate it up to speed -‐-‐ you’re adding kinetic energy. But once the car is at cruising speed, its kinetic energy is constant, and there’s no change in energy/time. Why do you then need to keep burning fuel to apply mechanical work to the engine? Wouldn’t that force your car to continue accelerating?
The answer is, your car is constantly experiencing losses of energy. You could use your engine only on accelerating if your car had no internal friction, slipped frictionlessly through the atmosphere, and had perfectly round, perfectly stiff wheels that made perfect contact with the ground at a single point. In real life though, the car loses energy in many ways. As the car moves, it must push air ahead of it, giving some of its kinetic energy to the air (“air resistance” or “aerodynamic drag”). The wheels deform on contact with the ground and so scrape along it and lose energy as they roll. (We call this “rolling resistance”. You have experienced it personally if you try switching between a road bike with skinny tires and then a mountain bike with fat knobby tires and observe how much slower you are on the fat-‐tired bike.) And no matter how well-‐oiled and well-‐greased your car is, the parts scrape against each other and lose energy to friction. You need to continue drawing power from your engine to compensate for all these ongoing losses. And, your actual power draw is in fact significantly more than the losses of kinetic energy, because the thermodynamic efficiency of a car engine is typically only around 25%. Since you throw ¾ of the chemical energy in fuel away as waste heat, you need to burn fuel equivalent to 4 times as much power as the car is losing to friction.
Side note: we use the terms ‘fuel economy’ or ‘fuel efficiency’ to describe the amount of energy a car uses in driving, but the term ‘efficiency’ is deceptive. Fuel usage differences aren’t mostly due to differences in the thermodynamic efficiency of engines. It’s just that heavier, blockier, and bigger cars necessarily require more work to keep in motion. Similarly, driving any car uses more energy than a motorcycle. Calling this ‘waste’ is a statement of values as much as physics.
In typical average driving for a small car, the energy losses in driving are split fairly equally between braking, rolling resistance, air resistance, and frictional losses. The slides posted online help walk you through the formulas for air resistance and rolling resistance. You may want to read also the posted document “Air_and_rolling_resistance.pdf” for the physics.
Note: Despite improved engineering of automobiles, fuel economy – the amount of energy needed to drive some distance – remained almost flat for decades. You can see this here: http://www.rita.dot.gov/bts/sites/rita.dot.gov.bts/files/publications/national_transportation_statistics/html/table_04_23.html
from the Dept. of Transportation. Over time, cars became more streamlined in profile, reducing their air resistance (see figure in the handout pdf), and engine technology improved, reducing engine losses. But cars also became heavier (note that both rolling resistance and the power needed to accelerate scale with the car’s mass), and we also demand that cars deliver higher performance, so their engines are no longer optimized for the kind of driving they actually do. Most technological improvements since the 1970s were thus counteracted by driving heavier and more powerful vehicles. Only in recent years, since new fuel economy standards were imposed, have fuel economy numbers begun to creep up.
Problems: Assume that you have a middle-‐of-‐the-‐road new car for some reasonable year (state it). You can use the DOT table if you want to estimate fuel economy for highway driving. Pick values for drag and rolling resistance coefficients (Cd Crr) from the figures in the slides. Make reasonable estimates about car weight and size. A typical passenger car will have mass ~ 1200 kg and cross sectional area ~ 3 m2. 60 mph (100 km/hour) is a reasonable speed for highway driving.
For optional problems – a passenger train weighs ~ 500 tons, a freight train ~ 3000 tons (with super-‐freights up to 5 x as much). Freight trains travel 50 mph max (80 km/hr); bullet trains travel 200 mph (320 km/hr). The mass of a bicycle + rider is dominated by that of the rider (the bike contributes ~ 10 kg), a reasonable racing speed is 25 mph ~ 40 km/hr
A. State your assumptions, including a bit higher fuel economy for highway driving. Give speeds in m/s.
B. Compute your gasoline consumption in gallons per hour for highway driving.
C. Now use your knowledge of the energy density of gasoline and convert units to state the car’s total power consumption in kW, during highway driving.
D. Divide the power in problem C by ~4 to get the mechanical work that the engine puts out, in kW. This term is often called the power “at the wheels”. Convert to horsepower for additional intuition.
E. Even my tiny little Honda Fit has a stated horsepower (the maximum power that the engine can provide to the wheels) of 130 hp. Why is this different from the answer in D?
F. Estimate your power loss to aerodynamic drag during highway driving, in kW.
G. Estimate your power loss to rolling resistance during highway driving, in kW.
H. Which of these is the more important factor to your fuel efficiency in highway driving?
I. (Optional) For your passenger car of the previous problems, assume you drop your speed by 1/3 to that of city driving. Which now matters more, aerodynamic drag or rolling resistance? Hint: do this simply, just use scale factors. The ratio of power loss to aerodynamic drag vs that to rolling resistance is Ratio = (Cd * ½ ρ A v3) / (Crr m g v) ….which scales as v3/v or v2. So just take the ratio of F to G (drag loss over rolling resistance loss), and then scale it by the appropriate factor.
J. For vehicles that drive slowly and do a lot of stopping and starting (postal trucks, golf carts), energy use is dominated by braking and rolling resistance, and aerodynamic drag is unimportant. They therefore do not look very streamlined in profile. Freight trains both drive fast and do not do a lot of braking, and yet they are also designed with blocky profiles. Explain -‐ why is aerodynamic drag not important for a freight train?
Optional problems for #2
K. (Optional) A bullet train, on the other hand, is streamlined in profile, suggesting that air resistance matters. Plug in numbers to demonstrate that this is the case.
L. (Optional) By how much is your gasoline consumption increased if you speed on the highway, driving 75 mph instead of 60 mph? (This difference is why you’re advised to drive as slowly as possible if you are low on gasoline.)
M. (Optional) The U.S. 55 mph speed limit was set in 1974 by Congress as part of the Emergency Highway Energy Conservation Act, intended to reduce gasoline use after the OPEC oil embargo. Was this a plausible way to reduce our dependence on foreign oil? (See handout to look at drag coefficients in 1974 – cars were more boxy then.)
N. (Optional) Formula One racing cars are intended to go very fast, in races where fuel consumption matters, but they aren’t designed like Porsche Boxsters – instead, they have high drag coefficients between 0.7 and 1. Aerodynamically, they are more “blocky” even than Hummers. This is a deliberate design choice (and the engineers are good). Why was it chosen?
O. (Optional) Formula One cars not only have a high Cd but also have high rolling resistance (see figures below). Tires for racing bikes, on the other hand, are designed for low rolling resistance. These must be rational choices, but why? Discuss the factors that led to these opposite choices for bike racing vs. auto racing.
Michelin Pro3 Race tire, 116 psi Pirelli PZero Formula 1 race tire: ~15 psi
P. (Optional) Go to a Metra station and observe the sanding that provides enough friction to let the train start. Photograph the sanding system in action if you can.