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11.22 Confinement of positive ions and electrons with a static electric and magnetic field

Some speed - temperature - voltage calculations

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 Mean speed of D2 molecules at 293 º K: ½ . m . v2  = 3/2 . k . T  (see: hyperphysics kinetic temperature ) k = Bolzmann constant = 1,38 E-23 J/K      ( k = R/Na,  R=8,31 Nm/mol, K    Na = 6,0 E23) m = massa D2 molecule =  2. 2.  1,67E-27 = 6,68E-27 kg v2 =  3/2 . 1,38 E-23 .293 / ( ½ . 6,68E-27 ) v  = 1,35 . 103 m/s -> kinetic energy = ½ . 6,68E-27 . (1,35 . 103 ) 2 = 6,09E-21 J => 6,09E-21 /(1,6E-19) = 0,038 eV v = 3. 104 m/s  of D+ ions. m(D+ ion) = 2. 1,67. 10-27 kg = 3,34 . 10-27 kg ½ . m . v2  = 3/2 . k . T -> T = ⅓ . m . v2 / k = ⅓ . 3,34 . 10-27 . (3.104 )2 / (1,38 . 10-23)  = 7,26 . 104 K ½ . m . v2 = ΔV. q    (ΔV = voltage difference needed to accelerate the ions; q = charge of a ion = 1,6 . 10-19 C ) ΔV = ½ . m . v2 / q = ½ . 3,34 . 10-27 . ( 3. 104 )2 / (1,6 . 10-19 ) =  9,4 V v = 3. 104 m/s  of electrons m(electron) = 9,1. 10-31 kg ΔV = ½ . m . v2 / q = ½ . 9,1. 10-31  . ( 3. 104 )2 / (1,6 . 10-19 ) = 0,0026 V v = 3. 105 m/s  of D+ ions. -> T = ⅓ . m . v2 / k = ⅓ . 3,34 . 10-27 . (3.105 )2 / (1,38 . 10-23)  = 7,26 . 106 K ½ . m . v2 = ΔV. q  ΔV = ½ . m . v2 / q = ½ . 3,34 . 10-27 . ( 3. 105 )2 / (1,6 . 10-19 ) =  900,4 V > kinetic energy = ½ . 3,34E-27 . (3 . 105 )2 = 1,5E-16 J => 1,5E-16 /(1,6E-19) = 939 eV v = 3. 105 m/s  of electrons ΔV = ½ . m . v2 / q = ½ . 9,1. 10-31  . ( 3. 105 )2 / (1,6 . 10-19 )  = 0,26 V v = 3. 106 m/s  of D+ ions. -> T = ⅓ . m . v2 / k = ⅓ . 3,34 . 10-27 . (3.106 )2 / (1,38 . 10-23)  = 7,26 . 108 K ½ . m . v2 = ΔV. q  ΔV = ½ . m . v2 / q = ½ . 3,34 . 10-27 . ( 3. 106 )2 / (1,6 . 10-19 ) =  90kV > kinetic energy = ½ . 3,34E-27 . (3 . 106 )2 = 1,5E-14 J => 1,5E-14 /(1,6E-19) = 93,9 keV

20 February 2014 ..     by  Rinze Joustra        www.valgetal.com