Chapter 1 Gas Power Cycle

Chapter 1
Gas Power Cycle
Content:
Air Standard Cycle,
Air Standard Carnot cycle
Reciprocating Engines,
Air Standard Otto Cycle (Spark Ignition)
Ideal Diesel Cycle (Compression Ignition)
Ideal Brayton Cycle (Gas turbine engines)
Gas cycles VS Vapor cycles
Gas cycles :
Working fluid remains in the gaseous throughout
the entire cycle applied in gasoline engine, diesel
engine, gas turbine engine
Source: Assoc.Prof.Dr.Sommai Prepream

Gas cycles VS Vapor cycles (cont’d)
Vapor cycles :
Working fluid exists in the vapor phase during
on part of the cycle and in the liquid phase
during another part such as in steam power
plant cycle and refrigeration cycles
The Carnot Cycle and Its Value in Engineering
Heat
Engine
Source, T H
Sink, T L
Q H
Q L
W net
The 4 processes of the carnot cycle: (Heat Engine)
� Reversible Isothermal heat transfer from high temp.
reservoir.
� Reversible adiabatic expansion.
� Reversible Isothermal heat transfer to low temp. reservoir
� Reversible adiabatic compression

Example 8.1 Show that the thermal efficiency of a Carnot
cycle operating between the temperature limits of T L AND
T H is solely a function of these two temperatures.
Solution
from
Wnet
QL
ηth
= = 1−
QH
QH
TH 1
2⎛
δQ
⎞
and S2
− S1
= ∫ ⎜ ⎟
1 ⎝ T ⎠rev
for rev. Isothermal process, T = const.
TL 4
1 2
1Q2
then S2
− S1
= Q
T ∫ δ =
1 rev T
or
1Q2
= T ( S2
− S1)
Therefore,
QH
= TH
( S2
− S1),
and QL
= TL
( S3
− S4
)
But S1
= S4,
and S2
= S3
QL
TL
( S3
− S4
) TL
ηth
= 1−
= 1−
= 1−
Q T ( S − S ) T
H
H
Actual Gas Cycle
2
1
H
T
Q H
Q L
s 1 =s 4 s2 =s 3
� Open Cycle (intake, discharge)
� Working fluid is not a pure
substance
� Heat input by COMBUSTION of
a fuel
� Involve friction
Source: Assoc.Prof.Dr.Sommai Prepream
Exhaust
gas
2
3
s
Fuel + air
mixture

Intake
valve
Bore
Air-Standard Assumption
� The working fluid is air which continuously
circulated in a close loop and behave as an
ideal gas
� All the processes are internally reversible
� The combustion process is replaced by a
heated-addition process from the external
source
� The exhaust process is replaced by a heat
rejection process which restores the
working fluid to its initial state
Over View on Reciprocating Engines
Exhaust
valve
Stroke
TDC
BDC
Top Dead Center (TDC) : Position of the piston
when it forms the smallest volume in diameter
Bottom Dead Center (BDC) : Position of the
piston when it forms the largest volume in diameter
Stroke : Length of piston travel
Bore : Diameter of the cylinder
Clearance Volume (Vc ) : minimum volume
formed in the cylinder when the piston is at TDC
Displacement Volume (Vd ) :Swept Volume (Vmax-Vmin )
Source: Assoc.Prof.Dr.Sommai Prepream
Compression Ratio (r v ) = (V max /V min ) = (V BDC /V TDC )
Mean Effective Pressure (MEP) : Fictious pressure that acted on the
piston during the entire power stoke
Net Work during the actual cycle (Wnet ):
Wnet = (MEP) x (Piston Area) x (Stroke) = (MEP) x (Displacement Volume)

P
Mean Effective Pressure, MEP Concept
Actual Processes
v min
W net
Equivalent
v max
TDC BDC
Source: Assoc.Prof.Dr.Sommai Prepream
Four Stroke Engine
v
P
MEP
Equivalent by MEP
v min
W net
v max
Wnet = (MEP) x (Displacement Volume)
= (MEP) x (Vmax-Vmin )
Intake Compression Power Exhaust
1. Intake Stroke piston moves from TDC to BDC,
drawing in fresh air-fuel mixture.
2. Compression Stroke piston moves from BDC to
TDC, compress air-fuel mixture.
3. Power Stroke piston at TDC, spark plug ignite
the air-fuel mixture. the combustion occur
very fast that, in theory, the piston still at
TDC. After that the piston is pushed to BDC.
4. Exhaust Stroke piston moves from BDC to TDC,
pushes the combustion gases out.
v

Two Stroke Engine Power
Compression
Intake &
Exhaust
1. Compression Stroke piston moves from
BDC to TDC, compress air-fuel
mixture.
2. Power Stroke piston at TDC, spark plug
ignite the air-fuel mixture. After the
piston is pushed to BDC. Meanwhile,
about half way, combustion gases are
discharged and fresh air-fuel mixture
is drawing in .
Actual and Ideal cycles in spark-ignition engine
Actual four-stoke spark-ignition
engine
Ideal Otto
cycle
Under the Air-Standard Assumption

P
Air Standard Otto Cycle
Ideal Otto cycle consists of four internally reversible processes:
Process 1-2 Isentropic Compression
Process 2-3 v = constant, heat added
Process 3-4 Isentropic expansion
Process 4-1 v = constant, heat rejection
3
2
v 2 =v 3
Pv k = c
Pv k = c
w in
w out
1
v 1 =v 4
TDC BDC
Source: Assoc.Prof.Dr.Sommai Prepream
4
v
T
1
2
v = const.
v = const.
q in
q out
s 1 =s 2 s3 =s 4
There are only 2-stroke of all 4-processes,
Analysis of Air Standard Otto Cycle
Constant volumeheat
transfer
1
st
v = const.
⇒ w = 0
Idealgas:
and
IsentropicProcessof
Idealgases
and
law : closedsystem
2
2
2
q
in
3
out
k
3
= u − u + w
3
=
2
4
3
3
1
= P
3
k
1v1
2
2
q = u − u
Pv
q
Pv
RT, du
= q = C (T −T
)
k
v
v
3
= P
2
=
q = q = C (T −T
)
1
3
k
2v2
C dT
P ⎛ 2 v ⎞ ⎛ 1 V ⎞ 1
= ⎜
⎟ = ⎜
⎟
P1
⎝ v2
⎠ ⎝V2
⎠
( k−1)
/ k
T ⎛ ⎞ ⎛ ⎞
2 P2
v1
= ⎜
⎟ = ⎜
⎟
T1
⎝ P1
⎠ ⎝ v2
⎠
2
4
= constant
k
v
k −1
P
T
3
2
v 2 =v 3
1
Pv k = c
2
s 1 =s 2
3
4
Pv k = c
v = const.
v =
const.
w in
q in
q out
s
w out
4
1
v 1 =v 4
3
4
s3 =s 4
v
s

Analysis of Air Standard Otto Cycle (cont’d)
Thermal efficiency
w
η th =
q
w
1
2
3
C
= c,
k =
C
P2v
2 − P1v
1
1w
2 =
1−
k
R(
T2
− T1
)
1w
2 =
1−
k
qL
4 q
or,
η th = 1−
= 1−
q q
Mean Effective Pressure
w
net
w =
Pv
k
net
= w + w
∫
net
in
Pdv
= MEP ( v − v )
4
H
1
p
v
2
2
1
3
P
T
3
2
v 2 =v 3
1
Pv k = c
2
s 1 =s 2
Pv k = c
v = const.
v =
const.
Spark-ignition Engine
1. The higher r v the higher thermal eff.
2. The higher r v cause Self-Ignition � engine knock
3. Higher Octane Number of fuel used retard the self-ignition
4. Typical r v of gasoline engine ~ 9.0 – 10.0
5. Thermal efficiency of actual spark ignition engine ~ 25-30%
v
w in
q in
q out
w out
4
1
v 1 =v 4
3
4
s3 =s 4
v
v
s

Example 8.2 An ideal Otto cycle has a compression ratio of 8. At the
begining of the compression process, the air is at 100 kPa and 17 o C,
and 800 kJ/kg of heat is transfered to air during the heat addition
proceed. Accounting for the variation of specific heats of air with
temperature, determine, (a) the maximum temperature and pressure
which occur during the cycle, (b) the net work out put, (c) the thermal
efficiency, and (d) the mean effective pressure of the cycle
Given:
rv = 8.0
P1 = 100 kPa and
T1 =17oC qH = 800 kJ/kg
variation of specific
heats
Determine:
a) T max
b) w net
c) η th
d) MEP
P
3
2
v 2 =v 3
Pv k = c
Pv k = c
4
1
v 1 =v 4
v
T
1
2
s 1 =s 2
v = const.
v = const.
(a) T max = T 3 :find state 2 and then 3 using Ideal gas eqn.
and Table A-17
Table A-17 ; T 1 = 290 K � u 1 = 206.91 kJ/kg, and v r = 676.1
w in
1-2 Isentropic proc. v r2/v r1 = v 2/v 1 , v 2/v 1 =1/r v = 1/8,
� v r2 = v r1/r v = 676.1/8.0 = 84.51
Table A-17 : at v r2 = 84.51 � T 2 = 652.4 K and u 2 = 475.11 kJ/kg,
P 2v 2 /T 2 = P 1v 1/T 1 ,
P 2 = P 1(v 1/v 2 )(T 2/T 1 ) = (100 kPa)(8.0)(652.4/290) = 1799.7 kPa
2-3 Constant volume heat added,
1 st law q 23 = w 23 + u 3 –u 2 ; w 23 = 0
q 23 = u 3 –u 2 ; 800 kJ/kg = u 3 – 475.11
� u 3 =1275.11 kJ/kg � table A-17: T 3 = 1575.1 K and v r3 = 6.108
P 3v 3/T 3 = P 2v 2 /T 2 ,
P 3 = P 2(v 2/v 3 )(T 3/T 2 ) = (1799.7 kPa)(1/8.0)(1575.1/652.4) = 543.4 kPa
T max =T 3 = 1575.1 K answer
w out
q in
q out
3
4
s3 =s 4
s

(b) w net = q H –q L , similar to q 23 ; -q L =q 41 = u 1 –u 4
3-4 Isentropic proc.
v r4/v r3 = v 4/v 3 , v 4/v 3 =r v = 8, � v r4 = v r3r v = 6.108*8.0 = 48.864
Table A-17 : at v r4 = 48.864 � T 2 = 795.6 K and u 4 =588.74 kJ/kg,
4-1 Constant volume heat rejected,
1 st law q 41 = w 41 + u 1 –u 4 ; w 41 = 0
q 41 = u 1 –u 4 = 206.91 - 588.74 = -381.83 kJ/kg
q L = -q 41 = 381.83 kJ/kg
w net = q H –q L = 800 – 381.83 = 418.17 kJ/kg answer
(c) η th = w net /q H = 418.83/800 = 0.523 or 52.3 % answer
(d) MEP = w net /(v 1-v 2 ) ; P 1v 1 = RT 1
� v 1 = 0.832 m 3 /kg , v 2 =v 1/8,
MEP = 574.4 kPa answer
Spark-ignition Engine
Diesel cycles: the ideal cycle for
compression-ignition engine
Ideal diesel cycle consists of four internally reversible processes:
Process 1-2 Isentropic Compression
Process 2-3 P = constant, heat added
Process 3-4 Isentropic expansion
Process 4-1 v = constant, heat rejection
P
2
q in
v 2 =v 3
3
Pv k = c
Pv k = c
4
1
v 1 =v 4
q out
v
T
1
2
P = const.
v = const.
q in
q out
s 1 =s 2 s3 =s 4
4
3
s

1st law : closed system
Ideal gas:
and
Analysis of Ideal Diesel Cycle
Constant pressure heat transfer
2
P = const.
⇒ w = P ( v − v )
and
2
2
4 1
3
3
3
2
= − q = C ( T −T
)
Isentropic Process of Ideal gases
3
1
4
3
2
3
1
k
1v1
4
3
Pv = RT, dh = C dT
q = q = C (T −T
)
in
2
2
q = u − u = −C
( T −T
)
q
q = u − u + w
out
k
Pv = P
k
p
3
v
= P
2
2
4
2
k
2v2
P ⎛ 2 v ⎞ ⎛ 1 V ⎞ 1
= ⎜
⎟ = ⎜
⎟
P1
⎝ v2
⎠ ⎝V2
⎠
( k −1)
/ k
T ⎛ ⎞ ⎛ ⎞
2 P2
v1
= ⎜
⎟ = ⎜
⎟
T1
⎝ P1
⎠ ⎝ v2
⎠
3
q = u − u + P ( v − v ) = h − h
v
2
3
4
1
= constant
k
p
k −1
2
1
3
2
P
T
2
q in
v 2 =v 3
2
1
s 1 =s 2
3
Pv k = c
P = const.
Pv k = c
q in
v = const.
Analysis of Ideal Diesel Cycle
Cut off ratio (r c ) is the ratio of the cylinder
volume after and before the combustion process
Thermal
w
η th =
q
from
at
net
H
r c
efficiency
q
= 1 −
q
V3
v3
= =
V v
T1
( T4
/ T1
− 1)
= 1 −
kT ( T / T − 1)
2
definition
processes
3
out
in
of
2
2
1 − 2 and 3 − 4
2
T4
− T1
= 1 −
k ( T − T )
r
c
and
k
1 ⎡ r ⎤
c − 1
η th = 1 − k −1
⎢ ⎥
r ⎣ k ( rc
− 1)
⎦
3
2
isentropic
P
T
2
q in
v 2 =v 3
2
3
Pv k = c
1
s 1 =s 2
Pv k = c
P = const.
q in
v = const.
q out
4
1
v 1 =v 4
q out
4
1
v 1 =v 4
4
3
s3 =s 4
q out
v
4
3
s3 =s 4
q out
v
s
s

Compression Ignition Engine
1. The thermal efficiency of Diesel cycle always higher than Otto cycle
( η th , diesel > ηth,
otto ).
2. The diesel engines operated at the higher rv and burn the fuel more
completely.
3. Typical rv of diesel engine ~ 12.0 – 23.0
4. In a gasoline engine, it’s air intake is carefully restricted and controlled by
the carburetor for a 15:1 air to fuel ratio. However, in the diesel engine,
the air intake is unrestricted.
5. Thermal efficiency of actual diesel engine ~ 35-40%.
Compressor
Typical
compression
ratios for
diesel
engines
Brayton Cycle: The Ideal Cycle for
Gas-turbine Engines
Combustion Chamber
Turbine
An open-cycle gas turbine engine
Win
v
qin
Heat Exchanger
Compressor Turbine
Heat Exchanger
qout
Wout
An closed-cycle gas turbine engine

Ideal Brayton Cycle
Brayton Cycle consists of four internally reversible processes:
Process 1-2 Isentropic Compression (in a compressor)
Process 2-3 P = constant, heat added
Process 3-4 Isentropic expansion (in a turbine)
Process 4-1 P = constant, heat rejection
qin
T
W Wout
s
in
= const.
22
4
2
3
2
4
3
qout
P = const. at process 2 -3
and 4 -1
and P = P and P = P
1
qin
P = const.
P = const.
qout
3
s 1 =s 2 s3 =s 4
Analysis of Ideal Brayton Cycle
The energy balance for a steady - flow process for Brayton cycle
and
( q − q ) + ( w − w ) = h
in
q = q = h − h = C (T −T
)
q
in
out
out
1
3
4
2
= q = h − h = C (T −T
)
1
Isentropic Process of Ideal gases at process1
- 2 and 3-
4
net
H
k
Pv = P
in
out
p
p
exit
3
− h
4
inlet
2
1
4 1
k k
1v1
= P2v
2 = constant
1
qout
( k −1) / k
⎞ 2
( k −1)
/ k
⎛ P ⎞ 3 T3
s 1 =s 2
s3 s3 =s 4 s
and
T ⎛ 2 P
=
T ⎜
1 P ⎟
⎝ 1 ⎠
= ⎜
P ⎟
⎝ 4 ⎠
=
T4
Thermal efficiency
w
η th =
q
qout
= 1 −
q
T4
− T1
T1
( T4
/ T1
− 1)
= 1 −
= 1 −
k ( T − T ) kT ( T / T − 1)
Pressure
P2
ratio ( rp
) = ; Then
P1
η th = 1 −
r
1
in
from isentropic at processes 1 − 2 and 3 − 4 and
k −1/
k
p
3
2
2
3
2
s
T
P
2
2
qin
qin
P = const.
P = const.
qin
3
s = const.
qout
3
1 4
qout
4
v

Gas Turbine Engine
1. The thermal efficiency of an ideal Brayton cycle depends on the
pressure ratio and k.
2. In commom designs, the pressure ratio should be in range of 11-16,
however, it can work in the range at r p= 5 – 20.
3. At the fixed turbine inlet temp. (T 3), the net work output increase with
the pressure ratio, reaches a maximum, and then starts to decrease.
4. Cycle efficiency can be improved through increasing the tutbine inlet
temp., increasing the efficiency of turbomachinery components,
modification of the basic cycle.