EGR 334 ThermodynamicsChapter 3: Section 11
Lecture 09: Generalized Compressibility Chart Quiz Today?
Today’s 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
3
Like cp and cv, today’s topic is about compressible gases….This method does not work for two phase mixtures such as water/steam. It only applies to gases.
Limitation:
pvZ
RT
where absolute pressure
absolute temperature
molar specific volume
p
T
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
4
Universal Gas Constant
Substance Chem. Formula
R (kJ/kg-K) R(Btu/lm-R)
Air --- 0.2870 0.06855
Ammonia NH3 0.4882 0.11662
Argon Ar 0.2082 0.04972
Carbon Dioxide CO2 0.1889 0.04513
Carbon Monoxide
CO 0.2968 0.07090
Helium He 2.0769 0.49613
Hydrogen H2 4.1240 0.98512
Methane CH4 0.5183 0.12382
Nitrogen N2 0.2968 0.07090
Oxygen O2 0.2598 0.06206
Water H2O 0.4614 0.11021
R can also be expresses on a per mole basis:
RR
M
where M is the molecular weight (see Tables A-1 and A-1E)
5Sec 3.11 : Compressibility
For low pressure gases it was noted from experiment that there was a linear behavior between volume and pressure at constant temperature.
The constant R is called the Universal Gas Constant.Where does this constant come from?
and the limit as P0
then RT
PvP
0
lim
The ideal gas model assumes low P molecules are elastic spheres no forces between molecules
6Sec 3.11 : Compressibility
To compensate for non-ideal behavior we can use other equations of state (EOS) or use compressibility
RT
PvZ
Define the compressibility factor Z,
Z1 whenideal gasnear critical pointT >> Tc or (T > 2Tc)
Step 1: Thus, analyze Z by first looking at the reduced variables
CR
CR
T
TT
P
PP
Pc = Critical Pressure
Tc = Critical Pressure
Fig03_12
Step 2: Using the reduced pressure, pr and reduced temperature, Tr 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 to a) state whether the substance behaves as an ideal gas, if Z ≈ 1 b) calculate the specific volume of the gas using
'Rc
c
vv
RTp
vv
M
The figures also let’s you directly read reduced specific volume where
RTv Z
p
where
RR
M
9Sec 3.11 : Compressibility
Summarize:
1) from given information, calculate any two of these:
or M
RR
RC
pp
p R
C
TT
T 'R
c
c
vv
RTp
2) Using Figures A-1, A-2, and A-3, read the value of Z
3) Calculate the missing property using
pvZ
RT pv
ZRT
where v
vM
(Note: pc and Tc can be found on
Tables A-1 and A-1E)
(Note: M for different gases can be found on 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
10
Example: (3.95) A tank contains 2 m3 of air at -93°C and a gage pressure of 1.4 MPa. Determine the mass of air, in kg. The local atmospheric pressure is 1 atm.
Sec 3.11 : Compressibility
V = 2 m3 T = -93°C pgage = 1.4 MPa
patm = 0.101 MPa
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Example: (3.95) Determine the mass of air, in kg
Sec 3.11 : Compressibility
V = 2 m3 T = -93°C = 180 Kp = pgauge + patm = 1.4 MPa + 0.101 MPa = 1.5 MPa = 15 bar
From Table A-1 (p. 816): For Air: 16) Tc = 133 K pc = 37.7 bar 15
0.4037.7
1801.35
133
RC
RC
pp
p
TT
T
Vmppv
ZRT RT
2
5 315 10 2 1 161.1
1000 10.95 8.314 18028.97
Nm
kJ kmolkmol K kg
m kJ Jm kg
J N mK
Z=0.95 View
Compressibility Figure
pV pV
mRZRT Z TM
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 << pc and / or T >> Tc
Tuu and h h T u T pv u T RT
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 ap V nb nRT
V
a attraction between particles
b volume of particles
Redlich–Kwong
Peng-Robinson
m m m
RT apV b TV V b
2 22m m m
RT apV b V bV b
virial 2 31 .....Z B T p C T p D T p
.....1
32
v
TD
v
TC
v
TBZ
B Two molecule interactionsC Three molecule interactions
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Example: (3.105) A tank contains 10 lb of air at 70°F with a pressure of 30 psi. Determine the volume of the air, in ft3. Verify that ideal gas behavior can be assumed for air under these conditions.
m = 10 lbT = 70°F p = 30 psi
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Example: (3.105) Determine the volume of the air, in ft3. Verify that ideal gas behavior can be assumed for air under these conditions.
Sec 3.12 : Ideal Gas
m = 10 lbT = 70°F = 530°R p = 30 psi= 2.04 atm
For Air, (Table A-1E, p 864)Tc = 239 °R and pc = 37.2 atm
2.040.055
37.2
5302.22
239
RC
RC
pp
p
TT
T
Z= 1.0 (Figure A-1)
pv pVZ
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
RM
mZ TmZRTV
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 pcr = 33.5 atm, Tcr = 126.2 K
At State 1, pr,1 = 200/33.5 = 5.97 and Tr,1 = 252.4/126.2 = 2.
According to compressibility factor chart , Z = 0.95 vr' = 0.34.
Following the constant vr' line until it intersects with the line at Tr,2 = 189.3/126.2 = 1.5 gives Pr,2 = 3.55.
Thus P2 = 3.55 x 33.5 = 119 atm.
Since the chart shows Z drops down to around 0.8 at State 2, so it would not be appropriate to treat it as an ideal gas law for this model.
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End of Slides for Lecture 09