Download Pure Substances

Pure Substances
P-T and P-V diagrams
Triple point and critical point
Sub cooled liquid
Saturated liquid, mixture of saturated liquid and vapor
Saturated vapor and superheated vapor states of a pure substance with water as
example.
Enthalpy of change of phase, dryness fraction, T-S and H-S diagrams.
Representation of various processes on these diagrams.
Steam tables and its use.
Throttling Calorimeter, Separating Calorimeter.
Throttling and Separating Calorimeter.
Introduction:
A pure substance is one that has a homogeneous and in variable chemical
composition. It may exist in more than one phase, but the chemical composition is
the same in all phases. Thus liquid water, a mixture of liquid water and water vapor,
and a mixture of ice and liquid water are all pure substance: every phase has the
same chemical composition. On the other and mixture of liquid air and gaseous air is
not a pure substance. Because of the composition of the liquid phase is different
from that of the vapor phase.
Some times a mixture of gases such as air is considered a pure substance as
long as there is no change of phase. Strictly speaking this is not true.
Vapor liquid solid phase equilibrium in a pure substance:
Fig 1: Constant pressure change from liquid to vapor phase for a pure
substance
Consider a system one kg of water contained in a piston cylinder
arrangement as shown in the figure. Suppose that the piston and weight maintain a
pressure of 0.1 MPa in the cylinder and that initial temperature be 20 0C. As the heat
is transferred to the water the temperature increases appreciably, the specific
volume increases slightly under constant pressure. When the temperature 99.6 0C,
additional heat transfer results in a change of phase. I.e. some of the liquid becomes
vapor. During this process both temperature and pressure remain constant where as
sp. Volume increases considerably. When the last drop of liquid has vaporized
further transfer of heat results in an increase in both temperature and sp. volume of
the vapor.
The term saturation temperature designates the temperature at which
vaporization takes place at a given pressure. Or the pressure is called saturation
pressure corresponding to the saturation temperature. Thus for water 99.6 0C the
saturation pressure is 0.1MPa, and for water at 0.1MPa the saturation temperature
is 99.60C. Thus there is a definite relation between saturation pressure and
saturation temperature.
Pressure
Vapor-pressure curve
Temperature
Fig 2: Vapor – Pressure curve of a pure substance
If the substance exists as liquid at the saturation temperature and pressure it
is called saturated liquid. If the temperature of the liquid is lower than the saturation
temperature for the existing pressure it is called either a sub cooled liquid or a
compressed liquid.
When the substance exists as a part liquid and part vapor at the saturation
temperature a dryness fraction comes into picture. It is also called as quality and it is
defined as the ratio of mass of vapor to the total mass. It is denoted by the symbol
’x’. Quality has meaning only when the substance is in saturated state. I.e at the
saturation pressure and temperature. The quality x is an intensive property.
If a substance exists as vapor at the saturation temperature it is called
saturated vapor or dry saturated vapor with x=1. When the vapor is at a temperature
greater than the saturation temperature at the saturation pressure, it is said to exist
as superheated vapor. After that the temperature increases as heat is added at
constant pressure.
Fixing of the State:
The state of a closed system at equilibrium is its condition as described by the
values of its thermodynamic properties. From the observations of many
thermodynamic systems it is known that not all these properties are independent of
one another, the state can be uniquely determined by giving the values of
independent properties. Values for all the other thermodynamic properties are
determined from this independent subset. A general rule known as the state
principle has been developed as a guide in determining the number of independent
properties to fix the state of the system. Based on the empirical evidence it has been
concluded that there is one independent property for each way a system’s energy
can be varied independently. Independent property can be associated with heat
transfer as one way of varying the energy and another independent property can be
counted for each relevant way the energy can be changed through work.
Simple compressible systems change in volume can have a significant
influence on the energy. The only mode of energy transfer by work that can occur as
a simple compressible systems undergoes quasi-equilibrium process. The
experience shows that it is useful for wide range of engineering applications the
properties of pure substances either independent or dependent.
Phases of a Pure Substance:
A phase is identified as having a distinct molecular arrangement that is
homogeneous throughout and separated from the others by easily identifiable
boundary surface. When studying phases or phase changes in thermodynamics one
does not need to be concerned with the molecular structure and behavior of different
phases. However it is very helpful tio have some understanding of the molecular
phenomenon involved in each phase, a brief discussion of phase transformation
follows.
Intermolecular bonds are strongest in solids and weakest in gases. One
reason is that molecules in solids are closely packed together whereas in gaseous
they are separated by relatively large distances.
There are many practical situations where 2 phases of a pure substance coexist in equilibrium.
Temperature – Volume (T-V) diagram for water:
Fig 3: T-V diagram for water
It is clear from the figure that constant pressure lines are ABCD, EFGH, IJKL
etc. the peak point of the figure indicated by N is the critical point of water. Thus the
critical pressure 22.089 MPa and corresponding critical temperature is 374.14 0C.
A constant pressure process at a pressure greater than the critical pressure is
represented by curve PQ. Thus water at 40 MPa, 20 0C is heated in a constant
pressure process, there will be never be two phases present at the state shown.
Instead there will be continuously in density at all the times and there will be only
one phase present. The question arises is: when do we have a liquid and when do
we have a vapor? The answer is that this is not a valid question at super critical
pressures. We simply term the substance as fluid. However rather arbitrarily at
temperatures below the critical temperatures we usually refer to it as a compressed
liquid and at temperatures above the critical temperatures as superheated vapor. It
should be noted that however at pressures above the critical pressures we never
have a liquid and vapor phase of pure substance existing in equilibrium.
Table1: Some Critical point data
Critical
Temperature, 0C
Water
374.14
Carbon dioxide
031.05
Oxygen
-118.35
Hydrogen
-239.85
Critical
Pressure, MPa
22.089
07.39
05.08
01.30
Critical Volume,
m3/kg
0.003155
0.002143
0.003438
0.032192
Consider another experiment with piston cylinder arrangement. Suppose that the
cylinder contains one kg of ice at –200C and one bar. When heat is transferred to the
ice the pressure remains constant the specific volume increases slightly and the
temperature increases until it reaches 0 0C, at which point the ice melts and
temperature remains constant. This state is called saturated solid state. For most
substances the specific volume increases during this melting process. But for water
specific volume of the liquid is less than the specific volume of the solid.
Water Vapor - An Ideal gas: Water vapor below 10kPa pressure vapor can be
treated as an ideal gas regardless of its temperature with negligible error. At higher
pressures, however, the ideal gas assumption yields un acceptable errors,
particularly in the vicinity of the critical point and the saturated vapor line. Therefore
in air-conditioning applications the water vapor in the air can be treated as an ideal
gas with essentially no error since the pressure of the water vapor is very low. In
steam power plant applications however the pressure involved are usually very high.
Therefore ideal gas relations should not be used.
Sublimation: If the initial pressure of the ice at –200C is 0.26 kPa, heat transferred
to the ice results in an increase in the temperature to –100C. At this point however
the ice passes directly from the solid phase to the vapor phase. This process is
known as sublimation. Further heat transfer results in superheating of the vapor
Triple Point: Consider the ice at 0.6113 kPa and temperature of –200C. Through
heat transfer let the temperature increase until it reaches 0.01C. At this point
however further heat transfer may cause some of the ice to become vapor and some
to become liquid. At this point it is possible to have three phases in equilibrium. This
point is called the triple point. Triple point is defined as the state in which all three
phases may be present in equilibrium. The pressure and temperature at the triple
point for a number of substance is given in following table.
Table 2:Triple Point Data
Temperature, 0C
Hydrogen
-259
Oxygen
-219
Nitrogen
-210
Carbon Dioxide
-56.4
Mercury
-39
Water
0.01
Zinc
419
Silver
961
Copper
1083
Pressure, kPa
7.194
0.15
12.53
520.8
0.00000013
0.6113
5.066
0.01
6.000079
Consider a solid state as shown in the figure, when the temperature increases
with constant pressure the substance passes directly from solid to vapor phase.
Along the constant pressure line EF the substance passes from solid to liquid phase
at one temperature and then from liquid to vapor phase at higher temperature.
Constant pressure line CD passes through the triple point and it is only at the
triple point the three phases exists together in equilibrium. At a pressure above
critical pressure such as GH line there is no sharp distinction between liquid and
vapor phases. The triple point temperature and critical temperature vary greatly from
substance to substance. For ex: critical temperature of helium is 5.3K. Therefore
absolute temperature of helium at ambient conditions is over 50 times greater than
the critical temperature. On the other hand water has a critical temperature 374.14
0C (647.29K) and at ambient conditions the temperature of water is less than ½ the
critical temperature.
Allotropic transformation
It should be pointed out that a pure substance can exist in a number of different solid
phases. A transition from one solid phase to another is called an allotropic
transformation. This can be well understood by the following figure.
Temperature
Fig 4: P-T Diagram for water.
Fig 5: P-T Diagram for iron
Independent properties of a pure substance:
One important reason for introducing the concept of pure substance is that
the state of a simple compressible pure substance is defined by two independent
properties. For ex: if the specific volume and temperature of a super heated steam
are specified the state of the steam is determined.
To understand the significance of the term independent property, consider the
saturate liquid and vapor state of a pure substance. These two states have the same
pressure and temperature but they are definitely not the same state. In a saturation
state therefore pressure and temperature are not independent properties. Two
independent properties such as pressure -specific volume, pressure and quality,
temperature-specific volumes are required to specify a saturation state of a pure
substance.
Common Properties of Pure Substances
Density
The density of any pure substance is its ratio of mass to unit volume.
Specific gravity
Closely related to density, specific gravity is a dimensionless ratio of the density of
the substance to the density of a standard--typically another substance, usually a
liquid, in which the substance in question is suspended. The usual standard of
specific gravity is water, in which case the specific gravity is numerically equivalent
to the density. But one can also define the specific gravity of a substance in, for
example, an organic liquid, such as benzene.
Melting point
The melting point of any pure substance is the temperature at which, under common
atmospheric pressure, that substance changes its state from solid to liquid. If the
substance is a liquid at 25 degrees Celsius, this temperature is usually called the
freezing point. Some substances, all of them compounds, will simply decompose
before they ever melt.
Boiling point
The boiling point of any pure substance is the temperature at which, under common
atmospheric pressure, that substance changes its state from liquid to gas. If the
substance is a gas at 25 degrees Celsius, this temperature is sometimes called the
condensation point.
Triple point
The triple point of any pure substance is that combination of temperature and
pressure at which all three phases of that substance coexist simultaneously.
Specific heat
The specific heat of any pure substance is the amount of heat required to raise the
temperature of a unit mass of that substance by one degree on a given temperature
scale.
Heat of fusion
The heat of fusion of any pure substance is the amount of heat required to change a
unit mass of that substance, once brought to the melting point, from solid to liquid.
Heat of vaporization
The heat of vaporization of any pure substance is the amount of heat required to
change a unit mass of that substance, once brought to the boiling point, from liquid
to gas.
Critical point
The critical point of any pure substance is a point on a three-dimensional graph of
temperature, pressure, and molar volume (ratio of volume to amount-of-substance)
beyond which that substance can exist only as a gas. Water, for example, has a
critical temperature beyond which it will exist only as dry steam. To make it liquid
again, one must cool it below this critical temperature and then apply the critical
pressure (the pressure coordinate of the critical point).
Equations of state for the vapor phase of a simple compressible
substance:
From the experimental observations it has been established that the equation
of state under low-density gases is given by PV=RuT. The ideal gas equation of state
and compressibility factor equation are good approximations at low-density
conditions. Therefore ideal gas equation of state is very convenient to use in
thermodynamic calculations. The question comes what is low density? Or what
range of density will the ideal gas equation of state hold with accuracy?
The analysis gives the pressure temperature deviations from ideal gas
behavior. To answer5 the question the concept of compressibility factor Z is
introduced and is defined by the relation Z= PV/RuT. For ideal gas Z=1 and the
deviation of Z from unity is a measure of the deviation of actual relation from the
ideal gas equation of state.
Fig 6: Compressibility of Nitrogen
It shows a skeleton compressibility charge for nitrogen. Three observations can be
made from this chart.
All the temperature Z
1 as P
0. i.e. as the pressure approaches zero the P-VT behavior closely approaches that predicted by the ideal gas equation of state.
Note also that at temperatures of 300K and above i.e. above room temperature the
compressibility factor is near unity up to a pressure of 10MPa. This means that the
ideal gas equation of state can be used for nitrogen over this range with
considerable accuracy.
Now suppose we reduce the temperature from 300K but keep the pressure constant
at 4MPa, the density will increase and we note a sharp decrease below unity in the
value of compressibility factor. Values of Z <1 mean that the actual density is greater
than would be predicted by ideal gas behavior.
As the temperature reduced from 300K and pressure remains constant at 400MPa
the molecules are brought closer together, there is an attractive force between the
molecules. The lower the temperature the greater is the intermolecular attractive
force. Thus more attractive force more density, which is greater than, would be
predicted by the ideal gas. (Ideal gas intermolecular attraction is negligible)
On comparison of various compressibility charts for other pure substances it is
observed that the diagrams are all similar in the characteristics described above for
nitrogen at least in qualitative sense. Quantitatively the diagram are all different sine
critical temperature, critical pressure of different substances vary over a wide range.
There is a way to put all these substances on a common basis called reduced
properties w.r.t. the values at critical point like
Reduced pressure = Pr = P / Pc and reduced temperature Tr = T / Tc
These equations state that the reduced property for a given state ids the value of
this property on this state divided by the value of the same property at the critical
point.
Tables of Thermodynamic properties:
Tables of thermodynamic properties of many substances are available and in
general these tables have the same form. The steam table is specially selected
because they are a vehicle for presenting thermodynamic tables and steam is used
extensively in power plants and industrial purposes. The table consists of four
separate tables namely:
1) Saturated steam table
2) Saturated water table
3) Temperature basis table and
4) Pressure basis table.
It contains values of enthalpy, internal energy, specific volume and entropy. If the
values of T and P are given other all values can be directly taken from the table.
Therefore we should first learn how to use the table. We note that all these are
independent variables.
To finding the correct table other nuisance of everyday use of the table in the
interpolation. I.e. when one of the stated values is not exactly equal to a value listed
in the table. Recently computerized tables are in use, which eliminates these
problems. But the students nevertheless must learn to understand the significance,
construction and limitations of the tables.
The actual steam table as presented in various books is a summary table based on
complicated curve fit to the behavior of water. Here we concentrate three properties
namely T,P and v and note that other properties listed namely u, h, s are presented.
It is further noted that the separation of phases in terms of values of P and T is
actually described by the relation illustrated in the pressure-temperature diagram.
The specific volume of a substance having given quality can be found out by the
definition of quality. The volume is the sum of volume of liquid and vapor.
V= Vliq + V vap.
In terms of masses: mv = mliq * vf + m vap* vg
By introducing the quality x we have v = (1-x) vf + x vg
We know that
vfg = vg - vf
We can write the specific volume equation for wet steam as v = vf + x vfg
In superheated region pressure and temperature are independent properties and
therefore for each pressure a large number of temperature is given and for each
temperature four thermodynamics properties are listed namely specific volume,
enthalpy, entropy and internal energy.
Temperature – specific volume diagram:
Figure shows a T-v plot for water with an indication of percent error in
assuming ideal gas behavior along the saturated vapor curve and also in several
area of superheated region. Generally a slight decrease and only a small error is
made if one same that the volume of a compressed liquid is equal to the specific
volume of the saturated liquid at the same temperature. It is generally accepted
procedure particularly when compressed liquid data are not available.
Fig 7: Temperature – specific volume diagram for water
In the modern scenario computerized table are available. The main program
operated with a visual interface in the window environment on a PC which is
generally user friendly. The program operates on DOS environment, which covers
not only the tables of water, but it covers most of the pure substances used in
engineering industries. The generalized chart with compressibility factor is also
included so it is possible to get the value of Z a more little accurately than reading
the graph. It is useful in the case of two-phase mixture, like saturated liquid and
vapor values are needed.
Thermodynamic Surfaces:
The matter discussed in this chapter can be well summarized by a consideration of a
pressure, specific volume, temperature surface (PVT surface). Two such surfaces
are shown in the figure for a substance such as water in which the specific volume
increases during freezing and the other in which a substance, its specific decreases
during freezing.
Fig 8 : PvT Surface -- Water
Fig 9 : PvT Surface – Water
In these diagrams the pressure, specific volume and temperature are plotted on a
mutually perpendicular coordinates and each possible equilibrium state is thus
represented by a point on the surface. This follows directly from the fact that a pure
substance has only two independent intensive properties. All points along a quasi
equilibrium surface lie on the PvT surface since such a process always passes
through equilibrium states.
The regions of the surfaces that represent a single surface- the solid, liquid and
vapor faces are indicated. These surfaces are curved. The two phase regions – the
solid-liquid, solid-vapor, liquid-vapor regions are ruled by surfaces. It is understood
that they are made up of straight lines parallel to the specific volume axis. This of
course follows from the fact that in the two-phase region lines of constant pressure
are also lines of constant temperature, although the specific volume may change.
The triple point actually appears as the triple line on the PVT surface, since the
pressure and temperature of the triple point are fixed but the specific volume may
vary depending upon the proportion of each phase.
It is also of interest to note that the pressure temperature and pressure volume
projections of these surfaces. We have already considered the P-T diagram for
water. It is on this diagram we observe the triple point. Various lines of constant
temperature are shown on the P-v diagram and corresponding constant temperature
sections are identically seen on the P-v-T surfaces. The critical isotherm as a point
of intersection at the critical point.
The equilibrium states of a simple, compressible substance can be specified in
terms of its pressure, volume and temperature. If any two of these state variables is
specified, the third is determined. This implies that the states of the substance can
be represented as a surface in a three dimensional PvT space. The PvT surface
above represents a substance which contracts upon freezing. Most substances do
so, the notable exception being water which expands upon freezing. A considerable
amount of information about the phases of matter can be illustrated with the PvT
surface.
The solid, liquid and gas(vapor) phases can be represented by regions on the
surface. Note that there are regions on the surface which represent a single phase,
and regions which are combinations of two phases. A point lying on a line between a
single-phase and a two-phase region represents a "saturation state". The line
between the liquid and the liquid-vapor regions is called the liquid-saturation line and
any point on that line represents a saturated-liquid state. A point on the boundary
between the vapor and the liquid-vapor regions is called a saturated-vapor state.
Note the critical state where the saturated-liquid and saturated-vapor lines meet. The
state variables of this unique point are denoted by P c, vc and Tc. If a substance is
above the critical temperature Tc, it cannot condense into a liquid, no matter how
high the pressure. This merging of the liquid and vapor states above the critical
temperature is a characteristic of all known substances. While a pure vapor state
can exist at a pressure lower than Pc, at pressures above Pc it is constrained to be a
vapor. States with pressures above P c are described as "supercritical states".
The remarkable "triple state" of matter where solid, liquid and vapor are in
equilibrium may be characterized by a temperature called the triple point. The triple
state is represented by a line parallel to the Pv plane with a characteristic pressure
for the substance but variable volume. The triple point temperature of water is
assigned the value 273.16 K and the triple state of water is used as the reference for
establishing the Kelvin temperature scale.
One notice that for a substance such as water which expands on freezing, the
freezing temperature decreases with an increase in pressure. For a substance that
contracts on freezing, the freezing temperature increases as the pressure increases.
Thus as the pressure of the vapor is increased along the constant temperature line a
substance that expands on freezing first becomes solid and then liquid. For
substance that contract on freezing, the corresponding constant temperature line
indicates that as the pressure of the vapor increases, it first becomes liquid and then
solid.
Fig 10 : PvT Surface for a Substance which Contracts Upon Freezing
Fig 11 : PvT Surface for a Substance which Expands Upon Freezing
T-s Diagram for a pure substance:
Consider the heating of 1 kg of ice at –50C to steam at 2500C. The pressure being
maintained at 1 atm. It is observed that the entropy of steam increases in different
regimes of heating namely
1) Entropy increase of ice to saturated freezing temperature
2) Entropy increase of ice as it melts into water.
3) Entropy increase of water as it is heated from 0 Oc to 100OC.
4) Entropy increase of water as it is vaporized at 100 OC absorbing latent heat of
vaporization.
5) Entropy increase of vapor as it is heated from 100 OC to 250OC.
Fig 12: T – s Diagram for water Constant pressure lines
These entropy changes are shown in T-S graph. It is a constant pressure process. If
during heating process the pressure had been maintained constant at 2 atm, a
similar curve would be obtained. If these states for different pressures are joined the
phase equilibrium diagram of a pure substance on the T-s coordinate would be
obtained as shown below.
Fig 13: T –s Diagram for water
Most often liquid vapor transformation only are of interest, the following figure the
liquid, the vapor and the transition zones only at a particular pressure, sf – specific
entropy of saturated water and sg - specific entropy of saturated vapor. The entropy
change of the system during the phase change from a liquid to vapor at that
constant pressure is sfg. (= sg – sf). The value of sfg decreases as pressure
increases. And becomes zero at the critical point.
Fig 14: Phase equilibrium diagram –T – s Diagram
H-S diagram (Mollier diagram) for a pure substance:
From the first and second law of thermodynamics following property relations are
obtained.
Tds = dh – v dp
and ( h / s )p = T
These equations form the basis of h-s diagram of a pure substance. The slope of an
isobar on the h-s coordinates is equal to the absolute temperature. If the
temperature remains constant the slope will remain constant. If the temperature
increases the slope of the isobar will increase.
Consider once again the heating of ice at –5OC to steam at 250OC the pressure
being maintained constant at 1 atm. The slope of the isobar of 1 bar on the h-s
coordinates first increases as the temperature of the ice increases from –5OC to
0OC. the slope then remains constant as ice melts to water at 0 OC. Te slope of
isobar again increases as the temperature of water rises from 0 OC to 100OC. the
slope again remains constant as water vaporizes at constant temperature. Finally
the slope of the isobar continues to increase as the temperature of steam increases
to 250OC and beyond (as shown in the figure below.
Fig 15: h –s diagram for water - constant pressure lines.
Similarly the isobars of different pressures can be drawn on h-s diagrams as shown
in the figure below.
Fig 16: h –s for water.
This figure shows the phase equilibrium diagram of a pure substance on the h-s
coordinate indicating the saturated solid line, saturated liquid line saturated vapor
line, the various phases and the transition zones.
Fig 17: h-s diagram for water:
This figure is the Mollier diagram indicating only the liquid and vapor phases. As the
pressure increases the saturation temperature increases, slope of the isobar
increases. Hence the constant pressure lines diverge from one another and the
critical isobar is at a tangent to the critical point. In the vapor region the states of
equal slopes at various pressures are joined by lines as shown, which are the
constant temperature lines. Here at a particular pressure h f is the specific enthalpy of
saturated water and hg is specific enthalpy of saturated vapor and h fg (= hg - hf) is
the latent heat of vaporization at that pressure. As the pressure increases hfg
decreases and at the critical pressure hfg becomes zero.
Dryness fraction and various equations:
Dryness fraction is defined as the ratio of mass of dry steam to total mass of steam.
It is denoted by x and is also called as quality of steam.
x = mv / (mv +ml ) where mv and ml are the masses of vapor and liquid respectively.
Let V be the toal volume of a liquid vapor mixture of quality x in which Vf volume of
saturated liquid and Vg volume of saturated vapor, the corresponding being m, mf,
mg respectively.
We have m = mf + mg and V= Vf +Vg
Therefore mv = mf * vf + mg*vg
v= (1-x) vf + x vg.
Similarly s = (1-x) sf + x sg
h= (1-x) hf + x hg
u= (1-x) uf + x ug.
Same equations are written as
v = vf + x vfg,
h = hf + x hfg
u = uf + x ufg
s = sf + x sfg
If the condition of the steam is superheated then we have Degree of superheat,
which is difference between the superheated temperature to the saturation
temperature.
 super = Tsuper –Tsaturation
The other properties are calculated as v super = vsat * Tsuper / Tsat
hsuper = hg + Cp (Tsuper-Tsat)
ssuper = sg + Cp log(Tsuper/Tsat)
PHASE DIAGRAMS OF PURE SUBSTANCES
This page explains how to interpret the phase diagrams for simple pure substances including a look at the special cases of the phase diagrams of water and carbon
dioxide. This is going to be a long page, because I have tried to do the whole thing
as gently as possible.
The basic phase diagram
What is a phase?
At its simplest, a phase can be just another term for solid, liquid or gas. If you have
some ice floating in water, you have a solid phase present and a liquid phase. If
there is air above the mixture, then that is another phase.
But the term can be used more generally than this. For example, oil floating on water
also consists of two phases - in this case, two liquid phases. If the oil and water are
contained in a bucket, then the solid bucket is yet another phase. In fact, there might
be more than one solid phase if the handle is attached separately to the bucket
rather than molded as a part of the bucket.
You can recognize the presence of the different phases because there is an obvious
boundary between them - a boundary between the solid ice and the liquid water, for
example, or the boundary between the two liquids.
Phase diagrams
A phase diagram lets you work out exactly what phases are present at any given
temperature and pressure. In the cases we'll be looking at on this page, the phases
will simply be the solid, liquid or vapor (gas) states of a pure substance.
This is the phase diagram for a typical pure substance.
These diagrams (including this one) are nearly always drawn highly distorted in
order to see what is going on more easily. There are usually two major distortions.
We'll discuss these when they become relevant.
If you look at the diagram, you will see that there are three lines, three areas marked
"solid", "liquid" and "vapour", and two special points marked "C" and "T".
The three areas
These are easy! Suppose you have a pure substance at three different sets of
conditions of temperature and pressure corresponding to 1, 2 and 3 in the next
diagram.
Under the set of conditions at 1 in the diagram, the substance would be a solid
because it falls into that area of the phase diagram. At 2, it would be a liquid; and at
3, it would be a vapour (a gas).
Moving from solid to liquid by changing the temperature:
Suppose you had a solid and increased the temperature while keeping the pressure
constant - as shown in the next diagram. As the temperature increases to the point
where it crosses the line, the solid will turn to liquid. In other words, it melts.
If you repeated this at a higher fixed pressure, the melting temperature would be
higher because the line between the solid and liquid areas slopes slightly forward.
So what actually is this line separating the solid and liquid areas of the diagram?
It simply shows the effect of pressure on melting point.
Anywhere on this line, there is an equilibrium between solid and liquid.
You can apply Le Chatelier's Principle to this equilibrium just as if it was a chemical
equilibrium. If you increase the pressure, the equilibrium will move in such a way as
to counter the change you have just made.
If it converted from liquid to solid, the pressure would tend to decrease again
because the solid takes up slightly less space for most substances.
That means that increasing the pressure on the equilibrium mixture of solid and
liquid at its original melting point will convert the mixture back into the solid again. In
other words, it will no longer melt at this temperature.
To make it melt at this higher pressure, you will have to increase the temperature a
bit. Raising the pressure raises the melting point of most solids. That's why the
melting point line slopes forward for most substances.
Moving from solid to liquid by changing the pressure:
Moving from liquid to vapor:
In the same sort of way, you can do this either by changing the temperature or the
pressure.
The liquid will change to a vapour - it boils - when it crosses the boundary line
between the two areas. If it is temperature that you are varying, you can easily read
off the boiling temperature from the phase diagram. In the diagram above, it is the
temperature where the red arrow crosses the boundary line.
So, again, what is the significance of this line separating the two areas?
Anywhere along this line, there will be an equilibrium between the liquid and the
vapour. The line is most easily seen as the effect of pressure on the boiling point of
the liquid.
As the pressure increases, so the boiling point increases.
The critical point
You will have noticed that this liquid-vapour equilibrium curve has a top limit that I
have labelled as C in the phase diagram.
This is known as the critical point. The temperature and pressure corresponding to
this are known as the critical temperature and critical pressure.
If you increase the pressure on a gas (vapor) at a temperature lower than the critical
temperature, you will eventually cross the liquid-vapor equilibrium line and the vapor
will condense to give a liquid.
This works fine as long as the gas is below the critical temperature. What, though, if
your temperature was above the critical temperature? There wouldn't be any line to
cross!
That is because, above the critical temperature, it is impossible to condense a gas
into a liquid just by increasing the pressure. All you get is a highly compressed gas.
The particles have too much energy for the intermolecular attractions to hold them
together as a liquid.
The critical temperature obviously varies from substance to substance and depends
on the strength of the attractions between the particles. The stronger the
intermolecular attractions, the higher the critical temperature.
Moving from solid to vapour:
There's just one more line to look at on the phase diagram. This is the line in the
bottom left-hand corner between the solid and vapour areas.
That line represents solid-vapour equilibrium. If the conditions of temperature and
pressure fell exactly on that line, there would be solid and vapour in equilibrium with
each other - the solid would be subliming. (Sublimation is the change directly from
solid to vapour or vice versa without going through the liquid phase.)
Once again, you can cross that line by either increasing the temperature of the solid,
or decreasing the pressure.
The diagram shows the effect of increasing the temperature of a solid at a (probably
very low) constant pressure. The pressure obviously has to be low enough that a
liquid can't form - in other words, it has to happen below the point labelled as T.
You could read the sublimation temperature off the diagram. It will be the
temperature at which the line is crossed.
The triple point
Point T on the diagram is called the triple point.
If you think about the three lines which meet at that point, they represent conditions
of:



solid-liquid equilibrium
liquid-vapour equilibrium
solid-vapour equilibrium
Where all three lines meet, you must have a unique combination of temperature and
pressure where all three phases are in equilibrium together. That's why it is called a
triple point.
If you controlled the conditions of temperature and pressure in order to land on this
point, you would see an equilibrium, which involved the solid melting and subliming,
and the liquid in contact with it boiling to produce a vapor - and all the reverse
changes happening as well.
If you held the temperature and pressure at those values, and kept the system
closed so that nothing escaped, that's how it would stay. A strange set of affairs!
Normal melting and boiling points
The normal melting and boiling points are those when the pressure is 1 atmosphere.
These can be found from the phase diagram by drawing a line across at 1
atmosphere pressure.
The phase diagram for water
There is only one difference between this and the phase diagram that we've looked
at up to now. The solid-liquid equilibrium line (the melting point line) slopes
backwards rather than forwards.
In the case of water, the melting point gets lower at higher pressures. Why?
If you have this equilibrium and increase the pressure on it, according to Le
Chatelier's Principle the equilibrium will move to reduce the pressure again. That
means that it will move to the side with the smaller volume. Liquid water is produced.
To make the liquid water freeze again at this higher pressure, you will have to
reduce the temperature. Higher pressures mean lower melting (freezing) points.
Now lets put some numbers on the diagram to show the exact positions of the
critical point and triple point for water.
Notice that the triple point for water occurs at a very low pressure. Notice also that
the critical temperature is 374°C. It would be impossible to convert water from a gas
to a liquid by compressing it above this temperature.
The normal melting and boiling points of water are found in exactly the same way as
we have already discussed - by seeing where the 1 atmosphere pressure line
crosses the solid-liquid and then the liquid-vapour equilibrium lines
The phase diagram shows that the water would first freeze to form ice as it crossed
into the solid area. When the pressure fell low enough, the ice would then sublime to
give water vapour. In other words, the change is from liquid to solid to vapour. I find
that satisfyingly bizarre!
Measurement of Steam Quality:
The state of a pure substance gets fixed if two independent properties are given.
Thus the pure substance is said to have two degrees of freedom.
Fig 18: T – s and h – s diagram
Figure shows the values of pressure and temperature would fix up the state. But
when the substance is in the saturation state or two phase region the measured
values of pressure and temperature could apply equally well to saturated liquid point
f and saturated vapor point g. or two mixtures of any quality points x 1, x2 or x3. of the
two properties, P and T only one is independent; the other is a dependent property.
If the pressure is given the saturation temperature gets automatically fixed for the
substance. In order to fix up the state of the mixture one more property such as
specific volume, enthalpy or composition of mixture or internal energy is required to
be known. Since it is relatively difficult measure the specific volume of the mixture
devices such as calorimeters are used for determining the quality or the enthalpy of
the mixture.
In the measurement of quality, the object is always to bring the state of the
substance from the two phase region to the single phase region or superheated
region where, the pressure and temperature are independent and measured to fix
the state, either by adiabatic throttling or e3lectric heating.
There are four types of calorimeter are common in use namely
1) Separating Calorimeter
2) Throttling Calorimeter
3) Combined separating and throttling calorimeter.
4) Electrical calorimeter.
Separating Calorimeter: the steam whose dryness fraction is to be determined
is very wet then separating calorimeter gives the quality of the steam. A known
quantity of steam is passed through a separating calorimeter as shown. The steam
is made to change direction suddenly, the water being denser than the dry steam is
separated out. The quantity of water, which is separated out, is measured at the
separator. The dry steam coming out of the separator is sent through a condenser
where it is condensed separately.
The dryness fraction of the steam is calculated by weighing the mass of the water
and mass of dry steam after condensation separately.
Throttling Calorimeter: sample of wet steam of mass m at pressure p1 is taken
from the steam main through a perforated sampling tube as shown in the figure.
Fig 19: Throttling calorimeter
Then it is throttled by the partially opened valve to a pressure p 2 measured by a
mercury manometer and temperature T 2 so that after throttling the steam is in the
superheated region. The process is shown on T-s and H-s diagram.
The steady flow energy equation gives the enthalpy after throttling as equal to
enthalpy before throttling. It is a irreversible process hence joined by a dotted line.
Thus the initial state of the steam is p 1 , t1 and its dryness fraction is x1 and the final
state of the superheated steam is p 2 x2.
Fig 20: T – s and h – s diagram for throttling calorimeter
Now h1 =h2
hf1+x1 hfg1 = h2
x1 =( h2 – hf1 ) / hfg1
With P2 and T2 being known, h2 can be found out from the superheated steam table.
The values of hf, hfg are taken from saturated steam table., thus quality of the wet
steam x1 can be calculated.
Combined separating and throttling calorimeter:
When the steam is very wet and the pressure after throttling is not low enough to
take the steam to the superheated region then a combined separating and throttling
calorimeter is used for the measurement of quality. Steam from the main is first
passed through a separator as shown in the figure, where some part of the moisture
separates due to sudden change in direction and falls by gravity and partially dry
vapor is then throttled and taken to the superheated region .
Fig 21: Schematic diagram for separating and throttling calorimeter.
As shown in the figure process 1-2 represents moisture separation from the wet
sample of steam at constant pressure P1 and process 2-3 represents throttling to
pressure P2 withP2 and T3 being measured, h3 can be found out from the
superheated steam table.
h3= h2 = hf1 + x2 hfg1
Therefore x2, the quality of the steam after partial moisture separation can be
evaluated. If m kg of steam is taken through the sampling tube in y seconds, m 1 kg is
separated and m2 kg is throttled and then condensed to water and collected, we
have
m=m1+m2.
The mass of dry vapor will be at state2 is x2m2.
Therefore the quality of the sample of the steam at state1 which ids x1 is given by
x1= x2m2 / (m1 + m2)
There is one more method of measurement of quality of wet steam by using electric
calorimeter as shown in the figure. The sample of steam is passed in steady flow
through an electric heater. Electrical energy input Q should be sufficient to taken the
steam to the superheated region where pressure P 2 and temperature T2 are
measured. If I is the current flowing through the heater in amperes and V the
voltage across the coil at steady state Q=VI. If m is the mass of steam taken in t
seconds under steady flow condition then the steady flow energy equation for heater
is given by m1h1 + Q = m1h2
Fig 22: Electric calorimeter
Hence h1 + Q / m1 = h2. With h2,Q and m1 =being known h1 can be computed. Thus
h1= hf1 + x1 hfg1.
Hence x1 can be calculated.
1) A steam boiler initially contains 5 m3 of steam and 5 m3 of water at 1 MPa.
Steam is taken out at constant pressure until 4 m3 of water is left. What is the
heat transferred during the process?
Solution:
Fig 23: Steam drum.
At 1 MPa vf = 0.001127 and vg= 0.1944 m3 / kg.
hg=2778.1 kJ/kg
uf = 761.68 , ufg = 1822 , ug = 2583.6 kJ/kg
The initial mass of saturated water and steam in the boiler = (Vf /vf) + (Vg / vg) =
[(5/0.001127) + (5/0.1944) ] = [(4.45 * 10 3) + (25.7) ] kg
Final mass of saturated water and steam = (4/0.001127) +(6/0.1944) = [ (3.55 *
103) + 30.8] kg
Mass of steam taken out of the boiler, ms = [4.45 * 10 3 + 24.7] - [ (3.55 * 103) + 30.8]
= 894.9 kg
Making an energy balance we have initial energy stored in saturated water and
steam + heat transfer from external source = final energy stored in saturated water
and steam + energy leaving the steam or
U1 + Q =U1 + ms*hg
Assuming that the steam taken out is dry.
Hence 4.45 * 103 * 761.68+27.7*2583.6+ Q = 3.55 * 103 *761.68 +30.8 *2583.6
+894.9 * 2778.1
Q = 2425000-685500+13176
Q = 1752676 kJ.
2) Steam flows in a pipeline at 1.5Mpa. After expanding to 0.1MPa in a
throttling calorimeter, the temperature is found to be 120 o C. Find the quality
of steam in the pipe line. What is the maximum moisture at1.5 MPa that can be
determined with this set-up if least 5 o C of superheat is required after
throttling for accurate readings?
Fig 24 : h –s Diagram
Solution.
At state 2 when p = 0.1 MPa and t = 120 o C by interpolation,
h2 = 2716.2 kJ/kg and
p = 1.5 MPa hf = 844.89 and hfg = 1947.3 kJ/kg
and h1 = h2 hf1+ x1 hfg1 = h2
844.89 + x1 *1947.3 = 2716.2
x1 = 1871.3 / 1947.3 = 0.963
Ans.
When p = 0.1MPa and t = 99.63 + 5 = 104.63 oC,
h3=2685.5 kJ/kg
Since h3=h4 2685.5 = 844.89 + x4 *1947.3
x4 = 1840.6 / 1947.3 = 0.948
The maximum moisture that can be determined with this set up is only 5.2% Ans
3) The following data were obtained with a separating and throttling
calorimeter:
Pressure in pipeline :1.5 MPa
Condition after throttling:0.1 MPa.110o C
During 5 min moisture collected in the separator:0.150 litre at 70 oC,
Steam condensed after throttling during 5 min
Find the quality of steam in the pipeline
Fig 25: h –s Diagram
Solution :
AT 0.1 MPa, 110 oC, h3 = 2696.2 kJ/kg
Now h3=h2 = hf2 + x2 hfg2
2696.2=844.89 + x2 1947.3
x2 = 1851.31/1947.3 = 0.955
If m1 = mass of moisture collected in the separator in 5 min and
m2= mass of steam condensed after throttling in 5 min then
x1 = (x2 m2) / (m1 + m2)
at 70oC vf = 0.001023 m3/kg
m1= 0.1462 kg
and m2 = 3.24 kg
Hence x1= 0.955*3.24 / (0.1462 +3.24) = 0.915 Ans.
4) A vessel having a volume of .4 m 3 contains 2.0 kg of liquid water and vapor
mixture in equilibrium at a pressure of 600 kPa. Calculate a) The volume and
mass of liquid b) The volume and mass of vapor.
Solution: The specific Volume is calculated first.
v =(0.4/2 ) = 0.2 m3/kg
The quality of steam can now be calculated
0.2 = 0.001101 + x * 0.3146, x = 0.6322
Therefore mass of liquid is 2.0*(1 - 0.6322) = 0.7356 kg
Mass of vapor is 2.0(0.6322) = 1.2644 kg
Volume of liquid is ml vf
= 0.7356(0.001101) =0.0008m3
Volume of vapor is mvvg = 1.2644(0.3157)=0.3992 kg
5) Steam at 1 bar and a dryness fraction of 0.523 is heated in a rigid vessel
until it becomes saturated vapor. Calculate the heat transferred per kg steam.
Solution:
From the steam table at 1 bar pressure, t s = 99.62oC,
vf =0.001043m3/kg,vfg=1.69296m3/kg,
uf =417.33 kJ/kg, ufg =2088.72kJ/kg,
Volume of one kg of given state of vapor = vf + x vfg
= 0.001043 + 0.523 * 1.69296 = 0.8864 m 3/kg
Enthalpy correspond to the state of steam,
u= uf + x ufg = 417.33 + 0.523 * 2088.72 =1509.73kJ/kg
v (0.8864 m3/kg ) correspond to the saturated condition of the steam, from steam
table, which is vg,
The pressure found to be 2 bar and u g =2529.49kJ/kg
Heat added = 2529.49– 1509.73 = 1019.76 kJ/kg Ans
.
6) A rigid vessel contains one kg of mixture of saturated water and saturated
steam at a pressure of 0.15MPa. When mixture is heated the state passes
trough the critical point. Determine,
a) Volume of the vessel
b) The mass of the liquid and vapor
c) The temperature of the mixture when the pressure rises to 3 MPa.
d) The heat transfer required to produce the final state.
Solution. Vc: Critical volume = 0.003155 m3/kg (From the steam table.)
v = vf + x X vfg
0.003155 = 0.001035 + x1 * 1.15828 ,
x1 = 0.00183
Hence, mass of vapor = x = 0.00183 kg
Mass of liquid = (1-x1) = 0.998 kg.
u1 = uf + x1 X ufg ;
u1 = 466.92 + 0.00183 X 2052072 = 470.68 kJ/kg
Saturation temperature correspond to 3 MPa, is 233.9 OC, the temperature of the
mixture
x2 = (0.003155 – 0.001216)/0.06546 = 0.02962
Heat transfer in constant volume process = u2 – u1,
u2 = 1004.76 + 0.02962 X 1599.34 = 1052.13 kJ/kg
Heat transfer = 1052.13 – 470.68 = 581.45 kJ/kg.
7) Steam initially at 0.3MPa, 250 OC is cooled at constant volume. Find
i) At what temperature will the steam become saturated vapor?
ii) What is the quality at 80OC? What is the heat transferred per kg of steam in
cooling from 250 OC to 80OC?
Solution:
At 300kPa, 250 OC, from the steam table , it is a superheated condition.
v = 0.79636 m3/kg , u = 2728.69 kJ/kg
vg = 0.79636 m3/kg , Ps = 225 kPa, ts = 124 OC
at 80 OC
0.79636 = 0.001029 + x 3.45612
x = (0.79636 - 0.001029) /3.45612 = 0.23
u2 = 520.45+0.23(2013.1) = 483.463 kJ/kg
Heat transfer = change in internal energy as the process is const. volume,
Heat transfer = u1 – u2 = 2728.69 – 983.463 = 1745.227 kJ/kg
8) State whether the following samples of steam are wet, dry or superheated:
Justify your answer.
I)
Pressure = 1 MPa absolute enthalpy = 2880 kJ/kg
II)
Pressure=500kPa absolute, volume =0.35m 3 /kg
III)
Temperature = 200oC Pressure = 1.2 MPa.
IV)
Temperature = 100oC, entropy =6.88kJ/kg K.
V)
Pressure= 10 kPa, enthalpy = 2584 kJ/kg.
Try as homework.
Problems for practice:
1) A vacuum pump is used to evacuate a chamber where some specimens are
dried at 500C. the pump rate of volume displacement is 0.5 m 3/s with an inlet
pressure of 0.1 kPa and temperature 323K. How much water vapor has been
removed over 30 minutes period?
2) A sample of steam from a boiler drum at 3 MPa is put through a throttling
calorimeter in which the pressure and temperature are found to be 0.1 MPa,
393K. Find the quality of the sample taken from the boiler.
3) Two streams of steam, one at 2MPa, 573K and the other at 2MPa, 673K
mixed in a steady flow adiabatic process. The rate of flow of two streams are
3 kg/min and 2 kg/min respectively. Evaluate the final temperature of the
emerging stream, if there is no pressure drop due to the mixing process.
What would be the rate of increase in the entropy of the universe?
4) Steam enters an engine at a pressure of 10 bar absolute and 523 K. it is
exhausted at 2 bar. The steam at exhaust is 0.9 dry. Find drop in enthalpy
and change in entropy.
5) Calculate the internal energy per kg of superheated steam at a pressure of 10
bar absolute and a temperature of 573 K. also find the change of internal
energy if this steam is expanded to 1.4 bar absolute and dryness fraction 0.8