Question

Water at 120°C with a quality of 0.33 has its temperature raised 65°C in a constant volume process. What is the new quality and pressure? Show the process in a P-v diagram.

Answer 1

final pressure = 0.7 MPa

quality = 20 degree super heat.

Step-by-step explanation:

Given data:

water temperature 120 degree celcius

quality
`x_1 = 0.33`

we know that from saturated water temperature table . for T =120 DEGREE

`VF_1 = 0.001060 m^3/kg`

`VG_1 = 0.89133 m^3/kg`

we know that
`v_1 = VF_1 + x_1 VFG_1`

= 0.001060 + 0.33(0.89133- 0.001060)

= 0.29485 m^3/kg

frinal temperature of water is
`T_2 = T + 65`

= 120 + 65 = 185 Degree

from saturated water temp. table

for T = 185 degree, VG = 0.17390 m^3/kg

`VG < v_1 =v_2`

from super heated table by interpolation

final pressure = 0.7 MPa

quality = 20 degree super heat.