How to convert your car to run on water
Guest
Posts: n/a
How to convert your car to run on water
Hey guys
I'm just letting you know about a new product I ordered yesterday, and
thought I'd let you know about it.
Run Your Car With Water Today - Double your gas mileage by running
your vehicle on a combination of water and gas. http://www.runyourcarwithwatertoday.com
It costs $49 to buy the document/plans, but hopefully I'll recover
that cost quickly with the gas saving. I haven't set it all up yet,
and I don't think it will save me 50% in gas, but I've read
independent sources and believe it's feasible to save around 10-20%.
I think the fact it's helping the environment by reducing emmissions,
spurred me into buying the product too.
Anyway, like I said, I thought I'd let you know, because I know a lot
of people are doing it tough with the increasing gas prices.
Have a good day,
Dave Tayler
I'm just letting you know about a new product I ordered yesterday, and
thought I'd let you know about it.
Run Your Car With Water Today - Double your gas mileage by running
your vehicle on a combination of water and gas. http://www.runyourcarwithwatertoday.com
It costs $49 to buy the document/plans, but hopefully I'll recover
that cost quickly with the gas saving. I haven't set it all up yet,
and I don't think it will save me 50% in gas, but I've read
independent sources and believe it's feasible to save around 10-20%.
I think the fact it's helping the environment by reducing emmissions,
spurred me into buying the product too.
Anyway, like I said, I thought I'd let you know, because I know a lot
of people are doing it tough with the increasing gas prices.
Have a good day,
Dave Tayler
Guest
Posts: n/a
Re: How to convert your car to run on water
On Mon, 16 Jun 2008 07:55:20 -0700 (PDT), "rick++"
<rick303@hotmail.com> wrote:
>Trade for some land in Florida!
Exactly. If that was real ( which it is not } how long do you figure
it would take for the gas companies to buy the patent.
<rick303@hotmail.com> wrote:
>Trade for some land in Florida!
Exactly. If that was real ( which it is not } how long do you figure
it would take for the gas companies to buy the patent.
Guest
Posts: n/a
Re: How to convert your car to run on water
Jim wrote:
> On Mon, 16 Jun 2008 07:55:20 -0700 (PDT), "rick++"
> <rick303@hotmail.com> wrote:
>
>> Trade for some land in Florida!
> Exactly. If that was real ( which it is not } how long do you figure
> it would take for the gas companies to buy the patent.
I checked the poster's website.
Evidently they're offering a kit that enables you to split water into
hydrogen and oxygen gases by electrolysis, using your car's battery for
the electricity. And then to partially power your car with those gases.
But they forgot the Law of Conservation of Energy: Your car's battery
is kept charged by running the (gasoline-powered) engine. And that's the
energy that will be used to do the electrolysis.
--
Steven L.
Email: sdlitvin@earthlinkNOSPAM.net
Remove the NOSPAM before replying to me.
> On Mon, 16 Jun 2008 07:55:20 -0700 (PDT), "rick++"
> <rick303@hotmail.com> wrote:
>
>> Trade for some land in Florida!
> Exactly. If that was real ( which it is not } how long do you figure
> it would take for the gas companies to buy the patent.
I checked the poster's website.
Evidently they're offering a kit that enables you to split water into
hydrogen and oxygen gases by electrolysis, using your car's battery for
the electricity. And then to partially power your car with those gases.
But they forgot the Law of Conservation of Energy: Your car's battery
is kept charged by running the (gasoline-powered) engine. And that's the
energy that will be used to do the electrolysis.
--
Steven L.
Email: sdlitvin@earthlinkNOSPAM.net
Remove the NOSPAM before replying to me.
Guest
Posts: n/a
Re: How to convert your car to run on water
David wrote:
> Hey guys
>
> I'm just letting you know about a new product I ordered yesterday, and
> thought I'd let you know about it.
>
> Run Your Car With Water Today - Double your gas mileage by running
> your vehicle on a combination of water and gas. http://www.runyourcarwithwatertoday.com
>
> It costs $49 to buy the document/plans, but hopefully I'll recover
> that cost quickly with the gas saving. I haven't set it all up yet,
> and I don't think it will save me 50% in gas, but I've read
> independent sources and believe it's feasible to save around 10-20%.
>
> I think the fact it's helping the environment by reducing emmissions,
> spurred me into buying the product too.
>
> Anyway, like I said, I thought I'd let you know, because I know a lot
> of people are doing it tough with the increasing gas prices.
>
> Have a good day,
>
> Dave Tayler
I call bullshit on this device until I see an independent third party
lab validate these results. The first person to say that big oil
companies are preventing independent labs from performing these tests
can stick it where the sun does not shine!
As taken from the discovery forum
(http://community.discovery.com/eve/f...m/2321969559):
Water-for-gas? Let's have a look, shall we?
The standard reaction for the electrolysis of water is:
2H2O → 2H2(g) + O2(g)
For this process, 4 moles of electrons take place, therefore the
standard free energy change is:
ΔG = -nFε
ΔG = -4*96487*1.229
ΔG = -474330 J
We can expect to extract -474 kJ from the products of the electrolysis
and use it to do work.
This is also borne out by looking at the approximate bond energies. 4
H-O bonds must be broken, at ≈460 kJ each, but energy is released when
the new bonds form. The O=O bond yields ≈498 kJ and the H-H bonds yield
≈433 kJ each. Our net free energy is:
-4(460000) + 2(433000) + 498000 = -476000 J
So we have to put ≈ 475 kJ into the system to separate the water, and we
can get ≈ 475kJ back out in utilizing the free energy.
According to the "Oxyhydrogen" people (more on this later), burning the
resulting gases gives off 576 kJ. If the TOTAL energy that can possibly
be released to do work is 475 kJ, how are they magically getting 576 kJ?
They're generally not.
Now, on to the electrolysis itself. The efficiencies for electrolysis of
water are reported as anywhere from 50% to 94%. So, in the best case, we
have to put 505 kJ in to be able to get 475 kJ out, and in the worst
case we need an input of 950 kJ to be able to get 475 kJ out. Even
before factoring in the other losses, we can see from the 50-94%
efficiencies, that we will have a net loss of 6-50% of the energy used
for electrolysis.
There are thermal losses from heat engines. From Carnot, the MAXIMUM
possible efficiency from our engine is 70-75%. The real value is about
25% due to friction and the fact that the combustion is not
spontaneously reversible.
So, with the maximum possible theoretical efficiency (which can NOT be
obtained in practice) we are now needing anywhere from 675 kJ to 1357
kJ. Even if we allow for the inflated output numbers, and ignore the
actual losses we'd encounter, you can NOT produce 576 kJ and use it to
keep a process needing 675 kJ to 1357 kJ on-going. And this is before we
try to extract any OTHER work from our engine -- we are solely using the
output power to try and drive the input power.
In reality, what you end up with is needing to output around 3150 kJ to
keep the cycle going, but you are only producing 475 kJ. This is why
water-4-gas did not work when Adam and Jamie tried it. The losses
completely overwhelm any energies "created". Water is not a fuel -- it
is the ashes of hydrogen that has been burned once before.
Now, I know the "newest" schemes do not try to run the car solely off
the hydrogen, but call for using it in a "hydro-assist" manner. The
claim is that the hydrogen makes the gasoline burn better.
The first thing to note is the losses from above. If we need 3150 kJ to
produce two moles of hydrogen gas, and can only get 475 kJ back from the
system, then we need to burn an ADDITIONAL 2675 kJ worth of gasoline
just to break even!! This isn't looking promising.
But, but, but... the claims are that the hydrogen is a "catalyst" and
makes the gasoline burn more efficiently.
So? That claim is just wrong. We know we can't affect the thermal
efficiency of the Carnot cycle by very much, so 75% is still going to be
"wasted". Even if the hydrogen did act as a "catalyst", there is no more
energy to be released -- 99% of the gasoline does undergo combustion. A
300% gain in efficiency would imply that we are now getting 399% of the
theoretically-retrievable energy that the gasoline contains (well,
actually more than 399% because we also need to cover the losses from
the electrolysis). This is just asinine and Carnot says otherwise.
Energy can not be created from nothing -- the gasoline can NOT give more
than 100% of what it has.
But, but, but... "it's not H2 gas that is produced, it's monatomic
hydrogen!"
Ok, let's look at that. The ½H2(g) → H(g) reaction is NOT exothermic
until you reach a temperature of about 4000K. At our temperatures you
need to INPUT another 800 kJ (4*200 kJ) to get the 2H2(g) into 4H(g).
Without even considering the thermal losses from the combustion cycle,
we now need an input of about 1600 kJ to our electrolysis system in
order to get the claimed 576 kJ of "oxyhydrogen" back out (and that
calculation was made in the 19th century and is taken from a 1911
encyclopaedic entry -- wonder why they don't use more modern sources and
numbers?). I'm still seeing a major net loss, aren't you?
But, but, but... "the hydrogen DOES improve the efficiency!"
Yes, it is reported that hydrogen can increase the lean limit from 1.7
to 1.85, and it is also reported that hydrogen can reduce the no-load
idle consumption of gasoline by up to 50% in small engines.
Let's look at these. First off, these are for volumes of hydrogen that
can't possibly be produced by these simple cells. And, the reported
lamdas of 1.7 and 1.85 are for natural gas combustion, not gasoline. The
lean limit for gasoline is a lamda of about 1.2. Until you get to the
lean misfire conditions, leaning your gasoline down WILL reduce
consumption. But you also get rough idle and loss of power. Except when
idling, you car isn't fully leaned (and even then it's not maxed out) --
in fact, the more demand you're putting on the engine, the richer the
computer (or even the carb in non-EFI engines) makes the mixture. By
overriding the computer and approaching the lean limit, you can reduce
consumption at idle even without hydrogen. But as you place demands on
the engine, it can NOT be run this lean. The faster you go, the more
engine power must be used to overcome drag, so at highway speeds, even
if you are not accelerating, you are not running a lean mixture. The
manufacturers do NOT lean the idle down as far as possible, because
excessive leaning can burn the spark plugs and pistons and lead to
detonation and preignition. Even if it was possible to reduce idle usage
of gasoline by up to 50%, the long-term engine damage and the percentage
of the time the engine is NOT idling dramatically reduce this savings.
Plus, you STILL need extra gasoline to produce the hydrogen in the first
place from all the above-mentioned losses.
"Ok, ok... so I can't get such phenomenal gains in mileage from
'hydro-assist.' But what about this PICC stuff?"
Ah, yes... The "pre-ignition catalytic converter". Let's take a look.
The claim is that the PICC will break the isooctane molecule of gasoline
down into smaller molecules that will "burn better." Oh, really? When we
looked at the energies involved in electrolysis, we noticed that
breaking bonds takes energy and forming bonds releases energy. It
doesn't matter what happens in-between. If our net result is 2C8H18 +
25O2 → 16CO2 + 18H2O, then we are breaking 37 C-H bonds, 14 C-C bonds
and 25 O=O bonds, but re-forming 32 C-O bonds and 36 H-O bonds. Any
bonds that form and break in-between are immaterial; we are left with a
net change in bonds of about 37800 kJ. But, and this is important, it
TAKES energy to make any smaller molecules. We have to put energy in, in
order to have the higher bond energies of any smaller molecules. The net
change from beginning to end, remains the same: whether you extract 37.8
MJ from the gasoline, or extract 57.8 MJ after putting 20 MJ into the
bonds of smaller molecules, is immaterial -- the net change is still
37.8 MJ.
And besides, the PICC sites claim they're making a "plasma". Plasmas do
not undergo normal chemical reactions. You won't get combustion in a
plasma, nevermind factoring in the large energies necessary to create
plasmas in the first place.
Every step of all these schemes consumes far more energy than it could
possibly release. There is no magic way to get such phenomenal gains in
fuel efficiency out of your existing car.
> Hey guys
>
> I'm just letting you know about a new product I ordered yesterday, and
> thought I'd let you know about it.
>
> Run Your Car With Water Today - Double your gas mileage by running
> your vehicle on a combination of water and gas. http://www.runyourcarwithwatertoday.com
>
> It costs $49 to buy the document/plans, but hopefully I'll recover
> that cost quickly with the gas saving. I haven't set it all up yet,
> and I don't think it will save me 50% in gas, but I've read
> independent sources and believe it's feasible to save around 10-20%.
>
> I think the fact it's helping the environment by reducing emmissions,
> spurred me into buying the product too.
>
> Anyway, like I said, I thought I'd let you know, because I know a lot
> of people are doing it tough with the increasing gas prices.
>
> Have a good day,
>
> Dave Tayler
I call bullshit on this device until I see an independent third party
lab validate these results. The first person to say that big oil
companies are preventing independent labs from performing these tests
can stick it where the sun does not shine!
As taken from the discovery forum
(http://community.discovery.com/eve/f...m/2321969559):
Water-for-gas? Let's have a look, shall we?
The standard reaction for the electrolysis of water is:
2H2O → 2H2(g) + O2(g)
For this process, 4 moles of electrons take place, therefore the
standard free energy change is:
ΔG = -nFε
ΔG = -4*96487*1.229
ΔG = -474330 J
We can expect to extract -474 kJ from the products of the electrolysis
and use it to do work.
This is also borne out by looking at the approximate bond energies. 4
H-O bonds must be broken, at ≈460 kJ each, but energy is released when
the new bonds form. The O=O bond yields ≈498 kJ and the H-H bonds yield
≈433 kJ each. Our net free energy is:
-4(460000) + 2(433000) + 498000 = -476000 J
So we have to put ≈ 475 kJ into the system to separate the water, and we
can get ≈ 475kJ back out in utilizing the free energy.
According to the "Oxyhydrogen" people (more on this later), burning the
resulting gases gives off 576 kJ. If the TOTAL energy that can possibly
be released to do work is 475 kJ, how are they magically getting 576 kJ?
They're generally not.
Now, on to the electrolysis itself. The efficiencies for electrolysis of
water are reported as anywhere from 50% to 94%. So, in the best case, we
have to put 505 kJ in to be able to get 475 kJ out, and in the worst
case we need an input of 950 kJ to be able to get 475 kJ out. Even
before factoring in the other losses, we can see from the 50-94%
efficiencies, that we will have a net loss of 6-50% of the energy used
for electrolysis.
There are thermal losses from heat engines. From Carnot, the MAXIMUM
possible efficiency from our engine is 70-75%. The real value is about
25% due to friction and the fact that the combustion is not
spontaneously reversible.
So, with the maximum possible theoretical efficiency (which can NOT be
obtained in practice) we are now needing anywhere from 675 kJ to 1357
kJ. Even if we allow for the inflated output numbers, and ignore the
actual losses we'd encounter, you can NOT produce 576 kJ and use it to
keep a process needing 675 kJ to 1357 kJ on-going. And this is before we
try to extract any OTHER work from our engine -- we are solely using the
output power to try and drive the input power.
In reality, what you end up with is needing to output around 3150 kJ to
keep the cycle going, but you are only producing 475 kJ. This is why
water-4-gas did not work when Adam and Jamie tried it. The losses
completely overwhelm any energies "created". Water is not a fuel -- it
is the ashes of hydrogen that has been burned once before.
Now, I know the "newest" schemes do not try to run the car solely off
the hydrogen, but call for using it in a "hydro-assist" manner. The
claim is that the hydrogen makes the gasoline burn better.
The first thing to note is the losses from above. If we need 3150 kJ to
produce two moles of hydrogen gas, and can only get 475 kJ back from the
system, then we need to burn an ADDITIONAL 2675 kJ worth of gasoline
just to break even!! This isn't looking promising.
But, but, but... the claims are that the hydrogen is a "catalyst" and
makes the gasoline burn more efficiently.
So? That claim is just wrong. We know we can't affect the thermal
efficiency of the Carnot cycle by very much, so 75% is still going to be
"wasted". Even if the hydrogen did act as a "catalyst", there is no more
energy to be released -- 99% of the gasoline does undergo combustion. A
300% gain in efficiency would imply that we are now getting 399% of the
theoretically-retrievable energy that the gasoline contains (well,
actually more than 399% because we also need to cover the losses from
the electrolysis). This is just asinine and Carnot says otherwise.
Energy can not be created from nothing -- the gasoline can NOT give more
than 100% of what it has.
But, but, but... "it's not H2 gas that is produced, it's monatomic
hydrogen!"
Ok, let's look at that. The ½H2(g) → H(g) reaction is NOT exothermic
until you reach a temperature of about 4000K. At our temperatures you
need to INPUT another 800 kJ (4*200 kJ) to get the 2H2(g) into 4H(g).
Without even considering the thermal losses from the combustion cycle,
we now need an input of about 1600 kJ to our electrolysis system in
order to get the claimed 576 kJ of "oxyhydrogen" back out (and that
calculation was made in the 19th century and is taken from a 1911
encyclopaedic entry -- wonder why they don't use more modern sources and
numbers?). I'm still seeing a major net loss, aren't you?
But, but, but... "the hydrogen DOES improve the efficiency!"
Yes, it is reported that hydrogen can increase the lean limit from 1.7
to 1.85, and it is also reported that hydrogen can reduce the no-load
idle consumption of gasoline by up to 50% in small engines.
Let's look at these. First off, these are for volumes of hydrogen that
can't possibly be produced by these simple cells. And, the reported
lamdas of 1.7 and 1.85 are for natural gas combustion, not gasoline. The
lean limit for gasoline is a lamda of about 1.2. Until you get to the
lean misfire conditions, leaning your gasoline down WILL reduce
consumption. But you also get rough idle and loss of power. Except when
idling, you car isn't fully leaned (and even then it's not maxed out) --
in fact, the more demand you're putting on the engine, the richer the
computer (or even the carb in non-EFI engines) makes the mixture. By
overriding the computer and approaching the lean limit, you can reduce
consumption at idle even without hydrogen. But as you place demands on
the engine, it can NOT be run this lean. The faster you go, the more
engine power must be used to overcome drag, so at highway speeds, even
if you are not accelerating, you are not running a lean mixture. The
manufacturers do NOT lean the idle down as far as possible, because
excessive leaning can burn the spark plugs and pistons and lead to
detonation and preignition. Even if it was possible to reduce idle usage
of gasoline by up to 50%, the long-term engine damage and the percentage
of the time the engine is NOT idling dramatically reduce this savings.
Plus, you STILL need extra gasoline to produce the hydrogen in the first
place from all the above-mentioned losses.
"Ok, ok... so I can't get such phenomenal gains in mileage from
'hydro-assist.' But what about this PICC stuff?"
Ah, yes... The "pre-ignition catalytic converter". Let's take a look.
The claim is that the PICC will break the isooctane molecule of gasoline
down into smaller molecules that will "burn better." Oh, really? When we
looked at the energies involved in electrolysis, we noticed that
breaking bonds takes energy and forming bonds releases energy. It
doesn't matter what happens in-between. If our net result is 2C8H18 +
25O2 → 16CO2 + 18H2O, then we are breaking 37 C-H bonds, 14 C-C bonds
and 25 O=O bonds, but re-forming 32 C-O bonds and 36 H-O bonds. Any
bonds that form and break in-between are immaterial; we are left with a
net change in bonds of about 37800 kJ. But, and this is important, it
TAKES energy to make any smaller molecules. We have to put energy in, in
order to have the higher bond energies of any smaller molecules. The net
change from beginning to end, remains the same: whether you extract 37.8
MJ from the gasoline, or extract 57.8 MJ after putting 20 MJ into the
bonds of smaller molecules, is immaterial -- the net change is still
37.8 MJ.
And besides, the PICC sites claim they're making a "plasma". Plasmas do
not undergo normal chemical reactions. You won't get combustion in a
plasma, nevermind factoring in the large energies necessary to create
plasmas in the first place.
Every step of all these schemes consumes far more energy than it could
possibly release. There is no magic way to get such phenomenal gains in
fuel efficiency out of your existing car.
Guest
Posts: n/a
Re: How to convert your car to run on water
"ChrisB" <ChrisB@somewhere.someplace.com> wrote in message
news:zoqdndwLOP3Sq8XVnZ2dnUVZ_qninZ2d@giganews.com ...
> David wrote:
>> Hey guys
>>
>> I'm just letting you know about a new product I ordered yesterday, and
>> thought I'd let you know about it.
>>
>> Run Your Car With Water Today - Double your gas mileage by running
>> your vehicle on a combination of water and gas.
>> http://www.runyourcarwithwatertoday.com
>>
>> It costs $49 to buy the document/plans, but hopefully I'll recover
>> that cost quickly with the gas saving. I haven't set it all up yet,
>> and I don't think it will save me 50% in gas, but I've read
>> independent sources and believe it's feasible to save around 10-20%.
>>
>> I think the fact it's helping the environment by reducing emmissions,
>> spurred me into buying the product too.
>>
>> Anyway, like I said, I thought I'd let you know, because I know a lot
>> of people are doing it tough with the increasing gas prices.
>>
>> Have a good day,
>>
>> Dave Tayler
>
>
> I call bullshit on this device until I see an independent third party lab
> validate these results. The first person to say that big oil companies
> are preventing independent labs from performing these tests can stick it
> where the sun does not shine!
>
> As taken from the discovery forum
> (http://community.discovery.com/eve/f...m/2321969559):
>
(snip)
You pointy-headed scientists obviously hate America! I took my Lincoln
Navigator, packed it full of lead bricks for added road-hugging weight, let
most of the air out of the Firestone tires, added two extra gas tanks,
supercharged it, turbocharged it and put the hydrogen doo-dad on it. Time
for a road trip! This is a smooth ride that respects the sensibilities of
real men. Real men like the executives at Exxon-Mobil. Have a nice day
Guest
Posts: n/a
Re: How to convert your car to run on water
In article <zoqdndwLOP3Sq8XVnZ2dnUVZ_qninZ2d@giganews.com>, ChrisB
<ChrisB@somewhere.someplace.com> wrote:
> David wrote:
> > Hey guys
> >
> > I'm just letting you know about a new product I ordered yesterday, and
> > thought I'd let you know about it.
> >
> > Run Your Car With Water Today - Double your gas mileage by running
> > your vehicle on a combination of water and gas.
> > http://www.runyourcarwithwatertoday.com
> >
> > It costs $49 to buy the document/plans, but hopefully I'll recover
> > that cost quickly with the gas saving. I haven't set it all up yet,
> > and I don't think it will save me 50% in gas, but I've read
> > independent sources and believe it's feasible to save around 10-20%.
> >
> > I think the fact it's helping the environment by reducing emmissions,
> > spurred me into buying the product too.
> >
> > Anyway, like I said, I thought I'd let you know, because I know a lot
> > of people are doing it tough with the increasing gas prices.
> >
> > Have a good day,
> >
> > Dave Tayler
>
>
> I call bullshit on this device until I see an independent third party
> lab validate these results. The first person to say that big oil
> companies are preventing independent labs from performing these tests
> can stick it where the sun does not shine!
>
> As taken from the discovery forum
> (http://community.discovery.com/eve/f...m/2321969559):
>
> Water-for-gas? Let's have a look, shall we?
>
> The standard reaction for the electrolysis of water is:
> 2H2O ? 2H2(g) + O2(g)
> For this process, 4 moles of electrons take place, therefore the
> standard free energy change is:
> ?G = -nF?
> ?G = -4*96487*1.229
> ?G = -474330 J
>
> We can expect to extract -474 kJ from the products of the electrolysis
> and use it to do work.
>
> This is also borne out by looking at the approximate bond energies. 4
> H-O bonds must be broken, at ‰460 kJ each, but energy is released when
> the new bonds form. The O=O bond yields ‰498 kJ and the H-H bonds yield
> ‰433 kJ each. Our net free energy is:
> -4(460000) + 2(433000) + 498000 = -476000 J
>
> So we have to put ‰ 475 kJ into the system to separate the water, and we
> can get ‰ 475kJ back out in utilizing the free energy.
>
> According to the "Oxyhydrogen" people (more on this later), burning the
> resulting gases gives off 576 kJ. If the TOTAL energy that can possibly
> be released to do work is 475 kJ, how are they magically getting 576 kJ?
> They're generally not.
>
> Now, on to the electrolysis itself. The efficiencies for electrolysis of
> water are reported as anywhere from 50% to 94%. So, in the best case, we
> have to put 505 kJ in to be able to get 475 kJ out, and in the worst
> case we need an input of 950 kJ to be able to get 475 kJ out. Even
> before factoring in the other losses, we can see from the 50-94%
> efficiencies, that we will have a net loss of 6-50% of the energy used
> for electrolysis.
>
> There are thermal losses from heat engines. From Carnot, the MAXIMUM
> possible efficiency from our engine is 70-75%. The real value is about
> 25% due to friction and the fact that the combustion is not
> spontaneously reversible.
>
> So, with the maximum possible theoretical efficiency (which can NOT be
> obtained in practice) we are now needing anywhere from 675 kJ to 1357
> kJ. Even if we allow for the inflated output numbers, and ignore the
> actual losses we'd encounter, you can NOT produce 576 kJ and use it to
> keep a process needing 675 kJ to 1357 kJ on-going. And this is before we
> try to extract any OTHER work from our engine -- we are solely using the
> output power to try and drive the input power.
>
> In reality, what you end up with is needing to output around 3150 kJ to
> keep the cycle going, but you are only producing 475 kJ. This is why
> water-4-gas did not work when Adam and Jamie tried it. The losses
> completely overwhelm any energies "created". Water is not a fuel -- it
> is the ashes of hydrogen that has been burned once before.
>
>
> Now, I know the "newest" schemes do not try to run the car solely off
> the hydrogen, but call for using it in a "hydro-assist" manner. The
> claim is that the hydrogen makes the gasoline burn better.
>
> The first thing to note is the losses from above. If we need 3150 kJ to
> produce two moles of hydrogen gas, and can only get 475 kJ back from the
> system, then we need to burn an ADDITIONAL 2675 kJ worth of gasoline
> just to break even!! This isn't looking promising.
>
> But, but, but... the claims are that the hydrogen is a "catalyst" and
> makes the gasoline burn more efficiently.
>
> So? That claim is just wrong. We know we can't affect the thermal
> efficiency of the Carnot cycle by very much, so 75% is still going to be
> "wasted". Even if the hydrogen did act as a "catalyst", there is no more
> energy to be released -- 99% of the gasoline does undergo combustion. A
> 300% gain in efficiency would imply that we are now getting 399% of the
> theoretically-retrievable energy that the gasoline contains (well,
> actually more than 399% because we also need to cover the losses from
> the electrolysis). This is just asinine and Carnot says otherwise.
> Energy can not be created from nothing -- the gasoline can NOT give more
> than 100% of what it has.
>
> But, but, but... "it's not H2 gas that is produced, it's monatomic
> hydrogen!"
>
> Ok, let's look at that. The ?H2(g) ? H(g) reaction is NOT exothermic
> until you reach a temperature of about 4000K. At our temperatures you
> need to INPUT another 800 kJ (4*200 kJ) to get the 2H2(g) into 4H(g).
> Without even considering the thermal losses from the combustion cycle,
> we now need an input of about 1600 kJ to our electrolysis system in
> order to get the claimed 576 kJ of "oxyhydrogen" back out (and that
> calculation was made in the 19th century and is taken from a 1911
> encyclopaedic entry -- wonder why they don't use more modern sources and
> numbers?). I'm still seeing a major net loss, aren't you?
>
> But, but, but... "the hydrogen DOES improve the efficiency!"
>
> Yes, it is reported that hydrogen can increase the lean limit from 1.7
> to 1.85, and it is also reported that hydrogen can reduce the no-load
> idle consumption of gasoline by up to 50% in small engines.
>
> Let's look at these. First off, these are for volumes of hydrogen that
> can't possibly be produced by these simple cells. And, the reported
> lamdas of 1.7 and 1.85 are for natural gas combustion, not gasoline. The
> lean limit for gasoline is a lamda of about 1.2. Until you get to the
> lean misfire conditions, leaning your gasoline down WILL reduce
> consumption. But you also get rough idle and loss of power. Except when
> idling, you car isn't fully leaned (and even then it's not maxed out) --
> in fact, the more demand you're putting on the engine, the richer the
> computer (or even the carb in non-EFI engines) makes the mixture. By
> overriding the computer and approaching the lean limit, you can reduce
> consumption at idle even without hydrogen. But as you place demands on
> the engine, it can NOT be run this lean. The faster you go, the more
> engine power must be used to overcome drag, so at highway speeds, even
> if you are not accelerating, you are not running a lean mixture. The
> manufacturers do NOT lean the idle down as far as possible, because
> excessive leaning can burn the spark plugs and pistons and lead to
> detonation and preignition. Even if it was possible to reduce idle usage
> of gasoline by up to 50%, the long-term engine damage and the percentage
> of the time the engine is NOT idling dramatically reduce this savings.
> Plus, you STILL need extra gasoline to produce the hydrogen in the first
> place from all the above-mentioned losses.
>
>
> "Ok, ok... so I can't get such phenomenal gains in mileage from
> 'hydro-assist.' But what about this PICC stuff?"
>
> Ah, yes... The "pre-ignition catalytic converter". Let's take a look.
>
> The claim is that the PICC will break the isooctane molecule of gasoline
> down into smaller molecules that will "burn better." Oh, really? When we
> looked at the energies involved in electrolysis, we noticed that
> breaking bonds takes energy and forming bonds releases energy. It
> doesn't matter what happens in-between. If our net result is 2C8H18 +
> 25O2 ? 16CO2 + 18H2O, then we are breaking 37 C-H bonds, 14 C-C bonds
> and 25 O=O bonds, but re-forming 32 C-O bonds and 36 H-O bonds. Any
> bonds that form and break in-between are immaterial; we are left with a
> net change in bonds of about 37800 kJ. But, and this is important, it
> TAKES energy to make any smaller molecules. We have to put energy in, in
> order to have the higher bond energies of any smaller molecules. The net
> change from beginning to end, remains the same: whether you extract 37.8
> MJ from the gasoline, or extract 57.8 MJ after putting 20 MJ into the
> bonds of smaller molecules, is immaterial -- the net change is still
> 37.8 MJ.
>
> And besides, the PICC sites claim they're making a "plasma". Plasmas do
> not undergo normal chemical reactions. You won't get combustion in a
> plasma, nevermind factoring in the large energies necessary to create
> plasmas in the first place.
>
>
>
> Every step of all these schemes consumes far more energy than it could
> possibly release. There is no magic way to get such phenomenal gains in
> fuel efficiency out of your existing car.
Nice explanation but I doubt the average reader can handle the physical
chemistry. An easier way is to jack up the back of the car will be
running down hill all the time. I also have a bridge to sell if anyone
is interested.
<ChrisB@somewhere.someplace.com> wrote:
> David wrote:
> > Hey guys
> >
> > I'm just letting you know about a new product I ordered yesterday, and
> > thought I'd let you know about it.
> >
> > Run Your Car With Water Today - Double your gas mileage by running
> > your vehicle on a combination of water and gas.
> > http://www.runyourcarwithwatertoday.com
> >
> > It costs $49 to buy the document/plans, but hopefully I'll recover
> > that cost quickly with the gas saving. I haven't set it all up yet,
> > and I don't think it will save me 50% in gas, but I've read
> > independent sources and believe it's feasible to save around 10-20%.
> >
> > I think the fact it's helping the environment by reducing emmissions,
> > spurred me into buying the product too.
> >
> > Anyway, like I said, I thought I'd let you know, because I know a lot
> > of people are doing it tough with the increasing gas prices.
> >
> > Have a good day,
> >
> > Dave Tayler
>
>
> I call bullshit on this device until I see an independent third party
> lab validate these results. The first person to say that big oil
> companies are preventing independent labs from performing these tests
> can stick it where the sun does not shine!
>
> As taken from the discovery forum
> (http://community.discovery.com/eve/f...m/2321969559):
>
> Water-for-gas? Let's have a look, shall we?
>
> The standard reaction for the electrolysis of water is:
> 2H2O ? 2H2(g) + O2(g)
> For this process, 4 moles of electrons take place, therefore the
> standard free energy change is:
> ?G = -nF?
> ?G = -4*96487*1.229
> ?G = -474330 J
>
> We can expect to extract -474 kJ from the products of the electrolysis
> and use it to do work.
>
> This is also borne out by looking at the approximate bond energies. 4
> H-O bonds must be broken, at ‰460 kJ each, but energy is released when
> the new bonds form. The O=O bond yields ‰498 kJ and the H-H bonds yield
> ‰433 kJ each. Our net free energy is:
> -4(460000) + 2(433000) + 498000 = -476000 J
>
> So we have to put ‰ 475 kJ into the system to separate the water, and we
> can get ‰ 475kJ back out in utilizing the free energy.
>
> According to the "Oxyhydrogen" people (more on this later), burning the
> resulting gases gives off 576 kJ. If the TOTAL energy that can possibly
> be released to do work is 475 kJ, how are they magically getting 576 kJ?
> They're generally not.
>
> Now, on to the electrolysis itself. The efficiencies for electrolysis of
> water are reported as anywhere from 50% to 94%. So, in the best case, we
> have to put 505 kJ in to be able to get 475 kJ out, and in the worst
> case we need an input of 950 kJ to be able to get 475 kJ out. Even
> before factoring in the other losses, we can see from the 50-94%
> efficiencies, that we will have a net loss of 6-50% of the energy used
> for electrolysis.
>
> There are thermal losses from heat engines. From Carnot, the MAXIMUM
> possible efficiency from our engine is 70-75%. The real value is about
> 25% due to friction and the fact that the combustion is not
> spontaneously reversible.
>
> So, with the maximum possible theoretical efficiency (which can NOT be
> obtained in practice) we are now needing anywhere from 675 kJ to 1357
> kJ. Even if we allow for the inflated output numbers, and ignore the
> actual losses we'd encounter, you can NOT produce 576 kJ and use it to
> keep a process needing 675 kJ to 1357 kJ on-going. And this is before we
> try to extract any OTHER work from our engine -- we are solely using the
> output power to try and drive the input power.
>
> In reality, what you end up with is needing to output around 3150 kJ to
> keep the cycle going, but you are only producing 475 kJ. This is why
> water-4-gas did not work when Adam and Jamie tried it. The losses
> completely overwhelm any energies "created". Water is not a fuel -- it
> is the ashes of hydrogen that has been burned once before.
>
>
> Now, I know the "newest" schemes do not try to run the car solely off
> the hydrogen, but call for using it in a "hydro-assist" manner. The
> claim is that the hydrogen makes the gasoline burn better.
>
> The first thing to note is the losses from above. If we need 3150 kJ to
> produce two moles of hydrogen gas, and can only get 475 kJ back from the
> system, then we need to burn an ADDITIONAL 2675 kJ worth of gasoline
> just to break even!! This isn't looking promising.
>
> But, but, but... the claims are that the hydrogen is a "catalyst" and
> makes the gasoline burn more efficiently.
>
> So? That claim is just wrong. We know we can't affect the thermal
> efficiency of the Carnot cycle by very much, so 75% is still going to be
> "wasted". Even if the hydrogen did act as a "catalyst", there is no more
> energy to be released -- 99% of the gasoline does undergo combustion. A
> 300% gain in efficiency would imply that we are now getting 399% of the
> theoretically-retrievable energy that the gasoline contains (well,
> actually more than 399% because we also need to cover the losses from
> the electrolysis). This is just asinine and Carnot says otherwise.
> Energy can not be created from nothing -- the gasoline can NOT give more
> than 100% of what it has.
>
> But, but, but... "it's not H2 gas that is produced, it's monatomic
> hydrogen!"
>
> Ok, let's look at that. The ?H2(g) ? H(g) reaction is NOT exothermic
> until you reach a temperature of about 4000K. At our temperatures you
> need to INPUT another 800 kJ (4*200 kJ) to get the 2H2(g) into 4H(g).
> Without even considering the thermal losses from the combustion cycle,
> we now need an input of about 1600 kJ to our electrolysis system in
> order to get the claimed 576 kJ of "oxyhydrogen" back out (and that
> calculation was made in the 19th century and is taken from a 1911
> encyclopaedic entry -- wonder why they don't use more modern sources and
> numbers?). I'm still seeing a major net loss, aren't you?
>
> But, but, but... "the hydrogen DOES improve the efficiency!"
>
> Yes, it is reported that hydrogen can increase the lean limit from 1.7
> to 1.85, and it is also reported that hydrogen can reduce the no-load
> idle consumption of gasoline by up to 50% in small engines.
>
> Let's look at these. First off, these are for volumes of hydrogen that
> can't possibly be produced by these simple cells. And, the reported
> lamdas of 1.7 and 1.85 are for natural gas combustion, not gasoline. The
> lean limit for gasoline is a lamda of about 1.2. Until you get to the
> lean misfire conditions, leaning your gasoline down WILL reduce
> consumption. But you also get rough idle and loss of power. Except when
> idling, you car isn't fully leaned (and even then it's not maxed out) --
> in fact, the more demand you're putting on the engine, the richer the
> computer (or even the carb in non-EFI engines) makes the mixture. By
> overriding the computer and approaching the lean limit, you can reduce
> consumption at idle even without hydrogen. But as you place demands on
> the engine, it can NOT be run this lean. The faster you go, the more
> engine power must be used to overcome drag, so at highway speeds, even
> if you are not accelerating, you are not running a lean mixture. The
> manufacturers do NOT lean the idle down as far as possible, because
> excessive leaning can burn the spark plugs and pistons and lead to
> detonation and preignition. Even if it was possible to reduce idle usage
> of gasoline by up to 50%, the long-term engine damage and the percentage
> of the time the engine is NOT idling dramatically reduce this savings.
> Plus, you STILL need extra gasoline to produce the hydrogen in the first
> place from all the above-mentioned losses.
>
>
> "Ok, ok... so I can't get such phenomenal gains in mileage from
> 'hydro-assist.' But what about this PICC stuff?"
>
> Ah, yes... The "pre-ignition catalytic converter". Let's take a look.
>
> The claim is that the PICC will break the isooctane molecule of gasoline
> down into smaller molecules that will "burn better." Oh, really? When we
> looked at the energies involved in electrolysis, we noticed that
> breaking bonds takes energy and forming bonds releases energy. It
> doesn't matter what happens in-between. If our net result is 2C8H18 +
> 25O2 ? 16CO2 + 18H2O, then we are breaking 37 C-H bonds, 14 C-C bonds
> and 25 O=O bonds, but re-forming 32 C-O bonds and 36 H-O bonds. Any
> bonds that form and break in-between are immaterial; we are left with a
> net change in bonds of about 37800 kJ. But, and this is important, it
> TAKES energy to make any smaller molecules. We have to put energy in, in
> order to have the higher bond energies of any smaller molecules. The net
> change from beginning to end, remains the same: whether you extract 37.8
> MJ from the gasoline, or extract 57.8 MJ after putting 20 MJ into the
> bonds of smaller molecules, is immaterial -- the net change is still
> 37.8 MJ.
>
> And besides, the PICC sites claim they're making a "plasma". Plasmas do
> not undergo normal chemical reactions. You won't get combustion in a
> plasma, nevermind factoring in the large energies necessary to create
> plasmas in the first place.
>
>
>
> Every step of all these schemes consumes far more energy than it could
> possibly release. There is no magic way to get such phenomenal gains in
> fuel efficiency out of your existing car.
Nice explanation but I doubt the average reader can handle the physical
chemistry. An easier way is to jack up the back of the car will be
running down hill all the time. I also have a bridge to sell if anyone
is interested.
Thread
Thread Starter
Forum
Replies
Last Post
Currently Active Users Viewing This Thread: 1 (0 members and 1 guests)
