Diesel Brayton Cycles

8/12/2019 Diesel Brayton Cycles

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Diesel / Brayton Cycles

ASEN 3113


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Diesel Cycle Invented by Rudolf

Christian Karl Diesel in 1893

First engine waspowered by powderedcoal

Achieved acompression ratio ofalmost 80

Exploded, almost killedDiesel

First working enginecompleted 1894 -generated 13 hp


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Diesel Engine

Also known asCompressionIgnition Engine

(CI) Can this engine

knock?

Difference fromOtto Cycle?


4/43


5/43

Early CI Engine Cycle and the Thermodynamic Diesel Cycle

AI

R

CombustionProducts

Fuel injected

at TC

IntakeStroke

Air

Air

BC

CompressionStroke

PowerStroke

ExhaustStroke

Q in Q out

CompressionProcess

Const pressureheat addition

Process

ExpansionProcess

Const volumeheat rejection

Process

ActualCycle

DieselCycle


6/43

Process a bIsentropic

compression

Process b c Constant

pressure heataddition

Process c dIsentropicexpansion

Process d aConstant volumeheat rejection

- a=1,b=2,etcfor

book

Air-Standard Diesel cycler

c v c

vb

v3

v2

( BOOK )

Cut-off ratio:


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m

V V P

m

Quu in 23223 )()(

AIR2 3 Constant Pressure Heat Addition

now involves heat and work

)()( 222333 v P uv P umQin

)()( 2323 T T chhmQ

pin

cr vv

T T

v RT

v RT

P 2

3

2

3

3

3

2

2

Q in

First Law Analysis of Diesel Cycle

Equations for processes 1 2, 4 1 are the same as those presentedfor the Otto cycle


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)()( 34m

W

m

Quu out

AIR

3 4 Isentropic Expansion

)()( 4343 T T cuumW

vout

note v 4=v 1 socr

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 v P

T v P c

3

4

3

4

3

33

4

44

11

4

3

3

4

k c

k

r r

vv

T T


9/43

23

14

11 hh

uu

mQ

mQ

in

out

cycle Diesel

Diese l const c V

1 1

r k 1

1

k

r c

k 1 r c 1

For cold air-standard the above reduces to:

Thermal Efficiency

11

1 k Ottor

recall,

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

So why is a Diesel engine usually more efficient?


10/43

Typical CI Engines15 < r < 20

When r c (= v 3/v2) 1 the Diesel cycle efficiency approaches theefficiency of the Otto cycle

Thermal Efficiency

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


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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

11

1

k in

c r V P

Q

k

k

r as Q

in 0, r

c 1

MEP W

net

V max V min

- compares performance

of engines of the samesize


12/43

Modern CI Engine Cycle and the Thermodynamic Dual Cycle

AI

R

CombustionProducts

Fuel injectedat 15 o beforeTDC

IntakeStroke

Air

Air

TC

BC

CompressionStroke

PowerStroke

ExhaustStroke

Q in Q out

CompressionProcess

Const pressureheat addition

Process

ExpansionProcess

Const volumeheat rejection

Process

ActualCycle

DualCycle

Q in

Const volumeheat addition

Process


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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

Q in

Q i n

Q o u t1

1

2

2

2.5

2.5

3

3

44

)()()()( 5.2325.25.2325.2 T T cT T chhuu

m

Q pv

in


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Thermal Efficiency

)()(11

5.2325.2

14

hhuu

uu

mQ

mQ

in

out

cycle Dual

1)1(11

1 1 c

k

c

k cconst

Dual r k

r

r v

11

1 k Otto r 1

1111 1c

k ck

const c Diesel

r r

k r V

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

2

3

5.2

3 andwhere P P

vvr c


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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 P 3.

Transformation of r c and into more natural variables yields

11111 1

11 k r V P Q

k k r k inc

1

31 P P

r k

For the same initial conditions P 1, V1 and the same compression ratio:

Diesel Dual Otto

For the same initial conditions P 1, V1 and the same peak pressure P 3 (actual design limitation in engines):

otto Dual Diesel


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Brayton Cycle

Introduced by GeorgeBrayton (an

American) in 1872

Used separateexpansion andcompression cylinder

Constant Combustionprocess


18/43

18

Brayton Cycle


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Other applications of Braytoncycle

Power generation - use gas turbines togenerate electricityvery efficient

Marine applications in large ships

Automobile racing - late 1960s Indy 500STP sponsored cars


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Schematic of simple cycle


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Idealized Brayton Cycle


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22

Brayton Cycle

1 to 2--isentropiccompression 2 to 3--constant pressure

heat addition (replacescombustion process) 3 to 4--isentropicexpansion in the turbine 4 to 1--constant pressureheat rejection to return airto original state


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Brayton cycle analysis

in

net

qw

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.


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24

wcomp h2 h1

1 to 2 (isentropic compression incompressor), apply first law

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



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25

2323in hhqq 2 to 3 (constant pressure heat addition -treated as a heat exchanger)


or ,hhw 34turb

43turb hhw

3 to 4 (isentropic expansion in turbine)


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26

,hhq 41out

14out hhq

4 to 1 (constant pressure heat rejection)

We know this is heat transfer out of thesystem and therefore negative. In book,theyll give it a positive sign and thensubtract it when necessary.



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Substituting:

compturbnet www net work:

)h(h)h(hw 1243net


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Thermal efficiency:


in

netq

w

)h(h )h(h)h(h 231243

)h(h)h(h1

23

14


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assume cold air conditions and manipulatethe efficiency expression:

)T(Tc)T(Tc1

23 p

14 p

1TT

1TTTT

123

14

2

1


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30

TT

p p

k k

2

1

2

1

1

;TT

p p

p p

k k

k k

4

3

4

3

1

1

2

1

Using the isentropic relationships,

Define:

4

3

1

2 p P

PPP

ratio pressurer



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4

3k 1k p

1

2

TT

r TT


Then we can relate the temperature ratios tothe pressure ratio:

Plug back into the efficiency

expression and simplify:

k 1k pr

11


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32



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

turb

comp

w

wBWR


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The Back-Work Ratio is the Fractionof Turbine Work Used to Drive the

Compressor


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EXAMPLE PROBLEM

The pressure ratio of an air standard Braytoncycle is 4.5 and the inlet conditions to thecompressor are 100 kPa and 27 C. The

turbine is limited to a temperature of 827 Cand mass flow is 5 kg/s. Determine

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

Assume constant specific heats.


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Draw diagram

P

v

1

2 3

4


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Start analysis

Lets get the efficiency:

k 1k pr

11

From problem statement, we know r p = 4.5

349.05.411 4.114.1


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Net power output:

Substituting for work terms:

Wnet mwnet m w turb wcompNet Power:

Wnet m (h3 h4 ) (h2 h1)

Wnet m c p (T3 T4) (T2 T1) Applying constant specific heats:


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Need to get T 2 and T 4 Use isentropic relationships:

TT

p p

k k

2

1

2

1

1

;k

1k

3

4

3

4

p p

TT

T1 and T

3 are known along with the

pressure ratios:

K 4614.5300T 1.40.42 T2 :

T4 : K 7.7150.2221100T 1.40.4

4

Net power is then: kW1120Wnet

Wnet m c p (T3 T4) (T2 T1)


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Back Work Ratio

43

12

turb

comp

hh

hh

w

wBWR

Applying constant specific heats:

42.07.7151100

300461

TT

TTBWR

43

12


41/43

Brayton Cycle

In theory, as the pressure ratio goes up,the efficiency rises. The limiting factor isfrequently 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., T 3


42/43

Brayton Cycle

Irreversibilities Compressor and turbine frictional effects -

cause increase in entropy

Also friction causes pressure drops throughheat exchangers Stray heat transfers in components Increase in entropy has most significance

w c = h 2 h 1 for the ideal cycle, which wasisentropic

w t = h 3 h 4 for the ideal isentropic cycle


43/43

Brayton Cycle

In order to deal with irreversibilities, weneed to write the values of h 2 and h 4 ash 2,s and h 4,s .

Then

s,43

act,43

s,t

a,tt hh

hh

w

w

act,21

s,21

a,c

s,cc hh

hhww

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Diesel Brayton Cycles