27 Sep 08
This gadget, invented by a professor at Temple University, does something with electric fields to make the gas or diesel fuel less viscous and so makes smaller drops in the fuel injector, it's apparently for real!
You have all seen those ads in the Pop magazines where they say increase your gas milage with our goodie, only 19.95, BUT WAIT, we will include a magic window cleaner at no extra cost... So finally something real!
http://www.sciencedaily.com/releases/2008/09/080925111836.htm
16 Oct 08
Originally posted by sonhouseless fuel in the injector makes less power. so whats the new idea, -save fuel by using less power?😕
This gadget, invented by a professor at Temple University, does something with electric fields to make the gas or diesel fuel less viscous and so makes smaller drops in the fuel injector, it's apparently for real!
You have all seen those ads in the Pop magazines where they say increase your gas milage with our goodie, only 19.95, BUT WAIT, we will include ...[text shortened]... st... So finally something real!
http://www.sciencedaily.com/releases/2008/09/080925111836.htm
16 Oct 08
Originally posted by eamon oDid you read the piece? It's not less fuel, the droplets are smaller, resulting in a more efficient burn so it uses less fuel for the same amount of power. The device lowers the viscosity of the fuel just before the injectors.
less fuel in the injector makes less power. so whats the new idea, -save fuel by using less power?😕
16 Oct 08
Originally posted by sonhouseno i didnt, thanks for explaining the idea.
Did you read the piece? It's not less fuel, the droplets are smaller, resulting in a more efficient burn so it uses less fuel for the same amount of power. The device lowers the viscosity of the fuel just before the injectors.
(personally, i feel it would be good if more people did that rather than
just posting a link to some massive article)
17 Oct 08
Originally posted by sonhouseA few points:
This gadget, invented by a professor at Temple University, does something with electric fields to make the gas or diesel fuel less viscous and so makes smaller drops in the fuel injector, it's apparently for real!
You have all seen those ads in the Pop magazines where they say increase your gas milage with our goodie, only 19.95, BUT WAIT, we will include ...[text shortened]... st... So finally something real!
http://www.sciencedaily.com/releases/2008/09/080925111836.htm
* The chemical energy being released by burning is several orders of magnitude higher than the energy used for breaking up droplets.
* Today sdiesel have a very efficient and clean burning process. If you would leave about 20 % chemical energy in your fuel you would observe:
+ dark clouds of soot
+ poisonous amounts of CO
* Organic molecules are electrically inert and won't be excited by an outer electrical field.
* And I want the original pubvlication in a scientific magazine.
Thank you for contributing
17 Oct 08
Originally posted by Ponderable* The chemical energy being released by burning is several orders of magnitude higher than the energy used for breaking up droplets.
A few points:
* The chemical energy being released by burning is several orders of magnitude higher than the energy used for breaking up droplets.
* Today sdiesel have a very efficient and clean burning process. If you would leave about 20 % chemical energy in your fuel you would observe:
+ dark clouds of soot
+ poisonous amounts of CO
* Orga ...[text shortened]...
* And I want the original pubvlication in a scientific magazine.
Thank you for contributing
True.
* Today sdiesel have a very efficient and clean burning process. If you would leave about 20 % chemical energy in your fuel you would observe:
+ dark clouds of soot
+ poisonous amounts of CO
Not quite true. The maximum efficiency of any engine is given by the following equation that stems from Carnot's theorem (http://en.wikipedia.org/wiki/Carnot_engine):
max efficiency = 1 - (Tc/Th)
Where Tc is the absolute temperature of the cold reservoir and Th is the absolute temperature of the hot reservoir. The operating temperature of a diesel engine is about 550 C (http://en.wikipedia.org/wiki/Diesel_engine#Early_fuel_injection_systems), so when discharging to ambient conditions the maximum efficiency is:
max efficiency = 1 - (298 K / (273 K + 550 K)) = 63.7%
So a perfect thermodynamically efficient diesel engine already leaves a significant amount of energy in the off gases.
* Organic molecules are electrically inert and won't be excited by an outer electrical field.
Most organic molecules in oily chemicals, especially the types used in fuel, are non-polar. However, viscosity is produced by electrostatic interactions between sub-atomic particles, manifesting in the resistance to shear stress. An electric field will most definitely have an effect on this property, whether the molecules are polar or not.
* And I want the original pubvlication in a scientific magazine.
Apparently, this article was published in Energy & Fuels in 2008. Here's the link to the published article at the bottom of the page: http://pubs.acs.org/cgi-bin/abstract.cgi/enfuem/asap/abs/ef8004898.html
I think you can purchase a PDF of the article there if you want to.
18 Oct 08
Originally posted by PBE6Am I right that that is a calculation based entirely on the assumption that a fixed amount of chemical energy is converted to heat and is a calculation of the efficiency of converting that produced heat into power?
Not quite true. The maximum efficiency of any engine is given by the following equation that stems from Carnot's theorem (http://en.wikipedia.org/wiki/Carnot_engine):
max efficiency = 1 - (Tc/Th)
Where Tc is the absolute temperature of the cold reservoir and Th is the absolute temperature of the hot reservoir. The operating temperature of a diesel engin ...[text shortened]... mically efficient diesel engine already leaves a significant amount of energy in the off gases.
If so, then how is it relevant to a process whose sole purpose is to increase the amount of fuel burnt? Or am I wrong about what the device claims to do?
18 Oct 08
Originally posted by twhiteheadHis calculation is for the maximum efficiency, so it's just a theoretical upper bound.
Am I right that that is a calculation based entirely on the assumption that a fixed amount of chemical energy is converted to heat and is a calculation of the efficiency of converting that produced heat into power?
If so, then how is it relevant to a process whose sole purpose is to increase the amount of fuel burnt? Or am I wrong about what the device claims to do?
18 Oct 08
2 edits
Originally posted by twhiteheadI was responding to Ponderable's post that if you leave 20% of the chemical energy in the fuel then you start producing too much soot and CO. I wanted to show that the amount of energy that doesn't get extracted from the fuel is much higher than 20%. The effects Ponderable mentioned seem more the result of incomplete combustion (formation of ash and particulate, formation of CO instead of CO2) than inefficient energy capture.
Am I right that that is a calculation based entirely on the assumption that a fixed amount of chemical energy is converted to heat and is a calculation of the efficiency of converting that produced heat into power?
If so, then how is it relevant to a process whose sole purpose is to increase the amount of fuel burnt? Or am I wrong about what the device claims to do?
The maximum efficiency of an ideal engine is the most efficient any engine can be. An ideal engine operating at those temperatures would produce about 64 J of useable energy for every 100 J extracted from the heat reservoir. Real engines don't do as well, because of multiple inefficiencies (friction, temperature gradients, fluid flow pattern inefficiencies, etc...). I understand that this device decreases the viscosity of the fluid through an applied electric field, thereby reducing the size of the fuel droplets as they leave the fuel injector resulting in a more complete/efficient combustion process. It reduces one of the local inefficiencies, resulting in better overall efficiency for the engine. However, it will never off-set the inefficiency so well that the engine's overall efficiency exceed about 64%.
18 Oct 08
So yes of course not all HEAT is converted into (useful) motion. This is an entropy effect and Carnot (the ideal machine) gives a efficiency factor of 63%.
But all the chemical energy is converted into HEAT even in today diesel machines. If this coonversion was ineffective then it wold result in CO, unburnt diesel and soot in the offgas.
I will look up the publication when I am at work (since we have a subscrption). Sometimes journalists reporting on something didn't get the details right.
19 Oct 08
3 edits
Originally posted by PBE6I have no objections to your claim that the maximum efficiency of converting the heat to power is a mere 64%
The maximum efficiency of an ideal engine is the most efficient any engine can be. An ideal engine operating at those temperatures would produce about 64 J of useable energy for every 100 J extracted from the heat reservoir.
However as Ponderable has already pointed out, that is almost irrelevant to the situation or possibly argues in favor of ponderables initial claim. Separate the engine into two parts, one part (1.) that converts chemical energy into heat and another (2.) that converts heat into power.
So if we start with 100 J of chemical energy and only x J gets converted to heat in 1. then only y J will be converted to power in 2. where y is less than 64% of x.
So in order to get an overall increase in output of 20% by increasing the efficiency of 1. we need an increase in efficiency in 1. by considerably more than 20% (more like 30-40% ).
This means that current engines must be burning only 60-70% of their fuel.
20 Oct 08
1 edit
Originally posted by twhiteheadBut you have to go with the bottom line. How do you figure a reported claim of 20% increase in milage?
I have no objections to your claim that the maximum efficiency of converting the heat to power is a mere 64%
However as Ponderable has already pointed out, that is almost irrelevant to the situation or possibly argues in favor of ponderables initial claim. Separate the engine into two parts, one part (1.) that converts chemical energy into heat and anoth ...[text shortened]... (more like 30-40% ).
This means that current engines must be burning only 60-70% of their fuel.
Are you saying the engine simply puts out 20% less power?
20 Oct 08
Originally posted by sonhouseLets say an engine currently burns x percent of its fuel in in order to produce y amount of heat of which only 50% actually drives the wheels.
But you have to go with the bottom line. How do you figure a reported claim of 20% increase in milage?
Are you saying the engine simply puts out 20% less power?
Then a 20% increase in x should result in a 20% increase in y, and a 20% increase in power at the wheels. So it looks like I was wrong, and we don't need to burn more than 20% more fuel.
But I still think that the efficiency of heat to power conversion is irrelevant, and that current engines burn more than 80% of their fuel, so the device remains somewhat suspect unless there is something I am missing.