? asked in Science & MathematicsPhysics · 1 decade ago

What change does the enthalpy undergo during Joule-Thompson throttling?

1 Answer

  • Anonymous
    1 decade ago
    Favorite Answer

    enthalpy remains constant through this process.

    0 = Qcv - Wcv +m[ (h1-h2) + (V1*V1 - V2*V2)/2 + g(z1 -z2) ]

    for a control volume enclosing a throttling device, the only work is flow work at locations where mass enters or exits the CV. => Wcv=0

    assume no heat transfer => Qcv=0

    assume change in specific kinetic energy is negligible

    so that, h2 = h1 for p2 <p1

    these assumptions are all included in the joule-thompson throttling process

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