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Internal Energy

In a coal bas...

Question

In a coal based thermal power plant the boiler operates at 250°C and 80 bar pressure and condenser operates at 0.10 bar and 29°C. The steam enters the turbine at 450°C and the expansion is 85% isentropic. Specific. heat ot steam at 80 bar and 250°C is 1.75 kJ/kg. The pumping efficiency is 90%. The following data table is given:

Actual dryness fraction in the turblne outlet is _____
0.89

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Solution

The correct option is A 0.89
$h_{1} = h_{2 (250^{o} C)} + C_{p} (Δ T_{s u p e r h e a t})$
$h_{1} = 2813 + 1.75 (450 - 250) = 3163 k J / k g$
$s_{1} = s_{g (250^{o} C)} + C_{p} I n \frac{T_{v a p}}{T_{s a t}}$
$= 6.180 + 1.75 (\frac{450 + 273}{250 + 273})$
$s_{1} = 6.7466 k J / k g K$
$S_{1} = S_{2}^{'} = (S_{f} - S_{f g})_{25^{o} C}$
$6.7466 = 0.661 + x (8.01 - 0.661)$
$x^{'} = 0.828$
Thus is the dryness fraction if the expansion would have been iseniropic but expansion is not 100% isenfropic.
$h_{2}^{'} = (h_{f} + x h_{f g})_{29^{o}} C$
$= 190 + 0.828 (2580 - 190)$
$h_{2}^{'} = 2169.156 k J / k g$
Now $η_{T} = 0.85 = \frac{h_{1} - h_{2}}{h_{1} - h_{2}}$
$0.85 = \frac{3163 - h_{2}}{3163 - 2169.156}$
$h_{2} = 2318.23 k J / k g$
$h_{2} = h_{f} + x h_{f g}$
$2318.33 = 190 + x (2580 - 190) = 0.69$

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