Question

1. Verify the state of water (compressed liquid, saturated mixture or superheated vapor) in the following conditions. For each condition, draw a sketch of the phase diagram (P-T), the T-v diagram (with the appropriate isobar) or P-v diagram (with the appropriate isotherm) to explain your answer. Calculate the quality if it is a saturated mixture.
   1. 120 ºC, 0.1334 MPa
   2. 0.36 MPa, 0.4 m³/kg
   3. 302 ºF, 8.0 ft³/lbm
   4. 0.2 MPa, 110ºC

Answer 1

Ans(a): At 120 degree Celsius and 0.1334 MPa the state is super-heated vapor and its shown below on the P-V diagram. This can be interpreted by looking at the saturated steam tables where pressure corresponding to 120 degree Celsius is 0.1985 MPa. This can be understood by the concept that water evaporates faster at lower pressure.

Ans(b). At 0.36 MPa and 0.4 m³ /kg the state is a saturated mixture and its shown below on the P-V diagram. This can be interpreted by looking at the saturated steam tables where Specific Volume of Saturated Steam corresponding to 0.36 MPa is 0.51051. The given specific volume is lower than the one matching with given steam pressure, hence quality of team is less than unity.

Ans(c). At 302 degree Fahrenheit, we can interpret from the steam table the corresponding volume of saturated vapor is 6.28 ft³/lbm and given volume is more than it. Hence our steam is super-heated since it has got more volume per mass.

Ans(d). At 0.2 MPa and 110ºC, the state is compressed liquid since pressure corresponding to given saturation temperature is 0.143376 which is lower than 0.2 MPa. The general rule is liquid requires higher temperatures to boil at higher pressures.