ES 202 Fluid and Thermal Systems Lecture 16: Property Tables (1/20/2003)

Lecture 16ES 202 Fluid & Thermal Systems2 Assignments Homework: –3-56, 3-57, 5-72 in Cengel & Turner Reading assignment –Tables A-4 to A-14 in Cengel & Turner

Lecture 16ES 202 Fluid & Thermal Systems3 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!

Lecture 16ES 202 Fluid & Thermal Systems4 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

Lecture 16ES 202 Fluid & Thermal Systems5 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

Lecture 16ES 202 Fluid & Thermal Systems6 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: where q is the intensive counterpart of any extensive property like internal energy, enthalpy, entropy and volume.

Lecture 16ES 202 Fluid & Thermal Systems7 T > T sat (P) T < T sat (P) T = T sat (P) 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  If T < T sat (P), compressed liquid.  If T = T sat (P), saturated liquid-vapor mixture.  If T > T sat (P), superheated vapor.

Lecture 16ES 202 Fluid & Thermal Systems8 P > P sat (T) P < P sat (T) P = P sat (T) 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  If P > P sat (T), compressed liquid.  If P = P sat (T), saturated liquid-vapor mixture.  If P < P sat (T), superheated vapor.

Lecture 16ES 202 Fluid & Thermal Systems9 v g (P) v f (P) 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 P sat = P 1)If v < v f (P sat ), compressed liquid. 2)If v = v f (P sat ), saturated liquid. 3)If v f (P sat )< v < v g (P sat ), saturated liquid-vapor mixture. 4)If v = v g (P sat ), saturated vapor. 5)If v > v g (P sat ), superheated vapor. (2) (4) (5) (1) (3) P sat

Lecture 16ES 202 Fluid & Thermal Systems10 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)

Lecture 16ES 202 Fluid & Thermal Systems11 Flow Chart Two-phaseCompressed liquidSuperheated vapor Determine phase of substance Determine qualityDirect look up Interpolate other properties

Lecture 16ES 202 Fluid & Thermal Systems12 Property Tables Compressed liquid water (A-7) Saturated water –temperature (A-4) –pressure (A-5) Superheated water (A-6)

Lecture 16ES 202 Fluid & Thermal Systems13 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)

Lecture 16ES 202 Fluid & Thermal Systems14 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: (weak function of T)

Lecture 16ES 202 Fluid & Thermal Systems15 Exercise with property tables