Answer:
a. T-V and P-V diagram are included
b. State 1: Specific volume = 0.0811753 m³/kg
State 2: Specific volume = 0.0811753 m³/kg
State 3: Specific volume = 0.0804155 m³/kg
c. Z = 51.1
d. Quality for state 1 = 100%
Quality for state 2 = 63.47%
Quality for state 3 = 100%
Explanation:
a. T-V and P-V diagram are included
b. State 1: Water vapor
P₁ = 3.0 MPa = 30 bar
T₁ = 300°C = 573.15
Saturation temperature = 233.86°C Hence the steam is super heated
Specific volume = 0.0811753 m³/kg
State 2:
Constant volume formula is P₁/T₁ = P₂/T₂
Specific volume = 0.0811753 m³/kg
T₂ = 200°C = 473.15
Therefore, P₂ = P₁/T₁ × T₂ = 3×473.15/573.15 = 2.4766 MPa
At T₂ water is mixed water and steam and the [tex]v_f[/tex] = 0.00115651 m³/kg
[tex]v_g[/tex] = 0.127222 m³/kg
State 3:
P₃ = 2.5 MPa
T₃ = 200°C
Isothermal compression P₂V₂ = P₃V₃
V₃ = P₂V₂ ÷ P₃ = 2.4766 × 0.0811753/2.5 = 0.0804155 m³/kg
Specific volume = 0.0804155 m³/kg
2) Compressibility factor is given by the relation;
[tex]Z = \dfrac{PV}{RT} = \dfrac{3\times 10^6 \times 0.0811753 }{8.3145 \times 573.15} = 51.1[/tex]
Z = 51.1
3) Gas quality, x, is given by the relation
[tex]x = \dfrac{Mass_{saturated \, vapor}}{Total \, mass} = \dfrac{v - v_f}{v_g - v_f}[/tex]
Quality at state 1 = Saturated quality = 100%
State 2 Vapor + liquid Quality
Gas quality = (0.0811753 - 0.00115651)/ (0.127222-0.00115651) = 63.47%
State 3: Saturated vapor, quality = 100%.
