Steam initially at a pressure of 15 bar and 0.95 dryness expands isentropically to 7.5 bar and is then throttled until it is just dry. Determine per kg of steam :

(i) Change in entropy ;

(ii) Change in enthalpy ;

(iii) Change in internal energy.

Using : (a) Steam tables

(b) Mollier chart.

Is the entire process reversible ? Justify your statement.

Question

Steam initially at a pressure of 15 bar and 0.95 dryness expands isentropically to 7.5 bar and is then throttled until it is just dry. Determine per kg of steam :

(i) Change in entropy ;

(ii) Change in enthalpy ;

(iii) Change in internal energy.

Using : (a) Steam tables

(b) Mollier chart.

Is the entire process reversible ? Justify your statement.

Accepted Answer

Using steam tables

Condition 1 : 15 bar, 0.95 dryness

[latex]h_{f_1}=844.7[/latex] kJ/kg ; [latex]t_{s_1}=198.3^{\circ} C[/latex], [latex]s_{f_1}=2.3145[/latex] kJ/kg K,

[latex]s_{g_1}=6.4406[/latex] kJ/kg K, [latex]v_{g_1}=0.132[/latex] m³/kg

[latex]h_1=h_{f_1}+x_1 h_{fg_1}=844.7+0.95 imes 1945.2=2692.64[/latex] kJ/kg

[latex]s_1=s_{f_1}+x_1\left(s_{g_1}-s_{f_1} ight)=2.3145+0.95(6.4406-2.3145)=6.2343[/latex] kJ/kg K.

Condition 2 : 7.5 bar

[latex]h_{f_2}=709.3[/latex] kJ/kg, [latex]t_{s_2}=167.7^{\circ} C[/latex], [latex]h_{fg_2}=2055.55[/latex] kJ/kg, [latex]s_{f_2}=2.0195[/latex] kJ/kg K

[latex]s_{g_2}=6.6816[/latex] kJ/kg K, [latex]v_{g_2}=0.255[/latex] m³/kg

Consider isentropic expansion 1-2 :

(i) Change in entropy = 0

i.e., Entropy at 1 = entropy at 2

∴ [latex]s_1=s_2[/latex]

[latex]6.2343=s_{f_2}+x_2\left(s_{g_2}-s_{f_2} ight)[/latex]

[latex]=2.0195+x_2(6.6816-2.0195)[/latex]

∴ [latex]x_2=\frac{6.2343-2.0195}{6.6816-2.0195}=0.9[/latex]

Now, enthalpy at point 2,

[latex]h_2=h_{f_2}+x_2 h_{f_{g_2}}=709.3+0.9 imes 2055.55=2559.29[/latex] kJ/kg.

(ii) Change in enthalpy [latex]=h_2-h_1[/latex]

= 2559.29 – 2692.64 = – 133.35 kJ/kg.

(–ve sign indicates decrease).

(iii) Change in internal energy :

Internal energy at point 1,

[latex]u_1=h_1-p_1 x_1 v_{g_1}[/latex]

= [latex]2692.64-15 imes 10^5 imes 0.95 imes 0.132 imes 10^{-3}=2504.54[/latex] kJ/kg

Internal energy at point 2,

[latex]u_2=h_2-p_2 x_2 v_{g_2}[/latex]

= [latex]2559.29-7.5 imes 10^5 imes 0.9 imes 0.255 imes 10^{-3}=2387.16[/latex] kJ/kg

∴ Change in internal energy

[latex]=u_2-u_1=2387.16-2504.54=-117.38[/latex] kJ/kg

(–ve sign indicates decrease)

Consider the throttling expansion 2-3 :

Entropy at point 2,

[latex]s_2=\left(s_1 ight)=6.2343[/latex] kJ/kg K

Entropy at point 3,

[latex]s_3=s_{f_3}+x_3\left(s_{g_3}-s_{f_3} ight)[/latex]

The pressure at point 3 can be read from h-s chart ([latex]p_3=0.06[/latex] bar) and the corresponding values of [latex]s f_3[/latex] and [latex]h_{f g_3}[/latex]$ from steam tables.

Condition 3. At 0.06 bar, [latex]x_3=1[/latex]. From steam tables,

[latex]s_{f_3}=0.521[/latex] kJ/kg K, [latex]s_{g_3}=8.330[/latex] kJ/kg K

∴ [latex]s_3=0.521+1 imes(8.330-0.521)=8.330[/latex] kJ/kg K

Change in entropy [latex]=s_3-s_2[/latex]

= 8.330 – 6.2343 = 2.0957 kJ/kg K

Change in enthalpy = 0

i.e., [latex]h_2=h_3[/latex]

Change in internal energy = 0

i.e., [latex]u_3=u_2[/latex]

Combining the results obtained from isentropic and throttling expansion, we get during the entire process :

(i) Change in entropy = 2.0957 kJ/kg K (increase).

(ii) Change in enthalpy = 133.35 kJ/kg K (decrease).

(iii) Change in internal energy = 117.38 kJ/kg (decrease).

Only the expansion of steam from point 1 to 2 (i.e., isentropic expansion) is reversible because of unresisted flow whereas the expansion from point 2 to point 3 (i.e., throttling expansion) is irreversible because of frictional resistance to flow. Increase of entropy also shows that expansion from point 2 to point 3 is irreversible.

(b) Using Mollier chart.

Refer to Fig. 4.77.

• Locate point 1 at an intersection of 15 bar pressure line and 0.95 dryness fraction line.

• Draw vertical line from point 1 intersecting 7.5 bar pressure line at point 2. Line 1-2 represents isentropic expansion.

• From point 2 draw a horizontal line intersecting at the saturation line at point 3. Line 2-3 then represents throttling expansion.

From Mollier chart :

[latex]h_1=2692[/latex] kJ/kg, [latex]h_2=2560[/latex] kJ/kg

[latex]s_1=s_2=6.23[/latex] kJ/kg, [latex]s_3=8.3[/latex] kJ/kg K

∴ (i) Change in entropy [latex]=s_3-\left(s_1 ext { or } s_2 ight)[/latex]

= 8.3 – 6.23 = 2.07 kJ/kg K (increase).

(ii) Change in enthalpy [latex]=h_2\left( ext { or } h_3 ight)-h_1[/latex]

= 2560 – 2692 = – 132 kJ/kg

= 132 kJ/kg (decrease).

Question 4.63: Steam initially at a pressure of 15 bar and 0.95 dryness exp...

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