Is this "efficiency" of a petroleum refinery meaningful?
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This article ( http://www.transportation.anl.gov/modeling_simulation/GREET/pdfs/energy_eff_petroleum_refineries-03-08.pdf ) concludes that a petroleum refinery is about 90% efficient, but uses the definition: Petroleum Refinery Energy Efficiency = energy in all petroleum products/(energy in crude input, other feedstock inputs, and process fuels) Efficiency is often a rough gauge of how much energy is wasted. In discussing the potential "efficiency" of a thermal power plant the amount of wasted heat is discussed and the "thermodynamic efficiency" is considered to be up to 60%: http://en.wikipedia.org/wiki/Thermal_power_station What is the "thermodynamic efficiency" of a petroleum refinery? Is this the proper measure for determining the well to wheel efficiency of an internal combustion engine vehicle?
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Answer:
We could consider the thermodynamic efficiency of components of the refinery, such as pumps.This would be defined as work done by component divided by energy input to that component. I don't think that is a very useful number. What you want for a well to wheels comparison is something that I would call the net refinery energy efficiency, defined as the net energy content of everything useful that comes out divided by the energy content of everything that goes in. The net energy content would be the total energy content of everything useful that comes out, minus the amount of energy brought in from external sources, including (as you suggest) any energy cost that arises from disposal of waste. The article you refer to seems to give a rather more restricted definition with some strange conventions, but I do not really know enough to comment further Either way, the reported efficiency will depend on local refinery practice. For example, gases that are flared do not count towards the output, but gases that are used presumably would (but see the detailed comments at the end of the article). On my definition, heat that is used for raising domestic hot water would be considered a useful outputs, but I do not know if it is included in the figures quoted in the article. What you really need to do is, with the help of an expert, a proper examination of the GREET model referred to in the article. But I suspect that give or take a few percent, the reported "refinery efficiencies" are good enough for your purpose. In a well to wheels comparison, you would need to include as factors the energy cost of bringing crude oil to the refinery, some realistic measure of refinery efficiency (I have done my best to suggest one) and distribution energy efficiency (charging as appropriate for the cost of getting the fuel from the refinery to the pump) for gasoline or diesel, just as you would need to include, for (say) an electric vehicle recharge from coal-burning power plant, an allowance for the energy charge of bringing coal to the power station, the thermodynamic efficiency (and here I really do mean thermodynamic efficiency) of the power plant (plus a little extra if the "waste" heat is put to some good use), and the efficiencies of electrical distribution and of re-charging the vehicle. All this before considering the efficiency with which the vehicle itself uses the supplied energy, which is of course far greater for an electrical vehicle than for one using internal combustion. "Every question in science should be made as simple as possible, and no simpler" (Einstein, attributed).
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Other answers
There is a big difference between thermodynamic efficiency and "refinery efficiency". Thermodynamic efficiency refers to the efficiency of an engine. Power plants are basically electricity generating engines, i.e., diesel, steam and gas turbines. A 60% thermodynamic conversion is VERY high. A regular car engine has a thermodynamic efficiency of less than 30%. The efficiency of a refinery, is how much crude oil is wasted in order to obtain distilled products. There are no engines converting fuel into mechanical power. The waste comes from evaporation, flaring, pumping and heating. ------------------ Ok, let me see if I got it now. You want to compare a refinery supplying liquid fuel to generate mechanical power, just like a power plant supplies electrical "fuel" for the same purpose? The thing is that most power plants do not use liquid fuels but - rather - coal, gas, nuclear, hydro and so. In addition, thermodynamically are much more efficient than car engines. When power plants use liquid fuels, it is in the form of very heavy No.6 oil. Electrical vehicle conversion losses are relatively low. Most losses come from pneumatic friction. IOW, it is a very tough comparison, given the many different variables. In the end, it is always spent in the form of heat. Sorry if this is not what you are looking for. --------------------------------------… OVERALL efficiency of ANY energy source is ZERO. Energy always ends up warming up the environment, no matter how efficient or commendable our purpose is. There is no point in discussing thermal efficiency of nuclear, solar, wind or hydro power plants, because fuel is inexpensive. Most of the cost comes from construction, operation and maintenance, which are much higher than for any other power plant. So, thermodynamic efficiency is a non-issue. Therefore, the concept of "efficiency" is applicable only to the RATE energy can be spent. .
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Interesting question. So, what you're wondering is how much energy is lost in the refining process in relation to the amount of energy used? Well, the refinery industry consumes quite a bit of energy; here's a profile of the refining industry: http://www1.eere.energy.gov/industry/petroleum_refining/pdfs/profile.pdf A table I think is most pertinent here is 1-11, "Petroleum Refining (NAICS 324110) Energy Consumeda—2005 (Trillion Btu)." It gives all of the energy sources the industry uses, from coal to purchased electricity, etc. Crude Oil ----- 0.0 Liquified Petroleum Gases ----- 16.0 Distillate Fuel Oil ----- 4.4 Residual Fuel Oil ----- 13.9 Refinery Gas ----- 1435.5 Marketable Petroleum Coke ----- 13.5 Catalyst Petroleum Coke ----- 526.5 Natural Gas ----- 701.4 Coal ----- 0.9 Purchased Electricity (including losses) ----- 384.2 Purchased Steam ----- 70.0 Other Productsb ----- 20.3 TOTAL ----- 3186.5 (Actually, there's a similar table in the link you gave). So, if we know the efficiencies of each, using the percentage values of each we can come up with a total thermodynamic efficiency. I'll post this now and see if I can do some more research.... Edit: Suddenly occurred to me that searching for efficiency values is pointless. The process itself heavily influences efficiency of the fuel used. Friction and insulation will make the efficiency smaller or greater, respectively; I don't know where to find (or how) the thermal efficiency of a refinery. I guess all I've given you is a total energy consumption, which was available in the article you saw anyways. I would guess the thermal efficiency would be low, esp. lower than 90%.
A Modest Proposal
The entire oil and gas industry's concept of efficiency fails to consider almost all of the true cost involved and the coal industry is not much better. Not only do they not take into consideration the geopolitical costs of securing the supply and the costs of addressing the environmental issues brought by fossil fuel use but if you consider the ultimate source of the energy is solar via photosynthesis, the plants that originally stored the energy as hydrocarbons had efficiencies well less than 1% (the theoretical maximum efficiency of photosynthesis is only 6.6% with most plants averaging less than 1% efficient), then only a small fraction of the ancient biomass got trapped by geologic forces which in themselves is a massive expenditure of energy to transform the biomass to the fossil reserves that we tap today. We can talk about the efficiency of a refinery all we want but from a global perspective (a sun to wheel perspective), it's still a fraction of a fraction of a fraction of 1% of energy from the sun collected over millions of years. If we looked at it from this global perspective then the cost of photoelectrics are not so prohibitive.
John W
I'm glad you clarified your target as well-to-wheel efficiency. You don't need to look at thermodynamic efficiency of the plant directly. For your purposes, the Petroleum Refinery Energy Efficiency is a good measure for this stage of the process. If you want it to be a tiny bit more accurate, you'd need to include other energy costs: any fuel the refinery uses to operate (electricity, natural gas), the energy cost of any additives, and if you want to get really extreme, the energy cost of the employees and the amortized energy cost of building the plant itself. But if you do all that work, you'll probably discover the difference to be within the error bounds of the Petroleum Refinery Energy Efficiency calculation. See the reference for the Tesla Motors calculated values. It shows their electric roadster far surpasses anything else. Try contacting [email protected] and asking for a paper publishing their well-to-wheel calculations. If you get that (or find it by web search), please pass it on to me. Thanks. Update: The well-to-wheel metric is to facilitate comparison of alternate ways to use crude oil to get cars to move. If you think the refinery efficiency is too optimistic, find a better measure from a credible source. Part of the benefit of electric cars is that you can power them from sources other than crude oil. They're ideal for variable sources like wind and solar. With a smart grid, it's easy to transfer surplus peak power availability into car batteries, or to cut back during peak demand. It's not clear exactly which methodology bothers you, or what alternative you're proposing. When Tesla cites electricity from natural gas, no refinery is required.
Frank N
you have all good answers. more technical than mine, to be sure. however, "well to wheel efficiency" is probably a measure of where our technological society is, than a measure of any specific thing. You might consider it something like "food efficiency = cost to deliver a calorie of food to your plate, starting with seed." clearly refinery efficiency is important, because it to some degree determines the price of fuel. I haven't checked, but I suspect that Valero is a bit less efficient because they specialize in "dirty" oil. They pay less for the oil they get, spend a bit more refining it, and probably get a bit less gasoline out of each barrel. Obviously their "refinery efficiency" would be lower, but their balance sheet would be equivalent. When you mix that in with whether the output is burned in cars, or trucks, or trains, or airplanes, you don't get something that's particularly useful. Over a period of decades, it should increase, but with the passing of peak oil, and the difficulty obtaining it, it might decline again. When you're comparing cars, you want to use an industry average for refining, rather than any specific refinery. As was mentioned, for the Tesla, then you would use the power plant efficiency, which is twice what internal combustion engine efficiency typically is. And then there is diesel efficiency, applied to cars, trucks, trains, etc. As i said first, to get a figure that's useful, you have to define the limits of what you want to check. In you original post, you included sources for refinery efficiency and also power plant efficiency. I might note that most thermal power plants in the US use either coal, or natural gas. Oil has become to price volatile for power companies to want to deal with. Coal is cheaper, and natural gas is cleaner. <<A comparison is made between electric and fueled vehicle system efficiencies but with this definition refinery waste does not seem to be considered. Is it an unequal comparison? >> For this, you probably want to use coal for the electric car, since that's the most prevalent power source today. And you'd want to use some average for refining. However, what's left out is the environmental damage that's left behind from the 2 power sources. One would hope that it won't be to long before solar panels become a major player. And electric cars become the standard.
linlyons
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