TPK-2 Minggu 3

TEKNIK PERMESINAN KAPAL II(Minggu – 3)

LS 1329 ( 3 SKS)Jurusan Teknik Sistem Perkapalan

ITS Surabaya

Gas Cycles

Carnot 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=η

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 engineOtto 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

4ds

2π

CVCVDVr +

=TDC

BDC

VV

=

Mean Effective Pressure

Mean 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 (andrejection) process;

Assuming constant specific heat:

QQ - 1 =

QQ - Q =

H

L

H

LHthη

T C m = Q vin ∆

1-TTT

1 - TTT

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

2

32

1

41

23v

14vthη

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

γγ

TT =

TT

1

4

2

3

Otto Cycle Derivation

TT-1 =

2

1thηLeading to

Differences between Otto and Carnot cycles

T

s

1

2

3

4

T

s

1

2

3

4

2

3

The compression ratio (rv) is a volume ratioand is equal to the expansion ratio in an ottocycle engine.

Compression Ratio

VV =

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 γ

γη

)r( = VV =

TT -1

v1

2-1

2

1 γγ

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

Otto Cycle Derivation

Summarizing

QQ - 1 =

QQ - Q =

H

L

H

LHthη T C m = Q v ∆

1-TTT

1 - TTT

-1 =

2

32

1

41

thη

)r( = VV =

TT -1

v1

2-1

2

1 γγ

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

-1vth γ

γη

TT =

TT

1

4

2

3

2

11TT th −=η

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 22°C 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 cycle

Use cold air cycle assumptions.

Draw cycle and label points

P

v

1

2

3

4

T1 = 295 K

P1 = 95 kPa

r = V1 /V2 = V4 /V3 = 9

Q23 = 8.6 kJ

Carry through with solution

kg 10 x 29.6RT

VPm 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

kgkJ0.855

kg6.29x10kJ8.6 3

+=−

K2304.7T3 =

Thermal Efficiency

11.41k 911

r11 −− −=−=η

585.0=η

Sample Problem – 2

Solution P

v

1

2

3

4

Diesel Cycle P-V & T-s Diagrams

Sample Problem – 3

Gasoline vs. Diesel Engine