If 1 barrell of oil (42 gallons, right?) could be refined by itself and the refiner could use only energy from that barrel of oil, how much (on average, realizing the output of refining varies depending upon alchemy) of that oil is consumed in the process?
I wonedered this - well more like how much oil is consummed in drilling, trucking, piping, shipping, pumping, refining, - basically how much is used to get it into my tank?
Also a barrel of oil is something like $30 (apx) which is 42 gal. Where can I buy an actually barrel and take it home?
Actually you can. Just go to the NW coast of Spain where that tanker broke up and you can take as much as you like for free. There are thousands of people scooping the oil by hand these days. I never knew people were so desperate for oil. 
The Oil Well to your car’s tank efficiency is around 88%. Read here - Page 6 The tank to wheel efficiency varies from 10 - 20, usually approximated to 16%. Those figures are for gasoline
Now your 1 gallon of crude is not only used to make gasoline, but the clothes you are wearing (Okay so maybe u’r wearing 100% cotton but what about the buttons ?) , the car body you use, the keyboard you are typing on, many parts of the computer, your CDs, … There’s so much products from CRUDE OIL around you right now that you’ll be surprised you are not overwhelmed. So talking about just energy efficiency in Crude refining terms make little sense when it comes to petrochemicals from Crude.
We’ve debated this before. Based on my 1st-order calculations in this thread, the total efficiency of a gasoline vehicle (which includes energy expended to make gasoline and tank-to-wheel efficiency) is around 108%.
Crafter - from your earlier post - “0.353 joules must be expended (by burning coal, oil, or whatever) to produce 2.11 joule of gasoline”
So you start out with (2.11 + 0.353) J of energy (Crude Oil) and end up with 2.11 J of useful energy :
Hence, the efficiency of this process = 100 x 2.11/(2.11+0.353) = 86%. I quoted a figure of 88%.
You can consider this to be the efficiency of Crude Oil Well to your Car Tank ( if you add the transportation energy usage too).
2.11 joule of gasoline, of which 0.38 joules will be used to turn the wheels of vehicle powered by an ICE
This is the fuel tank to wheel efficiency = 100 x (0.38/2.11) = 18%. I quoted a value of 16%.
The overall efficiency, that is Oil Well to Car Wheel efficiency in your case is :
Energy Taken from the Oil Well = 0.353 J + 2.11 J = 2.463J
Energy finall delivered to the wheel = 0.38J
So the efficiency = 100* 0.38/2.463= 15% (Which is also the same as multiplying 86% x 18%)and regret not 108%
Hope that helps.
andy_fl: To arrive at my answer you must combine the two items in italics. This gives you the following: 0.353 joules must be expended (by burning coal, oil, or whatever) to produce 0.38 joules at the wheel. This gives you an overall efficiency of 108%. Also note that I’m not violating any laws of physics here.
Perhaps we’re both right… I’m defining efficiency as the amount of energy expended vs. energy-at-the-wheel, while you’re defining it as thermodynamic efficiency. These are two different definitions with two different answers.
And by the way: you’re correct that I never added in the energy expended during transportation. That will probably bring my efficiency number down to around 100%.
Crafter, let me please try this one more time. I know no refinery which uses coal as a heat source and most use petroleum derivatives for heat source. And I’ll use all your numbers to make it easy :
You start with some amount of crude with 2.463J of energy.
Of this 2.463J of energy, you spend 0.353J to refine it to gasoline. So after refining you are left with 2.11J and 0.353J has been wasted as heat to the atmosphere.
Now lets assume that this 2.11J is transferred to your gas tank without any significant loss of energy (i.e. lets ignore the energy loss in transortation).
So out of the 2.463j originally from the oil well 2.11J lands up in your tank. Out of this 2.11 J of energy 1.73 J ends up as waste heat in the exhaust, heat in the radiator, and heating up the engine, transmission etc. and 0.38J actually shows up as power in the wheel which moves the car.
So going by what you took out of the well to the final desirable product - energy in the wheel (all other expenditure of energy is waste) the overall well to wheel efficiency is :
100x0.38/2.463 = 15%
I hope that was a typo otherwise, Im sure you’l soon get a lot of guys in tin foil clamouring to invest in your new startup.
No, it’s not a typo.
Let’s say you cut down a tree with a handsaw. Let’s say you expend 5,000 Joules in the process. You then burn the wood, which gives you 500,000 Joules of energy.
Assuming all you wanted was heat, the efficiency is 10,000%. Not bad.
We do the same thing with oil. We receive more energy from the oil than what was expended getting it out of the ground and processing it. This results in an efficiency of over 100%.
It’s a very simple concept to me, and I’m having a hard time figuring out why it’s difficult for some people to grasp…
No the efficiency is 99.0099%. You started off with (500,000 J + 5000J) but you lost 5000J and only 500,000J showed up as energy in your fireplace.
Since you are a electrical engineer, let me say this to you. Say, you expend 5,000 J/s in the process of transmitting power (transformer losses, line losses) from a power station. And 500,000 J/s actually reaches the consumers.
Assuming, all you wanted was the power to the consumers, What is the efficiency of the transmission process ?
It’s a very simple concept to me, and I’m having a hard time figuring out why it’s difficult for some people to grasp…
It is indeed a simple concept but you are misunderstanding it. Efficiency is not just a ratio - but a ratio of output as a percentage of input.
Andy_fl: As stated previously, you and I are simply defining “efficiency” in different ways.
When it comes calculating overall system efficiency of crude energy sources, I prefer to look at how much energy is acquired vs. how much energy is expended in order to aquire the energy.
There is nothing wrong with they way you’re calculating it; it is perfectly valid in a mathematical sense. But there’s a disadvantage to it: the final number provides no indication on whether or not it was worth the effort to acquire the energy source in the first place.
That’s why I prefer my method. If the efficiency is over 100%, then by definition it was worth it, and it can rightfully be called a “fuel.” If it’s under 100%, it is simply an energy transfer mechanism.
Am I making any kind of sense?
Sure it does. By andy_fl’s definition, if the efficiency is less than 0% then there is no net energy to be gained.
Actually, “nothing to gain” would be when andy_fl’s efficiency is between 0% and 50%. Anyway, I still prefer my definition, though “efficiency” might not be the proper term to use. How about “economic advantage”?
Its bad because the way you define efficiency isn’t mulplicitive. You can’t say that the Crude->Gas transformation is 500% efficient and the gas->motion is 600% efficient, therefore, its 30,000% efficient.
Perpetual motion sellers tend use similar definitions of efficiency to produce their large numbers.