Fundamentals of Thermodynamics
Chapter 10
Power and Refrigeration Systems
Gaseous Working Fluids
Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
10.1 Air-standard power cycles
‑ Working fluid : Fixed mass of air
‑ Combustion process : replaced by heat transfer from external s
ource
‑ Exhaust & intake process : replaced by heat transfer to the surr
ounding
‑ Internally reversible
‑ Constant specific heat
Internal Combustion Engine – Otto, Diesel, gas-turbine
External Combustion Engine – Steam, Stirling
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
10.2 The Brayton cycle
2개의 P = const
2개의 S = const
process를 구성
QL
hth = 1 QH
working fluid : gas ( single phase )
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
C p (T4 - T1 )
QL
hth = 1 = 1QH
C p (T3 - T2 )
= 1-
T1 (T4 / T1 - 1)
T2 (T3 / T2 - 1)
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
P3 P2
=
P4 P1
P2 æ T2 ö
=ç ÷
P1 è T1 ø
k
k -1
P3 æ T3 ö
=
=ç ÷
P4 è T4 ø
k
k -1
T3 T2
T T
T
T
=
® 3 = 4 ® 3 -1 = 4 -1
T4 T1
T2 T1
T2
T1
æ P1 ö
T1
\hth = 1 - = 1 - ç ÷
T2
è P2 ø
k -1
k
1
= 1-
( P2 / P1 )
k -1
k
1- 2¢ - 3¢ - 4 - 1 : TH,avg ­ ® h ­
1- 2¢ - 3¢¢ - 4¢¢ - 1: TH,avg ­ ® h ­, kg당 work 감소
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
ì Pv k = const
í
î Pv = RT
RT k
× v = const
v
Tv k -1 = const
k
æ RT ö
pç
÷ = const
è p ø
p1- k × T k = const
T ×P
1- k
k
= const
T2 æ P1 ö
=ç ÷
T1 è P2 ø
1- k
k
æ P2 ö
=ç ÷
è P1 ø
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k -1
k
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
실제 cycle
compressor가 40 ~ 80% 차지
hcomp =
h2 s - h1
h2 - h1
hturb =
h3 - h4
h3 - h4 s
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
Ex. 10.1
In an air-standard Brayton cycle, the air enters the compressor at 0.1 Mpa
and 15℃. The pressure leaving the compressor is 1.0 MPa, and the maxi
mum temperature in the cycle is 1100℃. Determine
1. The pressure and temperature at each point in the cycle.
2. The compressor work, turbine work, and cycle efficiency.
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
Ex. 10.2
Consider a gas turbine with air entering the compressor under the same c
onditions as in Example 10.1 and leaving at a pressure of 1.0 Mpa. The m
aximum temperature is 1100℃. Assume a compressor efficiency of 80 %,
a turbine efficiency of 85 %, and a pressure drop between the compressor
and turbine of 15 kPa. Determine the compressor work, turbine work, and
cycle efficiency.
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
10.3 The simple gas-turbine cycle with a regenerator
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
qH = C p (T3 - Tx )
wnet wt - wc
hth =
=
qH
qH
wt = C p (T3 - T4 )
Ideal case: Tx =T4 ® q H =w t
C p (T2 - T1 )
qH - wc
wc
hth =
= 1= 1qH
wt
C p (T3 - T4 )
k -1
é
ù
k
éæ T2 ö ù
P
æ 2 ö - 1ú
ê
1
ç
÷
T1 êëè T1 ø úû
T1 ê èç P1 ø÷
ú
= 1= 1k -1 ú
T3 é æ T4 ö ù
T3 ê
ê1 - æç P1 ö÷ k ú
êë1 - çè T3 ÷ø úû
ëê è P2 ø ûú
k -1
é
ù
k
k -1
P
æ
ö
ê1 - ç 1 ÷ ú
k
P2 ø ú
T æP ö
= 1- 1 ç 2 ÷ ê è
k -1 ú
T3 è P1 ø ê
ê1 - æç P1 ö÷ k ú
êë è P2 ø úû
= 1-
T1 æ P2 ö
ç ÷
T3 è P1 ø
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k -1
k
= 1-
T2
T3
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
regenerator efficiency
hreg
hx - h2
=
hx ' - h2
hreg
Tx - T2
=
Tx ' - T2
if C p = const.
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
Ex. 10.3
If an ideal regenerator is incorporated into the cycle of Example 10.1, det
ermine the thermal efficiency of the cycle.
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
10.4 Gas-turbine power cycle configurations
Ericsson Cycle
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2개의 const. T
2개의 const. P
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
Ex. 10.4
An air-standard power cycle has the same states given in Example 10.1. I
n this cycle, however, the compressor and turbine are both reversible, isot
hermal processes. Calculate the compressor work and the turbine work, a
nd compare the results with those of Example 10.1.
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
10.5 The air-standard cycle for jet propulsion
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
Ex. 10.5
Consider an ideal jet propulsion cycle in which air enters the compressor
at 0.1 Mpa and 15℃. The pressure leaving the compressor is 1.0 Mpa, an
d the maximum temperature is 1100℃. The air expands in the turbine to a
pressure at which the turbine work is just equal to the compressor work.
On leaving the turbine, the air expands in a nozzle to 0.1 Mpa. The proce
ss is reversible and adiabatic. Determine the velocity of the air leaving th
e nozzle.
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
10.6 The air-standard refrigeration cycle
CP (T1 - T4 )
qL
qL
h1 - h4
COP ( b ) =
=
=
»
wnet wC - wE h2 - h1 - ( h3 - h4 ) CP (T2 - T1 ) - CP (T3 - T4 )
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
P2 æ T2 ö
=ç ÷
P1 è T1 ø
k / ( k -1)
P3 æ T3 ö
=
=ç ÷
P4 è T4 ø
k / ( k -1)
T1 - T4
1
1
b=
=
=
1
T
/
T
T2
T
T2 - T1 - T3 + T4
3
2
2
-1
T1 1 - T4 / T1 T1
=
1
rp( k -1) / k
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
Ex. 10.6
Consider the simple air-standard refrigeration cycle of Fig. 10.12. Air ent
ers the compressor at 0.1 MPa and -20℃ and leaves at 0.5 MPa. Air enter
s the expander at 15℃. Determine
1. The COP for this cycle.
2. The rate at which air must enter the compressor to provide 1 kW of
refrigeration.
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
10.7 Reciprocating engine power cycles
Stroke : S = 2 Rcrank
Displacement : Vdispl = N cyl (Vmax - Vmin ) = N cyl Acyl S
Compression ratio : rv = Vmax / Vmin
Net work per cylinder:Wnet = mwnet = Pmeff (Vmax - Vmin )
RPM
RPM
&
Power : W = N cyl mwnet
= Pmeff Vdisp
60
60
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
10.8 The Otto cycle
2개의 S = const
2개의 V = const
process를 구성
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
mCv (T4 - T1 )
QH - QL
QL
h th =
= 1= 1QH
QH
mCv (T3 - T2 )
æ T4 - 1ö
1- k
ç
÷
æ v1 ö
T1 è T1 ø
T1
1
= 1= 1 - = 1 - ç ÷ = 1 - rv1- k = 1 - k -1
T2 æ T3 - 1ö
T2
rv
è v2 ø
ç T
÷
è 2 ø
Pv k = const
Pv = RT
Pv
1 1 = RT1
P2 v2 = RT2
R T1v1k -1 = R T2 v2 k -1
T2 æ v1 ö
=ç ÷
T1 è v2 ø
k -1
æ v4 ö
=ç ÷
è v3 ø
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k -1
T3
=
T4
Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
tetraethyl lead ® knocking 방지
옥탄가
iso - octan ( C8 H 18 ) 과 n - heptan ( C7 H 16 ) 의 질량비로 결정
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
*Deviation
1. Cv ­
T­
2. Incomplete Combustion
3. 배기 흡기를 위한 Work
4. Heat Loss
5. Irreversibilities
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
Ex. 10.7
The compression ratio in an air-standard Otto cycle is 10. At the beginnin
g of the compression stoke, the pressure is 0.1 MPa and the temperature is
15℃. The heat transfer to the air per cycle is 1800 kJ/kg air. Determine
1. The pressure and temperature at the end of each process of the cycle.
2. The thermal efficiency.
3. The mean effective pressure.
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
10.9 The Diesel cycle
hth = 1 = 1-
QL
QH
Cv ( T4 - T1 )
C P ( T3 - T2 )
T1 (T4 / T1 - 1)
= 1×
kT2 (T3 / T2 - 1)
(
Effect
i ) 1 - 2 - 3¢ - 4¢ - 1
ii ) Otto 1 - 2 - 3¢¢ - 4 - 1
iii ) Otto 1 - 2¢ - 3 - 4 - 1
h¯
h­
h¯
)
k
é
b
-1 ù
1
ú
hth = 1 - k -1 ê
rv ê k ( b - 1) ú
ë
û
b : cutoff ratio
v T
b= 3= 3
v2 T2
최고온도를 제한했을 때에는 h Diesel >hOtto
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
T - S 선도상에서 등압곡선과 등적곡선의 기울기
v = const
Tds = du + Pdv = Cv dT
¶T ö
T
=
÷
¶S øv Cv
P = const
Tds = dh - vdP = C p dT
¶T ö
T
=
÷
¶S ø p CP
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
Ex. 10.8
An air-standard diesel cycle has a compression ratio of 20, and the heat tra
nsferred to the working fluid per cycle is 1800 kJ/kg. At the beginning of t
he compression process, the pressure is 0.1 MPa and the temperature is 1
5℃. Determine
1. The pressure and temperature at each point in the cycle.
2. The thermal efficiency.
3. The mean effective pressure.
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
10.10 The Stirling cycle
2개의 T = const
2개의 V = const
process를 구성
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
10.11 The Atkinson and Miller cycles
Higher expansion ratio than compression ratio
For the compression amd expansion processes(s=const),
T2 æ v1 ö
=ç ÷
T1 è v2 ø
k -1
T4 æ v3 ö
,
=ç ÷
T3 è v4 ø
k -1
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
The heat rejection process gives
æ v4 ö
P = C : T4 = ç ÷ T1 , qL = h4 - h1
è v1 ø
The efficiency of the cycle becomes
qH - qL
qL
h4 - h1
h=
= 1= 1qH
qH
u3 - u2
C p (T4 - T1 )
T4 - T1
= 1= 1- k
Cv (T3 - T2 )
T3 - T2
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
The smaller compression ratio : CR1 = ( v1 / v3 )
The expansion ratio : CR = ( v4 / v3 )
æ
æ v4 ö
CR
k -1
T
=
T
CR
,
T
=
T
=
T1
ç 2 1 1
ç ÷ 1
4
v1 ø
CR1
è
ç
®
ç
k
CR
CR
k -1
ç T3 = T4CR k -1 =
T
CR
=
T1
1
ç
CR1
CR1
è
The efficiency of the cycle becomes
CR
-1
CR1
CR - CR1
h = 1- k
= 1- k
k
k
k
CR
CR
CR
1
- CR1k -1
CR1
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
Cycle in between the Otto cycle and Atkinson cycle
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
10.12 Combined-cycle power and refrigeration syste
ms
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Chapter 10. Power and Refrigeration Systems – Gaseous Working Fluids
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