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Thermodynamic Properties
August 26, 2010
Outline
Unit one – Properties of Pure
Substances
• Extensive, E, m, intensive, T, P, , and
specific, e = E/m, variables
•  = m/V = 1/v => density = 1/(sp. vol.)
• Look at P-v-T data for real substances
• Ideal gases PV = nRuT = mRT
Larry Caretto
Mechanical Engineering 370
Thermodynamics
– Engineering gas constant: R = Ru / M
– With v = V/m, Pv = RT
August 26, 2010
• Use of V or v = V/m gives difference between
PV = mRT and Pv = RT
2
Thermodynamic Variables
Thermodynamic System
• Extensive variables (volume, mass,
energy) depend on size of system
• Intensive variables (pressure,
temperature, density) do not
• Specific variables are ratio of an
extensive variable to mass (e.g. specific
volume, v = V/m)
• General notation, if E is an extensive
variable, e = E/m is specific variable
• Basic unit of analysis defined by boundaries
where all interactions occur
• State of system defined by properties
• Interaction of system with surroundings
– Can interchange energy and mass
– Open/closed systems: Have/Do not have mass
addition or removal
• Open system called control volume
– Isolated system: constant energy and mass
– All interactions across boundaries
3
4
Thermodynamic State
Ideal Gas Equations
• Two independent intensive properties
determine the state of a simple
compressible system
• Once the state of a system is determined all other properties are known
• Thermodynamic processes are a
continuous series of states known as a
quasi-equilibrium process
• From chemistry: PV = nRuT
• Ru = Universal gas constant = 8.314
kJ/kmol·K = 8.314 kPa · m3/kmol·K =
10.73 psia ·ft3/lbmol ·K
• Engineering gas constant, R = Ru/M
• Definition of mol, n = m / M PV = nR T
u
• PV = (m/M)(MR)T = mRT
PV = mRT
• P(V/m) = Pv = RT
Pv = RT
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ME 370 – Thermodynamics
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1
Thermodynamic Properties
August 26, 2010
Z = 1 is ideal gas
TR = T/Tcrit
Why Use Ideal Gas?
Figure 3-51 from Çengel,
Thermodynamics An Engineer
Approach, 2008
PR = P/Pcrit
• Simple equations
• Real gas behavior close to ideal gas for
low pressure (or high temperature)
• Want P low compared to Pc
• Example calculation: find specific
volume of water at 1,000oC and 10 kPa
• Compare to property table value of v =
58.758 m3/kg for water (page 918)
7
Finding R
8
Ideal Gas Calculation
• Table A-1, page 908: for water, R =
0.4615 kJ/kg·K = 0.4615 kPa·m3/kg·K
1kJ = 1 kN·m = 1 (kN/m2)·m3 = 1 KPa·m3
0.4615 kPa  m 3
(1273.15 K )
58.756 m 3
RT
kg  K


v
10 kPa
P
kg
• Error is only 0.0017% at this point
• Ideal gas often used for “permanent
gases” such as air
9
Ideal Gas Calculation II
P-v-T for Real Substances
• Three phases: gas, liquid and solid
• Look at gas and liquid
• Thought experiment
• A volume of 20 ft3 at a pressure of
300 psia contains 20 lbm of Helium;
what is the temperature?
• T = PV / (mR)
• R = 2.6809 psia•ft3/lbm•R (page 958)
T
PV

mR
300 psia 20
ft 3

2.6809 psia  ft 3
(20 lbm )
lbm  R
– Add heat to a liquid at constant pressure;
measure T and v
– After a while vapor starts to form
 111.9 R  347.7 o F
• Heat addition increases volume by converting
liquid to gas while T is constant (at constant P)
– Finally have only gas that expands readily
• Ideal gas has v/T = const (at constant P)
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ME 370 – Thermodynamics
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12
2
Thermodynamic Properties
August 26, 2010
Thought Experiment II
Thought Experiment III
• Repeat experiment with difference in
variables
• Hold temperature constant
• Add heat and measure pressure, P, and
specific volume v
• Have same liquid-vapor transition
• States 2 to 4 are liquid vapor transition
– P is constant if T is constant
– Liquid and vapor properties constant in
transition only relative proportions change
– Liquid and vapor have same properties (P,
vf = 1/f, vg = 1/g, and T) during transition;
only relative proportions change
Figures 3-6 to 3-10 from Çengel, Thermodynamics An
Engineering Approach, 2008
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14
P-v Diagram for Water
Thought Experiment Results
1.E+08
• A constant temperature line on P-v
T = 50 C
1.E+07
Pressure (Pascals)
– Looks like a hyperbola in the gas region
– Has constant pressure during the transition
from liquid to gas
– Is nearly vertical in the liquid region
• The points where the transition between
liquid and gas take place for different
temperatures: saturation curve
• Results of “experiment” on next slide
T = 100 C
T = 200 C
1.E+06
T = 300 C
Tcrit
1.E+05
Gas
Liquid
T = 500 C
T = 1000
C
Sat
1.E+04
Mixed
1.E+03
0.0001 0.001
15
0.01
0.1
1
10
100
1000
Specific Volume (cubic meters per kilogram)
Critical P-v Diagram for Water
Superheat P-v Diagram for Water
1.E+08
50,000
45,000
T = 200 C
T = 300 C
1.E+07
Tcrit
T = 500 C
T = 1000
C
Pressure (Pascals)
Pressure (Pascals)
Sat
40,000
Sat
35,000
T = 50 C
30,000
T = 100 C
T = 200 C
25,000
T = 300 C
20,000
Tcrit
15,000
T = 500 C
10,000
T = 1000
C
5,000
1.E+06
0.001
0.01
0.1
Specific Volume (cubic meters per kilogram)
1
0
0
50
100
150
200
Specific Volume (cubic meters per kilogram)
ME 370 – Thermodynamics
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Thermodynamic Properties
August 26, 2010
What We Have to Do
Finding Properties
• Find all intensive thermodynamic
properties when we are given only two
intensive properties
• When we have a mixture of liquid and
gas T and P are not independent
• For a single phase (liquid or gas) T and
P are independent (can specify state)
• In the mixed region we use vapor mass
fraction, called quality, x
mg
v  vf
v  v f • v and v are found
f
g

x

from either T or P
mg  m f v g  v f
v fg
– Two independent, intensive properties
define state
• Here we will discuss only four intensive
properties, P, v, T, and fraction of vapor
in mixed region called the quality, x
• Approach used here will be applied to
finding other properties: internal energy,
enthalpy, and entropy
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20
Saturation Properties
Quality Derived
• Mixed region volume is sum of liquid
and vapor volumes

mf
mg
m  mg
V mg

vg 
vf 
vg 
vf
m m
m
m
m
P
V  mg v g  m f v f
• Use quality: x = mg/(mg + mf) = mg/m
v
Sat
T = Tsat(P)
P = Psat(T)
Saturated liquid (f) and
saturated vapor (g)
properties can be found
if P or T are known
m  mg
V mg

vg 
v f  xvg  (1  x)v f  v f  xv g  v f 
m m
m
Use quality,
x, in mixed
region
• Will apply this later to other properties
vf
v
x
vg
v  vf
vg  v f
v  v f  xv g  v f 
21
22
Saturation Tables
• See steam tables, pp 914–923
• Saturation tables
– Table A-4 has T as lookup variable
– Table A-5 has P as lookup variable
– Same results in each table
• Other variables besides specific volume
have saturated liquid (f) and gas (g)
• Example: what are P and v if T = 220oC
and x = 0.25?
23
ME 370 – Thermodynamics
• Use Table A-4, Page 915, T = 220oC
• Find P = 2319.6 kPa, vf = 0.001190
m3/kg and vg = 0.086094 m3/kg
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Thermodynamic Properties
August 26, 2010
What is v for x = 0.25?
What if P and T are Given?
• Found from Table A-4 with lookup T
• If these are independent properties then
they define a state
• The “superheat” (gas) tables for water
(Table A-6, pp 918 – 921) have a set of
tables for P with rows of data for v
• Data are also given for other properties
• Basic question: Given T and P, how do
we know if we have a liquid or a gas?
– P = Psat(T = 220oC) = 2319.6 kPa
– vf(T = 220oC) = 0.001190 m3/kg
– vg(T = 220oC) = 0.086094 m3/kg
x
v  vf
vg  v f
 v  v f  x(vg  v f )  (1  x)v f  xvg
v  (1  x)v f  xvg  (1  0.25)
0.001190 m3
0.086094 m3
 (0.25)
kg
kg
v
0.022416 m 3
kg
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Finding State with Given T and P
Sat
T = T1 < T2
T = T2 >T1
Psat(T1)
Psat(T2)
P
Liquid when
T < Tsat(P) or
P > Psat(T)
Gas when
T > Tsat(P) or
P < Psat(T)
vf
27
v
vg
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General Rules
• When we are given T and P we have a
single phase. Is it gas or liquid?
– It is a gas if T > Tsat(P) or P < Psat(T)
– It is a liquid if T < Tsat(P) or P > Psat(T)
– “Superheat” tables have Tsat by P
– Can guess superheat and check
• Example: R134a at 0oC and P = 100 kPa, 400 kPa
• Guess superheat and find at 100 kPa and 0oC, v =
0.21630 m3/kg; at 400 kPa there is no 0oC entry
• Data from page 929
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ME 370 – Thermodynamics
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5
Thermodynamic Properties
August 26, 2010
Results
General Rules Continued
• v(0oC, 100 kPa) = 0.21630 m3/kg
• What is v(0oC, 100 kPa)? Page 929
• The example did not find a 0oC entry at
400 kPa because this is a liquid
– The header 0.4 MPa (Tsat = 8.91oC confirms
that T = 0oC < Tsat = 8.91oC
• Approximation for compressed liquid:
v(T,P)  vf(T) Note it’s vf(T)
• Why are there no data at T =
= 400 MPa (= 0.4 MPa)?
0oC
and P
– To get answer for 400 kPa and 0oC look at
saturation table for T = 0oC, A-11, page 926
– v(0oC,400 kPa)  vf(0oC) = 0.0007723 m3/kg
31
Two More Problems
Review Last Problem
• What is the specific volume for refrigerant
134a (R-134a) when the pressure is 90
psia and the temperature is 240oF
– R-134a tables in english units: pp. 976-979
– Guess that this is gas and look in
“superheated” tables on page 979
– v = 0.77796 ft3/lbm at given T and P
• What is P when T =
0.77796 ft3/lbm?
240oF
and v =
90 psia
Sat
T = Tsat(P)
P = Psat(T)
vf
34
Given T-v(or P-v); Find P(or T)
• Determine state: find vf and vg from
saturation tables for given T (or P)
vg
35
ME 370 – Thermodynamics
• Use general rule that gas occurs when P <
Psat(T) or T > Tsat(P)
• For this problem T = 240oF > Tsat(90 psia) = 72.83oF
• Also P = 90 psia < Psat(240oF) = 138.79 psia
• For liquid to use v  vf(T) we must be
given v-P: find T for which vf(T)= vgiven
• Compressed liquid tables for water
Gas
v
– R-134a tables in english units: pp. 978-81
– What if we could not guess that this is gas or
made a wrong guess?
– If v > vg it is a gas; use superheat table
– If v < vf it is a liquid; see below
– If vf < v < vg it is in mixed region. Can find x
P
Find vf and vg from T or P
Liquid if v < vf
Gas if v > vg
Otherwise mixed
Mixed
• What is the specific volume for refrigerant
134a (R-134a) when the pressure is 90
psia and the temperature is 240oF
33
Find State Given v and (T or P)
Liquid
32
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6
Thermodynamic Properties
August 26, 2010
Interpolation
Interpolation II
• Can use interpolation function on your
calculator if you have one
• Basic idea is that you have a table with
data pairs (x1, y1) and (x2, y2)
• You want to find a value of y for some
value of x between x1 and x2
• Simplest idea is to assume a straight
line between x1 and x2
y2
Interpolated y
y
y1
x
Interpolation III
x
Want y for this x
y2  y1 y  y1

x2  x1 x  x1
y  y1 
y2  y1
x  x1 
x2  x1
38
Example Problem
• Equation for a straight line between
data pairs (x1, y1) and (x2, y2) is
y2  y1
x  x1  or
x2  x1
x2
x1
37
y  y1 
• For a straight
line any two
points on the line
can give slope
y
• For H2O at P = 2 MPa, what is T when
v(m3/kg) = (a) 0.00115, (b) 0.01, (c)0.2?
• Table A-5, p 917 At 2000 kPa, vf =
0.001177 m3/kg, vg = 0.099587 m3/kg
y  y1 y2  y1

x  x1 x2  x1
– Thus (a) is liquid, (b) is mixed, (c) is gas
• Second form is easier to remember
• For compressed liquid state (a), correct
T is one for which vf(T)= 0.00115 m3/kg
– It says that the slope between any point
(x,y) and the point (x1, y1) is the same as
the slope between (x1, y1) and (x2, y2)
– Extract from Table A-4, next slide, shows
that this vf is between T = 195oC and T =
200oC
• First form used for computations
39
Data and Interpolation
Example: (b) Mixed, (c) Gas
• (b) for mixed region, T = Tsat(2 Mpa) =
212.38oC; can find quality
• We want to find T for v
= 0.00115 m3/kg
– Between 195oC and
200oC
v  vf
• For this interpolation x1
= 0.001149 m3/kg, x2 =
0.001157 m3/kg, y1 =
195oC and y2 = 200oC
200 o C  195o C
 .00115 m 3  .001149 m 3   195.6o C
T  195 C 
3
3
kg
kg 

 .001149 m
.001157 m
kg
kg
o
41
ME 370 – Thermodynamics
40
.01 m
3
kg

x
v g  v f .099587 m3
 .001177 m
kg
3
kg
 .001177 m
 0.0896  8.96%
3
kg
• (c) v = .2 m3/kg at P = 2 MPa found on p.
921 between T = 600oC and 700oC
T  600 o C 
700o C  600 o C
 .2 m3  .19962 m3   601.6o C
3
3
kg
kg 

 .19962 m
.22326 m
kg
kg
42
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Thermodynamic Properties
August 26, 2010
Compressed Liquid Table
More on Compressed Liquid
• Not a common table, use here to check
approximation that v(P,T)  vf(T)
• Table A-7, page 922, gives v(50 MPa,
100oC) = 0.0010201 m3/kg
• Table A-4, page 914, gives vf(100oC) =
0.001043 m3/kg
• Error is about 2% even for a pressure
difference of 49.9 MPa between the
actual pressure and saturation pressure
• Check approximation that v(P,T)  vf(T)
for previous problem: v = 0.00115 m3/kg
and P = 2 MPa
• Do not have data for compressed liquid
at 2 MPa in table on page 922
43
44
– First entry is at 5 MPa
• Have to use double interpolation and
interpolation with saturation points
– Interpolate between Psat and 5 MPa
Interpolations
Table
2 MPa
• We are looking for v = 0.00115 m3/kg
which is between 180oC and 200oC in
compressed liquid table at 5 MPa
• We have to find data at these
temperatures and 2 MPa
Compressed liquid data
shown here
Looking for v = 0.00115
m3/kg at P = 2 MPa
No data for P < 5 M Pa
Use saturated liquid data
for lower pressure
At 180oC, vf = 0.001127
m3/kg at P = 1.0021 MPa
At 200oC, vf = 0.001157
m3/kg at P = 1.5538 MPa
Interpolate to get v at 2
MPa at 180oC, 200oC
46
– Because 5 MPa is the first table entry we
use saturation as the lower data point
vt(T) = v(T, Psat)
v(T , 2 MPa)  v f (T ) 
v(T ,5 MPa)  v f (T )
5 MPa  Psat (T )
2 MPa  Psat (T )
45
Interpolations II
Interpolations III
• Apply formula for T = 180oC and 200oC
3
v(180 C , 2 MPa)  .001127 m
o
3
.001125 m
kg

3
 .001127 m
kg
kg
2 MPa  1.0021 MPa   .001126501 m
5 MPa  1.0021 MPa
kg
3
v( 200 C , 2 MPa)  .001157 m
o
3
kg
• Now interpolate between results for two
different T’s to find T at desired v
3

3
47
ME 370 – Thermodynamics
200o C  180o C
.00115648 m 3 .001126501 m3

kg
kg
 .00115 m3 .001126501 m 3 

  195.7 o C

kg
kg


 .001157 m
3
kg
kg
2 MPa  1.5538 MPa   .00115648 m
kg
5 MPa  1.5538 MPa
.001153 m
T  180o C 
• Previous result using v(T,P) ≈ vf(T) was
195.6oC
48
8