9. Entropy Change of Real and Idealized Processes#

During the last chapter we motivated the idea of entropy from a statistical perspective and showed that for any real process the entropy (and multiplicity) is always maximized. This explains why there is a directionality to processes, such as the transfer of heat from hot to cold. And for any real process, there is always entropy generation and thus the entropy of the universe is increasing. Using our statistical interpreation with an assumption that entropy is an extensive thermodynamic property, we were able to derive a thermodynamic definition of entropy that relates the entropy change to the heat transfer divided by temperature. For a reversible process, these quantities are equal, and for an irreversible, or real, process we must also account for the entropy generation. In this chapter, we will build off of our concept of the thermodynamic definition of entropy to help analyze processes, similar to how we used internal energy and enthalpy to analyze processes in earlier chapters.

Recall from last chapter in the thermodynamic discussion of entropy (Section 8.4 Thermodynamic Definition of Entropy), the following equation

(9.1)#\[dS = \frac{1}{T} dU + \frac{p}{T} dV\]

or in terms of specific properties

(9.2)#\[ds = \frac{1}{T} du + \frac{p}{T} dv\]

And upon rearrangement we see that

(9.3)#\[Tds = du + pdv\]

Also, recall that

(9.4)#\[h = u + pv\]

which becomes the following after differentiation

(9.5)#\[dh = du + pdv + vdp\]

And substituting (8.3) into this we also get

(9.6)#\[Tds = dh - vdp\]

And therefore, in addition to (8.2) we get another expression for entropy in terms of enthalpy

(9.7)#\[ds = \frac{1}{T} dh - \frac{v}{T} dp\]

9.1. Entropy Change of Ideal Gases#

From (8.2) and (8.7) we have two equations that can be used to determine entropy via integration with the appropriate equation of state. Perhaps the simplest is the ideal gas equation of state, from which we also prior derived relationships for enthalpy and entropy in terms of temperature only. See Section 6.4 Using Specific Heats with Ideal Gases, where we showed that

(9.8)#\[du = c_{\rm v} dT\]

and

(9.9)#\[dh = c_{\rm p} dT\]

Using these two equations and the ideal gas law (i.e., \(pv=RT\)), we can rearrange (8.2) and (8.7) to show that

(9.10)#\[ds = \frac{c_{\rm v}}{T} dT + \frac{R}{v} dv\]

and

(9.12)#\[ds = \frac{c_{\rm p}}{T} dT - \frac{R}{p} dp\]

which are now expressions that can be readily integrated because \(c_{\rm v}\) and \(c_{\rm p}\) are functions of temperature only for an ideal gas. Therefore

(9.12)#\[s_2-s_1 = \int_{1}^{2} \frac{c_{\rm v} }{T} dT + R ln \frac{v_2}{v_1}\]
(9.13)#\[s_2-s_1 = \int_{1}^{2} \frac{c_{\rm p} }{T} dT - R ln \frac{p_2}{p_1}\]

If it is assumed that specific heat is independent of temperature or linear with temperature, then (8.11) and (8.12) become

(9.14)#\[s_2-s_1 = c_{\rm v} ln \frac{T_2}{T_1} + R ln \frac{v_2}{v_1}\]
(9.15)#\[s_2-s_1 = c_{\rm p} ln \frac{T_2}{T_1} - R ln \frac{p_2}{p_1}\]

where c_{\rm v} or c_{\rm p} are the average values if it is assumed they are linear with temerature and not constant. Exact values for the integral are often tabulated in textbooks as a function of temperature in various forms, understanding that

(9.16)#\[\int_{T_1}^{T_2} \frac{c_{\rm p} }{T} dT = \int_{0}^{T_2} \frac{c_{\rm p} }{T} dT - \int_{0}^{T_1} \frac{c_{\rm p} }{T} dT\]

where the two right hand terms are now functions of a single temperature only and can be readily tabulated. Since we will be using Cantera to solve for property values this is not necessary to discuss in further detail, but in the case where only tables are avalable this is the approach often used. This is necessary because as can be seen from the equations above, entropy is not only a function of temperature, like enthalpy and internal energy and cannot therefore themselves be easily tabulated in a similar manner. Lets see how we can evaluate entropy changes of an ideal gas using specific heats and compare it to the output directly from Cantera.

9.1.1. Example - Evaluating Entropy Change of an Ideal Gas With and Without Specific Heats#

Consider oxygen that is heated from 300 K to 1500 K. Assume that during this process the pressure is decreased from 200 to 150 kPa and that oxygen behaves ideally. Calculate the change in specific entropy, s, using Cantera and with specific heats.

Solution - We know the initial temperature and pressure, so solving with Cantera should be straightforward. Using these two properties we can also solve directly using (8.15). We will obtain the specific heat from Cantera, assuming it is contstant and at 300 K.

import cantera as ct
#State 1
T1 = 300 #K
P1 = 200*1000 # Pa

species1 = ct.Oxygen()# define state 1
species1.TP = T1, P1

#State 2
T2= 1500 #K
P2 = 150*1000 # Pa

species2 = ct.Oxygen()# define state 2
species2.TP = T2, P2

#Using Cantera directly

ds=species2.SP[0]-species1.SP[0]

print(round(ds, 2), "J/kg using Cantera")
1724.55 J/kg using Cantera
import numpy as np
species1 = ct.Oxygen()
cp = species1.cp # cp at 300 K, the default temperature
ds = cp*np.log(T2/T1)-8.314/32*np.log(P2/P1)
print(round(ds, 2), "J/kg using specific heat equation")
1480.69 J/kg using specific heat equation

9.2. Isentropic Processes#