Diesel / Brayton Cycles ASEN 3113. Diesel Cycle Invented by Rudolf Christian Karl Diesel in 1893 First engine was powered by powdered coal Achieved a.

Diesel / Brayton Cycles

ASEN 3113

Diesel Cycle• Invented by Rudolf

Christian Karl Diesel in 1893

• First engine was powered by powdered coal

• Achieved a compression ratio of almost 80

• Exploded, almost killed Diesel

• First working engine completed 1894 - generated 13 hp

Diesel Engine

• Also known as Compression Ignition Engine (CI)

• Can this engine “knock”?

• Difference from Otto Cycle?

Thermodynamic Cycles for CI engines

• In early CI engines the fuel was injected when the piston reached TDC and thus combustion lasted well into the expansion stroke.

• In modern engines the fuel is injected before TDC (about 15o)

• The combustion process in the early CI engines is best approximated by a constant pressure heat addition process Diesel Cycle

• The combustion process in the modern CI engines is best approximated by a combination of constant volume and constant pressure Dual Cycle

Fuel injection startsFuel injection starts

Early CI engine Modern CI engine

Early CI Engine Cycle and the Thermodynamic Diesel Cycle

A I R

CombustionProducts

Fuel injectedat TC

IntakeStroke

Air

Air

BC

CompressionStroke

PowerStroke

ExhaustStroke

Qin Qout

CompressionProcess

Const pressure heat addition

Process

ExpansionProcess

Const volume heat rejection

Process

ActualCycle

DieselCycle

Process a b Isentropic compression

Process b c Constant pressure heat addition

Process c d Isentropic expansion

Process d a Constant volume heat rejection

- a=1,b=2,etc…for book

Air-Standard Diesel cycle

€

rc =vc

vb

=v3

v2

(BOOK)

Cut-off ratio:

( )m

VVP

m

Quu in 232

23 )()(−

−+=−

AIR23 Constant Pressure Heat Addition

now involves heat and work

)()( 222333 vPuvPum

Qin +−+=

)()( 2323 TTchhm

Qp

in −=−=

crv

v

T

T

v

RT

v

RTP ==→==

2

3

2

3

3

3

2

2

Qin

First Law Analysis of Diesel Cycle

Equations for processes 12, 41 are the same as those presented for the Otto cycle

)()( 34 m

W

m

Quu out+−=−

AIR

3 4 Isentropic Expansion

)()( 4343 TTcuum

Wv

out −=−=

note v4=v1 so cr

r

v

v

v

v

v

v

v

v

v

v=⋅=⋅=

3

2

2

1

3

2

2

4

3

4

r

r

T

T

P

P

T

vP

T

vP c⋅=→=3

4

3

4

3

33

4

44

11

4

3

3

4−−

⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

kc

k

rr

vv

TT

23

1411hh

uu

mQ

mQ

in

out

cycleDiesel −

−−=−=η

€

ηDieselconst cV

=1−1

rk−1

1

k⋅

rck −1( )

rc −1( )

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥

For cold air-standard the above reduces to:

Thermal Efficiency

1

11 −−= kOtto

rηrecall,

Note the term in the square bracket is always larger than one so for the same compression ratio, r, the Diesel cycle has a lower thermal efficiencythan the Otto cycle

So why is a Diesel engine usually more efficient?

Typical CI Engines15 < r < 20

When rc (= v3/v2)1 the Diesel cycle efficiency approaches the efficiency of the Otto cycle

Thermal Efficiency

Higher efficiency is obtained by adding less heat per cycle, Qin, run engine at higher speed to get the same power.

k = 1.3

k = 1.3

The cut-off ratio is not a natural choice for the independent variablea more suitable parameter is the heat input, the two are related by:

111

111 −⎟⎟

⎠

⎞⎜⎜⎝

⎛−−= kin

crVP

Qk

kr as Qin 0, rc1

€

MEP =Wnet

Vmax −Vmin

- compares performance of engines of the same size

Modern CI Engine Cycle and the Thermodynamic Dual Cycle

A I R

CombustionProducts

Fuel injectedat 15o before TDC

IntakeStroke

Air

AirTC

BC

CompressionStroke

PowerStroke

ExhaustStroke

Qin Qout

CompressionProcess

Const pressure heat addition

Process

ExpansionProcess

Const volume heat rejection

Process

ActualCycle

DualCycle

Qin

Const volume heat addition

Process

Process 1 2 Isentropic compressionProcess 2 2.5 Constant volume heat additionProcess 2.5 3 Constant pressure heat additionProcess 3 4 Isentropic expansionProcess 4 1 Constant volume heat rejection

Dual Cycle

Qin

Qin

Qout

11

2

2

2.5

2.5

33

44

)()()()( 5.2325.25.2325.2 TTcTTchhuum

Qpv

in −+−=−+−=

Thermal Efficiency

)()(11

5.2325.2

14

hhuu

uu

mQ

mQ

in

out

cycleDual −+−

−−=−=η

( )⎥⎦⎤

⎢⎣

⎡−+−

−−= − 1)1(

111 1

c

kc

kcconst

Dual rk

r

rv αα

αη

1

11 −−= kOtto

rη

( )( )⎥⎦

⎤⎢⎣

⎡

−−

⋅−= − 1

1111

1c

kc

kconst cDiesel

r

r

krV

η

Note, the Otto cycle (rc=1) and the Diesel cycle (=1) are special cases:

2

3

5.2

3 and where PP

vvrc ==

The use of the Dual cycle requires information about either:i) the fractions of constant volume and constant pressure heat addition (common assumption is to equally split the heat addition), orii) maximum pressure P3.

Transformation of rc and into more natural variables yields

⎥⎦

⎤⎢⎣

⎡

−−−⎟⎟

⎠

⎞⎜⎜⎝

⎛−−= − 1

1111

111 krVP

Q

k

kr

kin

cα

α1

31

P

P

rk=

For the same initial conditions P1, V1 and the same compression ratio:

DieselDualOtto ηηη >>

For the same initial conditions P1, V1 and the same peak pressure P3 (actual design limitation in engines):

ottoDualDiesel ηηη >>

Brayton Cycle

• Introduced by George Brayton (an American) in 1872

• Used separate expansion and compression cylinder

• Constant Combustion process

18

Brayton Cycle

Other applications of Brayton cycle

• Power generation - use gas turbines to generate electricity…very efficient

• Marine applications in large ships• Automobile racing - late 1960s Indy 500

STP sponsored cars

Schematic of simple cycle

Idealized Brayton Cycle

22

Brayton Cycle

•1 to 2--isentropic compression•2 to 3--constant pressure heat addition (replaces combustion process)•3 to 4--isentropic expansion in the turbine •4 to 1--constant pressure heat rejection to return air to original state

Brayton cycle analysis

in

net

q

w=η

compturbnet www −=

Efficiency:

Net work:

• Because the Brayton cycle operates between two constant pressure lines, or isobars, the pressure ratio is important.

•The pressure ratio is not a compression ratio.

24

€

wcomp = h2 − h1

1 to 2 (isentropic compression in compressor), apply first law

**When analyzing the cycle, we know that the compressor work is in (negative). It is standard convention to just drop the negative sign and deal with it later:

Brayton cycle analysis

25

2323in hhqq −==

2 to 3 (constant pressure heat addition - treated as a heat exchanger)

Brayton cycle analysis

or,hhw 34turb −=−

43turb hhw −=

3 to 4 (isentropic expansion in turbine)

26

,hhq 41out −=

14out hhq −=

4 to 1 (constant pressure heat rejection)

We know this is heat transfer out of the system and therefore negative. In book, they’ll give it a positive sign and then subtract it when necessary.

Brayton cycle analysis

Brayton cycle analysis

Substituting:

compturbnet www −=net work:

)h(h)h(hw 1243net −−−=

Thermal efficiency:

Brayton cycle analysis

in

net

q

w=η

)h(h

)h(h)h(h

23

1243

−−−−

=

)h(h

)h(h1

23

14

−−

−=η

Brayton cycle analysis

assume cold air conditions and manipulate the efficiency expression:

)T(Tc

)T(Tc1

23p

14p

−−

−=η

( )( )1TT

1TT

T

T1

23

14

2

1

−

−−=η

30

T

T

p

p

k

k2

1

2

1

1

=⎛

⎝⎜

⎞

⎠⎟

−

;T

T

p

p

p

p

k

k

k

k4

3

4

3

1

1

2

1

=⎛

⎝⎜

⎞

⎠⎟ =

⎛

⎝⎜

⎞

⎠⎟

− −

Using the isentropic relationships,

Define:

4

3

1

2p P

P

P

Pratiopressurer ===

Brayton cycle analysis

( )

4

3k1kp

1

2

T

Tr

T

T== −

Brayton cycle analysis

Then we can relate the temperature ratios to the pressure ratio:

Plug back into the efficiency expression and simplify:

( ) k1kpr

11

−−=η

32

Brayton cycle analysis

Brayton cycle analysis

An important quantity for Brayton cycles is the Back Work Ratio (BWR).

turb

comp

w

wBWR=

The Back-Work Ratio is the Fraction of Turbine Work Used to Drive the

Compressor

The Back-Work Ratio is the Fraction of Turbine Work Used to Drive the

Compressor

EXAMPLE PROBLEM

The pressure ratio of an air standard Brayton cycle is 4.5 and the inlet conditions to the compressor are 100 kPa and 27C. The turbine is limited to a temperature of 827C and mass flow is 5 kg/s. Determine

a) the thermal efficiencyb) the net power output in kWc) the BWR

Assume constant specific heats.

Draw diagram

P

v

1

2 3

4

Start analysis

Let’s get the efficiency:

( ) k1kpr

11

−−=η

From problem statement, we know rp = 4.5

( ) 349.05.4

11

4.114.1=−=

−η

Net power output:

Substituting for work terms:

€

˙ W net = ˙ m w net = ˙ m w turb − w comp( )

Net Power:

€

˙ W net = ˙ m (h3 −h4 )− (h2 −h1)( )

€

˙ W net = ˙ m cp (T3 −T4 )− (T2 −T1)( )

Applying constant specific heats:

Need to get T2 and T4

Use isentropic relationships:

T

T

p

p

k

k2

1

2

1

1

=⎛

⎝⎜

⎞

⎠⎟

−

;k

1k

3

4

3

4

p

p

T

T−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

T1 and T3 are known along with the pressure ratios:

( ) K4614.5300T 1.40.42 ==T2:

T4: ( ) K7.7150.2221100T 1.40.44 ==

Net power is then: kW1120Wnet =&

€

˙ W net = ˙ m cp (T3 −T4 )− (T2 −T1)( )

Back Work Ratio

43

12

turb

comp

hh

hh

w

wBWR

−−

==

Applying constant specific heats:

42.07.7151100

300461

TT

TTBWR

43

12 =−−

=−−

=

Brayton Cycle

• In theory, as the pressure ratio goes up, the efficiency rises. The limiting factor is frequently the turbine inlet temperature.

• The turbine inlet temp is restricted to about 1,700 K or 2,600 F.

• Consider a fixed turbine inlet temp., T3

Brayton Cycle

• Irreversibilities– Compressor and turbine frictional effects -

cause increase in entropy– Also friction causes pressure drops through

heat exchangers– Stray heat transfers in components– Increase in entropy has most significance

• wc = h2 – h1 for the ideal cycle, which was isentropic

• wt = h3 – h4 for the ideal isentropic cycle

Brayton Cycle

• In order to deal with irreversibilities, we need to write the values of h2 and h4 as h2,s and h4,s.

• Thens,43

act,43

s,t

a,tt hh

hh

w

w

−−

==η

act,21

s,21

a,c

s,cc hh

hh

w

w

−−

==η