The properties of water and steam used to do work can be determined from the steam table, given input of the temperature, pressure, specific volume, as well as other **thermodynamic ***variables*

The responses to the required values of pressure and specific volume are as follows;

Part I

(a) **Given **T = 140°C, v = 0.5 m³/kg, to find p

From the steam tables, we have, at 140°, = 0.00107976, = 0.508519, = 3.61501

Given that the < v < , the fluid has two phases and the experienced pressure =__ 3.61501 bar__

(b) **Given **that at 30 MPa, and 80°C the steam is in the superheated heated region

From the single phase region of the steam tables, we have;

v = __1.01553 × 10⁻³ m³/kg__

(c) **Given **that at p = 10 MPa = 100 bar, and T = 600°C, we have from the single phase region of the steam table

v = __3.83775 × 10⁻² m³/kg__

(d) At T = 80°C, and x = 0.4, we have;

v = 0.00102904 + 0.4 × (3.40527 - 0.00102904) ≈ 1.363

v = __1.363 m³/kg__

Part II

(a) At T = 440 °C, p = 20 MPa. from the single phase region of the steam table, we have;

v =__ 1·22459 × 10⁻² m³/kg__

(b) At T = 160°C, p = 20 MPa = 200 bar, the steam is in the superheated region, and we have;

v = __1.08865 × 10⁻³ m³/kg__

(c) At T = 40°C and p = 2 MPa = 20 bar, we have

v = 0.00100700 m³/kg = __1.007 × 10⁻³ m³/kg__

Learn more about the **steam tables** here;

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