Question

10 kg of water is placed in an enclosed volume of 3m³.Heat is added until the temperature is 250°C.For the final state; A) the pressure is? B) the mass of vapour is? C) the enthalpy of the vapour is?

Answer 1

a) We use the table of saturated water - temperature . In the quality region the pressure as $p = 3976.2 kPa$

b) To find the mass of the vapor we must determine the quality. We use the next equation-

$U = Uf + x(Ug - Uf)$

Then,

$0.3 m3/kg = 0.00125 m3/kg + x( 0.05008 - 0.00125 ) m3/kg$

$0.3 = 0.00125 + 0.04883x$

$0.04883x = 0.3 - 0.00125$

$0.04883x = 0.29875$

$x = 0.29875 (m3/kg) / 0.04883 (m3/kg)$

$x = 6.11816$

Using the relationship of-

$X = mg / m ,$

we find the vapor mass

$mg = xm = 6.11816 \times10 kg = 61.1816 kg$

To find the enthalpy of the vapor-

Using the relationships

$dinP = -\frac{d(AHvap)}{R}\frac{1}{T}$

$\Delta Нvap = (\frac{1}{3976.2}) \times (523?) \times8.314$

$\Delta Нvap = 571.93Ј$