Based on the Lugano Report, an estimate of the fuel composition may be attempted.
Known facts:
1. Fuel sample had a mass of 1 gram
2. Page 29: “From the analysis methods of the fuel we find that there are significant quantities of Li, Al, Fe and ICP-AES analysis we find there is about 0.011 grams of 7Li in the 1 gram fuel.”
3. Page 29: “… the information from ICP-AES that there is about 0.55 gram NI in the fuel.”
4. Page 28: “From all combined analysis methods of the fuel we find that there are significant quantities of Li, Al, Fe and H in addition to the Ni.”
5. Page 28: “… from the ICP-AES analysis which shows the mass ratio between Li and Al is compatible with a LiAlH4 molecule.”
6. Page 28: “…natural composition, i.e. 6Li 7% and 7Li 93%”
7. Page 28: “We remark in particular that hydrogen but no deuterium was seen by SIMS.”
Analysis
The average mass of the lithium atoms are 0.07*6 + 0.93*7 = 6.93 amu.
Aluminum atoms have a mass of 27 amu while hydrogen atoms have an average mass of 1.
So the molecular weight of the LiAlH4 must be 6.93 + 27 + 4 = 37.93 amu.
There for the amount of LiAlH4 must be 0.011 grams * 37.93 / 6.93 = 0.06 grams and the amount of aluminum must be 0.043 grams. The amount of hydrogen in the LiAlH4 must be 0.006 grams.
The iron mass must therefore be 1.0 grams (total) – 0.55 grams (Ni) – 0.043 grams (Al) – 0.011 grams (Li) – 0.006 grams (H) = 0.39 grams of iron.
Element % by Weight
Nickel 55.0
Iron 39.0
Aluminum 4.3
Lithium 1.1
Hydrogen (no Deuterium) 0.6
Total 100.0
LiAlH4 6.0
It is also likely that the LiAlH4 was prepared using hydrogen depleted of deuterium.
This website uses cookies to improve your experience. We'll assume you're ok with this, but you can opt-out if you wish. Cookie settingsACCEPT
Privacy & Cookies Policy
Privacy Overview
This website uses cookies to improve your experience while you navigate through the website. Out of these cookies, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. We also use third-party cookies that help us analyze and understand how you use this website. These cookies will be stored in your browser only with your consent. You also have the option to opt-out of these cookies. But opting out of some of these cookies may have an effect on your browsing experience.
Necessary cookies are absolutely essential for the website to function properly. This category only includes cookies that ensures basic functionalities and security features of the website. These cookies do not store any personal information.
Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. It is mandatory to procure user consent prior to running these cookies on your website.
Determining the Fuel Composition for an E-Cat Reactor (Steven Karels)
The following post was submitted by Steven Karels
Based on the Lugano Report, an estimate of the fuel composition may be attempted.
Known facts:
1. Fuel sample had a mass of 1 gram
2. Page 29: “From the analysis methods of the fuel we find that there are significant quantities of Li, Al, Fe and ICP-AES analysis we find there is about 0.011 grams of 7Li in the 1 gram fuel.”
3. Page 29: “… the information from ICP-AES that there is about 0.55 gram NI in the fuel.”
4. Page 28: “From all combined analysis methods of the fuel we find that there are significant quantities of Li, Al, Fe and H in addition to the Ni.”
5. Page 28: “… from the ICP-AES analysis which shows the mass ratio between Li and Al is compatible with a LiAlH4 molecule.”
6. Page 28: “…natural composition, i.e. 6Li 7% and 7Li 93%”
7. Page 28: “We remark in particular that hydrogen but no deuterium was seen by SIMS.”
Analysis
The average mass of the lithium atoms are 0.07*6 + 0.93*7 = 6.93 amu.
Aluminum atoms have a mass of 27 amu while hydrogen atoms have an average mass of 1.
So the molecular weight of the LiAlH4 must be 6.93 + 27 + 4 = 37.93 amu.
There for the amount of LiAlH4 must be 0.011 grams * 37.93 / 6.93 = 0.06 grams and the amount of aluminum must be 0.043 grams. The amount of hydrogen in the LiAlH4 must be 0.006 grams.
The iron mass must therefore be 1.0 grams (total) – 0.55 grams (Ni) – 0.043 grams (Al) – 0.011 grams (Li) – 0.006 grams (H) = 0.39 grams of iron.
Element % by Weight
Nickel 55.0
Iron 39.0
Aluminum 4.3
Lithium 1.1
Hydrogen (no Deuterium) 0.6
Total 100.0
LiAlH4 6.0
It is also likely that the LiAlH4 was prepared using hydrogen depleted of deuterium.