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kg of steam at a pressure of 17 bar and dryness 0.95 is heated at a constant pressure until it is completely dry. Determine: (i) Increase in volume (ii) Quantity of heat added.
1 Answer
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Given: m = 1 kg, P = 17 bar, x = 0.95

Solution:

Now from steam tables,

$h_1 = h_f + x xh_{fg} = 871.8 + 0.95 x 1921.5 = 2697.32 kJ/kg$

$h_2 = 2793.4 kJ/kg$

Quantity of Heat added = $m × (h_2 - h_1) = 96.08 kJ$

Now …

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