{"id":1837,"date":"2021-07-27T20:21:57","date_gmt":"2021-07-28T00:21:57","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/thermo1\/chapter\/6-7-determining-the-specific-entropy-of-a-state\/"},"modified":"2022-08-11T20:19:51","modified_gmt":"2022-08-12T00:19:51","slug":"6-7-determining-the-specific-entropy-of-a-state","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/thermo1\/chapter\/6-7-determining-the-specific-entropy-of-a-state\/","title":{"raw":"6.7 Specific entropy of a state","rendered":"6.7 Specific entropy of a state"},"content":{"raw":"
\r\n

6.7.1 Determining the specific entropy of pure substances by using thermodynamic tables<\/h2>\r\n

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<\/a>.\u00a0 In addition to the [latex]P-v[\/latex] and [latex]T-v[\/latex] diagrams, the [latex]T-s[\/latex] diagram is commonly used to illustrate the relation between temperature and specific entropy of a pure substance. Figure 6.7.1 <\/a>shows the [latex]T-s[\/latex] diagram <\/a> for water.<\/p>\r\n \r\n\r\n<\/div>\r\n\r\n[caption id=\"attachment_3365\" align=\"aligncenter\" width=\"600\"]\"T-s<\/a> Figure 6.7.1<\/strong> T-s diagram for water<\/em>[\/caption]\r\n\r\n

\r\n
\r\n

Example 1<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nFill in the table.\r\n\r\n\r\n\r\n\r\n\r\n
Substance<\/td>\r\nT, o<\/sup>C<\/td>\r\nP, kPa<\/td>\r\nv, m3<\/sup>\/kg<\/td>\r\nQuality x<\/em><\/td>\r\ns, kJ\/kg-K<\/td>\r\nPhase<\/td>\r\n<\/tr>\r\n
Water<\/td>\r\n250<\/td>\r\n<\/td>\r\n0.02<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n
R134a<\/td>\r\n-2<\/td>\r\n100<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSolution:<\/span><\/em>\r\n\r\nWater: T = 250 o<\/sup>C, [latex] v [\/latex] = 0.2 m3<\/sup>\/kg\r\n\r\nFrom Table A1<\/a>: T<\/em> = 250 o<\/sup>C, [latex] v_f [\/latex]\u00a0 = 0.001252 m3<\/sup>\/kg, [latex] v_g [\/latex]\u00a0 = 0.050083 m3<\/sup>\/kg\r\n\r\nSince [latex] v_f\u00a0 < v <\u00a0 v_g [\/latex], water at the given state is a two phase mixture; the saturation pressure is P<\/em>sa<\/em>t<\/sub> = 3976.17 kPa, and [latex]s_f[\/latex] = 2.7935 kJ\/kgK,\u00a0[latex]s_g[\/latex] = 6.0721 kJ\/kgK\r\n\r\nThe quality is\r\n

[latex] x = \\dfrac{v - v_f}{v_{g} - v_{f}} = \\dfrac{0.02 - 0.001252}{0.050083 - 0.001252} = 0.383936 [\/latex]<\/p>\r\nThe specific entropy is\r\n

[latex] \\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*}[\/latex]<\/p>\r\n \r\n\r\nR134a: T<\/em> = -2 o<\/sup>C, P<\/em> = 100 kPa\r\n\r\nFrom Table C1<\/a>: by examining the saturation pressures at 0 o<\/sup>C and - 5 o<\/sup>C, we can estimate that the saturation pressure for T<\/em> = -2 o<\/sup>C is about 270 kPa; therefore, R134a at the given state is a superheated vapour.\r\n\r\nFrom Table C2<\/a>,\r\n

P<\/em> = 100 kPa, T<\/em> = -10 o<\/sup>C, [latex] v [\/latex]= 0.207433 m3<\/sup>\/ kg, [latex]s[\/latex]<\/em> = 1.7986 kJ\/kgK<\/p>\r\n

P<\/em> = 100 kPa, T<\/em> = 0 o<\/sup>C, [latex] v [\/latex]= 0.216303 m3<\/sup>\/ kg, [latex]s[\/latex]<\/em> = 1.8288 kJ\/kgK<\/p>\r\nUse linear interpolation to find [latex] v [\/latex] and [latex]s[\/latex]<\/em> at T<\/em> = -2 o<\/sup>C.\r\n

[latex] \\because \\dfrac{v - 0.207433}{0.2160303 - 0.207433} = \\dfrac{s - 1.7986}{1.8288 - 1.7986} = \\dfrac{-2 - (-10)}{0 - (-10)} [\/latex]<\/p>\r\n

[latex] \\therefore v\u00a0 = 0.214529 \\ \\rm{m^3\/kg} [\/latex]\u00a0\u00a0\u00a0\u00a0 and\u00a0\u00a0\u00a0\u00a0\u00a0 [latex] s = 1.8228 \\ \\rm{kJ\/kgK} [\/latex]<\/p>\r\nIn summary,\r\n\r\n\r\n\r\n\r\n\r\n
Substance<\/td>\r\nT\r\n\r\no<\/sup>C<\/td>\r\nP\r\n\r\nkPa<\/td>\r\nv\r\n\r\nm3<\/sup>\/kg<\/td>\r\nQuality x<\/em><\/td>\r\ns\r\n\r\nkJ\/kg-K<\/td>\r\nPhase<\/td>\r\n<\/tr>\r\n
Water<\/td>\r\n250<\/td>\r\n3976.17<\/span><\/td>\r\n0.02<\/td>\r\n0.383936<\/span><\/td>\r\n4.0523<\/span><\/td>\r\ntwo-phase<\/span><\/td>\r\n<\/tr>\r\n
R134a<\/td>\r\n-2<\/td>\r\n100<\/td>\r\n0.214529<\/span><\/td>\r\nn.a.<\/span><\/td>\r\n1.8228<\/span><\/td>\r\nsuperheated vapour<\/span><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<\/div>\r\n
\r\n

Example 2<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nA rigid tank contains 3 kg of R134a initially at 0o<\/sup>C, 200 kPa. R134a is now cooled until its temperature drops to -20o<\/sup>C. Determine the change in entropy, [latex]\\Delta S[\/latex], of R134a during this process. Is [latex]\\Delta S=S_{gen}[\/latex]?\r\n\r\n \r\n\r\nSolution:<\/span><\/em>\r\n\r\nThe initial state is at T1<\/sub> = 0o<\/sup>C and P1<\/sub> =200 kPa. From Table C2<\/a> in Appendix C,\r\n

[latex]s_1[\/latex] = 1.7654 kJ\/kgK, [latex] v_1 [\/latex] = 0.104811 m3<\/sup>\/kg<\/p>\r\nThe tank is rigid; therefore, [latex] v_2 [\/latex] = [latex] v_1 [\/latex] = 0.104811 m3<\/sup>\/kg.\r\n\r\nFrom Table C1<\/a>, at T<\/em>2<\/sub> = -20o<\/sup>C:\r\n

[latex]v_f [\/latex] = 0.000736 m3<\/sup>\/kg,\u00a0\u00a0\u00a0 [latex] v_g [\/latex] = 0.147395 m3<\/sup>\/kg<\/p>\r\n

[latex]s_f [\/latex] = 0.9002 kJ\/kgK,\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex] s_g [\/latex] = 1.7413 kJ\/kgK<\/p>\r\nBecause [latex]\u00a0 v_f < v_2 < v_g[\/latex], the final state is a two-phase mixture.\r\n

[latex] x_2 = \\dfrac{v_2 - v_f}{v_g-v_f} = \\dfrac{0.104811 - 0.000736}{0.147395-0.000736}=0.70964 [\/latex]<\/p>\r\n

[latex] \\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*}[\/latex]<\/p>\r\nThe total entropy change is\r\n

[latex]\\Delta S = m (s_2 - s_1) = 3 \\times (1.4971 - 1.7654) = - 0.8049 \\ \\rm{kJ\/K}[\/latex]<\/p>\r\nIt is important to note that [latex]\\Delta S \\neq S_{gen} [\/latex] 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.\r\n\r\n<\/div>\r\n<\/div>\r\n

6.7.2 Determining the specific entropy of solids and liquids<\/h2>\r\n

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,<\/p>\r\n \r\n

[latex]s_2-s_1=C_pln\\dfrac{T_2}{T_1}[\/latex]<\/p>\r\n \r\n\r\nwhere\r\n

[latex]s[\/latex]: specific entropy, in kJ\/kgK<\/p>\r\n

[latex]C_p[\/latex]: specific heat, in kJ\/kgK. Note that for solids and liquids, [latex]C_p=C_v[\/latex]. <\/sub>Table G2<\/a> and Table G3<\/a> list the specific heats of selected solids and liquids, respectively.<\/span><\/p>\r\n

[latex]T[\/latex]: absolute temperature, in Kelvin<\/p>\r\n\r\n

6.7.3 Determining the specific entropy of ideal gases<\/h2>\r\nThe 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.\r\n

[latex]s_2-s_1=C_pln\\displaystyle\\frac{T_2}{T_1}-Rln\\frac{P_2}{P_1}[\/latex]<\/p>\r\n

[latex] s_2-s_1=C_vln\\displaystyle\\frac{T_2}{T_1}+Rln\\frac{v_2}{v_1}[\/latex]<\/p>\r\n \r\n\r\nwhere <\/span>\r\n

[latex]C_p[\/latex], [latex]C_v[\/latex] and [latex]R[\/latex] are the constant-pressure specific heat, constant-volume specific heat, and gas constant, respectively, in kJ\/kgK. Table G1<\/a> lists these properties of selected ideal gases.<\/p>\r\n

[latex]T[\/latex]: absolute temperature, in Kelvin<\/p>\r\n

[latex]P[\/latex]: pressure, in kPa<\/p>\r\n

[latex]s[\/latex]: specific entropy, in kJ\/kgK<\/p>\r\n

[latex]v[\/latex]: specific volume, in m3<\/sup>\/kg<\/p>\r\n \r\n

\r\n

Example 3<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nAir is compressed from an initial state of 100 kPa, 27o<\/sup>C to a final state of 600 kPa, 67o<\/sup>C. Treat air as an ideal gas. Calculate the change of specific entropy, [latex] \\Delta s [\/latex], in this process. Is [latex] \\Delta s = s_{gen}[\/latex]?\r\n\r\n \r\n\r\nSolution:<\/span><\/em>\r\n\r\nFrom Table G1<\/a>: Cp<\/sub> = 1.005 kJ\/kgK, R = 0.287 kJ\/kgK\r\n\r\n[latex] \\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*}[\/latex]\r\n\r\n \r\n\r\nIt is important to note that [latex]\\Delta s \\neq s_{gen} [\/latex] in general. The specific entropy decreases in this process, but the rate of entropy generation is always greater than zero in a real process.\r\n\r\n<\/div>\r\n<\/div>\r\n

6.7.4 Isentropic relations for an ideal gas<\/h2>\r\n \r\n\r\nIf a process is reversible and adiabatic, it is called an [pb_glossary id=\"674\"]isentropic process[\/pb_glossary]<\/strong> 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.\r\n\r\n \r\n

[latex]Pv^k= \\rm{constant}[\/latex]\u00a0\u00a0 and\u00a0\u00a0\u00a0 [latex]\\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)}[\/latex]<\/p>\r\nwhere\r\n

[latex]k=\\dfrac{C_p}{C_v}[\/latex]: specific heat ratio. The [latex]k[\/latex] values of selected ideal gases can be found in Table G1<\/a>.<\/p>\r\n

[latex]T[\/latex]: absolute temperature, in Kelvin<\/p>\r\n

[latex]P[\/latex]: pressure, in kPa<\/p>\r\n

[latex]v[\/latex]: specific volume, in m3<\/sup>\/kg<\/p>\r\n\r\n<\/div>\r\nIt is noted that the isentropic relation [latex]Pv^k = \\rm{constant} [\/latex]\u00a0<\/strong>for ideal gases is actually a special case of the polytropic relation [latex]Pv^n = \\rm{constant} [\/latex] \u00a0<\/strong>with [latex] n = k = \\dfrac{C_p}{C_v} [\/latex].\r\n\r\n \r\n

\r\n

Example 4<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nDerive the isentropic relation [latex]Pv^k= \\rm{constant}[\/latex]\r\n\r\n<\/div>\r\n

Solution:<\/em><\/span><\/p>\r\n\r\n

\r\n\r\nFor an ideal gas undergoing an isentropic process,\r\n

[latex] \\Delta s = s_2 - s_1 =C_pln\\displaystyle\\frac{T_2}{T_1}-Rln\\frac{P_2}{P_1}=0[\/latex]<\/p>\r\nSubstitute [latex] C_p = \\dfrac{kR}{k-1} [\/latex] in the above equation and rearrange,\r\n

[latex] \\because \\dfrac{k}{k-1}ln\\dfrac{T_2}{T_1} = ln\\dfrac{P_2}{P_1} [\/latex]<\/p>\r\n

[latex] \\therefore ln\\left(\\dfrac{T_2}{T_1}\\right)^{\\dfrac{k}{k-1}} = ln\\dfrac{P_2}{P_1} [\/latex]<\/p>\r\n

[latex] \\therefore \\dfrac{P_2}{P_1} = \\left(\\dfrac{T_2}{T_1}\\right)^{\\dfrac{k}{k-1}} [\/latex]<\/p>\r\nCombine with the ideal gas law, [latex] Pv = RT [\/latex],\r\n

[latex]\\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}}[\/latex]<\/p>\r\n

[latex] \\therefore \\dfrac{P_2}{P_1} = \\left(\\dfrac{v_1}{v_2}\\right)^k [\/latex]<\/p>\r\n

[latex]\\therefore Pv^k = \\rm{constant} [\/latex]\u00a0\u00a0 and\u00a0\u00a0 [latex]\\dfrac{P_2}{P_1} = \\left(\\dfrac{v_1}{v_2}\\right)^k = \\left(\\dfrac{T_2}{T_1}\\right)^{\\dfrac{k}{k-1}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

\r\n

Practice Problems<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\n[h5p id=\"46\"]\r\n\r\n<\/div>\r\n<\/div>","rendered":"
\n

6.7.1 Determining the specific entropy of pure substances by using thermodynamic tables<\/h2>\n

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<\/a>.\u00a0 In addition to the [latex]P-v[\/latex] and [latex]T-v[\/latex] diagrams, the [latex]T-s[\/latex] diagram is commonly used to illustrate the relation between temperature and specific entropy of a pure substance. Figure 6.7.1 <\/a>shows the [latex]T-s[\/latex] diagram <\/a> for water.<\/p>\n

 <\/p>\n<\/div>\n

\"T-s<\/a>
Figure 6.7.1<\/strong> T-s diagram for water<\/em><\/figcaption><\/figure>\n
\n
\n
\n

Example 1<\/p>\n<\/header>\n

\n

Fill in the table.<\/p>\n\n\n\n\n\n
Substance<\/td>\nT, o<\/sup>C<\/td>\nP, kPa<\/td>\nv, m3<\/sup>\/kg<\/td>\nQuality x<\/em><\/td>\ns, kJ\/kg-K<\/td>\nPhase<\/td>\n<\/tr>\n
Water<\/td>\n250<\/td>\n<\/td>\n0.02<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n
R134a<\/td>\n-2<\/td>\n100<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Solution:<\/span><\/em><\/p>\n

Water: T = 250 o<\/sup>C, [latex]v[\/latex] = 0.2 m3<\/sup>\/kg<\/p>\n

From Table A1<\/a>: T<\/em> = 250 o<\/sup>C, [latex]v_f[\/latex]\u00a0 = 0.001252 m3<\/sup>\/kg, [latex]v_g[\/latex]\u00a0 = 0.050083 m3<\/sup>\/kg<\/p>\n

Since [latex]v_f\u00a0 < v <\u00a0 v_g[\/latex], water at the given state is a two phase mixture; the saturation pressure is P<\/em>sa<\/em>t<\/sub> = 3976.17 kPa, and [latex]s_f[\/latex] = 2.7935 kJ\/kgK,\u00a0[latex]s_g[\/latex] = 6.0721 kJ\/kgK<\/p>\n

The quality is<\/p>\n

[latex]x = \\dfrac{v - v_f}{v_{g} - v_{f}} = \\dfrac{0.02 - 0.001252}{0.050083 - 0.001252} = 0.383936[\/latex]<\/p>\n

The specific entropy is<\/p>\n

[latex]\\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*}[\/latex]<\/p>\n

 <\/p>\n

R134a: T<\/em> = -2 o<\/sup>C, P<\/em> = 100 kPa<\/p>\n

From Table C1<\/a>: by examining the saturation pressures at 0 o<\/sup>C and – 5 o<\/sup>C, we can estimate that the saturation pressure for T<\/em> = -2 o<\/sup>C is about 270 kPa; therefore, R134a at the given state is a superheated vapour.<\/p>\n

From Table C2<\/a>,<\/p>\n

P<\/em> = 100 kPa, T<\/em> = -10 o<\/sup>C, [latex]v[\/latex]= 0.207433 m3<\/sup>\/ kg, [latex]s[\/latex]<\/em> = 1.7986 kJ\/kgK<\/p>\n

P<\/em> = 100 kPa, T<\/em> = 0 o<\/sup>C, [latex]v[\/latex]= 0.216303 m3<\/sup>\/ kg, [latex]s[\/latex]<\/em> = 1.8288 kJ\/kgK<\/p>\n

Use linear interpolation to find [latex]v[\/latex] and [latex]s[\/latex]<\/em> at T<\/em> = -2 o<\/sup>C.<\/p>\n

[latex]\\because \\dfrac{v - 0.207433}{0.2160303 - 0.207433} = \\dfrac{s - 1.7986}{1.8288 - 1.7986} = \\dfrac{-2 - (-10)}{0 - (-10)}[\/latex]<\/p>\n

[latex]\\therefore v\u00a0 = 0.214529 \\ \\rm{m^3\/kg}[\/latex]\u00a0\u00a0\u00a0\u00a0 and\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]s = 1.8228 \\ \\rm{kJ\/kgK}[\/latex]<\/p>\n

In summary,<\/p>\n\n\n\n\n\n
Substance<\/td>\nT<\/p>\n

o<\/sup>C<\/td>\n

P<\/p>\n

kPa<\/td>\n

v<\/p>\n

m3<\/sup>\/kg<\/td>\n

Quality x<\/em><\/td>\ns<\/p>\n

kJ\/kg-K<\/td>\n

Phase<\/td>\n<\/tr>\n
Water<\/td>\n250<\/td>\n3976.17<\/span><\/td>\n0.02<\/td>\n0.383936<\/span><\/td>\n4.0523<\/span><\/td>\ntwo-phase<\/span><\/td>\n<\/tr>\n
R134a<\/td>\n-2<\/td>\n100<\/td>\n0.214529<\/span><\/td>\nn.a.<\/span><\/td>\n1.8228<\/span><\/td>\nsuperheated vapour<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n
\n
\n

Example 2<\/p>\n<\/header>\n

\n

A rigid tank contains 3 kg of R134a initially at 0o<\/sup>C, 200 kPa. R134a is now cooled until its temperature drops to -20o<\/sup>C. Determine the change in entropy, [latex]\\Delta S[\/latex], of R134a during this process. Is [latex]\\Delta S=S_{gen}[\/latex]?<\/p>\n

 <\/p>\n

Solution:<\/span><\/em><\/p>\n

The initial state is at T1<\/sub> = 0o<\/sup>C and P1<\/sub> =200 kPa. From Table C2<\/a> in Appendix C,<\/p>\n

[latex]s_1[\/latex] = 1.7654 kJ\/kgK, [latex]v_1[\/latex] = 0.104811 m3<\/sup>\/kg<\/p>\n

The tank is rigid; therefore, [latex]v_2[\/latex] = [latex]v_1[\/latex] = 0.104811 m3<\/sup>\/kg.<\/p>\n

From Table C1<\/a>, at T<\/em>2<\/sub> = -20o<\/sup>C:<\/p>\n

[latex]v_f[\/latex] = 0.000736 m3<\/sup>\/kg,\u00a0\u00a0\u00a0 [latex]v_g[\/latex] = 0.147395 m3<\/sup>\/kg<\/p>\n

[latex]s_f[\/latex] = 0.9002 kJ\/kgK,\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 [latex]s_g[\/latex] = 1.7413 kJ\/kgK<\/p>\n

Because [latex]\u00a0 v_f < v_2 < v_g[\/latex], the final state is a two-phase mixture.\n\n\n

[latex]x_2 = \\dfrac{v_2 - v_f}{v_g-v_f} = \\dfrac{0.104811 - 0.000736}{0.147395-0.000736}=0.70964[\/latex]<\/p>\n

[latex]\\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*}[\/latex]<\/p>\n

The total entropy change is<\/p>\n

[latex]\\Delta S = m (s_2 - s_1) = 3 \\times (1.4971 - 1.7654) = - 0.8049 \\ \\rm{kJ\/K}[\/latex]<\/p>\n

It is important to note that [latex]\\Delta S \\neq S_{gen}[\/latex] 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.<\/p>\n<\/div>\n<\/div>\n

6.7.2 Determining the specific entropy of solids and liquids<\/h2>\n

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,<\/p>\n

 <\/p>\n

[latex]s_2-s_1=C_pln\\dfrac{T_2}{T_1}[\/latex]<\/p>\n

 <\/p>\n

where<\/p>\n

[latex]s[\/latex]: specific entropy, in kJ\/kgK<\/p>\n

[latex]C_p[\/latex]: specific heat, in kJ\/kgK. Note that for solids and liquids, [latex]C_p=C_v[\/latex]. <\/sub>Table G2<\/a> and Table G3<\/a> list the specific heats of selected solids and liquids, respectively.<\/span><\/p>\n

[latex]T[\/latex]: absolute temperature, in Kelvin<\/p>\n

6.7.3 Determining the specific entropy of ideal gases<\/h2>\n

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.<\/p>\n

[latex]s_2-s_1=C_pln\\displaystyle\\frac{T_2}{T_1}-Rln\\frac{P_2}{P_1}[\/latex]<\/p>\n

[latex]s_2-s_1=C_vln\\displaystyle\\frac{T_2}{T_1}+Rln\\frac{v_2}{v_1}[\/latex]<\/p>\n

 <\/p>\n

where <\/span><\/p>\n

[latex]C_p[\/latex], [latex]C_v[\/latex] and [latex]R[\/latex] are the constant-pressure specific heat, constant-volume specific heat, and gas constant, respectively, in kJ\/kgK. Table G1<\/a> lists these properties of selected ideal gases.<\/p>\n

[latex]T[\/latex]: absolute temperature, in Kelvin<\/p>\n

[latex]P[\/latex]: pressure, in kPa<\/p>\n

[latex]s[\/latex]: specific entropy, in kJ\/kgK<\/p>\n

[latex]v[\/latex]: specific volume, in m3<\/sup>\/kg<\/p>\n

 <\/p>\n

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Example 3<\/p>\n<\/header>\n

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Air is compressed from an initial state of 100 kPa, 27o<\/sup>C to a final state of 600 kPa, 67o<\/sup>C. Treat air as an ideal gas. Calculate the change of specific entropy, [latex]\\Delta s[\/latex], in this process. Is [latex]\\Delta s = s_{gen}[\/latex]?<\/p>\n

 <\/p>\n

Solution:<\/span><\/em><\/p>\n

From Table G1<\/a>: Cp<\/sub> = 1.005 kJ\/kgK, R = 0.287 kJ\/kgK<\/p>\n

[latex]\\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*}[\/latex]<\/p>\n

 <\/p>\n

It is important to note that [latex]\\Delta s \\neq s_{gen}[\/latex] in general. The specific entropy decreases in this process, but the rate of entropy generation is always greater than zero in a real process.<\/p>\n<\/div>\n<\/div>\n

6.7.4 Isentropic relations for an ideal gas<\/h2>\n

 <\/p>\n

If a process is reversible and adiabatic, it is called an isentropic process<\/a><\/strong> 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>\n

 <\/p>\n

[latex]Pv^k= \\rm{constant}[\/latex]\u00a0\u00a0 and\u00a0\u00a0\u00a0 [latex]\\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)}[\/latex]<\/p>\n

where<\/p>\n

[latex]k=\\dfrac{C_p}{C_v}[\/latex]: specific heat ratio. The [latex]k[\/latex] values of selected ideal gases can be found in Table G1<\/a>.<\/p>\n

[latex]T[\/latex]: absolute temperature, in Kelvin<\/p>\n

[latex]P[\/latex]: pressure, in kPa<\/p>\n

[latex]v[\/latex]: specific volume, in m3<\/sup>\/kg<\/p>\n<\/div>\n

It is noted that the isentropic relation [latex]Pv^k = \\rm{constant}[\/latex]\u00a0<\/strong>for ideal gases is actually a special case of the polytropic relation [latex]Pv^n = \\rm{constant}[\/latex] \u00a0<\/strong>with [latex]n = k = \\dfrac{C_p}{C_v}[\/latex].<\/p>\n

 <\/p>\n

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Example 4<\/p>\n<\/header>\n

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Derive the isentropic relation [latex]Pv^k= \\rm{constant}[\/latex]<\/p>\n<\/div>\n

Solution:<\/em><\/span><\/p>\n

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For an ideal gas undergoing an isentropic process,<\/p>\n

[latex]\\Delta s = s_2 - s_1 =C_pln\\displaystyle\\frac{T_2}{T_1}-Rln\\frac{P_2}{P_1}=0[\/latex]<\/p>\n

Substitute [latex]C_p = \\dfrac{kR}{k-1}[\/latex] in the above equation and rearrange,<\/p>\n

[latex]\\because \\dfrac{k}{k-1}ln\\dfrac{T_2}{T_1} = ln\\dfrac{P_2}{P_1}[\/latex]<\/p>\n

[latex]\\therefore ln\\left(\\dfrac{T_2}{T_1}\\right)^{\\dfrac{k}{k-1}} = ln\\dfrac{P_2}{P_1}[\/latex]<\/p>\n

[latex]\\therefore \\dfrac{P_2}{P_1} = \\left(\\dfrac{T_2}{T_1}\\right)^{\\dfrac{k}{k-1}}[\/latex]<\/p>\n

Combine with the ideal gas law, [latex]Pv = RT[\/latex],<\/p>\n

[latex]\\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}}[\/latex]<\/p>\n

[latex]\\therefore \\dfrac{P_2}{P_1} = \\left(\\dfrac{v_1}{v_2}\\right)^k[\/latex]<\/p>\n

[latex]\\therefore Pv^k = \\rm{constant}[\/latex]\u00a0\u00a0 and\u00a0\u00a0 [latex]\\dfrac{P_2}{P_1} = \\left(\\dfrac{v_1}{v_2}\\right)^k = \\left(\\dfrac{T_2}{T_1}\\right)^{\\dfrac{k}{k-1}}[\/latex]<\/p>\n<\/div>\n<\/div>\n

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Practice Problems<\/p>\n<\/header>\n

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