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.

Respuesta :

Answer:

final pressure = 0.7 MPa

quality = 20 degree super heat.

Explanation:

Given data:

water temperature 120 degree celcius

quality[tex] x_1 = 0.33[/tex]

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

[tex]VF_1 = 0.001060 m^3/kg[/tex]

[tex]VG_1 = 0.89133 m^3/kg[/tex]

we know that[tex]v_1 = VF_1 + x_1 VFG_1[/tex]

      = 0.001060 + 0.33(0.89133- 0.001060)

       = 0.29485  m^3/kg

frinal temperature of water is [tex]T_2 = T + 65[/tex]

                                                         = 120 + 65 = 185 Degree

from saturated water temp. table

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

[tex]VG < v_1 =v_2[/tex]

from super heated table by interpolation

final pressure = 0.7 MPa

quality = 20 degree super heat.

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