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The First Law ofThermodynamics &
Cyclic ProcessesMeeting 7Meeting 7
Section 4Section 4--11
Thermodynamic CycleThermodynamic Cycle• Is a series of
processes which form a closed path.
• The initial and the final states are coincident.
• All thermal engines work in a cyclic process.
11stst Law for a CycleLaw for a Cycle41 3
2
•The Energy change is zero due to the cyclic nature; • But not heat and work, they are path dependent functions. That is, the net work in a cyclic process has to equal to the net heat.
• Furthermore, the AREA inside the cycle represents the net Work or the net Heat
0∆Ecycle =
0 W Q ∆Eor cycle cyclecycle =−=
Qcycle = Wcycle
∫∫∫ δ−δ=cyclecyclecycle
WQdE0
1
Heat Engine Power CyclesHeat Engine Power Cycles
Hot body or source
Cold body or sink
System, or heat engine
Qin
Qout
Wcycle
A Thermal engines draws heat from a hot source and rejects heat to a cold source producing work
Heat Engine Efficiency: Heat Engine Efficiency: machines that produce workmachines that produce work
Wcycle
Hot body or source
Cold body or sink
System, or heat engine
QH
QLH
L
H
LH
H
LIQ
1
QQQ
QW
−=
−=
==η
Refrigerators and heat pumps: Refrigerators and heat pumps: machines that consume workmachines that consume work
Hot body or source
Cold body or sink
System
Qout
Qin
WcycleLH
L
IN
L
QQQ
WQ
−=
==η
LH
H
IN
H
QQQ
WQ
−=
==η
RE
FRIG
ER
AT
OR
HE
AT
PU
MP
TEAMPLAYTEAMPLAYA 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.
A
BWin = 45 Btu
Qin=40 BtuWout=60 Btu
1
2P
V
Th
Wη 0,4Q
= =
Ex4.14) Uma grande central de potência produz 1000MW de potência elétrica com uma eficiência térmica do ciclo de 40%. Qual é a taxa com que o calor é rejeitado ao ambiente por esta central?
H1000MWQ 2500MW
0,4= =
H L liq
L
Q W (Q Q ) W
Q 1500MW
= ⇒ − =
∴ =
∫ ∫
Wcycle
Hot body or source
Cold body or sink
System, or heat engine
QH
QL
CarnotCarnot Cycle for GasCycle for Gas• The Carnot cycle is a reversible cycle that is composed
of four internally reversible processes. – Two isothermal processes– Two adiabatic processes
Net Work Positive Net Work Negative,Reversed cycle.
•The area represents the net work or net heat.
Execution of the Carnot Cycle in a Closed SystemExecution of the Execution of the CarnotCarnot Cycle in a Closed SystemCycle in a Closed System
• 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
CarnotCarnot Cycle for VaporCycle for Vapor
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 the savings?
Considering you did this job, what would be reasonable charge for your services?
Question ….Question ….
$12,998/day$4,744,499/year
VaporVapor--cycle cycle Power PlantsPower Plants
BOILERTURBINE
PUMP
CONDENSER
qin
wout
qout
win1
3
42
We’ll simplify the power plantWe’ll simplify the power plant
1
23
4
•Low thermal efficiency•Compressor and turbine must handle two phase flows•The Carnot cycle is not a suitable model for vapor power cycles because it cannot be approximated in practice.
QL=-3500
Wb=-1
QH=+5000
Wt=?
Qp=-500
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
50001000
QW
η ===
MW 1001W1 W1000WWW TTbTliq =→−=→−=
Reversed Vapor Cycles: Reversed Vapor Cycles: Refrigerator and Heat PumpRefrigerator and Heat Pump
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 RefrigeratorInside The Household Refrigeratorcommon view engineering view
Reversed Reversed CarnotCarnot Vapor Cycle Vapor Cycle RefrigeratorRefrigerator
12
34
Expansion valve
1
4
3
2
Gas 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 SparkSpark--Ignition EnginesIgnition Engines
v
v
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
LTTmCTTmC
1QQ
1−−
−=−=η
( )( )1TTT
1TTT1
1232
141
H
L−−
−=−=η
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
2TT
TT
TT
VV
VV
TT
≡⇒=
=
=
−γ−γ
qout
qin
( )( )
( ) 2
1v
v VV r where
r
TT
TTTTTT
=−=η
−≡−−
−=η
−γ 1
2
1
232
141
11
1111
Thermal Efficiency of Ideal Otto Cycle
Thermal Efficiency of Ideal Thermal Efficiency of Ideal Otto CycleOtto 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
BraytonBrayton CycleCycle• This is another air standard cycle and it models
modern turbojet engines. Proposed by George Brayton in 1870!
Illustration of A Turbofan EngineIllustration of A Illustration of A Turbofan EngineTurbofan Engine
Other applications Other applications of of BraytonBrayton cyclecycle
• Power generation - use gas turbines to generate electricity…very efficient
• Marine applications in large ships
A Closed & OpenA Closed & Open--Cycles Cycles GasGas--Turbine EngineTurbine Engine
Aquecedor1000ºC
Condensador100ºC
TurbinaCompressor
1 2
4 3
Qh=+180kW
QL=-110kW
W>0W<0 0
0
h
liqT
liq
0,3918070
QW
η
70kW110180W
WQ:Lei1º
===
=−=
= ∫∫
Ex4.17) Uma central de potência com turbina a gás operando num ciclo fechado usa ar como fluido de trabalho, veja esquema da figura.
(a) Identifique as interações das transf. de calor e trabalho considerando o ar como sistema.
(b) (b) se 180KW são fornecidos ao aquecedor e 110KW são rejeitados no condensador, determine a potência líquida e a eficiência da central.
parte (a) parte (b)
BraytonBrayton CycleCycle• 1 to 2--adiabatic compression
in the compressor• 2 to 3--constant pressure heat
addition (replaces combustion process)
• 3 to 4--adiabatic expansion in the turbine
• 4 to 1--constant pressure heat rejection to return air to original state
BraytonBrayton CycleCycle• 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.
PressureRatio
32
1 4
PPP P
=
BraytonBrayton Cycle AnalysisCycle AnalysisLet’s assume cold air conditions and manipulate the efficiency expression:
p 4 1L
H p 3 2
C (T T )Q1 1 Q C (T T )
or
−η = − = −
−
( )( )1TT
1TTTT1
23
14
2
1
−−
−=η
1
2 2
1 1
T P ;T P
γ−γ
=
1
3 3
4 4
T PT P
γ−γ
=
Using the adiabatic relationships:
Pressure and temperature ratios
BraytonBrayton cycle analysiscycle analysis
4
3
1
2
PP
PP
=
1
4
2
3
1
2
4
3
1
2
4
3
TT
TTor
TT
TT
PP
PP since ==→
=
( )1 32p
1 4
TT rT T
γ− γ= =
BraytonBrayton Cycle AnalysisCycle AnalysisThen we can relate the temperature ratios to the pressure ratio:
Plug back into the efficiency expression and simplify:
( )1
th,Brayton 12 p
T 11 1T r γ− γ
η = − = −
BraytonBrayton CycleCycle
k1kpr
11 /)( −−=η
γ = 1.4
Thermal Efficiency of Thermal Efficiency of the Ideal the Ideal BraytonBrayton CycleCycle
Solução de Exercícios Cap 4
Ex4.13) 100 KJ de calor é adicionada a um ciclo de Carnot a 1000K. O ciclo rejeita calor a 300K. Quanto trabalho o ciclo produz e quanto calor ele rejeita?
1000K
300K
W
100KJ
Ciclo de CarnotW=?Qc=?
30KJQ
70)Q(100WQ
70KJW1000,7W0,7η1031
TT1
QWη
c
c
c
h
c
hc
=
=−→=
=×=→=
−=−==
∫ ∫
Ex4.16)
P2
T
s
P1>P2
550ºC
30ºC
0,638233031ηT =−=
Aquecedor1000ºC
Condensador100ºC
TurbinaCompressor
1 2
4 3
Qh=+180kW
QL=-110kW
W>0W<0 0
0
h
liqT
liq
0,3918070
QW
η
70kW110180W
WQ:Lei1º
===
=−=
= ∫∫
Ex4.17) Uma central de potência com turbina a gás operando num ciclo fechado usa ar como fluido de trabalho, veja esquema da figura.
(a) Identifique as interações das transf. de calor e trabalho considerando o ar como sistema.
(b) (b) se 180KW são fornecidos ao aquecedor e 110KW são rejeitados no condensador, determine a potência líquida e a eficiência da central.
parte (a) parte (b)
Rendimento Máximo -> Rendimento de Carnot
0,7112733731ηη Carnotmáx =−==
Ciclo Brayton
γ1γ
1
2
PP1η
−
−=
Conhecer Pressões
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
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