Answer to Question #79208 in Mechanical Engineering for Usman Nadeem

Question

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

Answer 1

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.