Question

B3. (20%) (Show all your calculation steps.) Water vapor initially at 240 °C, 1.0 MPa expands in a piston-cylinder assembly isothermally and without internal irreversibilities to a final pressure of 0.1 MPa. Evaluate the (a)work done, in kl/kg, (b) the change of entropy,, in kl/kg-K and (c) the change of internal energy, in kl/kg. Assume the system can be treated as a Van der Waals model.. [Given: the specific volume of the water vapor can be determined from the superheated steam table and they are: V1 = 0.2275 m'/kg and V, - 2.359 m /kg. The system can be treated as a single-phase condition.] Integration Equations: dx dx In *, -a xa and - G x Q2 X2

Answer 1

P = 1.0Mpa 1 1,- 200°C = 240+273 = 513k 2 P₂ = 0.1 MPa Given: - Process 1-2 : Isothermal expansion of gas (water vapor) and Reversible process. gas. the system Assume water vapour to be an ideal can be treated as Vander Wadis model. So pv=RT is valid for the process 1-2 P = Pressure v = specific volume R= Gas constant T= Temperature (in kelvin) Work done dw Pdv PV = MRT WE Spar P = MRT V W1-2= (MRT dv = MRT fen? -MRT M V2 *m = 1kg for per kg calculations. P2 V₂ = PV V2 Po 1.00 Vio o.10 V2 ση Isothermal 10. do

10.0 m= 1 kg PV = RT P = P2 U = RT W,-2 = = P. v, en 1₂ =(1-ox109( 0.2275) lu(100) N m3 x W2 kg (a) Work- done (W,-2) 523.8381 kJ kg The change of entropy for reversible process] S-8,- d₂-9, ds 189 T (c) for internal Iso thermal process, T = const. and for ideal gos, energy is the function of temperature only. AU= the change in So, LAU = o internal en from first law of thermodynamics (Neglecting AkE 4 APE) so= dut sw yo(7= Constant) (89) -2 = (40) + (W1-2) (89),-2 = Wi-z = 523.8381 kJ kg (b) (As), 2 (59) 1-2 523-8381 513 = 1.021127 kJ kg ok