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Ψ
T
1
( 29 Ψ
325K
(
)
=
70
J
kgair
K
⋅
=
Ψ
T1
70
J
kgair
K
⋅
=
Ψ
T2
Ψ
T1
nuR
bar
ln
1
r
c
⋅

=
Ψ
T2
832.688
J
kgair
K
⋅
=
From Figure 4.4
T
2
820K
=
T1 and T2 from Figure 43 gives
u
s.1
20
kJ
kgair
=
u
s.2
440
kJ
kgair
=
Compression Work
W
c1.2
u
s.2
u
s.1

420000
J
kg
⋅
→
=
To determine P2, use ideal gas equation
P
2
nuR
bar
T
2
⋅
v
2
=
need v.2, must find v.1 first
v
1
nuR
bar
T
1
⋅
P
1
=
v
1
0.47
m
3
kgair
=
v
2
v
1
r
c
=
P
2
nuR
bar
T
2
⋅
v
2
=
P
2
7065kPa
=
HOMEWORK #6 5.11 and 5.13
Helpful hints: use Cutoff ratio = 2.18Note that State 4 is at low temperatures (not on Fig 45), use
isentropic relationships w/ constant avg. specific heats to find T4, P4 using gamma=1.3. To find u4, must
use Fig 410 with Temperature to find us,4. Relate by U4=us,4 – 1320 kj/kg airFor part c, use Eq 517
kPa
10
3
Pa
=
kgair
kg
=
kJ
10
3
J
=
MJ
10
6
J
=
5.11  Constant Pressure AirFuel Cycle.
Initial Conditions
P
1
200kPa
=
T
1
325K
=
φ
0.4
=
x
b
0.025
=
r
c
14
=
β
2.18
=
γ
1.3
=
a. Determine temperature and pressure at states 2,3,4 and 5
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This note was uploaded on 01/25/2011 for the course ME 476 taught by Professor Broch during the Spring '10 term at Nevada.
 Spring '10
 Broch
 Combustion

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