the first law of thermodynamics & cyclic processes meeting 7 section 4-1

The First Law ofThermodynamics &

Cyclic Processes

Meeting 7Meeting 7

Section 4-1Section 4-1

Thermodynamic CycleThermodynamic Cycle

• Is a series of processes which form a closed path.

• The initial and the final states are coincident.

For What Thermodynamics For What Thermodynamics Cycles Are For?Cycles Are For?

• Thermal engines work in a cyclic process.

• A Thermal engines draws heat from a hot source and rejects heat to a cold source producing work

Heat Engine Power CyclesHeat Engine Power Cycles

Hot body or source

Cold body or sink

System, or heat engine

Qin

Qout

Wcycle

Heat Engine EfficiencyHeat Engine Efficiency

Wcycle

Hot body or source

Cold body or sink

System, or heat engine

QH

QL

H

L

H

LH

H

LIQ

Q

Q1

Q

QQ

Q

W

Refrigerators and heat Refrigerators and heat pumpspumpsHot body or source

Cold body or sink

System

Qout

Qin

WcycleLH

L

IN

L

QQ

Q

W

Q

LH

H

IN

H

QQ

Q

W

Q

RE

FR

IGE

RA

TO

RH

EA

T P

UM

P

Energy analysis of cyclesEnergy analysis of cycles

4

1

3

2

For the cycle, E1 E1 = 0, or

0ΔE cycle

0 W Q ΔE cycle cyclecycle

For cycles, we can write:For cycles, we can write:

Q Wcycle cycle

Qcycle and Wcycle represent net amounts

which can also be represented as:

Qcycle = Wcycle

TEAMPLAYTEAMPLAY• A closed system undergoes a cycle consisting

of two processes. During the first process, 40 Btu of heat is transferred to the system while the system does 60 Btu of work. During the second process, 45 Btu of work is done on the system.(a) Determine the heat transfer during the

second process.(b) Calculate the net work and net heat transfer

of the cycle.

TEAMPLAYTEAMPLAY

B

A

Win = 45 Btu

Qin=40 Btu

Wout=60 Btu1

2

Carnot CycleCarnot Cycle

• The Carnot cycle is a reversible cycle that is composed of four internally reversible processes.

– Two isothermal processes

– Two adiabatic processes

Carnot cycle for a gasCarnot cycle for a gasThe The area represents the net workarea represents the net work

TL

P-v Diagram of the P-v Diagram of the Reversed Carnot CycleReversed Carnot Cycle

P-v Diagram of the P-v Diagram of the Reversed Carnot CycleReversed Carnot Cycle

•

TL

The Carnot cycle for a gas might The Carnot cycle for a gas might occur as visualized below.occur as visualized below.

TL

QL

TH TLTL TH

Execution of the Carnot Execution of the Carnot Cycle in a Closed SystemCycle in a Closed SystemExecution of the Carnot Execution of the Carnot Cycle in a Closed SystemCycle in a Closed System

• (Fig. 5-43)

•This is a Carnot cycle involving two phases -- it is still two adiabatic processes and two isothermal processes.

•It is always reversible -- a Carnot cycle is reversible by definition.

TL

TL

TL

Vapor Power CyclesVapor Power Cycles

• We’ll look specifically at the Rankine cycle, which is a vapor power cycle.

• It is the primary electrical producing cycle in the world.

• The cycle can use a variety of fuels.

Question …..Question …..How much does it cost to operate a gas fired 1000 MW(output) power plant with a 35% efficiency for 24 hours/day for a full year if fuel cost are $2.00 per 106 Btu?

$467,952.27/day

$170,801,979/year

If you could improve the efficiency of a 1000 MW power plant from 35% to 36%, what would be a reasonable charge for your services? Let’s assume $2.00 per million BTUs fuel charge and 24 hr/day operation.

Question ….Question ….

$12,998.67/day

$4,744,499/year

Vapor-cycle Vapor-cycle Power PlantsPower Plants

BOILERTURBINE

PUMP

CONDENSER

q in

w out

qout

w in

1

3

42

We’ll simplify the power We’ll simplify the power plantplant

Carnot Vapor CycleCarnot Vapor Cycle

Low thermal efficiency

compressor and turbine must handle two phase

flows

Carnot Vapor CycleCarnot Vapor CycleCarnot Vapor CycleCarnot Vapor Cycle

•The Carnot cycle is not a suitable model for vapor power cycles because it cannot be approximated in practice.

Rankine CycleRankine CycleRankine CycleRankine Cycle• The model cycle for vapor power cycles is

the Rankine cycle which is composed of four internally reversible processes: constant-pressure heat addition in a boiler, isentropic expansion in a turbine, constant-pressure heat rejection in a condenser, and isentropic compression in a pump. Steam leaves the condenser as a saturated liquid at the condenser pressure.

Refrigerator and Heat Refrigerator and Heat Pump ObjectivesPump Objectives

•

The objective of a refrigerator is to remove heat (QL) from the cold medium; the objective of a heat pump is to supply heat (QH) to a warm medium

Inside The Household Inside The Household RefrigeratorRefrigerator

Ordinary Household Ordinary Household RefrigeratorRefrigerator

Gas Power CycleGas Power CycleGas Power CycleGas Power Cycle

• A cycle during which a net amount of work is produced is called a power cycle, and a power cycle during which the working fluid remains a gas throughout is called a gas power cycle.

Actual and Ideal Cycles in Actual and Ideal Cycles in Spark-Ignition EnginesSpark-Ignition Engines

v

v

Otto CycleOtto Cycleqin

qout

P-V diagram (Work)

T-S diagram (Heat Transfer)

Performance of cyclePerformance of cycle

H

L

H

net

Q

Q1

Q

w

Thermal Efficiency:

Need to know QH and QL

Ott

o C

ycle

Ott

o C

ycle

• Heat addition 2-3 QH = mCV(T3-T2)• Heat rejection 4-1 QL = mCV(T4-T1)

• or in terms of the temperature ratios

23V

14V

H

L

TTmC

TTmC1

Q

Q1

1TTT

1TTT1

Q

Q1

232

141

H

L

qout

qin

Ott

o C

ycle

Ott

o C

ycle

• 1-2 and 3-4 are adiabatic process, using the adiabatic relations between T and V

4

3

1

2

4

3

RATIO VOLUME SAME

1

3

41

2

1

1

2

T

T

T

T

T

T

V

V

V

V

T

T

2

11

2

1

V

V r ;

r

11

T

T1

qout

qin

1k2

1Otto,th r

11

T

T1

This looks like the Carnot efficiency, but it is not! TT1 1 and Tand T22 are not constant. are not constant.

Cycle performance with cold air Cycle performance with cold air cycle assumptionscycle assumptions

What are the limitations for this expression?

Thermal Efficiency Thermal Efficiency of Ideal Otto Cycleof Ideal Otto CycleThermal Efficiency Thermal Efficiency of Ideal Otto Cycleof Ideal Otto Cycle

• Under cold-air-standard assumptions, the thermal efficiency of the ideal Otto cycle is

where r is the compression ratio and k is the specific heat ratio Cp /Cv.

Effect of compression ratio on Otto Effect of compression ratio on Otto cycle efficiencycycle efficiency

k = 1.4

Otto CycleOtto CycleThe thermal efficiency of the Otto Cycle increases with the specific heat ratio k of the working fluid

Brayton CycleBrayton Cycle• This is another air standard cycle and it

models modern turbojet engines.

Brayton CycleBrayton Cycle

Proposed by George Brayton in 1870!

http://www.pwc.ca/en_markets/demonstration.htmlhttp://www.pwc.ca/en_markets/demonstration.html

jet engine with afterburner jet engine with afterburner for military applications.for military applications.

Schematic of A Schematic of A Turbofan EngineTurbofan Engine

Illustration of A Illustration of A Turbofan EngineTurbofan EngineIllustration of A Illustration of A Turbofan EngineTurbofan Engine

TurbopropTurboprop

compressor turbine

burner

Schematic of a Schematic of a Turboprop EngineTurboprop Engine

Other applications Other applications of Brayton cycleof 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

An Open-Cycle An Open-Cycle Gas-Turbine EngineGas-Turbine Engine

A Closed-Cycle A Closed-Cycle Gas-Turbine EngineGas-Turbine Engine

Brayton CycleBrayton CycleBrayton CycleBrayton Cycle• The ideal cycle for modern gas-turbine engines is the Brayton cycle, which is made up of four internally reversible processes: isentropic compression, constant pressure heat addition, isentropic expansion, and constant pressure heat rejection.

Turbojet Engine Basic Turbojet Engine Basic Components and T-s Components and T-s

Diagram for Ideal Turbojet Diagram for Ideal Turbojet CycleCycle

P-v and T-s Diagrams for the P-v and T-s Diagrams for the Ideal Brayton Cycle Ideal Brayton Cycle

Brayton CycleBrayton Cycle• 1 to 2--isentropic compression in the

compressor• 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 CycleBrayton Cycle

• 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.

Brayton cycle analysisBrayton cycle analysisLet’s assume cold air conditions and manipulate the efficiency expression:

)T(TC

)T(TC1

23p

14p

1TT

1TT

T

T1

23

14

2

1

;P

P

T

T k

1k

1

2

1

2

k

1k

2

1k

1k

3

4

3

4

P

P

P

P

T

T

Using the isentropic relationships,

Let’s define:

4

3

1

2p P

P

P

Pratiopressurer

Brayton cycle analysisBrayton cycle analysis

Brayton CycleBrayton Cycle• Because the Brayton cycle operates

between two constant pressure lines, or isobars, the pressure ratio is important.

• The pressure ratio is just that--a pressure ratio.

• A compression ratio is a volume ratio (refer to the Otto Cycle).

Brayton CycleBrayton Cycle• The pressure ratio is

• Also1

2

r

r

s1

2

1

2

P

P

P

P

P

P

1

2

r

r

s4

3

4

3

P

P

P

P

P

P

P

P

4

3

4

3k1kp

1

2

T

Tr

T

T

Brayton cycle analysisBrayton cycle analysisThen we can relate the temperature ratios to the pressure ratio:

Plug back into the efficiency expression and simplify:

k1kp2

1Braytonth r

11

T

T1 ,

Ideal Brayton CycleIdeal Brayton Cycle

k1kp

Braytonth r

11 ,

What does this expression assume?What does this expression assume?

Thermal Efficiency Thermal Efficiency of Brayton Cycleof Brayton Cycle

Thermal Efficiency Thermal Efficiency of Brayton Cycleof Brayton Cycle

• Under cold-air-standard assumptions, the Brayton cycle thermal efficiency is

where rp = Pmax/Pmin is the pressure ratio

and k is the specific heat ratio. The thermal efficiency of the simple Brayton cycle increases with the pressure ratio.

Brayton CycleBrayton Cycle

k1kpr

11

/)(

k = 1.4

Thermal Efficiency of Thermal Efficiency of the Ideal Brayton Cyclethe Ideal Brayton Cycle

Um sistema pistão cilindro contêm, inicialmente, três kg de

H2O no estado de líquido saturado com 0.6 MPa.

Calor é adicionado, vagarosamente, a água fazendo com

que o pistão se movimente de tal maneira que a pressão

seja constante.

Quanto de trabalho é realizado pela água?

Quanta energia deve ser transferida para a água de tal

maneira que ao final do processo ela esteja no estado de

vapor saturado?

Evaporação a pressão constante

Representaçãodo processo

Fronteirado

Sistema

processo a pressão const.

Primeira Lei:

1Q2 – 1W2 = U2 – U1

mas o trabalho 1W2 = Patm*(V2-V1),

Logo 1Q2 = (P2V2+U2)-(P1V1+U1) = H2-H1

Onde h2 é a entalpia do vapor saturado e h1 é a

entalpia do líquido. Na tabela 1-2 termodinâmico para 0.6MPa, tem-se que h2 = 2756,8 KJ/kg e h1 =

670,56 KJ/kg. Considerando 3kg de H2O, então o calor transferido será de 3*(2756-670) = 6259 KJ.

Resfriamento com Gelo Seco

0.5 kg de gelo seco (CO2) a 1 atm é colocado em cima de

uma fatia de picanha. O gelo seco sublima a pressão

constante devido ao fluxo de calor transferido pela picanha.

Ao final do processo todo CO2 está no estado de vapor (foi

completamente sublimado).

Determine a temperatura do CO2 e quanto de calor ele

recebeu da picanha.

Representaçãodo processoFronteira

do Sistema

Resfriamento com Gelo Seco – processo a pressão const.

Primeira Lei:

1Q2 – 1W2 = U2 – U1

mas o trabalho 1W2 = Patm*(V2-V1),

Logo 1Q2 = (P2V2+U2)-(P1V1+U1) = H2-H1

Onde h2 é a entalpia do vapor saturado e h1 é a

entalpia do sólido. No diagrama termodinâmico para Patm, tem-se que h2 = 340 KJ/kg e h1 = -220

KJ/kg. Considerando 0.5kg de CO2, então o calor

transferido será de 280 KJ. A temperatura de saturação do CO2 será de 175K (-98 oC)

SOLI

D

LIQUIDVAPOR

Solução de Exercícios Cap 4

Ex4.13)1000K

300K

W

100KJ

Ciclo de CarnotW=?Qc=?

30KJQ

70)Q(100WQ

70KJW

1000,7W0,7η

10

31

T

T1

Q

Wη

c

c

c

h

c

hc

Ex4.14)

1500MWQ

W)QQ(

WQ

2500MW0,4

1000MWQ

Q

W0,4η

c

liqch

h

hT

Ex4.15)

1000MW4000-5000WQ

Condensador

Turbina

Caldeira

Bomba

WT>0

Qc<0 (3500)

Wb<0

Qh>0(5000)

Qmeio<0 (500)

00

h

liqT 25

5000

1000

Q

Wη

MW 1001W1 W1000WWW TTbTliq

Ex4.16)

P2

T

s

P1>P2

550ºC

30ºC

0,63823

3031ηT

Ex4.17) Qh=+180kW

Aquecedor

Condensador

TurbinaCompressor

1 2

43

QL=-110kW

W>0W<0

1000ºC

100ºC

00

h

liqT

liq

0,39180

70

Q

Wη

70kW110180W

WQ:Lei1º

Rendimento Máximo -> Rendimento de Carnot

0,711273

3731ηη Carnotmáx

Ciclo Brayton

γ

1γ

1

2

P

P1η

Conhecer Pressões