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Deriving the kinetic energy flux in an effusion process
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[QUOTE="TSny, post: 6861630, member: 229090"] This doesn't look right. The rate of effusion of particles of a certain momentum should be proportional to ##p_x## rather than ##|\vec p|##. Suppose we define a distribution function ##n(p_x, p_y)## such that the number of particles per unit area which have ##x## and ##y## components of momentum in the range ##p_x## to ##p_x+dp_x## and ##p_y## to ##p_y+dp_y## is $$n(p_x,p_y)dp_xdp_y.$$ You can express ##n(p_x, p_y)## in terms of ##N##, ##A##, and the Maxwell-Boltzmann distribution. Show that the number of particles with momentum in this range that effuse through the opening per unit time is $$d\Phi_{particles} =l \left(\frac{p_x}{m}\right)\left[n(p_x,p_y)dp_xdp_y\right]$$ Then show that the flux of energy through the opening carried by these particles is $$d\Phi_\epsilon= l \left(\frac{p_x}{m}\right)\epsilon(p_x, p_y)\left[n(p_x,p_y)dp_xdp_y\right]$$ where ##\epsilon(p_x, p_y)## is the energy of a particle with momentum components ##p_x## and ##p_y##. The total flux of energy through the opening is then ##\Phi_\epsilon = \int d\Phi_\epsilon##. [/QUOTE]
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Deriving the kinetic energy flux in an effusion process
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