Source: https://www.blue.world/technology/
This is a sensible approach because waste heat is used to support reformulation. However, without knowing the kWh/l of methanol, it is not possible to tell if it makes economic sense. Grid power in our area is $0.10/kWh.
Gallon of methanol, $1.10, is 3 kg. Given it is CH{3}OH:
Bob Wilson
- High-Temperature PEM fuel cells
- Methanol reforming
Methanol reforming is a relatively simple process that converts a mix of methanol and water into a hydrogen-rich gas. Before the reforming can take place the fuel needs to go from liquid to gas form by evaporation, a process that requires energy and in a generic system would mean using primary fuel, leading to lower efficiency. In the combination with HT PEM the waste heat is of sufficient temperature to drive this process, meaning an energy-free process which leads to a superior overall efficiency.
The fundamental chemical process, which takes place in the reformer is: Ch3OH + H2O → CO2 + 3 H2
The hydrogen (H2) produced is subsequently used in the fuel cell to produce electricity.
The fundamental chemical process, which takes place in the reformer is: Ch3OH + H2O → CO2 + 3 H2
The hydrogen (H2) produced is subsequently used in the fuel cell to produce electricity.
This is a sensible approach because waste heat is used to support reformulation. However, without knowing the kWh/l of methanol, it is not possible to tell if it makes economic sense. Grid power in our area is $0.10/kWh.
Gallon of methanol, $1.10, is 3 kg. Given it is CH{3}OH:
C(14), O(16), H(1) -> 4H / (14C + 16O + 4H) :: mass ratios
4 / 24 = 16.7% hydrogen :: hydrogen content
3kg * 16.7% = 0.5 kg of hydrogen in 3 kg of methanol costing $1.10
~$2.20/kg of hydrogen - these are industrial hydrogen rates
4 / 24 = 16.7% hydrogen :: hydrogen content
3kg * 16.7% = 0.5 kg of hydrogen in 3 kg of methanol costing $1.10
~$2.20/kg of hydrogen - these are industrial hydrogen rates
Bob Wilson
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