- The table lists temperatures and specific volumes of water vapor at two pressures. Using the data provided here, estimate
(a) the specific volume at T = 240°C, P = 1.25 MPa.
(b) the Tin °C at P= 1.5 MPa, v = 0.1555 m3/kg.
Since data encountered do not fall exactly on the grid of values provided by property tables, and linear interpolation between adjacent table entries becomes necessary. (Watch “linear interpolation”) Compare the results to that from https://webbook.nist.govichemistry/fluid/ for the same conditions. Watch “Using NIST for Enthalpy Data” in the “Course Syllabus and Schedule” folder.
P = 1.0 MPa
T (°C) 3.4 (m3/kg)
200 0.20602
240 0.22756
280 0.24801
P = 1.5 MPa
T (°C) v (m3/kg)
200 0.13245
240 0.14831
280 0.16276
- For water, locate the state on a sketch of the T-v diagram. Then determine the specified property at the indicated state. (Class or Watch “identifying phases”) (i) T = 140°C, v = 0.5 m3/kg. Find P, in bar. (ii) P = 30 MPa, T = 100°C. Find ‘, in m3/kg. (iii) T = 80°C, x = 0.75. Find P, in bar, and s., in m3/kg.
