Two kg of steam with a quality of 30% is heated at a constant pressure of 300KPa until the
temperature reaches 600 °C. Calculate the work done by the steam
The work W done by the steam in an isobaric process is given by: W=p(V_2-V_1), (1) where p = 300 kPa – the pressure of the steam, V1 and V2 ¬are the volume, occupied by the steam at the beginning and the end of the process, resprctively. The volume occupied by the steam is given by: V=mv, (2) where m = 2 kg – the mass of the steam, v is the specific volume of the steam. Substitute (2) into (1): W=mp(v_2-v_1 ). (3) Initially the steam is saturated at 300kPa = 0.3MPa. So, its specific volume is given by: v_1=v_L+x(v_V-v_L ), (4) where x = 30% = 0.3 – the quality of the steam, vL and vV ¬are the specific volume of the saturated water and steam, respectively. From the steam table we can define the specific volume vL and vV ¬(see https://www.nist.gov/sites/default/files/documents/srd/NISTIR5078-Tab3.pdf): v_L=1/ρ_L =1/931.82=0.001073 m^3/kg, v_V=1/ρ_V =1/1.6508=0.60576 m^3/kg. Substitute into (4): v_1=0.001073+0.3∙(0.60577-0.001073)=0.1825 m^3/kg. From the steam table, the specific volume of the steam at the end of the process (p = 0.3 MPa, t = 600°C) is: v_2=1/ρ_2 =1/0.74550=1.3414 m^3/kg. Substitute into (3): W=2∙300∙(1.3414-0.1825)=695.3 kJ.