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freddie_mclair
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- TL;DR Summary
- Calculating the ideal ignition temperature for a D-T fusion reaction
I've been trying to calculate the ideal ignition temperature for a 50-50% Deuterium-Tritium (D-T) reaction. In the literature this value is ##4.4##keV and I'm getting ##5.2##keV. Here's how I'm carrying out my calculations.
This value can be calculated by making the alpha particle heating, ##S_{\alpha}=(1/4)n^2 E_{\alpha}\langle \sigma v \rangle##, equal to the Bremsstrahlung losses, ##S_B=C_B n^2 \sqrt{T}##:
$$S_{\alpha}=S_B \Leftrightarrow E_{\alpha}\langle \sigma v \rangle-4C_B \sqrt{T}=0$$
The value I get for ##C_B=(\sqrt{2}/3\pi^{5/2})(e^6/(\varepsilon_0^3 c^3 h m_e^{3/2})## is ##4.22\cdot10^{-29}## ##\sqrt{kg}\, m^4\, s^{-2}##, where ##e## is the electron charge, ##\varepsilon_0## the vacuum permittivity, ##h## Planck's constant and ##m_e## the electron mass. The alpha particle energy is ##E_{\alpha}=3.5##keV. For the reactivity, ##\langle \sigma v \rangle##, which is a function of ##T##, I have been using an equation derived analytically from "Plasma Physics and Fusion Energy book by J.P. Freidberg, 2007":
$$ \langle \sigma v \rangle = \frac{4 \sigma_m}{\sqrt{3}}\sqrt{\frac{2T_m}{m_r}} \left(\frac{T_m}{T}\right)^{2/3}e^{-3(T_m/T)^{1/3}+2} $$
Where ##\sigma_m = 5.03##barns and ##T_m = 296##keV. ##m_r## is the reduced mass of the deuterium and tritium, which is given by ##m_r=(m_D m_T)/(m_D + m_T)##, where ##m_D=2.0141u##, ##m_T=3.01605u## and ##u=1.660\cdot 10^{-28}##kg.
Here are all elements for calculating the ideal ignition temperature, but I feel that I'm messing up somewhere the conversions from keV to Joule or Kelvin.
Can someone help me to troubleshoot what I'm doing wrong? Or, do you indeed get the same value as me?
Thanks in advance!
This value can be calculated by making the alpha particle heating, ##S_{\alpha}=(1/4)n^2 E_{\alpha}\langle \sigma v \rangle##, equal to the Bremsstrahlung losses, ##S_B=C_B n^2 \sqrt{T}##:
$$S_{\alpha}=S_B \Leftrightarrow E_{\alpha}\langle \sigma v \rangle-4C_B \sqrt{T}=0$$
The value I get for ##C_B=(\sqrt{2}/3\pi^{5/2})(e^6/(\varepsilon_0^3 c^3 h m_e^{3/2})## is ##4.22\cdot10^{-29}## ##\sqrt{kg}\, m^4\, s^{-2}##, where ##e## is the electron charge, ##\varepsilon_0## the vacuum permittivity, ##h## Planck's constant and ##m_e## the electron mass. The alpha particle energy is ##E_{\alpha}=3.5##keV. For the reactivity, ##\langle \sigma v \rangle##, which is a function of ##T##, I have been using an equation derived analytically from "Plasma Physics and Fusion Energy book by J.P. Freidberg, 2007":
$$ \langle \sigma v \rangle = \frac{4 \sigma_m}{\sqrt{3}}\sqrt{\frac{2T_m}{m_r}} \left(\frac{T_m}{T}\right)^{2/3}e^{-3(T_m/T)^{1/3}+2} $$
Where ##\sigma_m = 5.03##barns and ##T_m = 296##keV. ##m_r## is the reduced mass of the deuterium and tritium, which is given by ##m_r=(m_D m_T)/(m_D + m_T)##, where ##m_D=2.0141u##, ##m_T=3.01605u## and ##u=1.660\cdot 10^{-28}##kg.
Here are all elements for calculating the ideal ignition temperature, but I feel that I'm messing up somewhere the conversions from keV to Joule or Kelvin.
Can someone help me to troubleshoot what I'm doing wrong? Or, do you indeed get the same value as me?
Thanks in advance!
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