Thermodynamic Cycles4

8/3/2019 Thermodynamic Cycles4

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Diesel / Brayton Cycles

ASEN 3113



Diesel Cycle

Invented by Rudolf Christian Karl Diesel in1893

First engine was powered bypowdered coal

Achieved a compression ratioof almost 80

Exploded, almost killedDiesel

First working enginecompleted 1894 - generated13 hp



Diesel Engine

Also known asCompressionIgnition Engine

(CI) Can this engine

³knock´?

Difference fromOtto 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

AI

R

CombustionProducts

Fuel injectedat TC

Intake

Stroke

Air

Air

BC

Compression

Stroke

Power

Stroke

Exhaust

Stroke

Qin Qout

Compression

Process

Const pressure

heat addition

Process

Expansion

Process

Const volume

heat rejection

Process

Actual

Cycle

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

r c!

vc

vb

!

v3

v2

( BOOK )

Cut-off ratio:



m

V V P

m

Q

uu

in 232

23 )()(

!

AIR

23 Constant Pressure Heat Addition

now involves heat and work

)()( 222333 v P uv P um

Qin!

)()( 2323 T T chhm

Q p

in !!

cr v

v

T

T

v

RT

v

RT P !!p!!

2

3

2

3

3

3

2

2

Qin

First LawAnalysis of Diesel Cycle

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

for the Otto cycle



)()( 34m

W

m

Quu out ! AIR

3 4 Isentropic Expansion

)()( 4343 T T cuum

W v

out !!

notev 4=v

1 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

v P

T

v P c!p!

3

4

3

4

3

33

4

44

11

4

3

3

4

¹ º ¸©

ª¨!¹¹

º ¸©©

ª¨!

k

c

k

r

r

v

v

T

T



23

1411hhuu

mQmQ

in

out

cycle Diesel

!!L

L Diesel con st cV !1

1

r k 1

1

k

r c

k 1 r c 1

«

-¬¬

»

½¼¼

For cold air-standard the above reduces to:

Thermal Efficiency

1

11

!k

Otto r

Lrecall,

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 efficiency

than the Otto cycle

So why is a Diesel engine usually more efficient?



Typical CI Engines

15 < r < 20

When r c

(= v3/v

2)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 variable

a more suitable parameter is the heat input, the two are related by:

111

111¹¹

º ¸©©

ª¨! k

inc

r V P Q

k k r as Qin 0, r c 1

ME P ! W net V max V min

- compares performance

of engines of the samesize



Modern CI Engine Cycle and the Thermodynamic Dual Cycle

AI

R

CombustionProducts

Fuel injectedat 15o beforeTDC

Intake

Stroke

Air

Air

TC

BC

Compression

Stroke

Power

Stroke

Exhaust

Stroke

Qin Qout

Compression

Process

Const pressure

heat addition

Process

Expansion

Process

Const volume

heat rejection

Process

Actual

Cycle

DualCycle

Qin

Const volume

heat addition

Process



Process 1 2 Isentropic compression

Process 2 2.5 Constant volume heat additionProcess 2.5 3 Constant pressure heat addition

Process 3 4 Isentropic expansion

Process 4 1 Constant volume heat rejection

Dual Cycle

Q in

Q in

Q out

11

2

2

2.5

2.5

33

44

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

m

Q pv

in !!



Thermal Efficiency

)()(115.2325.2

14hhuu

uu

mQ

mQ

in

out

cycle Dual

!!L

¼½»

¬-

«

!

1)1(

111

1 c

k

c

k

ccon st Dual

r k

r

r v EE

E

L

1

11

!

k Ottor

L ¼½

»¬-

«

! 1

1111

1c

k c

k con st c Diesel

r

r

k r V

L

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

2

3

5.2

3 andwhere P

P v

vr c !! E



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), or

ii) maximum pressure P3.

Transformation of r c and E into more natural variables yields

¼½»¬

-«

¹¹

º ¸©©

ª¨!

11111

111 k r V P

Q

k

k r k

inc

E

E1

31

P

P

r k

!E

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

Diesel Dual Otto LLL ""For the same initial conditions P1, V1 and the same peak pressure P3

(actual design limitation in engines):

otto Dual Diesel LLL ""





Brayton Cycle

Introduced by George

Brayton (an

American) in 1872

Used separateexpansion and

compression cylinder

Constant Combustion

process



18

Brayton Cycle



Other applications of Braytoncycle

Power generation - use gas turbines togenerate electricity«very efficient

Marine applications in large ships

Automobile racing - late 1960s Indy 500STP sponsored cars



Schematic of simple cycle



Idealized Brayton Cycle



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



Brayton cycle analysis

in

net

q

w!L

com pturbnet www !

Efficiency:

Net work:

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

The pressure ratio is not a compression ratio.



24

wcomp

! h2 h

1

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:




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)



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,they¶ll give it a positive sign and then

subtract it when necessary.





Substituting:

com pturbnet www

!

net work:

)h(h)h(hw 1243net !



Thermal efficiency:


in

net

qw!L

)h(h)h(h)h(h

23

1243

!

)h(h

)h(h1

23

14

!L




assume cold air conditions and manipulatethe efficiency expression:

)T(Tc

)T(Tc1

23 p

14 p

!L

1TT

1TT

T

T1

23

14

2

1

!L



30

TT

p p

k

k 2

1

2

1

1

! ¨ª© ¸

º¹

; TT

p p

p p

k

k

k

k 4

3

4

3

1

1

2

1

! ¨ª© ¸

º¹ ! ¨

ª© ¸

º¹

Using the isentropic relationships,

Define:

4

3

1

2 p

P

P

P

Pratio pressurer !!!




4

3k 1k p

1

2

TTr

TT !!


Then we can relate the temperature ratios tothe pressure ratio:

Plug back into the efficiency

expression and simplify:

k 1k

pr

11

!L



32





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

turb

com p

w

wBWR !



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

Compressor

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

Compressor



EXAMPLE PROBLEM

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

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

a) the thermal efficiency

b) 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:

k 1k pr

1

1

!L

From problem statement, we know rp = 4.5

349.05.4

114.114.1

!!

L



Net power output:

Substituting for work terms:

Ý Wnet ! Ý m w net ! Ý m w turb w com p Net Power:

Ý Wnet ! Ý m (h3h4 ) (h2 h1)

Ý Wnet ! Ý mc p (T3T4 ) (T2 T1)

Applying constant specific heats:



Need to get T2 and T4

Use isentropic relationships:

T

T

p

p

k

k

2

1

2

1

1

!

¨

ª©

¸

º¹

;

k

1k

3

4

3

4

p

p

T

T

¹¹ º

¸©©ª

¨!

T1 and T3 are known along with thepressure ratios:

K 4614.5300T1.40.4

2 !!T2:

T4: K 7.7150.2221100T1.40.4

4 !!

Net power is then: kW1120Wnet

!

Ý Wnet ! Ý mc p (T3T4 ) (T2 T1)



Back Work Ratio

43

12

turb

com p

hh

hh

w

wBWR

!!

Applying constant specific heats:

42.0

7.7151100

300461

TT

TTBWR

43

12!

!

!



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., T3



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

wc = h2 ± h1 for the ideal cycle, which wasisentropic

wt = h3 ± h4 for the ideal isentropic cycle



Brayton Cycle

In order to deal with irreversibilities, weneed to write the values of h2 and h4 ash2,s and h4,s.

Then

s,43

act,43

s,t

a,tt

hh

hh

w

w

!!L

act,21

s,21

a,c

s,c

chh

hh

w

w

!!L

Date post:	06-Apr-2018
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