Unit C- 2: List of Subjectsmercury.pr.erau.edu/~hayasd87/ES312/ES312_I_Notes_… · ·...

ES312 Energy Transfer Fundamentals

Unit C: Thermodynamic Analysis

ROAD MAP . . .

C-1: p-v-T Relations

C-2: Thermodynamic Property Tables

ES312 Energy Transfer Fundamentals

Unit C-2: List of Subjects

Thermodynamic Property Tables

Saturated Liquid and Saturated Vapor

Saturated 2-Phase (Liquid and Vapor) Mixture

Superheated Vapor

Compressed Liquid

THERMODYNAMICS PROPERTY TABLES

Appendix 1: PROPERTY TABLES AND CHARTS (SI UNITS)

Appendix 2: PROPERTY TABLES AND CHARTS (ENGLISH UNITS: “E”)

Water properties (for example)

Table A-4 (A-4E): Saturated water – Temperature table

Table A-5 (A-5E): Saturated water – Pressure table

Table A-6 (A-6E): Superheated water

Table A-7 (A-7E): Compressed liquid water

ENTHALPY

Enthalpy is a combined property: H U p V

Enthalpy per unit mass is: h u p v (kJ/kg)

Enthalpy is the combination of internal energy and energy added to the system by “flow work.”

Greek word “enthalpien” means “to heat.”

Enthalpy represents the “total heat content” of the system.

UNIT C-2PAGE 1 of 14

Thermodynamic Property Tables

Table A-4 (A-4E)

Table A-5 (A-5E)Table A-7 (A-7E)

Table A-6 (A-6E)

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SATURATED LIQUID AND SATURATED VAPOR

Saturated liquid and saturated vapor properties for water are given in tables A-4 (A-4E) / A-5 (A-5E).

Table A-4(A-4E): properties are listed under temperature (temperature table)

Table A-5(A-5E): properties are listed under pressure (pressure table)

For two-phase mixture, the subscript f is used to denote properties of a saturated liquid, and the

subscript g to denote the properties of saturated vapor.

Another subscript commonly used is fg, which denotes the difference between the saturated vapor

and saturated liquid values of the same property.

ENTHALPY OF VAPORIZATION

The quantity fgh is called the “enthalpy of vaporization”

This represents the amount of energy needed to vaporize a unit mass of saturated liquid at a given

temperature or pressure.

This decreases as the temperature or pressure increases, and becomes zero at the critical point.


Saturated Liquid and Saturated Vapor

Table A-4 (A-4E)

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Solution

A rigid tank contains 50 kg of saturated liquid water at 90 C. Determine the pressure in the tank and the

volume of the tank.

Analysis

The state of the saturated liquid water is shown on a T-v diagram in the figure. Since saturation

conditions exist in the tank, the pressure must be the saturation pressure at 90 C:

osat @ 90 Cp p 70.183 kPa (Table A-4: Saturated water – Temperature table)

The specific volume of the saturated liquid at 90 C is:

o @ 90 Cf v v30.001036 m /kg (Table A-4: Saturated water – Temperature table)

Therefore, the total volume of the tank becomes:

350 kg 0.001036 m /kgm V v30.0518 m


A rigid tank contains 50 kg of saturated liquid water at 90 C. Determine the pressure (in “kPa”) in the tank and the volume (in “m3”) of the tank.

EXERCISE C-2-1(Do-It-Yourself)

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Solution

A cylinder contains saturated water vapor. The temperature and the mass of vapor are to be determined.

Analysis

The state of the saturated water vapor is shown on a p-v diagram in the figure. Since the cylinder

contains saturated vapor at 50 psia, the temperature inside must be the saturation temperature at this

pressure:

sat @ 50 psiaT T o280.99 F (Table A-5E: Saturated water – Pressure table)

The specific volume of the saturated vapor at 50 psia is:

@ 50 psiag v v 38.5175 ft /lbm (Table A-5E: Saturated water – Pressure table)

Therefore, the mass of water vapor inside the cylinder becomes:

3

3

2 ft

8.5175 ft /lbmm V

v0.235 lbm


A piston-cylinder device contains 2 ft3 of saturated water vapor at 50 psia. Determine the temperature (in “F”) and the mass of the vapor (in “lbm”) inside the cylinder.


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Solution

Saturated liquid water is vaporized at constant pressure. The volume change and the energy transferred

are to be determined.

Analysis

(a) The process described is illustrated on a p-v diagram in the figure. The volume change per unit

mass during a vaporization process is vfg, which is the difference between vg and vf.

Reading these values from Table A-5 (Saturated water – Pressure table) at 100 kPa:

3 3 31.6941 m /kg 0.001043 m /kg 1.6931 m /kgfg g f v v v

Therefore,

30.2 kg 1.6931 m /kgfgm V v30.3386 m

(b) The amount of energy needed to vaporize a unit mass of a substance at a given pressure is the

enthalpy of vaporization at that pressure, which is:

2257.5 kJ/kgfgh from Table A-5 (Saturated water – Pressure table) at 100 kPa.

Therefore the amount of energy transferred to the water is:

0.2 kg 2257.5 kJ/kgfgmh 451.5 kJ


200 g of saturated liquid water is completely vaporized at a constant pressure of 100 kPa. Determine (a) the volume change (in “m3”) and (b) the amount of energy transferred to the water (in “kJ”).


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SATURATED TWO-PHASE MIXTURE

“Quality” of saturated mixture of liquid-vapor: is the ratio of the mass of vapor to the total mass of

the mixture.

vapor

total

mx

m where, total liquid vapor f gm m m m m

x = 0: Saturated liquid x = 1: Saturated vapor

AVERAGE PROPERTIES OF MIXTURE

A two-phase system can be treated as a homogeneous mixture.

The total volume and total mass of two-phase mixture can be determined as:

t f g V V V and t f gm m m

In terms of specific volume:

mV v => avgt f f g gm m m v v v (eqn. 1)

The average property (specific volume) can be given by:

t f gm m m => f t gm m m , this can be substituted into the eqn.1:

avgt t g f g gm m m m v v v => avg 1 f gx x v v v (eqn. 2)

Note that g f fg v v v , and the eqn. 2 yields: avg f fgx v v v (kJ/kg)

Similarly: avg f fgu u xu (kJ/kg) avg f fgh h xh (kJ/kg)

In general: avg f fgy y xy ( avgf fgy y y ) where, y is v, u, or h


Saturated 2-Phase (Liquid and Vapor) Mixture

The relative amounts of liquid and

vapor phases in a saturated mixture

are specified by the “quality” x

Quality is related to the horizontal distances on P-v and T-v diagrams

The value of v of a saturated liquid-vapor

mixture lies between the vf and vg values at

the specified temperature or pressure

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Solution

A rigid tank contains saturated mixture. The pressure and the volume of the tank are to be determined.

Analysis

(a) The state of the saturated liquid-vapor mixture is shown in the figure. Since the two phases coexist

in equilibrium, we have a saturated mixture, and the pressure must be the saturation pressure at the given

temperature:

osat @ 90 Cp p 70.183 kPa (Table A-4: Saturated water – Temperature table)

(b) At 90 C, we have specific volume of:

o @ 90 Cf v30.001036 m /kg (Table A-4: Saturated water – Temperature table)

o @ 90 Cg v32.3593 m /kg (Table A-4: Saturated water – Temperature table)

One way of finding the volume of the tank is to determine the volume occupied by each phase and then

add them:

3 38 kg 0.001036 m /kg 2 kg 2.3593 m /kgf g f f g gm m V V V v v34.73 m

Another way is to first determine the quality (x), then the average specific volume, and finally the total

volume:

2 kg0.2

10 kg

g

f

mx

m

3 3 30.001036 m /kg 0.2 2.3593 0.001036 m /kg 0.473 m /kgf fgx v v v

310 kg 0.473 m /kgm V v34.73 m


A rigid tank contains 10 kg of water at 90 C. If 8 kg of the water is in the liquid form and the rest is in the vapor form, determine (a) the pressure (in “kPa”) in the tank and (b) the volume of the tank (in m3”).


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Properties

From table A-4, at 220 C: vf = 0.001190 m3/kg and vg = 0.08609 m3/kg

Analysis

(a) Two-phase (liquid-vapor) mixture coexist in equilibrium, therefore this is a saturated liquid-vapor

mixture. The pressure of the tank is the saturation pressure at the given temperature:

p = Tsat@220 C = 2,319.6 kPa

(b) The total mass and the quality can be determined as:

3

3

1 3 1.8 m504.2 kg

0.001190 m /kg

f

f

f

m V

v

3

3

2 3 1.8 m13.94 kg

0.08609 m /kg

g

g

g

m V

v

504.2 kg 13.94 kg 518.1 kgt f gm m m => 13.94 kg

518.1 kg

g

t

mx

m 0.0269

(c) The density is determined from:

3 3 3 30.001190 m /kg 0.0269 0.08609 m /kg 0.001190 m /kg 0.003474 m /kgf g fx v v v v

3

1 1

0.003474 m /kg v

3287.8 kg/m


A 1.8 m3 rigid tank contains steam at 220 C. 1/3 of the volume is in the liquid phase and the rest is in the vapor form. Determine (a) the pressure (in “kPa”) of the steam, (b) the quality of the saturated mixture, and (c) the density (in “kg/m3”) of the mixture.

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SUPERHEATED VAPOR

In the region to the right of the saturated vapor line and at temperatures above the critical point

temperature, a substance exists as superheated vapor.

Since the superheated region is a single-phase region (vapor only), temperature and pressure are no

longer dependent properties and they can conveniently be used as the two independent properties in

the tables.

In table A-6 (A-6E), the properties are listed against temperature for selected pressures starting with

the saturated vapor data. The saturation temperature is given in parentheses following the pressure

data.

Compared to saturated vapor, superheated vapor is characterized by:

Lower pressures (p < psat at a given T)

Higher temperatures (T > Tsat at a given p)

Higher specific volumes (v > vg at a given p or T)

Higher internal energies (u > ug at a given p or T)

Higher enthalpies (h > hg at a given p or T)


Superheated Vapor

Table A-6 (A-6E)

At a specified pressure, superheated vapor

exists at a higher h than saturated vapor

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Solution

Determine the temperature of water at a state of p = 0.5 MPa and h = 2,890 kJ/kg.

Analysis

At 0.5 MPa, the enthalpy of saturated water vapor is:

hg = 2,748.1 kJ/kg.

Since h > hg, as shown in the figure, we have superheated vapor.

Under 0.5 MPa in table A-6:

By linear interpolation, the temperature can be determined as:

o o 2,890 kJ/kg 2,855.8 kJ/kg250 C 200 C

2,961 kJ/kg 2,855.8 kJ/kgT

o216.3 C


Determine the temperature (in “C”) of water at a state of pressure 0.5 MPa and enthalpy 2,890 kJ/kg.


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Properties

From table A-13, the properties of R-134a at the given state are:

p = 800 kPa & T = 120 C => u = 327.87 kJ/kg & v = 0.037625 m3/kg

Analysis

The total volume and internal energy can be determined as follows:

32 kg 0.037625 m /kgm V v30.0753 m

2 kg 327.87 kJ/kgU mu 655.7 kJ


A rigid vessel contains 2 kg of refrigerant 134a at 800 kPa and 120 C. Determine the volume (in “m3”) of the vessel and the total internal energy (in “kJ”).

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COMPRESSED LIQUID

Compressed liquid tables are not as commonly available: table A-7 (A-7E) is the only exception

(compressed liquid of water).

The format of table A-7 (A-7E) is very much like the format of the superheated vapor tables. One

reason for the lack of compressed liquid data is the relative independence of compressed liquid

properties from pressure.

In the absence of compressed liquid data, a general approximation is to treat compressed liquid as

saturated liquid at the given temperature:

at a given fy y T where, y is v, u, or h

Of these three properties, the enthalpy is most sensitive to variations in the pressure. Although the

approximation above will provide good results, one can correct the enthalpy by:

@ @ sat@f T f T Th h v p p

Compared to saturated liquid, compressed (or subcooled) liquid is characterized by:

Higher pressures (p > psat at a given T)

Lower temperatures (T < Tsat at a given p)

Lower specific volumes (v < vf at a given p or T)

Lower internal energies (u < uf at a given p or T)

Lower enthalpies (h < hf at a given p or T)

REFERENCE STATES AND VALUES

The absolute values of u, h, and s cannot be measured directly: hence often a convenient reference

state is chosen and zero value is assigned.

Water: state of saturated liquid at 0.01 C is taken as a reference state

Refrigerant R-134a: state of saturated liquid at 40 C is taken as a reference state


Compressed Liquid

A compressed liquid may be approximated as

a saturated liquid at the given temperature

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Solution

The exact and approximate values of the internal energy of liquid water are to be determined.

Analysis

At 80 C, the saturation pressure of water is 47.416 kPa, and since 5 MPa > psat, we have compressed

liquid, as shown in the figure.

(a) From the compressed liquid table (table A-7):

p = 5 MPa & T = 80 C => u = 333.82 kJ/kg

(b) From the saturation table (table A-4):

oat 80 Cfu u T => u = 334.97 kJ/kg

The % error can be computed as:

334.97 333.82100

333.82

0.34%


Determine the internal energy (in “kJ/kg”) of compressed liquid water at 80 C and 5 MPa, using (a) data from the compressed liquid table and (b) saturated liquid data. What is the error (in “%”) involved in the second case?


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Analysis

Initially, the cylinder contains compressed liquid (since p > psat@40 C) that can be approximated as a

saturated liquid at the specified temperature.

From table A-4: o 3

1 at 40 C 0.001008 m /kgf T v v & o

1 at 40 C 167.53 kJ/kgfh h T

(a) The mass is determined from:

3

1

3

1

0.050 m

0.001008 m /kgm V

v49.61 kg

(b) At the final state, the cylinder contains saturated vapor; the final temperature must be the saturation

temperature at the final pressure:

T = Tsat@200kPa = o120.21 C

(c) The final enthalpy is h2 = hg@200kPa = 2,706.3 kJ/kg. Therefore, the total enthalpy change is:

2 1 49.61 kg 2,706.3 kJ/kg 167.53 kJ/kgH m h h 125,943 kJ


A piston-cylinder device initially contains 50 liters of liquid water at 40 C and 200 kPa. Heat is transferred to the water at constant pressure until the entire liquid is vaporized.(a) What is the mass of the water (in “kg”)?(b) What is the final temperature (in “C”)(c) Determine the total enthalpy change (in “kJ”).(d) Show the process on a T-v diagram with respect to saturation lines.

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Unit C- 2: List of Subjectsmercury.pr.erau.edu/~hayasd87/ES312/ES312_I_Notes_… · ·...

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