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EGR 334 Thermodynamics
Chapter 3: Section 11
Lecture 09:Generalized Compressibility Chart
Quiz Today?
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Todays main concepts: Universal Gas Constant, R Compressibility Factor, Z. Be able to use the Generalized Compressibility to solve problems Be able to use Z to determine if a gas can be considered to be an
ideal gas. Be able to explain Equation of State
Reading Assignment:
Homework Assignment:
Read Chap 3: Sections 12-14
From Chap 3: 92, 93, 96, 99
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Like c p and c v, todays topic is about compressible gases. This method does not work for two phase mixtures such aswater/steam. It only applies to gases .
3
Limitation:
pv Z RT
where absolute pressureabsolute temperature
molar specific volume
pT
v
mol
f mol
8.314 kJ/kmol K
1.986 Btu/lb
1545 ft lb /lb
o
o
R R
R
and
Compressibility Factor, Z
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Universal Gas Constant
4
Substance Chem. Formula R (kJ/kg-K) R(Btu/lm-R)
Air --- 0.2870 0.06855
Ammonia NH 3 0.4882 0.11662Argon Ar 0.2082 0.04972
Carbon Dioxide CO 2 0.1889 0.04513
Carbon Monoxide CO 0.2968 0.07090
Helium He 2.0769 0.49613Hydrogen H 2 4.1240 0.98512
Methane CH 4 0.5183 0.12382
Nitrogen N 2 0.2968 0.07090
Oxygen O 2 0.2598 0.06206
Water H 2O 0.4614 0.11021
R can also be expresses on a per mole basis:
R R M
where M is the molecular weight (see Tables A-1 and A-1E)
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5Sec 3.11 : Compressibility
For low pressure gases it was noted from experiment that therewas a linear behavior between volume and pressure at constanttemperature.
The constant R is called the Universal Gas Constant.Where does this constant come from?
and the limit as P 0
then RT
Pv P 0lim
The ideal gas model assumeslow Pmolecules are elastic spheresno forces between molecules
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6Sec 3.11 : Compressibility
To compensate for non-ideal behavior we can use other equations of state (EOS) or use compressibility
RT Pv
Z
Define the compressibility factor Z,
Z 1 when
ideal gasnear critical pointT >> Tc or (T > 2T c)
Step 1: Thus, analyze Z by first
looking at the reduced variables
C R
C R
T
T T
P P
P Pc = Critical Pressure
Tc = Critical Pressure
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Step 2: Using the reduced pressure, p r and reduced temperature, T r determine Z from the Generalized compressibility charts.(see Figures A-1, A-2, and A-3 in appendix).
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Step 3: Use Z toa) state whether the substance behaves as an ideal gas, if Z 1
b) calculate the specific volume of the gas using
' Rc
c
vv
RT p
vv M
The figures also lets youdirectly read reducedspecific volume where
RT v Z
pwhere
R R M
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9Sec 3.11 : Compressibility
Summarize:
1) from given information,calculate any two of these:
or M R R
RC
p p
p R C
T T
T ' R
c
c
vv
RT p
2) Using Figures A-1, A-2, and A-3,read the value of Z
3) Calculate the missing property using
pv Z
RT pv
Z RT
where vv M
(Note: p c and T c can be found onTables A-1 and A-1E)
(Note: M for different gases can be foundon Table 3.1 on page 123.) mol
f mol
8.314 kJ/kmol K
1.986 Btu/lb
1545 ft lb /lb
o
o
R R
R
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Example: (3.95) A tank contains 2 m 3 of air at -93C and a gagepressure of 1.4 MPa. Determine the mass of air, in kg. The localatmospheric pressure is 1 atm.
Sec 3.11 : Compressibility
V = 2 m 3 T = -93Cpgage = 1.4 MPapatm = 0.101 MPa
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Example: (3.95) Determine the mass of air, in kg
Sec 3.11 : Compressibility
V = 2 m 3 T = -93C = 180 Kp = p gauge + p atm = 1.4 MPa + 0.101 MPa = 1.5 MPa = 15 bar
From Table A-1 (p. 816): For Air: 16) T c = 133 K p c = 37.7 bar
150.40
37.7180
1.35133
R C
RC
p p
pT
T T
V
m p pv Z RT RT
25 315 10 2 1 1
61.1
1000 10.95 8.314 18028.97
N m
kJ kmol kmol K kg
m kJ J m kg
J N m K
Z=0.95ViewCompressibility Figure
pV pV m
R ZRT Z T M
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12Sec 3.11.4 : Equations of State & Sec 3.12 : Ideal Gas Model
Ideal Gas pv RT
Equations of State: Relate the state variables T, p, V
Alternate Expressions pV mRT
pv mRT
When the gas follows the ideal gas law,Z = 1p > T c
T uu and h h T u T pv u T RT
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13Sec 3.11.4 : Equations of State & Sec 3.12 : Ideal Gas Model
Ideal Gas pv RT
Equations of State: Relate the state variables T, P, V
Van der Waals 2
2
n a p V nb nRT
V
a attraction between particles
b volume of particles
Redlich Kwong
Peng-Robinson
m m m
RT a p
V b TV V b
2 22m m m
RT a p
V b V bV b
virial 2 31 ..... Z B T p C T p D T p
.....1 32 v
T Dv
T C vT B
Z
B Two molecule interactions C Three molecule interactions
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Example: (3.105) A tank contains 10 lb of air at 70F with a pressure of 30 psi. Determine the volume of the air, in ft 3. Verify that ideal gasbehavior can be assumed for air under these conditions.
m = 10 lbT = 70Fp = 30 psi
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Example: (3.105) Determine the volume of the air, in ft 3. Verify thatideal gas behavior can be assumed for air under these conditions.
Sec 3.12 : Ideal Gas
m = 10 lbT = 70F = 530Rp = 30 psi= 2.04 atm
For Air, (Table A-1E, p 864)Tc = 239 R and p c = 37.2 atm
2.04 0.05537.2
5302.22
239
RC
RC
p p p
T T
T
Z= 1.0 (Figure A-1)
pv pV Z RT mRT
2
2 2
1
3
144
1545 28.97(10 )(1.0) / 53065.4
(30 )
f mol
mol m
f
ft lb
lb R
lbm lb
lb inin ft
lb RV ft
ViewCompressibility Figure
R
M mZ T mZRT V p p
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Example 3:Nitrogen gas is originally at p = 200 atm, T = 252.4 K.It is cooled at constant volume to T = 189.3 K.What is the pressure at the lower temperature?
SOLUTION:From Table A-1 for Nitrogen p cr = 33.5 atm, T cr = 126.2 K
At State 1, p r,1 = 200/33.5 = 5.97 and T r,1 = 252.4/126.2 = 2.
According to compressibility factor chart , Z = 0.95 v r' = 0.34.
Following the constant v r' line until it intersects with the line atTr,2 = 189.3/126.2 = 1.5
gives
Pr,2 = 3.55.
Thus P 2 = 3.55 x 33.5 = 119 atm.
Since the chart shows Z drops down to around 0.8 at State 2, so it would notbe appropriate to treat it as an ideal gas law for this model.
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End of Slides for Lecture 09
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