JJ 507 – THERMODYNAMIC 2

Prepared By :MUHAMMAD ZUHAIRY BIN ZULKIFLIB.Eng(Manufacturing), (Hons), UKMLecturer Jab. Kejuruteraan Mekanikal Politeknik Kota Kinabalu

UNIT 2

STANDARD AIR CYCLE

2.0 THE STANDARD AIR CYCLE2.1 Internal Combustion Engine

The internal combustion engine is an engine in which the combustion of a fuel occurs with an oxidizer (usually air) in a combustion chamber that is an integral part of the working fluid flow circuit.

In an internal combustion engine, the expansion of the high-temperature and high-pressure gases produced by combustion apply direct force to some component of the engine. This force is applied typically to pistons, turbine blades, or a nozzle. This force moves the component over a distance, transforming chemical energy into useful mechanical energy.

Examples of internal combustion cycles are :I. The open cycle gas turbine unitII. The petrol engine, III.The diesel engineIV.The gas engine.

In the petrol engine a mixture of air and petrol is drawn into the cylinder, compressed by the piston, then ignited by an electric spark. The hot gases expand, pushing the piston back, and are then swept out to exhaust, and the cycles recommences with induction of a fresh charge of petrol and air.

In the diesel the oil is sprayed under pressure into the compressed air at the end of the compression stroke, and combustion is spontaneous due to the high temperature of the air after compression.

Figure 2.1 General arrangement of an internal combustion engine

2.2 THE OTTO CYCLE

The Otto cycle is the ideal air standard cycle for the petrol engine, the gas engine and the high-speed oil engine. The cycle is shown on a p-v diagram in Figure 2.2

  Process 1 to 2 is isentropic compression

Process 2 to 3 is reversible constant volume heating

Process 3 to 4 is isentropic expansion

Process 4 to 1 is reversible constant volume cooling

Figure 2.2 p-v diagram of Otto cycle

Figure 2.3 T-s diagram of Otto cycle

Heat supplied QS= mcv (T3 - T2)

Heat rejected, QR= mcv (T4 – T1)

Applying the First Law of Thermodynamics,

Wnet = QS - QR

= mcv (T3 - T2) - mcv (T4 – T1)

s

net

QW

)()]()[(

23

1423

TTmcTTTTmc

v

v

)()(1

23

14

TTTT

The air standard efficiency of the Otto cycle may be defined as:

2

1v V

Vr

volumeclearance volumeclearance meswept volu

To give a direct comparison with an actual engine the ratio of the specific volume, V1/V2 is taken to be the same as the compression of the actual engine,

Compression ratio,

1

2

TT 1

2

1

VV 1

vr1vr

Since process 1 - 2 is isentropic compression then

= or T2 =T1 =

4

3

TT 1

3

4

VV 1

vr 1vr

Similarly, process 3 - 4 is isentropic expansion and

= = or T3 =T4

114

14

)()(1

vrTTTT

1

11 vr

So,

EXAMPLE 1 An engine working on the Otto cycle has a

compression ratio 10/1. At the beginning of the compression stroke the temperature of the air is 300 C. After the addition of the heat at constant volume the temperature of the air rises to 11000 C. Find the heat supplied, heat rejected and the net work done per kilogram of air in the cylinder. What is the efficiency of the cycle? For air assume γ = 1.4 and cv = 0.718 kJ/kgK.

SOLUTION:Data: rv =10 , T1 = 30 + 273 = 303 K , T3 = 1100 + 273 =

1373 K

1

2

TT 1

2

1

VV

1vr

1vr

=

T2 =T1

= 303 (101.4-1)

=

= 761.1K

4

3

TT 1

3

4

VV

1

3

vrT

1vr

=

T4 =

=

= 546.6K

14.1101373

QS= cv (T3 - T2) = 0.718(1373 - 761.1) = 439.3 kJ/kg

QR= cv (T4 – T1) =0.718(546.6 - 303) = 174.9 kJ/kg Wnet = QS - QR

= mcv (T3 - T2) - mcv (T4 – T1)

= 439.3 – 174.9 = 264.4 kJ/kg

s

net

QW

1

11 vr

3.4394.264

14.110

11

Cycle efficiency,

= 0.602 @ 60.2%

Or

= 0.602 @ 60.2%

EXAMPLE 2

Calculate the ideal air standard cycle efficiency based on the Otto cycle for a petrol engine with a cylinder bore of 50 mm, a stroke of 75 mm and a clearance volume of 21.3 cm3. For air assume γ = 1.4

SOLUTION:

Clearance volume, V2 = 21.3 cm3

Total cylinder volume, V1 = Swept volume + Clearance volume

= 147.2 + 21.3 = 168.5 cm3

Swept volume

3

3

2

2.147

147200

75504

42

cm

m

LD

2

1v V

Vr

3.21168.5

1

11 vr

14.1911.711

= 7.911

= 1 – 0.437 = 0.563 @ 56.3%

EXAMPLE 3When working on the Otto cycle with air as the working fluid, an engine has an air standard efficiency of 54.5% and rejects heat at the rate 520 kJ/kg of air used. The pressure and temperature at the beginning of compression are 0.98 bar and 66o C respectively. Determine:

a) The compression ratio of the engineb) The net work done per kilogram of airc) The pressure and temperature at the end of

compressiond) The maximum pressure and temperature in

the cycle

SOLUTION: Data: η = 0.545, QR = 520 kJ/kg, T1= 66 + 273 =

339 K, P1 = 0.98 bar, D = 0.072 m, L = 0.085 m

1

11 vr

14.1

11545.0 vr

7.16rv

a) For air, γ = 1.4

b)

623kJ/kg 5201143RSnet QQW

s

net

QW

s

RS

QQQ

545.0

s

S

QQ 520545.0

kgkJQS /1143

1

2

TT 1

2

1

VV 1

vr1y

vr

2

22

1

11 T

VP

TV

P

12

2112 TV

TVPP

339745)16.7(98.02 P

c) For isentropic compression,

T2 =T1

For a perfect gas such as air;

= =

= 339 (7.161.4-1)= 745 K

= 15.4 bar

3

33

2

22 T

VP

TV

P

2

323 T

TPP

48.3bar745

23374.15

d) Maximum pressure and temperature in the cycle occur after heat addition (point 3)

QS = cv (T3 - T2)

1143 = 0.718(T3 - 745) T3 = 2337 K

; but V2 = V3

2.3 THE DIESEL CYCLE

Diesel work on the idea of spontaneous ignition of powdered coal, which was blasted into the cylinder by compressed air. Oil became the accepted fuel used in compression-ignition engines and the oil was originally blasted into the cylinder in the same way that Diesel had intended to inject the powdered coal.

Process 1 to 2 is isentropic compression

Process 2 to 3 is reversible constant pressure heating

Process 3 to 4 is isentropic expansion

Process 4 to 1 is reversible constant volume cooling

Figure 2.4 p-v diagram of Diesel cycle

Figure 2.5 T-s diagram of Diesel cycle

Heat supplied QS= mcp (T3 - T2)

Heat rejected, QR= mcv (T4 – T1)

Applying the First Law of Thermodynamics,

Wnet = QS - QR

= mcp (T3 - T2) - mcv (T4 – T1)

s

net

QW

)()()(

23

1423

TTmcTTmcTTmc

p

vp

)()(

123

14

TTcTTc

p

v

)()(11

23

14

TTTT

The air standard efficiency of the Diesel cycle may be defined as:

2

1v V

Vr

2

3c V

Vr

)1(11 1

cv

c

rrr

Compression ratio,

Combustion ratio (cut-off ratio),

So,

EXAMPLE 4

An air standard diesel cycle has a compression ratio of 19/1 and the heat supplied to the working fluid is 1800 kJ/kg. At the beginning of the compression process the pressure and temperature are 1.0 bar and 25o C respectively. For air assume γ = 1.4, cp = 1.005 kJ/kgK and cv = 0.718 kJ/kgK. Determine:a) The temperature at the remaining point in the

cycleb) The air standard thermal efficiency

SOLUTION:Data: rv = 19, QS= 1800 kJ/kgK, P1 = 1 bar =105 N/m2, T1 = 25 + 273 =298 K

1

2

TT 1

2

1

VV

1vr

1vr T2 =T1

= 298 (191.4-1) = 968 K

Process 2 -3 is at constant pressure during which QS = 1800 kJ/kg is supplied

QS = cp (T3 – T2)

1800 = 1.005(T3 – 968) T3 = 2759 K

a) = =

4

3

TT 1

3

4

VV 1

3

1

VV

35

1

11 885.0

10)298)(287(1 m

PmRTV

312 045.0

19885.0

19mVV

3

33

2

22

TVP

TVP

3

2

323 128.0

9682759)045.0( m

TTVV

Process 3 – 4 is isentropic expansion,

= = since V1 = V4

Assuming 1 kg of working fluid (air),

but P2 = P3

4

2759T

11.4

0.1280.855

Hence,

=

T4 = 1291 K

18001087

s

net

QW

)()(11

23

14

TTTT

)9682759(4.1)2981291(11

= 0.604 @ 60.4%

or

= 0.604 @ 60.4%

b) Heat supplied, Qs = 1800 kJ/kg

Heat rejected, QR = cv (T4 – T1) = 0.718(1291 -298) = 713 kJ/kg

Wnet = QS – QR = 1800 -713 = 1087 kJ/kg

2.4 MEAN EFFECTIVE PRESSURE

The mean effective pressure is defined as the height of a rectangle having the same length and area as the cycle plotted on a p - v diagram. This is illustrated for an Otto cycle in Figure 2.6

Figure 2.6 Mean effective pressure for the Otto cycle

The rectangle ABCDEA is the same length as the cycle 12341, and area ABCDEA is equal to area 12345. Then the mean effective pressure, pm, is the height AB of the rectangle. The work output per kilogram of air can therefore be written as W =area ABCD = pm (v1 – v2)

)( 21 VVW

pm

; (V1 – V2) is swept volume of the cylinder Mean effective pressure,

EXAMPLE 5

Calculate the mean effective pressure if the heat supplied, QS and the cycle efficiency are 260 kJ/kg and 68.2 % respectively. Compression ratio, rv = 18/1 and inlet pressure 1.01 bar and temperature 20o C. R= 287 J/kgK

SOLUTION:

sQW

W = ηQs

= 0.682(260) = 177 kJ/kg

18VVVV 1

121

kgmVV /786.0 321

)( 21 VVW

pm

786.010177 3x

=225190 N/m2

= 2.25 bar

1V1817

118P17RT

)18(1.01x103)17(287)(295

18;2

VV1

1

11;

PmRTV