Lesson 1 (Properties of Pure Substance)

8/3/2019 Lesson 1 (Properties of Pure Substance)

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Thermodynamics 2 Lesson 1

LESSON 1

PROPERTIES OF PURE SUBSTANCE

DEFINITIONS AND USEFUL INFORMATION

Pure substance is a substance that is homogeneous in composition and homogeneous and invariable in

chemical aggregation (Faires, 1978).

A substance is said to be pure substance if its chemical compositions do not change even in solid,

liquid, or gaseous phase. Example of which is water that has two atoms of hydrogen and one atom of

oxygen as it is in solid, liquid or gaseous state.

State of pure substance or system refers to its condition as quantified or identified through the

thermodynamic properties of the substance.

Phase is a quantity of substance that is homogeneous throughout.

Relevant Reversible Work Modes refers only to the important work modes for the system in question

and irreversible work modes are not considered.

Reversible work mode refers to the amount of energy added in a forward process that can be removed

by the reversed process.

Additional Thermodynamic Properties1. Enthalpy is a thermodynamic property with the following definitions:

For a closed system, it means the sum of internal energy and the product of pressure andspecific volume, and given by the relation:

pvuh +=

Where, h = enthalpy, kJ/kg, Btu/lb p = pressure, kPaa, psia

v = specific volume, m3/kg

For an open system, enthalpy means the sum of internal energy and the flow energy or flowwork, and given by the relation:

fEUH +=

Where, H = total enthalpy, kJ, kW, Btu, Btu/min

U = total internal energy, kJ, kW, Btu, Btu/min

Ef= flow energy or work, kJ. KW, Btu, Btu/min

2. Entropy is a thermodynamic property first introduced by Clausius in 1865 and could be defined asfollows:

Entropy is the measure of the microscopic disorder of the molecules of a substance.

It is a thermodynamic property that remains constant in an adiabatic reversible process. The change of entropy in an irreversible process is the measure of the unavailable energy. Entropy is also defined as the measure of the irreversibility of the system or substance. The change of entropy for a reversible process is given by the relation:

==

rev

12T

dQsss

Details of entropy werre discussed in the Second law of thermodynamics.

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Explanation for the Change of Phase in Figure 1.1o At state 1, water is at solid phase at, say, - 20 C under atmospheric pressure of 101.325 kPa. Heat

is added to the solid water to undergo the process from state 1 to state 2 without melting. This

process is heating of solid with temperature increases from 20oC to 0

oC.

o At state 2, water is still at solid phase at 0 oC. This is the time in which melting process is about tostart. With the heat addition continues, melting process is from state 2 to state 3. During the

melting process, temperature remains the same at 0

o

C. The temperature for process 2 to 3 isknown as melting temperature or melting point.

o State 3 is the end of melting process but the temperature remains 0 oC. Water at this point is atliquid phase. Then heat addition continues to perform heating process from state 3 to state 4.

During this process, temperature increases to, say, 60 oC.

o At state 4, water is called subcooled liquid and the temperature is called subcooled liquidtemperature. Water heated but still no occurring of evaporation. With the heat addition continues,

process 4 to 5 is performed with the increase of temperature from 60 oC to 100 oC. This process is

still heating process.

o At state 5, water is called as saturated liquid and at this point evaporation or boiling process isabout to start at 100 oC. With the heat addition continues, process 5 to 6 is performed without

changing the temperature.

o At state 6, water is a mixture liquid and vapor and is called as wet vapor or wet steam. At his pointthe temperature remains the same at 100

oC. The proportion of the liquid and vapor components

depends on time being considered from state 5 to state 6. The amount of vapor content may bedetermined if the quality of vapor or steam is given. With the heat addition continues, evaporation

process is performed from state 6 to state 7 with the same temperature at 100 oC.

o At state 7, boiling or evaporation process ended with the temperature still 100 oC. At this point, allliquid are converted into vapor or steam and is called as saturated vapor/steam or dry and saturated

vapor/steam. With the heat addition continues, process 7 to 8 is performed with an increase of

temperature from 100oC to, say, 150

oC.

Temperature for processes 5 to 6 and 6 to 7 is called boiling temperature or boiling point andtechnically termed in thermodynamics as saturation temperature and the existing pressure is

also called as saturation pressure.

o At state 8, the condition of water is called superheated vapor/steam and the temperature issuperheated vapor temperature at, say, 150 oC. The temperature at this point is higher than the

saturation temperature. Any point above point 7 at a given temperature higher than the saturation

temperature and following the same existing pressure is called as superheated vapor or steam. The difference of superheated temperature and the saturation temperature or boiling

temperature is called degrees of superheat(oSH).

The difference of saturation temperature and the subcolled liquid temperature is calleddegrees subcooled(

oSC).

Note:

If the existing pressure is increased, melting and boiling temperatures are also increased. Withfurther increased of an existing pressure, a critical point is attained, a point in which saturated

liquid and saturated vapor conditions are coexisted in equilibrium.

If the existing pressure is decreased, melting and boiling temperatures are also decreased. With thefurther decreased of an existing pressure, a triple point will be attained, a point in which melting

and boiling temperatures are the same. Existing pressure lower than the triple point, sublimation

will take place.

OTHER DEFINITIONS Existing pressure or applied pressure refers to the pressure of the system in which change of phase at

constant pressure is considered.

Solid water is a condition of water in which the temperature is below the freezing temperature

corresponding to an existing pressure.

Subcooled liquid is a liquid condition in which the given temperature is higher than the melting

temperature and lower than the saturation or boiling temperature corresponding to an existing pressure.

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Compressed liquid is a liquid condition in which the pressure is higher than saturation pressure

corresponding to the given temperature.

Saturated liquid is a liquid condition in which boiling or evaporation is about to begin and with the

temperature is equal to the saturation or boiling temperature corresponding to an existing pressure.

Wet vapor or wet steam is the mixture of saturated liquid and saturated vapor with the temperature

equal to the saturation temperature corresponding to an existing pressure.

Saturated vapor/steam or dry and saturated vapor/steam is vapor condition in which evaporation is

ended and condensation is about to begin with the temperature equal to the saturation temperature

equal to the saturation temperature corresponding to an existing pressures.

Superheated vapor or steam is a vapor in which the temperature is higher than the saturation

temperature corresponding to an existing pressure.

Heat of fusion is the amount of heat required to change the phase from solid to liquid or vice versa.

Latent heat or enthalpy of evaporation is the amount of heat required to convert saturated liquid to

saturated vapor; it is the amount of heat required to change saturated liquid to saturated vapor at

constant temperature without changing the pressure.

Sensible heat is the amount of heat required to change the temperature without changing the pressure.

Critical point is a point in which liquid and vapor are coexisted in equilibrium.

Critical pressure is the existing pressure at the critical point. Critical temperature is the temperature at

the critical point corresponding to an existing critical pressure.

o For water, critical pressure, pc = 22.09 MPaa, and critical temperature, tc = 374.14oC.

Triple point is a point in which the melting and boiling or saturation temperature are the samecorresponding to the existing pressure.

o For water, at triple point: p = 0.6113 kPaa and t = 0.01oC

Sublimation is the process in which solid will directly change from solid to vapor at a temperature

corresponding to an existing pressure.

o For water, sublimation process will occur at p = 0.260 kPaa and t = - 10 oC. Super critical condition is a condition above the critical point (p > 22.09 kPaa for water). This is a

condition where there is no liquid or vapor phase of pure substance existing in equilibrium.

The thermodynamic properties considered in the study of pure substance are:

P = Pressure, MPaa, psia T = Temperature, oC, oF v = Specific volume, m3/kg, ft3/lb u = Specific Internal energy, kJ/kg, Btu/lb

h = Specific Enthalpy, kJ/kg/ Btu/lb s = Specific Entropy, kJ/kg-oK, Btu/lb-oR

DIAGRAMS Diagrams hereto are graphical representation of the relation of thermodynamic properties for the

change of phase at constant pressure such as pV, TV, Ts, pT, and ph diagrams.

Figure 1.2. Tv Diagram

o

SH

oSC

Vvgvf

f x

sc

shSuperheated temperature, tsh

Saturation temperature, tsat

Subcooled temperature, tsc

Pressure, p, line

Saturated LiquidCurve

cp

Saturated VaporCurve

T

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From Figure 1.2 above,

Point sc is subcooled liquid condition Point f is saturated liquid condition Point x is wet vapor condition Point g is saturated vapor condition Point sh is superheated vapor condition Point cp is critical point condition oSH = tsh tsat oSC = tsat tsc

Figure 1.3. pV Diagram

V

p

f

gx sh

sc

Existing pressure or

Saturation pressure, p, line

Superheated

Temperature, tsh

SaturationTemperature, tsat

cp

Subcooled liquid

Temperature, tsc

Figure 1.4. Ts Diagram

ptp

p

tsb

S-L

L-V

oSC

oSH

Sublimation

Triple Point

gf x

sh

sc

t

tsh

tsc

tmp

ttp

tc cppc

Saturated VaporCurve

Saturated Liquidcurve

T

psb

s

From Figure 1.4 above

tmp = melting point temperature ttp = triple point temperature tsb = sublimation temperature ptp = triple point pressure psb = sublimation pressure pc = critical point pressure tc = triple point temperature p = existing or applied pressure

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Figure 1.5. hs Diagram

p

tsat

g

f

sh

sc

tsh

tsct

cp

h

s

Figure 1.6. pT Diagram

Triple Point

Critical Point

Vapor

Solid

Liquid

EvaporationMelting

Sublimation

Sublimation Curve

Evaporation

Curve

Fusion

Curve

T

p

From figures above

f stands for saturated liquid g stands for saturated vapor fg stands for evaporation or process from f to g Evaporation (fg) the process of boiling, liquid to vapor phase Condensation (gf) the process of opposite direction, from vapor to liquid Saturated liquid curve is the locus of all states of saturated liquid. Saturated vapor curve is the locus of all states of saturated vapor. Triple point is the state in which three phases may all be present in equilibrium. Enthalpy of sublimation, hig,- is the heat of sublimation or the amount of heat required to

sublimate the substance.

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Quality of Vapor/Steam; xo Quality of vapor/steam or dryness factor is the ratio of the mass of vapor content in the mixture to

that of the mass of the mixture of vapor and liquid.

%100mm

mx

gf

g

+=

Moisture Content; yo Moisture content is amount of saturated liquid in the mixture of saturated vapor and saturated

liquid; it is the ratio of the mass of saturated liquid to that of the mass of the mixture of saturated

liquid and saturated vapor.

%100mm

my

gf

f

+= Note: %100yx =+

Figure 1.7. Quality of vapor

mf mg

Q

Where, mf= mass of saturated liquid content, kg, lb.

mg = mass of saturated vapor content, kg, lb.

x = quality of vapor or steam, or dryness factor, %

y = moisture content, %

THERMODYNAMIC PROPERTIES OF EACH STATE

For Subcooled or Compressed Liquid

To obtain the properties v, u, h, & s, p and t must be given. Properties are obtained from steam tables if the given substance is water. Values of properties for subcooled liquid must be lower than the properties of saturated

liquid as indicated in the following relations:

215 x 10

-3m

3/kg

Therefore, the condition is superheated vapor. Ans.

h. For p = 1.5 MPaa & oSH = 152.01 oCSince the given temperature is

oSH, then from steam tables, at 1.5 MPaa,

t = tsat +o

SH = 198.32 + 152.01 = 350.33o

C > tsatTherefore, the given condition is superheated steam. Ans.

i. For p = 5 MPaa & oSC = 65.43 oCSince the given temperature is

oSC, then from steam tables, at 5 MPaa,

t = tsat -oSH = 263.99 65.43 = 198.65

oC < tsat

Therefore, the given condition is subcooled liquid. Ans.

j. For p = 0.35 Mpaa and v = 850 x 10 -3 m3/kgFrom steam tables, at p = 0.35 MPaa,

vg = 524.3 x 10-3

m3/kg

v > vg = 850 x 10-3

m3/kg > 524.3 x 10

-3m

3/kg

Therefore, the given condition is superheated vapor/steam. Ans.

Ex. # 1.3] For water as a pure substance and using steam tables, determine the:a. p & u if t = 200 oC and h = 852.45 kJ/kgb. t & h if p = 150 kPaa and u = 1000 kJ/kgc. u & h if p = 800 kPaa and t = 300 oCd. u & h if p = 5 000 kPaa and t = 200 oCe. t & h if p = 300 kPaa and v = 0.850 m3/kg

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Solution:

a. For the values of p & u if t = 200 oC & h = 852.45 kJ/kg

fg

tsat

tsh

p

T

s

Superheated

Vapor

x

sc

WetVapor

Saturated

Vapor

Saturated

Liquid

Subcooled

Liquid

From Steam tables, at t = 200oC, hf= 852.45 kJ/kg.

Since h = hf= 852.45 kJ/kg, water is at saturated liquid condition.

Therefore, from steam tables, at t = 200oC,

p = psat = 1.5538 MPaa and u = uf= 850.65 kJ/kg ans.

b. For the values of t & h if p = 0.15 MPaa and u = 1000 kJ/kgAt p = 0.15 MPaa, uf= 466.94 kJ/kg and ug = 2519.7 kJ/kg

Since uf< u < ug, water is at wet vapor condition.

From steam tables, at p = 0.15 MPaa,

uf= 466.94 kJ/kg ufg = 2052.7 kJ/kghf= 467.11 kJ/kg hfg = 2226.5 kJ/kg

tsat = 200oC

For quality x, f

fg

u u 1000 466.94x 100% 100% 25.97%

u 2052.7

= = =

Then, ( )f fgh h x h 467.11 0.2597 2226.5 1045.33kJ / kg= + = + = ans

ans.C37.111tto

sat ==

c. For the values of u & h if p = 0.80 MPaa and t = 300 oCAt p = 0.80 MPaa, tsat = 170.43

oC

Since tsat < t, the water is superheated steam.

From steam tables, at p = 0.80 MPaa & t = 300oC,

u = 2797.2 kJ/kg and h = 3056.5 kJ/kg ans.

d. For the values of u & h if p = 5 MPaa and t = 200 oCAt p = 5 MPaa, tsat = 263.99

oC.

Since tsat > t, the water is subcooled liquid.

From steam tables (table 4), at p = 5 MPaa and t = 200oC,


e. For the values of u & h if p = 0.30 MPaa and v = 850 x 10 -3 m3/kgAt p = 0.30 MPaa, vg = 605 x 10

-3m

3/kg.

Since v > vg, water is superheated steam.

From steam tables (table 3), at p = 0.30 MPaa and v = 850 x 10-3

m3/kg, and

Using interpolation,

t,oC V 10

3m

3 /kg u, kJ/kg h, kJ/kg

280 843.8 2775.4 3028.6

T 850 u h

290 859.6 2791.1 3048.9

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Interpolating,t 280 850 843.8 u 2775.4 h 3028.6

290 280 859.6 843.8 2791.1 2775.4 3048.9 3028.6

= = =

t = 283.92oC


Ex. # 1.4] An 100-liter tank containing a mixture of saturated liquid water and saturated steam at 370

o

C. a)Find the mass of each phase if their volumes are equal. b) Find the volume occupied by each phase if theirmasses are equal.

Given: 100-liter tank with saturated water and steam

t = 370oC

Required:

a. mf& mg if Vf= Vg

Vg

Liquidt = 370

oC

Va or

b. Vf& Vg if mf= mg

Solution:

a. For the values of mf& mg if Vf= VgmgFrom steam tables at t = 370 oC,

vf= 2.213 x 10-3

m3/kg Vf mfvg = 4.925 x 10

-3m

3/kg

Considering the total volume,3T

gf m05.02

VVV ===

Therefore, kg59.22kg/m10x213.2

m05.0

v

Vm

33

3

f

ff === ans

kg15.10kg/m10x925.4

m05.0

v

Vm

33

3

g

g

g === ans.

b. For the values of Vf& Vg if mf= mgFrom steam tables at t = 370 oC,

vf = 2.213 x 10-3

m3/kg

vg = 4.925 x 10-3

m3/kg

Considering the mass,gf

f g

f g

VVm m

v v= =

Then, ggg

fgf V44934.0V

925.4

213.2

v

vVV =

=

= eq. 1

Liquid

Vapor

t = 370oC

Vg mg

Vf mf

Considering the volume, 10.0VVV Tgf ==+

gf V10.0V = eq. 2

Equating eq. 1 and eq. 2, gg V10.0V44934.0 =

3g m069.0

44934.1

10.0V == ans

Then, ans3

gf m031.0069.010.0V10.0V ===

Ex. # 1.5] Determine the volume occupied by 2.25 kg steam at 8 MPaa and 60 % quality.

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Given: Wet steam

m = 2.25 kgp = 8 MPaa

Saturated

LiquidSaturated

Va or

x = 60 %

Required:

The volume occupied by the wet vapor.

Solution:

From steam tables, at p = 8 MPaa,

vf= 1.3842 x 10-3

m3/kg and vg = 23.52 x 10

-3m

3/kg

For the specific volume of wet vapor,

( ) ( )[ ] 3fgf 103842.152.2360.03842.1vvxvv +=+=

v = 14.66568 x 10-3

m3/kg

Then, for the volume occupied by the 2.25 kg wet vapor,

( ) liters33m033.010x66568.1425.2vmV 33 ====

Ex. # 1.6] A rigid container with a volume of 2 000 liters contains 4-kg mixture of saturated water andsteam at 120

oC. The mixture is slowly heated until the liquid content is completely vaporized. Determine

the a) quality of the vapor at the initial condition; and b) temperature of water after heating.

Given: A rigid tankVT = 2 000 liters = 2 m

3m = 4 kg t1 = 120

oC

x2 = 100 %

Required:

a. The initial quality of vaporb. The final temperature

V1 = V2

2

1

t2

t1 =120oC

Figure:T

Rigid

ContainerVapor

Liquid

QVSolution:

From steam tables, at t1 = 120oC,

vf1 = 1.0603 x 10-3

m3/kg

vg1 = 891.9 x 10-3

m3/kg

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Solving for the initial specific volume of the mixture,

kg/m10x500kg/m5.0kg4

m2

m

Vv 333

31

1====

a. For the quality of steam at initial condition

( ) ( ) %01.56%1000603.19.891

0603.1500%100

v

vvx

1fg

1f11 =

=

=

b. For the final temperature after heatingSince the container is rigid, the volume does not change during the process.

v2 = v1 = vg2 = 500 x 10-3

m3/kg

From steam table (table 1), at v2 = 500 x 10-3

m3/kg and x2 = 100 %

T, C vg, m3/kg

140 508.9 x 10-3

t 500 x 10-3

141 495.6 x 10-3

Interpolating,t 140 500 508.9

141 140 495.6 508.9

=

Then, ( ) o500 508.9

t 141 140 140 140.67 C495.6 508.9

= + =

Ex. # 1.7] A piston-cylinder arrangement contains 100 liters of saturated water and 900 liters of saturated

steam in equilibrium at 1 MPaa. Heat is added at constant pressure until the temperature becomes 300oC.

Determine a) the initial temperature; b) the total mass; c) the quality of vapor at initial condition; d) thefinal volume; and e) the work done during the process.

Given: A piston-cylinder device

Vf1

Vg1

Q

Vf1 = 100 liters = 0.10 m3

Vg1 = 900 liters = 0.90 m3

p1 = p2 = 1 MPaat2 = 300

oC

Required:

a. Initial temperatureb. Total massc. Quality of steam at initial conditiond. Final Volumee. Work done during the process

Solution:

a. For the initial temperature

From steam tables, at 1 MPaa,

t1 = tsat = 179.91oC

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b. For the total mass

At 1 MPaa, vf1 = 1.1273 x 10-3

m3/kg and vg1 = 194.44 x 10

-3m

3/kg

Then,3

f1f1 3 3

f1

V 0.10 mm 88.71 kg

v 1.1273 x10 m / kg

= = =

3g1

g1 3 3

g1

V 0.90 mm 4

v 194.44 x10 m / kg= = = .63 kg

Therefore, the total mass will be

kg34.9363.471.88mmmm 21g1f1 =+==+=

c. For the quality of vapor

g1

1

1

m 4.63x 100 % 100% 4.96 %

m 93.34

= = =

d. For the final volume

At point 2, p2 = 1 MPaa and t2 = 300oC

v2 = 257.9 x 10-3

m3/kg

Then, ( )3 3 32 2 2V m v 93.34kg 257.9x10 m / kg 24.07 m= = =

e. For the work done during the process

( )122

1

2

1VVpdVpdVpW ===

( )W 1000 24.07 1.0 23070 kJ= =

IMPORTANT POINTS TO REMEMBER

Saturation temperature is a temperature at which vaporization of liquid or condensation of vapor

takes place at a given or existing pressure. Saturation pressure is an existing pressure during the vaporization or condensation of the substance

at a temperature known as saturation temperature. Sub-cooled liquid is the liquid that has a temperature lower than the saturation temperature

corresponding to the given or existing pressure. Compressed Liquid, synonymous with sub-cooled liquid, is the liquid that has a pressure higher than

the saturation pressure corresponding to the existing temperature.

Saturated liquid is the liquid that has a temperature equal to the boiling point corresponding to theexisting or given pressure.

Vapor is a substance existing in the gaseous phase but relatively near its saturation temperature. Saturated Vapor is the vapor that has temperature equal to the saturation temperature corresponding

to an existing pressure. Degree of superheat,

oSH, is the difference between the temperature of superheated vapor and the

saturation temperature for an existing pressure. Degrees Sub-cooled,

oSC, is the difference between the saturation temperature for the given pressure

and the actual sub-cooled liquid temperature.

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Wet Vapor is the mixture of saturated vapor and the saturated liquid. Quality of wet vapor, x, is the percentage by weight of vapor in the mixture of saturated vapor and

saturated liquid. Percent Moisture, y, is the percentage by weight of liquid in the mixture of saturated liquid and

saturated vapor. Latent Heat of Evaporation is the amount of energy required to convert saturated liquid to saturated

vapor or from saturated vapor to saturated liquid with the temperature remains constant.

Sensible heat is the amount of heat required to change the temperature of a substance without

changing the phase. Critical point is a point that represents the highest pressure and highest temperature at which liquid

and vapor can coexist in equilibrium.

PROBLEMS

Prob. # 1] Determine the properties and draw the Tv, pV, Ts, & hs diagrams of water for the following

conditions:

a) t = 102 oC & saturated liquidb) p = 700 kPaa & x = 100 %c) t = 70 C & x = 80 %d) p = 0.5 MPaa & h = 2895.6 kJ/kge) p = 0.356 MPaa & t = 321.67 oC

Prob. # 2] Specify the condition (subcooled liquid, saturated liquid, wet vapor, etc.) of water for the

following conditions. Justify your answer and show pV and Ts diagrams for each item.a. p = 900 kPaa & v = 1.0245 x 10-3 m3/kgb. t = 76 C & u = 3542.2 kJ/kgc. t = 100 oC and p = 2.5 Mpaad. p = 1.2 MPaa & oSH = 234.6 oCe. p = 1 Mpaa & oSC = 100 oC

Prob. # 3] A pumping discharging water at 1 MPaa and 165oC. Determine the specific volume and internal

energy of the water at the pump discharge.

Prob. # 4] A 500-liter tank contains saturated mixture of water at 300 kPaa. If the saturated liquid occupies10 % of the volume, determine a) the the quality of vapor, and b) the total mass of water in the tank.

Prob. # 5] A piston-cylinder device initially contains 300 liters of water at 420 kPaa and 110oC. Heat is

added to the water at constant pressure until the liquid is entirely vaporized. Determine a) the mass of

water; b) the final temperature; c) the final volume of water; d) the work done during the process; and e)

the amount of heat added.

Lesson 1 (Properties of Pure Substance)

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Transcript of Lesson 1 (Properties of Pure Substance)