Steam at a pressure of 15 bar and 250°C is expanded through a turbine at first to a pressure of 4 bar. It is then reheated at constant pressure to the initial temperature of 250°C and is finally expanded to 0.1 bar. Using Mollier chart, estimate the work done per kg of steam flowing through the

Question

Steam at a pressure of 15 bar and 250°C is expanded through a turbine at first to a pressure of 4 bar. It is then reheated at constant pressure to the initial temperature of 250°C and is finally expanded to 0.1 bar. Using Mollier chart, estimate the work done per kg of steam flowing through the turbine and amount of heat supplied during the process of reheat. Compare the work output when the expansion is direct from 15 bar to 0.1 bar without any reheat. Assume all expansion processes to be isentropic.

Accepted Answer

Refer Fig. 12.26.

Pressure, [latex]p_1=15[/latex] bar ;

[latex]p_2=4[/latex] bar ;

[latex]p_4=0.1[/latex] bar.

Work done per kg of steam,

W = Total heat drop

[latex]=\left[\left(h_1-h_2 ight)+\left(h_3-h_4 ight) ight][/latex] kJ/kg ...(i)

Amount of heat supplied during process of reheat,

[latex]h_{ ext {reheat }}=\left(h_3-h_2 ight)[/latex] kJ/kg ...(ii)

From Mollier diagram or h-s chart,

[latex]h_1=2920[/latex] kJ/kg, [latex]h_2=2660[/latex] kJ/kg

[latex]h_3=2960[/latex] kJ/kg, [latex]h_2=2335[/latex] kJ/kg

Now, by putting the values in eqns. (i) and (ii), we get

W = (2920 – 2660) + (2960 – 2335)

= 885 kJ/kg.

Hence work done per kg of steam = 885 kJ/kg.

Amount of heat supplied during reheat,

[latex]h_{ ext {reheat }}=(2960-2660)= 3 0 0[/latex] kJ/kg.

If the expansion would have been continuous without reheating i.e., 1 to 4′, the work output is given by

[latex]W_1=h_1-h_4^{\prime}[/latex]

From Mollier diagram,

[latex]h_{4^{\prime}}=2125[/latex] kJ/kg

∴ [latex]W_1=2920-2125=795[/latex] kJ/kg.

Question 12.18: Steam at a pressure of 15 bar and 250°C is expanded through ...

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