Question 12.18: Steam at a pressure of 15 bar and 250°C is expanded through ...
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.
Refer Fig. 12.26.
Pressure, p_1=15 bar ;
p_2=4 bar ;
p_4=0.1 bar.
Work done per kg of steam,
W = Total heat drop
=\left[\left(h_1-h_2\right)+\left(h_3-h_4\right)\right] kJ/kg …(i)
Amount of heat supplied during process of reheat,
h_{\text {reheat }}=\left(h_3-h_2\right) kJ/kg …(ii)
From Mollier diagram or h-s chart,
h_1=2920 kJ/kg, h_2=2660 kJ/kg
h_3=2960 kJ/kg, h_2=2335 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,
h_{\text {reheat }}=(2960-2660)= 3 0 0 kJ/kg.
If the expansion would have been continuous without reheating i.e., 1 to 4′, the work output is given by
W_1=h_1-h_4^{\prime}From Mollier diagram,
h_{4^{\prime}}=2125 kJ/kg
∴ W_1=2920-2125=795 kJ/kg.
