Editing Joule–Thomson effect (section)

==Proof that the specific enthalpy remains constant==
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In thermodynamics so-called "specific" quantities are quantities per unit mass (kg) and are denoted by lower-case characters. So ''h'', ''u'', and ''v'' are the [[enthalpy#Specific enthalpy|specific enthalpy]], specific internal energy, and specific volume (volume per unit mass, or reciprocal density), respectively. In a Joule–Thomson process the specific [[enthalpy]] ''h'' remains constant.<ref>See e.g. M.J. Moran and H.N. Shapiro "Fundamentals of Engineering Thermodynamics" 5th Edition (2006) John Wiley & Sons, Inc. page 147</ref> To prove this, the first step is to compute the net work done when a mass ''m'' of the gas moves through the plug. This amount of gas has a volume of ''V''<sub>1</sub> = ''m'' ''v''<sub>1</sub> in the region at pressure ''P''<sub>1</sub> (region 1) and a volume ''V''<sub>2</sub> = ''m'' ''v''<sub>2</sub> when in the region at pressure ''P''<sub>2</sub> (region 2). Then in region 1, the "flow work" done ''on'' the amount of gas by the rest of the gas is: W<sub>1</sub> = ''m'' ''P''<sub>1</sub>''v''<sub>1</sub>. In region 2, the work done ''by'' the amount of gas on the rest of the gas is: W<sub>2</sub> = ''m'' ''P''<sub>2</sub>''v''<sub>2</sub>. So, the total work done ''on'' the mass ''m'' of gas is
:<math>W = mP_1 v_1 - mP_2 v_2.</math>

The change in internal energy minus the total work done ''on'' the amount of gas is, by the [[first law of thermodynamics]], the total heat supplied to the amount of gas.
:<math> U - W = Q </math>

In the Joule–Thomson process, the gas is insulated, so no heat is absorbed. This means that
:<math>\begin{align}
     (mu_2 - mu_1) &- (mP_1 v_1 - mP_2 v_2) = 0 \\
   mu_1 + mP_1 v_1 &= mu_2 + mP_2 v_2 \\
     u_1 + P_1 v_1 &= u_2 + P_2 v_2
\end{align}</math>

where ''u''<sub>1</sub> and ''u''<sub>2</sub> denote the specific internal energies of the gas in regions 1 and 2, respectively. Using the definition of the specific enthalpy ''h = u + Pv'', the above equation implies that
:<math>h_1 = h_2</math>

where h<sub>1</sub> and ''h''<sub>2</sub> denote the specific enthalpies of the amount of gas in regions 1 and 2, respectively.