In a steam engine the steam at the beginning of the expansion process is at 7 bar, dryness fraction 0.98 and expansion follows the law $p v^{1.1}=$ constant, down to a pressure of 0.34 bar. Calculate per kg of steam :

(i) The work done during expansion ;

(ii) The heat flow to or from the cylinder walls during the expansion.

Question

In a steam engine the steam at the beginning of the expansion process is at 7 bar, dryness fraction 0.98 and expansion follows the law $p v^{1.1}=$ constant, down to a pressure of 0.34 bar. Calculate per kg of steam :

(i) The work done during expansion ;

(ii) The heat flow to or from the cylinder walls during the expansion.

(i) The work done during expansion ;

(ii) The heat flow to or from the cylinder walls during the expansion.

Accepted Answer

Initial pressure of steam, [latex]p_1=7 ext { bar }=7 imes 10^5[/latex] N/m²

Dryness fraction, [latex]x_1=0.98[/latex]

Law of expansion, [latex]p v^{1.1}=[/latex] constant

Final pressure of steam, [latex]p_2=0.34 ext { bar }=0.34 imes 10^5[/latex] N/m².

At 7 bar : [latex]v_g=0.273[/latex] m³/kg

∴ [latex]v_1=x_1 v_g=0.98 imes 0.273=0.267[/latex] m³/kg

Also, [latex]p_1 v_1{ }^n=p_2 v_2{ }^n[/latex]

i.e., [latex]\frac{v_2}{v_1}=\left(\frac{p_1}{p_2} ight)^{1 / n}[/latex]

∴ [latex]\frac{v_2}{0.267}=\left(\frac{7}{0.34} ight)^{\frac{1}{1.1}} \quad ext { or } \quad v_2=0.267\left(\frac{7}{0.34} ight)^{\frac{1}{1.1}}=4.174[/latex] m³/kg.

(i) Work done by the steam during the process :

[latex]W=\frac{p_1 v_1-p_2 v_2}{n-1}=\frac{7 imes 10^5 imes 0.267-0.34 imes 10^5 imes 4.174}{(1.1-1)}[/latex]

= [latex]\frac{10^5}{0.1}(1.869-1.419)=10^5 imes 4.5[/latex] N-m/kg

i.e., Work done [latex]=\frac{10^5 imes 4.5}{10^3}= 4 5 0[/latex] kJ/kg.

At 0.34 bar : [latex]v_{g 2}=4.65[/latex] m³/kg, therefore, steam is wet at state 2 (since [latex]v_2

Now, [latex]v_2=x_2 v_{g 2}, ext { where } x_2=[/latex] dryness fraction at pressure [latex]p_2[/latex] (0.34 bar)

[latex]4.174=x_2 imes 4.65 \quad ext { or } \quad x_2=\frac{4.174}{4.65}=0.897[/latex]

The expansion is shown on a p-v diagram in Fig. 4.69, the area under 1-2 represents the work done per kg of steam.

(ii) Heat transferred :

Internal energy of steam at initial state 1 per kg,

[latex]u_1=\left(1-x_1 ight) u_{f 1}+x_1 u_{g_1}=(1-0.98) 696+0.98 imes 2573=2535.46[/latex] kJ/kg

Internal energy of steam at final state 2 per kg,

[latex]u_2=\left(1-x_2 ight) u_{f 2}+x_2 u_{g 2}[/latex]

= (1 – 0.897) 302 + 0.897 × 2472 = 2248.49 kJ/kg

Using the non-flow energy equation,

[latex]Q=\left(u_2-u_1 ight)+W[/latex]

= (2248.49 – 2535.46) + 450 = 163.03 kJ/kg

i.e., Heat supplied = 163.03 kJ/kg.

Question 4.58: In a steam engine the steam at the beginning of the expansio...

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