Post on 19-Jan-2015
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CHAPTER
5MEC 451Thermodynamics
Air Standard Cycle
Lecture Notes:MOHD HAFIZ MOHD NOHHAZRAN HUSAIN & MOHD SUHAIRIL Faculty of Mechanical EngineeringUniversiti Teknologi MARA, 40450 Shah Alam, Selangor
For students EM 220 and EM 221 only
thnet
in
W
Q
th CarnotL
H
T
T, 1
Upon derivation the performance of the real cycle is often measured in terms of its thermal efficiency
The Carnot cycle was introduced as the most efficient heat
engine that operate between two fixed temperatures TH
and TL. The thermal efficiency of Carnot cycle is given by
Review – Carnot Cycle
2
The ideal gas equation is defined as
mRTPVorRTPv
where P = pressure in kPav = specific volume in m3/kg (or V = volume in
m3)R = ideal gas constant in kJ/kg.Km = mass in kgT = temperature in K
Review – Ideal Gas Law
3 3
The Δu and Δh of ideal gases can be expressed as
)( 1212 TTCuuu v
)( 1212 TTChhh P
Δu - constant volume processΔh - constant pressure process
4
Process Description Result of IGL
isochoricconstant volume (V1 =
V2)
isobaricconstant pressure (P1 =
P2)
isothermalconstant temperature
(T1 = T2)
polytropic -none-
isentropicconstant entropy (S1 =
S2)
According to a law of constantnVP
2
2
1
1
T
P
T
P
2
2
1
1
T
V
T
V
2211 VPVP
1
2
1
1
2
2
1
n
nn
T
T
V
V
P
P
Review – Thermodynamics Processes
5
R = 0.2871 kJ/kg.KCp = 1.005 kJ/kg.K
Cv = 0.718 kJ/kg.K
k = 1.4
where R = ideal gas constant
Cp = specific heat at constant pressure
Cv = specific heat at constant volume
k = specific heat ratio
Review – Properties of Air
6
IC Engine – combustion of fuel takes place inside an engine’s cylinder.
Introduction
7
Air continuously circulates in a closed loop.
Always behaves as an ideal gas.
All the processes that make up the cycle are
internally reversible.
The combustion process is replaced by a heat-
addition process from an external source.
Air-Standard Assumptions
8
A heat rejection process that restores the working
fluid to its initial state replaces the exhaust process.
The cold-air-standard assumptions apply when the
working fluid is air and has constant specific heat
evaluated at room temperature (25oC or 77oF).
No chemical reaction takes place in the engine.
Air-Standard Assumptions
9
Top dead center (TDC), bottom dead center (BDC), stroke, bore, intake valve, exhaust valve, clearance volume, displacement volume, compression ratio, and mean effective pressure
Terminology for Reciprocating Devices
10
The compression ratio r of an engine is defined as
rV
V
V
VBDC
TDC
max
min
The mean effective pressure (MEP) is a fictitious pressure that, if it operated on the piston during the entire power stroke, would produce the same amount of net work as that produced during the actual cycle.
MEPW
V V
w
v vnet net
max min max min11
Otto CycleThe Ideal Cycle for Spark-Ignition Engines
12
The processes in the Otto cycle are as per following:
Process
Description
1-2 Isentropic compression
2-3 Constant volume heat addition
3-4 Isentropic expansion
4-1 Constant volume heat rejection
1
2
3
4Qout
Qin
PvgConstant
v1v2 v
P
s
T
Qout
Qin
1
2
3
4
(a) P-v diagram (b) T-s diagram 13
Related formula based on basic thermodynamics:
1
2
1
1
2
2
1
n
nn
T
T
V
V
P
P
1
2
1
1
2
2
1
n
nn
T
T
V
V
P
P
3 2in vQ mC T T
4 1out vQ mC T T
14
Thermal efficiency of the Otto cycle:
thnet
in
net
in
in out
in
out
in
W
Q
Q
Q
Q Q
Q
Q
Q
1
Apply first law closed system to process 2-3, V = constant.
Thus, for constant specific heats
Q U
Q Q mC T T
net
net in v
,
, ( )
23 23
23 3 2
,23 ,23 23
3
,23 ,23 ,23
2
0 0
net net
net other b
Q W U
W W W PdV
15
Apply first law closed system to process 4-1, V = constant.
Thus, for constant specific heats,
Q U
Q Q mC T T
Q mC T T mC T T
net
net out v
out v v
,
, ( )
( ) ( )
41 41
41 1 4
1 4 4 1
The thermal efficiency becomes
th Ottoout
in
v
v
Q
Q
mC T T
mC T T
,
( )
( )
1
1 4 1
3 2
,41 ,41 41
1
,41 ,41 ,41
4
0 0
net net
net other b
Q W U
W W W PdV
16
th Otto
T T
T T
T T T
T T T
,
( )
( )
( / )
( / )
1
11
1
4 1
3 2
1 4 1
2 3 2
Recall processes 1-2 and 3-4 are isentropic, so
Since V3 = V2 and V4 = V1,
3 32 4
1 4 1 2
T TT Tor
T T T T
11
32 1 4
1 2 4 3
kkTT V V
andT V T V
17
The Otto cycle efficiency becomes
th Otto
T
T, 1 1
2
Since process 1-2 is isentropic,
where the compression ratio is r = V1/V2 and
th Otto kr, 11
1
1
2 1
1 2
1 1
1 2
2 1
1
k
k k
T V
T V
T V
T V r
18
An Otto cycle having a compression ratio of 9:1 uses air as
the working fluid. Initially P1 = 95 kPa, T1 = 17°C, and V1 =
3.8 liters. During the heat addition process, 7.5 kJ of heat
are added. Determine all T's, P's, th, the back work ratio
and the mean effective pressure.
Example 5.1
Solution:Data given:
1
1
2
23
1
1
290
9
7.5
95
3.8
T K
VV
Q kJ
P kPa
V Litres
19
Example 5.1
10.42 1
21 2
11.42 1
21 2
290 9 698.4
95 9
Pr 1 2
Pr 2 3 .
2059
1 :
k
k
stnet net
T VT K
T V
P VP kPa
P V
law Q W
ocess isentropic compression
ocess Const volume heat addition
3
0
23 3 2
1 1 1 1
23 123 23
1
0.2871 290: 0.875
95
1727
v
mkg
kJkg
U
Q mC T T
IGL Pv RT v
Q vq Q
m V
20
Example 5.1
3 2
3 223 3 2
3 2
3 3
3
10.434
4 33 4
1.4344 3
3 4
:
0.718 698.4 9.15
3103.7
1/ 9 1288.8
1/ 9 422
Pr 3 4 exp
v
k
k
ocess isentr
Back to IGL ButV V
P Pq C T T
T T
T P MPa
T K
VTT T
opic ansi
KT V
VPP P kPa
P
on
V
21
Example 5.1
41 4 1
41 4 1
0.718 1288.8 290
717.
Pr 4 1 .
1
v
v
kJkg
Q mC T
ocess Const volume heat rejec
T
q C T
tion
T
Then:
23 41
,
1009.6
0.585 58.5%
net in out
kJkg
netth Otto
in
W q q
q q
W
q
22
Example 5.1
What else?
max min max min
1 2 1 2 1
11
12
exp 34
1 /
1009.61298
1 0.875 1 1/ 9
net net
net net
net
r
vcomprbw
ans
W wMEP
V V v v
w w
v v v v v
wkPa
v
Cw ur
w u
2 1
v
T T
C
3 4
0.225 22.5%
T T
23
Supplementary Problems 5.1
1. An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 27°C, and 750 kJ/kg of heat is transferred to air during the constant-volume heat-addition process. Taking into account the variation of specific heats with temperature, determine (a) the pressure and temperature at the end of the heat addition process, (b) the net work output, (c) the thermal efficiency, and (d) the mean effective pressure for the cycle.
[(a) 3898 kPa, 1539 K, (b) 392.4 kJ/kg, (c) 52.3 percent,(d ) 495 kPa]2. The compression ratio of an air-standard Otto cycle is 9.5. Prior to the isentropic compression process, the air is at 100 kPa, 35°C, and 600 cm3. The temperature at the end of the isentropic expansion process is 800 K. Using specific heat values at room temperature, determine (a) the highest temperature and pressure in the cycle; (b) the amount of heat transferred in, in kJ; (c) the thermal efficiency; and (d) the mean effective pressure.
[(a) 1969 K, 6072 kPa,(b) 0.59 kJ, (c) 59.4 percent, (d) 652 kPa] 24
The processes in the Diesel cycle are as per following:
Diesel Cycle
25
cv rratiooffCutv
vandrrationCompressio
v
v,,
2
3
2
1
Diesel Cycle
26
Related formula based on basic thermodynamics:
1
2
1
1
2
2
1
n
nn
T
T
V
V
P
P
1
2
1
1
2
2
1
n
nn
T
T
V
V
P
P
3 2in PQ mC T T
4 1out vQ mC T T
27
Thermal efficiency of the Diesel cycle
th Dieselnet
in
out
in
W
Q
Q
Q, 1
Apply the first law closed system to process 2-3, P = constant.
Thus, for constant specific heats
Q U P V V
Q Q mC T T mR T T
Q mC T T
net
net in v
in p
,
,
( )
( ) ( )
( )
23 23 2 3 2
23 3 2 3 2
3 2
,23 ,23 23
3
,23 ,23 ,23
2
2 3 2
0 0
net net
net other b
Q W U
W W W PdV
P V V
28
Apply the first law closed system to process 4-1, V = constant
Q U
Q Q mC T T
Q mC T T mC T T
net
net out v
out v v
,
, ( )
( ) ( )
41 41
41 1 4
1 4 4 1
Thus, for constant specific heats
The thermal efficiency becomes
th Dieselout
in
v
p
Q
Q
mC T T
mC T T
,
( )
( )
1
1 4 1
3 2
,41 ,41 41
1
,41 ,41 ,41
4
0 0
net net
net other b
Q W U
W W W PdV
29
PV
T
PV
TV V
T
T
P
P
4 4
4
1 1
14 1
4
1
4
1
where
Recall processes 1-2 and 3-4 are isentropic, so
PV PV PV PVk k k k1 1 2 2 4 4 3 3 and
Since V4 = V1 and P3 = P2, we divide the second equation by the first equation and obtain
Therefore,
34
4 2
k
kc
VPr
T V
, 1
111
1
kc
th Diesel kc
r
r k r
30
An air-standard Diesel cycle has a compression ratio of 18 and a cut-off ratio of 2.5. The state at the beginning of compression is fixed by P = 0.9 bar ant T = 300K. Calculate:
i. the thermal efficiency of the cycle,
ii. the maximum pressure, Pmax, and
iii. The mean effective pressure.
Example 5.2
Solution:
Data given:
1
2
3
2
18
2.5
V
V
V
V
31
Example 5.2
10.42 1
21 2
3 322 3 3 2
2 3 2
4 1 2
3 2 3
4
Pr 1 2
Pr 2 3
300 18 953.3
2383.
.
Pr 3 4 exp
3
. 18 1/ 2.5 7.2
k
ocess isentropic compression
ocess Const pressure heat addition
ocess isentropic ansio
T VT K
T V
V VVP P T T K
T T V
V V V
V V V
T
n
1
0.434
3 4
2383.3 1/ 7.2 1082k
VT K
T V
32
Example 5.2
23 3 2 3 2
41 4 1 4 1
1437.15
561.48
875.67
kJin P in p kg
kJout P out p kg
kJnet in out kg
Q Q mC T T q C T T
Q Q mC T T q C T T
w q q
What we need?
,
max 2 3
1
2 22 max
1 1
1
0.6093 60.93%
5148
875.67969.1
1 1/ 0.9566 1 1/18
netth diesel
in
k
k
net
wi
q
ii P P P
P TP kPa P
P T
wiii MEP kPa
V r
33
Supplementary Problems 5.2
1. An ideal diesel engine has a compression ratio of 20 and uses air as the working fluid. The state of air at the beginning of the compression process is 95 kPa and 20°C. If the maximum temperature in the cycle is not to exceed 2200 K, determine (a) the thermal efficiency and (b) the mean effective pressure. Assume constant specific heats for air at room temperature.
[ (a) 63.5 percent, (b) 933 kPa]
2. An ideal diesel cycle has a compression ratio of 16 to 1. The maximum cycle temperature is 1700C and the minimum cycle temperature is 15C. Calculate:i. the specific heat transfer to the cycleii. the specific work of the cycleiii. the thermal efficiency of the cycle
34
Dual cycle gives a better approximation to a real engine. The heat addition process is partly done at a constant volume and partly at constant pressure. From the P-v diagram, it looks like the heat addition process is a combination of both Otto and Diesel cycles.
Dual Cycle
35
The same procedure as to Otto and Diesel cycles can be applied to Dual cycle. Upon substitutions, the thermal efficiency of Dual cycle becomes
111
11
kvcpp
kcp
th rrckr
rr
Dual Cycle
36
At the beginning of the compression process of an air-standard dual cycle with a compression ratio of 18, the temperature is 300 K and the pressure is 1 bar. The pressure ratio for the constant volume part of the heating process is 1.5 to 1. The volume ratio for the constant pressure part of the heating process is 1.2 to 1. Determine (a) the thermal efficiency and (b) the mean effective pressure. (WRONG SOLUTION!!)
Example 5.3
Solution:
1 1
2 2
41
3
1
18 1.5
300 1.2
1
V P
V P
VT K
V
P bar
Data given:
37
10.42 1
21 2
3 322 3 3 2
2 3 2
4 1 2
3 2 3
4
Pr 1 2
Pr 2 3
300 18 953.3
2383.
.
Pr 3 4 exp
3
. 18 1/ 2.5 7.2
k
ocess isentropic compression
ocess Const pressure heat addition
ocess isentropic ansio
T VT K
T V
V VVP P T T K
T T V
V V V
V V V
T
n
1
0.434
3 4
2383.3 1/ 7.2 1082k
VT K
T V
38
11 1
5 34 4 45 4 4
4 5 5 3 5
0.41
181715.94 1.2
5
Pr 4 5
84.85
exp
kk k
ocess isentropic ansio
T VV V VT T T
V
n
T V V V
K
Information needed?
51 5 1
23 34 3 2 4 3
204.52
629.65
kJout v kg
in v p
kJkg
Q Q C T T
Q Q Q C T T C T T
m
m m
39
Answer the questions ?
11
118
204.521 1 0.675 67.5%
629.65
1
425.13
0.8613 1
522.63
net in out outth
in in in
net
r
W Q Q Qa
Q Q Q
Wb MEP
v
kPa
40
Indicated power (IP) Brake power (bp) Friction power (fp) and mechanical efficiency,
m
Brake mean effective pressure (bmep), thermal efficiency and fuel consumption
Volumetric efficiency, v
Criteria of Performance
41
Defined as the rate of work done by the gas on the piston as evaluated from an indicator diagram obtained from the engine using the electronic engine indicator.
2
LANnpIP i
For four-stroke engine,
And for two-stroke engine,
LANnpIP i
Indicated Power
42
ip = work done per cycle × cycle per minute
n is the number of cylinders.
Indicated Power
43
constantdiagramoflength
diagramofareanetip
Indicated mean effective pressure, pi given by:,
For one engine cylinder Work done per cycle = pi
A L Where A = area of piston
L = length of stroke
timeunitpercycleALPip i Power output = (work done per cycle) x (cycle
per minute)
For four-stoke engines, the number of cycles per unit time is N/2 and for two-stroke engines the number of cycles per unit time is N, where N is the engine speed.
volumentdisplaceme
cycleperdoneworkip
Brake power is a way to measure the engine power output.
The engine is connected to a brake (or dynamometer) which can be loaded so that the torque exerted by the engine can be measured.
The torque is obtained by reading off a net load, w at known radius, r.
Wr
Brake Power
44
Therefore
Nbp 2
22
LANnPLANnpIPbp bim
m
Brake power is also can be expressed as
Then the brake mean effective pressure (Pb) is
imb PP
45
Friction Power
46
The difference between the Ip and bp is the friction power (fp). It is the power that overcome the frictional resistance of the engine parts.
bpIPfp
Power input to the shaft is usually bigger than the indicated power due to frictional losses or the mechanical efficiency.
powerindicated
powerbrakemech
Mechanical Efficiency
47
Brake Mean Effective Pressure
48
From the definition of Brake power IPBP m
Since 2
LANnpIP i for 4 stroke engine
and 22
LANnPLANnpbp bim
Since and Pi are difficult to obtain, they may be combined and replaced by a brake mean effective pressure, Pb
m
m
Equating this equation to another definition of bp: NTLANnPb 22
TLAn
Pb
4 So:
Its observed that bmep is proportional to torque.
Brake Thermal Efficiency
49
The power output of the engine is obtained from the chemical energy of the fuel supplied. The overall engine efficiency is given by the brake thermal efficiency,
m
vnetf
p
fe
pbp Qm
b
P
b
,power equivalent fuel
power brake
givenpower
power brake
mf = mass flow fuel , Qnet,v = net calarofic value of the fuel.
sfc is the mass flow rate of fuel consumed per unit power output and is a criterion of economical power production.
bp
msfc f
Specific Fuel Consumption
50
Volumetric efficiency is only used with four-stroke cycle engine, which have a distinct induction process.
The parameter used to measure the effectiveness of an engine’s induction process is the volumetric efficiency.
sV V
V
Volumetric Efficiency
51
An engine operating at 2400 rpm consumes 12 ml of fuel (s.g. 0.85) in 60 second. The engine indicates a load of 30 N on the pony brake system and the brake’s torque arm is 20 cm. Determine (a) the brake power, (b) the mass flow rate of fuel, and (c) the specific fuel consumption.
Example 5.4
Solution:
52
A four-cylinder petrol engine has a bore of 57 mm and a stroke of 90 mm. Its rated speed is 2800 rpm and it is tested at this speed against a brake which has a torque arm of 0.356 m. The net brake load is 155 N and the fuel consumption is 6.741 l/h. The specific gravity of the petrol used is 0.735 and it has a net calorific value of 44,200 kJ/kg. The engine is tested in an atmospheric condition at 101.325 kPa and 15 oC at air-fuel ratio of 14.5/1. Calculate for this speed, the engine torque, the bmep, the brake thermal efficiency, the specific fuel consumption and the volumetric efficiency of the engine.
Example 5.4
Solution:
53
2
LANnpIP i
Nbp 2
Real Case
54