Transcript Lecture 06: Specifying Properties

EGR 334 Thermodynamics
Chapter 3: Section 1-5
Lecture 06:
Evaluating Properties
Quiz Today?
Today’s main concepts:
Number of properties needed to set simple compressible system.
State properties: Temperature, Pressure, Specific Volume
Phase and Quality.
Given two properties of a pure compressible substance, find
other state properties.
• Using the Thermodynamic Property tables.
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Reading Assignment:
• Read Chap 3: Sections 6-8
Homework Assignment:
From Chap 3: 5, 7, 10, 29
Sec 3.1.1: Phase and Pure Substance
3
Pure Substance: a substance that is uniform and invariable in chemical composition.
May each of these be considered a pure substance?
Helium gas
in a tank?
Air
Ice
Yes
Oil
Water
Honey
A contents of a
glass containing
water and ice?
Yes
Atmospheric
air?
Yes…maybe
A jar of honey,
water, oil, and
ice topped
with air?
No
Sec 3.1.1: Phase and Pure Substance
4
Phase: Matter which is homogeneous throughout in both its chemical
composition and physical structure.
Bottle Contents : Air (N2, O2, ect.)
Homogeneous chemical composition?
Homogeneous physical structure?
Bottle : Glass
Homogeneous chemical composition?
Homogeneous physical structure?
Cap : Aluminum?
Homogeneous chemical composition?
Homogeneous physical structure?
Sec 3.1.1: Phase and Pure Substance
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Phase: Matter which is homogeneous throughout in both its chemical
composition and physical structure.
- Gaseous Phase
- Liquid Phase
- Solid Phase
Pure substances can exist in multiple phases.
Pure substances can undergo changes in Phase.
liquid  gas: vaporization
solid  liquid: melting
gas  liquid: condensation
liquid  solid: freezing
solid  gas: sublimation
gas  solid: deposition
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The intensive state of a closed system at equilibrium is its condition as described
by the values of its intensive thermodynamic properties.
In this class we will be dealing with simple compressible systems.
Examples of Simple Compressible Systems:
--Standard Air (Oxygen-Nitrogren mix)
--Ideal Gases
--Superheated Water Vapor
For a simple compressible system, specifying any ___2_____
independent intensive thermodynamic properties will fix the
other intensive thermodynamics properties of the system.
Intensive Properties include:
Pressure
Temperature
Specific Volume
Enthalpy
Density
Internal Energy
Entropy
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Most important skill for today’s lecture:
• Given any two intensive properties of H20, be
able to specify other intensive properties.
Given any two of
Temperature
Specific Volume
and Pressure
...find the missing intensive property value.
p  p (T , v)
or
T  T ( p, v )
or
v  v(T , p )
Sec 3.2: P-v-T relation
3D p-v-T surface model:
A representation of how
p, v, and T take on specific
values depending upon their
intensive properties and
phase changes.
The different surfaces of
the model represent
different phases of a pure
substance (like H20) which
depend only on the
pressure, temperature,
and specific volume of the
state.
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Sec 3.2: P-v-T relation
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Phase Diagram from the 3D p-v-T surface model:
If you pull off the Pressure-Temperature projection you create a plot
given as the Phase Diagram.
Critical Point
Pressure
•
•
Triple point
Sec 3.2: P-v-T relation
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Pressure
p-V Diagram from the 3D p-v-T surface model:
The Pressure-Specific Volume projection
of the surface model is a useful tool
for showing processes involving
ice-water-water vapor system.
Sec 3.2: P-v-T relation
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Projections of p-v-T surface:
Triple Point--the state at which solid, liquid, and gas coexist
Critical Point-- the point where saturated liquid and saturated vapor lines meet.
Isotherms--Lines of constant temperature
Isotherms
Phase diagram
p-V Diagram
Sec 3.2: P-v-T relation
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T-v diagram from the 3D p-v-T surface model:
If you pull off the Temperature-Specific Volume projection you create
a plot that can be useful for showing
lines of constant pressure.
•
isobar: line of constant pressure
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You should be able to recognize:
Temperature
i) Saturated Liquid Line (identified values of Tf , vf, and pf )
ii) Saturated Vapor Line (identified values of Tg , vg, and pg
iii) Critical Point (inflection point at top of dome)
iv) On p-v diagram, lines of constant temperature run from high left to low right.
v) On T-v diagram, lines of constant pressure run from low left to high right.
vi) While traversing the liquid-vapor phase (the area under the dome),
both temperature and pressure remain constant for changes in spec. volume.
Sec 3.3: Phase Change
Consider a Constant Pressure Process
•j
•i
water
sub-cooled liquid
(compressed liquid)
water
saturated
liquid
Run
Animation
vapor
water
vapor
water
two phase
liquid vapor
two phase
liquid-vapor
sat.
vapor
saturated
vapor
super
vapor
super heated vapor
(steam)
Sec 3.3: Phase Change
Two Phase, Liquid-Vapor Mix
Can have different mixtures of liquid
and vapor (10% liquid, 90% vapor)
Distinguish mixture of
liquid/vapor using quality.
x
mvapor
mliquid  mvapor
x = 1, saturated vapor (g)
x = 0, saturated liquid (f)
Can also be expressed
as a percentage (%)
Sec 3.5: Saturation
Two Phase, Liquid-Vapor Mix
Quality (x) is also a property. It is a way to express the relative amount
of a substance that contains two different phases of material.
For example, in a given vessel, we know that the two volumes must add
to the total volume
VT  VL  VV
 ml
vT  
 mT

 mv 
v f  
v g

 mT 
with
1 x 
ml
mT
;
x
mv
mT
v  1  x  v f  xvg
v  v f  x  vg  v f

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Quantitative Thermodynamic Properties:
Given any two intensive properties, can you
a) determine the phase (liquid, saturated, mixture, solid, or gas?)
b) determine the other intensive property values at that state.
Methods:
1) Read off a p-v or a T-v diagram
2) Look up on Steam Tables: (Appendix and Handout)
i) Table A.2: Prop. of Saturated H20: Temperature Table
ii) Table A.3: Prop. of Saturated H20: Pressure Table
iii) Table A.4: Prop. of Superheated Water Vapor
iv) Table A.5: Prop. of Compressed Water
v) Table A.6: Prop. of Saturated H20: Solid-vapor table
3) Use a Computer Steam Application
i) IT (download from www.wiley.com/college/moran)
ii) http://www.dofmaster.com/steam.html
Link to IT
dofmaster
*
pressure
Link to IT
download
(p, v, T)
spec. vol.
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Examples of
Property Diagrams
Given two properties, locate
and read other values.
Molliere chart
Pyschrometric chart
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Saturated Steam Table
Given T= 25 deg C
and saturated vapor
Read p and v.
Given two property values,
look up other intensive values.
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Example 1: Determine property values
Determine phase or phases of H20 at the following conditions and sketch
p-v and T-v diagrams showing the positions of each of the following
Identify the specific volume, v, if possible:
a) p = 5 bar
T = 151.9 deg C.
b) p = 5 bar
T = 200 deg. C.
c) p = 2.5 MPa
T = 200 deg C.
d) p = 2.8 bar
T = 160 deg C.
e) p = 1 bar
T = -12 deg C.
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Example 1: Determine property values
Determine phase or phases of H20 at the following conditions and sketch
p-v and T-v diagrams showing the positions of each of the following:
Identify the specific volume, v, if possible:
a) p = 5 bar
T = 151.9 deg C.
from table A.3: at 5 bar the saturated temperature is 151.9 which means
it could be anywhere on the liq-vapor line as a two phase material.
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Example 1: Determine property values
Determine phase or phases of H20 at the following conditions and sketch
p-v and T-v diagrams showing the positions of each of the following
Identify the specific volume, v, if possible:
a) p = 5 bar
T = 151.9 deg C.
b) p = 5 bar
T = 200 deg. C.
c) p = 2.5 MPa
T = 200 deg C.
d) p = 2.8 bar
T = 160 deg C.
e) p = 1 bar
T = -12 deg C.
23
Example 1: Determine property values
Determine phase or phases of H20 at the following conditions and sketch
p-v and T-v diagrams showing the positions of each of the following:
b) p = 5 bar
T = 200 deg. C.
from table A.3: at 5 bar the temp. of 200 > 151.9 which means the
substance is vapor or superheated vapor…need to consult table A.4
from table A.4 at 5 bar and T = 200 deg C….v = 0.4249 m3/kg
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Example 1: Determine property values
Determine phase or phases of H20 at the following conditions and sketch
p-v and T-v diagrams showing the positions of each of the following
Identify the specific volume, v, if possible:
a) p = 5 bar
T = 151.9 deg C.
b) p = 5 bar
T = 200 deg. C.
c) p = 2.5 MPa
T = 200 deg C.
d) p = 2.8 bar
T = 160 deg C.
e) p = 1 bar
T = -12 deg C.
25
Example 1: Determine property values
c) p = 2.5 MPa T = 200 deg C.
from table A.3 at 2.5 MPa = 25 bar the temp of 200 is 200< 224 which means
that the substance is compressed liquid. For compressed liquid we assume that
the spec. vol, will be similar to vf = 1.1973x10-3. This can be checked on Table
A.5 for compressed liquids which gives 1.1555x10-3 m3/kg
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Example 1: Determine property values
Determine phase or phases of H20 at the following conditions and sketch
p-v and T-v diagrams showing the positions of each of the following
Identify the specific volume, v, if possible:
a) p = 5 bar
T = 151.9 deg C.
b) p = 5 bar
T = 200 deg. C.
c) p = 2.5 MPa
T = 200 deg C.
d) p = 2.8 bar
T = 160 deg C.
e) p = 1 bar
T = -12 deg C.
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Example 1: Determine property values
d) p = 2.8 bar
T = 160 deg C.
from Table A.2 at Tsat = 160, psat = 6.178 > 2.8 which means the
substance will be vapor or superheated…so refer to table A.4.
from Table A.4 at p = 1.5 and T = 160….v = 1.317
at p = 3 and T = 160….v = 0.651
Interpolation is required:
v  1.317
2.8  1.5

0.651  1.317 3.0  1.5
v  0.7398
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Example 1: Determine property values
Determine phase or phases of H20 at the following conditions and sketch
p-v and T-v diagrams showing the positions of each of the following
Identify the specific volume, v, if possible:
a) p = 5 bar
T = 151.9 deg C.
b) p = 5 bar
T = 200 deg. C.
c) p = 2.5 MPa
T = 200 deg C.
d) p = 2.8 bar
T = 160 deg C.
e) p = 1 bar
T = -12 deg C.
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Example 1: Determine property values
e) p = 1 bar
T = -12 deg C.
from table A.3 at psat of 1 bar, Tsat = 99.63 C > -12 which means the
substance will be compressed liquid or solid. Noticing that Table A-5
doesn’t handle pressures this low, if you assumed that the material was
liquid, you would use the saturated spec. vol. at the given temperature of
-12 deg. However, you recognize that -12 degrees is below the freezing
point of water and therefore you expect that this is in solid phase (ice).
Table A-6 also has information on saturated solid-vapor data.
Assume v = vi for a T of -12 deg. C.
v = 1.0888 x 10-3 = 0.0010888 m3/kg
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Linear Interpolation:
yunknown  y1 xknown  x1

y2  y1
x2  x1
yunknown
 y2  y1 
 y1  
 ( xknown  x1 )
 x2  x1 
*
y2
yunknown
y1
*
x1
xknown
x2
Sec 3.5: Evaluating P, v, T
You try it: What is the specific volume of a water at 40 bar and 140 oC?
Linear Interpolation will be needed.
Given the following:
@ 140°C & 25 bar, v1 = 1.0784 m3/kg
@ 140°C & 50 bar, v2 = 1.0768 m3/kg
Find the spec. vol at p = 40 bar.
v  v1 
v2  v1
( p  p1 )
p2  p1
v  1.0784 
(1.0768  1.0784)
(40  25)  1.0774
(50  25)
Sec 3.5: Evaluating P, v, T
Example 2:
This time try it for the same pressure with different temperatures.
Given:
40 bar and 180 oC, the specific volume of water is 1.1248 m3/kg.
40 bar and 140 oC, the specific volume of water is 1.0774 m3/kg.
What is the specific volume of water at 40 3bar and 150 oC?
@ 180°C & 40 bar, v1 = 1.1248 m /kg
v2  v1
v  v1 
(T  T1 )
T2  T1
v  1.0774 
(1.1248  1.0774)
(150  140)  1.0893
(180  140)
m3 / kg
Sec 3.5.2 : Saturation Tables
Example:
A cylinder-piston assembly initially contains water at 3 MPa
and 300oC. The water is cooled at constant volume to 200
oC, then compressed isothermally to a final pressure of 2.5
MPa. Sketch the process on a T-v diagram and find the
specific volume at the 3 states.
Sec 3.5.2 : Saturation Tables
Example:
TV diagram
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State 1:
p1 = 3 MPa= 30 bar and
T1 = 300 deg C.
i) Start with superheated vapor table A-4
ii) Interpolate between 280 and 320 deg C.
vb  va
v1  va 
(T1  Ta )
Tb  Ta
v1  0.0771 
(0.0850  0.0771)
(300  280)
(320  280)
v1  0.0811 m3 / kg
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State 2:
v2 = v1 = 0.0811 m3/kg and T2 = 200 deg C.
Recognize that this is in the liquid-vapor mixture range.
Refer to Table A-2
i) Locate vg and vf at T = 200 deg C.
v f  1.1565  103
vg  0.1274
v2  0.0811
ii) Determine quality, x2
v  v f  x (v g  v f )
x

v  vf
vg  v f
0.0811  0.0011565
 0.633  63.3%
0.1274  0.0011565
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State 3:
p3 = 2.5 MPa = 25 bar and T2 = T3= 200 deg C.
Recognize that this is in the compressed liquid.
Refer to Table A-5
i) Locate p3=25 bar and T3=200 deg C
ii) Read the value of v directly.
v3  1.1555 103 m3 / kg
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End of Slides for Lecture 06