Transcript ES 202 Lecture 1 - Rose

ES 202
Fluid and Thermal Systems
Lecture 16:
Property Tables
(1/20/2003)
Assignments
• Homework:
– 3-56, 3-57, 5-72 in Cengel & Turner
• Reading assignment
– Tables A-4 to A-14 in Cengel & Turner
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Announcements
• Homework set #5 due by 5 pm today in my office
• Comments on Exam 1
– Problem 1: difference between static and stagnation pressure, straws
serve as frictional pipe
– Problem 2: hydrostatic pressure distribution in vertical and horizontal
directions, centroid, center of pressure, moment analysis
– Problem 3: mechanical energy balance, major loss, minor loss
– Session 5: max = 91, min = 31, average = 61
– Session 6: max = 90, min = 44, average = 67
• Mid-term grade based on Exam 1 score on an absolute scale
• Keep your heads up, we still have much to learn and improve!
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Road Map of Lecture 16
• Quiz on Lecture 15
• Phase determination in various regions
• Data interpolation
– linear
– bi-linear
• Compressed Liquid Approximation
• Practice with property tables
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Quiz on Lecture 15
• True/False Temperature and pressure are independent in the
two-phase region.
– False
– However, they are independent of each other in the compressed liquid and
superheated vapor regions.
• Apart from temperature, name at least three other independent,
intensive properties in the two-phase region.
– u, h, s, v, x
• Give the definition of quality
– mass fraction of vapor in mixture
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x
ES 202 Fluid & Thermal Systems
mg
m f  mg
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Quiz on Lecture 15 (Cont’d)
• In the two-phase region, most intensive properties can be
expressed as a mass-weighted/volume-weighted average of the
saturated liquid and saturated vapor values.
– mass-weighted average
– in terms of equation:
q  (1  x)q f  xqg
 q f  x(qg  q f )
 q f  x q fg
where q is the intensive counterpart of any extensive property like
internal energy, enthalpy, entropy and volume.
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Phase Determination (Case 1)
• Case 1: Given P and T
– Look up saturation table
– Compare given P and T against saturation values in the table
 In pressure table,
 Recall constant pressure line on
T-v diagram
T > Tsat(P)
T = Tsat(P)
 If T < Tsat(P),
compressed liquid.
T < Tsat(P)
 If T = Tsat(P),
saturated liquid-vapor mixture.
 If T > Tsat(P),
superheated vapor.
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Phase Determination (Case 1 Cont’d)
• Case 1: Given P and T
– Look up saturation table
– Compare given P and T against saturation values in the table
 In temperature table,
 Recall constant temperature line
on P-v diagram
P > Psat(T)
P = Psat(T)
 If P > Psat(T),
compressed liquid.
P < Psat(T)
 If P = Psat(T),
saturated liquid-vapor mixture.
 If P < Psat(T),
superheated vapor.
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Phase Determination (Case 2)
• Case 2: Given P (or T) and v (or u, h, s)
– Look up saturation table
– Find saturated liquid and vapor values for v (or u, h, s) at Psat = P
1) If v < vf(Psat),
compressed liquid.
(1)
2) If v = vf(Psat),
(2)
(3)
(4)
Psat
saturated liquid.
(5)
3) If vf(Psat)< v < vg(Psat),
saturated liquid-vapor mixture.
4) If v = vg(Psat),
saturated vapor.
5) If v > vg(Psat),
superheated vapor.
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vf(P)
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vg(P)
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Summary
• Compressed liquid (quality is undefined, any two intensive
thermodynamic properties suffice)
• Saturated liquid ( x = 0 )
• Saturated liquid-vapor mixture ( 0 < x < 1 )
• Saturated vapor ( x = 1 )
• Superheated vapor (quality is undefined, any two intensive
thermodynamic properties suffice)
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Flow Chart
Determine phase of substance
Compressed liquid
Direct look up
Two-phase
Superheated vapor
Determine quality
Direct look up
Interpolate other properties
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Property Tables
• Compressed liquid water (A-7)
• Saturated water
– temperature (A-4)
– pressure (A-5)
• Superheated water (A-6)
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Data Interpolation
• The property tables only tabulate discrete values for pressure or
temperature as the independent property.
• If you are interested in values which do not fall on the tabulated data
points, interpolation within the “sandwich” interval will be necessary.
• Since the property tables report data at small intervals, linear
interpolation should be adequate for most purposes.
– Example: specify T (not tabulated) and x in two-phase region
• If both independent, intensive thermodynamic properties do not fall on
the tabulated data points, bi-linear interpolation is necessary to
completely specify the thermodynamic states.
– Example: specify P and T in superheated vapor region (both not
tabulated)
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Compressed Liquid Approximation
• If you find
– the substance is a compressed (subcooled) liquid;
– the compressed liquid table is unavailable or inadequate,
you may invoke the compressed liquid approximation:
u(T , P)  u f (T )
v(T , P)  v f (T )
(weak function of T)
s(T , P)  s f (T )
h(T , P)  u f (T )  Pvf (T )  hf (T )  P  Psat (T )v f (T )
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Exercise with property tables
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