Answer to Question #230730 in Mechanical Engineering for Pavan

Question

A vessel contains 2.5m³ of steam at a pressure of 25 bar. Determine the mass of the steam under following conditions.

a. Steam is wet having dryness fraction 0.85.

b. Steam is dry saturated.

c. Steam is superheated at constant pressure to 310⁰C of superheated steam as 2.21 kJ/kgK.

Answer 1

V=2.5m³

P=25bar=2500kPa

T_s=226.1°C, V_s=0.077 both from steam tables of a saturated steam at 25 bar

a. Specific volume of wet steam of dryness fraction,x, neglecting volume of water =xVs

=0.85*0.077=0.06545m³/kg

Mass of steam=$\frac{volume\ of\ steam}{specific\ volume\ of\ steam,V_s}$

=$\frac{2.5}{0.0645}$

=38.1971kg

b. Mass of dry saturated steam=$\frac{volume\ of\ steam}{specific\ volume\ of\ steam}$

=$\frac{2.5}{0.077}$

=32.4675kg

c. V_sup=$V_s*\frac{T_{sup}}{T_s}$

T_sup=310+226.1+273=809.1K

T_s=226.1+273=499.1K

V_sup=0.077*$\frac{809.1}{499.1}$ =0.1248

Mass of steam=$\frac{volume\ of\ steam}{specific\ volume\ of\ steam}$

=$\frac{2.5}{0.1248}$

=20.03kg