Calculate the following specific volumes: a. Carbon dioxide: 10◦C, 80% quality b. Water: 4 MPa, 90% quality c. Nitrogen: 120 K, 60% quality

a) v = 0.006153 m^3 / kg

b) v = 0.08781 m ^3 / kg

c) v = 0.005563 m^3 / kg

Explanation:

All states are two-phase with quality given. The overall specific

volume is given by

v = vf + x vfg = (1 - x) vf + x vg

a) CO2

10◦C, 80% quality

v = 0.001161 + x * 0.00624 = 0.006153 m^3 / kg

b) H2O

4 MPa, 90% quality

v = 0.001177 (1 - x) + x*0.09845 = 0.08781 m ^3 / kg

c) Nitrogen

120 K, 60% quality

v = 0.001915 + x*0.00608 = 0.005563 m^3 / kg