me2121 - me2121e slides chapter 3 (2014)
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Mechanical Engineering,NUS
• Thermodynamic Properties of Pure Substances– A pure substance is one where its properties
remain invariant (or no change) when it undergoes state changes, from gaseous to liquid or to solid phase.
ME2121/ TM1121- Thermodynamics
Chapter 3 – 1/ 12
Water, H2O
Ice (Solid) liquid
Water, H2O
Melting (NMP)
O
H H
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• State Diagrams– Water,
– Isobaric Process,(A-B-C-D)
ME2121/ TM1121- Thermodynamics
Chapter 3 – 2/ 12
v
TCritical point
1 bar10 bar
Pcr= 221 bar
Liquid & VapourVapour region
Liquid
AB
C
Fig. 3.1 2-D State diagram
374 o C
L
L+V
V
vf = vg
g-linef-line
D
State A State B State C-D State D
Liquid phase
State C
Two-phase mixture
subcooled Saturated (T=Tsat)
Gaseousphase
L
V
Qin
100oCTA <Tsat
T>Tsat
100 bar
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• 3-D state diagrams
ME2121/ TM1121- Thermodynamics
Chapter 3 – 3/ 12
liquidice
load
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ME2121/ TM1121- Thermodynamics
Chapter 3 – 3/ 12
A thermodynamic state on the surface require two independent properties to locate it, e.g, P=P(v,T).
Constant pressure
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ME2121/ TM1121- Thermodynamics
Chapter 3 – 3/ 12
3-D and 2-D representations of a substance that contracts on freezing
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• Variables in the liquid+vapour region– Quality or dryness fraction of steam
- Specific volume of atwo-phase mixture
ME2121/ TM1121- Thermodynamics
Chapter 3 – 4/ 12
fg
g
mmm
vapourofmassliquidofmassvapourofmassx
fgv
fgf
gf
T
gg
T
ff
T
T
ggffmixture
vvxv
xvvxvorm
vmm
vmv
mV
givesmbyand
vmvmV
)(
)1(
fgfg
vtocomparedsmall
f
vv
vvv
x
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• Other derived properties– enthalpy
– internal energy
- Entropy
- Quality
ME2121/ TM1121- Thermodynamics
Chapter 3 – 5/ 12
fgfgf xhhxhhxh )1(
fgfgf xuuxuuxu )1(
fgfgf xssxssxs )1(
fg
f
fg
f
fg
f
uuu
hhh
sss
x
At a given P, T, v
Example 3.1Find the states of steam for the following given conditions:
(i) t = 34o C, v = 26.6 m3/kg, (ii) P = 0.8 bar, t = 100o C, (iii) P = 2.7 bar, x = 0.5.
Refer to the Table of saturated steam and water (pg.2), the column 1 is the temperature of mixtures in oC, and column 3 is
the specific volume at dry saturated vapour.
At v =26.6 m3/kg, note that t of steam/water is 34oC. Thus, we conclude that the state point
sits on the dry saturatedvapour line or the “g” line, shown below.
t
vV=26 m3/kg
34oCg-linef-line
Critical pt
At P =0.8 bar, t =100o C,
Look up page 3 of Table where column 1 is the saturated pressure, and at 0.8 bar, the corresponding temperature is only 93.5oC.
This temperature is found to be less than 100oC (given), thus, we conclude that the steam is super-heated (that is, temperature of state higher than the saturated temperature).
t
v=2.087m3/kg
100oC
g-linef-lineCritical pt P=0.8
bar
Tsat=93.5o
C
Superheated state
At P =2.7 bar, x=0.5,
vf = 0.00107 m3/kg (pg. 10 of Table)., vg =0.6686 m3/kg (pg.4),
Note that 2.7 bar is very much lower than Pcr =221 bar, thus, vf <<< vg., and Tsat = 130oC.
By definition, any other properties could be found;
In this region (L+V), P & T are no longer independent.
t
vNote: at low P, vf << vg,,
thus vi ~ x(vg) +(1-x)vf
g-linef-line
Critical pt P=2.7 bar
X=0.5
L+V
=0
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• Interpolation technique- data in tables are sparsely tabulated. -To interpolate for v at pressure (P) when values of v1 and v2 are available from tables at P1 and P2.
ME2121/ TM1121- Thermodynamics
Chapter 3 – 6/ 12
g-line
f-line
Critical pt P2
L+V
P1P
vv2 v1
t 12
12
11)( vv
PPPP
vPv
v (m3/kg) P1 =20 bar P=25 bar P2 =30 bar
t = 250o C 0.1115 m3/kg 0.0706 m3/kg
t =275o C =(0.1115+0.1255)/2=0.1185
=(0.1185+0.0759) / 2=0.0972
=(0.0706+0.0812)/2=0.0759
t =300o C 0.1255 0.0812
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• Gaseous Phase– The equation of state (EOS),– Two forms: (i) gravimetric , (ii) molar form
or
– Compressibility factor
ME2121/ TM1121- Thermodynamics
Chapter 3 – 7/ 12
)tan(lim
tconsgasaRTmPV
)./314.8
tan(lim
KmolJ
tconsgasuniversalRTnPV
o
gasidealforRTPvZ 0.1
Z=Pv/RT1.0
Pr = P/ Pcr
Tr = 1.0
Tr= 1.5
Tr = 10
5 10
Tr = T/Tcr
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• Empirical equations for real gases Losses incurred when gas molecules can be described by Virial Equations of the form
It can represent many other forms of EOS. If written in terms of pressure,
where B2,p = B2,v /RoT, and B2,P =(B3,v –B2,v2)/(RoT)2.
ME2121/ TM1121- Thermodynamics
Chapter 3 – 8/ 12
2,3,2
2,3,2
)()(1
,)/(
)()(1
vTB
vTB
vR
TP
vnVvolumemolaroftermsinorV
TBV
TBZ
TRPV
vvo
vv
o
2,3,21 PBPBZ
TRPV
ppo
Example
• van del Waals (1837), accounted for the intermolecularattractions or pressure ( ) and the presence or effectof gas molecules (b) on the “free volume”
Comparing with the Virial Eq.,
And expanding the van der Waals as
then we see that B2,v(T) =and B3,v (T) =
TRbvvaP o
2
P
L+VT
TC
Critical pt,
02
2
CC TT VP
VP
2
)(
2
)(,3,2
11vb
RTab
vvR
TP
TBTBvv
2b
RTab
2
,3,2 )()(1
vTB
vTB
vR
TP vvo
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• Example 3.1– Derive, from first principles, the ideal gas law, PV/T = constant?– From experiments, the expansion and compressibility factors
are– and
– The volume of ideal gas in a system is V=V(P,T) and the change
–– Substitute for
Ln V - ln T+ln P = ln (constant),or
PV/T = constant. (QED)
ME2121/ TM1121- Thermodynamics
Chapter 3– 9/ 12
TTV
V P
11
PPV
V T
11
dPPVdT
TVdV
TP
0
PdP
TdT
VdVor
PdPV
TdTV
VdPVdTdV
Example 3.3 (notes)
Calculate the van der Waals parameters (aand b) for the following gases with the experimental values at the critical point:
Write the van der Waals equation in another form as
Gases Pc (bar) Tc (K)
Benzene 49.1 562
Water 221.2 647.3
023
Pabv
Pavb
PTR
v o
And at the critical point, T=Tc, P=Pc,
023
ccc
co
Pabv
Pavb
PTRv
At the critical point, the three roots (solutions) of v, as given mathematically by a cubic equation, must converge i.e.,
03 cvv
After expanding, we have
033 3223 ccc vvvvvv
Comparing equations for the roots (solution) of v,
The root of vo:
The root of v1:
The root of v2:
Thus,
Thus, substitute for a and b, we
have
3cc
vPab
23 cc
vPa
cc
c vbP
RT 3
cc Pva 23
33 2
3c
cc
cc vPvPvb
c
coc P
TRv
83
Gases Pc (bar)or 105 Pa
Tc (K)
[Jm3 / mol2]. [m3/mol]
Benzene 49.1 562
=1.88 1.19 E-4
Water 221.2 647.3 0.552 3.04 E-5.
cPcToR
a2
6427
c
co
PTR
b8
Gas molecule effect on the free volumeEffect of intermolecular forces
5101.49
2)562(3143.86427
xxa
5101.4985623143.8
xxxb
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• Steam Tables by Rogers & Mayhew
ME2121/ TM1121- Thermodynamics
Chapter 2 – 10/ 12
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• Saturated liquid properties
ME2121/ TM1121- Thermodynamics
Chapter 3 – 11/ 12
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• Superheated properties
ME2121/ TM1121- Thermodynamics
Chapter 3 – 12/ 12