Modern Engineering Thermodynamics

13.20 Miller Cycle 511
Then, from Figure 13.52b,
V c = V 1 = V 7s = V 4 = 9:72 × 10 −5 m 3
Also,
V 6s = V 2s = V 3 = V d + V c = 8:75 × 10 −4 m 3 + 9:72 × 10 −5 m 3 = 9:72 × 10 −4 m 3
and
V 5 = V 7s × CR = ð9:72 × 10 −5 m 3 Þ × 8:00 = 7:78 × 10 −4 m 3
Now, we can compute the temperature and pressure at each point in the cycle by starting at state 5 in Figure 13.52b, the
closing of the intake valve, where we know the pressure and temperature.
State 5 in Figure 13.52b
This is where the intake valve closes and the following information is given in the problem statement: p 5 = 200 kPa and
T 5 = 40.0ºC = 313 K.
State 6 in Figure 13.52b
The process from states 5 to 6 is isentropic, so
k
V
p 6s = p 5
5 = ð200: kPaÞ 7:78 ×
10−4 m 3
V 6s
9:72 × 10 4 = 148 kPa
m 3
and
k−1
V
T 6s = T 5
5 = ð313 KÞ 7:78 ×
10−4 m 3
V 6s
9:72 × 10 −4 = 648 K
m 3
State 7s in Figure 13.52b
The process from states 5 to 7s is also isentropic, so
and
State 1 in Figure 13.52b
The mass of air in the cylinder is
p 7s = p 5 ðV 5 /V 7s Þ k = p 5 ðCRÞ k = ð200: kPaÞð8:00Þ 1:35 = 3310 kPa
T 7s = T 5 ðV 5 /V 7s Þ k−1 = T 5 ðCRÞ k−1 = ð313 KÞð8:00Þ 0:35 = 648 K
m air = p 6sV 6s
RT 6s
= ð148 kPaÞð9:72 × 104 m 3 Þ
ð0:287 kJ/kg . KÞð648 KÞ
Since we have an air–fuel ratio of 15.0, the mass of fuel in the cylinder is m fuel =
The heat produced by the combustion process is then
= 1:73 × 10 − 3 kg
m air
AF + 1 = 1:73 × 10 − 3 kg
= 1:08 × 10 − 4 kg
15:0 + 1
Q comb = m fuel × Fuel heating value = ð1:08 × 10 −4 kgÞð43300 kJ/kg-fuelÞ = 4:69 kJ:
Also, Q comb = m air c v-air (T 1 – T 7s ) = (1.73 × 10 –3 kg)(T 1 – 648 K) = 4.69 kJ. Solving for T 1 gives T 1 = 3940 K. Then, since the
process from state 7s to 1 is a constant volume process, p 1 = p 7s (T 1 /T 7s ) = (3310 kPa)(3940 K/624 K) = 20.2 MPa.
State 2s in Figure 13.52b
The process from 1 to 2s is isentropic, so
k
V
p 2s = p 1
1 = ð20:2 × 10 3 kPaÞ 9:72 × 10−5 m 3 1:35
V 2s
9:72 × 10 −4 = 901 kPa
m 3
and
k−1
V
T 2s = T 1
1 = ð3920 KÞ 9:72 ×
10−5 m 3
V 2s
9:72 × 10 −4 = 1760 K
m 3
(Continued )