03 Cycles Aux
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Transcript of 03 Cycles Aux
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Independent Properties
Simple Compressible Pure Substance
(Absence of motion, gravity, surface, magnetic, electrical effects)
States are defined by 2 independent properties
Saturated liquid, saturated vapor are in diff state but at same P, T
In a saturated state, Press & Temp are dependent
Superheated vapor: any two of P, v, T define the state
(P, v), (P, T), (v, T)
Saturated state: the following would define the state but not (P,T)
(P, v), (v, T)
(T, x), (P, x)
Example
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P-V-T Behavior of Low- and Moderate-Density Gases
Ideal
gas eqn
(See next slide)
cr
cr P
P
PT
T
T==
, Generalized compressibility chart
ZRTPvRTPvZ == OR Compressibility factor: Unity for ideal gas
22dbcbvv
abv
RTP++
=
Ideal gas can beassumed at low andmoderate densities(See next slide)
)(
,
)KJ/mol3145.8(
,
MRR
RTPvmRTPV
Mmn
R
TRvPTRnPV
=
==
=
=
==
Reduced properties
Cubic eqn of state
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T-v Diagram of water (Opt)
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N2
Compressibility Factor for N2 (Opt)
Ex. 3.8, 9
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==
=
=
==
2
1
2
121 PdVWW
PdVWdVAdx
PAdxFdxW
++
::
::
WdV
WdV
Work done during the process from state 1 to state 2
Integral can be:1. Graphically or experimentally evaluated OR2. Analytical evaluated when the P-V relation is
known
Work done at the Piston
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Work = Areain a P-V diagram
=
2
121 PdVW
Graphical Approach to Find Work
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During any cycle a system (control mass)undergoes, the cyclic integral of heat is equal to
the cyclic integral of work.
= WQ
The First Law of Thermodynamics for a
Control Mass Undergoing a Cycle
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Consider different processes between two states
= WQ 121
,
BCA
The First Law of Thermodynamics for a
Change in State of a Control Mass
depends only on the initial and finalstates not on the path.
212112 WQEE
WQdE
=
= Energy = +in out
( )WQ
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Energy is present in various forms
Net change in energy of the system is exactly
equal to energy transfer as work or heat
PEKEUE
E
++=
++= EnergyPotentialEnergyKineticEnergyInternal
mgZPEVVmKE == ,2
1 rr
212112
2
1
2
21212 )(
2)(
2
WQZZmgVVmUUEE
mgdZVVm
ddUdE
=++=
+
+=
rr
WQPEdKEddUdE =++= )()(
Conservation of Energy
m
Uu
U
=
InternalEnergy
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Consider a constant-pressure process withnegligible kinetic energy and potential energy
Enthalpy
12
111222
11221221
1221
211221
)()(
)(
HHVPUVPU
VPVPUUQ
VVPW
WUUQ
=
++=
+=
=
+=
Pvuh
PVUH
+=
+=
Thermodynamic Property Enthalpy
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Recap 1st law of thermodynamics
Definition of specific heats at const volume/press
For solids and liquids, specific heat reduced to:
Little volume change
VdPdHQ
PdVdUWdUQ
=
+=+=
PPP
P
vvv
v
T
h
T
H
mT
Q
mC
T
u
T
U
mT
Q
mC
=
=
=
=
=
=
11
11
CdTdudh
vdPduPvddudh
++= )(
Specific heat atconst volume
Specific heat atconst pressure
pv CCC ==
Const-Vol, Const-Press Specific Heats
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In general, ),( vTuu =
However, for a low-density gas(Confirmed by steam table)
)(~ Tuu
It is known for an ideal gas)(Tuu =
dTmCdU
dTCdudT
duC
v
v
v
0
0
0
=
=
=
Similarly, for an ideal gas)(Thh =
dTmCdH
dTCdhdT
dhC
p
p
p
0
0
0
=
=
=
Specific heats for an ideal gas:
RTuPvuh +=+=
U, H, Cv, Cp of Ideal Gases
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Specific heats for an ideal gas are also functionsof temperature only like internal energy andenthalpy
)(),( 0000 TCCTCC ppvv ==
RCC
RdTdTCdTC
RdTdudh
RTuh
vp
vp
+=
+=
+=
+=
00
00
Specific heats for an ideal gas are related as
follows:
Specific Heats of Ideal Gases
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The inequality of Clausius:
It is a consequence of the 2nd law ofthermodynamics
The equality holds for reversible cyclesThe inequality holds for irreversible cycles
0 TQ
The Inequality of Clausius: Second Law
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Consider reversible cycles between 1 and 2
+
==
1
2
2
10
BA T
Q
T
Q
T
Q
+
==
1
2
2
10 BC T
Q
T
Q
T
Q
For a cycle along A and B:
For a cycle along C and B:
=
2
1
2
1CA T
Q
T
Q
We can thus define a property entropy like
revT
QdS
revT
Q
: path independent
=
2
112
revT
QSS
Entropy A Property of a System
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Entropy: Thermodynamic propertyTherefore, once entropy is obtained for areversible process, the same entropy can be
used for an irreversible as long asthermodynamic states are the same
Zero entropy
1. All pure substances in the (hypothetical)ideal-gas state at absolute zero temperaturehave zero entropy
2. However, it is not practically easy to definezero entropy (or reference state) so anarbitrary state is chosen for zero entropy. Water: Saturated liquid at 0.01C Refrigerants: Saturated liquid at -40C
Entropy A Property of a System
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Extensive property S Intensive property s(specific entropy)
At saturation ( )
fgf
gf
xss
xssxs
+=
+= 1
Temperature-entropy diagramEnthalpy-entropy diagram (Mollier diagram)
Entropy of a Pure Substance
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0
0
41
434
334
23
212
112
=
=
=
=
=
=
SS
T
Q
T
QSS
SS
T
Q
T
QSS
Lrev
Hrev
Consider a heat engine on the Carnot cycle
Rev. isothermal
Isentropic
Rev. isothermal
Isentropic
1-a-b-2-1area
1-4-3-2-1area==
H
netth
Q
W
TH th and TL thSimilar argument for a refrigerator
(see text)
Entropy Change in Reversible Processes
*Isentropic = Rev. adiabatic
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Consider a reversible heat-transfer process
12: The change of state from saturated liquid tosaturated vapor at constant pressure
Heat transfer = area 1-2-b-a-1
T
h
T
q
T
qsss
fg
rev
fg ==
== 21
2
112
23: The change of state fromsaturated vapor to superheatedvapor at constant pressure
Heat transfer = area 2-3-c-b-2
==3
2
3
232 Tdsqq
Entropy Change in Reversible Processes
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Lets start with the Gibbs equation.
PdvduTds +=
+=
+=
==
2
11
2012
0
0
ln
gas,idealFor
vvRdT
TCss
dvv
RdT
T
Cds
v
R
T
PdTCdu
v
v
v
=
=
2
1 1
20
12
0
ln
P
PRdT
T
Css
dPPRdT
TCds
p
p
Similarly,
Method of integration1. Constant specific heat2. Functional form known
3. Tabulated Standardentropy
=T
T
p
T dTT
Cs
0
00
Entropy Change of An Ideal Gas
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Entropy Change of An Ideal Gas
const
or2
1
1
2
1
2
1
1
2
1
1
2
1
2
1
2
1
20
=
=
=
=
=
k
kk
k
k
C
R
Pv
v
v
P
P
v
v
T
T
P
P
T
T
P
P
T
T p
In an isentropic process, 0ln2
11
20
12 == PP
RdTT
Css
p
When the specific heat is constant, this equation becomes:
0lnln1
2
1
20 =
PPR
TTCp
k
k
kC
CC
C
R
p
vp
p
111
0
00
0
==
=
Ratio of specific heats is defined as
0
0
v
p
C
Ck=
Ideal gas equation
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Reversible Polytropic Process for an Ideal Gasnnn
VPVPPV 2211const ===Polytropic process:
nn
n
n
VV
PP
TT
V
V
P
P
=
=
=
1
1
2
1
1
2
1
2
1
2
1
2
In a polytropic process for an ideal gas,
n
TTmR
n
VPVP
dV
V
PdVWn
=
=
==
1
)(
1
1const
121122
2
1
2
1
21
for n1