Question

A mass of 5.5 kg of saturated liquid-vapor mixture of water is contained in a piston-cylinder at 101.325 kPa. Initially, 2.3 kg of the water is in the liquid phase and the rest is in the vapor phase. Heat is now transferred to the water, and the piston, which is resting on a set of stops, starts moving when the pressure inside reaches 300 kPa. Heat transfer continues until the total volume increases by 21 percent. Determine:
a) the initial temperature (oC),
b) the final temperature (oC),
c) the specific volume of water when the piston first starts moving (m3/kg), and
d) the work done during this process (kJ).

T1 (oC)	Format : 94.62
T3 (oC)	Format : 227.2
v2 (m3/kg)	Format : 0.945
Wb (kJ)	Format : 242.2

Answer 1

Pressure Voleme (P-V) of the system is as follows- 23 :582 + We have 1 given total mass = m=5.5kg mass of liquid= mg=2-3 kg so that mass of vaboo (mg)-5.5- 2:3 = 3.2 kg . So that dryness fraction at state 1 is - 3.2 X = = 0:582 2.3+ 3.2 Pressure of state 1 is = Pi = 101.325 kPa so that we can write following properties from saturated water tables (steam table) at 101.325 kPa Saturation temperature Ti= Tsat = gg.goc Specific volume of saturated liquid u = 1.04 34 x10-3 mol kg Specific in temal energy of safrated liquid Uf = 418.885 kJ/kg

specific volume of saturated vapor Ug = 1-6733 m3/kg specific internal energy of saturated vapor Ug = 2505.9848 k slka. Parto initial temperature is wa the Ti= gg. goc Past - (6) Specific roleme al state 1 is given as Q = p + s (lg-Uf) = 1.0434X16-3 + 0.582 (1-6733 - 1.0434X10-3) W = 0.9743 m/kg so that initial volume in the cylinder is Vi = ml, = 5.5x0.9743= 5.359 m3 Voleme at state 3 is increases by 21%. so that V3 = 1.21 Vi = 1.21 X5. 359. V3 = 6.484 m3 so that final Specific voleme is Uz - Va 6=484 = 1199 m3/kg 5.5

- - We have given pressure al starte 3 as P3= 3ookpa . So that final temperature T3 is given from steam table at P3 and ils The final temperature is | T3 = 495.0°C Past - (6) Pressure al state 2 = B = A = 300 kPa Voleme al statte 2 = V2 = V, = 5.359 m3 so that specific voleme at this state is L2= V25.359 69 = 0.974 m®/kg So that specific volume of water when the piston first starts moving is given as 02= 0.974 m3/kg 1. Past-(d) The work done during this process is given as Wb= Pz (V3-V2) = 300 x( 6. 484 – 5.359) TWb= 337.5 kJ