TH E RM ODYN AM I CS-I : Ai r Sta Standa ndard Cycle Cycless and and H eat Engi nes nes
A I R
S T A N DA R D
chapter V -
1
of 17
C Y C LE S A N D
H EA T E NG INE S
The chemical energy of the fuel is converted to thermal energy (heat), which in turn is converted into mechanical energy, or work. The work per unit time is the power propelling the car. The fuel used in internal combustion engines is a hydrocarbon mixture, such as gasoline, diesel fuel, alcohol, or gas.
Heat engine is a device that continuously converts heat to work (or power). The word ‘continuous’ is of critical importance in this definition and needs further elaboration; it means that the engine will continue to operate as long as the heat energy input is maintained. The ability of the device to convert heat to work does not necessarily make the device a heat engine. A heat engine is a device that operates in a cycle.
I. Air Standard Standard Otto Cycle
The Otto cycle cycle is a positive-displacement, spark ignition, open cycle, internal-combustion use today. The cycle was clearly described for the first time (1862) by a Frenchman, Beau de Rochas. Rochas . Fourteen years later, Nicholous A. Otto Otto (1832-1891), a German, independently
MI T-Schoo T-School of Mech Mecha anical nical E ngineeri ng
invented the same cycle, and going further than Beau de Rochas, he built an engine to operate on it. The ideal cycle consists of two isentropic and two constant volume processes. It is customary to analyze the ideal cycle as though there were no suction and exhaust strokes and as though the working substance were air only, such an analysis is referred to as an air-standard analysis. Processes: 1. Starting with the piston at the bottom dead center, compression proceeds isentropically from state 1 to 2. 2. Heat is added at constant volume from state 2 to state 3. 3. Expansion occurs isentropically from state 3 to 4. 4. Heat is rejected at constant volume from state 4 to 1.
TH E RM ODYN AM I CS-I : Ai r Standard Cycles and H eat Engi nes
chapter V -
2
of 17
V 1 V 4 and V 2 V 3
INTAKE STROKE: The intake valve is open; exhaust valve is initially open, then it closes and the piston moves down, bringing fresh air/fuel mixture into the cylinder. COMPRESSION STROKE: Both intake and exhaust valves are closed, and the air/fuel mixture is compressed by the upward piston movement POWER STROKE: Both intake and exhaust valves are closed; spark ignition and combustion occur, with the resultant pressure increase forcing the piston downward. EXHAUST STROKE: The exhaust valve is open, the intake valve is closed, and the upward movement of the piston forces the products of combustion from the engine.
S 1 S 2 and Two and Four Stroke Engine Four Stroke comprises induction (intake), compression, expansion (power) and exhaust strokes. The cycle is completed in two revolutions of the crankshaft. Two Stroke cycle combines compression in the cylinder with induction (intake) below piston; and expansion (power) in the cylinder with exhaust below the piston (crankcase). Ideally the two stroke engine would develop a larger power output than the four stroke engine of the same size at the same speed. In practice the efficiency of the two-stroke engine is less than that of the four stroke engine, though it can be improved by supercharging.
r k
V 1
S 3 S 4
Compression Ratio
V 2
r e
V 4
r P
P 3 , P 2
Expansion Ratio
V 3 Pressure ratio
(the ratio of pressure during isometric heat addition)
V 1 V 2
V 4
r e r k
V 3
V C V 2 V 3
where:
Clearance volume
c
is the percent clearance
V C c V D
V D V 1 V 2
V D V 1 V 2 1
V 2
V D c V D
r k 1
r k
1 c
c
Process 1 – 2 : Isentropic Compression T 2 P 2 T 1 P 1
k 1 k
V 1 V 2
k 1
k 1
T 2 T 1 r k
MI T-School of Mechanical E ngineeri ng
k 1
r k
TH E RM ODYN AM I CS-I : Ai r Standard Cycles and H eat Engi nes
chapter V -
of 17
3
W 1 2 U 12 mC V T 2 T 1
Q R Q41 mC V T 1 T 4
Q12 0
Q R mC V T 1 T 1r P Q R m C V T 1 1 r P
Process 2 – 3 : Isometric Heat Addition
T 3
P 3
T 2
P 2
Net Work
r P
W NET PdV W 12 W 23 W 34 W 41 W NET mC V T 2 T 1 0 mC V T 4 T 3 0
k 1
T 3 T 2 r P T 1 r P r k
W NET mC V T 1 T 2 T 3 T 4
W 23 0
W NET mC V T 1 1 r k
Q23 mC V T 3 T 2
k 1
W NET mC V T 1 r k Process 3 – 4 : Isentropic Expansion T 4 P 4 T 3 P 3
k 1 k
V 3 V 4
k 1
1 r k
k 1
W NET mC V T 1r k
k
k 1
k 1
1
r P 1 mC V T 1 1 r P r P 1 mC V T 1 r P 1 k 1
1
k 1
W NET m C V T 1 r P 1 r k Thermal Efficiency TH
W NET
P 4 TH
m C V T 1 r P 1 r k
Q A
P 1
m C V T 1r k
r k
k 1
r k
1
k 1
k 1
1
1
r P 1
1
r k
k 1
1 1 k 1 100% r k
TH
PVT-Relationship
P 2 T 1 P 1
k 1
Q R
A
W NET mC V T 1r k
Q41 mC V T 1 T 4
T 2
Q Q
W NET
W 41 0
k 1
r P
or
Process 4 – 1 : Isometric Heat Rejection T 4
k 1
r P 1 r P 1
Q43 0
r P r k
W NET m C V T 1 r P 1 r k
1 T 4 T 3 k 1 T 1 r P r k W 34 U 34 mC V T 4 T 3
T 1
k 1
Mean Effective Pressure
V 1 V 2
T 2 T 1
P 2 P 1 V 1 V 2
T 3 T 4
P 3 P 4 V 4 V 3
k 1
P 3 T 4 P 4 T 3
T 4 T 1
P 4 P 1 V 3 V 2
k 1 k
V 4 V 3
k 1
T 3
P MEP
T 2
P 2
k 1
V 1
r P 1
Heat Rejected
MI T-School of Mechanical E ngineeri ng
k 1
T 1r k k 1
k 1
1
V V 1 V 2 1
P MEP
V 4
m C V T 1 r P 1 r k
m C V T 1 r P 1 r k
Heat Added
Q A m C V T 1 r k
V D
V 1 1
P 3
Q A Q23 mC V T 3 T 2 Q A mC V T 1r P r k
P MEP
W NET
1
r k
k 1
V 1 k 1 1 P 1C V r P 1 r k 1 r 1 R k r k
k 1
P 1 r k r P 1 r k
k 1 r k 1
1
TH E RM ODYN AM I CS-I : Ai r Standard Cycles and H eat Engi nes
chapter V -
4
of 17
(2-3) Reversible, constant pressure heat addition (Ignition)
(3-4) Isentropic expansion to initial volume (Power Stroke)
(4-1) Reversible constant-volume heat rejection (Exhaust)
Advantages and Disadvantages of the Diesel Cycle Advantages Diesel engines are the most efficient of all heat engines. This follow from the fact that they are able to use high compression ratios and because they are able to use excess air to ensure thorough combustion of the fuel.
Because Diesel engines do not require electrical equipment to operate and do not need a precise air-fuel ratio for combustion, they are very reliable and will operate under adverse conditions. For example, moisture tends to cause malfunction of the high tension electrical gear in the spark-ignition engines but this is not a problem in diesel engines.
Effect of Compression Ratio on Thermal Efficiency
Diesel engines use excess air and do not require fuel additive and therefore produce less combustion pollutants. Also, the fuel is less volatile and less fuel vaporization to the atmosphere occurs.
Diesel engines are the safest of all the heat engines. Diesel fuel is less volatile than petrol and does not have the explosive problems associated with gaseous fuels. Disadvantages
They have lower specific power output than spark-ignition engines. This is mainly because combustion is slower and this limits the operating speed. Lower speed means lower power output because of the basic relationship between power and speed. Also the higher compression ratio requires a heavier flywheel and starting equipment.
Cylinder temperatures vary between 20K and 2000K so 1.2 < k < 1.4 k = 1.3 most representative
They cost more than spark- ignition engines. This is because of the greater mass of the engine (more materials) and the higher cost of fuel-injection and starting equipment.
II. Air Standard Diesel Cycle
The Diesel cycle is a positive displacement, open cycle, internal, compression ignition engine invented by Rodulf Diesel in 1860s. The Diesel engine differs from the gasoline powered Otto cycle by using a higher compression of the fuel to ignite the fuel rather than using a spark plug ("compression ignition" rather than "spark ignition"). In the diesel engine, air is compressed adiabatically with a compression ratio typically between 15 and 20. This compression raises the temperature to the ignition temperature of the fuel mixture which is formed by injecting fuel once the air is compressed. The ideal air-standard cycle is modeled as a reversible adiabatic compression followed by a constant pressure combustion process, then an adiabatic expansion as a power stroke and an isometric exhaust. A new air charge is taken in at the end of the exhaust, as indicated by the processes the diagram. The four steps of the air-standard Diesel Cycle are outlined below:
(1-2) Isentropic Compression (Compression Stroke)
MI T-School of Mechanical E ngineeri ng
DIESEL ENGINE CYCLINDER
TH E RM ODYN AM I CS-I : Ai r Standard Cycles and H eat Engi nes
chapter V -
V D V 1 V 2 V D
5
of 17
1
V 2
r k 1
c V D
r k
1 c
c
V V 4 3 V 2 V 3 V 2 V 4
r k r C r e
Process 1 – 2 : Isentropic Compression T 2 P 2 T 1 P 1
PV DIAGRAM OF DIESEL CYCLE
k 1
V 1 V 2
k
k 1 k 1
r k
k 1
V 1 V 4 and P 2 P 3
T 2 T 1 r k
W 12 U 12 mC V T 2 T 1
Q12 0 Process 2 – 3 : Isobaric Heat Addition T 3 T 2
V 3 V 2
r C k 1
T 3 T 2 r C T 1 r C r k
W 23 P V 3 V 2 mR T 3 T 2 Q23 mC P T 3 T 2
Process 3 – 4 : Isentropic Expansion TS DIAGRAM OF DIESEL ENGINE
S 1 S 2 and r k
r C
V 1 V 2
V 4
S 3 S 4
Compression Ratio
V 2
V 3
P 4 T 3 P 3
T 4
k 1 k
V 3 V 4
k 1
r V C 2 V 1
k 1
r C r k
k 1
k 1 r C k 1 k 1 r k C T 4 T 3 k 1 T 1 r C r k k 1 T 1r C r k r k W 34 U 34 mC V T 4 T 3
Q43 0
Cut-off Ratio
V 2 Process 4 – 1 : Isometric Heat Rejection T 1
(the ratio of volume during isobaric heat addition)
r e
V 1 V 2
V 4 V 3
V 4 V 3
V 1 , V 3
c
P 4
Q41 mC V T 1 T 4
Clearance volume
is the percent Clearance
V C c V D
P 1
W 41 0
r e r k
V C V 2 V 3 , where:
T 4
Expansion Ratio
Heat Added
Q A Q23 mC P T 3 T 2 mC P T 1r C r k
Q A m C P T 1 r k
V D V 1 V 2 Heat Rejected
MI T-School of Mechanical E ngineeri ng
k 1
r C 1
k 1
T 1r k k
1
TH E RM ODYN AM I CS-I : Ai r Standard Cycles and H eat Engi nes
Q R Q41 mC V T 1 T 4 mC V T 1 T 1r C Q R m C V T 1 1 r C
k
k
chapter V -
6
of 17
r C 1 r C k 1 k 1 r k 1 k 1
P MEP
P 1 r k kr k
Net Work III. Air Standard Dual-Combustion Cycle
W NET PdV W 12 W 23 W 34 W 41 W NET mC V T 2 T 1 mR T 3 T 2 mC V T 4 T 3 0
W NET mC V T 2 mC V T 1 mRT 3 mRT 2
mC V T 4 mC V T 3 W NET m C V T 1 C P T 2 C P T 3 C V T 4 W NET m C P T 3 T 2 C V T 4 T 1
The assumption of either isometric or isobaric combustion is an oversimplification; neither occurs in practice. However, a more accurate representation of the combustion process in high-speed internal combustion engines can be obtained by assuming there is an initial sharp pressure increase (at constant volume) followed by essentially constant pressure combustion.
W NET mC V k T 3 T 2 T 4 T 1
W NET mC V k T 1r C r k
k 1
W NET mC V T 1 kr k
k 1
W NET m C V T 1 kr k or
W NET
T 1r k k 1 T 1r C k T 1
r C 1 r C k 1
k 1
r C 1 r C k 1
Q Q A Q R
W NET m C P T 1 r k
k 1
(1-2) Isentropic compression
(2-3) Reversible, constant volume heat addition
(3-4) Reversible, constant pressure heat addition (4-5) Isentropic expansion
(5-1) Reversible constant volume heat rejection
r C 1 m C V T 1 1 r
k C
r C 1 C V r C k 1 k 1 k W NET m C V T 1 kr k r C 1 r C 1
W NET mT 1 C P r k
The theoretical cycle bases on the combination of isometric and isobaric combustion is called dual cycle or mixed cycle. Calculation follows the same method of stepwise movement around the cycle.
k 1
Thermal Efficiency
r C 1 r C k 1 TH k 1 Q A m C P T 1 r k r C 1 k 1 k kr k r C 1 r C 1 TH k 1 k r k r C 1 m C V T 1 kr k
W NET
k 1
k 1 r C 1 TH 1 k 1 k r 1 100% r k C
V 1 V 5 ;
Mean Effective Pressure P MEP
W NET V D
mC V T 1 kr k
k 1
r C 1 r C k 1
V V 1 V 2 1
V 1
P MEP
mC V T 1 kr k
k 1
r C 1 r C k 1
V 1 1
P MEP
P 1C V kr k
PV DIAGRAM OF DUAL CYCLE
k 1
1
r k
r C 1 r C k 1 r k 1 r k
R
MI T-School of Mechanical E ngineeri ng
V 2 V 3 and
P 3 P 4
TH E RM ODYN AM I CS-I : Ai r Standard Cycles and H eat Engi nes
chapter V -
7
of 17
Q23 mC V T 3 T 2 Process 3 – 4 : Isobaric Heat Addition
T 4
T 3
V 4 V 3
r C k 1
T 4 T 3 r C T 1 r P r C r k
W 34 P V 4 V 3 mR T 4 T 3 Q34 mC P T 4 T 3 TS DIAGRAM OF DUAL CYCLE
S 1 S 2 and
V 1
r k
V 2
V 5
,
r C
V 4
r e
V 5
P 5 T 4 P 4 T 5
k
V 4 V 5
k 1
V r 3 C V 1
k 1
r C r k
k 1
r C k 1 k r k 1 T 1 r P r C k
Pressure ratio ,
k 1
T 5 T 4
V 2
W 45 U 45 mC V T 5 T 4
Q45 0
Cut-off Ratio
V 3
V 4
V 1 V 4
V V 4 1 V 2 V 3 V 4 V 1
r k
S 4 S 5
Compression Ratio
P 3 , P 2
r P
Process 4 – 5 : Isentropic Expansion
1 c
c
,
Expansion Ratio
Process 5 – 1 : Isometric Heat Rejection T 1 T 5
P 1 P 5
W 51 0
r k r C r e
Q51 mC V T 1 T 5
Heat Added V D V 1 V 2 V 5 V 3
and
Q A Q23 Q34 mC V T 3 T 2 mC P T 4 T 3
Q A mC V T 3 T 2 k T 4 T 3 Process 1 – 2 : Isentropic Compression
P 2 T 1 P 1
k 1
T 2
k
V 1 V 2
T 1r P r k k 1 T 1r k k 1 Q A mC V k 1 k 1 k T 1r P r C r k T 1r P r k
k 1
r k
k 1
k 1
T 2 T 1 r k
Q A mC V T 1r k
k 1
r P 1 kr P r C 1
W 12 U 12 mC V T 2 T 1
Q12 0
Heat Rejected
Q R Q51 mC V T 1 T 5 mC V T 1 T 1r P r C
k
Process 2 – 3 : Isometric Heat Addition T 3 T 2
P 3 P 2
Q R m C V T 1 1 r P r C
r P k 1
T 3 T 2 r P T 1 r P r k W 23 0
MI T-School of Mechanical E ngineeri ng
k
Net Work
W NET PdV W 1 2 W 23 W 3 4 W 45 W 51
TH E RM ODYN AM I CS-I : Ai r Standard Cycles and H eat Engi nes W NET mC V T 2 T 1 mR T 4 T 3 mC V T 5 T 4
W NET mC V T 2 mC V T 1 mRT 4 mRT 3
chapter V -
8
of 17
Comparison of Otto, Diesel and Dual Cycle For the same inlet conditions P 1, V1 and the same compression ratio:
mC V T 5 mC V T 4 W NET m C V T 1 C V T 2 C p C v T 3 C P T 4 C V T 5 W NET mC V T 1 T 2 k 1T 3 kT 4 T 5 W NET mC V T 1 T 2 kT 3 T 3 kT 4 T 5 W NET mC V T 3 T 2 k T 4 T 3 T 5 T 1 k 1 k 1 T 1r P r k T 1r k W NET mC V k 1 k 1 k k T 1r P r C r k T 1r P r k T 1r P r C
W NET m C V T 1 r k
k 1
r P 1 kr P r C 1
Q Q
or
W NET
k
Q R
A
W NET m C V T 1r k
r P r C 1
k 1
r P 1 kr P r C 1
m C V T 1 1 r P r C k W NET m C V T 1 r k
k 1
r P 1 kr P r C 1
r P r C 1 k
Thermal Efficiency
TH
W NET Q A TH
TH
r P 1 kr P r C 1 r P r C k 1 k 1 m C V T 1r k r P 1 kr P r C 1
m C V T 1 r k
r k
k 1
r P 1 kr P r C 1 r P r C k 1 k 1 r k r P 1 kr P r C 1
k 1
k 1 r P r C 1 100% 1 k 1 r k r P 1 kr P r C 1
Otto
Mean Effective Pressure
Dua l Dies el
For the same inlet conditions P 1, V1 and the same peak pressure P3 (actual design limitation in engines): P MEP
W NET V D
m C V T 1 r k
k 1
r P 1 kr P r C 1 r P r C k 1
V V 1 V 2 1
V 1
P MEP
k 1
P 1 C V r k
r P 1 kr P r C 1 r P r C k 1
R1
P MEP
P 1r k r k
k 1
1
r k
r P 1 kr P r C 1 r P r C k 1 k 1r k 1
MI T-School of Mechanical E ngineeri ng
TH E RM ODYN AM I CS-I : Ai r Standard Cycles and H eat Engi nes
Dies el Dua l otto
chapter V -
9
of 17
ENGINE PERFORMANCE OF I.C.E.
Actual Performance of Internal Combustion Engine
PISTON-CYLINDER ARRANGEMENT OF INTERNAL COMBUSTION ENGINE ( ICE )
IV. Stirling Cycle
The Stirling cycle is an external-heat, closed two stroke cycle, that uses gas as the working substance. It was invented by Robert Stirling in about 1816 and has historical importance as one of the first practical heat engines to use air rather than steam as the working substance. The basic principle of the engine is that one part of a cylinder (or a separate cylinder) is kept hot by the application of heat source (usually by external
MI T-School of Mechanical E ngineeri ng
TH E RM ODYN AM I CS-I : Ai r Standard Cycles and H eat Engi nes
chapter V -
10
of 17
combustion of the fuel), and another part of the cylinder (or a separate cylinder) is kept cool by the use of cooling fins or water cooling, and is the heat sink. The Stirling cycle is a regenerative cycle, which uses a regenerator to alternatively absorb and give back heat as the gas flows over it
PV DIAGRAM OF STIRLING CYCLE
TS DIAGRAM OF STIRLING CYCLE
V 1 V 4 ,
V 2 V 3 ,
and
T 1 T 2 T H
T 3 T 4 T L
Heat Added Q A Q1 2 W 1 2 P 1V 1 ln
V 2 V 1
mRT H ln
V 2 V 1
Heat Rejected Q R Q34 W 34 P 3V 3 ln
V 4 V 3
mRT L ln
V 1 V 2
Net Work W NET Q A Q R mRT H ln
V 2 V 1
mRT L ln V 2 V 1
W NET mRT H T L ln
Thermal Efficiency
TH
W NET Q A
TH
MI T-School of Mechanical E ngineeri ng
V 2 V 1
mRT H T L ln
mRT H ln
T 1 L T H
V 2 V 1
100%
V 2 V 1
TH E RM ODYN AM I CS-I : Ai r Standard Cycles and H eat Engi nes
chapter V -
11
of 17
(similar to Carnot cycle)
Mean Effective Pressure P MEP
W NET V D
T T V mR H L ln 2 V 2 V 1 V 1
V. Ericsson Cycle
An Ericsson Cycle Engine is also known as a "hot-air engine," and is named after John Ericsson (1803-1889). It is often compared to a Stirling engine.
PV DIAGRAM OF ERICSSON CYCLE
The cycle of the Ericsson engine are as follows: Air at the bottom of the cylinder (E) is heated, thus expanding and forcing the piston (A) upward. At this time the displacer (B) is driven downward to the bottom of the cylinder. Since the displacer is of a smaller diameter than the cylinder, the hot air rushes around the displacer to the cool end of the engine (F). Once in the top end of the cylinder, the hot air begins to contract, sucking the piston downward. Now the displacer moves upward, forcing all the cool air from the top end of the cylinder into the bottom end. Here the air is heated and the cycle begins again. TS DIAGRAM OF ERICSSON CYCLE
Heat Added Q A Q1 2 W 1 2 P 1V 1 ln
V 2 V 1
mRT H ln
V 2
mRT L ln
V 1
V 1
Heat Rejected Q R Q34 W 34 P 3V 3 ln
V 4 V 3
V 2
Net Work W NET Q A Q R mRT H ln
V 2
mRT L ln
V 1
V 2 V 1
W NET mRT H T L ln
Thermal Efficiency
TH
W NET Q A
TH
V 2 V 1
mRT H T L ln
mRT H ln
T 1 L T H
V 1
100%
(similar to Carnot cycle)
MI T-School of Mechanical E ngineeri ng
V 2
V 2 V 1
TH E RM ODYN AM I CS-I : Ai r Standard Cycles and H eat Engi nes
chapter V -
Mean Effective Pressure P MEP
W NET V D
12
kJ
w NET wCOMP w EXP 279.8 855.6
T T V mR H L ln 2 V 2 V 4 V 1
w NET 575.8
of 17
kg
k J k g
(f) the heat supplied per kg of air SAMPLE PROBLEMS
q A C V T 3 T 2 0.7176
5.1) For a theoretical Otto cycle operating on a operating on a compression ratio of 8.5:1, taking in air at 15° C and 101.3 kPa and having a maximum cycle temperature of 1800°C, calculate: (a) the temperature and pressure after compression k 1
P T 2 T 1 2 P 1
V 1 V 2
k
k 1
T 1r k k 1
T 1
1.41
T 2 15 273 K 8.5
q A 1,001.11
677.89 K or 404.89 C TH
P 3 P 2
T 3 T 2
P 2
T 3 T 2
1800 273 677.89
(h) the mean effective pressure P MEP P MEP 101 .3 kPa
3.058
2,026.73kPa 3.058 6,197.74 kPa V 3 V 4
k 1
T 4 T 3
V T 3 2 V 1
k 1
1 T 3 r k
v1
k 1
P 4
and
P 1
P 4 P 1
T 4 T 1
101.3 kPa
kJ kg K
P
wNET v1 v 2
T 1 880.71
15 273
kJ kg
(d) the theoretical work of expansion w EXP C V T 4 T 3 0. 7176
k g K
w EXP 855.6
1.4 1
1
1.4 18.5 1
800 kPa
v1 r k
15 273 K
kg K 101.3kPa
0.816m
3
0.816 m
3
kg
3
0.096 m kg
8.5
575.8 kJ
kg
0.816 0.096 m
800 kPa
3
kg
309.78 kPa
677.89 288 K 279.8
kJ
1
k 1 r k 1
8.53.058 1 8.5
0.287 kJ
v2
P MEP
(c) the theoretical work of compression per kg of air; wCOMP C V T 2 T 1 wCOMP 0.7176
R T 1
880.71 K or 607.71C T 4
k 1
P 1 r k r P 1 r k
or
1.41
1 8.5
T 4 1800 273 K
k g
1 1 1 k 1 100% 1 1.41 100% 57.5% 8.5 r k
(b) the temperature and pressure after expansion
k J
or
k
P 3
2073 677.89 K
(g) the theoretical cycle efficiency w NET 575.8 100% 57.5% TH q A 1,001.11
V k 1 101.3kPa8.51.4 P 2 P 1 1 T 1r k V 2 P 2 2,026.73 kPa
r P
kJ kg K
5.2) An air standard Otto cycle uses 0.1 kg of air and has a 17% clearance. The intake conditions are 98 kPa and 37°C, and the energy release during combustion is 1600 kJ/kg. Determine: (a) the compression ratio 1 c 1 0.17 r k 6.8824 c 0.17 (b) the pressure and temperature at the four cycle state points
880.71 2073 K
V 4 V 1
k J
0.1kg 0.287
k g
V 4 V 1
(e) the net work per kg of air V 3 V 2
MI T-School of Mechanical E ngineeri ng
mRT 1 P 1
kJ
37 273 K
kg K
0.0908 m3
98kPa
V 2 r k
0.0908m
6.8824
3
0.0132 m3
TH E RM ODYN AM I CS-I : Ai r Standard Cycles and H eat Engi nes
chapter V -
13
of 17
or T 2 T 1r k
k 1
1.41
310 K 6.8824
670.6 K or 397.6C
q A C V T 3 T 2 T 3
T 3
q A C V
T 2
1600 kJ 0.7176 kJ
kg
q A
1 r k
T 3
T 2
k 1
T 4 T 3
670.6
4.325
1.4 1
mRT 2
P 2
V 2
P 3
mRT 3 V 3
0.1kg 0.287
P 1V 1
m
RT 1
3
2900.25 K
kg K
T 2 T 1r k
6305.85 kPa
3
V 2
0.0908m
T 3
V D V 1 V 2 0.0908 0.0132 m 3
0.0776 m
0.0486 kg
Q A mC P
k 1
V 1 r k
30512.54421.41 838.84 K 3
T 2
(d) the work
W NET mC V T 1 T 2 T 3 T 4 kJ 0.10kg 0.7176 kg K 310 670.6 2900.25 1340.72 K W NET 86.03 kJ
mRT 3 P 3
V T 4 T 3 3 V 4
28.5kJ
kJ 0.0486kg 1.0047 kg K T 3 1,422 .52 K
0.04850.287 1423.72 3450
k 1
1 1 100% 53.7% 1 k 1 100% 1 1.4 1 6 . 8824 r k
1
V 3 V 2
0.00575 0.00339
W NET m C V T 1 kr k W NET 0.0486 kg 0.7176
1.412.5442
3
192.46 K
209.71kPa
1.6962
1.4 1
k 1
r C 1 r C k 1
kJ
305 K
kg K
1.6962 1 1.6962
W NET 16.86 kJ
6.8824 4.325
MI T-School of Mechanical E ngineeri ng
0.00575m
(b) the work
k 1r k 1
16.8824 1.41 1 P MEP 98 kPa 1.4 16.8824 1 P MEP 1108.7 kPa
838.84
1.4
k 1
3
1.41
0.00575 1422.52 0.0425
k
r C
(f) the mean effective pressure.
0.00339 m
V 0.00575 P 4 P 3 3 3450kPa 0.0425 V 4
(e) the cycle efficiency
P 1 r k r P 1 r k
0.0425m 12.5442
3
V 3
P MEP
kg K
(c) the displacement volume
TH
3
100kPa 0.0425m kJ 305 K 0.287
kJ
kJ 1340.72 K 0.1kg 0.287 K kg mRT 4 423.77 kPa P 4 3
W NET
1
P k 3450 1.4 r k 2 12.5442 V 2 P 1 100
670.6 K kg K 1458.05 kPa
0.0132m
V 4
1108 .7 kPa
V 1
kJ
0.0132m
0.0776
1
T 4 1340 .72 K or 1067 .72C
86.03kJ
(a) P, V and T at each state point V 1 V 4 0.0425 m 3 , T 1 305 K , P 1 100 kPa and P 2 P 3 3450 kPa
1 2900 .25K 6.8824
0.1kg 0.287
5.3) An air standard Diesel cycle receives 28.5 kJ/cycle of heat while operating at 300 RPM. At the beginning of compression, pressure is 100 kPa, temperature is 305K and volume is 0.0425m 3. At the beginning of heat addition, the pressure is 3450 kPa. Determine:
670 .6
kg K
2900.25
V D
T 2
C V
T 3 2900 .25 K or 2627 .25C
r P
W NET
P MEP
(c) the power
cycle
1.4
1
TH E RM ODYN AM I CS-I : Ai r Standard Cycles and H eat Engi nes
W NET
Power
t
chapter V -
T 2 T 1r k
W NET N
T 3
kJ rev 1 min cycle Power 16.86 300 min 60 s 1rev cycle Power 84.3 kW two stroke
Q A mC P
P MEP P MEP 100 kPa
k 1 P 1 r k kr k r C
1 k 1r k 1
12.5442 1.4 12.5442
1.4 1
k r C
1
1.6962 1 1.6962
1.4
1
P MEP 431 k Pa TH
P MEP
16.86kJ
0.0425 0.00339m3
TH
Btu 0.1082 lb 0.24 R lb T 3 2546 .27 R
T 3 T 2
2546.27
1506.53
1506 .53
1.6902
k 1 r 1 1 k 1 C 100% r k k r C 1
1.69021.4 1 1 100 % 59% 1 1.4 1 12.4176 1.41.6902 1
(b) the power if the engine runs at 300RPM W NET Q A TH 27
k 1 r 1 1 k 1 C 100 % r k k r C 1
1.6962 1.4 1 1 100 % 59.1% 1 1.4 1 12.5442 1.41.6962 1
Btu rev
15.93
Btu rev
ft lb
300
or 12,396 .09
ft lb rev
rev min HP
min 33,000 ft lb rev Power 112 .7 HP
(c) the mean effective pressure
r C 1 r C k 1 k 1r k 1 k 1
P MEP
W 16.86 NET 100 % 100 % 59.1% 28.5 Q A
0.59
Power W NET N 12,396.09
or TH
27 Btu
431kPa
(e) the thermal efficiency of the cycle TH
V 2
TH
1.4 112.5442 1
or
V 3
of 17
55012.41761.41 1506.53R
T 2
r C
(d) the mean effective pressure
k 1
14
P MEP 14.7 psia
P 1 r k kr k
12.4176 1.4 12.4176
1.4 1
1.6902 1 1.6902
1.4
1
1.4 112.4176 1 P MEP 62.4 psi
5.4) A one cylinder Diesel engine operates on the air-standard cycle and receives 27 Btu/rev. The inlet pressure is 14.7 psia, the inlet temperature is 90°F, and the volume at the bottom dead center is 1.5 ft 3. At the end of compression the pressure is 500 psia. Determine: (a) the cycle efficiency (b) the power if the engine runs at 300RPM (c) the mean effective pressure Solution (a) the cycle efficiency P 1 14.7 psia , T 1 550 R , V 1 V 4 1.5 ft 3 P 2 500 psia
and Q A 27 BTU
rev
5.5) The compression ratio of an air standard dual cycle is 12, and at the beginning of compression the pressure is 100 kPa, the volume is 1.2 liters, and the temperature is 37°C. During the heat addition processes, 0.4 kJ is transferred at constant volume and 1.0 kJ at constant pressure. Determine (a) the cycle thermal efficiency (b) the work of the cycle (c) the mean effective pressure Solution (a) the cycle thermal efficiency K , V 1 V 5 0.0012 m 3 P 1 100 kPa , T 1 310
Q AV 0.4 kJ , 1
Q AP 1.0 kJ and r k 12
1
P k 500 1.4 r k 2 12.4176 V 2 P 1 14.7 lb in 2 3 14.7 2 144 2 1.5 ft in ft P 1V 1 0.1082 lb m ft lb RT 1 53.34 550 R lb R V 1
MI T-School of Mechanical E ngineeri ng
m
P 1V 1 RT 1
3
100kPa 0.0012 m kJ 0.287 310 K
0.00135 kg
kg K
Q AV mC V T 3 T 2 T 3
Q AV mC V
T 2
TH E RM ODYN AM I CS-I : Ai r Standard Cycles and H eat Engi nes
T 2 310 K 121.41 837 .6 K
Q AP mC P T 4 T 3 T 4 T 3
T 4
Q AV mC V
Q AP mC P
T 2
Q AP mC P
chapter V -
T 3
T 4
r C
T 3
1987 .8 1250 .5
r P
kJ kg K
T 3 T 1 r P r k
1250 .5 1987 .8 K
T 4
q AP C P
1.5896
T 5 T 1 r P r C 310 1.493 1.5896 1.4 k
P 2
k P 1r k
k 1
0.00135 k g 1.0047
T 2 540 R 91.41 1300 .4 R
837.6 1250 .5 K kJ 0.00135 kg 0.7176 k g K T 1250 .5 r P 3 1.493 837 .6 T 2 1.0k J
of 17
addition at P=C is 100 BTU/lb. Determine the thermal efficiency and the mean effective pressure.
0.40kJ
T 3
15
T 3
P 3 P 2
470
318 .61
1.4752
540 R 1.4752 91.4 1 1918 .4 R 100 Btu
0.24 Btu
r C
885 .6 K
14.7 91.4 318.61 k Pa
T 4 T 3
lb
1918 .41 R 2335 .1 R
lb R 2335 .1
1918 .41
1.2172
k
T 5 T 1 r P r C 540 R 1.4752 1.2172 1.4 1048 .9 R 1.4 1 1.493 1.5896 1 100 % 60.2% TH 1 1.4 1 12 1.493 1 1.41.493 1.5896 1
q AV C V T 3 T 2 0.1714
Btu lb R
q AV 105.93 Btu q R C V T 1 T 5 0.1714
(b) the work of the cycle
Q R mC V T 1 T 5 0.00135 kg 0.7176
kJ
310 885.6
kg K
w NET
Q 0.4 1.0 0.558kJ
0.842 kJ
or
TH
W NET TH Q A 0.6021 0.4kJ
0.842 k J
k 1
540 1048 .9
q 105 .93 100 87.23 Btu lb
118 .7 Btu
lb
w 118 .7 NET 100 % 100 % 57.6% q 105 .93 100 A
or
or W NET m C V T 1 r k
lb
lb R q R 87.23 Btu lb
Q R 0.558 kJ
W NET
Btu
1918 .4 1300 .4
r P 1 k r P r C 1 r P r C k 1
TH
k r P r C 1 1 1 k 1 r P 1 kr P r C 1 r k
1.4 1 1.4752 1.2172 1 100% 1 1.4 1 9 1.4752 1 1.41.4752 1.2172 1 kJ 1.4 310 K 121.41 1.493 1 1.41.4931.5896 1 1.493 TH 1.5896 57.6% 1 0.842 kJ
TH
W NET 0.00135kg 0.7176
kg K
(c) the mean effective pressure P MEP P MEP 100kPa
k 1 P 1r k r k
r P 1 kr P r C 1 k 1r k 1
k r P r C
v1
1
RT 1 P 1
53.34
ft lb lb R
1.4 112 1
v2
or W NET V 1 V 2
0.842kJ W NET 765 kPa 3 V 1 0 . 0012 m 3 V 1 0.0012 m r k 12
P MEP
w NET v D
14.1862 9 r k
v1
118 .7
540 R
3
14.1862 ft lb 2 lb in 14.1 2 144 2 in ft
12 121.41 1.493 1 1.41.4931.5896 1 1.493 1.5896 1.4 1
P MEP 765 k Pa
P MEP
ft 3
3
lb 1.5762 ft
Btu lb
14.1862 1.5762
lb
ft lb ft 2 778 . 16 Btu 144 in2 ft 3
lb P MEP 50.8 psi
or 5.6) An air standard dual cycle is characterized by the following : P1=14.1 psia, T 1=80°F, P3=470psia, and r k=9; heat
MI T-School of Mechanical E ngineeri ng
991.41 1.4752 1 1.41.4752 1.2172 1 1.4752 1.2172 1.4 1 P MEP 14.1 psi 1.4 19 1 P MEP 50.8 psi
TH E RM ODYN AM I CS-I : Ai r Standard Cycles and H eat Engi nes
chapter V -
16
of 17
1550°C and 20°C, respectively. Determine the compression ratio of the engine (Ans 9.145:1) Problems: 01) An air standard Otto cycle has a compression ratio of 8:0 and has air condition at the beginning of compresson of 100 kPa and 25°C. The heat added is 1400 kJ/kg. Determine (a) the four cycle state points: (b) the thermal efficiency; (c) the mean effective pressure. (Ans 56.5%, 1057 kPa) 02) An engine operates on air standard Otto cycle with a compression ratio of 9 to 1. The pressure and temperature at the end of the compression stroke are 800 kPa and 700°C. Determine the net cycle work per kg if the pressure at the end of heat addition is 3.0MPa. 03) An engine operates on air standard Otto cycle. The pressure and temperature at the beginning of isentropic compression are 120 kPa and 35°C. The peak pressure and temperature are 4.8 MPa and 2500°C. Determine (a) the net cycle work in kJ/kg; (b) the cycle efficiency. (Ans 713.5 kJ/kg, 44.9%) 04) An air standard Otto cycle has the following cycle states, where state 1 is at the beginning of the isentropic compression: P1=101kPa, T1=333K, V1=0.28m3, T3=2000K, r k=5. Determine (a) the remaining cycle state points; (b) the thermal efficiency; (c) the head added; (d) the heat rejected. (Ans 47.5%, 290kJ, -152.4kJ) 05) A four cylinder spark ignition engine with a compression ratio of 8 has pistons with a bore of 9 cm and a stroke of 10 cm. The air pressure at the beginning of compression is 98 kPa, and the temperature is 37°C. The engine may be modeled by the air standard Otto cycle. The maximum cycle temperature is 1700°K. If the engine produces 75 kW of power, determine (a) the heat supplied per cylinder, (c) the RPM required. 06) An air standard Otto cycle has a compression ratio of 7.5. The maximum and minimum cycle temperatures are 1600° and 300°K, and the minimum pressure is 100 kPa. Determine (a) the cycle efficiency (b) the change of entropy during heat addition; (c) the change of availability per unit mass during the expansion process. 07) An engine working on the Otto cycle has a theoretical cycle efficiency of 70% of that of a Carnot-cycle engine working between the same upper and lower cycle temperatures of
MI T-School of Mechanical E ngineeri ng
08) An engine working on the Otto cycle has a heat input of 460kJ/kg. The air conditions at the beginning of compression are 0.95 bar, 27 °C. The temperature at the end of the expansion is 211°C. Calculate the compression ratio; maximum pressure and temperature during the cycle; and the cycle efficiency. C P=1.10 kJ/kg-K (Ans 7.78; 27.08 bar, 1100 K, 56%) 09) In an air standard Otto cycle the compression ratio is 9:1. Compression begins at 0.95 bar and 17°C. The maximum cycle temperature is 1100°C and C V=0.718 kJ/kg-K. Calculate the heat supplied per kg air, maximum cycle pressure and temperature, the cycle efficiency and mean effective pressure. Also calculate the efficiency of a Carnot cycle working between the same temperature limits. (Ans 484.7 kJ, 40.50 bar, 571 K, 58.5%, 3.63 bar, 78.9%) 10) An air standard Otto cycle has an initial temperature of 100°F, a pressure of 14.7 psia, and a pressure at the end of compression of 356 psia. The pressure at the end of heat addition is 1100 psia. Determine (a) the compression ratio; (b) the thermal efficiency; (c) the percentage clearance; (d) the maximum temperature. (Ans 9.74, 59.7%, 11.4%) 11) An air standard Diesel cycle has a compression ratio of 14. The maximum and minimum cycle temperatures are 2940°R and 540°R, and the minimum pressure is 14.7 psia. Determine (a) the cycle efficiency; (b) the change of entropy during heat addition; (a) the mean effective pressure. 12) An air standard Diesel cycle has a compression ration of 20 and a cutt-off ratio of 3. Inlet pressure and temperature are 100 kPa and 27 °C. Determine (a) the heat added per kg; (b) the net work per kg. (Ans 1994.6kJ/kg, 1209.2 kJ/kg) 13) In an air standard Diesel cycle, the compression ratio is 17. The cutoff ratio is 2.5:1. The air conditions at the beginning of compression are 101 kPa and 300°K. Determine (a) the thermal efficiency; (b) the heat added per kg of air; (c) the mean effective pressure. (Ans 60%, 1404.1 kJ/kg, 1050 kPa) 14) A four-cylinder compression-ignition engine with a compression ratio of 18 has pistons with a bore of 9 cm and a stroke of 10 cm. the air pressure at the beginning of compression is 98 kPa, and the temperature is 37°C. The engine may be modeled by the air-standard Diesel cycle. The maximum cycle temperature is 1700°K. If the engine produces 75 kW of power,
TH E RM ODYN AM I CS-I : Ai r Standard Cycles and H eat Engi nes
determine (a) the heat supplied per cylinder; (b) the thermal efficiency, (c) the RPM required. 15) In an air standard dual cycle, the isentropic compression starts at 100 kPa and 300°K. The compression ratio is 13 the maximum temperature is 2750 °K, and the maximum pressure is 6894 kPa. Determine (a) the cycle work per kg; (b) the heat added per kg; (c) the mean effective pressure. (Ans 1040.3 kJ/kg, 1705.5 kJ/kg, 1308 kPa) 16) A dual cycle engine has a compression ratio of 14. The air state at the beginning of compression is 100 kPa and 300°K. Fifteen hundred kJ/kg of heat is added during the heat-addition process, with one third at constant volume and two thirds at constant pressure. Find the thermal efficiency and mean effective pressure of the engine. 17) An Internal combustion engine runs on an ideal cycle in which the heat rejection is at constant volume, but the heat is supplied such that dP/dv = C. The heat is supplied over 1/8 x expansion stroke. Compression ratio is 6:1. Compression commences at 0.95 bar, 20°C. Maximum cycle pressure is 40 bar. Calculate the cycle efficiency, mean effective pressure and maximum temperature of the cycle. (Ans 45.1%, 13.55 bar, 3343K)
MI T-School of Mechanical E ngineeri ng
chapter V -
17
of 17
