6.7 Specific entropy of a state

Claire Yu Yan

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 P_sat = 3976.17 kPa, and $s_{f}$ = 2.7935 kJ/kgK, $s_{g}$ = 6.0721 kJ/kgK

The quality is

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

The specific entropy is

$\begin{aligned} s & = s_{f} + x (s_{g} - s_{f}) \\ = 2.7935 + 0.383936 \times (6.0721 - 2.7935) = 4.0523 k J / k g K \end{aligned}$

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.

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

$∴ v = 0.214529 m^{3} / k g$ and $s = 1.8228 k J / k g K$

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 0^oC, 200 kPa. R134a is now cooled until its temperature drops to -20^oC. Determine the change in entropy, $Δ S$ , of R134a during this process. Is $Δ S = S_{g e n}$ ?

Solution:

The initial state is at T₁ = 0^oC 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 = -20^oC:

$v_{f}$ = 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.

$x_{2} = \frac{v_{2} - v_{f}}{v_{g} - v_{f}} = \frac{0.104811 - 0.000736}{0.147395 - 0.000736} = 0.70964$

$\begin{aligned} s_{2} & = s_{f} + x_{2} (s_{g} - s_{f}) \\ = 0.9002 + 0.70964 \times (1.7413 - 0.9002) = 1.4971 k J / k g K \end{aligned}$

The total entropy change is

$Δ S = m (s_{2} - s_{1}) = 3 \times (1.4971 - 1.7654) = - 0.8049 k J / K$

It is important to note that $Δ S \neq S_{g e n}$ 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,

$s_{2} - s_{1} = C_{p} l n \frac{T_{2}}{T_{1}}$

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.

$s_{2} - s_{1} = C_{p} l n \frac{T_{2}}{T_{1}} - R l n \frac{P_{2}}{P_{1}}$

$s_{2} - s_{1} = C_{v} l n \frac{T_{2}}{T_{1}} + R l n \frac{v_{2}}{v_{1}}$

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, 27^oC to a final state of 600 kPa, 67^oC. Treat air as an ideal gas. Calculate the change of specific entropy, $Δ s$ , in this process. Is $Δ s = s_{g e n}$ ?

Solution:

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

$\begin{aligned} Δ s & = s_{2} - s_{1} = C_{p} l n \frac{T_{2}}{T_{1}} - R l n \frac{P_{2}}{P_{1}} \\ = 1.005 l n \frac{273.15 + 67}{273.15 + 27} - 0.287 l n \frac{600}{100} = - 0.3885 k J / k g K \end{aligned}$

It is important to note that $Δ s \neq s_{g e n}$ 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.

$P v^{k} = c o n s t a n t$ and $\frac{P_{2}}{P_{1}} = {(\frac{v_{1}}{v_{2}})}^{k} = {(\frac{T_{2}}{T_{1}})}^{k / (k - 1)}$

where

$k = \frac{C_{p}}{C_{v}}$ : 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 $P v^{k} = c o n s t a n t$ for ideal gases is actually a special case of the polytropic relation $P v^{n} = c o n s t a n t$ with $n = k = \frac{C_{p}}{C_{v}}$ .

Example 4

Derive the isentropic relation $P v^{k} = c o n s t a n t$

Solution:

For an ideal gas undergoing an isentropic process,

$Δ s = s_{2} - s_{1} = C_{p} l n \frac{T_{2}}{T_{1}} - R l n \frac{P_{2}}{P_{1}} = 0$

Substitute $C_{p} = \frac{k R}{k - 1}$ in the above equation and rearrange,

$∵ \frac{k}{k - 1} l n \frac{T_{2}}{T_{1}} = l n \frac{P_{2}}{P_{1}}$

$∴ l n {(\frac{T_{2}}{T_{1}})}^{\frac{k}{k - 1}} = l n \frac{P_{2}}{P_{1}}$

$∴ \frac{P_{2}}{P_{1}} = {(\frac{T_{2}}{T_{1}})}^{\frac{k}{k - 1}}$

Combine with the ideal gas law, $P v = R T$ ,

$∴ \frac{P_{2}}{P_{1}} = {(\frac{T_{2}}{T_{1}})}^{\frac{k}{k - 1}} = {(\frac{P_{2} v_{2}}{P_{1} v_{1}})}^{\frac{k}{k - 1}}$

$∴ \frac{P_{2}}{P_{1}} = {(\frac{v_{1}}{v_{2}})}^{k}$

$∴ P v^{k} = c o n s t a n t$ and $\frac{P_{2}}{P_{1}} = {(\frac{v_{1}}{v_{2}})}^{k} = {(\frac{T_{2}}{T_{1}})}^{\frac{k}{k - 1}}$

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Introduction to Engineering Thermodynamics Copyright © 2022 by Claire Yu Yan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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

6.7.2 Determining the specific entropy of solids and liquids

6.7.3 Determining the specific entropy of ideal gases

6.7.4 Isentropic relations for an ideal gas

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