Question

Determine the temperature, volume, and quality for one kg of water at H=3500kJ/kg, P =0.1MPa. (Using steam tables)

Answer 1

mass of water: 1 kg

H = 3500KJ/kg

P = 0.1 MPa = 100 KPa = 1 bar

consider the water is at saturated state

from the pressure steam tables

at 1 bar T = 99.6 ^oC

volume = 1.694 m³/kg

as we have 1 kg of water

volume = 1.694 m³

quality of steam

H = Hf + xHfg

from the steam tables

Hf = 417 kj/kg

Hfg = 2258 kj/kg

H = 3500 kj/kg

x is the fraction of saturated mixture that is steam.

Hence, 3500 = 417 + (x 2258)

x =1.36 (means the fraction of steam is excess)

the value of x should be less than 1 for saturated water

hence, this is a superheated steam

see the superheated steam table, at 0.1 Mpa or 1 bar and 99.6 ^oC

Specific Volume of Superheated Steam = 0.880277 m³

Specific Heat of Superheated Steam = 2.66198 KJ/ kg-K

Specific Enthalpy of Superheated Steam = 2657.94 kJ/kg