How to Cause Nuclear Reactions at Low Energy and Why Should You Care Edmund Storms, Ph.D. Santa Fe, NM These slides are from this video lecture: http://www.youtube.com/view_play_list?p=3b79262131ca1bcf
Profs. Pons and Fleischmann (Courtesy Univ. of Utah)
What is Cold Fusion? Cold Fusion = popular name Low Energy Nuclear Reaction = LENR = General description Chemically Assisted Nuclear Reaction = CANR = LENR Condensed Matter Nuclear Science = CMNS = Field of study
Work is Being Done at Laboratories in Many Countries 1. Japan, 2. Italy, 3. Russia, 4. Ukraine, 5. Israel, 6. U.S.A., 7. China, 8. France
What Reactions Occur? Detected d + d > 3 He(0.82 MeV) + n(2.45 MeV) d + d > p(3.02 MeV) + t(1.01 MeV) d + d > 4 He + (23.8 MeV) (No gamma) n*d + M > fusion or fission (transmutation) Not Detected d + t > n(14.01mev) + 4 He(3.5 MeV) d + p > 3 He + gamma (5.5 MeV)
ITER REACTOR
How does Cold Fusion Compare to Hot Fusion? Cold Fusion Occurs in a special solid materials at low applied energy. Produces mainly helium-4 and heat Small, simple devices can make useful power Involves fusion of deuterium and many other nuclear reactions. Has been studied for 18 years and is not understood. Is a clean energy source. Is difficult to replicate. Hot Fusion Occurs in a plasma or upon application of high energy. Produces neutrons, tritium and heat. A very large reactor is required to make useful power. Involves fusion of tritium and deuterium. Has been studied for 60 years and is well understood. Is a radioactive energy source. Is difficult to make useful.
Kinds of Evidence Anomalously large heat production, Helium production consistent with heat production, Occasional tritium production, Production of anomalous isotopes, Consistent patterns of behavior, Energetic particle and gamma emission.
How is Cold Fusion Initiated? Electrolysis Ambient Gas Liquid Plasma Gas Plasma Proton Conductor Sonic Implantation Laser Light
Representation of Calorimeter Methods Single Wall Double Wall Open method Closed method Single and Double Wall Isoperbolic EP = T*C - V*I, (<0.1 W) Seebeck EP = Vs*C - V*I, (<0.02W) Flow EP = T*(F*Cp + C) - V*I, (±0.05 W)
Flow calorimeter schematic (This slide not included in video)
Simplified flow calorimeter schematic showing only the cooling water in the outer jacket (This slide not included in video)
How Much Power is Produced? All 157 values, 0.005-183 W, 12.7 W mean 120 100 100 80 80 COUNT COUNT 120 60 B All Values A Electrolytic Values 60 40 40 20 20 0 0 0 40 80 120 WATT 160 200 0 5 10 15 20 25 WATT Power based on 157 reported studies 30 35 40
Recent Example of Significant Power using Electrolysis (Dardik et al., Energetics Technologies, Israel) 17 h Average Pout ~20 watts Pout=KΔT Average Pinet=Iin*Vin PPinet dis ~0.74 watts Excess Power of up to 34 watts; Average ~20 watts for 17 h
Example of Significant Power using Gas Loading (Arata and Zhang, Osaka Univ., Japan)
Consistent Patterns for Heat Production When Pd is used during Electrolysis Effect of average D/Pd ratio Effect of applied current Effect of batch of Pd Effect of temperature Effect of electrolysis time Effect of cracking of the metal
Effect of D/Pd ratio on excess power (McKubre et al., SRI) C1 Parabolic 0.875 555 6 5 4 3 2 1 0 0.88 0.89 0.90 0.91 0.92 0.93 0.94 Atomic ratio (D/Pd) 0.95 0.96 0.97 0.98
Summary of Heat Production (McKubre et al., SRI) D/Pd = 0.70 to 0.90 D/Pd = 0.90 to 0.95 D/Pd = 0.95 to 1.05 17 no heat 9 no heat, 6 heat 15 heat
Example of a Consistent Pattern (McKubre et al., SRI) Light & Heavy Water Cells; Excess Power vs. Current Density P14/D2O Linear P13/H2O 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0.1 0.2 0.3 0.4 0.5-0.1 Electrochemical Current Density (A/cm 2 ) 0.6
Effect of Current on Excess Power EFFECT OF CURRENT DENSITY ON EXCESS POWER Produced by Palladium 2.0 Variable Temperature 1.8 Constant Temperature McKubre et al., (wire, flow) Storms (plate, ²T) Bush (plate, ²T) 1.6 Aoki et al. (plate, flow) Kunimatsu (rod, ²T) 1.4 Kainthla et al. (wire, ²T) 1.2 1.0 0.8 0.6 0.4 0.2 0.0 0.0 0.1 0.2 0.3 0.4 CURRENT DENSITY, A/cm2 0.5 0.6 0.7
Effect of Applied Current on Excess Power 100 2 10 1.1 Santhanam et al. Miles et al. Oriani et al. Eagleton and Bush Kainthla et al. Fleischmann and Pons.01.001.0001 0 200 400 Guruswamy and Wadsworth Hutchinson et al. Liaw et al. Lewis and Skold Appleby et al. Noninski and Noninski 600 800 1000 1200 CURRENT (ma/cm )2 1400 1600
Example of relationship between helium and heat production
Growth of heat and helium (McKubre et al., SRI) 9 180 8 160 ppmv SC2 3 line fit for 4He 7 140 6 120 Gradient 5 100 4 80 3 60 2 40 1 20 0 0 0 5 10 Time (Days) 15 20 Excess Energy (kj) Differential
Grown of Tritium during Electrolysis (Bockris et. al, Texas A&M) 20000 Active Cell Inactive Cell 15000 10^8 atoms/sec 10000 Current Increase 5000 10^7 atoms/sec 0 0 50 100 TIME, hr 150
Growth of Tritium during Gas Discharge (Claytor et al., LANL)
Extra Elements Commonly Found 8 Cu Pd cathode, electrolysis Zn NUMBER OCCASIONS REPORTED 7 Pd Pt Fe 6 Ti Al 5 Mg Pb 4 Ag 3 2 1 0 2 10 18 26 34 42 50 ATOMIC NUMBER 58 66 74 82
Iwamura et al., Mitsubishi, Japan Caused the following reactions by passing D2 through Pd-CaO-Pd layers. 4D + 133Cs55 = 141Pr59 +? 4D + 88Sr38 = 96Mo42 +? 6D + 138Ba56= 150Sm62 +? Measured target loss and product gain using XPS, SIMS, XANES, ICP-MS, and X-ray Fluorescence. Basic process replicated in Italy (INFN-LNF) and Japan (Osaka Univ.).
Active Region (Iwamura et al., Mitsubishi Heavy Industries, Japan) D2 gas permeation through the Pd complex D2 gas 重水素ガス D Permeation 重陽子 Cross Section of Pd Complex 独自開発ナノ構造反応膜 Surface 最表面 CaO 2nm Pd 40nm D Flux 重陽子透過 真空 Vacuum Pd基板 Pd substrate
Decrease of Cs and Emergence of Pr 14 2 Number of atoms (/10 /cm ) (Iwamura et al., Mitsubishi Heavy Industries, Japan) Cs D Pd Pd CaO Cs 1st Pr 1st Cs 2nd Pr 2nd 16 14 12 10 8 6 4 2 0 0 20 40 60 80 Time (h) 100 120
Conversion of Cs to Pr (Iwamura et al., Mitsubishi Heavy Industries, Japan) Electrochemical Addition Ion Implantation Conversion rateη (%) 100 80 60 0.3(1/SCCM) 40 20 0 0.00 1.00 2.00 3.00 4.00 Average D2 gas permeation rate:fl SCCM)
Production of Pr from Cs (Iwamura et al., Mitsubishi Heavy Industries, Japan) X 1 2 3 4 5 6 Y 13-4 1 2 3 4 100μm 5 6 100μm Much Pr detection Pr detection No Pr
Mass Correlation between Given and Detected Elements (Iwamura et al., Mitsubishi Heavy Industries, Japan)
Why is LENR Hard to Replicate? Active material is present only in very small regions, Active material is a complex alloy with an unknown structure and composition, Active material rarely forms. (The challenge is to make the alloy on purpose and in large amount.)
What Do We Know? * Nuclear reactions involve 1, 2, 4, and 6 deuterons that add to a variety of nuclei * Nuclear reactions involve both H and D * Nuclear reactions occur only in special solid environments (NAE) * Nuclear reactions produce little if any radiation or energetic particles detected outside of apparatus. * Significant radiation is detected inside of apparatus along with the nuclear products. * Nuclear reactions can occur at rates that make significant heat energy. * Nuclear reactions can be initiated many different ways.
Requirements of a Theory The nuclear charge of the deuteron or proton must be lowered, but not by neutron formation. The deuterons must form a cluster of 2, 4, or 6 d after this lowering process has occurred. The reactions must produce very little gamma radiation and few radioactive isotopes. The process must occur spontaneously without significant energy being applied. The mechanism must occur only under very special conditions in unique solid materials.
Conclusions The LENR phenomenon is real and has the potential to be an ideal energy source. Significant energy can be produced using several methods and with simple devices. The effect requires use of a special solid material. The challenge now is to identify this special material and manufacture it in large amount.
For more information read the book The Science of Low Energy Nuclear Reaction available at www.worldscientific.com and consult the website www.lenr-canr.org