Answer:
Explanation:
tank volume = 1.8 , T = 220°C, = 1/3, = 2/3,
a) The tank contain liquid-vapor mixture, therefore the pressure of the steam in the tank is gotten by the saturated pressure at 220°C
From Table A-4 'saturated water temperature table' we have:
P = 2319.6 KPa ≈ 2320 KPa
b) x = = 2/3 ÷ (1/3 + 2/3)
x = 0.67 = 67%
c) Dmix = Dv * xv + Dl * xl
Dv = 0.6 kg/m3, Dl = 1000 kg/m3
Dmix = 0.6(2/3) + 1000(1/3)
Dmix = 0.4 + 333
Dmix = 333.4 kg/m3
Dmix = 333 kg/m3
A) The pressure of the steam is; P_sat = 2319.6 kPa
(b) The quality of the saturated mixture is; 26.9%
(c) The density of the mixture is; ρ = 287.86 kg/m³
We are given;
Total volume; V = 1.8 m³
Temperature of rigid tank; T = 220°C
Volume in liquid phase; V₁ = ¹/₃(1.8) = 0.6 m³
Volume in vapour phase; V₂ = ²/₃(1.8) = 1.2 m³
From the steam table 4 attached at T = 220°C, we have;
A) Pressure of steam; P_sat = 2319.6 kPa
B) Specific volume for saturated liquid; υ_f = 0.001190 m³/kg
Specific volume for saturated vapour; υ_g = 0.086094 m³/kg
Mass of saturated liquid is gotten from the formula;
m_f = V₁/υ_f
m_f = 0.6/0.001190
m_f = 504.2 kg
Mass of saturated vapour;
m_g = V₂/υ_g
m_g = 1.2/0.086094
m_g = 13.94 kg
Total mass is;
m_t = m_f + m_g
m_t = 504.2 + 13.94
m_t = 518.14 kg
Quality of steam is;
x = m_g/m_t
x = 13.94/518.14
x = 0.02690
x = 26.9%
C) Formula for density is;
Density is; ρ = Total mass/total volume
ρ = 518.14/1.8
ρ = 287.86 kg/m³
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