Methane is cooled from a temperature of 128 degrees
Fahrenheit to 90 degrees Fahrenheit at a pressure of 1346 psia. Calculate the change in specific entropy
using generalized ‘entropy correction due to lack of ideal behavior’ chart.
Critical pressure of methane is 673 psia, critical temperature
is 344 degrees Rankin.
Solution:
Entropy correction is plotted as a group (S*
minus S) versus reduced pressure and reduced temperature. For our situation, initial condition
corresponds to reduced temperature of 2 and reduced pressure of 2. Using Figure 107 of Hougen
and Watson, this correction is found to be 0.53 calories per mole per degree
Kelvin. At final condition, reduced temperature
changes to 1.6, and associated correction is read as 0.9 calories per mole per
degree Kelvin.
Entropy change for an isobaric process is given as sum of
two terms. First one corresponds to
ideal gas, and second one is correction term.
Please note that for ideal gases entropy change for a
isobaric process is given by integral T1 to T2 of Cp
over T dT. We
know that CP is given in the format of constant a plus bT Entropy change for an ideal gas reduces to a
log (T2 over T1) plus b time (T2 minus T1). This value is found to be minus 1.637 Btus per pound mole per degree Rankine.
Plugging in the values for second group, entropy change for
real gas is (0.9 minus 0.53) or 0.37 calories per mole per degree Kelvin or 0.37
Btus per pound mole per degree Rankin. Entropy change is (minus 1.637) minus 0.37 or
minus 2.0 Btus per pound mole per degree Rankin.