Solutions to ASE 3333 - Test No. 2: Thermodynamics Problems - Prof. Pasquale Cinnella, Exams of Aerospace Engineering

Download Solutions to ASE 3333 - Test No. 2: Thermodynamics Problems - Prof. Pasquale Cinnella and more Exams Aerospace Engineering in PDF only on Docsity! ASE 3333 - Test No. 2 - Solutions (last revised 10/19/2011) Problem 1: mass 5kg kJ 1000J Using the EES sofware, we find the entropy for saturated water vapor at 3 MPa (pressure and quality=1 known): s_1 6.186 kJ kg K  At the end of the condensation, the pressure is the same, and the quality has gone to zero. The new entropy is: s_2 2.645 kJ kg K  Then the entropy change is: Δs s_2 s_1 3.541 kJ kg K  ΔS mass Δs 17.705 1 K kJ Problem 2: γ_he 5 3 1.667 V_1 2.5 L V_2 0.25L 2.5 10 4  m 3  T_1 20 273.15( ) K 293.15 K Volume ratio: ratio V_2 V_1  ratio 0.1 Using isentropic formulas: T_2 T_1 ratio γ_he 1( )  1.361 10 3  K Then: T_3 T_2 700 K 2.061 10 3  K After the expansion, we have (using isentropic formulas again): T_4 T_3 ratio γ_he 1  443.96 K Problem 3: Using the values of net work and work ratio, we can find the values for turbine and compressor work: w_net 150 kJ kg  backwork 0.4 η 0.85 w_t w_c w_net= w_c w_t backwork= w_t w_net 1 backwork( ) 250 kJ kg  w_c w_t w_net 100 kJ kg  Next, we can adjust those values, based on the isentropic efficiencies: w_t_new w_t η 212.5 kJ kg  w_c_new w_c η 117.647 kJ kg  Therefore: w_net_new w_t_new w_c_new 94.853 kJ kg  Problem 4: γ 1.4 c_P 1 kJ kg K  T_in 25 273.15( ) K 298.15 K P_in 101 kPa Ratio 10 Bleed 0.1 T_max_a 1200 K T_max_b 1600 K