Question

the enthalpy of 1kg of steam at 70bar pressure is 2680KJ/kg.what is the condition of steam??​

Answer 1

From the steam tables, at 70 bar

$T_{s}$ = 285.8°C, $h_{f}$ = 1267.4 kJ/kg, $h_{fg}$ = 1506 kJ/kg

$h_{g}$ = 1267.4 + 1506 = 2773.5 kJ/kg

Since the given enthalpy is lesser than the enthalpy in a dry saturated state,

it is a wet steam

h = $h_{f}$ + x$h_{fg}$

2680 = 1267.4 + x(1506)

1412.6 = 1506x

x = 0.938 (dryness fraction)