💸 Will this tech be cheap? 💸
How to Assess ClimateTech P.2
Welcome to My Essays
Hi. I’m Ahmed. I’m an analyst with a deep curiosity about the world.
I explore ideas. I break them down. I share what I learn in essays like this.
I hope you find them interesting.
What I’m Sharing Today
I am continuing to share what I learned about evaluating climate technologies.
The goal is to figure out the best way to stop climate change.
What’s in This Essay?
This framework answers four questions:
Should I spend time on this technology?
Will it get cheap enough?
If it’s cheap, will it scale?
If it can scale, how should I scale it?
If you missed my thoughts on the first question, check out this essay.
Today, I am focusing on the second question.
How do you know a technology can become cheap?
How can you tell if it will be cheap enough to beat the fossil fuel alternative?
The Origin Of This Essay
One of the most common beliefs in technology is that producing more makes a product cheaper.
Many people treat this idea as if it were a law of nature.
This belief leads to a dangerous conclusion.
“I can take any technology and succeed if I get enough subsidies to reach high-volume production.”
People assume that once they reach high-volume production, the technology will work. This conclusion is wrong.
The main driver of success for any technology is not innovation or production volume.
What matters most are the fundamental laws of physics and chemistry.
These laws set the absolute minimum cost for any technology.
Since these laws never change, your lowest possible cost will not change unless you change your process.
Optimizing your process or buying in bulk can help you get closer to this minimum cost.
However, they will not reduce the minimum cost. In most climate industries, price is king.
In industries like steel or cement, everyone has the exact same product. The only way to win in the market is price.
If your minimum cost for a zero-emission solution is higher than the fossil fuel alternative, you are unlikely to be successful.
This essay is about what I learned about calculating the minimum cost of any technology.
To find the minimum cost, we will use the fundamental principles of physics and chemistry.
Tool 1: Enthalpy of Formation
Enthalpy of formation helps us find the minimum energy needed to make 1 unit of a product.
The minimum energy is a key factor in determining the minimum product cost.
The steps for finding the minimum energy are listed below.
Step 1: Identify the Chemical Reaction
Say we want to make hydrogen (H₂) from water (H₂O). Here is the chemical reaction: 2H₂O→ 2H₂+O₂
Note: Your chemical equation must have an equal amount of each element on both sides.
If not, the minimum energy calculation will be incorrect.
Step 2: Find the Enthalpy of Each Molecule in the Reaction
The enthalpy of most elements and molecules is available in databases like this one.
H₂O has an enthalpy of formation of –285.8 kJ/mol.
H₂ has an enthalpy of formation of 0 kJ/mol.
O₂ has an enthalpy of formation of 0 kJ/mol.
kJ means kilojoule, which is a unit of energy. Mol refers to mole, which is a unit of mass.
We use kJ/mol to standardize energy changes per mole of substance.
This makes it easier to apply these values directly to stoichiometric calculations (Tool 3).
Step 3: Multiply the Number of Moles by the Enthalpy of Each Molecule
Recall our reaction is: 2H₂O→ 2H₂+O₂
The chemical equation above says that we need 2 moles of H₂O (water) to make 2 moles of H₂ (hydrogen) and 1 mole of O₂ (oxygen).
Enthalpy of 2H₂O: –285.8 kJ/mol × 2 moles = –571.6 kJ
Enthalpy of 2H₂: 0 kJ/mol × 2 moles = 0 kJ
Enthalpy of O₂: 0 kJ/mol × 1 mole = 0 kJ
Enthalpy = (sum of enthalpy of outputs) – (sum of enthalpy of inputs).
Enthalpy = (0 kJ of 2H₂ + 0 kJ of O₂) – (–571.6 kJ of 2H₂O) = 571.6 kJ.
The calculations above show that we need 571.6 kJ to make 2 moles of H₂ and 1 mole of O₂.
A positive kJ value means the reaction is endothermic, meaning it absorbs energy.
A negative kJ value means the reaction is exothermic, meaning it releases energy.
A reaction that absorbs energy requires you to apply energy to maintain it.
A reaction that releases energy does not require additional energy to sustain it.
The 571.6 kJ for 2 moles of hydrogen (or 285.8 kJ for 1 mole of hydrogen) and 1 mole of oxygen does not provide useful information on its own.
To make this data meaningful, we need molar mass (Tool 2) and stoichiometry (Tool 3).
Tool 2: Molar Mass
Molar mass tells us the mass of one mole of an element or molecule.
You can find the molar mass of any element or molecule by doing a quick search on Google.
1 mole of H₂O = 18.02g (or grams). 2 moles of H₂O = 36.04g.
1 mole of H₂ = 2.02g. 2 moles of H₂ = 4.04g.
1 mole of O₂ = 32g.
The mass of inputs and outputs must be the same because of the law of conservation of mass.
This law states that matter cannot be created or destroyed.
Every chemical equation assumes 100% mass conversion efficiency, also called mass yield.
In reality, the actual mass yield will be less than 100%.
Your mass does not disappear, but some of it does not get converted into the desired output.
Molar mass values alone are not very helpful.
To make molar mass and enthalpy useful, we need stoichiometry (Tool 3).
Tool 3: Stoichiometry
Stoichiometry is the process of making the enthalpy of formation and molar mass useful.
Below are the steps.
Step 1: Identify the Product You Want to Make
Recall the chemical reaction of splitting water to produce hydrogen and oxygen: 2H₂O→ 2H₂+O₂
For this example, we want to make hydrogen.
Step 2: Identify What Mass We Want for Our Product
Hydrogen companies sell hydrogen by dollars per kilogram ($/kg).
For this calculation, we will use 1 kilogram as the reference amount.
Step 3: Find How Many Moles You Need to Get the Mass You Want
Remember that the mass of 1 mole of H₂ is 2.02g. We want to turn that mass into 1 kg.
Remember that the mass of 1 mole of H₂ is 2.02 grams.
We want to convert this mass into 1 kilogram.
First, convert the target mass (1 kilogram) into grams: 1 kilogram = 1000 grams.
Next, divide 1000 grams by 2.02 grams (mass of 1 mole of H₂). 1000/2.02= 496 moles
Now we know that to produce 1 kilogram of H₂, we need 496 moles of H₂.
Adjust the chemical equation:
Original: 2H₂O→ 2H₂+O₂
Adjusted: 496H₂O→496H₂+248O₂
Note: I don’t recommend running stoichiometry calculations by hand.
With multi-step reactions, doing everything manually increases the likelihood of mistakes.
I use spreadsheets to help organize the calculations.
Once the formulas are set up, I can run my analysis faster and with fewer errors. Here is an example spreadsheet.
Putting the 3 Tools Together
To make 1 kilogram of H₂, we need 496 moles.
Remember, the energy required to produce 1 mole of H₂ is 285.83 kJ/mole.
Multiply: 496 moles × 285.83 kJ/mole = 141,772 kJ per kilogram of H₂.
Convert kJ to kWh: 141,772 kJ ÷ 3600 = 39.4 kWh per kilogram of H₂.
Assume industrial electricity costs are 6.26 cents per kWh (Texas, September 2024).
Energy cost: 39.4 kWh × $0.0626 per kWh = $2.46 per kilogram of H₂.
Compare this to fossil fuel hydrogen costs: $0.5–$1.7 per kilogram.
Clearly, hydrogen from water is more expensive.
Note: This analysis assumes no capital costs, labor, land, maintenance, or financing.
Real-world costs would make green hydrogen even less competitive.
Another Example Using the 3 Analysis Tools
Let’s analyze making natural gas (CH₄) from CO₂ and H₂.
Producing 1 MMBtu of natural gas requires 9.6 kilograms of H₂ and 77 kilograms of CO₂.
Assume renewable energy costs 1 cent per kWh.
Cost of H₂: 39.4 kWh per kilogram × $0.01 per kWh = $0.39 per kilogram.
Total cost: $0.39 per kilogram × 9.6 kilograms = $3.74 per MMBtu.
Compare this to natural gas extracted from the ground: $2.20 per MMBtu (October 2024).
This shows that the technology is unlikely to be competitive.
Similar to the previous example, this analysis assumes no capital costs, labor, land, maintenance, or financing.
I also didn’t account for the cost of CO₂.
I only considered the energy cost of hydrogen, and it’s clear that this approach will not work.
A Few Final Words
I hope you found this essay interesting! If you know someone who might enjoy it, feel free to share.
Got feedback? Email me at theahmedhassan1@gmail.com.
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Until next time,
Ahmed
Thanks for sharing! I enjoyed the systematic analysis and the step-by-step explanation.