From 1b0394a3222751ac76c00d56db43bc079737e935 Mon Sep 17 00:00:00 2001 From: c2v2zw4k Date: Mon, 22 May 2023 03:14:04 +0000 Subject: [PATCH] Upload files to 'assets/docs/nuclear/equations' --- assets/docs/nuclear/equations/E-EFF-EPF-MM | 79 ++++++++++++++++++++++ 1 file changed, 79 insertions(+) create mode 100644 assets/docs/nuclear/equations/E-EFF-EPF-MM diff --git a/assets/docs/nuclear/equations/E-EFF-EPF-MM b/assets/docs/nuclear/equations/E-EFF-EPF-MM new file mode 100644 index 0000000..9becb67 --- /dev/null +++ b/assets/docs/nuclear/equations/E-EFF-EPF-MM @@ -0,0 +1,79 @@ +Uranium-235: 200 MeV per fission +Uranium-233: 197 MeV per fission +Plutonium-239: 210 MeV per fission + +The amount of energy needed to reach 15,111 megatons is: +E = 15.111 × 10^6 × 4.184 × 10^12 Joules, where 4.184 × 10^12 Joules +is the conversion of 1 megaton to Joules. + +Now we can calculate the number of fissions needed to +each element, considering the efficiency of 0.147: + +For Uranium-235: +N_fissions_Uranium-235 = E / (Efficiency × Energy per fission) +N_fissions_Uranium-235 = 15.111 × 10^6 × 4.184 × 10^12 / (0.147 × 200) +N_fissions_Uranium-235 ≈ 5,351 × 10^17 fissions + +For Uranium-233: +N_fissions_Uranium-233 = E / (Efficiency × Energy per fission) +N_fissions_Uranium-233 = 15.111 × 10^6 × 4.184 × 10^12 / (0.147 × 197) +N_fissions_Uranium-233 ≈ 5,442 × 10^17 fissions + +For Plutonium-239: +N_fissions_Plutonium-239 = E / (Efficiency × Energy per fission) +N_fissions_Plutonium-239 = 15.111 × 10^6 × 4.184 × 10^12 / (0.147 × 210) +N_fissions_Plutonium-239 ≈ 5057 × 10^17 fissions + +Finally, we can calculate the amount of mass of each element needed, +considering the number of fissions and the molar mass of each element: + +For Uranium-235: +m_Uranium-235 = N_fissions_Uranium-235 × Molar mass of Uranium-235 / Avogadro's number + +The molar mass of Uranium-235 is approximately 235 g/mol. + +For Uranium-233: +m_Uranium-233 = N_fissions_Uranium-233 × Molar mass of Uranium-233 / Avogadro's number + +The molar mass of Uranium-233 is approximately 233 g/mol. + +For Plutonium-239: +m_Plutonium-239 = N_fissions_Plutonium-239 × Molar Mass of Plutonium-239 / Avogadro's Number + +The molar mass of Plutonium-239 is approximately 239 g/mol. + +Now we can calculate the masses: + +For Uranium-235: +m_Uranium-235 = 5.351 × 10^17 × 235 / 6.022 × 10^23 +m_Uranium-235 ≈ 21.04 kg + +For Uranium-233: +m_Uranium-233 = 5.442 × 10^17 × 233 / 6.022 × 10^23 +m_Uranium-233 ≈ 21.43 kg + +For Plutonium-239: +m_Plutonium-239 = 5.057 × 10^17 × 239 / 6.022 × 10^23 +m_Plutonium-239 ≈ 19.97 + +Now we can calculate the required masses of each element +to reach 15,111 megatons of energy: + +For Uranium-235: +m_Uranium-235 = 5.351 × 10^17 × 235 / 6.022 × 10^23 +m_Uranium-235 ≈ 21.04 kg + +For Uranium-233: +m_Uranium-233 = 5.442 × 10^17 × 233 / 6.022 × 10^23 +m_Uranium-233 ≈ 21.43 kg + +For Plutonium-239: +m_Plutonium-239 = 5.057 × 10^17 × 239 / 6.022 × 10^23 +m_Plutonium-239 ≈ 19.97 kg + +Therefore, to reach 15,111 megatons of energy, +It would take approximately: + + 21.04 kg of Uranium-235 + 21.43 kg of Uranium-233 + 19.97 kg of Plutonium-239