basic thermodynamics manual

8/3/2019 Basic Thermodynamics Manual

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Basic Thermodynamics manual

Chapter 1

THERMAL PROPERTIES OF LIQUIDS AND SOLIDS

GENERALIZATION

Materials are often subjected to temperature variations in various industrial applications.

Their physical properties such as dimensions, densities, viscosity, etc. vary significantly with

temperatures. Specific terms such as specific heats, latent heat of fusion, latent heat of

vaporization, etc. are used in describing the physical change of the material with temperature.

An understanding of how the physical nature of materials changes with temperature is essential.

GENERAL OBJECTIVES

The student will learn about:

(a) the definition of specific heat, latent heat of fusion and latent heat of vaporization.

(b) the units of heat energy.

(c) the determination of heat required to raise or lower the temperature of a substance.

(d) the determination of the specific heat of a substance by experimental data.

(e) the determination of the heat flow required to raise or lower the temperature of a flowing

substance.

1. EXPANSION OF SOLIDS AND LIQUIDS

The physical dimensions of solids and liquids change as they are subjected to any

change of temperature. The thermal energy in the molecules of a solid or liquid increases as it

is heated. The molecules become more energetic and move more vigorously. This process is

reversed when cooling takes place.

A. Linear and Volume expansion of solids

For solids, the change in length of the material, l is :

l = lo t eq.(1-1)

where

lo

[m] is the original length of the solid

[/Co or /K] is the coefficient of linear expansion of the solid

t is the change in temperature from t1

to to

[Co].

The new length becomes :

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lt = lo (1 + t) eq.(1-2)

Extending into 3 dimensions, the new volume of any solid material is

Vt = Vo (1 + t)3 eq.(1-3)Where:

Vo

[m3] is the original volume of the solid.

EXAMPLE1

The horizontal trusses in the roof of the Coliseum are constructed from individual carbon steelbeams, each 20 m long. It takes 3 beams end to end to make one truss. If the temperature

inside of the Coliseum varies from 5oC to 23oC, how much allowance must be made for the

changes in length of each truss?

B. Volume expansion of liquids

Since liquid does not have its own definite shape, one can only consider the bulk expansion of

the liquid. The new volume of any liquid material is

Vt = Vo (1 + t) eq.(1-4)where

[/Co or /K] is the coefficient ofvolume expansion of the liquid.

EXAMPLE2

PHASE 1


So lu t io n D a ta

L .. s tee lL o t 1 t o L o.60 m

=L 0.013068 mt o

.5 d e g C

t 1.23 d e g C

s tee l..1. 21 1 0 5 d e g C 1 (p . 17 D B


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A 2 liter carbon steel spherical bottle is completely filled with ethyl alcohol at 20C. If the

bottle is heated to 45C, how much ethyl alcohol will overflow ?

Solution Data

V bottle_at_20.2 l

V alcohol.. alcohol V alcohol_at_20 t f t o

V alcohol_at_20.2 l

=V alcohol 5.35 102

l

V bottle+

....V bottle_at_20 1. steel t f t o

3

V bottle_at_20

=Vbottle

1.82 103

l

steel..

1.21 10

5

degC1

V overflow V alcohol V bottle alcohol..1.07 10 3 degC 1

(p.18 databook)=V overflow 5.17 10

2l

t o.20 degC

t f.45 degC

2. HEAT TRANSFER OF SOLIDS AND LIQUIDS

The temperature of an object rises as thermal energy is supplied to it. Similarly, its

temperature will fall if thermal energy is taken away from it. The amount of thermal energy

absorbed or lost can be calculated as follows :

Q = mc t eq.(1-5)Where:

Q [J] is the amount of heat absorbed or lost.

m [kg] is the mass of the object in discussion. t [Co] is the absolute difference between the initial and final temperature.c [or ] is the specific heat of the object. The specific heat of solids and liquids are

given in tables.

EXAMPLE3

Determine the heat required to heat 5 liters of water in an electric kettle from 20oC to its boiling

point temperature in Edmonton.

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Solution DataQ ..m water c t 1 t o m water

...5 10 3 998.2 kg (p.20 Databk

t o.20 degC

=Q 1629.7 kJt 1

.98 degC

c .4.1863kJ

.k g d eg C(23-3 SI datab k)

When a solid material is heated, the molecules become more energetic and vibrate so much that

they become loosely bound to each other. The material changes from solid phase to liquidphase. The temperature at which this takes place is called the melting point of the solid. The

amount of heat required to change 1 kg of solid at melting point to liquid is called the latent

heat of fusion. In the process of melting, any heat absorbed by the material is used for melting

and no heat is used to raise the temperature. The temperature thus remains at the melting point

until all material is melted.

The latent heat of fusion for a solid can be calculated by

Qf

= mLf

eq.(1-6)

Where: Qf[J] is the total heat for fusion.

Lf[] is the latent heat of fusion per kg.

m is the total mass of solid.

EXAMPLE4

Determine the heat which must be removed to freeze 7.5 kg of Benzene initially at 40oC.

DataSolution

mbenzene

.7.5 kgQ ..m benzenec t o t 1 .m benzeneL f

t o.40 degC

=Q 1398.8 kJt 1

.5.533 degC

c .1.7147kJ

.k g deg C(p.23-3 SI datab k

L f.127.4

kJ

kg (p.18 d atabk)

If the same liquid is further heated, its temperature rises again. More energy is absorbed by the

molecules and consequently each molecule can be totally free from each other At this stage, the

liquid turns into vapor. Any supplied energy is used for phase conversion and thus the

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temperature remains constant. The temperature is called the boiling point. The amount of

heat required to change 1 kg of liquid at boiling point to vapor is called the latent heat ofvaporization.

The latent heat of vaporization for a liquid is expressed as

Qv

= mLv

eq.(1-7)

Where: Qv

[J] is the total heat for vaporization.

Lv[] is the latent heat of vaporization per kg.

m is the total mass of liquid.

EXAMPLE 5

Calculate the amount of heat required to vaporize 2 kg of benzene at 15C.

Solution Data

Q ..m benzenec t boiling t o.m benzeneL v m benzene

.2 kg

t o.15 degC

=Q 1010.7 kJ

tboiling

.80.07 degC

(p.23-2 SI databk)

c .1.7147kJ.kg degC

(p.23-3 SI databk

L v.393.78

kJ

kg

(p.23-4 SI databk

3. HEAT TRANSFER IN MIXTURES

When two non-chemically reacting material at different temperatures are mixed together

in an isolated system, their thermal energies are exchanged until their temperatures achieve an

equilibrium value. By the principle of conservation of energy, the amount of heat gained by one

material is equal to the amount of heat lost by the other (provided there is no external losses).

This can be expressed mathematically as

Qgained

= Qlost

eq.(1-8)

or

m1c

1(t

f- t

i1) = m

2c

2(t

i2- t

f) eq.(1-9)

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where: m1is the heat gaining mass.

m2 is the heat losing mass.

c1, c2 are the specific heats of the corresponding masses.

tfis the final temperature of the mixture.

ti1,is the initial temperature of the mass gaining heat

ti2isthe initial temperature of the mass losing heat.

EXAMPLE6

A carbon steel container has a mass of 165 kg and is at a temperature of 40oC. 50 kg of ethyl

alcohol at an initial temperature of 10oC is poured into the container. What will be the finaltemperature of the container and the alcohol ?

Solution Data

..m h c h t h_o t final..m c c c t final t c_ o Hot:

m h.165 kg

t final

..m h c h t h_o..m c c c t c_ o

.m h c h.m c c c

c h.0.448

kJ.k g d eg C

(p.17 db )

t h_o.40 degC

=t final 21.6 degCCold:

m c.50 kg

t c_o.10 degC

c c.2.3564

kJ.k g d eg C(p.23-3 SI db )

EXAMPLE 7

A heat exchanger uses high pressure steam to heat up n-Nonane. The initial heat energy of the

flowing steam is 900 . After going through the exchanger, the steam condenses to water which

carries only 300 . If the n-Nonane has an initial temperature of 25C and reaches a final

temperature of 80 oC, what is the flow rate of the n-Nonane ?.

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steam@900kJ/sec

water@300kJ/sec

n-Nonane@25C,

?kg/sec

n-Nonane

@80o

The mass flow rate of n-Nonane is 4.992 kg/sec.

PROBLEMS

1. A brass bar 2 cm square in cross section measures 87.75 cm long at room temperature

(21oC). It is placed in an oven at 95oC where its length is measured to be 87.88 cm.

What is the coefficient of linear expansion of the material of the bar?

2. The horizontal trusses in the roof of the Coliseum are constructed from individual

(carbon) steel beams each 15 m long at 5C. It takes 3 beams end to end to make one

truss. If the temperature inside of the Coliseum varies from 5oC to 23oC, how much

allowance must be made for the changes in length of each truss?

3. A stainless steel measuring tape is graduated to read correctly at 20oC. An

instrumentation technologist uses the tape to make a measurement outdoors when the

temperature is -18oC. If the tape indicates 315.4 cm (read at -18C), does the number

show the true length ? If not, what is the true length of the object measured? What is the

measurement error ?

PHASE 1


Solution: Data

..q m c t f t i ( )h Q Steam: h .900 kJ

sec

q mh Q

.c t f t i

water:Q .300

kJ

sec

n-Nonane:=q m 4.992

kg

sec c .2.1855kJ

.kg degC(p.23-2 SI db

t i.25 degC t f

.80 degC


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4. The North-South section of the Syncrude 360 kPaa steam line is 120 m long (carbon

steel). It was welded together when the average outdoor temperature was 10oC. How

much would the line expand when the line was put into service? The temperature of thesteam is 140oC.

5. A 0.5 liter carbon steel spherical bottle is completely filled with ethyl alcohol at 20C.

If the bottle is heated to 35C, how much ethyl alcohol will overflow ?

6. Determine the heat required in each of the following cases.

a. To raise the temperature of 17.5 kg of Carbon Tetrachloride from 10o

C to 35o

C.

b. To raise the temperature of 1.00 Imp gallon of Methyl Alcohol at 15oC to its

boiling point temperature.

c. To heat 2 liters of water in an electric kettle from 20oC to its boiling point

temperature in Edmonton.

d. To heat 20,000 bbl per hour of heavy crude oil (specific heat 2.14 ) from 95oC to

260oC in a fired heater if the heater is 78% efficient. (Barrels are at 15oC).

7. Determine the heat to be added/removed in each of the following cases.

a. To freeze 10.5 kg of Benzene initially at 38oC.

b. To vaporize 2.10 US gal of Ethyl Alcohol initially at 22oC.

c. To thaw out 2 Imp gal of Ethylene Glycol at its freezing point, and raise its

temperature to 15oC. (The gallons were measured out at 20oC.)

8. A 45.4 kg (100 lb ) block of ice at an initial temperature of -40oC is to be vaporized at

an atmospheric pressure of 101.3 kPaa. Calculate the amount of heat required in each of

the following cases.

a. To raise the temperature of the ice from -40oC to 0oC. The specific heat of ice is

2.102 .

b. To melt the ice at 0 oC.

c. To raise the temperature of the water produced from 0oC to 100oC.

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d. To vaporize the water to steam at 100oC.

e. What is the total heat required to convert the ice at -40oC into steam at

atmospheric pressure ?

9. 8.00 kg of water is to be heated from 4oC to 62oC in an iron kettle which has a mass of

1.80 kg.

a. How much heat energy is required to heat the water alone ?

b. How much energy is required to heat the kettle alone ?

c. How much energy is needed to heat both the water and the kettle ?

d. If the heatup is to be completed in 8 minutes, how many watts of power must the

heater put out ?

10. a. A carbon steel container has a mass of 152 kg and is at a temperature of 60oC.

45.3 kg of ethyl alcohol at an initial temperature of 5oC is poured into the

container. What will be the final temperature of the container and the alcohol ?

b. To determine the specific heat of an unknown liquid, the following procedure

was used. 3.00 kg of the liquid was poured into a copper container which has a

mass of 0.870 kg. The equilibrium temperature of the container and the liquid

was measured to be 21.3oC. A bar of copper, mass 2.55 kg, was heated to a

temperature of 88.3oC, and then dropped into the liquid in the container. The

resulting temperature was measured to be 31.6oC. What is the specific heat of

the liquid ? What is the liquid ?

c. To determine the specific heat of a metal, a procedure similar to that in 9(b) is

used. 4.50 kg of water is placed in a copper container which has a mass of 0.580kg. The equilibrium temperature is measured to be 23.8oC. A 3.75 kg bar of the

metal is heated to a temperature of 67.2oC, then immersed in the water. The

resulting temperature is 25.7oC. What is the specific heat of the metal? What is

the metal ?

11. A heat exchanger uses high pressure steam to heat up benzene. The initial heat energy of the

flowing steam is 700 . After going through the exchanger, the steam condenses to water

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which carries only 210 . The benzene has an initial temperature of 25C and flows at at rate

of 6 , what is the final temperature of the exiting benzene ?.

steam @700 kJ/sec

water@210kJ/sec

benzene@25C,6 kg/sec

benzene@ ?C

Chapter 2

VAPOUR PRESSURE

GENERALIZATION

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The flow of fluids in pipes such as water pipes or instrumentation air supply lines are

frequently subjected to a change in pressure and temperature. Such changes may cause liquidsto vaporize or vapours to condense. Flashing and cavitation in valves are due to such phase

changes in liquids. The change in phase of a fluid depends on its vapour pressure which is

temperature dependent. It is necessary to understand the different pressure and temperature

conditions which govern the phase changes of liquids.

GENERAL OBJECTIVES

The student will gain a knowledge about :

(a) the concept of vapour pressure.

(b) the absolute and gauge pressure.(c) how to find vapour pressure at a given temperature from tables.

(d) the relationship between pressure at boiling points and vapour pressure of liquids

1. OCCURRENCE

Whether a substance is in its liquid or vapour state, its molecules are always in motion. It is

the collision of the molecules of the substance with the inner walls of a containing vessel that

creates the pressure against these surfaces.

If a closed vessel is completely evacuated, and then partly filled with a sample liquid, the

molecules near the liquid surface leave the liquid and migrate up into the space above the

liquid to form vapour. This will continue until the space above the liquid saturates with

vapour molecules, at which time the reverse effect occurs. Some vapour molecules in the

space return to the liquid. Eventually, an equilibrium condition is reached at which the

number of molecules leaving the liquid per unit time exactly matches the number of

molecules returning from the vapour to the liquid.

Since molecular collision creates pressure, there will be a measurable pressure in the space.

This pressure is the vapour pressure of the substance.

2. UNITS OF VAPOUR PRESSURE

Vapour pressure is always reported in absolute pressure units (kPaa, psia, etc). Otherwise,

a different table of vapour pressure values for liquids would be required for every different

location.

3. VARIATION OF VAPOUR PRESSURE WITH TEMPERATURE

The vapour pressure of a liquid increases considerably as the temperature of the liquid

increases. It is because more liquid molecules become energetic and are ready to leave the

liquid forming vapour. The vapour - temperature relationship is not linear. Thus, any vapour

pressure reading for a liquid is meaningless unless the temperature of the liquid is also given.

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4. THE RELATIONSHIP BETWEEN THE PRESSURE AT BOILING POINT ANDTHE VAPOUR PRESSURE OF THE LIQUID

If the temperature of the liquid is raised to the point where the vapour pressure of the liquid

becomes equal to the surrounding pressure, the liquid boils. Consequently, a liquid can be

caused to boil either by raising its temperature, or by lowering the ambient pressure. This

explains, for example, why water will boil at a lower temperature at higher elevations where

the atmospheric pressure is lower.

Liquids having relatively high vapour pressures will evaporate faster if the container is left

open. This is caused by the higher molecular mobility in liquids which have higher vapour

pressures.

For pure liquids, the relationships between temperature and vapour pressure are well known

and are predictable. Hence, it is possible to build a thermometer which is activated by the

vapour pressure of a liquid, and which is used, with a suitable correlation, to indicate

temperature.

5. VARIATION OF ATMOSPHERIC PRESSURE WITH ELEVATION

Since the density of air decreases with height, the air pressure at higher altitudes is

correspondingly lowered. The variation of atmospheric pressure with altitudes can be

calculated by considering various factors such as wind speed, temperature, altitude and

humidity. However, it can be approximated (within 20% error) by the expression :

Ps = Pe (1 + 0.000127E)

Where:

Ps is the pressure at sea level (101.325 kPa).

Pe is the pressure at elevation E.E is the elevation above sea level in meters.

6. HUMIDITY

The humidity of atmospheric pressure is a typical example of the presence of vapour

pressure. Any volume of atmospheric air at a specific temperature contains a certain amount

of water vapour. When the temperature of the air drops to a particular temperature, the water

vapour in the air will condense back to tiny droplets of water. The air is said to be saturated

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with water. The air temperature at which this occurs is called the dew point. The vapour

pressure of the water at saturation is called the saturation vapour pressure.

The humidity of air at any temperature is measured in terms of this saturation vapour

pressure. The percent humidity of air is defined as

%humidity =

v a p o u r p r e s s u r e a t

s a t u r a t e d v a p o u r p x 100%

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EXAMPLE1The room temperature is 22oC and the relative humidity is found to be 45%. The barometric

pressure is 95 kPa.a. Refer to the steam tables to find the saturated vapour pressure of water at 22 o(SI

databook p 24-38) .

IP 22 C

LRV i 20 C URV i 25C LRV o 2.339kPaa URV o 3.169 kPaa

SPAN i URV i LRV i

SPAN o URV o LRV o

y .

IP LRV i

SPAN iSPAN o LRV o

=y 2.671 kPaa

b. Find the partial pressure of the water vapour in the room.

Since humidity is defined as the ratio of water vapour pressure to saturated vapourpressure,that is,

Humidity ( )in_%water_vapour_pressure

saturated_vapour_pressure

p H2O.

0.45 y=p H2O 1.202 kPaa

c. At what temperature would the water vapour start to condense if the room air was cooleddown. This temperature is called the dew point temperature or just the dew point .

From the SI databook p 24-38, the vapour pressure of the room is found to lie betweenthose at 5C and 10C.

IP p H2O

LRV i 0.8721 kPaa URV i 1.2276kPaa LRV o 5C URV o 10C

SPAN i URV i LRV i

SPAN o URV o LRV o

y .IP LRV i

SPAN iSPAN o LRV o

=y 9.639C

That is, if the air is cooled to 9.639C, the water vapour in the air will turn to liquid water.In other words, the air is saturated with water vapour at 9.639C

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d. The room contains an air compressor which produces instrument air at 800 kPaa. Whatis the partial pressure of the water vapour in the high pressure air line from the compressor?

By direct proportions, at 95 kPaa, the vapour pressure is 1.202 kPaa; therefore, at 800kPaa the vapour pressure is given as

p H2O.800

951.202 kPaa

=p H2O 10.122 kPaa

e. Find the dew point temperature of the instrument air.

From the SI databook p 24-38, it is found that the temperature of the calculated pressureobtained in (d) is between 45C and 50C.

IP p H2O LRV i 9.593kPaa URV i 12.349kPaa

LRV o 45C URV o 50 C

SPAN i URV i LRV i

SPAN o URV o LRV o

y .IP LRV i

SPAN iSPAN o LRV o

=y45.96

C

f. What is the highest air pressure allowable in the air line at 22oC so that condensationcould not occur?As the air is compressed, the water vapour pressure will rise along with the air pressure.The maximum water vapour pressure which can be attained at 22C without condensationis 2.671 kPaa. It is given that at 95 kPaa air pressure, the water vapour pressure is1.202 kPaa.

By direct proportions, the highest air pressure is

p highest.95

2.671

1.202kPaa

=p highest 211.10 kPaa

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EXAMPLE 2

PHASE 1


A barometer can be made by inverting a glass tube completely filled

with a liquid. The liquid column will fall to a certain height at which the

atmospheric pressure can sustain. The empty space above the liquid

surface is ideally in vacuum. However, due to temperature changes,

some of the liquid always vaporizes and fills the vacuum space. A

vapour pressure is exerted on the column of liquid so that the height of

the column does not represent the actual atmospheric pressure.

Corrections to the reading of the column has to be made in order to

account for this.

If the liquid used in the column is mercury, find(a) the error in pressure

(b) the height of the mercury column if the atmospheric pressure is 100

kPaa.

SolutionData

(a) The error is caused by the vapour pressure of mercury. At

25C the vapour pressure of mercury is 252 mPaa which is the

error introduced to the atmospheric pressure.

(b) The height of the mercury column only represents anatmospheric pressure of

p atm.100 kPaa

p vapour_Hg.252 mPaa

Hg .13534kg

m3

p apparent p atm p vapour_Hg(databook p.19)

=p apparent 100 kPaag .9.81

m

sec2

Since p=gh, the height of the mercury column is

hp apparent

.Hg

g

=h 0.753 m


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EXAMPLE3

The fact that liquids at their prevailing temperature will boil and create vapour if their pressure

is reduced to a low enough value can lead to problems in pumps, valves, and flow elements.

This effect is called flashing. If the vapour formed recondenses downstream, the effect is called

cavitation. Both vaporization and cavitation can cause damage to plant equipment and piping.

For the following questions, use the average barometric pressure at the Northern Alberta

Institute of Technology (NAIT).

a) Acetone at 40oC enters a valve. If the inlet pressure is 30 psig, what will be the

maximum allowable pressure drop across the valve if flashing is to be avoided ?

b) The pressure at the suction of a methyl alcohol pump is -60 kPag. What is themaximum temperature which the alcohol can have before the pump begins to draw

vapour as well as liquid ?

o u on a a

NAIT average pressure3. a.

pat.93.5kPa

pinlet..30psi

.6.89

kPapsi

patvapour pressure of acetone at 40 deg C

=pinlet 300.3

kPa absolute pv.8kPa (databook p.29)

the permissible pressure drop ispsuction.

.60kPa=pinletpv 232.3

kPa

3b. ppumppat

.0kPa

=ppump 33.5kPaabsolute

the temperature that results in that methyl alcohol vapour pressure is about

T .39degC (from databook p.29)

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PROBLEMS

1. Determine the gauge pressure or temperature, as required, in the following cases. The

atmospheric pressure is reported to be 93.65 kPaa.

a) Butane in a cigarette lighter is in the liquid phase. If the temperature is 22oC,

what is the pressure inside the lighter?

b) Propane is normally stored in liquid form. What will be the pressure of the

propane vapour at 20oC?

c) If a propane cylinder has a maximum pressure rating of 2000 kPag, what is the

maximum temperature which it can tolerate?

d) What pressure will be necessary to liquefy carbon dioxide at a temperature of

15oC?

e) At what temperature would Freon 11 boil under normal atmospheric pressure at

the location?

2. The boiling point (temperature) of liquids depends on the prevailing pressure. For

water, this temperature/pressure relationship is plotted with the graphs of vapourpressure vs. temperature for various liquids. It is also shown in the tables for steam,

and for the physical properties of water. Use both data sources to determine the answers

to the following questions.

a) At what temperature will water boil if the barometer reads 689 mm of mercury

with a temperature of 20oC?

b) At what temperature would water boil at the top of Mt. Robson, 12972 feet

above sea level? The air pressure at the top of Mt. Robson is 62.7 kPaa (as given

by the chart for geopotential altitude versus pressure scale from Environmental

Canada).

c) If water were used in barometers instead of mercury, what would be the error in the

barometer reading due to vapour pressure at a temperature of 23oC?

d) What will be the error due to vapour pressure when the barometer contains mercury?

(The vapour pressure of mercury at 23oC is 0.0015 mm mercury)

3. The fact that liquids at their prevailing temperature will boil and create vapour if their

pressure is reduced to a low enough value can lead to problems in pumps, valves, and

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flow elements. This effect is called flashing. If the vapour formed condenses again

downstream, the effect is called cavitation. Both vaporization and cavitation can cause

damage to plant equipment and piping. For the following questions, use the averagebarometric pressure at Edmonton.

a) Acetone at 60oC enters a valve. If the inlet pressure is 20 psig, what will be the

maximum allowable pressure drop across the valve if flashing is to be avoided ?

b) The pressure at the suction of a methyl alcohol pump is -74.5 kPag. What is the

maximum temperature which the alcohol can have before the pump begins to

draw vapour as well as liquid ?

4. A barometer is made by inverting a column of liquid in a

glass tube as shown. If the liquid was acetone at 25C andthe local atmospheric pressure was 97 kPaa, find

(a) the error in pressure introduced by the vapour

pressure of acetone

(b) the height of the column of acetone.

5. The room temperature is 33C and the relative humidity is found to be 75%. The

barometric pressure is 100 kPaa. Find

(a) the saturated vapour pressure of water at 33C

(b) the partial pressure of the water vapour in the room

(c) the temperature at which water vapour would start to condense if the

room air was cooled down

(d) the partial pressure of the water vapour in the high pressure air line from

the compressor if an air compressor is used to compress the room air into

instrument air at 1000 kPaa

(e) the dew point temperature of the instrument air

(f) the highest air pressure allowable in the air line at 33C so that

condensation could not occur.

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Chapter 3

GAS LAWS

GENERALIZATION

Gases behave quite differently from liquids due to their compressible nature. Their

physical behaviors were investigated. Theories and laws were deduced to predict how gases

change with temperature and pressure. Such laws provide guidance in determining the proper

control of gases used or produced in industrial plants. An understanding of how gases behave at

various temperatures and pressures is crucial in the process of plant operation or production.

GENERAL OBJECTIVES

The student will learn how to:

(a) explain the basic theory of gases and its relationship to the ideal gas law and general gas

law.

(b) determine volume, pressure, temperature or number of moles for gases under various

conditions.

(c) determine the density of gases with the general gas law.

(d) state the conditions for STP, API or "contract" conditions.

(e) explain the importance of calculating the volume or volume flow at STP or API conditions.

(f) calculate the reduced pressure and reduced temperature and then determine the gascompressibility factors by graph.

1. DEFINITION OF A MOLE

A mole of substance is the amount of substance that contains 6.022*1023 elementary units of

the substance. An elementary unit can be an atom or a molecule. The mass of one mole of

substances is expressed in units of g/mole or kg/kmol. For example, one mole of Carbon-12

(C12) contains 6.022*1023 carbon atoms and has a mass of 12 grams ; one mole of Hydrogen

(H2) contains 6.022*1023 molecules and has a mass of 2 grams.

2. BASIC THEORY OF GASES

The basic kinetic-molecular theory of gases is the basis for microscopic analysis of gas

behavior. The basic theory assumes:

(a) gases are made up of minute particles called molecules whose dimensions are very

small as compared with the average distances between them.

PHASE 1



21/75

(b) the molecules of gases are in constant rapid motion colliding with each other and with

the walls of their containers in a perfectly random fashion.

(c) all molecular collisions areperfectly elastic, resulting in no decrease in the total kinetic

energy.

(d) the average kinetic energy of the molecules of a gas is proportional to the absolute

temperature.

(e) equal volumes of all gases at the same pressure and temperature contain the same

number of molecules: i.e., 22.414 m3 of gas at 101.325 kPaa and at 273.15o K contain

6.022 x 1026 molecules (1 kilomole) of gas and have a mass(kg) equal to its molecular

mass.

(f) that there are no intermolecular forces between molecules.

(g) that the volume occupied by the molecules is negligible compared to the volume of the

container.

Statistical mechanics could be used to illustrate that the gas molecules in a container of

volume (V) will set up an absolute pressure (p) that is proportional to the number of

molecules (n kilomoles) in the container and their average kinetic energy or absolute

temperature (T).

3. IDEAL GASES

The gas equation which relates p, V, T of an ideal gas is:

pV=nRT eq. (3-1)

where:

V = volume of gas in m3

p = absolute pressure of gas in Pascals

T = absolute temperature of gas in K

R = universal gas constant = 8314.510

n = number of kilomoles of gas

Variations of this equation are:

PHASE 1



22/75

(a) Boyle's Law1

p1 V1 = p2 V2 = constant (constant T,n) eq. (3-2)

(b) Charles' Law2

= = constant (constant p,n) eq. (3-3)

EXAMPLE1

A storage tank attached to an air compressor has a volume of 0.65 m3 . It contains air at a

pressure of 150 psig. The temperature is 20 oC. The barometer reading at the location is 690

mm of mercury.

a. How many kilomoles of air are there in the tank?

b. What is the density of the air in the tank?

c. What is the mass of the air in the tank?

d. How many API cubic meters of air are in the tank?

1 Boyle's Law - the product of the pressure and volume of a given mass of gas is constant if the temperature and

the number of moles do not change.2 Charles' Law - the volume of a gas at constant pressure is proportional to its absolute temperature if the pressure

and the number of moles do not change.

PHASE 1



23/75

So lu t io n Datapressureconversion

p a tmos.. Hg g h h .0 . 6 9 m

=p a tmos 9 1 . 7 0 9 k Pa Hg.1 3 5 4 6 . 3

k g

m3

(p .19 da taboop g a u g e

....1 5 0lb

in24 . 4 4 8

N

lb

.1 in

..2 .5 4 1 0 2 m

2

g .9 . 8 1 1 6 9m

sec2=p g a u g e 1 . 0 3 4 1 0

3k Pa

V 1.0 . 6 5 m3

p ab s p g a u g e p a tmos

T 1.2 0 de gC .2 7 3 K

=p ab s1 . 1 2 6 1 0

3

k Pa =T 1 2 9 3 K

R .8 . 3 1 4 4k J

.kmolKa. n

.p ab s V 1.R T 1

=n 0 .3 k mo l M .2 8 . 9 6 2 5k g

k m o l(p.23-2 Sda ta book

p std.1 0 1 . 3 2 5k Pa

b .

.p ab s M

.R T 1= 1 3 . 3 8 5

k g

m3 T st d.1 5 de gC .2 7 3 K

c. mass.

V 1 =mass8 .7

k g =T st d2 8 8

K

d .V std

..n R T st d

p st d=V st d 7 . 0 9 9 m

3

4. REAL GASES

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The ideal gas equation assumes no interactive forces between molecules. Such forces, called

Van der Waals forces do exist and become more significant as the average kinetic energy ofthe molecules becomes less (lower temperature) and/or the molecules are forced closer

together (higher pressures). This can be illustrated by the fact that all gases can be liquefied at

a unique temperature and pressure. Liquefaction occurs only when the cohesive forces which

tend to bind the molecules together counteract their kinetic energy to a sufficient extent to

keep the molecules confined to a relatively small volume. These interactive forces between

gas molecules cause the gas equation to deviate from the ideal one.

The pVT behavior of real gases can be analyzed by using the general equation:

pV = ZnRT eq. (3-4)

In this equation, Z, the compressibility factor shows deviation from ideality. Z= 1.0 for an

ideal gas and a departure from ideality will be measured by the deviation of Z from unity.

Real gases at low pressure and high temperature have a Z value close to 1. More

specifically Z will be close to 1.0 whenever the temperature of the gas is much greater than

its critical temperature and its pressure is much lower than its critical pressure.

The extent of the deviations from ideality depends on the gas absolute temperature and

pressure and on the gas critical temperature and critical pressure. The gas critical

temperature3 (TC) is the highest temperature at which a gas can still be liquefied and the gas

critical pressure (pC) is the pressure necessary to liquefy the gas at T

C.

One of the most convenient ways to find Z for any gas is to use available gas compressibility

(Z) graphs. To use these graphs we find the reduced pressure (pR) and the reduced

temperature (TR) and then use these to find Z from the available graphs (SI Engineering

Data Book p. 23-12 to 23-13).

pR = where: p is the absolute gas pressure

pC

is the gas critical pressure

TR = where: T is the absolute gas temperature

TC is the gas critical temperatureZ = f(T

R, p

R)

3The critical temperature and pressure of a substance have to do with the change of state of the substance between its gas andliquid phases.

The critical temperature Tc is the more fundamental one. If the temperature of a gas is greater than Tc, then no amount of

applied pressure will be able to cause it to change to its liquid state.

The critical pressure pc is the pressure corresponding to Tc on the T-p equilibrium curve (i.e., the vapor pressure curve). This

point is called the "Critical Point" (NB: the critical temperature establishes the critical point). There is also a critical density

c corresponding to the critical point.

Critical temperature, pressure, and density are definable only for pure substances, not for mixtures.

PHASE 1



25/75

EXAMPLE2

A 45 litre cylinder of oxygen stored at a temperature of 18 oChas a gauge pressure of 15MPag.

The barometric pressure is 98 kPaa.

a. What is the gas law deviation factor Z at the cylinder pressure and temperature?

b. How many kilomoles of oxygen are in the cylinder?

c. What is the density of the oxygen in the cylinder?

d. What is the mass of the oxygen in the cylinder?

e. How many API m3 of oxygen are in the cylinder?

Data

o u on a a

T .273K .18degCa. TRT

Tc

=TR 1. 5

=T 255 K

pR

ppc

patmos.98kPa=pR 2.99

p..1510

3 kPa patmosZ 0.84

(from SI data book p.23-12)=p 1.5110

4 kPab. n

.p V..Z R T

=n 0.37

kmolV ..4510

3 m3

Tc.154.5

K (p.23-2 SIdb)c.

.p M.

.Z R T

= 269.66

kg

m3 pc.504

kPa (p.23-2 SIdb)

d. mass .n M =mass 12.13

kgM .31.99

kgkmol

(p.23-2 SIdb)

e. Zstd 0.9992Tstd

.288K pstd.101.32

kPa

Vstd

...Zstdn R Tstd

pstd

=Vstd 8.96

m3

FINDING Z FACTOR BY REDLICH-KWONG METHOD

Another way of finding the Z factor is to use the Redlich-Kwong method4. In this method, the

Z factor is solved from the equation:

Z3- Z2 - (B2 + B -A)Z - AB = 0 eq. (3-5)

where Z is the compressibility factor

A = 0.42748*, B = 0.086647 *

EXAMPLE3

4 Flow measurement Engineering Handbook, R.W. Miller, 2nd Edition, 1983. p. 2-18

PHASE 1



26/75

(same as example 2 except that Redlich Kwong's method was used for finding the Z factor)

A 45 litre cylinder of oxygen stored at a temperature of -18o

C has a gauge pressure of 15MPag.The barometric pressure is 98 kPaa.

a. What is the gas law deviation factor Z at the cylinder pressure and temperature ?

b. How many kilomoles of oxygen are in the cylinder ?


d. What is the mass of the oxygen in the cylinder ?

e. How many API m3 of oxygen are in the cylinder ?

o u on a a

a. TR

T

Tc =TR 1. 5 T.273 K

.18degC

=T 255KpR

ppc

=pR 2.99

patmos.98kPa

Guess p ..15103 kPa patmosZ 0.9

=p 1.51104 kPa

For part a, using Redlich-Kwong method,V ..4510

3 m3

A .0.4274

pR

TR

2.5Tc

.154.5

K (p.23-2 SIdb)=A 0.3

pc .504

kPa (p.23-2 SIdb)

B .0.08664

pRTR

=B 0.15

M .31.998

kgkmol (p.23-2 SIdb)

Using the calculator to solve Z from the equation, we get:

Z3 Z

2 .B2 B A Z .AB 0

=Z .8

b. n .pV..ZRT=n 0.37

kmol

c.

.pM..ZRT

= 2 3.8kg

m3

d. mass .nM =mass 11.87 kg

e. Zstd 1.0Tstd

.288K pstd.101.32

kPa

Vstd

...ZstdnRTstd

pstd=

Vstd 8.7

m

3

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27/75

This method is used explicitly for non-hydrocarbon products. However, when the Z value is

high enough, namely, greater than 0.8, both the graph method or the Redlich-Kwong method

give quite close values.

5. AVOGADRO'S HYPOTHESIS

Equal volumes of different ideal gases at the same pressure and temperature contain the same

number of molecules.

6. GAS LAW CALCULATIONS

(a) Mass of gas

The mass of gas in a container cannot change without a leak; therefore, it is independent

of p, V, T, and Z. It is the product of n (in kilomoles) and the molecular mass ( M ) of

the gas.

i.e.: Mass = nM eq. (3-6)

(b) Density of gas.

density = = eq. (3-7)

(c) Volume of gas at STP conditions

The normal standard temperature and standard pressure are 0 0C (273.15K) and

101.325 kPaa respectively. The volume of 1 kmol of gas at these conditions, using the

real gas law, can be worked out to be :

VSTP = 22.414 ZSTP (m3 STP) eq. (3-8)

(d) Volume of gas at API conditions

The API (American Petroleum Institute) conditions refer to a base pressure of 101.325

kPaa and a base temperature of 15 oC (288.15 K). In this course, API conditions will

always be used as the standard reference. If you ever encounter the word standard

in this manual, it will mean API conditions. The simplified real gas law is :

VAPI = 23.645 ZAPI (m3 API) eq. (3-9)

The value of ZAPI at 15C is given in the SI data book p.23-3.

(e) Process of iteration

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In certain real gas law problems, the gas equation might involve two unknowns. An

example of this is to find the pressure of a sealed tank of gas. In such a case, thepressure, p, cannot be solved explicitly since the Z factor is pressure dependent. Iteration

has to be used. The iteration process involves an initial guess of one parameter (namely,

Z) and find the other parameter (namely, p). Once the new value of p is found, a more

accurate value of Z can be found. As soon as a more accurate value of Z is found, p can

be recalculated again. This process is repeated until the values of p repeats itself or

become very close.

EXAMPLE 4

A 20 litre gas cylinder contains 2 kg of methane at 13 oC. What will be the gauge pressure of

the gas if the cylinder is stored in J32 which has an atmospheric pressure of 92.03 kPaa? (Note:

this problem requires a trial and error approach. Start with Z = 1 and adjust Z with each trial

until the calculated value of Z matches the estimated value.)

V

TR

M

mZp

= =

31020

13)(273.158314.51

16.043

2.00Z

+ = Z14.8301255106 Pa

pc= 4599 kPa

Tc= 190.56 K

Z P=Z14830.12 kPa

4599

pp R =

190.56

286.15TR =

Znew

1.00 14830.12 kPaa 3.2246 1.5016 0.797099

0.797099 11821.08 2.5703 1.5016 0.809113

0.809113 11999.26 2.6091 1.5016 0.807879

0.807879 11980.96 2.6051 1.5016 0.808003

0.808003 11982.78 2.6055 1.5016 0.807991

0.807991 11982.61 2.6055 1.5016 0.807992

0.807992 11982.62 kPaa

pg = 11982.62 kPaa + 92.03 kPa = 12 074.65 kPag

PHASE 1



29/75

PROBLEMS

1. A balloon is filled with air. At a pressure of 100 kPaa and a temperature of 18 oC, its

volume is 0.522 m3. Calculate its new volume under the following different conditions.

a. It is placed in a vacuum chamber and the pressure is reduced to 32 kPaa (the

temperature remains at 18 oC).

b. It is placed in a pressurized chamber and the pressure is increased to 550kPaa at

the same temperature.

c. It is heated in an oven to a temperature of 95 oC, at a pressure of 100,000 kPaa.

d. It is cooled in a refrigerator to -70 oC, at 100 kPaa.

e. It is cooled to -40 oC, and its pressure is increased to 150 kPaa.

f. It is heated to 50 oC, and its pressure is decreased to 75kPaa.

2. A room is 4.0 m wide by 7.6 m long by 2.8 m high. A barometer located in the room

reads 701.3 mm mercury and the temperature is 23 oC.

a. How many kilomoles of air are there in the room?

b. What is the mass of the air in the room?

c. What is the density of the air in the room? Calculate the density of air using the

ratio or the formula.

d. What would the volume of the air be if it were at a temperature of 15

o

C and101.325 kPaa? This is the number of API meters (m3 API).

3. A storage tank attached to an air compressor has a volume of 0.455 m3 . It contains air

at a pressure of 105 psig. The temperature is 20 oC. The barometer reading at the

location is 707.2 mm of mercury.

a. How many kilomoles of air are there in the tank?

b. What is the density of the air in the tank?

c. What is the mass of the air in the tank?

d. How many API cubic meters of air are in the tank?

4. 6.28 sm3 (standard cubic meters at API conditions) of nitrogen are compressed into a

tank at 54 oC. The volume of the tank is 0.0625 m3.

a. How many kilomoles of nitrogen are in the tank?

b. What will be the absolute pressure of the nitrogen in the tank?

c. What will a pressure gauge on the tank read if the Edmonton weather office

reports an atmospheric pressure of 92.91 kPaa?

PHASE 1



30/75

d. What is the density of the nitrogen in the tank?

5. For nitrogen, calculate the reduced temperature and reduced pressure in each of thefollowing cases. Then determine the gas law deviation factor (compressibility factor) Z,

from the Gas Compressibility charts and the Redlich-Kwong method.

a. A pressure of 100 kPaa, and a temperature of 20 oC.

b. 100 kPaa and 300 oC.

c. 5.00 MPaa and 20 oC.

d. 45.0 MPaa and 0 oC.

e. 6.0 MPaa and 130 oC.

6. Calculate the reduced temperature and the reduced pressure for the following gases at

the specified p and T, then determine the gas law deviation factor (compressibility

factor) Z.

a. Air at a pressure of 7.5 MPaa, and a temperature of 55 oC.

b. Hydrogen at 15.0 MPaa and 200 oC.

c. Propane at 5.0 MPaa and 200 oC.

d. Methane at 10.0 MPaa and 50 oC.

e. Carbon Monoxide at 30.0 MPaa and 10 oC.

7. A 32 litre cylinder of oxygen stored at a temperature of 25 oC has a gauge pressure of

21.7 MPag. The barometric pressure is 91.63 kPaa.

a. What is the gas law deviation factor Z at the cylinder pressure and temperature ?

b. How many kilomoles of oxygen are in the cylinder ?


d. What is the mass of the oxygen in the cylinder ?

e. How many m3 API of oxygen are in the cylinder ?

8. A tank contains 25 kg of nitrogen at a temperature of 10 oC and a pressure of 5800

kPag. The barometer is steady at the average Edmonton value.

a. What is the value of Z at the tank pressure and temperature? What is the gas law

deviation factor Z at API pressure and temperature.

b. How many kilomoles of gas are in the tank?

c. Calculate the volume of the tank.

d. What is the density of the gas in the tank?

e. How many API m3 of nitrogen are in the tank?

PHASE 1



31/75

9. A 35 litre gas cylinder contains 4.25 kg of methane at 25 oC. What will be the gauge

pressure of the gas if the cylinder is stored in J32 which has an atmospheric pressure of

92 kPaa? (Note: this problem requires a trial and error approach. Start with Z = 1 andadjust Z with each trial until the calculated value of Z matches the estimated value.)

10. Propane gas flows in a pipeline at a rate of 55 . The line pressure and temperature are

measured to be 1025 kPag and 115oC respectively. The atmospheric pressure is 95 kPaa.

a. What is the gas law deviation factor of the propane at pipeline conditions?

b. What will be the density of the propane in the pipeline?

c. What is the flow rate in ?

d. What is the mass flow rate of the propane?

e. What is the flow rate of the propane in API ?

11. The compressed air main in a refinery consists of 100 m of 6 inch Schedule 40 steel

pipe. 22 of dry air are compressed into the line at 825 kPag and 20 oC. The air leaves

the line at the far end at 700 kPag and 15 oC. The atmospheric pressure is 92.7 kPa.

a. What is the flow rate of the air in kmol/s?

b. What will be the density of the air at the inlet and the outlet of the main?

c. What property of the (air) stream remains constant all along the main?

Chapter 4

THERMAL PROPERTIES OF GASES

GENERALIZATION

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When thermal energy is supplied to a gas, the energy is absorbed by the gas molecules

of which the kinetic energy increases correspondingly. Consequently, the temperature of the

gas increases as a result of an increase in internal energy. The energy per kilogram of the gascan be increased under constant volume or constant pressure. The ratio of these two types of

energies is defined as the isentropic constant, which is used in orifice flow metering and sizing.

GENERAL OBJECTIVE

The student will be able to:

(a) explain the energy equation for heat transfer in terms of the two specific heat constants (c Vand cP)

(b) derive the relationship for cP knowing cV, R and M.

1. SPECIFIC HEAT AT CONSTANT VOLUME (cV)

The quantity of heat necessary to change the temperature of m kg of gas enclosed in a fixed

volume from T1 to T2 is:

Q = cV m ( T2 - T1 ) eq.(4-1)

where

cVis the amount of heat necessary to raise the temperature of 1 kg of constant volume

gas by 1Co or 1 Ko. It is measured in units of oCkg

J

or oKkg

J

. Note that the

notation Co and Ko refer to a temperature range, not the actual temperatures. Actual

temperatures are measured in oC (degrees celcius) or K (kelvins)

Since the volume does not change, no external work is done by this system. All the heat

added/removed is gained /lost to the internal energy in the gas.

2. SPECIFIC HEAT AT CONSTANT PRESSURE (cP)

The heat necessary to change the temperature of m (kg) of gas contained under a constant

pressure (P) from T1 to T2 is :

H= cP m (T2 - T1) eq.(4-2)

where

PHASE 1



33/75

cP is the amount of heat necessary to raise the temperature of 1 kg of gas at constant

pressure by 1 Co or 1 Ko. It is measured in units ofoCkg

J

or oKkg

J

..

In order to keep the gas pressure constant, the volume of the container must change to

accommodate for gas expansion due to a rise in gas temperature. As a result, external work is

done by the additional heat. The external work done is p V where V is the change involume due to the temperature change.

Fig. 4.1

The total heat supplied, H is (assuming no heat losses)

H= Q + external work at constant pressure

cPm t = cVm t + p V

cP - cV = Tm

Vp

Using the gas law to replaceTm

Vp

by , the equation becomes

cP- cV = eq.(4-3)

EXAMPLE 1:

For monatomic helium M = 4.0026 , cp = 5193.1 oCkg

J

(p.23-3 SI databook)

PHASE 1



34/75

cv = 5193.1 oCkg

J

- = 3115 oCkg

J

Note that in the SI databook, only cp is given.

The ratio, is defined as the isentropic constant, k or . This constant will be used in the orificeflow metering section.

EXAMPLE2

Look up the value of the molar mass, and the specific heat at constant pressure cP( oCkg

J

) for

chlorine. From this data, calculate the value of the specific heat at constant volume cV, and the

ratio of the specific heats k)(cc

V

P =

Solution Data

C v C pR

MM .70.9054

kg

kmol(SI databook ,p.23-2

C p.476

J.kg K=C v 358.74

J.kg K

k

C p

C v =k 1.33

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EXAMPLE 3

The temperature of a 70 m3 steel tank contains air at 20 oC. The barometer is steady at the

average Edmonton value.

a. How much heat is required to raise the temperature of the air to 30 o C, assuming that the tank

is open to the atmosphere?

b. How much heat is required to raise the temperature of the air to 30 oC, assuming that the tank

is completely sealed?Solution Data

For ideal gas, Z 1 M .28.9625 kgkmol

(p.23-2 SIdb)

Finding specific heat at constant volume:p atmos

.93.44 kPa

V .70 m3

m air

..p atmos V M

..Z R T

T .20 degC .273 K

=m air 77.76 kg

=T 293 K

C v C pR

MC

p

.1004J

.kg K (p.23-2 SIdb)=C v 716.93

J

.kg K t 2.30 degC

a. Q ..m air C p ( )t 2 t 1 t 1.20 degC

=Q 780. 73 kJ

b. H ..m air C v ( )t 2 t 1

=H 557.5 kJ

PHASE 1



36/75

PROBLEMS

1. For the following gases, look up the value of the molecular mass (weight), and the

specific heat at constant pressure cP( oCkg

J

). From this data, calculate the value of the

specific heat at constant volume cV, and the ratio of the specific heats k)(c

c

V

P=

a. Air

b. Methane

c. Ethane

d. Propane

2. A storage room is 4 m wide by 7 m long by 3.2 m high. Its temperature is 18 oC. The

barometer is steady at the average Edmonton value.

a. Determine the specific heat of air at constant pressure, and at constant volume.

b. How much heat is required to raise the temperature of the air to 28 oC, assuming

that the room is open to the atmosphere?

c. How much heat is required to raise the temperature of the air to 28 oC,

assuming that the room is completely sealed (keeping the volume constant)?

3. An air cylinder with a movable piston is used to perform work at constant pressure. The

pressure in the cylinder is 50 MPaa at 15 C. The initial volume of the cylinder is 0.5 L.

It is heated from 15C to 100C. The volume of the air expands in order to keep the

pressure constant. Consider the air to be an ideal gas.

(a) How much work does the piston do?(b) If the diameter of the piston is 10 cm, how much does the piston move from its

original position?

4. Repeat Question #3 with air being a real gas.

(a) Find the compressibility factor of air at 100oC

(b) How much work does the piston do?

(c) If the diameter of the piston is 10 cm, how much does the piston move from its

original position?

PHASE 1



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Chapter 5

MIXTURES OF GASES

GENERALIZATION

In many situations, the gas products are mixtures of gases. Such mixtures possess

different properties than the pure ones. The density behaviour of a mixture of gases is similar to

that of a pure gas. The pressure and temperature characteristics of a mixture can be measured in

the same way as those of a pure gas. However, specific gravity, compressibility and specific

heat must be calculated by certain mixture laws.

GENERAL OBJECTIVES

The student will be able to:

(a) explain and use Dalton's law of partial pressure.

(b) explain the significance of mass, volume and mole fraction and convert from one to the

other.

(c) calculate the compressibility factor for a gas mixture.

(d) solve for two unknowns in a gas problem by iteration.

(e) determine the quantities and pressures of pure gases required to make a calibration gas at a

particular pressure and composition.

1. MOLE FRACTION, VOLUME FRACTION AND MASS FRACTION

Under identical conditions of temperature (> 0 oC) and pressure (< 10 atmospheres), equal

volumes of any gas mixture contain equal number of molecules. That is, one kilomole of any

gas mixture contains 6.022 x 1026 molecules and has a mass in kilograms which is equal to its

molar mass, M.

The composition of a gas mixture can be expressed by either specifying the mole fraction,

volume fraction or the mass fraction of each gas component.

a. Mole fraction

Mole fraction is the most fundamental. It is defined as the ratio of the number of moles of a

gas component, n1, in a gas mixture to the number of total moles, nt, of the gas mixture.

x B1 = eq.(5-1)

PHASE 1



38/75

EXAMPLE 1

At a dance there is a group of 20 football players and another group of 15 cheerleaders. Find

the molar fraction of each.

nfootballplayers = 20

ncheerleaders = 15

ntotal = 35

xfootballplayers = 20/35 = 0.571 = 57.1%

xcheerleaders = 15/35 = 0.429 = 42.9 %

In other words 57.1% of the persons at the dance are football players

b. Mass fraction

Mass fraction is defined as the ratio of the mass of a gas component, m1, in a gas mixture to

the total mass, mt, of the gas mixture.

WB1 =t

1

m

meq.(5-2)

In real gas situations, the mass fraction and the mole fraction differ by the ratio of their molar

masses. It can be shown that

B1

1

t

t

1

1

tB1 W

M

M

m

m

M

Mx == eq.(5-3)

EXAMPLE 2

At a dance there are 20 football players and 15 cheerleaders. Each player weighs 85 kg and

each cheerleader weighs 55 kg. Find the mass fraction of each group.

mplayers =20 * 85 kg/player

= 1700 kg

Mcheerleaders = 15 * 55 kg/cheerleader

= 825 kg

Mt = 1700 kg + 825 kg = 2525 kg

Wplayers = 1700 kg / 2525 kg = 0.673 = 67.3 %

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Wcheerleaders = 825 kg / 2525 kg = 0.327 = 32.7 %

In other words although the football players make up 57.1% of the number of persons theymake 67.3% of the total mass.

c. Volume fraction

Volume fraction is defined as the ratio of the volume of a gas component, V1, in a gas

mixture to the total volume, Vt, of the gas mixture.

B1 = t

1

V

V

eq.(5-4)

The volume V1 is the volume of component 1 at the total pressure of the mixture.

In ideal gas situations, the volume fraction, B1, and the mole fraction, x B1, are identical sinceall Z's are 1. However, in real gas situations, the volume fraction and the mole fraction differ by

the ratio of their compressibility factors. It can be shown, using the real gas law, that

t

1

1

t

B1 V

V

Z

Zx = =

1

t

Z

Z B1 eq.(5-5)

EXAMPLE 3

At a dance there are 20 football players and 15 cheerleaders. Each player weighs 85 kg and

each cheerleader weighs 55 kg. The density of a muscular young man is 999 kg/m3, and the

density of a cheerleader is 995 kg/m3. Find the volume fraction of each group.

= m/V

V = m/

Vplayers = (20 * 85kg/player)/999 kg/m3= 1.7017 m3

Vcheerleaders = (15 * 55 kg/cheerleader)/995 kg/m3= 0.8291 m3

Vt = 1.7017 m3 + 0.8291 m3 = 2.5308 m3

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players = 1.7017 m3 / 2.5308 m3 = 0.6724 = 67.24 %

cheerleaders = 0.8291 m3 / 2.5308 m3 = 0.3276 = 32.76 %

Note that in the case of a gas mixture the volume fraction and the mass fraction will generally

be very different from one another.

2. PHYSICAL PROPERTIES OF A GAS MIXTURE

(1) MOLAR MASS

The molar mass of a pure gas is a physical property associated with the gas. However, the

molar mass of a gas mixture does not exist, but it can be computed based on the molar

masses of individual gas components. The computed value provides us a convenient way of

finding other physical properties such as the critical pressure or volume for the gas mixture.

Suppose a gas mixture is made up of several pure gases which have molar masses, M1, M2,

M3, etc. The corresponding portion of each individual gas is measured by the number ofmoles, namely, n1, n2, n3, etc.

The total number of moles of mixed gas, nt = n1 + n2 + n3 +...

The total mass of gas mixture, mt = n1 M1 + n2M2 + n3M3 + ...

Let the molar mass of the mixture be Mmix, then, the total mass of the gas mixture

mt = ntMmix

Comparing the expressions for the total mass of the mixture, we have

ntMmix = n1M1 + n2M2 + n3M3 + ...

Dividing the expression by nt,

Mmix = M1 + M2 + M3 + ...

or Mmix = xB1M1 + xB2M2 + xB3M3 + ... eq.(5-6)

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Furthermore, the mass fraction of individual gas component can be stated as:

W B1 =mix

M

1M

x B1 eq.(5-7)

(2) SPECIFIC HEAT AT CONSTANT PRESSURE

The specific heat of a gas mixture at constant pressure can be approximated as the sum of

the products of the mole fraction and specific heat of each individual component. (p.2-103,

Miller)

cpmix = x B1 cp1 + x B2 c p2 + x B3 c p3 + ... eq.(5-8)

and the specific heat at constant volume can be obtained from the equation :

cvmix = cpmix -mixM

R eq.(5-9)

3. DALTON'S LAW OF PARTIAL PRESSURES

The pressure exerted independently by any single gas component in a mixture of gases is

called its partial pressure. Dalton's law of partial pressures states that "Each gas in a gaseous

mixture exerts a partial pressure equal to the pressure which it would exert if it were the only

gas present in the same volume. The total pressure of the mixture is the sum of the partial

pressures of all the component gases." This law may be used to calculate the total pressure

(ptotal) or the total volume (Vtotal) of a gas mixture.

ptotal = p1 + p2 + p3 +. . . eq.(5-10)

where p1

, p2, p

3

are partial pressures which individual gases would exert if contained alone in

the same volume and at the same temperature as the mixture.

Vtotal= V1 + V2 + V3 + . . . eq.(5-11)

where V1, V2, V3, are volumes which individual gases would occupy if present alone at the

given temperature and at the total pressure of the mixture.

The partial pressure and partial volume of a gas component can be found by using the mole

fractions. Namely,

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p1 = x B 1 pmix eq.(5-12)

V1 = x B 1 Vmix eq.(5-13)

4. PSEUDO-CRITICAL PROPERTIES OF GAS MIXTURES

In section 23 of the SI data book (p.23-10 to 23-30), a discussion of the Z factor is presented.

The computation of the compressibility factor, Z of a gas mixture is summarized as :

(1) first determine the pseudo-critical pressure (pC) and the pseudo-critical temperature (TC)

of the gas mixture

pC = x B 1 pC1+ x B 2 pC2 + x B 3 pC3 + . . . eq.(5-14)

TC = x B 1 TC1+ x B 2 TC2 + x B 3 TC3 + . . . eq.(5-15)

where

pC1, pC2, pC3 are critical pressures of each component.

TC1, TC2, TC3 are critical temperatures of each component.

(2) use the calculated pc, Tc values to determine the pseudo reduced pressure (pr) and the

pseudo reduced temperature (Tr).

Pseudo-reduced pressure, pr = where p is the absolute pressure of the mixture.

Pseudo-reduced temperature, Tr= where T is the absolute temperature of the mixture.

(3) From the compressibility graphs given in Sec. 23 (p.23-12) of the SI Engineering Data

Book, look up the value ofZ(pr, Tr) for hydrocarbons. For pure gases, use the Redlich-

Kwong method to actually calculate the Z factor.

5. GAS LAW CALCULATIONS

The general gas equation applies to all gas law calculations.

pV = ZnRT eq.(5-16)

Since some gas mixtures are not "ideal" at standard conditions, check Z and use it in your

calculations if it deviates from 1.000 by more than 0.001.

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EXAMPLE4

Calculate the mole percent of each of the components in the following gas mixtures. The

compositions have been provided on a mass percent basis as follows:

Nitrogen 40%

Carbon Dioxide 50%

Carbon Monoxide 10%.

Record the data in a table and calculate the weighted average of the molar mass:

gas Mass % M

N2 40 28.0134

CO2 50 44.01

CO 10 28.01

Mix 100 36.0196

where: 36.0196 kg/kmol = 0.40*28.0134 + 0.5*44.01 + 0.10*28.01

Using x B1 =1

Mmix

MWB1 calculate the composition based on moles

gas Mass % M Rel molN2 40 28.0134 51.43

CO2 50 44.01 40.92

CO 10 28.01 12.86

Mix 100 36.0196 105.20

Note that the sum of the mol % is not 100. Therefore we must normalise the values by

multiplying each by the normalisation factor20.105

100

gas Mass % Mrelativemol

Relative mol*20.105

100

Answer

N2 40 28.0134 51.43 48.87

CO2 50 44.01 40.92 38.91

CO 10 28.01 12.86 12.23

Mix 100 36.0196 105.20 100.0

You can check your answer by using W B1 =mix

M

1M

x B1 to

convert back to the original mass composition

gas mol % M Mass %

N2 48.87 28.0134 40.0

CO2 38.91 44.01 50.0CO 12.23 28.01 10.0

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Mix100.0 34.25 100.0

EXAMPLE5

A gas mixture consists of 70% methane, 20 % ethane and 10 % propane by mol fraction. Find

the proportions by volume fraction.

Write the data in tabular form:

gas molar% M Tc Pc

CH4 70 16.04 190.6 4599

C2H6 20 30.07 305.4 4880C3H8 10 44.1 369.8 4240

Mix 100

Calculate the weighted averages

gas molar% M Tc Pc

CH4 70 16.04 190.6 4599

C2H6 20 30.07 305.4 4880

C3H8 10 44.1 369.8 4240

Mix 100 21.65 231.5 4619.3

Calculate the Zs using Redlich-Kwonggas molar% M Tc Pc Z

CH4 70 16.04 190.6 4599 0.9201

C2H6 20 30.07 305.4 4880 0.2395

C3H8 10 44.1 369.8 4240 0.1942

Mix 100 21.65 231.5 4619.3 0.8336

Calculate the Vol % using B1= B1mix

1 xZ

Z

gas molar% M Tc Pc Z rel vol

CH4 70 16.04 190.6 4599 0.9201 77.26C2H6 20 30.07 305.4 4880 0.2395 5.75

C3H8 10 44.1 369.8 4240 0.1942 2.33

Mix 100 21.65 231.5 4619.3 0.8336 85.34

Sum not 100

Normalise by multiplying by 100/85.34

gas molar% M Tc Pc Z rel vol vol%

CH4 70 16.04 190.6 4599 0.9201 77.26 90.54

C2H6 20 30.07 305.4 4880 0.2395 5.75 6.73

C3H8 10 44.1 369.8 4240 0.1942 2.33 2.73

Mix 100 21.65 231.5 4619.3 0.8336 85.34 100.00

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Done!

EXAMPLE6

A gas mixture consists of 90.54 % methane, 6.73 % ethane and 2.73 % propane by volume

fraction. Find the proportions by mol fraction.

Write the data in tabular form

gas vol% M Tc Pc

CH4 90.54 16.04 190.6 4599

C2H6 6.73 30.07 305.4 4880

C3H8 2.73 44.1 369.8 4240

Mix 100.00

Calculate the weighted averages

gas vol% M Tc Pc

CH4 90.54 16.04 190.6 4599

C2H6 6.73 30.07 305.4 4880

C3H8 2.73 44.1 369.8 4240

Mix 100.00 17.75 203.2 4608.1

Calculate the Zs using Redlich-Kwong

gas vol% M Tc Pc Z

CH4 90.54 16.04 190.6 4599 0.9201C2H6 6.73 30.07 305.4 4880 0.2395

C3H8 2.73 44.1 369.8 4240 0.1942

Mix 100.00 17.75 203.2 4608.1 0.8980

Calculate the mol % using x B1= B11

mix

Z

Z

gas vol% M Tc Pc Z Rel mol

CH4 90.54 16.04 190.6 4599 0.9201 88.36

C2H6 6.73 30.07 305.4 4880 0.2395 25.24

C3H8 2.73 44.1 369.8 4240 0.1942 12.62

Mix 100.00 17.75 203.2 4608.1 0.8980 126.22

Sum is not 100

Normalise by multiplying by 100/126.22

gas vol% M Tc Pc Z Rel mol mol %

CH4 90.54 16.04 190.6 4599 0.9201 88.36 70.00

C2H6 6.73 30.07 305.4 4880 0.2395 25.24 20.00

C3H8 2.73 44.1 369.8 4240 0.1942 12.62 10.00

Mix 100.00 17.75 203.2 4608.1 0.8980 126.22 100.00

Done!

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EXAMPLE7

A gas mixture consists of 70% Methane, 18% Ethane, 2% Hydrogen Sulfide, and 10%Nitrogen, by mole fraction. 2 kg mass of this mixture is stored in a cylinder at a temperature of

25oC and a pressure of 10000 kPag. The barometer is steady at the average Edmonton value.

a. What will be the molar mass, the pseudo critical pressure, and the pseudo critical

temperature of the mixture?

b. What will be the compressibility factor (gas law deviation factor) of the mixture at its

storage P and T?

c. What is the volume of the cylinder?

d. How many API cubic meters of the gas mixture are contained in the cylinder?

e. What is the density of the gas in the cylinder at the conditions stated in the question?

SOLUTION

a & b. Write the data in tabular form and calculate the weighted averages

gas % mol M Tc Pc

CH4 70 16.043 190.56 4599

C2H6 18 30.07 305.41 4880

H2S 2 34.082 373.37 8963

N2 10 28.0134 126.21 3398

Mix 100 20.12568 208.4542 4616.76

Calculate Z of the mixture

gas % Vmol M Tc Pc Z

CH4 70 16.043 190.56 4599

C2H6 18 30.07 305.41 4880

H2S 2 34.082 373.37 8963

N2 10 28.0134 126.21 3398

Mix 100 20.12568 208.4542 4616.76 0.7864

c.32-

7m101.918

10009.1

29851.8314

126.20

2.000.7864

p

TRM

m*Z

V =

=

=Pa

KKkmol

J

kmolkg

kg

d. Calculate Zstd using Redlich-Kwong Zstd = 0.9972

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3

3stdm2.343

10325.101

28851.8314

126.20

2.009972.0

p

TRM

m*Z

V =

=

=Pa

KKkmol

J

kmol

kg

kg

e.332-

104.3101.917

2.00

V

m

m

kg

m

kg=

==

EXAMPLE8

A gas mixture consists of 70% Methane, 18% Ethane, 2% Hydrogen Sulfide, and 10%

Nitrogen, by volume fraction. 2 kg mass of this mixture is stored in a cylinder at a temperature

of 25oC and a pressure of 10000 kPag. The barometer is steady at the average Edmonton value.



b. What will be the compressibility factor (gas law deviation factor) of the mixture at

its storage P and T?



e. What is the density of the gas in the cylinder at the conditions stated in the question?

SOLUTION

a . Write the data in tabular form and calculate the weighted averagesgas vol% M Tc Pc

CH4 70.00 16.04 190.56 4599

C2H6 18.00 30.07 305.41 4880

H2S 2.00 34.082 373.37 8963

N2 10.00 28.0134126.21 3398

Mix 100.00 20.1236208.4544616.76

Calculate Z of the mixture and of each of the components

gas vol% M Tc Pc Z

CH4 70.00 16.04 190.56 4599 0.8525

C2H6 18.00 30.07 305.41 4880 0.3460

H2S 2.00 34.082 373.37 8963 0.1838

N2 10.00 28.0134126.21 3398 0.9888

Mix 100.00 20.1236208.4544616.760.7881

Calculate the mole fractions using xB1= 1

t

Z

Z B1

gas vol% M Tc Pc Z Rel mol

CH4 70.00 16.04 190.56 4599 0.8525 64.71

C2H6 18.00 30.07 305.41 4880 0.3460 41.00H2S 2.00 34.082 373.37 8963 0.1838 8.57

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N2 10.00 28.0134126.21 3398 0.9888 7.97

Mix 100.00 20.1236208.4544616.760.7881 122.26

Note that the sum is not 100Because the sum is greater than 100 so we must normalize the mol % by multiplying by100/121.74

gas vol% M Tc Pc Z rel mol mol %

C 70.00 16.04 190.56 4599 0.8525 64.71 52.93

CH4 18.00 30.07 305.41 4880 0.3460 41.00 33.54

C2H6 2.00 34.082 373.37 8963 0.1838 8.57 7.01

N2 10.00 28.0134126.21 3398 0.9888 7.97 6.52

Mix 100.00 20.1236208.4544616.760.7881 122.26 100.00

b. Find the weighted averages of M and Tc and Pc and finally find the Z of the mixturegas vol% M Tc Pc Z rel mo mol % M Tc Pc Z

CH4 70.00 16.04 190.56 4599 0.8525 64.71 52.93 16.04 190.56 4599

C2H6 18.00 30.07 305.41 4880 0.3460 41.00 33.54 30.07 305.41 4880

H2S 2.00 34.082 373.37 8963 0.1838 8.57 7.01 34.08 373.37 8963

N2 10.00 28.0134126.21 3398 0.9888 7.97 6.52 28.01 126.21 3398

Mix 100.00 20.1236208.4544616.760.7881 122.26 100.00 21.0021212.0094318.570.7641

c. 337

m1087.1710009.1

29851.8314

0021.21

2.000.7641

p

TRM

m*Z

V =

=

=Pa

KKkmol

J

kmol

kg

kg

d. Calculate Zstd using Redlich-Kwong Zstd = 0.9968

3

3stdm245.2

10325.101

28851.8314

0021.21

2.009968.0

p

TRM

m*Z

V =

=

=

Pa

KKkmol

J

kmolkg

kg

e.333

9.111m1087.17

2.00

V

m

m

kgkg=

==

EXAMPLE9

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A calibration gas sample consisting of 80% Oxygen and 20% Nitrogen by volume is to be

prepared. Two 10 litre bottles are evacuated, then charged from high pressure component

storage cylinders. One of the bottles is filled with Oxygen, the other with Nitrogen. Then thetwo bottles are connected together to provide 2 bottles of the required mix. There is to be

enough gas mixture in each bottle so that the bottle has a "full" pressure of 350 kPaa at 20oC.

a. The initial pressure in the 10 liter bottle of Oxygen.

b. The initial pressure in the other 10 liter bottle containing Nitrogen.

c. The total mass of the gases in the two sample bottles.

d. The number of API cubic meters of calibration gas produced.

PHASE 1



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assume Z = 1.0

Examining th e cal gas bottle we see there are 20 L @ 35 0 kPa

The partial pressure of oxyg en is p O2..0.8 350 kPa =p O2 280 kPa

The p artial pressure of nitrogen is p N2..0. 2 350 kPa =p N2 70 kPa

The initial v olume o f each gas was 10 L, thus the in itial pressures become

p O2.

.20 L

.10 Lp O2a.

=p O2 560 kPa

b.p N2

..20 L.10 L

p N2

=p N2 140 kPa

c. M N2.28.0134

kg

kmolM O2

.31.9988kg

kmol(p.23 -2 SIdb)

T .293 K

n O2

..p O2 10 L

.R T

=n O2 0.002 kmol

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m O2 .n O2 M O2

=m O2 0.074 kg

n N2

..p N2 10 L

.R T

=n N2 5.747 104

kmol

m N2.n N2 M N2

=m N2 0.016 kg

mtotal

mO2

mN2

=m total 0.09 kg

d. T std.288 K

p std.101.325 kPa

V std

..n O2 n N2 R T std

p std

=V std 67.906 L

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PROBLEMS

1. Determine the molar mass, the pseudo critical pressure, and the pseudo critical

temperature, for the following gas mixtures. The compositions are mole percent.

a. Nitrogen 79%, Oxygen 21%.

b. Fluorine 18%, Chlorine 82%. (Fluorine : Tc =-129C, pc=5573 kPa)

c. Helium 70%, Oxygen 12%, Nitrogen 18%.

d. Methane 83%, Ethane 16%, Propane 1%.

e. Nitrogen 85%, Carbon Dioxide 5%, Oxygen 5%, Carbon Monoxide 5%.

2. Calculate the mole percent of each of the components in the following gas mixtures.

The compositions have been provided on a mass percent basis.

a. Nitrogen 60%, Carbon Dioxide 30%, Carbon Monoxide 10%.

b. Helium 50%, Oxygen 30%, Carbon Dioxide 20%.

c. Methane 78%, Ethane 18%, Propane 4%.

3. A gas mixture consists of 58% Methane, 7% Ethane, 2% Hydrogen Sulfide, and 33%

Nitrogen, by volume. 2 kg mass of this mixture is stored in a cylinder at a temperature of

25 oC and a pressure of 15000 kPag. The barometer is steady at the average Edmonton

value.

a. What will be the molar mass , the pseudo critical pressure, and the pseudo

critical temperature of the mixture?

b. What will be the compressibility factor (gas law deviation factor) of the mixture

at its storage P and T?



e. What is the density of the gas in the cylinder at the conditions stated in the

question?

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4. A gas mixture consists of 62% Methane, 8% Ethane, and 30% Propane, by mole. Two

point five (2.5) kg of this mixture is to be stored in a cylinder, which has a volume of 28L. The temperature is 30oC.



b. What absolute pressure must be applied in order to store the 2.5 kg of the

mixture in the cylinder?

5. A calibration gas sample consisting of 60% Oxygen and 40% Nitrogen by volume is to

be prepared. Two 10 liter bottles are evacuated, then charged from high pressurecomponent storage cylinders. One of the bottles is filled with Oxygen, the other with

Nitrogen. Then the two bottles are connected together to provide 2 bottles of the

required mix. There is to be enough gas mixture in each bottle so that the bottle has a

"full" pressure of 250 kPaa at 20oC.

a. The initial pressure in the 10 liter bottle of Oxygen.

b. The initial pressure in the other 10 liter bottle containing Nitrogen.

c. The total mass of the gases in the two sample bottles.

d. The number of API cubic meters of calibration gas produced.

6. Natural Gas as it comes from the well has a typical composition of 82% Methane, 14%

Ethane, 2% Propane, and 2% Hydrogen Sulfide, by mole fraction. It is piped to a gas

treatment plant at a pressure and temperature of 3.35 MPag and 45 oC, at the rate of 150

actual . The atmospheric pressure is 99.75 kPaa.

a. What is the flow rate of the gas in ?

b. If the component gases are separated at the treatment plant, what will be the

mass flow rate of each component?

c. As the Methane is produced, it is piped off to be sold as fuel gas. What will be

the flow rate of the Methane in API ?

d. What will be the draw off rate of the Methane in actual if its pressure and

temperature are 253 kPag and 18oC?

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7. Natural gas has a typical composition of methane 72%, ethane 19%, and propane 9% by

mole. Determine the following quantities.

a The specific heat at constant pressure (cP). and at constant volume (cV).

b The heat required to raise the temperature of 150 kmol of the natural gas from

0oC to 50oC at constant volume.

c The heat which must be removed from the 150 kmol of the gas in order to cool it

back down to 0oC at constant pressure.

Chapter 6

STEAM

GENERALIZATION

High pressure and temperature steam is widely used in many industrial production

plants such as power plants, chemical plants, pulp mills or heavy crude oil processing plants.

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Various types of steam may be used throughout a plant according to the required temperature

and pressure. Proper understanding and operation of steam generation and maintenance are

essential.

GENERAL OBJECTIVES

The student will learn how to:

(a) explain the concept of generating steam

(b) define specific volume and enthalpy at saturated and unsaturated conditions

(c) define saturated, superheated and wet steam

(d) define the quality factor (x) for wet steam

(e) explain the pressure versus specific volume curve for water

(f) perform calculations for heat transfer in boilers, heat exchangers and turbines using thesteam table in the SI Engineering data book.

THE BASIC CONCEPT

Steam is used to move thermal energy from a boiler through pipes to a device which will

extract the energy such as a heat exchanger or steam turbine. It is similar to wires carrying

electrical energy from a generator to heaters or motors. The analogy can be carried further.

In electricity the rate of energy flow is given by

Power = V Iwhere V is the potential difference in volts and I is the current in amps. One volt isdefined as 1 joule per coulomb (J/C) and one amp is a flow rate of 1 coulomb per second

(C/s). (Recall that the coulomb is a unit of electric charge which equals combined charge of

6.25*1018 electrons.)

In steam generation the analogous formula is given by

Power = H qm

where H is the change of thermal energy measured in kilojoules per kilogram (kJ/kg) andis called the change of ENTHALPY. The term qm is the rate of steam flow in kilograms per

second (kg/s).

2. CONCEPT OF STEAM GENERATION

When water boils, steam is generated. Fig.1 shows the amount of heat content, called specific

enthalpy, absorbed by 1 kg of water at 101.325 kPaa and from 0C to 300C.

PHASE 1



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Starting at 0C, the water temperature rises to 100C at 101.325 kPaa atmospheric pressure.

The amount of heat can be calculated by the heat gain equation, Q = mc t. The first rising

section of the straight line represents the amount of heat absorbed during the process ofreaching the boiling condition.

When the water reaches its boiling point, any further supply of heat does not raise the

temperature of the water but will only change the phase of the water from liquid to vapour,

namely,

basic thermodynamics manual

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