6. Entropy and the Second Law of Thermodynamics

# 6.7 Specific entropy of a state

## 6.7.1 Determining the specific entropy of pure substances by using thermodynamic tables

The specific entropy of a pure substance can be found from thermodynamic tables if the tables are available. The procedures are explained in Section 2.4.  In addition to the $P-v$ and $T-v$ diagrams, the $T-s$ diagram is commonly used to illustrate the relation between temperature and specific entropy of a pure substance. Figure 6.7.1 shows the $T-s$ diagram for water.

Example 1

Fill in the table.

 Substance T, oC P, kPa v, m3/kg Quality x s, kJ/kg-K Phase Water 250 0.02 R134a -2 100

Solution:

Water: T = 250 oC, $v$ = 0.2 m3/kg

From Table A1: T = 250 oC, $v_f$  = 0.001252 m3/kg, $v_g$  = 0.050083 m3/kg

Since $v_f < v < v_g$, water at the given state is a two phase mixture; the saturation pressure is Psat = 3976.17 kPa, and $s_f$ = 2.7935 kJ/kgK, $s_g$ = 6.0721 kJ/kgK

The quality is

$x = \dfrac{v - v_f}{v_{g} - v_{f}} = \dfrac{0.02 - 0.001252}{0.050083 - 0.001252} = 0.383936$

The specific entropy is

\begin{align*} s &= s_f + x(s_g - s_f) \\&= 2.7935 + 0.383936 \times (6.0721 -2.7935) = 4.0523 \ \rm{kJ/kgK} \end{align*}

R134a: T = -2 oC, P = 100 kPa

From Table C1: by examining the saturation pressures at 0 oC and – 5 oC, we can estimate that the saturation pressure for T = -2 oC is about 270 kPa; therefore, R134a at the given state is a superheated vapour.

From Table C2,

P = 100 kPa, T = -10 oC, $v$= 0.207433 m3/ kg, $s$ = 1.7986 kJ/kgK

P = 100 kPa, T = 0 oC, $v$= 0.216303 m3/ kg, $s$ = 1.8288 kJ/kgK

Use linear interpolation to find $v$ and $s$ at T = -2 oC.

$\because \dfrac{v - 0.207433}{0.2160303 - 0.207433} = \dfrac{s - 1.7986}{1.8288 - 1.7986} = \dfrac{-2 - (-10)}{0 - (-10)}$

$\therefore v = 0.214529 \ \rm{m^3/kg}$     and      $s = 1.8228 \ \rm{kJ/kgK}$

In summary,

 Substance T oC P kPa v m3/kg Quality x s kJ/kg-K Phase Water 250 3976.17 0.02 0.383936 4.0523 two-phase R134a -2 100 0.214529 n.a. 1.8228 superheated vapour

Example 2

A rigid tank contains 3 kg of R134a initially at 0oC, 200 kPa. R134a is now cooled until its temperature drops to -20oC. Determine the change in entropy, $\Delta S$, of R134a during this process. Is $\Delta S=S_{gen}$?

Solution:

The initial state is at T1 = 0oC and P1 =200 kPa. From Table C2 in Appendix C,

$s_1$ = 1.7654 kJ/kgK, $v_1$ = 0.104811 m3/kg

The tank is rigid; therefore, $v_2$ = $v_1$ = 0.104811 m3/kg.

From Table C1, at T2 = -20oC:

$v_f$ = 0.000736 m3/kg,    $v_g$ = 0.147395 m3/kg

$s_f$ = 0.9002 kJ/kgK,        $s_g$ = 1.7413 kJ/kgK

Because $v_f < v_2 < v_g$, the final state is a two-phase mixture.

$x_2 = \dfrac{v_2 - v_f}{v_g-v_f} = \dfrac{0.104811 - 0.000736}{0.147395-0.000736}=0.70964$

\begin{align*} s_2 &= s_f + x_2 (s_g-s_f) \\&= 0.9002 + 0.70964 \times (1.7413-0.9002)=1.4971 \ \rm{kJ/kgK} \end{align*}

The total entropy change is

$\Delta S = m (s_2 - s_1) = 3 \times (1.4971 - 1.7654) = - 0.8049 \ \rm{kJ/K}$

It is important to note that $\Delta S \neq S_{gen}$ in general. The total entropy of R134a decreases in this cooling process, but the entropy generation is always greater than zero in a real process.

## 6.7.2 Determining the specific entropy of solids and liquids

The specific entropy of a solid or a liquid depends mainly on the temperature. The change of specific entropy in a process from states 1 to 2 can be calculated as,

$s_2-s_1=C_pln\dfrac{T_2}{T_1}$

where

$s$: specific entropy, in kJ/kgK

$C_p$: specific heat, in kJ/kgK. Note that for solids and liquids, $C_p=C_v$. Table G2 and Table G3 list the specific heats of selected solids and liquids, respectively.

$T$: absolute temperature, in Kelvin

## 6.7.3 Determining the specific entropy of ideal gases

The specific entropy of an ideal gas is a function of both temperature and pressure. Here we will introduce a simplified method for calculating the change of the specific entropy of an ideal gas in a process by assuming constant specific heats. This method is reasonably accurate for a process undergoing a small temperature change.

$s_2-s_1=C_pln\displaystyle\frac{T_2}{T_1}-Rln\frac{P_2}{P_1}$

$s_2-s_1=C_vln\displaystyle\frac{T_2}{T_1}+Rln\frac{v_2}{v_1}$

where

$C_p$, $C_v$ and $R$ are the constant-pressure specific heat, constant-volume specific heat, and gas constant, respectively, in kJ/kgK. Table G1 lists these properties of selected ideal gases.

$T$: absolute temperature, in Kelvin

$P$: pressure, in kPa

$s$: specific entropy, in kJ/kgK

$v$: specific volume, in m3/kg

Example 3

Air is compressed from an initial state of 100 kPa, 27oC to a final state of 600 kPa, 67oC. Treat air as an ideal gas. Calculate the change of specific entropy, $\Delta s$, in this process. Is $\Delta s = s_{gen}$?

Solution:

From Table G1: Cp = 1.005 kJ/kgK, R = 0.287 kJ/kgK

\begin{align*} \Delta s &= s_2 - s_1 = C_pln\displaystyle\frac{T_2}{T_1}-Rln\frac{P_2}{P_1} \\&= 1.005 ln\dfrac{273.15 + 67}{273.15 + 27} - 0.287ln\dfrac{600}{100} = -0.3885 \ \rm{kJ/kgK} \end{align*}

It is important to note that $\Delta s \neq s_{gen}$ in general. The specific entropy decreases in this process, but the rate of entropy generation is always greater than zero in a real process.

## 6.7.4 Isentropic relations for an ideal gas

If a process is reversible and adiabatic, it is called an isentropic process and its entropy remains constant. An isentropic process is an idealized process. It is commonly used as a basis for evaluating real processes. The concept of isentropic applies to all substances including ideal gases. The following isentropic relations, however, are ONLY valid for ideal gases.

$Pv^k= \rm{constant}$   and    $\displaystyle\frac{P_2}{P_1}=\displaystyle\left(\displaystyle\frac{v_1}{v_2}\right)^k=\left(\frac{T_2}{T_1}\right)^{k/(k-1)}$

where

$k=\dfrac{C_p}{C_v}$: specific heat ratio. The $k$ values of selected ideal gases can be found in Table G1.

$T$: absolute temperature, in Kelvin

$P$: pressure, in kPa

$v$: specific volume, in m3/kg

It is noted that the isentropic relation $Pv^k = \rm{constant}$ for ideal gases is actually a special case of the polytropic relation $Pv^n = \rm{constant}$  with $n = k = \dfrac{C_p}{C_v}$.

Example 4

Derive the isentropic relation $Pv^k= \rm{constant}$

Solution:

For an ideal gas undergoing an isentropic process,

$\Delta s = s_2 - s_1 =C_pln\displaystyle\frac{T_2}{T_1}-Rln\frac{P_2}{P_1}=0$

Substitute $C_p = \dfrac{kR}{k-1}$ in the above equation and rearrange,

$\because \dfrac{k}{k-1}ln\dfrac{T_2}{T_1} = ln\dfrac{P_2}{P_1}$

$\therefore ln\left(\dfrac{T_2}{T_1}\right)^{\dfrac{k}{k-1}} = ln\dfrac{P_2}{P_1}$

$\therefore \dfrac{P_2}{P_1} = \left(\dfrac{T_2}{T_1}\right)^{\dfrac{k}{k-1}}$

Combine with the ideal gas law, $Pv = RT$,

$\therefore \dfrac{P_2}{P_1} = \left(\dfrac{T_2}{T_1}\right)^{\dfrac{k}{k-1}} = \left(\dfrac{P_{2}v_{2}}{P_{1}v_{1}}\right)^{\dfrac{k}{k-1}}$

$\therefore \dfrac{P_2}{P_1} = \left(\dfrac{v_1}{v_2}\right)^k$

$\therefore Pv^k = \rm{constant}$   and   $\dfrac{P_2}{P_1} = \left(\dfrac{v_1}{v_2}\right)^k = \left(\dfrac{T_2}{T_1}\right)^{\dfrac{k}{k-1}}$

Practice Problems