Transcript General Compressibility
EGR 334 Thermodynamics Chapter 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:
•
Read Chap 3: Sections 12-14 Homework Assignment: From Chap 3: 92, 93, 96, 99
Limitation:
Like c p and c v , 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.
Compressibility Factor, Z
Z
pv RT
where
p
absolute pressure
T
absolute temperature
v
molar specific volume and
R
1.986 Btu/lb
mol f mol
o
o R R
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Universal Gas Constant 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)
Substance
Air Ammonia Argon Carbon Dioxide Carbon Monoxide Helium Hydrogen Methane Nitrogen Oxygen Water
Chem. Formula
-- NH 3 Ar CO 2 CO He H 2 CH 4 N 2 O 2 H 2 O
R (kJ/kg-K)
0.2870
0.4882
0.2082
0.1889
0.2968
2.0769
4.1240
0.5183
0.2968
0.2598
0.4614
R(Btu/lm-R)
0.06855
0.11662
0.04972
0.04513
0.07090
0.49613
0.98512
0.12382
0.07090
0.06206
0.11021
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Sec 3.11 : Compressibility The constant R is called the Universal Gas Constant.
Where does this constant come from? For low pressure gases it was noted from experiment that there was a linear behavior between volume and pressure at constant temperature. then lim
P
0
Pv T
R
and the limit as P 0 5 The ideal gas model assumes low P molecules are elastic spheres no forces between molecules
Sec 3.11 : Compressibility To compensate for non-ideal behavior we can use other equations of state (EOS) or use compressibility Define the compressibility factor Z,
Z
Pv RT
Z 1 when ideal gas near critical point T >> T c or (T > 2T c ) Step 1: Thus, analyze Z by first looking at the reduced variables
P R
P P C T R
T T C
P c = Critical Pressure T c = Critical Pressure 6
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).
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 where
v
Z RT p v
v M R
R M
The figures also let’s you directly read reduced specific volume where
v
'
R
v RT c p c
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Sec 3.11 : Compressibility Summarize: 1) from given information, calculate any two of these:
p R
p p C T R
T T C v
'
R
v RT c p c
(Note: p c and T c can be found on Tables A-1 and A-1E) 2) Using Figures A-1, A-2, and A-3, read the value of Z 3) Calculate the missing property using
Z
pv RT
or
Z
pv RT
(Note: M for different gases can be found on Table 3.1 on page 123.) where
v
v M R
R M R
1.986 Btu/lb mol f mol
o
o R R
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Sec 3.11 : Compressibility Example: (3.95) A tank contains 2 m atmospheric pressure is 1 atm.
3 of air at -93°C and a gage pressure of 1.4 MPa. Determine the mass of air, in kg. The local V = 2 m 3 T = -93°C p gage p atm = 1.4 MPa = 0.101 MPa 10
Sec 3.11 : Compressibility Example: (3.95) Determine the mass of air, in kg V = 2 m 3 T = -93°C = 180 K p = p gauge + p atm = 1.4 MPa + 0.101 MPa = 1.5 MPa = 15 bar 11 From Table A-1 (p. 816): For Air: 16) T c = 133 K p c = 37.7 bar
p R
p p C T R
T T C
15 37.7
0.40
180 133 1.35
Z=0.95 View Compressibility Figure
Z
pv RT
p RT m
0.95
8.314
m
pV ZRT
Z kJ
5
N m
2
m
3
kmol
28.97
kg
180
K
1
kJ R pV M
1000
J T
1
J
1 61.1
kg
Sec 3.11.4 : Equations of State & Sec 3.12 : Ideal Gas Model
Equations of State:
Relate the state variables T, p, V Ideal Gas
pv
RT
Alternate Expressions
pV
mRT pv
mRT
When the gas follows the ideal gas law,
u
Z = 1 p << p c
u
and / or T >> T c and
h
pv
RT
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Sec 3.11.4 : Equations of State & Sec 3.12 : Ideal Gas Model 13
Equations of State:
Relate the state variables T, P, V Ideal Gas Van der Waals
pv
RT
Redlich–Kwong
p p
2
n a V
2
V V RT
b m
nb
nRT
a
a
b volume of particles attraction between particles
TV m
V m
b
Peng-Robinson
p
V RT m
b
V m
2
a
2
bV m
b
2 virial
Z Z
1
B
v
C
v
2 B C 2 3 .....
D v
3 .....
Two molecule interactions Three molecule interactions
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 ft 3 . Verify that ideal gas behavior can be assumed for air under these conditions.
m = 10 lb T = 70°F p = 30 psi 14
Sec 3.12 : Ideal Gas Example: (3.105) Determine the volume of the air, in ft 3 . Verify that ideal gas behavior can be assumed for air under these conditions.
m = 10 lb T = 70°F = 530°R p = 30 psi= 2.04 atm For Air, (Table A-1E, p 864) T c = 239 °R and p c = 37.2 atm View Compressibility Figure
T R Z p R p
2.04
0.055
p C
37.2
Z= 1.0 (Figure A-1)
T T C
530 2.22
239
V pv RT
pV mRT
(10
lb m
)(1.0) 1545
V lb mol f
R
/
mZRT p
1
lb mol
28.97
lb m
(30
lb f in
2 ) 144
in
2
ft
2
mZ
p
530
R
M
T
65.4
ft
3 15
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' T r,2 line until it intersects with the line at = 189.3/126.2 = 1.5 gives P r,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 not be 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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