the first law of thermodynamics & cyclic processes meeting 7 section 4-1
TRANSCRIPT
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The First Law ofThermodynamics &
Cyclic Processes
Meeting 7Meeting 7
Section 4-1Section 4-1
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Thermodynamic CycleThermodynamic Cycle
• Is a series of processes which form a closed path.
• The initial and the final states are coincident.
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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
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Heat Engine Power CyclesHeat Engine Power Cycles
Hot body or source
Cold body or sink
System, or heat engine
Qin
Qout
Wcycle
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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
Q
W
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Refrigerators and heat Refrigerators and heat pumpspumpsHot body or source
Cold body or sink
System
Qout
Qin
WcycleLH
L
IN
L
Q
W
Q
LH
H
IN
H
Q
W
Q
RE
FR
IGE
RA
TO
RH
EA
T P
UM
P
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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
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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
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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.
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TEAMPLAYTEAMPLAY
B
A
Win = 45 Btu
Qin=40 Btu
Wout=60 Btu1
2
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Carnot CycleCarnot Cycle
• The Carnot cycle is a reversible cycle that is composed of four internally reversible processes.
– Two isothermal processes
– Two adiabatic processes
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Carnot cycle for a gasCarnot cycle for a gasThe The area represents the net workarea represents the net work
TL
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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
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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
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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)
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•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
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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.
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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
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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
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Vapor-cycle Vapor-cycle Power PlantsPower Plants
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BOILERTURBINE
PUMP
CONDENSER
q in
w out
qout
w in
1
3
42
We’ll simplify the power We’ll simplify the power plantplant
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Carnot Vapor CycleCarnot Vapor Cycle
Low thermal efficiency
compressor and turbine must handle two phase
flows
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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.
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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.
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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
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Inside The Household Inside The Household RefrigeratorRefrigerator
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Ordinary Household Ordinary Household RefrigeratorRefrigerator
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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.
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Actual and Ideal Cycles in Actual and Ideal Cycles in Spark-Ignition EnginesSpark-Ignition Engines
v
v
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Otto CycleOtto Cycleqin
qout
P-V diagram (Work)
T-S diagram (Heat Transfer)
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Performance of cyclePerformance of cycle
H
L
H
net
Q
Q1
Q
w
Thermal Efficiency:
Need to know QH and QL
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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
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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
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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?
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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.
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Effect of compression ratio on Otto Effect of compression ratio on Otto cycle efficiencycycle efficiency
k = 1.4
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Otto CycleOtto CycleThe thermal efficiency of the Otto Cycle increases with the specific heat ratio k of the working fluid
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Brayton CycleBrayton Cycle• This is another air standard cycle and it
models modern turbojet engines.
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Brayton CycleBrayton Cycle
Proposed by George Brayton in 1870!
http://www.pwc.ca/en_markets/demonstration.htmlhttp://www.pwc.ca/en_markets/demonstration.html
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jet engine with afterburner jet engine with afterburner for military applications.for military applications.
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Schematic of A Schematic of A Turbofan EngineTurbofan Engine
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Illustration of A Illustration of A Turbofan EngineTurbofan EngineIllustration of A Illustration of A Turbofan EngineTurbofan Engine
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TurbopropTurboprop
compressor turbine
burner
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Schematic of a Schematic of a Turboprop EngineTurboprop Engine
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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
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An Open-Cycle An Open-Cycle Gas-Turbine EngineGas-Turbine Engine
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A Closed-Cycle A Closed-Cycle Gas-Turbine EngineGas-Turbine Engine
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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.
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Turbojet Engine Basic Turbojet Engine Basic Components and T-s Components and T-s
Diagram for Ideal Turbojet Diagram for Ideal Turbojet CycleCycle
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P-v and T-s Diagrams for the P-v and T-s Diagrams for the Ideal Brayton Cycle Ideal Brayton Cycle
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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
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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.
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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
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;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
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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).
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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
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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 ,
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Ideal Brayton CycleIdeal Brayton Cycle
k1kp
Braytonth r
11 ,
What does this expression assume?What does this expression assume?
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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.
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Brayton CycleBrayton Cycle
k1kpr
11
/)(
k = 1.4
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Thermal Efficiency of Thermal Efficiency of the Ideal Brayton Cyclethe Ideal Brayton Cycle
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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
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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.
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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
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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)
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SOLI
D
LIQUIDVAPOR
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Solução de Exercícios Cap 4
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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
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Ex4.14)
1500MWQ
W)QQ(
WQ
2500MW0,4
1000MWQ
Q
W0,4η
c
liqch
h
hT
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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
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Ex4.16)
P2
T
s
P1>P2
550ºC
30ºC
0,63823
3031ηT
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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
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Rendimento Máximo -> Rendimento de Carnot
0,711273
3731ηη Carnotmáx
Ciclo Brayton
γ
1γ
1
2
P
P1η
Conhecer Pressões