The First Law of
Thermodynamics &
Cyclic Processes
Meeting 7
Section 4-1
Thermodynamic Cycle
• Is a series of
processes which
form a closed
path.
• The initial and
the final states
are coincident.
For What Thermodynamics 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 Cycles
Hot body or source
Qin
System,
or heat
engine
Qout
Cold body or sink
Wcycle
Heat Engine Efficiency
Hot body or source

QH
System,
or heat
engine
QL
Cold body or sink
Wcycle
WLIQ
QH

QH  QL

QH
QL
 1
QH
Qout
Wcycle
System
Qin
Cold body or sink
QL


WIN
QL

QH  QL
QH


WIN
QH

QH  QL
HEAT PUMP
Hot body or source
REFRIGERATOR
Refrigerators and heat pumps
Energy analysis of cycles
4
3
1
2
For the cycle, E1 E1 = 0, or
ΔE cycle  0
ΔE cycle  Q cycle  Wcycle  0
For cycles, we can write:
Qcycle = Wcycle
Qcycle and Wcycle represent net amounts
which can also be represented as:
Q  W
cycle
cycle
TEAMPLAY
• 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.
TEAMPLAY
Win = 45 Btu
2
B
Qin=40 Btu
1
A
Wout=60 Btu
Carnot 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 gas
The area represents the net work
P-v Diagram of the
Reversed Carnot Cycle
•
The Carnot cycle for a gas might
occur as visualized below.
TL TH
TH TL
QL
TL
Execution of the Carnot
Cycle 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.
TL
•It is always reversible
-- a Carnot cycle is
reversible by definition.
TL
TL
Vapor 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 …..
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
Question ….
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.
$12,998.67/day
$4,744,499/year
Vapor-cycle
Power Plants
We’ll simplify the power
plant
3
BOILER
wout
TURBINE
qin
4
2
CONDENSER
win
1
PUMP
qout
Carnot Vapor Cycle
Low thermal
efficiency
compressor
and turbine
must handle
two phase
flows
Carnot Vapor Cycle
•The Carnot cycle is not a
suitable model for vapor
power cycles because it cannot
be approximated in practice.
Rankine Cycle
• The model cycle for vapor power cycles is
the Rankine cycle which is composed of four
internally reversible processes: constantpressure 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
Pump 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
Refrigerator
Ordinary Household
Refrigerator
Gas 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
Spark-Ignition Engines
v
v
Otto Cycle
qin
qout
P-V diagram
T-S diagram
(Work)
(Heat Transfer)
Performance of cycle
Thermal Efficiency:
w net
QL

 1
QH
QH
Need to know QH and QL
Otto Cycle
• Heat addition 2-3 QH = mCV(T3-T2)
• Heat rejection 4-1 QL = mCV(T4-T1)
mCV T4  T1 
QL
  1
 1
QH
mCV T3  T2 
• or in terms of the temperature ratios
QL
T1 T4 T1  1
  1
 1
QH
T2 T3 T2  1
qin
qout
Otto Cycle
• 1-2 and 3-4 are adiabatic process, using
the adiabatic relations between T and V
 1
 1
 V4 
T3
T2  V1 
T2 T3


 
 



T1  V2 
V3 
T4
T1 T4




SAME VOLUME RATIO
T1
V1
1
  1
 1
; r
T2
V2
r  1
qin
qout
Cycle performance with cold air
cycle assumptions
th,Otto
T1
1
 1   1  k 1
T2
r
This looks like the Carnot efficiency, but it
is not! T1 and T2 are not constant.
What are the limitations for this expression?
Thermal Efficiency
of 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
cycle efficiency
k = 1.4
Otto Cycle
The thermal efficiency of the Otto Cycle increases
with the specific heat ratio k of the working fluid
Brayton Cycle
• This is another air standard cycle and it
models modern turbojet engines.
Brayton Cycle
Proposed by George Brayton in 1870!
http://www.pwc.ca/en_markets/demonstration.html
jet engine with afterburner
for military applications.
Schematic of A
Turbofan Engine
Illustration of A
Turbofan Engine
Turboprop
burner
compressor
turbine
Schematic of a
Turboprop Engine
Other applications
of 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
Gas-Turbine Engine
A Closed-Cycle
Gas-Turbine Engine
Brayton Cycle
• The ideal cycle for modern gasturbine 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
Components and T-s
Diagram for Ideal Turbojet
Cycle
P-v and T-s Diagrams for the
Ideal Brayton Cycle
Brayton 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 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 analysis
Let’s assume cold air conditions and
manipulate the efficiency expression:
 1 
C p (T4  T1 )
C p (T3  T2 )
T1 T4 T1  1
  1
T2 T3 T2  1
Brayton cycle analysis
Using the isentropic relationships,
T2  P2 
  
T1  P1 
k 1
k
T4  P4 
  
;
T3  P3 
k 1
k
 P1 
  
 P2 
Let’s define:
P2 P3
rp  pressure ratio 

P1 P4
k 1
k
Brayton 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 Cycle
• The pressure ratio is
Pr2
P2  P2 
 
 
P1  P1  s
Pr1
• Also
Pr3
P3  P3 
P2
 
 

P4  P4  s Pr4
P1
Brayton cycle analysis
Then we can relate the temperature ratios
to the pressure ratio:
T3
T2
k  1  k
 rp

T1
T4
Plug back into the efficiency expression
and simplify:
th,Brayton
T1
1
 1   1  k  1  k
T2
rp
Ideal Brayton Cycle
What does this expression assume?
 th, Brayton  1 
1
k 1  k
rp
Thermal Efficiency
of 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 Cycle
k = 1.4
  1
1
rp
( k 1) / k
Thermal Efficiency of
the Ideal Brayton Cycle
Evaporação a pressão constante
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?
Fronteira
do
Sistema
Representação
do processo
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.
Fronteira
do
Sistema
Representação
do processo
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)
LIQUID
VAPOR
Solução de Exercícios
Cap 4
Ex4.13)
1000K
100KJ
W
300K
Ciclo de Carnot
W=?
Qc=?
Tc
3
W
 1
 1
ηc 
10
Th
Qh
ηc  0,7  W  0,7 100
W  70KJ
 Q   W  (100 Q )  70
c
Qc  30KJ
Ex4.14)

W
ηT  0,4 

Q
h
1000MW

Qh 
 2500MW
0,4
  W

Q


 Q
 )W

(Q
h
c
liq
  1500MW
Q
c
Ex4.15)
WT>0
Turbina
Qh>0 Caldeira
(5000)
Condensador Qc<0 (3500)
Bomba
Wb<0
Qmeio<0 (500)
  W
  5000 - 4000  1000MW
Q
 
Wliq
1000
ηT 

 25 0 0
Qh 5000
Wliq  WT  Wb  1000 WT  1  WT  1001MW
Ex4.16)
T
P1>P2
P2
550ºC
30ºC
s
303
ηT  1 
 0,63
823
Ex4.17)
Qh=+180kW
1
Aquecedor
1000ºC
Compressor
2
Turbina
W<0
W>0
4
100ºC
Condensador
QL=-110kW
1º Le i :  Q   W
Wliq  180 110  70kW
ηT 
Wliq
Qh
70

 0,390 0
180
3
Rendimento Máximo -> Rendimento de Carnot
ηmáx  ηCarnot
373
 1
 0,71
1273
Ciclo Brayton
 P2 
η  1   
 P1 
γ 1
γ
Conhecer Pressões