Question

A well-insulated 1.8 m³ rigid tank contains steam at 220°C. One third of the volume (not the mass) is in the liquid phase and the rest is in the vapor phase. Determine a) the pressure of the steam, b) the quality of the saturated mixture, and c) the density of the mixture. Use the property tables included in the textbook appendices to solve this problem.

Answer 1

Answer: a) 2320kPa b) 0.0269 c) 287.8kg/m^3

Step-by-step explanation:

A)

2 phases co exist in equilibrium therefore we have a saturated liquid-vapor mixture. The pressure of the steam is the saturated pressure at the given temperature and the pressure in the tank is the saturated pressure at the specific temperature, therefore we have,

P = Tsat at 220 degrees = 2320kpa

B) Total mass of quantity

Mf = Vf / vf = (1/3 x (1.8m^3)) / 0.001190 m^3/kg = 504.2kg

Mg = Vg /vg = (2/3 x (1.8m^3)) / 0.08609m^3/kg = 13.94kg

Mt = Mf + Mg = 504.2 + 13.94 = 518.1kg

X = Mg/Mt = 13.94 / 518.1 = 0.0269

C) v = vf + X (vg - vf) = 0.001190 + 0.0269 (0.08609 - 0.001190) = 0.003474 m^3/kg

Therefore Pressure (P) = 1 / v = 1/ 0.003474 = 287.8 kg/m^3