What change does the enthalpy undergo during Joule-Thompson throttling?
- Anonymous1 decade agoFavorite 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