General Compressibility

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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
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•
•
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Reading Assignment:
• Read Chap 3: Sections 12-14
Homework Assignment:
From Chap 3: 92, 93, 96, 99
3
Limitation:
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.
Compressibility Factor, Z
Z 
pv
RT
where
p  ab so lu te p ressu re
T  ab so lu te tem p eratu re
v  m o lar sp ecific vo lu m e
and
 8 .3 1 4 k J/k m o l  K

o
R   1 .9 8 6 B tu /lb m o l  R
o
1 5 4 5 ft  lb /lb

R
f
m ol

4
Universal Gas Constant
R can also be expresses on a per mole basis:
R
R 
where M is the molecular weight (see Tables A-1 and A-1E)
M
Substance
Chem. Formula
R (kJ/kg-K)
R(Btu/lm-R)
Air
---
0.2870
0.06855
Ammonia
NH3
0.4882
0.11662
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
H2O
0.4614
0.11021
Argon
Water
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
 R
T
and the limit as P0
The ideal gas model assumes
low P
molecules are elastic spheres
no forces between molecules
5
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,
Pv
Z 
RT
Z1 when
ideal gas
near critical point
T >> Tc or (T > 2Tc)
Step 1: Thus, analyze Z by first
looking at the reduced variables
PR 
TR 
P
PC
T
TC
Pc = Critical Pressure
Tc = Critical Pressure
6
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).
8
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
RT
v  Z
p
where
v 
v
R 
M
R
M
The figures also let’s you
directly read reduced
specific volume where
v 'R 
v
R Tc
pc
Sec 3.11 : Compressibility
9
Summarize:
1) from given information,
calculate any two of these:
pR 
p
TR 
pC
T
TC
v
v 'R 
R Tc
pc
(Note: pc and Tc 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
 8 .3 1 4 k J/k m o l  K

o
R   1 .9 8 6 B tu /lb m o l  R
o
1 5 4 5 ft  lb /lb
 R
f
m ol

Sec 3.11 : Compressibility
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.
V = 2 m3
T = -93°C
pgage = 1.4 MPa
patm = 0.101 MPa
10
Sec 3.11 : Compressibility
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Example: (3.95) Determine the mass of air, in kg
V = 2 m3
T = -93°C
= 180 K
p = 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
pR 
TR 
Z 
p
pC
T
pv
 0 .4 0
3 7 .7

TC
RT
m 

15
180
pc = 37.7 bar
View
Compressibility Figure
Z=0.95
 1 .3 5
133

p V m 
m 
ZRT
RT
1 5  1 0
pV
5

N
m
 0 .9 5   8 .3 1 4  kmkJo l  K
2
2 m3
km o l
2 8 .9 7 kg

pV

Z R
1kJ
1 8 0  K
M
T
1J
1000 J 1N  m
 6 1 .1 kg
Sec 3.11.4 : Equations of State & Sec 3.12 : Ideal Gas Model
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Equations of State: Relate the state variables T, p, V
Ideal Gas
pv  RT
Alternate Expressions
pV  m RT
p v  m RT
When the gas follows the ideal gas law,
Z=1
p << pc and / or T >> Tc
u  u T

and h  h  T
  u T  
pv  u T

RT
Sec 3.11.4 : Equations of State & Sec 3.12 : Ideal Gas Model
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Equations of State: Relate the state variables T, P, V
Ideal Gas
pv  RT
2

n a
Van der Waals  p 
 V  n b   n R T b  volume of particles
2 
V 

a  attraction between particles
a
Redlich–Kwong p  R T 
Vm  b
T V m V m  b 
Peng-Robinson
p 
virial Z  1  B  T

RT
Vm  b
p  C T


a
Vm  2bVm  b
2
2
p  D  T  p  .....

B T  C T  D T 
Z  1


 .....
2
3
v
v
v
2
3
B  Two molecule interactions
C  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 lb
T = 70°F
p = 30 psi
Sec 3.12 : Ideal Gas
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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.
For Air, (Table A-1E, p 864)
Tc = 239 °R
and pc = 37.2 atm
m = 10 lb
T = 70°F = 530°R
p = 30 psi= 2.04 atm
pR 
TR 
Z 
p
2 .0 4

pC
T
TC
pv
RT
V 
View
Compressibility Figure
 0 .0 5 5
3 7 .2

530
Z= 1.0 (Figure A-1)
 2 .2 2
239

pV

mZRT
V 

p
mRT

(1 0 lb m )(1 .0 ) 1 5 4 5
(3 0
ft  lb f
lb m o l  R
lb f
2 )
in

mZ R
M
p
  2 8 .9 7   5 3 0 R 
1 lb m o l
/
lb m
1 4 4 in
ft
T
2
2
 6 5 .4 ft
3
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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
At State 1,
pcr = 33.5 atm,
pr,1 = 200/33.5 = 5.97
and
Tcr = 126.2 K
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

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