Question

Example # 01 4-43 100-kg of R-134a at 200 kPa are contained in a ps ton-cylinder device whose volume is 12.322m.The piston is now moved until the volume is one-half its ornginal size. This is done such that the pressure of the R-134a does not change Determine the final temperature and the change in the total internal energy of the R-134a

Answer 1

The initial specific volume is v_1 = V/m = 12.322m^3/100kg = 0.12322m^3/kg the initial stae is super heated and the internal energy at this state is p_1 = 200kPa v_1 = 0.12322 m^3/kg the final specific volume is v_2 = v_1/2 = 0.12322m^3/kg/2 = 0.06161m^3/kg this is is a constant pressure process. the final state is determined to be saturated mixture whose temperature is T_2 = T_sat@200kpa = -10.09 degree centigrade the internal energy at the final state is x_2 = V_2-v_f/v_fg = (0.06161-0.0007533)m^3/kg/(0.099867-0.0007533)m^3/kg = 0.6140 u_2 = u_f + x_2u_fg = 38.28 + (0.6140)(186.21) = 152.61 kJ/kg hence,the change in the internal energy is delta u = u_2-u_1 = 152.61- 263.08 = -110.47kj/kg