6.7 Specific entropy of a state

Fill in the table.

Solution:

Water: T = 250 ^oC, [latex]v[/latex] = 0.2 m³/kg

From Table A1: T = 250 ^oC, [latex]v_f[/latex]  = 0.001252 m³/kg, [latex]v_g[/latex]  = 0.050083 m³/kg

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

The quality is

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

The specific entropy is

[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]

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, [latex]v[/latex]= 0.207433 m³/ kg, [latex]s[/latex] = 1.7986 kJ/kgK

P = 100 kPa, T = 0 ^oC, [latex]v[/latex]= 0.216303 m³/ kg, [latex]s[/latex] = 1.8288 kJ/kgK

Use linear interpolation to find [latex]v[/latex] and [latex]s[/latex] at T = -2 ^oC.

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

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

In summary,

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, [latex]\Delta S[/latex], of R134a during this process. Is [latex]\Delta S=S_{gen}[/latex]?

Solution:

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

[latex]s_1[/latex] = 1.7654 kJ/kgK, [latex]v_1[/latex] = 0.104811 m³/kg

The tank is rigid; therefore, [latex]v_2[/latex] = [latex]v_1[/latex] = 0.104811 m³/kg.

From Table C1, at T₂ = -20^oC:

[latex]v_f[/latex] = 0.000736 m³/kg,    [latex]v_g[/latex] = 0.147395 m³/kg

[latex]s_f[/latex] = 0.9002 kJ/kgK,        [latex]s_g[/latex] = 1.7413 kJ/kgK

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

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

[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]

The total entropy change is

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

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.

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, [latex]\Delta s[/latex], in this process. Is [latex]\Delta s = s_{gen}[/latex]?

Solution:

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

[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]

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.