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    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

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    Profs. Pons and Fleischmann(Courtesy Univ. of Utah)

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    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

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    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

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    What Reactions Occur?Detected

    d + d > 3He(0.82 MeV) + n(2.45 MeV)

    d + d > p(3.02 MeV) + t(1.01 MeV)

    d + d > 4He + (23.8 MeV) (No gamma)

    n*d + M > fusion or fission (transmutation)Not Detected

    d + t > n(14.01MeV) + 4He(3.5 MeV)

    d + p > 3He + gamma (5.5 MeV)

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    ITER REACTOR

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    How does Cold Fusion Compare to Hot Fusion?

    Cold Fusion

    Occurs in a special solidmaterials at low applied energy.

    Produces mainly helium-4 andheat

    Small, simple devices can makeuseful power

    Involves fusion of deuteriumand many other nuclearreactions.

    Has been studied for 18 yearsand is not understood.

    Is a clean energy source.

    Is difficult to replicate.

    Hot Fusion

    Occurs in a plasma or uponapplication of high energy.

    Produces neutrons, tritium andheat.

    A very large reactor is requiredto make useful power.

    Involves fusion of tritium anddeuterium.

    Has been studied for 60 yearsand is well understood.

    Is a radioactive energy source.

    Is difficult to make useful.

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    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.

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    How is Cold Fusion Initiated? Electrolysis

    Ambient Gas

    Liquid Plasma

    Gas Plasma

    Proton Conductor

    Sonic Implantation

    Laser Light

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    Representation of Calorimeter Methods

    Open method

    Closed method

    Single and Double WallIsoperbolic

    EP = T*C - V*I, (

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    Flow calorimeter schematic(This slide not included in video)

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    Simplified flow calorimeter schematic showing only

    the cooling water in the outer jacket

    (This slide not included in video)

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    How Much Power is Produced?

    Power based on 157 reported studies

    0

    20

    40

    60

    80

    100

    120

    0 40 80 120 160 200

    WATT

    COUNT

    All 157 values, 0.005-183 W, 12.7 W mean

    0

    20

    40

    60

    80

    100

    120

    0 5 10 15 20 25 30 35 40

    BA

    COUNT

    WATT

    All Values

    Electrolytic Values

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    Recent Example of Significant Power using

    Electrolysis(Dardik et al., Energetics Technologies, Israel)

    Pout=KT

    Pinet=Iin*Vin P dis

    Excess Power of up to 34 watts; Average ~20 watts for 17 h

    17 h

    Average Pinet ~0.74 watts

    Average Pout ~20 watts

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    Example of Significant Power using Gas

    Loading(Arata and Zhang, Osaka Univ., Japan)

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    Consistent Patterns for Heat ProductionWhen 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

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    Effect of D/Pd ratio on excess power(McKubre et al., SRI)

    0

    1

    2

    3

    4

    5

    6

    0.88 0.89 0.90 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98

    Atomic ratio (D/Pd)

    C1 Parabolic 0.875 555

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    Summary of Heat Production

    (McKubre et al., SRI)

    D/Pd = 0.70 to 0.90 17 no heat

    D/Pd = 0.90 to 0.95 9 no heat, 6 heat

    D/Pd = 0.95 to 1.05 15 heat

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    Example of a Consistent Pattern

    (McKubre et al., SRI)

    Light & Heavy Water Cells; Excess Power vs. Current Density

    -0.1

    0.0

    0.1

    0.2

    0.3

    0.4

    0.5

    0.6

    0.1 0.2 0.3 0.4 0.5 0.6

    Electrochemical Current Density (A/cm 2)

    P14/D2O Linear P13/H2O

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    Effect of Current on Excess Power

    0.70.60.50.40.30.20.10.00.0

    0.2

    0.4

    0.6

    0.8

    1.0

    1.2

    1.4

    1.6

    1.8

    2.0

    Storms (plate, T)

    Kainthla et al. (wire, T)

    McKubre et al., (wire, flow)

    Kunimatsu (rod, T)

    Aoki et al. (plate, flow)

    Bush (plate, T)

    CURRENT DENSITY, A/cm2

    Constant TemperatureVariable Temperature

    EFFECT OF CURRENT DENSITY ON EXCESS POWER

    Produced by Palladium

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    Effect of Applied Current on Excess Power

    160014001200100080060040020000.0001

    .001

    .01

    .1

    1

    10

    100

    Miles et al.

    Liaw et al.

    Hutchinson et al.

    Kainthla et al.

    Fleischmann and Pons

    Oriani et al.

    Appleby et al.

    Santhanam et al.

    Noninski and Noninski

    Lewis and Skold

    Guruswamy and Wadsworth

    Eagleton and Bush

    CURRENT (mA/cm )

    2

    2

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    Example of relationship between helium and heat

    production

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    Growth of heat and helium(McKubre et al., SRI)

    0

    1

    2

    3

    4

    5

    6

    7

    8

    9

    0 5 10 15 20

    Time (Days)

    0

    20

    40

    60

    80

    100

    120

    140

    160

    180

    ExcessEnergy

    (kJ)

    ppmV SC2

    3 line fit for 4He

    Differential

    Gradient

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    Grown of Tritium during Electrolysis

    (Bockris et. al, Texas A&M)

    1501005000

    5000

    10000

    15000

    20000

    Active Cell

    Inactive Cell

    TIME, hr

    Current Increase

    10^8 atoms/sec

    10^7 atoms/sec

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    Growth of Tritium during Gas Discharge

    (Claytor et al., LANL)

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    Extra Elements Commonly Found

    0

    1

    2

    3

    4

    5

    6

    7

    8

    2 10 18 26 34 42 50 58 66 74 82

    Pd cathode, electrolysis

    NUMBER

    OCCASION

    SREPORTED

    ATOMIC NUMBER

    Al

    Cu Zn

    Ag

    Pb

    Fe

    Ti

    Mg

    Pd Pt

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    Iwamura et al., Mitsubishi, Japan

    Caused the following reactions by passing D2through 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.).

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    Active Region(Iwamura et al., Mitsubishi Heavy Industries, Japan)

    D2 gas permeation through the Pd complex

    Pd

    CaO

    Pd

    Pd

    CaO

    Pd

    Pd

    CaO

    Pd

    Pd substrate

    CaO

    Pd 40nm

    2nm

    D Flux

    Vacuum

    D2 gas

    D PermeationSurface

    Cross Section of Pd Complex

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    Decrease of Cs and Emergence of Pr

    (Iwamura et al., Mitsubishi Heavy Industries, Japan)

    0 20 40 60 80 100 1200

    2

    4

    6

    8

    1012

    14

    16Cs 1st

    Pr 1st

    Cs 2nd

    Pr 2nd

    Numberofatoms(/1

    014

    /cm

    2)

    Time (h)

    DCs

    PdCaO

    Pd

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    Conversion of Cs to Pr

    (Iwamura et al., Mitsubishi Heavy Industries, Japan)

    Average D2 gas permeation rate:FLSCCM)

    Electrochemical Addition

    Ion Implantation

    0

    20

    40

    60

    80

    100

    0.00 1.00 2.00 3.00 4.00

    Conversionrate(%)

    (1/SCCM)3.0

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    Production of Pr from Cs

    (Iwamura et al., Mitsubishi Heavy Industries, Japan)

    6

    5

    4

    3

    2

    1

    654321

    6

    5

    4

    3

    2

    1

    654321X

    Y

    100m

    100m

    Much Pr detection

    Pr detection

    No Pr

    13-4

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    Mass Correlation between Given and Detected

    Elements

    (Iwamura et al., Mitsubishi Heavy Industries, Japan)

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    Why is LENR Hard to Replicate?

    Active material is present only in very smallregions,

    Active material is a complex alloy with an

    unknown structure and composition, Active material rarely forms.

    (The challenge is to make the alloy onpurpose and in large amount.)

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    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 ofapparatus.

    * Significant radiation is detected inside ofapparatus along with the nuclear products.

    * Nuclear reactions can occur at rates that

    make significant heat energy.* Nuclear reactions can be initiated many

    different ways.

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    Requirements of a Theory

    The nuclear charge of the deuteron or proton mustbe lowered, but not by neutron formation.

    The deuterons must form a cluster of 2, 4, or 6 dafter this lowering process has occurred.

    The reactions must produce very little gammaradiation and few radioactive isotopes.

    The process must occur spontaneously withoutsignificant energy being applied.

    The mechanism must occur only under veryspecial conditions in unique solid materials.

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    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.

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    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