Question

For H2O, determine the specific volume at the indicated state, in m3/kg.

(a) T = 480°C, p = 20 MPa.

(b) T = 160°C, p = 20 MPa.

(c) T = 40°C, p = 2 MPa.

Answer 1

Answer -

a) T=480^o C , P = 20 MPa

At P = 20 MPa, saturation temperature T_sat = 365.81 ^oC

As T = 480 ^oC , therefore given state is superheated vapour at 20 MPa

From the property table of superheated water vapour at given state,

Specific volume, v = 0.01399 m³/ kg

b) T = 160^oC , P= 20MPa

The given state is compressed liquid water as T_sat = 365.81^oC

Therefore, from the property table of compressed liquid water af given state,

v = 0.0010899 m³/ kg (by linear interpolation)

c) T = 40^o C, P = 2 MPa

At given condition, the state is compressed liquid water

Usuall we approximate compressed liquid behavior evaluated at the given temperature as v = v_{f@T= 40^oC}

Therefore, from the property table of saturation water at given temperature,

i.e., v = v_f@T=40^oC = 0.0010078 m³/ kg