Question

Is this process possible and why?

73 Air at 500 kPa, 980 K enters a turbine operating at steady state and exits at 200 kPa, 680 K. Heat transfer from the turbine occurs at an average outer surface temperature of 320 K at the rate of 40 kJ per kg of air flowing. Kinetic and potential energy effects are negligible. For air as an ideal gas with c, 1.5 kJ/kg K, determine (a) the rate power is developed in kJ per kg of air flowing,and (b) the rate of entropy production within the turbine, in kJ/K per kg of air flowing

Answer 1

Heat transfer from turbine Q_Lost = -40KJ/kg @320K = T_avg C_p = 1.5 KJ/kg.k a) steady flow energy equation: q-w = Delta h + Delta KE + Delta PE Neglecting Delta KE & Delta PE material implication q-w = h_2-h_1 = C_p(T_2 - T_1) material implication w = -40 + 1.5(980-680) material implication w = 410 KJ/kg Therefore Rate of power produced = 410 KJ/kg of air. b) Entropy production S_gen = C_p ln T_2/T_1 - R ln P_2/P_1 - Q/T_avg material implication S_gen = 1.5 ln(680/980) - 0.287 ln(200/500)-(-40)/320 = -0.548 + 0.263 + 0.125 S_gen = -0.16 KJ/kg.K Hence, the system is not possible. Because the entropy generation cannot be negature.