1. EXAMPLE PROBLEMS
FOR MIDTERM I

2. 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 = kJ/kg-K and Cp = kJ/kg-K for the whole cycle)
Calculate;
The work done during compression.
b) The work done during expansion
c) The mean effective pressure

3. 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 = kJ/kg-K
Cp = 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

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

8. 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.

9. 1 2 3 4 P v

11. 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

12. SOLUTION
v1 =
0,9184 m3/kg
v2 =
0, 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