EXAMPLE PROBLEMS FOR MIDTERM I

PROBLEM 1 For an air standard ideal Otto cycle the pressure and temperature at the beginning of compression are 90 kPa and 300 K respectively. The temperature at the end of heat input at constant volume is 2300 K. The compression ratio is 9.5. (R = 0.287 kJ/kg-K and Cp = 1.005 kJ/kg-K for the whole cycle) Calculate; The work done during compression. b) The work done during expansion c) The mean effective pressure

The logic of the solution will be from TOP to BOTTOM 1 2 3 4 P v GIVEN P1 = 90 kPa T1 = 300 K R = 0.287 kJ/kg-K Cp = 1.005 kJ/kg-K R = 9.5 T3 = 2300 K REQUIRED Wc = ? SOLUTION The logic of the solution will be from TOP to BOTTOM The solution will be from BOTTOM to TOP

k = 1.4 T2 = 738 K T4 = 935 K Cv = 0.718 kJ/kg-K v1 = 0.956667 m3/kg v2 = 0.1007 m3/kg vs = 0.8560 m3/kg Wcomp = -314 kJ/kg Wexp = 980 kJ/kg Wnet = 666 kJ/kg Pe = 778 kJ/kg

Problem 3 The temperature and pressure of the reactants at the beginning of compression in an otto cycle is 25 C and 90 kPa respectively. The temperature and pressure of the reactants at the end of compression is 320 C and 1500 kPa. The fuel used is gasoline (C7.76H13.1) and the excess air coefficient is 0.9. Find the heat lost per kg of reactants, during compression. The mean molar specific heat of the reactants during compression. b) The mean specific heat of the reactants per kg of mixture, during compression c) The change of the internal energy of the reactants during compression. d) The work done on the reactants during compression e) The heat lost by the reactants during compression.

1 2 3 4 P v

Air-standard dual cycle r = 15 Pmax = 10 Mpa Qv = 0,2 Qin Qp = 0,8 4 REQUIRED hth = ? GIVEN Air-standard dual cycle r = 15 Pmax = 10 Mpa Qv = 0,2 Qin Qp = 0,8 P1 = 0,1 MPa T1 = 320 K 44 1 2 3 4 5

SOLUTION v1 = 0,9184 m3/kg v2 = 0,061227 m3/kg T2 = 945 K P2 = 4431 kPa T3 = 1707 K Qv = 546 kJ/kg Qp = 2185 kJ/kg T4 = 3882 K v4 = 0,1114 m3/kg d = 8,24 T5 = 1670 K Qout = 1356 kJ/kg hth = 0,50