Water vapor initially at 3.0 MPa and 300°C (state 1) is contained within a piston- cylinder. The water is cooled at constant volume until its temperature is 200°C (state 2). The water is then compressed isothermally to a state where the pressure is 2.5 MPa (state 3).a. Locate states 1, 2, and 3 on a T-v and P-v diagram. b. Determine the specific volume at all three states. c. Calculate the compressibility factor Z at state 1 and comment. d. Find the quality (if applicable) at all three states.

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 = 0.00115651 m³/kg

= 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;

Z = 51.1

3) Gas quality, x, is given by the relation

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%.

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