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originally posted by: deadeyedick
true or false? according to physics it takes 119.282 watts to produce one liter of hho atm
I'm not sure what exactly you're trying to do
So you put 15.9MJ in and you get 15.9MJ out if there are zero losses anywhere and if the result of the combustion is liquid water, which are obviously totally unrealistic assumptions.
just to set one thing straight.
how much energy is used to light a 60 watt bulb for 1 hour. and watts per hour is the correct term.
a 60 watt light bulb on for 1 hour uses 60 watt hours of energy.
how much if used if it was on for 15 minutes?
it still uses 60 watts, it doesn't magically drop to 15 watts.
originally posted by: bigx001
just to set one thing straight.
how much energy is used to light a 60 watt bulb for 1 hour. and watts per hour is the correct term.
a 60 watt light bulb on for 1 hour uses 60 watt hours of energy.
3.658kWh to seperate 1liter of water
so actually if that figure is correct it would be 1956.52174 J/s to produce 1l hho = ? continous watts
A watt is a rate?[/quote[
Exactly, just like miles per hour is a rate. A Watt is a Joule per second.
The main problem i have is that a watt is defined as a joule/second and not only does that convey a rate but also a time frame at which the rate is happening. I think much of this confusion is not only from my understanding but also from the need to screw people every month on their electric bill.
The rate is independent from the time frame. It's like saying that you're driving your car at a constant speed of 60 mph... that rate is completely independent of how long you drove the car far. When you take the rate and time frame, that's how you know how long you've driven. When you take your power and the time frame, that's how you know how much energy you've used or generated.
When you first asked the question I didn't get that but from your follow-up question I got that idea, so it's a good line of questioning, well done.
originally posted by: DenyObfuscation
Trying to compare required input vs the output and get an idea of how much is lost due to inefficiency.
Your electrolyzer can be in STP conditions, but if you have no other source of heat or energy besides electricity, that 2.7MJ will also have to come from electricity. Remember the HHV value of hydrogen of 141.83 kJ/g which is the energy content between hydrogen gas and liquid water? Convert the joules to kWh and you get 0.0394kWh/g, or 39.4 kWh/kg, where this EU report says potentially all that energy might come from electricity (which would include the 2.7MJ in our example):
About the bottom red arrow in the graphic you posted, that ~ 2.7 MJ figure is the energy provided by environment in STP conditions?
So to rephrase that in the numbers from our example, if all energy input is provided by electricity (which is the default assumption in the efficiency measure described above), the required energy is 15.9 MJ. If you have a waste heat source available, you can reduce the 15.9MJ requirement by up to 2.7MJ, since that will be provided by the waste heat, instead of electricity.
Here, we describe ‘efficiency’ as energy input in kWh per kg of hydrogen output. For commercial technologies (alkaline, PEM, AEM) this energy is supplied in electrical form, with a theoretical minimum electrical energy input of 39.4 kWh/kg H2 (HHV of hydrogen), if water is fed at ambient pressure and temperature to the system and all energy input is provided in the form of electricity. The required electrical energy input may be reduced below 39.4 kWh/kg H2 if suitable heat energy is provided to the system. High temperature electrolysis, such as PEM steam electrolysis and particularly solid oxide electrolysis could have lower operating costs if the electrolyser were co-located with a low cost or waste heat source, than if all the energy were provided through electricity.
As I've been trying to explain in my calculations here, that number is incomplete. It's the change in energy between gaseous water and hydrogen, not between liquid water and hydrogen. It relates to the explanation above that if you have a source of waste heat you can get close to that figure if the waste heat provides an additional 0.751kWh, but the total energy requirement if the only input is electricity is 3.659kWh + 0.751kWh = 4.410 kWh, per liter of liquid water.
originally posted by: deadeyedick
3.658kWh to seperate 1liter of water
so actually if that figure is correct
I'm getting screwed on my electric bill, but it's because of the aftermath of the fraud of Enron etc, and not because of any confusion on units. There are people who understand all this stuff and don't find it that confusing, but if you're not used to dealing with this stuff, I can understand how it might seem confusing.
I think much of this confusion is not only from my understanding but also from the need to screw people every month on their electric bill.
so 1J/s would be 1JX1second
Here is the problem i see in a watt
1w=1j=1sec
by assigning a time value to the definition of a joule you are also assigning that same time value to a watt
without foregoing logic we can determine that 1 watt = 1 sec bu definition
the only way to get into diminsional view of it is to forget that a watt has been defined as a second
if we forget that then we can arrive at the false understanding put forth today.
yes i am hopeless because i stuck to the basic math principles
if you do a simple search you will find that many agree with me and have determined just as i said a kWh is a false but it does remain steady as a rate for billing.
You can find much info that shows the need for such definitions was derrived from the need to charge people money.
I also admit that by all current standards i am wrong in what i am saying and you all are right