02 Basic Thermodynamics

7/29/2019 02 Basic Thermodynamics

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UNIT 1

THERMAL PROPERTIES OF LIQUIDS AND SOLIDS

Introduction

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:

1. The definition of specific heat, latent heat of fusion and latent heat of

vaporization.

2. The units of heat energy.

3. The determination of heat required to raise or lower the temperature of a substance.

4. The determination of the specific heat of a substance by experimental data.

5. The determination of the heat flow required to raise or lower the temperature of

a flowing substance.

LESSON 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 = lot 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].

Integrated Training Program / Phase A Basic Thermodynamics Page 2 of 46

Copyright 2004 International Human Resources Development Corporation


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S o l u t i o n D a t a

L .. s t e e lL o t 1 t o L o.6 0 m

= L 0 . 0 1 3 0 6 8mt o

.5 d e g C

t 1.2 3 d e g C

s t e e l..1 . 2 11 0 5 d e g C1 ( p . 1 7

The new length becomes:

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 steel beams, 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 bulkexpansion 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 of volume expansion of the liquid.




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Example2

A 2 liter carbon steel spherical bottle is completely filled with ethyl alcohol at 20C. Ifthe 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=V bottle 1.82 10

3l

steel..1.21 10 5 degC 1

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




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LESSON 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 = mct 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.

Solution Data

Q ..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.kg degC

(23-3 SI databk)

When a solid material is heated, the molecules become more energetic and vibrate somuch that they become loosely bound to each other. The material changes from solid

phase to liquid phase. 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)




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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.

D a t aS o l u t i o n

m b e n z e n e.7 . 5

k gQ ..m b e n z e n ec t o t 1 .m b e n z e n eL ft o

.4 0 d e g C=Q 1 3 9 8 . 8k J

t 1.5 . 5 3 3d e g C

c .1 . 7 1 4 7k J

.k g d e g C( p . 2 3 - 3 S I d

L f.1 2 7 . 4

k J

k g ( p . 1 8 d a t a b k )

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 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 of vaporization.

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.



J

kg

Jkg


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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

t boiling.80.07 deg

(p.23-2 SI databk)

c .1.7147kJ

.kg degC

(p.23-3 SI databk

L v.393.78kJ

kg

(p.23-4 SI databk




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LESSON 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 are no external losses). This can be expressed mathematically as

Qgained = Qlost eq.(1-8)

or

m1c1(tf- ti1) = m2c2(ti2 - tf) eq.(1-9)

where: m1 is the heat gaining mass.

m2 is the heat losing mass.

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

t fis the final temperature of the mixture.

t i1, is the initial temperature of the mass gaining heat

t i2 is the 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 final temperature 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:

mh

.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 cc h

.0.448kJ.kg degC

(p.17 db)

t h_o.40 degC

=t final 21.6 degC

Cold:

m c.50 kg

t c_o.10 degC

c c.2.3564

kJ.

kg degC(p.23-3 SI db




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Solution: Data

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

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

Example 7

A heat exchanger uses high pressure steam to heat up n-Nonane. The initial heat energyof 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?.

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.




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UNIT 2

VAPOUR PRESSURE

Introduction

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

liquids to 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 :

1. The concept of vapour pressure.

2. The absolute and gauge pressure.

3. How to find vapour pressure at a given temperature from tables.

4. The relationship between pressure at boiling points and vapour pressure of

liquids

LESSON 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.




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LESSON 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 forevery different location.

LESSON 3

Variation Of Vapour Pressure with Temperature

The vapour pressure of a liquid increases considerably as the temperature of the liquidincreases. It is because more liquid molecules become energetic and are ready to leavethe liquid forming vapour. The vapour - temperature relationship is not linear. Thus,any vapour pressure reading for a liquid is meaningless unless the temperature of theliquid is also given.

LESSON 4

The Relationship between the pressure at boiling point at the vapourpressure of the LIQUID

If the temperature of the liquid is raised to the point where the vapour pressure of theliquid becomes equal to the surrounding pressure, the liquid boils. Consequently, aliquid 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 athigher elevations where the atmospheric pressure is lower.

Liquids having relatively high vapour pressures will evaporate faster if the container isleft open. This is caused by the higher molecular mobility in liquids which have highervapour pressures.

For pure liquids, the relationships between temperature and vapour pressure are wellknown and are predictable. Hence, it is possible to build a thermometer which isactivated by the vapour pressure of a liquid, and which is used, with a suitablecorrelation, to indicate temperature.




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LESSON 5

Variation of atmospheric pressure with elevation

Since the density of air decreases with height, the air pressure at higher altitudes iscorrespondingly lowered. The variation of atmospheric pressure with altitudes can becalculated by considering various factors such as wind speed, temperature, altitude andhumidity. 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.

LESSON 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 certainamount 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 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 =

vapour pressure at a specific temperatu

saturated vapour pressure at dew poin x 100%




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Example1

The room temperature is 22oC and the relative humidity is found to be 45%. The barometricpressure is 95 kPa.

a. Refer to the steam tables to find the saturated vapour pressure of water at 2o(SIdatabook p 24-38) .

IP 22 C

LRV i 20C URV i 25

C LRVo 2.339kPaa URV o 3.169

kPaa

SPAN iURV i

LRV i

SPAN oURV o

LRVo

y .

IP LRVi

SPAN iSPAN o LRVo

=y 2.671kPaa

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.0.45y

=p H2O 1.202kPaa

c. At what temperature would the water vapour start to condense if the room air was codown. This temperature is called thedew point temperature or just thedew 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 0.8721kPaa URV i 1.227

kPaa LRVo 5C URV o 10

C

SPAN iURV i

LRV i

SPAN oURV o

LRVo

y .IP LRV i

SPAN i

SPAN oLRVo

=y 9.639C

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




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

By direct proportions, at 95kPaa, the vapour pressure is 1.202kPaa; 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 SIdatabook p 24-38, it is found that the temperature of the calculated pressure obtained 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

=y 45.96 C

f. What is the highest air pressure allowable in the air lineat 22oC so that condensation could not occur? As the air is compressed, the watervapour pressure will rise along with the air pressure.

The maximum watervapour pressure which can be attained at 22C without condensation is 2.671 kPaa. It is given that at95 kPaa air pressure, the watervapour pressure is

1.202 kPaa.By direct proportions, the highest air pressure is

p highest.952.671

1.202kPaa

=p highest 211.10 kPaa




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Example 2



A barometer can be made by inverting a glass tube completely filledwith 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 an

atmospheric pressure of

p atm.100 kPaa

p vapour_Hg.252 mPaa

Hg.13534kg

m

3

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 theirpressure 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 the

maximum temperature which the alcohol can have before the pump begins todraw vapour as well as liquid ?

o u on a a

NAIT average pressure3. a.

pat.93.5kPa

p inlet..30psi

.6.89

kPa

psipat

vapour pressure of acetone at 40 deg

=p inlet 300.3

kPa absolute pv.68kPa (databook p.29)

the permissible pressure drop is psuction..60kPa=p inlet

pv 232.3

kPa

3b. ppumppat

.60kPa

=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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UNIT 3

GAS LAWS

Introduction

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:

1. Explain the basic theory of gases and its relationship to the ideal gas law and

general gas law.

2. Determine volume, pressure, temperature or number of moles for gases under

various conditions.

3. Determine the density of gases with the general gas law.4. State the conditions for STP, API or "contract" conditions.

5. Explain the importance of calculating the volume or volume flow at STP or API

conditions.

6. Calculate the reduced pressure and reduced temperature and then determine the

gas compressibility factors by graph.

LESSON 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.




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LESSON 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.

(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).




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LESSON 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 KR = universal gas constant = 8314.510

n = number of kilomoles of gas

Variations of this equation are:

(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)

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.




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Example1

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

S o l u t i o n D a t ap re ss u rec o n v e rsi on

p a tm o s.. H g g h h .0 .6 9m

=p a tm o s 9 1 .7 0 9k P a H g.1 3 5 4 6 .3

k g

m3

(p .1 9 d a t a b op g a u g e

....1 5 0l b

i n24 .4 4 8

N

l b

.1 i n

..2 .5 41 0 2 m

2

g .9 .8 1 1 6 9m

s e c2=p g a u g e 1 .0 3 41 0

3k P a

V 1.0 .6 5m3

p a b s p g a u g e p a tm o s

T 1 .2 0 d eg C .2 7 3 K=p a b s 1 .1 2 61 0

3k P a

=T 1 2 9 3 K

R .8 .3 1 4 4k J

.k m o lKa . n

.p a b s V 1.R T 1

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

k m o l(p .2 3 -2d a ta b o o

p s t d.1 0 1 .3 2 5k P a

b .

.p a b s M

.R T 1= 1 3 .3 8 5

k g

m3 Ts t d

.1 5 d e g C .2 7 3 K

c . m a ss . V 1 =m a ss 8 . 7 k g =T s t d 2 8 8 K

d .V s t d

..n R T s t d

p s t d=V s t d 7 .0 9 9 m

3




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LESSON 4

Real Gases

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 of the 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 andpressure 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 TC.

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).

3The critical temperature and pressure of a substance have to do with the change of state of the

substance between its gas and liquid 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 temperatureestablishes the critical point). There is also a critical density

ccorresponding to the critical point.

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




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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 temperature

Z = f(TR, pR)

Example2

A 45 litre cylinder of oxygen stored at a temperature of 18 oC has a gauge pressure of15MPag. 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?

o u on a a

T.K

.1 degCa. TR

TTc

=TR 1.

=T K

pR

ppc

patmos.kPa=p .

p..1 1

3 kPa patmosZ 0.84

(from SI data book p.23-12)=p 1. 11

4 kPa. n

.p V..

=n .

mo V ..1

3 m3

Tc .1 .

K (p.23-2 SIdb)c. .p M

..Z R T= .

kg

m3

pc.kPa (p.23-2 SIdb)

. mass.n =mass 12.13

gM .1.

kgkmol

(p.23-2 SIdb)

e. Zstd . Tstd

.K pst .1 1.

kPa

Vstd

...Zst n R Tst pstd

=Vstd 8.96

m



p

pc

T

TC


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Finding Z Factor by Redlich-Kwong Method

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

(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 -18 oC 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?

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?



PrT

r2.5

PrT

r


24/46

o u on a a

a. TR

T

Tc

=TR

. T.K .degC

=T KpR

ppc

=pR 2.99

patmos.98kPa

Guess p..1510

3 kPa patmosZ . =p 1.5110

4 kPaFor part a, using Redlich-Kwong method,

V ..45103 m

3

A .0.4274

pR

TR2.5

Tc.154.5

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

pc.504

kPa (p.23-2 SIdb)

B .0.08 4

pRTR

=B 0.15

M

..

kgkmol (p.23-2 SIdb)

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

Z3 Z

2 .B2 B A Z .AB

=Z . b. n

.p V..Z R T

=n 0.37

kmol

c.

.p M..Z R T

= 2 3.8kg

m3

d. mass.n M =mass . kg

e. Zstd 1.0Tstd

.288K pstd.101.32

kPa

Vstd

...Zstdn R Tstdpstd

=Vstd . m

3

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.




25/46

LESSON 5

Avogadro's Hypothesis

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

same number of molecules.

LESSON 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.




26/46

(e) Process of iteration

In certain real gas law problems, the gas equation might involve two unknowns. Anexample of this is to find the pressure of a sealed tank of gas. In such a case, the

pressure, 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

= =

3

1020

13)(273.158314.51

16.043

2.00Z

+ = Z14.8301255106 Pa

pc= 4599 kPa

Tc= 190.56 K

Z P=Z14830.12 kPa4599

ppR =

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 0.8078790.807879 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




27/46

UNIT 4

THERMAL PROPERTIES OF GASES

Introduction

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 gas can 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:

1. Explain the energy equation for heat transfer in terms of the two specific heat

constants (cV and cP)

2. Derive the relationship for cP knowing cV, R and M.

LESSON 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.




28/46

LESSON 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

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 of oCkg

J

or o

Kkg

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 pV where V is the

change in volume 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

cPmt = cVmt + pV

cP - cV =Tm

Vp




29/46

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)

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 . This constant will be used in the

orifice flow metering section.

Example2

Look up the value of the molar mass, and the specific heat at constant pressure c P (

o

Ckg

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)(c

c

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 pC v

=k 1.33




30/46

Example 3

The temperature of a 70 m3 steel tank contains air at 20 oC. The barometer is steady atthe average Edmonton value.

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

that the tank is open to the atmosphere?

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

that the tank is completely sealed?

Solution Data

For ideal gas, Z 1M .28.9625

kg

kmol(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

M

C 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




31/46

UNIT 5

MIXTURES OF GASES

Introduction

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

different properties than the pure ones. The density behavior 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:

1. Explain and use Dalton's law of partial pressure.

2. Explain the significance of mass, volume and mole fraction and convert from

one to the other.

3. Calculate the compressibility factor for a gas mixture.

4. Solve for two unknowns in a gas problem by iteration.

5. Determine the quantities and pressures of pure gases required to make acalibration gas at a particular pressure and composition.

LESSON 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)




32/46

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

m

eq.(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)




33/46

Example 2

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

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

they make 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, areidentical since all 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)




34/46

Example 3

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

3

= 0.8291 m

3

Vt = 1.7017 m3 + 0.8291 m3 = 2.5308 m3

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.




35/46

LESSON 2

Physical Properties of A Gas Mixture

a- 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, M 1,

M2, M3, etc. The corresponding portion of each individual gas is measured by thenumber of moles, 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 + xB3 M3 + ... eq.(5-6)

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

W B1 =

mixM

1M

x B1 eq.(5-7)




36/46

b- Specific Heat at Constant Pressure

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

LESSON3

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, p3 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 themole fractions. Namely,

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

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




37/46

LESSON 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 of Z(pr, Tr) for hydrocarbons. For pure gases, use

the Redlich-Kwong method to actually calculate the Z factor.




38/46

LESSON 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.

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 % MN2 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 =

1M

mixM

WB1 calculate the composition based on moles

gas Mass % M Rel mol

N2 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 factor

20.105

100




39/46

gas Mass % M relative mol 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 =

mixM

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 50.0

CO 12.23 28.01 10.0

Mix 100.0 34.25 100.0

Example 5

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 4880

C3H8 10 44.1 369.8 4240

Mix 100




40/46

Calculate the weighted averages

gas molar% M Tc PcCH4 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-Kwong

gas molar% M Tc Pc Z

CH4 70 16.04 190.6 4599 0.9201

C2H6 20 30.07 305.4 4880 0.2395C3H8 10 44.1 369.8 4240 0.1942

Mix 100 21.65 231.5 4619.3 0.8336

Calculate the Vol % using B1= B1mix

1x

Z

Z

gas molar% M Tc Pc Z rel vol

CH4 70 16.04 190.6 4599 0.9201 77.26

C2H6 20 30.07 305.4 4880 0.2395 5.75C3H8 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 6.73C3H8 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

Done!




41/46

Example 6

A gas mixture consists of 90.54 % methane, 6.73 % ethane and 2.73 % propane byvolume 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 averagesgas 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 0.9201

C2H6 6.73 30.07 305.4 4880 0.2395

C3H8 2.73 44.1 369.8 4240 0.1942Mix 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




42/46

Example 7

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 thecylinder?

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 4880H2S 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

29851.8314126.20

2.000.7864

p

TRM

m*Z

V =

=

=Pa

KKkmol

J

kmolkg

kg


43/46

d. Calculate Zstd using Redlich-Kwong Zstd = 0.9972

3

3stdm2.343

10325.101

28851.8314126.202.009972.0

p

TRM

m*Z

V =

=

=Pa

KKkmol

J

kmolkg

kg

e.332-

104.3101.917

2.00

V

m

m

kg

m

kg=

==

Example 8

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.

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 . Write the data in tabular form and calculate the weighted averages

gas vol% M Tc Pc

CH4 70.00 16.04 190.56 4599C2H6 18.00 30.07 305.41 4880

H2S 2.00 34.082 373.37 8963

N2 10.00 28.0134 126.21 3398

Mix 100.00 20.1236 208.454 4616.76

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 28.0134 126.21 3398 0.9888

Mix 100.00 20.1236 208.454 4616.76 0.7881




44/46

Calculate the mole fractions using xB1=

1

t

Z

Zx 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 41.00

H2S 2.00 34.082 373.37 8963 0.1838 8.57

N2 10.00 28.0134 126.21 3398 0.9888 7.97

Mix 100.00 20.1236 208.454 4616.76 0.7881 122.26

Note that the sum is not 100 Because the sum is greater than 100 so we must normalize

the mol % by multiplying by 100/121.74

gas vol% M Tc Pc Z rel mol mol %

C 70.00 16.04 190.56 4599 0.8525 64.71 52.93CH4 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 28.0134 126.21 3398 0.9888 7.97 6.52

Mix 100.00 20.1236 208.454 4616.76 0.7881 122.26 100.00

b. Find the weighted averages of M and Tc and Pc and finally find the Z of the mixture

gas 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 28.0134 126.21 3398 0.9888 7.97 6.52 28.01 126.21 3398

Mix 100.00 20.1236 208.454 4616.76 0.7881 122.26 100.00 21.0021 212.009 4318.57 0.7641

c.

d. Calculate Zstd using Redlich-Kwong Zstd = 0.9968

3

3stdm245.2

10325.101

28851.8314

0021.21

2.009968.0

p

TRMm*Z

V =

=

=Pa

KKkmol

J

kmolkg

kg

e.333

9.111m1087.17

2.00

V

m

m

kgkg=

==



33

7m1087.17

10009.1

29851.83140021.21

2.000.7641

p

TRM

m*Z

V =

=

=Pa

KKkmol

J

kmolkg

kg


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Example 9

A calibration gas sample consisting of 80% Oxygen and 20% Nitrogen by volume is tobe 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 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 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.

assume Z=1.0

Examiningthe cal gas bottlewesee there are 20L@350kPa

The partial pressureof oxygenis p O2..0.8 350 kPa =p O2 280 kPa

The partial pressureof nitrogenis p N2..0.2 350 kPa =p N2 70 kPa

The initial volumeof eachgas was 10L, thus the initial 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. MN2.28.0134

kg

kmolMO2

.31.9988kg

kmol(p.23-2SIdb)

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

m total m O2 m N2

=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

02 Basic Thermodynamics

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