Hydrogen vs. Battery

[ohmman]

The product itself is zero emission. Rest depends on where the H2 is coming from. Just like with electricity.

That's a weak argument, and you know better. How about this - I'll provide you the energy mix for Tesla's Supercharger at the Nurburgring right here, and you provide me the same transparency for this yet to be used H2 generator that you're trying to sell as being cleaner than Tesla's solution.

Ball's in your court.

 

[nwdiver]

Current electrolyser product, can be deployed anywhere:
https://nelhydrogen.com/assets/uploads/2017/01/Nel_Electrolyser_brochure.pdf
elecrolyser, includes transformer and rectifier:
200 bar 3.8 - 4.4 kWh/m3; H2: 39.4kWh/kg 0.08988 kg/m3 = 80.5 - 93.2% efficiency

Maybe this is the source of your confusion? H2 isn't 39.4kWh/kg. It's 33.33kWh/kg. The higher number you used is the 'higher heating value' which assumes that the latent heat of vaporization is scavenged from the exhaust. Maybe a little helpful in the winter but costly and not of much use at any other time of the year. The density of H2 is 0.082

There's a lot of wiggle room to pick and choose which numbers are used but the upper limit of efficiency is ~75% not 93%. It's a little odd that Nel likes to tout how efficient their electrolysers are but no where I could find is there a 'Our electrolyses are XX% efficient'. They make you guess which assumptions they used and do the math...

 

[nwdiver]

The product itself is zero emission. Rest depends on where the H2 is coming from. Just like with electricity. EVs are called ZEVs as well.

I can sync the charging of my BEV to use surplus wind or solar that would have been wasted. The energy I used could not have displaced fools fuel... that's what 'surplus' means. It just would have been wasted. Every gram of H2 a FCV uses could have displaced ~2g of natural gas since we still need ~10B kg/yr of H2 for industry. Until the amount of H2 from CH4 is ~0 100% of the fuel for FCVs is natural gas.

 

[mtndrew1]

I bet if you walked in with $8k, they would take it on the spot.

I don’t think you’re far off the mark. There are a dozen or so Mirais on cars dot com that have been there for months with asking prices steadily sinking.

The only things driving the pitiful numbers of new H2 vehicle deliveries are heavily subsidized leases, massive government kickbacks, and a used Civic’s worth of free fuel.

Nobody actually wants these cars for anything near full freight.

 

[acoste]

Maybe this is the source of your confusion? H2 isn't 39.4kWh/kg. It's 33.33kWh/kg. The higher number you used is the 'higher heating value' which assumes that the latent heat of vaporization is scavenged from the exhaust. Maybe a little helpful in the winter but costly and not of much use at any other time of the year. The density of H2 is 0.082

There's a lot of wiggle room to pick and choose which numbers are used but the upper limit of efficiency is ~75% not 93%. It's a little odd that Nel likes to tout how efficient their electrolysers are but no where I could find is there a 'Our electrolyses are XX% efficient'. They make you guess which assumptions they used and do the math...

Their H2 has a purity of 99.9998% so that's very close to the HHV.

 

[acoste]

That's a weak argument, and you know better. How about this - I'll provide you the energy mix for Tesla's Supercharger at the Nurburgring right here, and you provide me the same transparency for this yet to be used H2 generator that you're trying to sell as being cleaner than Tesla's solution.

Ball's in your court.

You won. I can get H2 from Koblenz, but that's steam reformed. However in theory I could still transport zero emission sourced H2 from farther.

 

[nwdiver]

Their H2 has a purity of 99.9998% so that's very close to the HHV.

???? Explain what you think the relevance of that is.... I'm not talking about purity. I'm talking about kWh electricity in vs kWh H2 out. Higher vs lower heating value has nothing to do with purity.

 

[ohmman]

You won. I can get H2 from Koblenz, but that's steam reformed. However in theory I could still transport zero emission sourced H2 from farther.

Fair enough. Of course, the transportation would also have to be H2-based using the same zero emission sourced H2.. etc. Or, more realistically, it could be hauled using a BEV.

 

[acoste]

???? Explain what you think the relevance of that is.... I'm not talking about purity. I'm talking about kWh electricity in vs kWh H2 out. Higher vs lower heating value has nothing to do with purity.

Hydrogen - Wikipedia

Combustion of Hydrogen gas

2 H2(g) + O2(g) → 2 H2O(l) + 572 kJ (286 kJ/mol)

The enthalpy of combustion is −286 kJ/mol

Weight of H2 molecule ~= 2g/mol


Energy released when burning 1kg of pure H2 = −286 kJ/mol * 1kg / (2g/mol) = - 143MJ/kg >>> this is the HHV. The best case when we have a perfect H2 gas with a purity of 100%.

143 MJ energy = 39.7kWh energy (this is energy, not electricity; just a different unit)

Since H2 has the highest energy density, if we add other gasses like moisture (H20) or methane CH4 to the mix, the heating value will drop.


Combustion of methane
CH4 + 2 O2 → CO2 + 2 H2O (ΔH = −891 kJ/mol, at standard conditions)
methane weight 16g/mol

Energy released when burning 1kg of methane = −891 kJ/mol * 1kg / (16g/mol) = - 56MJ/kg

 

[nwdiver]

Hydrogen - Wikipedia




Weight of H2 molecule ~= 2g/mol


Energy released when burning 1kg of pure H2 = −286 kJ/mol * 1kg / (2g/mol) = - 143MJ/kg >>> this is the HHV. The best case when we have a perfect H2 gas with a purity of 100%.

143 MJ energy = 39.7kWh energy (this is energy, not electricity; just a different unit)

Since H2 has the highest energy density, if we add other gasses like moisture (H20) or methane CH4 to the mix, the heating value will drop.


Combustion of methane
CH4 + 2 O2 → CO2 + 2 H2O (ΔH = −891 kJ/mol, at standard conditions)
methane weight 16g/mol

Energy released when burning 1kg of methane = −891 kJ/mol * 1kg / (16g/mol) = - 56MJ/kg

Ok.... none of that has anything to do with LHV of 33.33kWh/kg vs HHV of 39kWh/kg..... Purity has NOTHING to do with HHV vs LHV...

The numerical difference between the LHV and HHV of a fuel is roughly equivalent to the amount of latent heat of vaporization that can be practically recovered in a secondary condenser per unit of fuel burned.
​

Do FCVs have a 'secondary condenser'? Which number is used to calculate the efficiency of a FCV? If you're using 33kWh/kg then you need to use 33kWh/kg to calculate electrolysis efficiency. FCVs assume 33kWh/kg....

 

[acoste]

Ok.... none of that has anything to do with LHV of 33.33kWh/kg vs HHV of 39kWh/kg..... Purity has NOTHING to do with HHV vs LHV...

The numerical difference between the LHV and HHV of a fuel is roughly equivalent to the amount of latent heat of vaporization that can be practically recovered in a secondary condenser per unit of fuel burned.
​

Do FCVs have a 'secondary condenser'? Which number is used to calculate the efficiency of a FCV? If you're using 33kWh/kg then you need to use 33kWh/kg to calculate electrolysis efficiency. FCVs assume 33kWh/kg....

Ok. Let's add the energy loss of evaporating water molecules.

Produced gas purity: 99.9998%

That is 999,998 mol of H2 and 2 mol of H2O

We burn the H2

Energy released when burning 999,998 mol of H2 of pure H2 = −286 kJ/mol * 999,998 mol = - 285,999,428 kJ

Energy needed to evaporate 2 mol of H2O = 2 * 40.8 kJ/mol = 81.6 kJ
(Latent Heat of Vaporization/Condensation)


So by burning the gas above the energy released equals - 285,999,428 kJ + 81.6 kJ = -285,999,346.4 kJ

Total weight of gas = 999,998 mol of H2 * 2g/mol + 2 mol of H2O * 18g/mol = 2000.032 kg

285,999,346.4 kJ / 2000.032 kg = 142.997 MJ/kg = 39.7214 kWh/kg

 

[nwdiver]

Ok. Let's add the energy loss of evaporating water molecules.

Produced gas purity: 99.9998%

That is 999,998 mol of H2 and 2 mol of H2O

We burn the H2

Energy released when burning 999,998 mol of H2 of pure H2 = −286 kJ/mol * 999,998 mol = - 285,999,428 kJ

Energy needed to evaporate 2 mol of H2O = 2 * 40.8 kJ/mol = 81.6 kJ
(Latent Heat of Vaporization/Condensation)


So by burning the gas above the energy released equals - 285,999,428 kJ + 81.6 kJ = -285,999,346.4 kJ

Total weight of gas = 999,998 mol of H2 * 2g/mol + 2 mol of H2O * 18g/mol = 2000.032 kg

285,999,346.4 kJ / 2000.032 kg = 142.997 MJ/kg = 39.7214 kWh/kg

How is the energy of water vapor harvested in a FCV? That's my point... You're making this more complicated that it is. 33.33kWh/kg is used to calculate the efficiency of FCVs because the other ~6kWh cannot be used. You can't assume 39kWh/kg for the output from electrolysis and 33.33kWh/kg for the input to a FCV....

HHV is generally a boiler metric because there is such a thing as a condensing boiler. To my knowledge there is no such thing as a condensing FCV (at least one that uses the heat)....

You can have electroysis that's ~80% efficient and a Mirai that gets 57mpge OR electrolysis that <75% efficient and a Mirai that gets ~67mpg. But you can't have electrolysis that's ~80% efficient AND a Mirai that gets 67mpge. Seems 33kWh/kg is the number that's used to calculate mpge....

 

[acoste]

You are getting lost. We are calculating the efficiency of the electrolyser, remember? There is no fuel cell there.

33.33kWh/kg is used to calculate the efficiency of FCVs because the other ~6kWh cannot be used.

Why? And if that's the case, that should be reflected in the efficiency of the fuel cell. Unrelated to the electrolyser.

 

[nwdiver]

You are getting lost. We are calculating the efficiency of the electrolyser, remember? There is no fuel cell there.

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Why? And if that's the case, that should be reflected in the efficiency of the fuel cell. Unrelated to the electrolyser.

If you're using two different numbers how is the result accurate??? If your definition of H2 is 39kWh/kg for electrolysis but changes to 33kWh/kg for a fuel cell then you're magically gaining ~20% in your well-to-wheels analysis. That's why you can't see how terrible FCVs are....

The Miria gets 67mpge because the 'gallon equivalent' of H2 is assumed to be 33kWh/kg. If it was 39kWh/kg it would be rated at 57mpge.

So which number do you want to go with? 33kWh and the Mirai gets 67mpge or 39kWh and the Miari gets 57mpge.....

 

[acoste]

If you're using two different numbers how is the result accurate??? If your definition of H2 is 39kWh/kg for electrolysis but changes to 33kWh/kg for a fuel cell then you're magically gaining ~20% in your well-to-wheels analysis. That's why you can't see how terrible FCVs are....

The Miria gets 67mpge because the 'gallon equivalent' of H2 is assumed to be 33kWh/kg. If it was 39kWh/kg it would be rated at 57mpge.

So which number do you want to go with? 33kWh and the Mirai gets 67mpge or 39kWh and the Miari gets 57mpge.....

https://www.sciencedirect.com/topics/engineering/fuel-cell-efficiency


The fuel cell efficiency is defined as a ratio between the electricity produced and the hydrogen consumed.

Sometimes, instead of hydrogen’s higher heating value (HHV), ΔH = 286 kJ/mol, the lower heating value (LHV) is used (ΔHLHV = 241 kJ/mol). The difference between the higher and lower heating values is the heat of product water condensation. Because the product water may leave the fuel cell in either form, that is, as liquid or as vapor, both values are correct; however, the type of heating value used to calculate the efficiency must be specified.


Ok. This has nothing to do with the energy density of the H2 going in. It is about the evaporation.

So if the fuel cell efficiency is given in LHV, then it is an optimistic value.

Let's say rated LHV efficiency is 60% = 20kWh / 33.33kWh. That means I fed the cell with 1 kg of Hydrogen with 39kWh/kg energy density and It produced 20kWh electric energy. So the fuel cell's real efficiency is closer to 51%.

 

[nwdiver]

Because the product water may leave the fuel cell in either form, that is, as liquid or as vapor, both values are correct;

No... they're not. Whether water leaves as a liquid or vapor the energy is lost in a fuel cell. It only matters if the objective it heat energy. The objective for a fuel cell is ELECTRICAL energy.

Regardless... everyone appears to use the LHV for calculating mpge so that's the number that needs to be assumed as the energy content for both the output from electrolysis AND the input to a fuel cell....

 

[acoste]

No... they're not. Whether water leaves as a liquid or vapor the energy is lost in a fuel cell. It only matters if the objective it heat energy. The objective for a fuel cell is ELECTRICAL energy.

Regardless... everyone appears to use the LHV for calculating mpge so that's the number that needs to be assumed as the energy content for both the output from electrolysis AND the input to a fuel cell....

This LHV efficiency makes no sense unless the heat energy is utilized as well.

 

[nwdiver]

This LHV efficiency makes no sense unless the heat energy is utilized as well.

.... that’s HHV that assumes heat energy is used. LHV for cars, HHV for boilers.

 

[jelloslug]

I don’t think you’re far off the mark. There are a dozen or so Mirais on cars dot com that have been there for months with asking prices steadily sinking.

The only things driving the pitiful numbers of new H2 vehicle deliveries are heavily subsidized leases, massive government kickbacks, and a used Civic’s worth of free fuel.

Nobody actually wants these cars for anything near full freight.

There was a Mirai that some used car dealership in New Jersey bough recently thinking they were getting some screaming deal only to find out that the closest fueling station was a few thousand miles away. The ad was pulled quickly...

 

[Curious George]

Current H2 availability in California.

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