MOTOR BAKAR

( 3 SKS)

Thermodynamic Principles All internal combustion

Open cycle, heated engine Gasoline (Otto) engine

Spark ignitionCompresses air-fuel mixture

Diesel engineCompressed ignitionCompresses air only

INTERNAL COMBUSTION ENGINE:

AN ENGINE THAT PRODUCES POWER BY BURNING FUEL INSIDE A COMBUSTION CHAMBER WITHIN THE ENGINE

Gas CyclesCarnot Cycle

T2

T1

s1 s2

Work W

1

2 3

4

1-2 - ADIABATIC COMPRESSION (ISENTROPIC)2-3 - HEAT ADDITION (ISOTHERMAL)3-4 - ADIABATIC EXPANSION (ISENTROPIC)4-1 - WORK (ISOTHERMAL)

Heat Q

Carnot Cycle

Carnot cycle is the most efficient cycle that can be executed between a heat source and a heat sink.

However, isothermal heat transfer is difficult to obtain in reality--requires large heat exchangers and a lot of time.

2

1

TT-1

Carnot Cycle

Therefore, the very important (reversible) Carnot cycle, composed of two reversible isothermal processes and two reversible adiabatic processes, is never realized as a practical matter.

Its real value is as a standard of comparison for all other cycles.

Gas cycles have many engineering applications Internal combustion engine

Otto cycleDiesel cycle

Gas turbines Brayton cycle

RefrigerationReversed Brayton cycle

Some nomenclature before starting internal combustion engine cycles

More terminology

Terminology

Bore = d Stroke = s Displacement volume =DV = Clearance volume = CV Compression ratio = r

4ds2

CVCVDVr

TDC

BDCVV

Mean Effective PressureMean Effective Pressure (MEP) is a fictitious pressure, such that if it acted on the piston during the entire power stroke, it would produce the same amount of net work.

minmax VVWMEP net

The net work output of a cycle is equivalent to the product of the mean effect pressure and the displacement volume

Real Otto cycle

Real and Idealized Cycle

Otto Cycle P-V & T-s Diagrams

Pressure-Volume Temperature-Entropy

Otto Cycle Derivation

Thermal Efficiency:

For a constant volume heat addition (and rejection) process;

Assuming constant specific heat:

QQ - 1 =

QQ - Q =

H

L

H

LHth

T C m = Q vin

1-TTT

1 - TTT

-1 =)T - T( C m)T - T( C m - 1 =

2

32

1

41

23v

14vth

T C m = Q v Rej

For an isentropic compression (and expansion) process:

where: γ = Cp/Cv

Then, by transposing,

TT =

VV =

VV =

TT

4

3

3

41-

2

11-

1

2

TT =

TT

1

4

2

3

Otto Cycle Derivation

TT-1 =

2

1thLeading to

Differences between Otto and Carnot cycles

The compression ratio (rv) is a volume ratio and is equal to the expansion ratio in an otto cycle engine.

Compression RatioVV =

VV = r

3

4

2

1v

1 + vv = r

vv + v =

volume Clearancevolume Total = r

cc

sv

cc

ccsv

where Compression ratio is defined as

Otto Cycle Derivation

Then by substitution,

)r(1 - 1 = )r( - 1 = 1-v

-1vth

)r( = VV =

TT -1

v1

2-1

2

1

The air standard thermal efficiency of the Otto cycle then becomes:

Otto Cycle Derivation

Summarizing

QQ - 1 =

QQ - Q =

H

L

H

LHth T C m = Q v

1-TTT

1 - TTT

-1 =

2

32

1

41

th

)r( = VV =

TT -1

v1

2-1

2

1

)r(1 - 1 = )r( - 1 = 1-v

-1vth

TT =

TT

1

4

2

3

2

11TT th

where

and then

Isentropic behavior

Otto Cycle Derivation

Heat addition (Q) is accomplished through fuel combustion

Q = Lower Heat Value (LHV) BTU/lb, kJ/kg

Q AF m =Q

fuelain cycle

Otto Cycle Derivation

T C m = Q vin also

Effect of compression ratio on Otto cycle efficiency

Sample Problem – 1The air at the beginning of the compression stroke of an air-standard Otto cycle is at 95 kPa and 22C and the cylinder volume is 5600 cm3. The compression ratio is 9 and 8.6 kJ are added during the heat addition process. Calculate:

(a) the temperature and pressure after the compression and heat addition process(b) the thermal efficiency of the cycleUse cold air cycle assumptions.

Draw cycle and label points

T1 = 295 KP1 = 95 kPa

r = V1 /V2 = V4 /V3 = 9

Q23 = 8.6 kJ

Carry through with solution

kg 10 x 29.6RTVPm 3-

1

11

Calculate mass of air:

Compression occurs from 1 to 2:

ncompressio isentropic VVTT

1

2

112

k

11.42 9K 27322T

K 705.6T2 But we need T3!

Get T3 with first law:

23v23 TTmcQ Solve for T3:

2v

3 TcqT K705.6

kgkJ0.855kg6.29x10kJ8.6 3

K2304.7T3

Thermal Efficiency

11.41k 911r

11

585.0

Sample Problem – 2

Solution

Diesel Cycle P-V & T-s Diagrams

Sample Problem – 3

Gasoline vs. Diesel Engine