Part A1
CHAPTER 1
Rankine and Brayton
Cycles
Here is what you will be able to do when you complete each objective:
1. Explain heat engines and their application to a steam power plant.
2. Explain the Rankine Cycle using a steam temperature-entropy diagram.
3. Evaluate a Rankine Cycle power plant in terms of efficiency, work ratio,
specific steam consumption, isentropic efficiency and efficiency ratio.
4. Explain the Rankine Cycle improvements that can be incorporated into a
power plant.
5. Explain the Brayton Cycle and its application to a gas turbine.
6. Explain the Brayton Cycle using pressure-volume and temperature-entropy
diagrams.
7. Evaluate a Brayton Cycle power plant in terms of temperatures, work output,
and efficiency.
8. Explain the Brayton Cycle improvements that can be incorporated into a
power plant.
9. Describe the design, layout, and advantages of a gas turbine / steam turbine
combined cycle plant.
10. Explain the total energy concept as it applies to a power plant.
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OBJECTIVE 1
Explain heat engines and their application to a steam power plant.
HEAT ENGINES
A power plant can be described in thermodynamic terms as a heat engine that
operates on a variety of thermodynamic cycles. The primary cycles involved in a
power plant are the Rankine Cycle which applies to a steam power plant and the
Brayton Cycle which describes a gas turbine application.
Any device used to convert heat into mechanical work is referred to as a heat
engine. A heat engine is a system that operates in a repeatable thermodynamic
cycle and performs net work from a supply of heat.
All heat engines have common characteristics. A conversion of energy from heat
to mechanical work must take place. The operations of heat supply and rejection
must be repeated to maintain the work output. There must be a relationship
between the quantities of heat involved and the amount of work done.
These characteristics lead to the following general statements that apply to all
heat engines:
1. All heat engines must have a source of heat, a working fluid and a
sink.
2. To produce continuous work, the processes of heat supply and heat
rejection are carried out cyclically.
3. The working fluid must be capable of returning to its original state at the
end of each cycle.
4. During the cycle of operations, the work done is equivalent to the
difference between the heat supplied and the heat rejected, reduced by
any inherent losses. Theoretical heat engines and thermodynamic cycles
are assumed to have no losses.
These statements are illustrated in Figure 1.
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FIGURE 1
Diagrammatic
Representation of a
Heat Engine
IDEAL CYCLES
Each practical engine in its present day form is a development of one of the
theoretical cycles proposed by scientists in the past. Figure 2 illustrates the
applications of the most common ideal cycles.
Reference to these cycles shows the basis of operation of each engine and
calculations based on these theoretically ideal cycles show the output work and
maximum efficiency obtainable.
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FIGURE 2
Thermodynamic
Cycles for a Heat
Engine
The Carnot Cycle is not a practical cycle, but since it results in the maximum
possible efficiency for any thermodynamic cycle, it is useful as a benchmark for
other cycles. Because power plants can operate on the Rankine or Brayton cycles,
these are the focus of this module.
HEAT ENGINE APPLIED TO A STEAM PLANT
The Rankine cycle used in a steam plant is classified as a closed cycle because the
working fluid circulates entirely within the plant components. The conceptual
heat engine shown in Figure 1 can be translated into the steam plant as illustrated
in Figure 3.
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FIGURE 3
Heat Engine Applied to
a Steam Plant
The boiler acts as the hot body where fuel is burned and heat is added to water
to turn it into steam. Work is extracted from the hot steam using a turbine. Heat
is then removed from the steam in the condenser after which a small amount of
work is added to pump the water to boiler feed pressure.
General statements that apply to heat engines are as follows:
1. The source of heat is the boiler fuel, the working fluid is the
steam/water circulating through the system and the sink (or place of
heat rejection) is the steam condenser.
2. The processes of heat supply and rejection are carried out cyclically upon
each portion of the working fluid as it progresses around the system.
3. The working fluid returns to its original state (as boiler feedwater) at the
completion of each circuit through the plants.
4. Finally, the maximum amount of work which can be achieved by such a
plant cannot be more than the difference between the heat supplied to,
and the heat rejected from, the working fluid used.
A standard version of a steam plant is provided in Figure 4.
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FIGURE 4
Steam Plant
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OBJECTIVE 2
Explain the Rankine Cycle using a steam temperature-entropy
diagram.
COMPARISON OF CARNOT AND RANKINE CYCLE
The Carnot cycle, the most idealized version of the heat engine, requires that
heat transfer from the heat source to the heat sink occur at a constant
temperature. This condition results in the most efficient cycle possible although,
technically, the Carnot cycle is not feasible.
The Rankine cycle is used as the standard for steam plants because it takes into
consideration some practical limitations which the Carnot cycle ignores. The
result is that the calculated Rankine efficiency of a plant shows a lower, but more
realistic, figure than the Carnot theoretical maximum.
A temperature - entropy (TS) diagram for steam, as shown in Figure 5, illustrates
the situation when a Carnot cycle diagram is superimposed upon it.
FIGURE 5
Carnot Cycle on
Temperature-Entropy
Diagram for Steam
The Carnot cycle demands that all heat supply be carried out at a constant top
temperature. In the case of a steam plant, this condition does not allow
superheating because the temperature rises during heat addition. Instead, the
Rankine cycle calls only for heat addition to be carried out at constant maximum
pressure.
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Further, the Carnot cycle requires a true adiabatic compression stage without any
heat addition or removal. This is also called an isentropic stage because it occurs
at constant entropy. When this principle is applied to a steam plant cycle, it does
not permit complete condensation of the steam in the condenser.
The compression here is shown to begin before condensation is complete in the
condenser. There is no physical way that partially condensed steam can be taken
from a condenser and handled in a feed pump.
The Rankine cycle allows changes that accommodate the practical requirements
of a steam plant as Figure 6 shows.
FIGURE 6
Rankine Cycle on
Temperature-Entropy
Diagram for Steam
Heat addition occurs at constant pressure into the superheat region.
Condensation is allowed to continue to completion and compression then takes
place in a feed pump back to the top pressure in the cycle.
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OBJECTIVE 3
Evaluate a Rankine Cycle power plant in terms of efficiency, work
ratio, specific steam consumption, isentropic efficiency and
efficiency ratio.
RANKINE CYCLE CALCULATIONS
There are various basic calculations applicable to a vapour cycle. They allow the
calculation of the overall efficiency of the cycle, the amount of work available,
the amount of steam required, and the isentropic efficiency of the turbine. To
keep things simple, these calculations are defined in terms of the basic Rankine
cycle without any of the improvements described in the next objective. However,
the same calculations can be applied to a modified Rankine cycle with minor
changes.
FIGURE 7
Basic Rankine Cycle
CYCLE EFFICIENCY
The primary concern for any cycle is its efficiency. The efficiency of any
thermodynamic cycle can be determined by the ratio of the net mechanical work
extracted from the cycle and the heat supplied. The heat converted into work is
equal to the heat supplied minus the heat rejected to the heat sink. This results in
the formula:
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work done
heat supplied
heat supplied - heat rejected
thermal efficiency =
heat supplied
thermal efficiency =
The efficiency of a steam plant is determined from the heat the boiler supplies to
the steam and the work extracted from the steam turbine due to the expansion of
the steam. The work done by the boiler feed pump is comparatively small so that
it can be ignored.
The heat supplied to the steam is the difference between the enthalpy of the
feedwater h1 and the steam as it leaves the boiler h2 or h2 – h1. Disregarding the
work of the feed pump, h1 is equivalent to hf and the heat supplied is equal to h2 –
hf.
The work done is likewise the difference between the enthalpy at the boiler exit
h2 and the enthalpy h3 at the exit of the turbine or h2 – h3.
The efficiency is thus the ratio
efficiency =
work done
heat supplied
η=
h2 − h3
h2 − h f
STEAM CONSUMPTION
The next important calculation is the one that determines the amount of steam
required to produce a certain amount of work output. This is known as specific
steam consumption. The amount of steam needed directly determines the size of
the components, especially the size of the boiler and turbine, and is a very useful
comparative value.
The steam consumption can be derived from the amount of work produced
which is normally given in kJ/kg of steam. The specific steam consumption is
stated in units of measure as kg/kWh and can be found using the equation:
EQUATION 3.1
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EQUATION 3.2
kg ⎤
⎡ kg
⎡s⎤
⎢⎣ kJ or kWs ⎥⎦ × 3600 ⎢⎣ h ⎥⎦
3600 ⎡ kg ⎤
=
W ⎢⎣ kWh ⎥⎦
specific steam consumption =
=
1
W
3600
W
where W = work in kJ/kg.
ISENTROPIC EFFICIENCY
The expansion process in the turbine as well as the recompression by the pump
are not entirely reversible in practice and thus their efficiency can never be 100%.
The efficiency is known as the isentropic efficiency because it entails an increase
in entropy. Referring to Figure 8, actual expansion occurs from point 2 to point
3 as compared to ideal isentropic expansion from point 2 to point 3΄. Likewise,
actual compression occurs from point 4 to point 5 instead of the ideal case from
point 4 to point 5΄ (Figure 9).
FIGURE 8
Expansion with
Isentropic Losses
Isentropic efficiency can be defined as:
EQUATION 3.3
isentropic efficiency =
actual enthalpy drop
isentropic enthalpy drop
isentropic efficiency =
h2 − h3
h2 − h3′
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WORK RATIO
In the Rankine cycle, not only is work extracted, but some work is also needed
for the process to pump the water back to boiler feed pressure. In other words,
the positive work from the turbine is partially offset by the negative work of the
pump. The measure used to indicate this action is called the work ratio rw and is
defined as the ratio of the net work to the positive work done in the cycle.
Referring to Figure 7, the work ratio is expressed as:
work ratio =
net work
positive work
rw =
EQUATION 3.4
W23 − W45
W23
Because irreversibilities decrease the positive work and increase the negative
work, the work ratio is a strong indicator of the effect that irreversibility has on
the efficiency of a system. Therefore, a high cycle efficiency, together with a high
work ratio, is a reliable indicator that the real power plant will have good overall
efficiency.
EFFICIENCY RATIO
The final measure is the efficiency ratio. This is the ratio of the actual efficiency
obtained and the ideal cycle efficiency and is thus an indicator of the total impact
of the reversibilities in the whole cycle. The ideal cycle efficiency is calculated
using isentropic processes compared to the actual efficiency (see Figure 9).
FIGURE 9
Rankine Cycle with
Isentropic Losses
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Using Figure 9:
ideal efficiency =
W23′ − W45′
Q5′2
where Q = the difference in enthalpy between the two designated points.
W − W45
actual efficiency = 23
Q52
Because the work of the pump is minor when compared to the turbine work and
the difference in pump efficiency is also quite minor, Q52 is nearly equal to Q5’2.
The efficiency ratio ends up being close to the isentropic efficiency or
actual efficiency
ideal efficiency
W23 − W45
Q52
=
W23′ − W45′
Q5′2
EQUATION 3.5
efficiency ratio =
efficiency ratio =
W23 − W45
W23′ − W45′
efficiency ratio =
W23
= isentropic efficiency
W23′
Example 1
A steam plant operates between pressures of 3000 kPa and 4 kPa. Isentropic
efficiency of the expansion and compression processes is 0.80. Using the basic
Rankine cycle process as in Figure 9, calculate:
a)
b)
c)
d)
e)
ideal efficiency
actual efficiency
efficiency ratio
work ratio for the actual cycle
specific steam consumption for the actual cycle
Solution
a) From the Steam tables, h2 = 2804.2 kJ/kg
A1 • Chapter 1 • Rankine and Brayton Cycles
For the ideal cycle, h3΄ is at the same entropy as h2 and this calculation can be
used to find the quality of the steam at 4 kPa and from there the enthalpy at
point 3΄.
s2 = s3′ = 6.1869 kJ/kgK
s3 = s f + X 3 s fg
6.1869 kJ/kgK = 0.4226 kJ/kgK + X 3 × 8.052 kJ/kgK
6.1869 kJ/kgK − 0.4226 kJ/kgK
8.052 kJ/kgK
= 0.716
X3 =
h3′ = h f + X 3 h fg
= 121.46 kJ/kg + (0.716 × 2432.9 kJ/kg)
= 1863.4 kJ/kg
The expansion work from the turbine is therefore
W = h2 − h3′
= 2804.2 kJ/kg − 1863.4 kJ/kg
= 940.8 kJ/kg
The compression work W45΄ can be found from the equation
W45′ = v( p5 − p4 )
= 0.001(3000 kPa − 4 kPA)
= 3 kJ/kg
This result can be used to calculate the enthalpy at point 5΄.
h4 = h f (at 4 kPa) = 121.46 kJ/kg
h5′ = h4 + W45′
= 121.46 kJ/kg + 3 kJ/kg
= 124.5 kJ/kg
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The ideal efficiency is:
ηi =
h2 − h3
h2 − h5′
2804.2 kJ/kg − 1863.4 kJ/kg
2804.2 kJ/kg − 124.5 kJ/kg
940.8 kJ/kg
=
2679.7 kJ/kg
= 0.351 or 35.1% (Ans.)
=
b) For the actual cycle, the expansion work will be:
actual expansion work = W23 × isentropic efficiency
= 940.8 kJ/kg × 0.8
= 752.6 kJ/kg
W45′
isentropic efficiency
3 kJ/kg
=
0.8
= 3.8 kJ/kg
actual compression work =
The enthalpy at point 5 now becomes:
h5′ = h4 + W45
= 121.46 kJ/kg + 3.8 kJ/kg
= 125.3 kJ/kg
actual efficiency =
W23 − W45
Q52
725.6 kJ/kg − 3.8 kJ/kg
2804.2 kJ/kg − 125.3 kJ/kg
721.8 kJ/kg
=
2678.9 kJ/kg
= 0.269 or 26.9% (Ans.)
=
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c) The efficiency ratio is:
actual efficiency
ideal efficiency
0.269
=
0.351
= 0.766 or 76.6% (Ans.)
efficiency ratio =
d) The work ratio for the actual cycle is:
rw =
W23 − W45
W23
752.6 kJ/kg − 3.8 kJ/kg
752.6 kJ/kg
= 0.995 (Ans.)
=
e) The specific steam consumption is:
3600 s/h
W
3600 s/h
=
752.6 kJ/kg
= 4.78 kg/kWh (Ans.)
s.s.c. =
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OBJECTIVE 4
Explain the Rankine Cycle improvements that can be incorporated
into a power plant.
There are several improvements that can be made to the Rankine cycle to
increase the cycle efficiency. They can achieve efficiency in some combination of
three basic aims:
•
Increasing the enthalpy of the steam by raising its temperature
•
Increasing the amount of energy which is recovered from the steam to
do useful work so that less energy is wasted in the heat sink (i.e. the
condenser)
•
Increasing the amount of energy which is transferred from the
combustion gases to the steam, so that less energy is lost in the stack
Even a small gain in efficiency will have large benefits to a plant because of the
resulting reduction in fuel cost and/or increase in the energy output. In each
case, there is also a cost associated and, the balance between cost and benefit
must be assessed before any improvements are made in the initial plant design or
in later redesigns.
Raising the working pressure of a boiler to increase the pressure of the steam
increases efficiency by increasing its enthalpy. For this reason, the evolution of
steam power plants has included a trend towards ever higher steam pressures up
to the supercritical pressures that are now sometimes used. The technology and
metallurgy of boilers and downstream equipment have been the limiting factors.
However, this is not strictly a Rankine cycle improvement since the relationships
between pressure and enthalpy, or between temperature and entropy, are
unchanged. Additional heat energy added to the boiler furnace is simply reflected
in additional heat energy in the steam. The P-V and T-S diagrams are
repositioned on their respective scales, but their shape will remain similar. On the
other hand, increasing the working temperature of the steam creates a distinct
change in the cycle parameters. That is because the temperature of the steam
remains constant as latent heat is added, as can be seen in Figures 7 and 9, until
the steam begins to be superheated. From that point on, the sensible heat that is
added produces a major reshaping of the T-S diagram.
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SUPERHEATING
The first improvement is superheating which entails continuing the boiler
heating process so that steam is heated into the superheat region with a
temperature higher than the saturated temperature. The general layout of a steam
plant with a superheater is shown in Figure 10 along with the resulting
temperature-entropy diagram. The superheater is located after the boiler’s steam
drum in the water / steam flow path to raise the temperature without raising the
boiler pressure.
FIGURE 10
Rankine Cycle with
Superheat
Compare the temperature-entropy diagram in Figure 10 with those in Figures 7
and 9 which show the Rankine cycle without superheating. The gain in entropy
as the steam is superheated can be clearly seen as the superheat cycle diagram
extends to the right well beyond the superimposed curve that is delineated by
saturated steam and water. This gain in entropy accompanies a gain in enthalpy,
indicating that more energy has been transferred to the steam. The additional
energy can be recovered in the form of additional useful work done. Figure 11
shows two temperature-entropy diagrams superimposed on each other with the
only difference being that one of them, labelled p1, has a higher working
temperature than the other, labelled p2. The temperature increase, as a result of
additional superheating, produces a significant gain in entropy (and thus,
enthalpy and work done.)
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FIGURE 11
Rankine Cycle
Superheat
Temperatures
The benefit of a superheater is that more work can be extracted from a specific
amount of steam. For a given amount of work, the specific steam consumption
is, therefore, lower and the size of all of the downstream components is also
reduced. This is evident in Figure 12 which shows graphically that efficiency
increases and specific steam consumption is reduced as the steam temperature is
increased.
FIGURE 12
Steam Temperature
Gains
There is another reason why superheating is desirable. As seen in Figure 7,
without superheating, the steam is quite wet when it exits the turbine. Wet steam
is undesirable in a steam turbine because it causes corrosion and erosion.
Superheating results in a higher dryness fraction as is clear from Figure 10.
Disadvantages of superheating include:
•
The increased capital investment required to build a new plant or to
modify an existing one because of the additional heating surface, piping,
temperature control system, floor space, and, possibly, building height.
•
The maintenance, inspection, and repair costs of the increased heating
surface.
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•
The requirement for a desuperheating or attemperation system to
prevent overheating of the superheater, steam piping, and downstream
equipment.
•
Added resistance to flow of the boiler combustion gases which requires
greater capacity from the boiler draft fans.
•
A risk of gas pass plugging where solid fuels are used as the ash tends to
accumulate on the superheater elements. This, in turn, requires extra
sootblowing equipment and added diligence toward the sootblowing
operation.
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The degree of superheat is limited by the metallurgy of the superheater tubing. It
is rare for a plant to have superheat temperatures above 565°C; and 593°C
remains a practical limit for nearly all new plants.
REHEATING
In a reheat cycle, expansion takes place in two turbines with constant pressure
reheating in between. The reheating normally occurs back to the original
superheat temperature but at a lower pressure. This condition is achieved with a
bank of reheater tubes either inside the boiler or in a separately fired reheater
furnace. The modified reheat cycle is illustrated in Figure 13 with the
accompanying temperature-entropy diagram.
FIGURE 13
Rankine Cycle with
Reheat
It can be seen in Figure 13 that reheating, like superheating, increases steam
entropy (and enthalpy) as the temperature is raised.
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The main reason for reheat is to avoid too wet a condition at the exit of the
turbine; this has become more of a concern as boiler pressures have increased.
There is a significant reduction in the specific steam consumption which
translates into smaller equipment such as a smaller boiler. Efficiency gains are
not usually the primary reason to use reheat.
The disadvantages and metallurgical limitations of reheating are the same as for
superheating. The large, high-pressure, forced circulation boilers now commonly
used in central power generating stations would not be possible without
reheating because too much condensation would occur in the steam as it passed
through the turbine.
REGENERATIVE FEEDWATER HEATING
It is a common practice with large steam turbines to extract or bleed some of the
steam from the turbine for use in a shell-and-tube feedwater heater. During this
procedure, heat energy that would otherwise be lost in the condenser can be used
to raise the temperature of the boiler feedwater prior to its entering the boiler.
The heat addition step of the Rankine cycle is then carried out with a smaller
temperature increase and less sensible heat is required to achieve saturation
temperature. In this way, the efficiency of the process more nearly approaches
the ideal Carnot cycle which requires constant temperature heat addition, as
shown in Figure 5.
Figure 14 shows the layout of a steam plant with a single feedwater heater and
the accompanying temperature-entropy diagram.
FIGURE 14
Regenerative
Feedwater Heating
Note that y kg of steam, representing the bled steam going to the heater, is
returned with a substantially higher temperature to the feedwater stream, while
more entropy is extracted (i.e. more work is done) from the stream that continues
through the turbine (1 – y kg).
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Modern power plants often utilize six or more feedwater heaters with the
feedwater flow in series through them. The steam for each one is bled from a
different location on the turbine, so that the feedwater is heated by a succession
of ever higher steam pressures in each successive heater. Very large plants may
use as many as 12 heaters, sometimes with two complete trains of heaters parallel
to each other.
The disadvantages of regenerative feedwater heating are as follows:
•
Increases in plant capital and maintenance costs which can be
considerable
•
A considerable increase in plant complexity
•
A resulting pressure drop in the feedwater requiring larger capacity boiler
feedwater and condenser extraction pumps
•
Specific steam consumption is increased
ECONOMISER AND AIR PREHEATER
There are further efficiencies to be obtained by making use of the heat contained
in the flue gases since they exit the boiler at a reasonably high temperature.
As shown in Figure 15, the economiser preheats water from the condenser and is
located after the superheater in the flue gas stream, so that the effectiveness of
the superheater is not reduced. As with regenerative feedwater heating, the
process efficiency more nearly approaches the ideal Carnot cycle as the feedwater
more nearly approaches its saturation temperature.
FIGURE 15 (a) (b)
Rankine Cycle with (a)
Economiser and (b) Air
Preheater
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An additional way of extracting heat from the combustion gases is to preheat the
intake air with an air preheater. Again, this provides a relatively free increase in
combustion efficiency because less fuel is required to heat the air.
Disadvantages of economizers and air preheaters are:
•
•
An increase in the complexity and cost of the boiler.
A requirement for more boiler draft fan and feedwater pump capacity.
OTHER CONSIDERATIONS
Many plants utilize steam that is bled or extracted from the turbine for purposes
outside the Rankine cycle flow. This steam may be for process heating in a
process or manufacturing plant, for space heating, or be sold to outside
customers for their own use. The resulting impact on thermal efficiency is
dependent on the efficiency of the usage of the bled steam. If more heat energy
is extracted from the bled steam than could have been recovered as useful work
in the turbine, then there is a net gain in efficiency; this is often the case.
Disadvantages include:
•
An increase in the specific steam consumption.
•
If the resulting condensate is not returned to the cycle, then additional
raw water is treated and used for makeup. This process affects both the
cost of water treatment and the plant capacity that must be available for
treating water.
•
There is a possibility that the overall thermal efficiency may be reduced if
the bled steam is not efficiently used. This is especially true if the
condensate is not returned because any heat energy remaining in it will be
lost.
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OBJECTIVE 5
Explain the Brayton Cycle and its application to a gas turbine.
HEAT ENGINE APPLIED TO A GAS TURBINE
The gas turbine is the other major type of heat engine used in a power plant. The
conceptual heat engine shown in Figure 1 can be redesigned into a gas turbine as
shown in Figure 16.
FIGURE 16
Heat Engine Applied to
a Gas Turbine
The gas turbine operates on the Brayton cycle which is an open cycle; that is, air
is taken in from the atmosphere and then exhausted back at the end of the
process. The intake air is first compressed. Fuel is added and combusted to
increase the air temperature. The heated air is expanded through a turbine and
then returned to the atmosphere.
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The general statements that apply to all heat engines are:
1. The source of heat is the fuel, the working fluid is the air, and the sink
(or place of heat rejection) is the atmosphere.
2. The processes of heat supply and rejection are carried out cyclically upon
each portion of the working fluid as it progresses around the system.
3. The working fluid returns to its original state (as exhausted air) at the
completion of each circuit by mixing with the large amount of
atmospheric air.
4. Finally, the maximum amount of work which could be achieved by such
a plant cannot be more than the difference between the heat supplied to,
and the heat rejected from the working fluid used.
A practical layout of a gas turbine is given in Figure 17.
FIGURE 17
A Gas Turbine
A1 • Chapter 1 • Rankine and Brayton Cycles
OBJECTIVE 6
Explain the Brayton Cycle using pressure-volume and temperatureentropy diagrams.
BRAYTON CYCLE
The Brayton cycle consists of an air compressor and turbine operating in series
with a combustor in between as previously illustrated in Figure 17. Because
energy is not stored at any point, it is the steady flow type.
Air is drawn into the compressor from the atmosphere and compressed
adiabatically to high pressure with a decrease in volume and an increase in
temperature.
It is then discharged to the combustor where fuel is added and burned which
gives a heat addition at constant pressure and a consequent small increase in
volume.
The working fluid is then passed to the turbine where it does work by expanding
adiabatically down to atmospheric pressure again. It is finally exhausted to the
atmosphere to complete the heat rejection process.
Figure 18 shows the Brayton cycle on the pressure-volume (PV) and
temperature-entropy (T-S) diagrams. In summary:
Stage 1 - 2
Adiabatic compression of the air in the compressor.
Stage 2 – 3
Heat addition at constant pressure by burning of the fuel
in the combustor.
State 3 – 4
Adiabatic expansion of the working fluid through the
turbine.
Stage 4 - 1
Exhaust to atmosphere at constant pressure.
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FIGURE 18
Brayton Cycle Shown
on P-V and T-S
Diagrams
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OBJECTIVE 7
Evaluate a Brayton Cycle power plant in terms of temperatures,
work output, and efficiency.
BRAYTON CYCLE CALCULATIONS
The calculations for a Brayton cycle are similar to those for the Rankine cycle
although there are some differences. One difference is that gas turbines are open
cycle (although a closed cycle is possible, it is rarely used) and the intake
temperature and pressure and exhaust pressure are atmospheric. The other major
difference is that the combustor and turbine contain not only air, but also the
products of combustion. Usually, this addition of mass from the fuel and
differences in properties of combusted products are ignored and properties for
air are assumed throughout the gas turbine.
EFFICIENCY OF BRAYTON CYCLE
The efficiency of a Brayton cycle is obtained from the basic equation for thermal
efficiency:
thermal efficiency =
=
heat supplied - heat rejected
heat supplied
mC p (T3 − T2 ) − mC p (T4 − T1 )
mC p (T3 − T2 )
η = 1−
T4 − T1
T3 − T2
It can be seen that this is the same equation as was derived for the Otto cycle.
However, the version based on compression ratio is a bit different because, with
the Brayton cycle, the ratio of pressures is relevant whereas with the Otto cycle
the pressure ratio is based on the volume ratio. Referring back to Figure 18, the
pressure ratio is defined as:
rp =
p2 p3
=
p1 p4
EQUATION 7.1
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A1 • First Class • SI Units
The temperatures are related to the pressures by the equation:
T2 ⎛ p2 ⎞
=⎜ ⎟
T1 ⎝ p1 ⎠
T3 ⎛ p3 ⎞
=⎜ ⎟
T4 ⎝ p4 ⎠
T3 − T2 = rp
γ −1
γ
γ −1
γ
γ −1
γ
and T2 = T1 ( rp )
γ −1
γ
and T3 = T4 ( rp )
γ −1
γ
(T4 − T1 )
This can be substituted into equation (7.1) which results in the efficiency being
dependent on the pressure ratio:
EQUATION 7.2
γ −1
1 γ
η = 1− ( )
rp
In practice, pressure ratios for gas turbines are mostly between 10 and 15
although some engines exceed 20. High pressure ratios are difficult to achieve
efficiently due to increased compressor temperatures. There is also a
metallurgical limit to combustion and turbine components that limits the
efficiency that can be obtained to about 40% with 30-35% being typical.
Consider the T-S diagram in Figure 15. T3 on the TS diagram is the limiting value
of temperature at inlet to the turbine which the turbine blading can withstand.
This temperature T3 is reached in two steps:
T1 - T2 which is the temperature rise due to compression
T2 - T3 which is the temperature rise due to burning of the fuel in the
combustor
If the compressor pressure ratio was maximum and hence T1 to T2 maximum,
the cycle on TS diagram Figure 19, would be abcd which would give high
efficiency, but obviously small output work. Alternatively, if the compressor
pressure ratio was minimum and hence T1 to T2 minimum, the cycle would be
aefg which would again give low efficiency together with small work output.
A1 • Chapter 1 • Rankine and Brayton Cycles
33
FIGURE 19
Limits of the Brayton
Cycle
It can be said that if the temperature T1 (temperature of the atmosphere) and the
turbine inlet temperature T3 are fixed, there will be a certain compressor pressure
ratio which gives the maximum work output and the best practical efficiency of
the gas turbine.
Reciprocating engines operating on the Otto and Diesel cycles are limited in
power output by the mass of the working fluid which they can handle; whereas,
the gas-turbine with its constant flow is not limited in this way.
WORK OUTPUT
The work done by the compressor is supplied by the turbine and must be
deducted from the turbine output to arrive at the net output from the plant.
From Figure 18 the work done by the compressor is shown as area 1, 2, b, a, the
turbine output is shown as area 4, 3, b, a and the net work of the cycle is
therefore given by the area 1, 2, 3, 4, which is the Brayton cycle diagram.
The net work output is the difference between the turbine work and compressor
work or
W = W34 − W12
EQUATION 7.3
= mC p (T3 − T4 ) − mC p (T2 − T1 )
W = C pT (
1
rp
γ −1
γ
− 1) − C pT3 (1 − rp
γ −1
γ
)
34
A1 • First Class • SI Units
Example 2
A gas turbine working on the air standard Brayton cycle operates between
pressures of 100 kPa and 1000 kPa. The inlet air temperature is 20°C and the
maximum temperature reached in the cycle is 1090°C. γ = 1.4.
Calculate:
a) The temperature at the end of the compression and of expansion
b) The heat supplied, heat rejected, and net work per kg of air
c) The efficiency
Solution
a) The temperature at the end of compression is:
From data given
P1
P2
T1
=
=
=
=
=
=
T3
100 kPa
1000 kPa
20 + 273
293 K
1090 + 273
1363 K
T2 = T1 ( rp )
γ −1
γ
1.4 −1
= 293K × (10 ) 1.4
= 565K or 292°C (Ans.)
The temperature at the end of expansion is as follows:
For the adiabatic expansion stage 3 - 4 it can be shown that
0.4
T3 ⎛ P3 ⎞ 1.4
=⎜ ⎟
T4 ⎝ P4 ⎠
But since P2 = P3 and P1 = P4
0.4
T3 ⎛ P2 ⎞ 1.4 ⎛ T2 ⎞
=⎜ ⎟ =⎜ ⎟
T4 ⎝ P1 ⎠
⎝ T1 ⎠
Given
T3
T2
T1
= 1363 K
= 565 K
= 293 K
A1 • Chapter 1 • Rankine and Brayton Cycles
T4 = T3 ×
T1
T2
293 K
565 K
= 707 K or 434°C (Ans.)
= 1363 K ×
b) Where m
=
cp
1 kg
= 1.005 kJ/kgK (Specific heat at constant pressure for air)
Heat supplied = mC p (T3 − T2 )
= 1 kg × 1.005 kJ/kgK (1363 K − 565 K )
= 802 kJ (Ans.)
Heat rejected = mC p (T4 − T1 )
= 1 kg ×1.005 kJ/kgK ( 707 K − 293 K )
= 416 kJ (Ans.)
net work per kg of air = heat supplied − heat rejected
= 802 kJ − 416 kJ
= 386 kJ (Ans.)
c)
work done
heat supplied
386 kJ
=
802 kJ
= 0.4813 or 48.13% (Ans.)
efficiency =
Checking this from the efficiency equation given earlier for the Brayton Cycle:
1
rp
η = 1− ( )
γ −1
γ
100 kPa
= 10
1000 kPa
1.4−1
1 1.4
η = 1− ( )
10
= 1 − 0.518
= 48.2%
where rp =
35
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A1 • First Class • SI Units
OBJECTIVE 8
Explain the Brayton Cycle improvements that can be incorporated
into a power plant.
BRAYTON CYCLE IMPROVEMENTS
To improve the efficiency of the basic gas turbine cycle, four approaches (or a
combination of them) can be implemented.
The four possibilities for improving the gas turbine cycle are inlet cooling,
intercooling, regeneration, and reheat. For various compatibility reasons, they are
normally used individually and, at the moment, no gas turbine exists that uses all
of these methods. As simple cycle gas turbines are improving in efficiency, these
cycle improvements are becoming less necessary and combined cycle (utilizing
waste heat for other purposes) is becoming more prevalent. Only inlet cooling
remains as a commonly applied technology for new plant design.
INLET COOLING
Inlet air cooling for a gas turbine will be described in more depth in a later
module. To understand its significance, refer to Figure 15. The line from
points 1 to 2 represents the compression of the inlet air. On the temperatureentropy diagram, if the inlet temperature is reduced (point 1), the area that is
circumscribed by the Brayton Cycle is increased; this represents an increase in
the useful work done. Reference to Formula 7.1 also shows a positive effect of
reducing the inlet temperature.
η = 1−
T4 − T1
T3 − T2
It can be seen that reducing T1 (the inlet air temperature) increases η , the cycle
efficiency. Thus, inlet cooling both increases the work that can be done and the
efficiency of the process.
Cooling of the inlet air can be achieved through a closed refrigeration system, by
using water sprays or using a heat exchange process with a lower temperature
fluid. Some form of inlet cooling is common on large gas turbine installations.
A1 • Chapter 1 • Rankine and Brayton Cycles
37
REGENERATION
The most common cycle improvement in the past has been the regenerative
cycle, or regeneration, where exhaust heat is used to increase the temperature of
compressed air before combustion. This process is accomplished by installing a
heat exchanger in the exhaust to preheat the air between the compressor and the
combustors as shown in Figure 20.
FIGURE 20
Regeneration
FIGURE 21
Regeneration T-S
Diagram
Figure 21 illustrates the effect of regeneration on a temperature-entropy diagram.
Comparing this to the basic Brayton cycle shown with dotted lines, it can be seen
that regeneration produces a loss in available work during the compression stage
(1 - 2), but that this loss is more than matched by a larger gain in the expansion
stage (3 - 4.)
This process approach was quite common until recently because it allowed the
efficiency of the gas turbine to be improved by 15-20%. Disadvantages are the
increased capital cost and the fact that there are increased pressure losses with
the newer high pressure ratio compressors. Instead, many installations now use
the exhaust heat for combined cycle or cogeneration.
38
A1 • First Class • SI Units
INTERCOOLING
In some gas turbines, the compression of the inlet air is done in two stages by
using a dual shaft arrangement with the air being cooled between the stages in a
heat exchanger (intercooler) as shown in Figure 22. Because isothermal
compression (compression without an increase in temperature) takes less work
than adiabatic compression (compression where no heat is removed so that the
air temperature increases), more of the turbine power is available for the output
load. Another advantage of intercooling is that the specific volume of the air is
reduced which allows a smaller machine size. The effect on a temperatureentropy diagram is similar to the effect of regeneration as shown in Figure 21,
but the effect of intercooling is less pronounced.
FIGURE 22
Intercooling
However, the beneficial effect of intercooling decreases with pressure ratio. A
high pressure ratio also means that losses through the intercooler become more
significant. Using an intercooler makes more sense if it is combined with
regeneration because much more of the exhaust heat will be recovered resulting
in improved overall cycle efficiency.
The intercooler is a shell and tube heat exchanger similar in construction to the
regenerator. Cooling water passes through the tubes while the air passes over the
outside of the tubes. In some cases the air may pass through tubes surrounded
by water.
REHEATING
In addition to compressing the air in two stages and intercooling between these
stages, the gas turbine may also be arranged to expand the hot gases in two
stages, with the gases being reheated between the stages. The gases are expanded
first in a high pressure turbine and then in a second set of combustion chambers
before entering and expanding through a low pressure turbine (see Figure 23).
A1 • Chapter 1 • Rankine and Brayton Cycles
39
FIGURE 23
Reheat
The effect of this reheating is to increase the energy content of the gases; as a
result the thermal efficiency of the cycle will be somewhat improved. However,
the major purpose of reheat is to increase the specific work output. The effect on
the temperature-entropy diagram can be seen in Figure 24 which clearly shows
the increase in the area within the cycle diagram representing an increase in work
available.
FIGURE 24
Reheat T-S Diagram
There are some gas turbines available today that use the reheat cycle, but they are
fairly rare.
40
A1 • First Class • SI Units
OBJECTIVE 9
Describe the design, layout, and advantages of a gas turbine /
steam turbine combined cycle plant.
COMBINED CYCLE
One aspect of the gas turbine is that a large amount of heat is available in the
exhaust with typical exhaust temperatures reaching 400°C-600°C. This heat can
be at least partially recovered in a waste heat recovery system with the restriction
that the final temperature not be reduced below the dewpoint to avoid corrosion.
If the waste heat is used to provide additional generation capability, the result is
referred to as combined cycle.
With waste heat recovery, the thermal efficiency can be increased from a typical
simple cycle gas turbine efficiency of 30-40% to a total plant efficiency of 6070%.
The most important use of the combined cycle is in large baseload power
generation applications that can exceed 1000 MW of total power. The exhaust
gases are routed to a Heat Recovery Steam Generator (HRSG) or Once-Through
Steam Generator (OTSG) that supplies steam to a steam turbine which may
drive a separate generator or, in some cases, be attached directly to the same
generator as the gas turbine. These two configurations are shown in Figure 25.
FIGURE 25 (a)
Combined Cycle
Configurations
(a) Single Generator
with Steam Turbine on
Common Shaft
A1 • Chapter 1 • Rankine and Brayton Cycles
41
FIGURE 25 (b)
Combined Cycle
Configurations
(b) Separate
Generators with
Common Steam
Turbine
The HRSG may be unfired; that is; no extra heat is added or it may be fired. In
which case, an additional burner is installed in the inlet duct, just before the
HRSG, to increase the temperature of the exhaust gas. This duct burner has an
advantage because it can compensate for reductions in gas turbine output if it is
run at part load and increase the output of the HRSG. It also prevents corrosion
in the back passes of the HRSG due to the flue gas temperature falling below the
acid dewpoint.
Early combined cycle installations used a single-pressure HRSG. Modern HRSGs
may consist of a double-pressure or triple-pressure configuration to enable the
maximum extraction of heat and the greatest resultant efficiency. The choice of
HRSG is dependent on the temperature of the exhaust and a triple-pressure
HRSG is most suitable for gas turbines with a high firing temperature above
550°C.
COGENERATION
Where the waste heat produces steam or hot water for other uses such as
heating, cooling or general steam application, it is called cogeneration or
combined heat and power (CHP).
Cogeneration for gas turbines is increasingly found in distributed power
applications. One or more gas turbines producing power have their exhaust fed
to a common HRSG which also has supplementary firing so that the amount of
42
A1 • First Class • SI Units
steam can be varied. There is a diverter valve on each engine in case steam is not
required.
Steam power plants may also incorporate additional heat exchangers after the
turbine to provide additional steam or water heating for industrial or residential
purposes. This action boosts their efficiency to higher levels than are normally
achieved.
A1 • Chapter 1 • Rankine and Brayton Cycles
OBJECTIVE 10
Explain the total energy concept as it applies to a power plant.
As the need for energy efficiency increases, power plant designers are constantly
improving the overall efficiency of power plants by optimizing the use of the
energy that is consumed.
Using all of the available energy in a plant to do useful work or to supply heating
and other needs, and minimizing or eliminating wastage of energy, requires an
integrated approach to assessing all the energy sources and uses in the plant. This
approach is called the total energy concept. Its application begins with a
structured and detailed energy audit for the plant; and this process will be
described in a later module. All available energy sources are considered, including
recovery of waste heat. All available energy consumption is also considered,
including low-level needs such as lighting and space heating.
Attention to the total energy package includes the use of waste heat for power
generation or the production of steam for industrial or residential heating
purposes. Combined cycle operations and cogeneration were described in the last
learning objective. On a smaller scale, heat recovery is often possible from plant
and process equipment. For example, an engine used for power generation often
generates a significant stream of cooling water that has been heated in the
engine’s cooling jacket or lubricating oil cooler. This heated water may be usable
for heat recovery at another location.
43
44
A1 • First Class • SI Units
CHAPTER QUESTIONS
1. State the general heat engine characteristics of a steam plant.
____________________________________________________________
____________________________________________________________
____________________________________________________________
2. Draw a T-S diagram for a basic Rankine cycle without improvements along
with a corresponding layout of the major components.
3. Calculate the efficiency of a Rankine cycle with boiler stop valve steam
conditions of 7000 kPa and 800 K (enthalpy is 3476 kJ/kg). The turbine
exhaust conditions are 3.75 kPa and 300K with 10% wetness (enthalpy is
2309 kJ/kg). The enthalpy of feedwater at entry to the boiler is 110 kJ/kg,
assuming no feed heating.
____________________________________________________________
4. Define work ratio and calculate it for a situation where the turbine work is
900 kJ/kg and the feedwater pump uses 50 kJ/kg.
____________________________________________________________
5. Describe these possible improvements to the Rankine cycle and indicate the
primary benefit for each one:
a) Superheating
b) Reheat
c) Economiser and air preheater
A1 • Chapter 1 • Rankine and Brayton Cycles
6. Describe the Brayton cycle and illustrate with P-V and T-S diagrams.
____________________________________________________________
____________________________________________________________
____________________________________________________________
____________________________________________________________
7. A gas turbine working on the air standard Brayton cycle operates between
pressures of 100 kPa and 900 kPa. The inlet air temperature is 15°C and the
maximum temperature reached in the cycle is 1100°C. Calculate:
a) the temperature at the end of the compression and at the end of
expansion
b) the heat supplied, heat rejected and net work per kg of air.
c) the efficiency
45
46
A1 • First Class • SI Units
8. Describe these possible improvements to the Brayton cycle and indicate the
primary benefit for each one:
a) regeneration
b) intercooling
c) reheat
Part A1
CHAPTER 2
Thermodynamics of
Steam
Here is what you will be able to do when you complete each objective:
1. Describe the basis for non-flow processes of vapours.
2. Explain the constant volume process for steam and calculate heat
supplied, work done and internal energy.
3. Explain the constant pressure process for steam and calculate heat
supplied, work done and internal energy.
4. Explain the constant temperature process for steam and calculate heat
supplied and work done.
5. Calculate steam entropy given the steam conditions.
6. Explain the significance of a Temperature-Entropy diagram for steam.
7. Explain the reversible adiabatic process for steam and calculate work
done and internal energy.
8. Explain the significance of a Mollier chart for steam.
47
A1 • Chapter 2 • Thermodynamics of Steam
49
OBJECTIVE 1
Describe the basis for non-flow processes of vapours.
PROPERTIES OF VAPOURS
The solutions for practical problems dealing with vapour processes, whether
non-flow (closed cycle) or uniform-flow (steady-flow open cycle) systems, are
complicated beyond those of perfect gases because the working fluid may be wet,
dry, or superheated at any given stage in the process.
Steam is the most common vapour encountered in power engineering work
followed by refrigerants such as ammonia and Freon.
NON-FLOW PROCESSES OF VAPOURS
In non-flow processes using vapours as the working medium, the various types
of energy entering into the process consist of heat energy Q, internal energy U
and the external work W.
The energy equation involved in an expansion or a compression of this type is
the fundamental energy equation:
heat added = change in internal energy + external work done
Q = (U 2 - U1 ) + W
EQUATION 1.1
The external work performed during the expansion or compression of a vapour
is calculated, as in the case of gases, from the area under the expansion or
compression curve in a pressure-volume diagram.
The internal energy is also related to the enthalpy with the equation:
h = U + PV
EQUATION 1.2
50
A1 • First Class • SI Units
where h
p
v
= enthalpy
= pressure
= volume
If the enthalpy is known, the internal energy can be calculated from this
equation.
PRESSURE-VOLUME DIAGRAMS
A PV (Pressure-Volume) diagram for any vapour can be constructed as shown in
Figure 1.
FIGURE 1
A Pressure-Volume
Diagram
When a liquid is heated at any one constant pressure, there is one fixed
temperature at which boiling takes place. The higher the pressure of the liquid,
the higher the temperature at which boiling occurs.
It is also found that the volume occupied by 1 kg of a boiling liquid at a higher
pressure is slightly larger than the volume occupied by 1 kg of the same liquid
when it is boiling at a lower pressure.
A series of boiling points plotted on a PV diagram appear as a sloping line called
the saturated liquid line as shown in Figure 1. The points P, Q and R represent
the boiling points of a liquid at pressures Pp, Pq and Pr respectively.
When a liquid at boiling point (saturation temperature) is given further heat at
constant pressure, it changes phase from liquid to vapour. The heat supplied
during this change is the latent heat of vaporization.
There is a definite value of specific volume for the vapour at each pressure
considered. A series of points R΄, Q΄ and P΄ indicate these volumes. The line
joining these and similar points is called the saturated vapour line.
A1 • Chapter 2 • Thermodynamics of Steam
The word saturation as used here refers to energy saturation. For example,
liquid saturated with sensible heat is shown to be on the saturated liquid line. A
slight addition of heat to this boiling liquid changes some of it into a vapour; it is
no longer a liquid but is now a wet vapour. Similarly, when a substance just on
the saturated vapour line is cooled slightly, droplets of liquid begin to form, and
the saturated vapour becomes a wet vapour. A saturated vapour is usually called
dry saturated to emphasize the fact that no liquid is present in the vapour in this
state; at the same time, it is saturated with latent heat.
The PV diagram can be used to demonstrate thermodynamic processes and
working cycles, bearing in mind that areas on a PV diagram represent work
done.
51
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A1 • First Class • SI Units
OBJECTIVE 2
Explain the constant volume process for steam and calculate heat
supplied, work done and internal energy.
CONSTANT VOLUME PROCESS
Work done is a function of the change in volume of the fluid. In a constant
volume process, there is no change in volume and the work done is zero. The
energy equation (1.1) can be written as:
EQUATION 2.1
Q = (U 2 - U1 ) + W
Q = (U 2 - U1 )
Any heat supplied or removed results in a change in internal energy only.
Figure 2 illustrates a constant process on a pressure-volume diagram.
FIGURE 2
Constant Volume P-V
Diagram
Figure 2(b) shows a perfect gas, and Figure 2(a) shows a steam process.
Example 1
One kilogram of steam at a pressure of 101.3 kPa and 50 percent dry receives
heat under constant volume raising the pressure to 200 kPa absolute.
A1 • Chapter 2 • Thermodynamics of Steam
Find:
a)
b)
c)
d)
the volume of the steam
the quality of the steam at the end of the process
the work done
the change in internal energy
Solution
Values from the steam tables:
At 101.3 kPa
vg1 = 1.673 m3/kg
Uf1 = 418.94 kJ/kg
hf1 = 419.1 kJ/kg
Ufg1 = 2087.6 kJ/kg
At 200 kP
avg2 = 0.8854 m3/kg
Uf2 = 504.49 kJ/kg
hf2 = 504.7 kJ/kg
Ufg2 = 2025 kJ/kg
hfg =
2256.9 kJ/kg
hfg2 =
2201.6 kJ/kg
a) Because the steam is 50 percent dry, its volume at the beginning of the
process is:
V1 = 0.50 ×1.673 m3 /kg × 1 kg
= 0.837 m 3 (Ans.)
This is also the volume at the end of the process because it is at constant volume.
b) If the steam at the end of the process was dry and saturated, its volume
would be 0.8854 m3. However, its actual volume is 0.837 m3, so the quality
X2 of the steam at the end of the process is:
0.837
0.8854
= 0.945 or 94.5% dry (Ans.)
X2 =
c) Because the volume is constant, the work done during the process is:
W =0
d) The heat required for a constant volume process is the difference between
the internal energies as per equation (2.1). To find the internal energies, we
can either use the Steam Tables or calculate the internal energies from
equation (1.2).
53
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A1 • First Class • SI Units
Using the Steam Tables, the internal energy U1 at 101.3 kPa and a quality of 0.50
is calculated as:
U1 = u f + X 1u fg
= 418.94 kJ/kg + 0.50 × 2087.6 kJ/kg
= 1462.7 kJ/kg
Similarly, at the final condition of 200 kPa and a quality of 0.945, the internal
energy U2 is:
U 2 = u f + X 2u fg
= 504.49 kJ/kg + 0.945 × 2025 kJ/kg
= 2418.1 kJ/kg
The heat required is then:
Q = (U 2 - U1 )
= 2418.1 kJ/kg − 1462.7 kJ/kg
= 955.4 kJ/kg (Ans.)
The internal energy at the beginning of the process can also be found from
equation (1.2) once the enthalpy has been determined.
At the end of the process:
h2 = h f 2 + X 2 h fg 2
= 504.7 kJ/kg + 0.945 × 2201.6 kJ/kg
= 2585.2 kJ/kg
h2 = U 2 + PV
2
U 2 = h2 − PV
2
= 2585.2 kJ/kg − (200 kPa × 0.837 m3 /kg)
= 2417.8 kJ/kg
At the beginning of the process:
h1 = h f 1 + X 1h fg1
= 419.1 kJ/kg + 0.50 × 2256.9 kJ/kg
= 1547.6 kJ/kg
A1 • Chapter 2 • Thermodynamics of Steam
h1 = U1 + PV
1
U1 = h1 − PV
1
= 1547.6 kJ/kg − (101.3 kPa × 0.837 m3 /kg)
= 1462.8 kJ/kg
Q = (U 2 - U1 )
= 2417.8 kJ/kg − 1462.8 kJ/kg
= 955 kJ/kg (Ans.)
55
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A1 • First Class • SI Units
OBJECTIVE 3
Explain the constant pressure process for steam and calculate heat
supplied, work done and internal energy.
CONSTANT PRESSURE PROCESS
Because the work is being done at constant pressure, it can be written as:
EQUATION 3.1
W = P(V2 − V1 )
The energy equation then becomes:
Q = (U 2 - U1 ) + P(V2 − V1 )
Since enthalpy is defined as:
EQUATION 3.2
h = U + PV
the equation can be rearranged as expressed in terms of enthalpy as follows:
EQUATION 3.3
Q = (U 2 + PV2 ) - (U1 + PV1 )
Q = h2 − h1
where h1 and h2 are the enthalpies of the vapour at the initial and final conditions.
Figure 3 illustrates a constant pressure process on a pressure-volume diagram.
FIGURE 3
Constant Pressure P-V
Diagram
A1 • Chapter 2 • Thermodynamics of Steam
Example 2
One kilogram of steam at a pressure of 1000 kPa and 50 percent dry is heated
under constant pressure until it becomes dry and saturated.
Find:
a)
b)
c)
d)
the volume of the steam at the final and initial conditions
the work done during the process
the heat required for the process
the internal energy of the steam at the final and initial conditions
Solution
a) From steam tables, the volume of 1 kg of dry saturated steam at 1000 kPa is:
V2 = 0.19444 m 3 (Ans.)
Since the steam at the initial condition is 50 percent dry, its volume will be:
V1 = 0.19444 m3 × 0.5
= 0.09722 m 3 (Ans.)
Note that the volume of liquid water is small enough that it can be ignored in the
calculation.
b) The work done during the expansion is obtained from equation (3.1):
W = P (V2 − V1 )
= 1000 kPa (0.19444 m3 − 0.09722 m3 )
= 97.22 kNm (Ans.)
c) The heat required is found from equation (3.3):
h2 = 2776.2 kJ/kg from steam tables
h1 = h f + X 1h fg
= 762.81 kJ/kg + 0.50 × 2015.3 kJ/kg
= 1770.5 kJ/kg
Q = h2 − h1
Q = 2776.2 kJ − 1770.5 kJ
= 1005.7 kJ (Ans.)
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A1 • First Class • SI Units
d) The internal energy can either be obtained directly from the Steam Tables or
calculated.
Using the Steam Tables, the internal energy U2 at 1000 kPa and dry conditions is
read directly as 2583.6 kJ/kg (Ans.).
U1 = u f + X 1u fg
= 761.68 kJ/kg + 0.50 × 1822 kJ/kg
= 1672.7 kJ/kg (Ans.)
The internal energy can also be calculated from equation (3.2):
h = U + PV
U1 = h1 − PV1
= 1770.5 kJ/kg − (1000 kPa × 0.09722 m3 /kg)
= 1673.3 kJ/kg (Ans.)
U 2 = h2 − PV2
= 2776.2 − (1000 × 0.19444)
= 2581.8 kJ/kg (Ans.)
A1 • Chapter 2 • Thermodynamics of Steam
59
OBJECTIVE 4
Explain the constant temperature process for steam and calculate
heat supplied and work done.
CONSTANT TEMPERATURE PROCESS
A constant temperature, or isothermal, process requires that heat be added
during expansion and removed during compression, and this makes it impractical
for most plant applications. The student will already be familiar with isothermal
processes for perfect gases, but a vapour process such as steam is slightly
different. It is illustrated in a pressure-volume diagram in Figure 4.
FIGURE 4
Isothermal PressureVolume Diagram
It can be seen in Figure 4 that the starting point for the process (point 1) is
assumed to be in the region of wet steam. Isothermal expansion of wet steam is
also a constant pressure process until the saturated vapour line is crossed and the
steam becomes superheated. From that point (point A) onward, pressure falls as
volume increases.
Calculation of heat supplied requires use of the entropy of the steam, which will
be discussed in the next learning objective. For now, values for entropy can be
obtained from the steam tables and used to calculate heat supplied as follows:
Q = T ( s2 − s1 )
where T = absolute temperature and s = entropy.
EQUATION 4.1
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A1 • First Class • SI Units
The relationship between heat supplied and work done is defined by the
difference between the internal energy of the steam before and after expansion,
so that:
Q + W = u2 − u1
EQUATION 4.2
Example 3
One kilogram of steam at a pressure of 4000 kPa and 80 percent dry is expanded
under constant temperature until it reaches a pressure of 2000 kPa.
Find:
a) The heat removed during the process
b) The work done during the process
Note: The steam tables provided do not include internal energy values for
superheated steam. Take this value to be 2681 kJ/kg.
Solution
a) From the steam tables, the initial entropy of the steam is:
sv = s f + 0.8s fg
= 2.7964 kJ/kgK + 0.8 × 3.2737 kJ/kgK
= 5.4154 kJ/kgK
The temperature throughout the process is the saturation temperature at 4000
kPa, which is (250.4 + 273) K = 523.4 K (say 523 K ).
The final entropy of the steam at 2000 kPa and superheated to 250°C can be
found in the steam tables and is 6.5453 kJ/kgK .
The heat removed during the process is found from equation (4.1):
Q = T ( s2 − s1 )
= 523K(6.5453 kJ/kgK − 5.4154 kJ/kgK)
= 523(1.1299 kJ/kg)
= 590.94 kJ/kg (Ans.)
A1 • Chapter 2 • Thermodynamics of Steam
b) From the steam tables, the initial internal energy of the steam is:
uv = u f + 0.8u fg
= 1082.31 kJ/kg + 0.8 × 1520.0 kJ/kg
= 2298.31 kJ/kg
The work done during the process is found from equation (4.2):
Q + W = u2 − u1
W = u2 − u1Q
= 2681 kJ/kg − 2298.31 kJ/kg + 590.94 kJ/kg
= 973.63 kJ/kg or 973.63 kNm/kg (Ans.)
Note the reversal of the “–” sign to a “+” sign for Q. This is because heat was
removed during expansion; therefore, Q has a negative value when calculating
work done.
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A1 • First Class • SI Units
OBJECTIVE 5
Calculate steam entropy given the steam conditions.
EXPLANATION OF ENTROPY
When a thermodynamic process is drawn on a PV diagram, the area under the
curve defines the amount of work that is done. A thermodynamic process can
also be thought of in terms of the heat that is supplied or removed by the various
processes that make up the cycle. In a similar fashion, a graph can be drawn that
has temperature on the y-axis and a property called entropy on the x-axis. The
area under the curve of the graph of a reversible process or cycle is the amount
of heat (see Figure 5).
FIGURE 5
Explanation of Entropy
The symbol for entropy is S or dS for a change in entropy and the units are
kJ/kgK.
CALCULATION OF ENTROPY FOR STEAM
To measure the increase in entropy of 1 kg of superheated steam from water at
0°C to some higher temperature, the areas beneath the curve drawn on Figure 6
must be calculated.
Referring to Figure 6, the entropy of water at 0°C (273 K) is shown as zero. To
raise its temperature to T1, the amount of heat represented by the area O, T0, T1,
S1 must be determined.
A1 • Chapter 2 • Thermodynamics of Steam
63
FIGURE 6
Diagram for
Calculation of Entropy
of Steam
This area will be the sum of all small strips defined as TdS .
Heat is related to entropy by the equation:
dQ = TdS
dS =
EQUATION 5.1
dQ
T
While in the liquid state,
Q = Cw dT
and
dS = Cw
dT
T
where Cw is the specific heat of water.
If we take the area from the y-axis to the saturation point, the result is the
equation:
dT
T0 T
= cw (ln T1 − ln T0 )
S f = cw ∫
T1
S f = cw ln
T1
T0
EQUATION 5.2
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A1 • First Class • SI Units
Because the point of zero entropy is taken at 0°C, the equation can be changed
to:
EQUATION 5.3
S f = cw ln
T1
273
To convert the water into steam at the saturation temperature T1, the latent heat
hfg is added. This process is done at constant temperature; thus the entropy
increase Sfg is given by:
EQUATION 5.4
S fg =
S fg =
latent heat
temperature of saturation
h fg
Tsat
Further heating produces superheated steam and the increase of entropy during
this heat addition is calculated in a similar manner to the sensible heat addition
except that, in this case, the specific heat of superheated steam must also be
known.
EQUATION 5.5
Ssup = csup ln
Ts
Tsat
The entropy of a vapour at temperature T is, therefore, the sum of the three
terms:
EQUATION 5.6
S = cw ln
T
T1 h fg
+
+ csup ln s
Tsat
273 Tsat
Example 4
Calculate from first principles the entropy of 1 kg of steam at 3000 kPa and
350°C and check the results from the steam tables. Take the specific heat of
water and superheated steam to be 4.32 kJ/kgK and 2.55 kJ/kgK respectively.
Solution
From the Steam Tables, the saturation temperature at 3000 kPa is 233.9°C and
the latent heat of evaporation at this pressure is 1795.7 kJ/kg.
A1 • Chapter 2 • Thermodynamics of Steam
Converting the temperatures to absolute yields:
Tsat = 273 + 233.9
= 506.9 K
T = 273 + 350
= 623 K
The entropy is calculated from equation (5.6):
S = cw kJ/kgK ln
h
T1
T
+ fg + csup ln
o
Tsat
273 C Tsat
= 4.32 kJ/kgK × ln
506.9 K 1795.7 kJ/kg
623 K
+
+ 2.55 kJ/kgK × ln
o
273 C
506.9 K
506.9 K
= 4.32 kJ/kgK × 0.6188 + 3.543 kJ/kgK + 2.55 kJ/kgK × 0.2062
= 2.673 kJ/kgK + 3.5425 kJ/kgK + 0.5258 kJ/kgK
= 6.741 kJ/kgK
A check from the Steam Tables provides the value for entropy as:
entropy/kg at 3000 kPa and 350°C = 6.73 kJ/kgK
The difference is mainly due to the inaccuracy entailed in the assumption that the
specific heat of water remains constant. The value from the Steam Table is the
correct result.
The above calculation has been carried out with the sole purpose of
demonstrating the relationship between entropy, heat and temperature to
provide a clearer idea of the meaning of entropy.
In all future calculations involving entropy, the Steam Tables will be used.
65
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A1 • First Class • SI Units
OBJECTIVE 6
Explain the significance of a Temperature Entropy diagram for
steam.
TEMPERATURE ENTROPY DIAGRAMS
A temperature/entropy (T-S) diagram can be drawn for any working fluid. A
simplified form of the T-S diagram for steam appears in Figure 7.
FIGURE 7
TS Diagram for a
Vapour
The ordinate (vertical axis) T, is measured in degrees Celsius (°C) or Kelvin (K).
The abscissa (horizontal axis) S is Entropy (kJ/kgK).
The liquid line is produced by plotting the temperature of liquid water against its
calculated entropy content. Zero entropy content occurs in water at 0°C. The
steam tables are based on the premise that zero enthalpy occurs in water at 0°C.
A1 • Chapter 2 • Thermodynamics of Steam
67
Addition of a small amount of heat to 1 kg of water at 0°C results in a
temperature increase. The increase in entropy is calculated from the ratio of the
heat added (kJ) and the absolute temperature at which it was added (K):
ΔS =
ΔQ
T
The entire T-S diagram can be constructed for any working fluid using this
calculation.
The saturation line is produced by plotting the entropy content against the
corresponding saturation temperature of 1 kg of dry saturated steam for a series
of pressures.
The area beneath the liquid and saturation line represents water not yet fully
converted to steam. This is the wet region. Lines drawn parallel to the
saturation line, but within the wet region, are lines of quality.
These represent the condition of 1 kg of steam which is not completely dry, i.e.
which has a percentage of wetness. Lines drawn parallel to the saturation line,
but in the superheat region, are lines of constant superheat. The pressure at
which the steam is being held can be found from the lines of constant pressure.
A section of a temperature - entropy diagram for steam is shown in Figure 8. The
various shaded areas represent the heat added to water at 0°C to vaporize it
completely at the pressure P1.
FIGURE 8
Temperature – Entropy
Diagram of Steam
The area ABCD represents the heat added to the water to bring it to saturation
temperature corresponding to the pressure P1. The area of a TS diagram
represents heat. This area ABCD may be called the sensible heat hf. Further
heating produces evaporation at the constant temperature T1 corresponding to
the pressure P1 and is represented by the area under the line CE. When
vaporization is complete, the latent heat, or the heat of vaporization (hfg), is
represented by the area DCEF. If, after all the water is vaporized, more heat is
added, the steam becomes superheated, and the additional heat required is
represented by an area EGHF.
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A1 • First Class • SI Units
Temperature-Entropy Diagram for a Steam Power Plant
Figure 9 illustrates the heat process occurring when the feedwater received in the
boilers of a power plant, at 40°C, is heated and converted into steam at a
temperature of 200°C and then gives up heat in doing work.
FIGURE 9
Temperature – Entropy
Diagram for Steam
Power Plant
When the feedwater first enters the boiler, its temperature must be raised from
40°C to 200°C before any steaming begins. The quantity of heat added to the
water is indicated by the area MNCD. This area represents approximately the
difference between the enthalpies of the water, using the steam tables; this is
shown to be (852 kJ/kg – 168 kJ/kg) or about 684 kJ/kg. The horizontal or
entropy scale shows the difference in entropies of the water to be (2.33 - 0.57
kJ/kgK) or about 1.76 kJ/kgK.
The curve NC, the liquid line, is constructed by plotting, from the steam tables,
the values of the entropy of the water for a number of different temperatures
between 40°C and 200°C.
A1 • Chapter 2 • Thermodynamics of Steam
When water at 200°C is converted into steam at that temperature, the curve
representing the change is a constant temperature line, and therefore horizontal,
represented by CE. Provided the evaporation has been complete, the heat added
in this process is the latent heat, or the heat of evaporation (hfg), at 200°C which
is 1941 kJ/kg. This quantity of heat, causing the water to change to steam, is
indicated by the area DCEF. The change in entropy during evaporation is 4.10
kJ/kgK.
Enthalpy fg at 200DC
S fg =
Absolute Temperature
1941 kJ/kg
=
(273 + 200)K
= 4.10 kJ/kgK
The total entropy of steam completely evaporated at 200°C is, therefore:
S = 0.57 kJ/kgK + 1.76 kJ/kgK + 4.1 kJ/kgK
= 6.43 kJ/kgK
To represent this final condition of the steam, the point E is plotted as shown in
the Figure 9. The point E is shown located on the curve RS, which is determined
by plotting a series of points calculated the same as E, but for different pressures.
The area MNCEF represents the total heat added to 1 kg of feedwater at 40°C to
produce steam at 200°C. Area OBCEF represents the total heat required to form
1 kg of steam at 200°C.
If the steam generated in the boiler is allowed to expand adiabatically
(isentropically) in an engine cylinder or in a turbine nozzle to a temperature of
40°C, this expansion is represented by the line EG. GN represents the
condensation of the exhaust steam.
The area MNGF represents the heat rejected to the condenser Q2
Q2 = T2 ( S E - S N )
= 313 K × ( S E - S N ) kJ/kgK
and the area NCEG or (Q1 – Q2) represents the heat which has been consumed
by being converted into work in the engine.
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A1 • First Class • SI Units
OBJECTIVE 7
Explain the reversible adiabatic process for steam and calculate
work done and internal energy.
ADIABATIC PROCESSES
No heat is added or subtracted during an adiabatic process. The energy equation
for an adiabatic non-flow process is:
EQUATION 7.1
Q = (U 2 - U1 ) + W = 0
W = (U1 - U 2 )
Thus, the work done during a non-flow adiabatic process is equal to the change
in internal energy of the vapour. The change in internal energy can be
determined by the change in the quality of the vapour, and this change can be
found from the entropy.
In a reversible or isentropic type of adiabatic process there is no change in
entropy and the process can be represented by a straight vertical line on a T-S
diagram. Because the entropy remains constant, the following equations hold
true:
S1 = S2
In the case of steam:
EQUATION 7.2
s f 1 + X 1s fg1 = s f 2 + X 2 s fg 2
Figure 10 illustrates a reversible adiabatic process on a pressure-volume diagram.
A1 • Chapter 2 • Thermodynamics of Steam
71
FIGURE 10
Reversible Adiabatic
P-V Diagram
Example 5
In a non-flow process one kilogram of steam at 700 kPa and 200°C expands
adiabatically to 120 kPa. Determine the amount of work done during the
expansion.
Solution
Because the process is adiabatic, the entropy can be used to determine the quality
of the steam at the end of the expansion.
From the Steam Tables, the entropy of superheated steam at 700 kPa and 200°C
is 6.89 kJ/kgK. At 120 kPa for wet steam, the values of saturated and dry
entropy are 1.361 kJ/kgK and 5.938 kJ/kgK.
s1 = s f 2 + X 2 s fg 2
6.89 kJ/kgK = 1.361 kJ/kgK + X 2 × 5.938 kJ/kgK
6.89 kJ/kgK − 1.361 kJ/kgK
5.938 kJ/kgK
= 0.931 or 93.1%
X2 =
The internal energy can be calculated from equation (3.2) using values for h1 and
V1 obtained from the Steam Tables
h = U + PV
U1 = h1 − PV1
= 2844.8 kJ/kg − (700 kPa × 0.2999) m3 /kg
= 2634.9 kJ/kg
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A1 • First Class • SI Units
The enthalpy after expansion can be calculated using the quality of steam found
earlier:
h2 = 439.4 kJ/kg + 0.931× 2244.1 kJ/kg
= 2528.7 kJ/kg
and
U 2 = h2 − PV2
= 2528.7 kJ/kg − (120 kPa ×1.329) m3 /kg
= 2369.2 kJ/kg
The work done is the difference between the internal energies:
W = (U1 - U 2 )
= 2634.9 kJ/kg − 2369.2 kJ/kg
= 265.7 kJ/kg (Ans.)
A process such as that in the previous example is said to be isentropic or
reversible. However, in actual practice no process is reversible. Instead, an actual
process is irreversible because of internal friction or turbulence which require
energy for overcoming or producing them. Because of this process, there is less
energy transformed into useful work and a corresponding increase in the internal
energy of the working substance compared to that of the theoretical reversible
process.
A1 • Chapter 2 • Thermodynamics of Steam
73
OBJECTIVE 8
Explain the significance of a Mollier chart for steam.
MOLLIER DIAGRAM
The Mollier Chart, or H-S diagram, for steam is of much greater value for
calculation purposes than the T-S diagram whose principal value is in
demonstrating cycles or processes.
A simplified form of a Mollier diagram is shown in Figure 11. The ordinate H is
measured in kJ and the abscissa S in entropy units. It shows essentially the same
form as the T-S diagram with a saturated liquid line and a saturated vapour line.
The area beneath these lines is again the wet region and the area above the
saturated vapour line is the superheated area.
FIGURE 11
H-S Diagram for a
Vapour
Lines of constant pressure, temperature and dryness are marked on Figure 11 to
give a general idea of how they appear on the full diagram.
The student should now consult the separate booklet of Steam Tables and study
the Enthalpy-Entropy diagram. It is a section only of the diagram in Figure 11 as
the portion approaching the origin or zero point is cut off.
The remainder has been magnified considerably with the lines of pressure,
temperature and dryness clearly marked.
74
A1 • First Class • SI Units
Note that the lines of constant dryness have become quality lines. Additional
lines of superheat have been added at 100°C, 200°C, 300°C, etc., above
saturation temperature.
In steady flow problems, the change of enthalpy is of prime importance and,
once the final and initial states for a process have been plotted, this quantity may
be read directly from the chart. Large scale H-S charts for steam have been
published provide results which are accurate for many engineering purposes.
The chart is useful when solving problems of enthalpy change during an
isentropic process (reversible adiabatic, at constant entropy). Such a process is
represented by a straight vertical line. If the initial pressure and condition are
known, the enthalpy drop can be read by dropping a perpendicular to the final
pressure.
One other process is easily followed on a Mollier diagram, namely a throttling
process in which enthalpy remains constant. This process is represented on
the diagram by a straight horizontal line.
Suppose, for example, that steam enters a throttling calorimeter at a pressure of
1000 kPa and is throttled down to a pressure of 101.325 kPa and a resulting
temperature of 120°C. To find the quality of the steam entering the calorimeter,
the Mollier diagram may be used as follows:
Find the intersection of the constant temperature 120°C line with the 101.3 kPa
pressure line. From this point follow a horizontal line (line of constant enthalpy)
to the left until it intersects the 1000 kPa pressure line. This point of intersection
is found to lie on a constant moisture line of approximately 3 percent. Thus the
entering steam has a quality of 97 percent.
A1 • Chapter 2 • Thermodynamics of Steam
CHAPTER QUESTIONS
1. Describe the three types of energy for non-flow processes that involve
vapours, and state the fundamental energy equation.
•
_________________________________________________________
•
_________________________________________________________
•
_________________________________________________________
2. What is the energy equation for a constant volume process, and why is it
different from the fundamental energy equation?
____________________________________________________________
____________________________________________________________
3. Derive the energy equation for a constant pressure process as expressed in
terms of enthalpy.
____________________________________________________________
4. Explain the significance of the areas in a PV diagram and a TS diagram.
____________________________________________________________
____________________________________________________________
5. Calculate from first principles the entropy of 1 kg of steam at 2000 kPa and
500°C and check the results from the steam tables. Take the specific heat of
water and superheated steam to be 4.32 and 2.55 kJ/kgK respectively.
____________________________________________________________
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A1 • First Class • SI Units
6. Draw the T-S diagram for a vapour showing the various lines of constant
properties.
7. What is the energy equation for a reversible adiabatic process and how is it
derived?
____________________________________________________________
____________________________________________________________
8. What two processes can be most easily read from a Mollier diagram?
•
_________________________________________________________
•
_________________________________________________________
Part A1
CHAPTER 3
Steady-flow Process
Calculations
Here is what you will be able to do when you complete each objective:
1. Describe the steady-flow energy equation and calculate the work done
in a steady-flow process.
2. Calculate the power consumed in a steady-flow process.
3. Explain the principle of conservation of energy and supersaturation as
they apply to a nozzle and calculate nozzle inlet and outlet velocities.
4. Calculate the initial dryness fraction of steam in a throttling process.
5. Determine, using a Mollier Chart, the quality, enthalpy, and entropy of
steam entering a calorimeter.
6. Calculate energy transfer, work done, and power produced in a steam
turbine.
7. Calculate the heat lost, surface area, required cooling water flow, and
heat transfer coefficient in a steam condenser.
8. Define and calculate availability and effectiveness in the context of the
steady-flow processes.
77
A1 • Chapter 3 • Steady-flow Process Calculations
79
OBJECTIVE 1
Describe the steady-flow energy equation and calculate the work
done in a steady-flow process.
STEADY-FLOW ENERGY EQUATION
Many heat engine applications consist of thermodynamic processes that require a
flow of a gas or vapour in and out of the system. A steady-flow process, as in the
case of steam flowing through a turbine, is one in which the total energy flows
uniformly into the process and leaves in a similar manner. Here, not only the
internal energy and work done, but also those energies and work associated with
a fluid in motion i.e. kinetic and potential energy and flow work must be
considered.
Consider a steady-flow process as indicated in Figure 1. In this process, 1 kg of
the working fluid is flowing through the system boundary at Section 1 under a
condition of constant pressure (P1) and at a higher elevation than Section 2. A
quantity of heat energy (Q) is added to the working fluid between Sections 1 and
2 and work equal to W is done by the working fluid. The working fluid leaves the
system under constant pressure (P2) at Section 2.
FIGURE 1
A Steady-flow Process
80
A1 • First Class • SI Units
The energy in kJ/kg of the working fluid entering the system at Section 1
consists of the sum of the following energies:
a) The potential energy (Z1) due to its elevation above the datum plane
b) The internal energy (Ul)
c) The work of flow (P1V1)
a. where P = pressure
V = volume
⎛ v12 ⎞
d) The kinetic energy ⎜ ⎟
⎝ 2 ⎠
where
v = velocity
The energy in kJ/kg of the working substance exiting the system at Section 2
consists of the sum of the following energies:
a) The potential energy (Z2) due to its elevation above the datum plane
b) The internal energy (U2)
c) The work of flow (P2V2)
⎛v 2 ⎞
d) The kinetic energy ⎜ 2 ⎟
⎝ 2 ⎠
The complete process may then be expressed using these symbols and adding
terms for heat and work as follows:
Z1 + U1 + PV
1 1+
v12
v2 2
+ Q = Z 2 + U 2 + PV
+
+W
2 2
2
2
The terms, internal energy and work of flow, always appear together in a steadyflow process and their sum is enthalpy (h) where
h = U + PV
The equation may then be written:
Z1 + h1 +
v12
v2
+ Q = Z 2 + h2 + 2 + W
2
2
The potential energy term (Z) is constant throughout the process in almost all
practical applications so that the equation now becomes the steady-flow energy
equation:
EQUATION 1.1
h1 +
v12
v2
+ Q = h2 + 2 + W
2
2
A1 • Chapter 3 • Steady-flow Process Calculations
In words, this equation reads:
enthalpy + kinetic energy + heat supplied = enthalpy + kinetic energy + work done
at entry
at entry
at exit
at exit
Remember that each term in this equation represents the energy of 1 kg of the
working fluid. For problems involving a greater mass flow of working fluid, each
side of the equation must be multiplied by the mass flow rate.
Example 1
Dry air enters a compressor with a velocity of 6 m/s and an enthalpy of 230
kJ/kg and leaves with a velocity of 12 m/s and an enthalpy of 475 kJ/kg.
Assuming adiabatic compression and negligible change in potential energy, how
many kNm of work are required to compress 1 kg of air?
Solution
Because the compression process is adiabatic, there is no heat supplied and the
steady-flow energy equation (1.1) becomes:
h1 +
v12
v2
= h2 + 2 + W
2
2
v2 v 2
W = h1 − h2 + 1 − 2
2
2
2
2
v1
6
= J/kg
2
2
= 18 J/kg
= 0.018 kJ/kg
v2 2 122
=
J/kg
2
2
= 72 J/kg
= 0.072 kJ/kg
= −245 kJ/kg or - 245 kN m/kg (Ans.)
As is often the case, these kinetic energy valves are small enough to be
insignificant and are ignored.
81
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A1 • First Class • SI Units
Therefore,
W = h1 − h2
= 230 kJ/kg − 475 kJ/kg
= −245 kJ/kg
The work done is 245 kNm/kg or air.
A1 • Chapter 3 • Steady-flow Process Calculations
OBJECTIVE 2
Calculate the power consumed in a steady-flow process.
CALCULATING POWER
The steady-flow energy equation can be used to calculate the power produced by
a process by first determining the work done and then deriving the power from
the rate of flow.
Example 2
Air flowing steadily through a rotary compressor is raised from 100 kPa and
16°C to 350 kPa and 115°C. The compressor coolant removes 8 kJ per kg of air
and the inlet and outlet air velocities are 10 m/s and 15 m/s respectively. If the
change of enthalpy of the air amounts to100 kJ/kg and the value of R is taken as
0.287 kJ/kgK, find the work done on the air and the power required to drive the
compressor if the airflow at inlet is 6 m3/min and mechanical efficiency of the
compressor is 90 percent. If the specific volume of air is 1.29 kg/m3, what is the
inlet area?
Solution
The energy balance equation for a steady-flow system is:
h1 +
v12
v2
+ Q = h2 + 2 + W
2
2
v2 v 2
W = h1 − h2 + 1 − 2 + Q
2
2
The difference in enthalpy is negative because work is being done on the
process:
h1 − h2 = −100 kJ/kg
The heat removed by cooling is also negative so that:
Q = −8 kJ/kg
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A1 • First Class • SI Units
For each kg of air:
102 ×10−3 kJ/kg 152 ×10−3 kJ/kg
−
− 8 kJ/kg
2
2
= −100 + 0.05 − 0.1125 − 8
= −108 kJ
W = −100 kJ/kg +
The terms for kinetic energy are small enough to be insignificant and are ignored.
Under inlet conditions, the flow in kg of air per minute is:
m=
PV
RT
100 kPa × 6 m3 /min
0.287 kJ/kgK × (16 + 273) K
= 7.23 kg/min or 0.1206 kg/s
=
The work done per second is thus:
W/s = 0.1206 kg/s × 108 kJ/kg
= 13.02 kJ/s (Ans.)
The power required to drive compressor is:
100
90
= 14.47 kW (Ans.)
power = 13.01 kJ/s ×
Because the flow and the velocity at the inlet are known, the area can be
calculated as:
mass flow (kg/s) = velocity (m/s) × area (m 2 ) × specific volume (kg/m3 )
m
area =
Vv
0.1206 kg/s
=
10 m/s × 1.29 kg/m3
= 0.0093 m 2 or 93cm 2 (Ans.)
A1 • Chapter 3 • Steady-flow Process Calculations
85
OBJECTIVE 3
Explain the principle of conservation of energy and supersaturation
as they apply to a nozzle and calculate nozzle inlet and outlet
velocities.
NOZZLE CALCULATIONS
Nozzles are found in steam and gas turbines and are a special case of steady state
flow processes. The function of the nozzle is to produce a jet of high velocity
that can be directed at the blades of a turbine. The expansion of the fluid is
converted to an increase in kinetic energy. The process is considered to be
adiabatic because there is very little opportunity for the fluid to exchange heat
energy with the surroundings during the time available.
The inlet and outlet conditions for a nozzle can be determined by applying the
principle of conservation of energy as shown in Figure 2.
FIGURE 2
Steady-flow Process
for a Nozzle
As the inlet and outlet of a nozzle are at the same elevation, there is no
difference in potential energy. Also, because the outlet velocity is much greater
than the inlet velocity, the inlet velocity may be ignored thus eliminating the
kinetic energy term at inlet conditions. Furthermore, the terms Q and W are both
zero and may also be eliminated. Hence, the steady-flow energy equation
becomes:
h1 = h2 +
v2 2
2
h1 − h2 =
v2 2
2
enthalpy drop = kinetic energy gain (neglecting v1 )
EQUATION 3.1
86
A1 • First Class • SI Units
If the drop in enthalpy is known, the exit velocity can be calculated by
reorganizing equation (3.1) into the form:
v2 = 2(h1 − h2 )
EQUATION 3.2
Because enthalpy is usually given in kJ/kg and the above formula assumes J/kg,
the enthalpies in equation (3.2) need to be multiplied by 103; now the formula
can be rewritten as:
EQUATION 3.3
v2 = 2 × 103 × (h1 − h2 )
= 2 × 103 × h1 − h2
v2 = 44.72 h1 − h2
Losses due to friction and other fluid losses mean that the efficiency of the
nozzle will not be 100% which needs to be considered. This efficiency can be
introduced into the equation so that it reads:
EQUATION 3.4
v2 = 44.72 ( h1 − h2 ) × nozzle efficiency
Example 3
Steam at a pressure of 1500 kPa and 240°C undergoes a reversible adiabatic
expansion through a nozzle to 550 kPa absolute. Assuming a nozzle efficiency of
96 percent, calculate the exit velocity of the steam.
Solution
Using the Mollier Chart, locate point (A) at the intersection of the 1500 kPa
pressure line and the 240°C temperature line. The enthalpy at this point is 2899
kJ/kg. Because the expansion is reversible, point (B) is located by dropping a
vertical constant entropy line from point (A) to the 550 kPa pressure line. The
enthalpy at this point is 2693 kJ/kg and the moisture is 2.75 percent.
The velocity is obtained from equation (3.4):
v2 = 44.72 ( h1 − h2 ) × nozzle efficiency
= 44.72 (2899 kJ/kg − 2693 kJ/kg) × 0.96
= 44.72 197.76
= 44.72 ×14.06
= 628.8 m/s (Ans.)
A1 • Chapter 3 • Steady-flow Process Calculations
SUPERSATURATION
Expansion of steam in a nozzle occurs very rapidly and, even though the steam is
well below its saturation temperature, there isn’t sufficient time for condensation
to occur as the steam passes through the nozzle. Instead of condensate forming
gradually as the steam pressure drops, it is formed suddenly at a single point in
the flow path. The point at which condensation occurs may be within the nozzle
or after the steam leaves the nozzle. Nozzles are manufactured to eliminate
damage from erosion.The condensation usually occurs at some point
downstream. Until that point, the steam is supersaturated, meaning that it
remains dry even though it is below its saturation temperature.
If the supersaturated steam comes into contact with an object, then its velocity
drops and condensation occurs very quickly. This can happen when the steam
encounters a measuring instrument such as a flowmeter or thermowell. When
such instruments are placed downstream of a nozzle, it is very important to
consider the potential erosion effects of supersaturation.
87
88
A1 • First Class • SI Units
OBJECTIVE 4
Calculate the initial dryness fraction of steam in a throttling process.
THROTTLING PROCESS
When steam expands by passing through a very small opening such as an orifice
or a partly open valve in a steam line, the pressure is considerably reduced. This
process is known as throttling or wiredrawing and may take place in an
apparatus such as a throttling calorimeter as shown in Figure 3.
FIGURE 3
Simple Throttling
Calorimeter
A1 • Chapter 3 • Steady-flow Process Calculations
89
An energy equation for the throttling process can be developed making the
following assumptions:
1. One kilogram of fluid leaves the apparatus for each kilogram of fluid that
enters the apparatus.
2. The apparatus is well insulated and heat transfer to the cooler
surroundings may be neglected.
3. No mechanical work is done between the entrance and the exit.
4. The changes in the velocity or kinetic energy and in the potential energy
of the fluid are both so small that they can be neglected.
Then, using the steady-flow energy equation, the terms with insignificant values
can be cancelled out:
h1 +
v12
v2
+ Q = h2 + 2 + W
2
2
h1 = h2
Thus, the enthalpy at entrance equals the enthalpy at exit. Also, as the pressure at
exit is lower than that at entrance, the same enthalpy content at the lower
pressure results in drier steam or even superheated steam.
If dry saturated steam at ordinary boiler pressures is throttled to atmospheric
pressure, the steam will become superheated because its enthalpy is greater than
the enthalpy of dry saturated steam at atmospheric pressure. Also, if steam
having 2 or 3 percent moisture is throttled from ordinary boiler pressures to
atmospheric pressure, its enthalpy may be high enough to produce superheated
steam at the lower pressure.
When a throttling calorimeter is used, the pressure and temperature of the steam
after throttling can be measured and its enthalpy can then be found from the
steam tables. Then the initial quality X1 of the steam can be determined as
follows:
h f 1 + X 1h fg1 = h1 = h2
X1 =
h2 − h f 1
h fg1
where
h2 = enthalpy of superheated steam in the calorimeter
h f 1 = enthalpy of saturated liquid at the initial pressure
h fg1 = enthalpy of evaporation at the initial pressure
X 1 = quality of wet steam at the initial pressure
EQUATION 4.1
90
A1 • First Class • SI Units
Example 4
Steam at 2000 kPa is reduced in pressure in a throttling calorimeter to 101.3 kPa.
If its temperature after throttling is 150°C, find the quality of the initial steam.
Solution
From superheated steam tables:
h2 at 101.3 kPa and 150°C = 2776.3 kJ/kg
h f 1 + X 1h fg1 = 2776.3 kJ/kg
From tables, at 2000 kPa:
h f 1 = 908.8 kJ/kg and
h fg1 = 1890.7 kJ/kg
2776.3 kJ/kg − 908.8 kJ/kg
1890.7 kJ/kg
= 0.988 or 98.8% (Ans.)
X1 =
A1 • Chapter 3 • Steady-flow Process Calculations
OBJECTIVE 5
Determine, using a Mollier Chart, the quality, enthalpy, and entropy
of steam entering a calorimeter.
CALORIMETER USING A MOLLIER CHART
To solve the example in the previous objective using the Mollier Chart, locate
point (B) at the intersection of the 101.3 kPa pressure line and the 150°C
constant temperature line. This point is in the superheat region of the chart
above the saturation line.
Point (A) is then found by projecting a horizontal constant enthalpy line from
point (B) to intersect with the 2000 kPa pressure line. The quality at point (A) is
about 0.99.
If the quality of the high pressure steam is too low to produce superheated steam
in the calorimeter, the throttling calorimeter cannot be used, and some other type
of apparatus such as the separating calorimeter must be used.
In the separating calorimeter, the water is removed from the steam using
mechanical separation which depends for its action on abruptly changing the
direction of flow of the steam. Since the water is several hundred times as dense
as the steam, it will be deposited because of its greater inertia. The amount of
steam leaving the calorimeter after removal of the moisture, divided by the sum
of the steam leaving and the moisture separated, will equal the quality of the
entering steam.
Example 5
Steam at 2000 kPa is throttled in a calorimeter to a pressure of 150 kPa and a
temperature of 150°C. Using the Mollier chart, determine the quality, the
enthalpy, and the entropy of the 2000 kPa steam.
Solution
From the Mollier chart, h2 is in the superheated region at 150 kPa and 150°C and
has an enthalpy of approximately 2770 kJ/kg. The entropy at this point is about
7.42 kJ/kgK.
91
92
A1 • First Class • SI Units
Because throttling is at constant enthalpy, a horizontal line is drawn until it
reaches a pressure of 2000 kPa. The quality at this point is 0.98 and the entropy
is about 4.3 kJ/kgK.
A1 • Chapter 3 • Steady-flow Process Calculations
93
OBJECTIVE 6
Calculate energy transfer, work done, and power produced in a
steam turbine.
STEADY-FLOW PROCESSES OF VAPOURS
The flow of vapour through a nozzle, a turbine, an orifice, or a throttling
calorimeter is an example of adiabatic processes which may in theory be either
reversible or irreversible.
The energy equation of a steady-flow process per kg of the working substance is
(neglecting the potential energy term):
h1 +
v12
v2
+ Q = h2 + 2 + W
2
2
In a steam turbine, the change in velocity from one stage to the next is negligible,
so the equation reduces to:
W = h1 − h2 + Q
If the process is adiabatic, the equation further reduces to:
W = h1 − h2
Allowing for thermal efficiency η, there is a certain amount of loss of enthalpy
which is not converted to practical work. The formula becomes:
W = η (h1 − h2 )
Thus, it can be seen that in an adiabatic process during which the change of
potential energy and velocity is small, the work done equals the difference
between the enthalpies at the beginning and the end of the process. The value of
the enthalpy at the end of the process can be determined from the steam quality
which, in turn, in a reversible process, can be determined from the constant value
of entropy that exists during such a process.
EQUATION 6.1
94
A1 • First Class • SI Units
Example 6
a) One kilogram of dry and saturated steam at 1100 kPa expands
adiabatically and reversibly in a steam turbine and exhausts at 101.3 kPa.
Assume that η = 0.65. If the changes in velocity and potential energy are
small enough to be neglected, what amount of work is done?
b) If the steam flow is 50 000 kg/h, what is the output power?
Solution
a) The entropy is constant; therefore, using the steam tables, the quality X2 at
the end of the process can be found:
entropy before = entropy after
6.550 kJ/kgK = 1.307 kJ/kgK + 6.049 kJ/kgK X 2
6.550 − 1.307
6.049
= 0.867
X2 =
enthalpy at start of process h1 = 2781.7 kJ/kg
enthalpy at end of process h2 = 419.1 kJ/kg + 0.867 × 2256.9 kJ/kg
= 2375.8 kJ/kg
W = η (h1 − h2 )
= 0.65(2781.7 kJ/kg − 2375.8 kJ/kg
= 0.65 × 405.9 kJ/kg
= 263.8 kJ/kg (Ans.)
If, in the above example, the process was irreversible, as it would be in
actual practice, there would be an increase in entropy during the process
which would increase both the quality and the enthalpy.
The solution can also be found using the Mollier Chart. A point (A) is
located at the intersection of the 1100 kPa pressure line with the saturation
line. From this point a vertical constant entropy line is dropped to intersect
with the 101.3 kPa pressure line giving point (B).
From the Mollier Chart, the enthalpy at (A) is 2780 kJ and the enthalpy at
(B) is 2380 kJ, giving a heat drop of 400 kJ. Also point (B) is on the 13
percent constant moisture line giving a quality of 0.87. These values
compare closely to those found by calculation and using steam tables.
A1 • Chapter 3 • Steady-flow Process Calculations
b)
50 000 kg/h × 263.8 kJ/kg
3600 s/h
= 3663.9 kJ/s
= 3663.9 kW or 3.66 MW (Ans.)
P=
95
96
A1 • First Class • SI Units
OBJECTIVE 7
Calculate the heat lost, surface area, required cooling water flow,
and heat transfer coefficient in a steam condenser.
STEAM CONDENSER
The application of the steady-flow energy equation to a steam condenser results
in a simplified equation because no work is done in a condenser. The inlet and
outlet velocities are low enough that they are not taken into consideration. The
equation can be written as:
Q = h2 − h1
EQUATION 7.1
The velocity in the condenser is even smaller than the inlet and outlet velocities
and, for all practical purposes, there are no friction or other losses and the
process is reversible. Therefore, there is no pressure drop due to friction, and the
pressure can be assumed to be constant.
To determine the amount of heat transfer across a heat exchanger such as a
condenser, it is necessary to know the overall heat transfer coefficient for the
wall and fluid layers on both sides of the cooling tubes. The amount of heat
transfer is also dependent on the mean temperature difference.
The equation for the amount of heat transferred is:
EQUATION 7.2
Q=
λ At ΔTm
d
where
λ = overall heat transfer coefficient in W/m°C
A = area in m 2
t = time in seconds
ΔTm = mean temperature difference in °C
d = wall thickness in m
A1 • Chapter 3 • Steady-flow Process Calculations
Example 7
Steam with a dryness of 0.901 and a pressure of 101.3 kPa flowing at a rate of 2
kg/s is condensed to dry steam. Calculate the amount of heat that needs to be
removed, the area of the condenser tubes, and the flow of water required to
accomplish this if the cooling water enters at 20°C and leaves at 40°C. The
overall heat transfer coefficient is 1050 W/m°C and the specific heat of water is
4200 J/kg°C.
Solution
enthalpy at entry to the condenser h1 = 419.1 kJ/kg + 0.901× 2256.9 kJ/kg
= 2452.6 kJ/kg
enthalpy at exit of the condenser h2 = 419.1 kJ/kg
The heat that needs to be removed by the condenser is:
Q = h1 − h2
= 2452.6 kJ/kg − 419.1 kJ/kg
= 2033.5 kJ/kg
With a flow of 2 kg/s, the required rate of heat removal is:
heat removal rate = heat rate × steam flow
= 2033.5 kJ/kg × 2 kg/s
= 4067 kJ/s (Ans.)
From the Steam Tables, the temperature of the steam is 100°C. The heat transfer
between the steam and the cooling water occurs by means of conduction. The
increase in temperature of the cooling water is (40°C - 20°C) = 20°C and the
wall thickness is 1 cm. The required area for cooling is, therefore:
Q=
λ At ΔT
d
Qd
A=
λt ΔT
4067 kJ/s ×103 J/kJ × 0.01 m
=
1050 W/mDC × 1 s × 20DC
= 1.937 m 2 (Ans.)
97
98
A1 • First Class • SI Units
The amount of cooling water needed is dependent on the specific heat of the
water and the temperature increase and can be found using the equation:
heat transfer (J/s) = cooling water flow (kg/s) × specific heat (J/kg°C)
× average temperature increase (°C)
(4067 × 103 ) J/s
cooling water flow =
4200 J/kg o C × 20o C
= 48.42 kg/s (Ans.)
A1 • Chapter 3 • Steady-flow Process Calculations
OBJECTIVE 8
Define and calculate availability and effectiveness in the context of
the steady-flow processes.
AVAILABILITY
So far, heat and work have been studied without examining how they are used
and how effectively they are applied. The ability to extract work and make
efficient use of energy is a function of the environment and conditions of the
process.
The measure of a system to produce useful work is called availability or exergy.
Availability is not the same as the amount of energy in a system but instead can
be thought of as the quality of the energy.
For example, consider a container of fuel in a large space of air. This system has
a large potential to produce work if the fuel is used in an engine. Now, burn the
fuel in a burner and allow it to heat the whole space. The air and combustion
products are now at a higher temperature. The energy of both systems is exactly
the same, but the availability of the energy to produce work is severely limited in
the second case when compared to the first.
Another example involves the potential energy at the top of a waterfall. Once the
water reaches the bottom of the waterfall, the decrease in potential energy has
been converted into a very small increase in temperature and lots of noise. Again,
the energy in both situations is the same, but the availability or usefulness of the
energy at the top of the waterfall is much greater than at the bottom.
Availability is defined as the maximum amount of work that can be extracted
from a system before it returns to the same conditions as its surrounding
environment. Once the system returns to the same conditions as its
surroundings, no more work can be done and the system is in what is called a
dead state.
99
100
A1 • First Class • SI Units
In a steady-flow process, the product of entropy and final temperature represents
energy that is unavailable for practical work, and the enthalpy of the incoming
working fluid represents the total energy in the system. The energy that is
available at any given point in the system can, therefore, be calculated as follows:
EQUATION 8.1
E
where
V2
+ Zg - T0S
2
available energy in kJ/kg of working fluid
enthalpy of the working fluid in kJ/kg
velocity of the working fluid in m/s
height of the system above a fixed datum in m
the gravity constant (9.81 m/s/s)
final system temperature in K (i.e. the temperature of the
system’s heat sink)
entropy of the working fluid in kJ/kg K
= h +
E
h
V2
Z
g
T0
=
=
=
=
=
=
S
=
V2
represents energy due to velocity, and Zg represents potential
2
energy. In actual practice, these terms are frequently ignored because they are
either insignificant or because they do not vary significantly from one point in a
system to another point. The working formula then becomes:
The term
EQUATION 8.2
E
= h - T 0S
The difference in available energy between any two points represents the
maximum amount of practical work that can be done per kg of working fluid.
This value is called the specific availability or specific exergy of the system. It is
generally applied between a given point and the heat sink (such as a turbine’s
condenser) as a measure of the efficiency of all or part of the system.
EQUATION 8.3
Specific availability = E1 - E0
= (h1 - T0S1) - (h0 - T0S0)
(8.3)
A1 • Chapter 3 • Steady-flow Process Calculations
101
Example 8
Dry saturated steam at 950 kPa is supplied to a back pressure turbine which
discharges dry saturated steam at 50 kPa. What is the specific availability of the
turbine?
Solution
Using Formula 8.2, specific availability = (h1 - T0S1) - (h0 - T0S0)
From the Steam Tables, at 950 kPa:
h1 = 2776.1 kJ/kg
S1 = 6.6041 kJ kg K
At 50 kPa:
T0
h0
S0
=
=
=
=
81.33 °C = (81.33 + 273) K
354.33 K
2645.9 kJ/kg
7.5939 kJ kg K
specific availability = (2776.1 kJ/kg − 354.33 K × 6.6041 kJ/kgK)
− (2645.9 kJ/kg − 354.33 K × 7.5939 kJ/kgK)
= (2776.1 − 2340) kJ/kg − (2645.9 − 2690.7) kJ/kg
= 436.1 kJ/kg + 44.8 kJ/kg
= 480.9 kJ/kg (Ans.)
EFFECTIVENESS
There is also a different interpretation of the Second Law of Thermodynamics
that leads to a further understanding of the value of availability. It is possible to
define and calculate the usefulness of the amount of work produced as compared
to the minimum required to achieve a result. This is known as effectiveness:
net output work
maximum reversible work
net output work
=
availabiltiy
effectiveness =
Note that this is a different way of expressing efficiency than is normally used
where efficiency is the ratio of output energy or work as compared to the input
energy.
EQUATION 8.4
102
A1 • First Class • SI Units
A good way of comparing the two is to look at an electric heater. Electric heaters
are normally quoted as being about 90% efficient because they convert about
90% of the electrical energy to heat. However, the resultant amount of heat
produced is now of a much lower quality in terms of its ability to do any more
work. The effectiveness is about 2.5%. The high quality energy of the electricity
used in the heater could, therefore, be much more effectively utilized to produce
heat by a system such as a heat pump.
Example 9
What is the system effectiveness of the turbine in Example 8?
Solution
net output work
effectiveness
= h1 - h0
= 2776.1 kJ/kg - 2645.9 kJ/kg
= 130.2 kJ/kg
net output work
availability
130.2 kJ/kg
=
480.9 kJ/kg
= 0.271 or 27.1% (Ans.)
=
The effectiveness in Example 9 has a relatively low value because the flow in this
system is largely non-reversible. Compare it to the more standard measure of
thermal efficiency:
output energy
input energy
2776.1 kJ/kg − 2645.9 kJ/kg
=
2776.1 kJ/kg
130.2 kJ/kg
=
2776.1 kJ/kg
= 0.047 or 4.7%
η=
The efficiency is very low because much of the energy contained in the steam has
not been used. This is to be expected for a back pressure turbine. The efficiency
would be substantially higher if the discharge was into a condenser or if the final
usage of the heat energy remaining in the discharged steam could be accounted
for as part of the system.
A1 • Chapter 3 • Steady-flow Process Calculations
CHAPTER QUESTIONS
1. Describe the four types of energy for steady-flow processes and state the
steady-flow energy equation.
•
_________________________________________________________
•
_________________________________________________________
•
_________________________________________________________
•
_________________________________________________________
2. A nozzle undergoes an enthalpy drop of 500 kJ/kg. Assuming that the entry
velocity is negligible and the nozzle efficiency is 97%, calculate the exit
velocity.
____________________________________________________________
3. Steam at 1500 kPa is reduced in pressure in a throttling calorimeter to 100
kPa. If its temperature after throttling is 150°C, find the quality of the initial
steam.
____________________________________________________________
4. One kilogram of dry and saturated steam at 1200 kPa expands adiabatically
and reversibly in a steam turbine and exhausts at 10 kPa. If the changes in
velocity and potential energy are small enough to be neglected, what amount
of work is done?
____________________________________________________________
5. Calculate the rate of heat transfer in MJ/h for a steam condenser if the
average temperature difference is 60°C and the area of the tubes is 20 m3.
Assume an overall heat transfer coefficient of 175 W/m°C.
____________________________________________________________
6. Define availability and effectiveness.
____________________________________________________________
____________________________________________________________
103
Part A1
CHAPTER 4
Thermodynamics of
Perfect Gases
Here is what you will be able to do when you complete each objective:
1. Review the behaviour of perfect gases.
2. Explain Joule’s law and its significance.
3. Calculate the heat added or rejected by a mass of perfect gas under
changing temperature and pressure conditions.
4. Explain the isothermal cycle using a pressure-volume diagram and
calculate heat rejected and work done using a perfect gas as the
working fluid.
5. Explain the reversible adiabatic cycle using a pressure-volume diagram
and calculate work done, final volume, and final temperature using a
perfect gas as the working fluid.
6. Calculate work done in a polytropic cycle using a perfect gas as the
working fluid.
7. Calculate the efficiency of a polytropic process using a perfect gas as
the working fluid.
8. Explain the Gibbs-Dalton law and calculate the work done and heat flow
per kilogram when a gas mixture is expanded.
105
A1 • Chapter 4 • Thermodynamics of Perfect Gases
OBJECTIVE 1
Review the behaviour of perfect gases.
PROPERTIES OF GASES
The working fluid, or medium, through which a heat engine converts heat into
work may be either a gas or a vapour. The laws governing the conditions of these
two substances differ.
For this reason, the study of thermodynamics is divided into two parts: one
concerning the behaviour of gases and one concerning the behaviour of vapours.
Similarly, heat engine cycles are divided into gas cycles and vapour cycles. This
module is concerned with gases and their behaviour. Another module deals with
vapours.
A gas may be defined as a fluid which remains in a gaseous state throughout the
pressure and temperature changes of the thermodynamic operations under consideration. Examples are oxygen, hydrogen, nitrogen, air, and carbon dioxide.
A vapour is a fluid which is easily transformed into a liquid by a moderate
reduction in temperature or an increase in pressure. Examples are steam and
ammonia.
Properties of working fluids may be divided into intensive and extensive
groups. Intensive properties are not dependent upon quantity, for example,
pressure or temperature. Extensive properties are dependent on quantity, for
example, mass or volume.
The state of a thermodynamic system may be defined by two or more of these
properties. The changes taking place during a thermodynamic operation may be
demonstrated graphically by plotting these properties against one another, for
example, pressure against volume, temperature against entropy, enthalpy against
entropy, or pressure against enthalpy.
Experiments carried out on the expansion and compression of perfect gases
resulted in the following gas laws. A perfect gas is one which follows Boyle’s
law, Charles’ law and Gay-Lussac’s law. These laws relate the changes in
pressure, volume, and temperature that occur as each one is varied individually.
107
108
A1 • First Class • SI Units
BOYLE’S LAW
Boyle’s law states that if a given mass of gas is compressed or expanded at
constant temperature, the absolute pressure varies inversely with the volume.
Expressed mathematically, this is:
PV = constant
Another way of expressing Boyle’s law is to compare two different states, in
which case the equation becomes:
PV
1 1 = PV
2 2
A common term for the fact that the process occurs at constant temperature is
isothermal. This is sometimes a reasonable assumption if the process of
expansion or compression is slow enough for the temperature to remain
constant.
CHARLES’ LAW
Charles’ law states, under constant pressure, the volume of a given mass of gas
varies directly with the absolute temperature.
Expressed mathematically, this is:
V
= constant
T
An alternate expression comparing two different states is:
V1 V2
=
T1 T2
A1 • Chapter 4 • Thermodynamics of Perfect Gases
GAY-LUSSAC’S LAW
Gay-Lussac’s law states that under constant volume, the absolute pressure of a
given mass of gas varies directly with the absolute temperature.
Expressed mathematically, this is:
P
= constant
T
An alternate expression comparing two different states is:
P1 P2
=
T1 T2
GENERAL GAS LAW
The three laws given above can seldom be used as they stand because it is very
rare that any one of the three variables (P, V, or T) remains constant during a
thermodynamic operation.
Boyle's law, Charles' law, and Gay-Lussac's law can be combined into a equation
known as the General Gas Law:
PV
= constant
T
The alternate version that compares two states is:
PV
PV
1 1
= 2 2
T1
T2
109
110
A1 • First Class • SI Units
The constant can be expressed depending on the mass of gas considered and the
characteristics of the gas. The equation can thus be written as:
PV
= mR
T
Where:
m
R
= mass of gas
= characteristic constant for the particular gas
This is most commonly written as follows:
PV = mRT
Where:
P
V
T
m
R
=
=
=
=
=
pressure in kPa, absolute
volume, m3
absolute temperature, K
mass, kg
characteristic gas constant, kJ/kg K
Example 1
If the volume of air at 0°C and atmospheric pressure is 0.773 m3/kg, calculate
the value of R.
Solution
Using equation PV = mRT :
PV = mRT
PV
R=
mT
101.3 kPa × 0.773 m3 /kg
R=
1 kg × ( 0°C + 273) K
101.3 kPa × 0.773 m3 /kg
1 kg × 273 K
R = 0.2868 kJ/kgK (Ans.)
R=
A1 • Chapter 4 • Thermodynamics of Perfect Gases
Example 2
If 1 kg of air occupies 0.313 m3 at a temperature of 95°C, find the corresponding
pressure.
Solution
Using the value of R for air calculated in example 1:
PV = mRT
mRT
P=
V
1 kg × 0.2868 kJ/kg K × (95 + 273) K
P=
0.313 m3
1 kg × 0.2868 kJ/kg K × 368 K
P=
0.313 m3
105.54
P=
0.313 m3
P = 337.2 kPa (Ans.)
111
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A1 • First Class • SI Units
OBJECTIVE 2
Explain Joule’s law and its significance.
ENERGY RELATIONSHIPS IN A NON-FLOW PROCESS
The formation of energy equations used to solve problems involving
thermodynamic processes is based on the principle of conservation of energy.
This principle states that energy can neither be created nor destroyed, but may be
transformed into other forms without loss.
A non-flow process is one in which the working medium is confined, such as in
a cylinder behind a movable piston. Heat application causes an expansion of the
medium. Hence, the energies involved in such processes are those associated
with a body at rest. An example of this process is the cycle of events in a spark
ignition engine as shown in Fig. 1.
FIGURE 1
Non-Flow Process
If a quantity of heat energy is imparted to the working medium in such a process,
the following effects are produced:
1. The moving molecules speed up, thereby, increasing the internal energy.
2. The increased speed of the molecules changes the size of the body as a
whole. This requires that something must be displaced to make room for
the new volume. This is a work process that the working medium
performs on its surroundings, thus producing external work.
A1 • Chapter 4 • Thermodynamics of Perfect Gases
These effects may be stated as:
heat added = change in internal energy + external work done
Q = (U 2 − U1 ) + W
This equation is known as the fundamental or General Energy Equation and is
the first law of thermodynamics in equation form for non-flow processes.
Any of the terms may be negative. For example, if heat is rejected or extracted,
there is a decrease in internal energy. If work is done on the gas by compression
then W will be negative.
INERNAL ENERGY
Internal energy (U) may be defined as the energy contained within a substance
due to its molecular activity. In general, the internal energy of a substance is
made up of two parts:
1. The internal kinetic energy due to molecular velocity which manifests
itself by temperature.
2. The internal potential energy arising from molecular dispersion.
Joule discovered that for a perfect gas, the internal energy depends only on its
absolute temperature and is independent of changes in pressure and volume.
This is known as Joule’s law.
It is often necessary to relate the heat added to a process and the resulting work
that is done. Since the difference between the two is a change in internal energy,
Joule's law becomes significant in relating the change in internal energy to a
change in absolute temperature.
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A1 • First Class • SI Units
OBJECTIVE 3
Calculate the heat added or rejected by a mass of perfect gas under
changing temperature and pressure conditions.
CONSTANT VOLUME PROCESS
The general energy equation that relates heat, internal energy, and work done can
be used to describe and calculate changes for a number of different reversible
non-flow processes. The first one is for a situation where heat is supplied to a
perfect gas without a change in volume. The result is an increase in temperature
but without work being done because the volume does not change.
The energy equation can be rewritten as:
heat added = change in internal energy + external work done
mcv (T2 − T1 ) = (U 2 − U1 ) + 0
Where:
cv = specific heat of the gas at constant volume
(U 2 − U1 ) = mcv (T2 − T1 )
A1 • Chapter 4 • Thermodynamics of Perfect Gases
Example 3
One kilogram of air having an initial temperature of 15°C is heated at constant
volume until the final temperature is 40°C. Using a value for cv of 0.718 kJ/kg K,
calculate:
a) External work done
b) Heat required
Solution
a) Since the process happens at constant volume, the external work done is
zero.
b) Heat required:
The heat required results from a change in internal energy only, and thus
depends only on temperature as per the following equation,
Heat added = mcv (T2 − T1 ) :
heat added = mcv (T2 − T1 )
heat added = 1 kg × 0.718 kJ/kg K × (313 − 288) K
heat added = 1 kg × 0.718 kJ/kg K × 25 K
heat added = 17.95 kJ (Ans.)
CONSTANT PRESSURE PROCESS
When a process happens at constant pressure, there is a change in temperature
and in volume, and therefore, there is work done. The work may be done
externally as the gas volume increases, or internally as the gas volume decreases.
This can be expressed by the equation:
work done = P(V2 − V1 )
By substituting the general gas law from equation PV = mRT , it can be
described in terms of temperature as follows:
work done = P (V2 − V1 ) = P (
work done = mR (T2 − T1 )
mRT2 mRT1
−
)
P
P
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A1 • First Class • SI Units
The heat supplied is found from the equation:
heat added = mc p (T2 − T1 )
Where:
c p = specific heat of the gas at constant pressure
Note that this equation is similar to the previous one for constant volume except
that the coefficient of heat transfer (specific heat) is now at constant pressure
instead of at constant volume.
The energy equation can now be written as:
heat added = change in internal energy + external work done
mc p (T2 − T1 ) = (U 2 − U1 ) + mR(T2 − T1 )
or:
(U 2 − U1 ) = mc p (T2 − T1 ) − mR(T2 − T1 )
Example 4
One kilogram of air, having an initial temperature of 15°C, is heated at constant
pressure until the final temperature is 40°C. Using a value for cp of 1.005 kJ/kg K
and a value for R of 0.287 kJ/kg K, find:
a) external work
b) heat required
c) change in internal energy
Solution
a) external work:
Because the process happens at constant pressure, the external work done is
obtained from equation work done = mR(T2 − T1 ) :
work done = mR(T2 − T1 )
work done = 1 kg × 0.287 kJ/kg K × (313 − 288) K
work done = 1 kg × 0.287 kJ/kg K × 25 K
work done = 7.175 kJ (Ans.)
A1 • Chapter 4 • Thermodynamics of Perfect Gases
b) heat required:
Heat required is obtained using equation Heat added = mc p (T2 − T1 ) :
heat added = mc p (T2 − T1 )
heat added = 1 kg × 1.005 kJ/kg K × (313 − 288) K
heat added = 1 kg × 1.005 kJ/kg K × 25 K
heat added = 25.125 kJ (Ans.)
c) change in internal energy:
heat added = change in internal energy + external work done
change in internal energy = heat added − external work done
change in internal energy = 25.125 kJ − 7.175 kJ
change in internal energy = 17.95 kJ (Ans.)
JOULE’S LAW
Joule’s law relates internal energy to absolute temperature. Also, if no external
work is done, the gas volume remains constant, and the specific heat is therefore
cv. Combining these two principles, we derive the following equation:
U = mc v T
Thus, the internal energy of a known mass of a perfect gas can be calculated for
any given temperature.
RELATIONSHIP OF SPECIFIC HEATS
Referring to examples 3 and 4, in each case, 1 kg of a perfect gas was heated
from the same initial temperature to the same final temperature. Therefore, the
change in internal energy was the same. However, there is a difference in the
amount of heat applied. This difference is due to the amount of external work
done in the constant pressure process, whereas no work was done in the
constant volume process. Therefore, comparing the two processes:
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118
A1 • First Class • SI Units
Heat required at constant press.=Heat required at constant vol.+Work done at constant press.
mc p (T2 − T1 ) = mcv (T2 − T1 ) + mR (T2 − T1 )
c p = cv + R
R = c − cv
The relationship between R, cp, and cv is important for many calculations
involving perfect gases. Another important relationship is the ratio of specific
heats, referred to as gamma (γ).
γ=
cp
cv
γ will be used in a later learning objective when dealing with adiabatic expansion
of a perfect gas.
POLYTROPIC PROCESSES
In practical applications, a true constant volume or constant pressure process is
rare since there is almost always some change in both volume and pressure
(called a polytropic process). The relationship between volume and pressure for a
perfect gas is more accurately described by modifying Boyle’s law, as follows:
PV n = constant
Where n is determined by the nature of the gas and by the conditions of the
process.
For a constant volume process, n = ∞ , while for a constant pressure process,
n = 0 . In most cases, the actual value of n will fall between 0 and ∞ . For
example, in a later objective, for an adiabatic process, n = γ .
The term for external work in a polytropic process must also be modified as
follows:
work done =
mR
(T2 − T1 )
n −1
The heat supplied is as follows:
heat added = mcn (T2 − T1 )
A1 • Chapter 4 • Thermodynamics of Perfect Gases
The energy equation can now be written as:
heat added = change in internal energy + external work done
mR
mcn (T2 − T1 ) = (U 2 − U1 ) +
(T2 − T1 )
n -1
(U 2 − U1 ) = mcn (T2 − T1 ) - (T2 − T1 )
Example 5
One kilogram of air, having an initial temperature of 15ºC, is heated until the
final temperature is 40oC. Using cn = 0.92 kJ/kg K , find the heat required.
Solution
Using equation, heat added = mcn (T2 − T1 ) , the heat required is as follows:
heat added = mcn (T2 − T1 )
heat added = 1kg × 0.92 kJ/kg K × ( 313 - 288 ) K
heat added = 1kg × 0.92 kJ/kg K × 25K
heat added = 23 kJ (Ans.)
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120
A1 • First Class • SI Units
OBJECTIVE 4
Explain the isothermal cycle using a pressure-volume diagram and
calculate heat rejected and work done using a perfect gas as the
working fluid.
ISOTHERMAL EXPANSION AND COMPRESSION
In an isothermal expansion or compression, the temperature of the working
substance is kept constant throughout the process. The form of the isothermal
curve on pressure-volume coordinates follows Boyle’s law:
PV = constant
This equation, describing a rectangular hyperbola, is a special case of the general
equation:
PV n = constant
Where the exponent, n = 1 , is represented by Fig. 2.
FIGURE 2
Work Done During
Isothermal Expansion
and Compression
A1 • Chapter 4 • Thermodynamics of Perfect Gases
The external work performed during an expansion from 1 to 2 (co-ordinates
P1V1 to P2V2), is shown graphically by the area under the curve between 1 and 2
(see Fig. 2). The two vertical lines close together in the figure are the limits of a
narrow area and indicate an infinitesimal volume change dV.
Work done during this small change of volume is:
dW = PdV
For a finite change of volume of any size, as from V1 to V2, the work done (W
(kN·m)) is the area under the curve, or:
V2
W = ∫ PdV
V1
For this integration, it is necessary to express P in terms of V. Assume that P and
V are values of pressure and volume for any point on the curve of expansion of
a gas. The equation is:
PV = C
C
P=
V
Where C = a constant
Substituting for the value of P :
C
dV
V1 V
V2 dV
W = C∫
V1 V
W =∫
V2
Which, when integrated, gives:
W = C (log e V2 − log e V1 )
Note that log e is also commonly shown as ln .
Since the initial conditions of the gas are P1 and V1 :
PV = C
PV = PV
1 1
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122
A1 • First Class • SI Units
Substituting this value for C:
W = PV
1 1 (log e V2 − log e V1 )
W = PV
1 1 log e
V2
V1
The units of measure for work are kN m.
Using the general gas law from equation PV = mRT , the work done can also be
expressed in terms of mass m in kg, so that the work done can be shown as
V
W = mRT log e 2
V1
Using Boyle’s law from equation PV
1 1 = PV
2 2 , it can also be expressed in terms of
pressure:
P
W = mRT log e 1
P2
Often, the ratio
V2
is called the ratio of expansion and is represented by r .
V1
Making this substitution in the equation W = mRT log e
V2
, the equation
V1
becomes:
W = mRT log e r
The above calculations of work done refer to an expansion from P1V1 to P2V2.
If, on the other hand, the work done is the result of compression from P2V2 to
P1V1, the curve of compression would be from 2 to 1. The area under the curve
would represent the work done during compression in a similar manner to that
for expansion, except that the expression would have a negative value, that is,
work is to be done upon the gas to decrease its volume.
The isothermal expansion or compression of a perfect gas causes no change in
its internal energy since the temperature T is constant. During such an expansion,
the gas must take in an amount of heat just equal to the work it does. Conversely,
during an isothermal compression it must reject an amount of heat just equal to
the work spent upon it.
A1 • Chapter 4 • Thermodynamics of Perfect Gases
Example 6
Air, having a pressure of 700 kPa abs. and a volume of 1 m3, expands
isothermally to a volume of 4 m3. Calculate:
a) External work of the expansion
b) Heat required to produce the expansion
c) Pressure at the end of expansion
Solution
a) External work of the expansion
Using the equation for work done, W = PV
1 1 log e
W = PV
1 1 log e
V2
:
V1
V2
V1
W = 700 kPa ×1 m3 × log e
4m3
1m3
W = 700 kPa ×1.3863 m3
W = 970.41 kNm (Ans.)
b) Heat required to produce the expansion
Since the heat added equals the increase in internal energy plus external work,
and since the temperature remains constant (requiring no heat to increase the
internal energy), the change in internal energy equals zero, and the heat added
equals the work done.
heat added = external work
heat added = 970.41 kJ (Ans.)
c) Pressure at the end of expansion
Using Boyle’s law, equation PV
1 1 = PV
2 2 , the pressure at the end of the
expansion will be:
PV
1 1 = PV
2 2
P2 =
PV
1 1
V2
700 kPa × 1 m3
4 m3
P2 = 175 kPa (Ans.)
P2 =
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A1 • First Class • SI Units
OBJECTIVE 5
Explain the reversible adiabatic cycle using a pressure-volume
diagram and calculate work done, final volume, and final
temperature using a perfect gas as the working fluid.
ADIABATIC EXPANSION AND COMPRESSION
In adiabatic expansion or compression, the working substance neither receives
nor rejects heat as it expands or compresses. A curve showing the relation of
pressures to volumes in such processes is called an adiabatic curve (see Fig. 3).
In any adiabatic process, the substance neither gains nor loses heat by
conduction, radiation, or internal chemical action. Hence, the work a gas does in
an adiabatic expansion is all done at the expense of its internal energy, and the
work done upon a gas in an adiabatic compression all goes to increase its
internal energy. Ideally, adiabatic action could be secured by a gas expanding, or
being compressed, in a cylinder which is a perfect non-conductor of heat. The
compression of a gas in a cylinder is approximately adiabatic when the process is
performed very rapidly. Conversely, when the process is done so slowly that the
heat has time to dissipate by conduction, the compression is more nearly
isothermal.
FIGURE 3
Isothermal and
Adiabatic Expansion
Lines
The pressure-volume diagram in Fig. 3 shows a comparison between four
standard forms of expansion. Each begins at point 1 with co-ordinates P1V1.
1. Constant volume expansion from point 1 to point a.
2. Constant pressure expansion from point 1 to point b.
3. Isothermal expansion (constant temperature) from point 1 to point c.
c
4. Adiabatic expansion (where n = γ = p ) from point 1 to point d.
cv
A1 • Chapter 4 • Thermodynamics of Perfect Gases
To derive the pressure-volume relation for a gas expanding adiabatically,
consider the fundamental equation:
heat added = increase in internal energy + external work done
Q = mcv (T2 − T1 ) + ∫ PdV
In the adiabatic expansion, no heat is added to or taken away from the gas by
conduction or radiation; therefore, Q becomes zero.
increase in internal energy = external work
mcv (T1 − T2 ) = ∫ PdV
To evaluate the work done, it is necessary to express P in terms of V in the
integration
∫ PdV
The law of the curve representing an adiabatic expansion is:
PV γ = C
C
P= γ
V
Where C = a constant
Therefore, the work done for an adiabatic expansion is:
C
dV
Vγ
dV
work done = C ∫ γ
V
work done = ∫
work done = C ∫ V −γ dV
work done =
PV
1 1 − PV
2 2
γ −1
It can also be expressed in terms of mRT using the general gas law to give us the
equation:
work done =
mR(T1 − T2 )
γ −1
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A1 • First Class • SI Units
Another method of calculating work done is as follows. Since an adiabatic
process involves no net addition or subtraction of heat, any external work done
is at the expense of the internal energy of the gas.
Therefore:
heat added = change in internal energy + external work done
0 = (U 2 − U1 ) + W
W = (U 2 − U1 )
From the equation for a perfect gas, (U 2 − U1 ) = mcv (T2 − T1 ) :
(U 2 − U1 ) = mcv (T2 − T1 )
But W = (U 2 − U1 )
Therefore, W = mcv (T2 − T1 )
PRESSURE, VOLUME, TEMPERATURE RELATIONSHIPS
The relationships between pressure, volume, and temperature which hold during
the expansion or compression of a gas can be demonstrated as follows, using the
nature of the PV curve from equation PV γ = C and the general gas law equation
PV
PV
1 1
= 2 2.
T1
T2
Using equation
PV
PV
1 1
= 2 2 and inserting equation PV γ = C :
T1
T2
T2 PV
= 2 2
T1 PV
1 1
From equation PV γ = C :
P2 V1γ
=
P1 V2γ
P2 ⎡ V1 ⎤
=⎢ ⎥
P1 ⎣V2 ⎦
γ
γ
T
PV
P ⎡V ⎤
Substituting 2 = ⎢ 1 ⎥ into equation 2 = 2 2 :
T1 PV
P1 ⎣V2 ⎦
1 1
A1 • Chapter 4 • Thermodynamics of Perfect Gases
γ
T2 ⎡ V1 ⎤ V2
=⎢ ⎥ ×
T1 ⎣V2 ⎦ V1
γ
T2 ⎡ V1 ⎤ ⎡ V1 ⎤
= ⎢ ⎥ ×⎢ ⎥
T1 ⎣V2 ⎦ ⎣ V2 ⎦
T2 ⎡ V1 ⎤
=⎢ ⎥
T1 ⎣V2 ⎦
−1
γ −1
In a similar fashion, the equation for pressure and volume can be rewritten as:
P1 V2γ
=
P2 V1γ
P1 ⎡V2 ⎤
=⎢ ⎥
P2 ⎣ V1 ⎦
γ
1
⎡ P ⎤γ V
and ⎢ 1 ⎥ = 2
V1
⎣ P2 ⎦
or
V2 ⎡ P2 ⎤
=⎢ ⎥
V1 ⎣ P1 ⎦
−
1
γ
−
1
PV PV
V ⎡P ⎤ γ
Substituting for the volume in 2 = ⎢ 2 ⎥ , into 1 1 = 2 2 :
T1
T2
V1 ⎣ P1 ⎦
−1
T2 P2 ⎡ P2 ⎤ γ
= ×⎢ ⎥
T1 P1 ⎣ P1 ⎦
−1
1
T2 ⎡ P2 ⎤ ⎡ P2 ⎤ γ
= ⎢ ⎥ ×⎢ ⎥
T1 ⎣ P1 ⎦ ⎣ P1 ⎦
T2 ⎡ P2 ⎤
=⎢ ⎥
T1 ⎣ P1 ⎦
T2 ⎡ P2 ⎤
=⎢ ⎥
T1 ⎣ P1 ⎦
1−
1
γ
γ −1
γ
127
128
A1 • First Class • SI Units
Collecting these results produces the following equation:
T2 ⎡ V1 ⎤
=⎢ ⎥
T1 ⎣ V2 ⎦
T2 ⎡ P2 ⎤
=⎢ ⎥
T1 ⎣ P1 ⎦
γ −1
γ −1
γ
A1 • Chapter 4 • Thermodynamics of Perfect Gases
OBJECTIVE 6
Calculate work done in a polytropic cycle using a perfect gas as the
working fluid.
POLYTROPIC PROCESSES
Those processes which do not follow either isothermal or adiabatic laws are
termed polytropic and follow the general law:
PV n = constant
n
n
Or PV
1 1 = PV
2 2
Where n is not unity and not equal to the ratio
cp
(the ratio of the specific heats
cv
of the working gas). The great majority of practical applications of gas processes
fall under this heading in that heat may be added or removed during the process,
and the gas temperature does not remain constant.
c
The value of n is usually greater than 1.0 but less than the ratio p .
cv
The work done during a polytropic expansion is calculated as shown earlier for
an adiabatic process and is given by the same equation except that γ (gamma) is
replaced by n .
work done =
PV
1 1 − PV
2 2
n −1
It can also be expressed in terms of mRT using the general gas law to give the
equation:
work done =
mR(T1 − T2 )
n −1
129
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A1 • First Class • SI Units
PRESSURE, VOLUME, TEMPERATURE RELATIONSHIPS
The relationships between pressure, volume, and temperature are the same as for
the adiabatic case, again substituting n for γ.
T2 ⎡ V1 ⎤
=⎢ ⎥
T1 ⎣V2 ⎦
T2 ⎡ P2 ⎤
=⎢ ⎥
T1 ⎣ P1 ⎦
n −1
n −1
n
Example 7
A 7 litre volume of air at a pressure of 300 kPa and a temperature of 150°C is
expanded polytropically (n = 1.32) to a pressure of 100 kPa. Calculate:
a) Final volume
b) Work transfer from the air
c) Final temperature
Solution
a) Final volume
n
n
The final volume is obtained using equation PV
1 1 = PV
2 2 :
n
n
PV
1 1 = PV
2 2
V2 n =
n
PV
1 1
P2
300 kPa × 71.32
L
100 kPa
300 kPa × 13.048
V2 n =
100 kPa
n
V2 = 39.144 L
V2 n =
1
V2 = 39.1441.32 L
V2 = 16.09 litres (Ans.)
A1 • Chapter 4 • Thermodynamics of Perfect Gases
b) Work transfer from the air:
The work done is obtained using equation work done =
PV
1 1 − PV
2 2
:
n −1
PV
1 1 − PV
2 2
n −1
300 kPa × 0.007 m3 − 100 kPa × 0.01609 m3
work done =
1.32 − 1
2.10 − 1.609
work done =
1.32 − 1
0.491
work done =
0.32
work done = 1.534 kNm (Ans.)
work done =
It is important to note that in part a, it was possible to use any consistent unit
of measure because the units of measure for pressure cancel out. However,
in part b, the volume had to be converted to m3 to result in the correct unit
of measure for the work done.
c) Final temperature:
T ⎡V ⎤
The final temperature may be obtained using either equations 2 = ⎢ 1 ⎥
T1 ⎣V2 ⎦
T2 ⎡ P2 ⎤
=⎢ ⎥
T1 ⎣ P1 ⎦
n −1
n
since both sets of pressure and volume are known.
T ⎡P ⎤
Using equation 2 = ⎢ 2 ⎥
T1 ⎣ P1 ⎦
n −1
n
:
n −1
or
131
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A1 • First Class • SI Units
⎡P ⎤
T2 = ⎢ 2 ⎥
⎣ P1 ⎦
n −1
n
× T1
⎡ 100 kPa ⎤
T2 = ⎢
⎣ 300 kPa ⎥⎦
1.32 −1
1.32
× (150 + 273) K
0.32
⎡ 100 kPa ⎤ 1.32
× (150 + 273) K
T2 = ⎢
⎣ 300 kPa ⎥⎦
T2 = 0.33330.2424 × 423 K
T2 = 0.7662 × 423 K
T2 = 324.1 K
T2 = ( 324.1 − 273) °C
T2 = 51.1°C (Ans.)
A1 • Chapter 4 • Thermodynamics of Perfect Gases
OBJECTIVE 7
Calculate the efficiency of a polytropic process using a perfect gas
as a working fluid.
EFFICIENCY OF A POLYTROPIC PROCESS
The efficiency of a thermodynamic process is defined by the basic equation:
output
input
work done
efficiency =
heat supplied
efficiency =
In order to differentiate from other types of efficiency, this is known as thermal
efficiency since it is the result of thermodynamic processes. Mechanical
efficiency, which deals with losses such as friction and windage, is the other type
of efficiency that normally has to be taken into account in heat engines.
For a compression process, the heat that results in an increase in temperature is
essentially wasted, while the energy provided to increase the pressure is useful
energy. Thus, depending on the mode of compression (constant volume or
pressure, adiabatic or polytropic), it is possible to calculate the work done using
the equations developed for each one of these. The heat supplied can likewise be
determined from the relevant equations.
In the case of polytropic compression, the work done is defined by equation
PV − PV
mR (T1 − T2 )
Work done = 1 1 2 2 or Work done =
. The heat supplied is
n −1
n −1
the sum of the internal energy and the work done. The internal energy is given by
the expression mcv (T2 − T1 ) .
Therefore, the thermal efficiency is:
work done
heat supplied
work done
efficiency =
change in internal energy + work done
efficiency =
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A1 • First Class • SI Units
Example 8
A 1 kg mass of air at a pressure of 100 kPa and a temperature of 25°C is
compressed to 250 kPa. Assuming a polytropic coefficient of 1.3, values for cv
of 0.718 kJ/kg K and R , 0.287 kJ/kg K, calculate the efficiency of the
compression process.
Solution
T ⎡P ⎤
The final temperature is obtained using equation 2 = ⎢ 2 ⎥
T1 ⎣ P1 ⎦
T2 ⎡ P2 ⎤
=⎢ ⎥
T1 ⎣ P1 ⎦
⎡P ⎤
T2 = ⎢ 2 ⎥
⎣ P1 ⎦
n −1
n
:
n −1
n
n −1
n
× T1
1.3−1
⎡ 250 kPa ⎤ 1.3
T2 = ⎢
× (25°C + 273) K
⎣ 100 kPa ⎥⎦
T2 = 2.50.2308 × 298 K
T2 = 1.2355 × 298 K
T2 = 368.18 K
T2 = ( 368.18 − 273) °C
T2 = 95.18°C
Therefore,
the work done
mR(T1 − T2 )
:
work done =
n −1
is
calculated,
using
the
equation:
A1 • Chapter 4 • Thermodynamics of Perfect Gases
mR (T1 − T2 )
n −1
1 kg × 0.287 kJ/kg K × (298 − 368) K
work done =
1.3 − 1
1 kg × 0.287 kJ/kg K × (−70) K
work done =
1.3 − 1
1 kg × 0.287 kJ/kg K × −70 K
work done =
1.3 − 1
1 kg × 0.287 kJ/kg K × −70 K
work done =
1.3 − 1
−20.09
work done =
0.3
work done = −66.97 kJ
work done =
Negative (compression) work was done on the gas.
Change in internal energy, using equation U 2 − U1 = mcv (T2 − T1 ) , is:
U 2 − U1 = mcv (T2 − T1 )
U 2 − U1 = 1 kg × 0.718 kJ/kg K × (368 − 298) K
U 2 − U1 = 1 kg × 0.718 kJ/kg K × (70) K
U 2 − U1 = 50.26 kJ
The efficiency, using equation efficiency =
work done
is:
internal energy + work done
work done
internal energy + work done
66.97 kJ
efficiency =
50.26 kJ + 66.97 kJ
66.97 kJ
efficiency =
117.23 kJ
efficiency = 0.5713
efficiency = 57.13% (Ans.)
efficiency =
135
136
A1 • First Class • SI Units
OBJECTIVE 8
Explain the Gibbs-Dalton law and calculate the work done and heat
flow per kilogram when a gas mixture is expanded.
DALTON’S LAW
When two or more gases are present together without a chemical reaction taking
place, they are referred to as a mixture. Air, a combination of oxygen, nitrogen,
water vapour, and other miscellaneous gases, is an example of a mixture. It is
possible to derive the thermodynamic properties of a mixture from those of its
individual component gases.
Dalton’s law, dealing with pressures of a mixture, states that the total pressure of
a mixture is equal to the sum of the partial pressures of its constituents. The
partial pressure of each constituent gas equals the pressure that the gas would
have if it filled the same volume at the same temperature by itself.
Using two containers of the same volume and temperature, one with gas A and
the other with gas B, the general gas law for each container is:
PAV = mA RT
and PBV = mB RT
If both of the gases are placed in one container, the total mass will be the sum of
the two masses:
PV = mRT
PV = (mA + mB ) RT
PV = ( PA + PB )V
P = PA + PB
This is known as Dalton’s law of partial pressures. In its general form, it applies
to any number of constituents and can be written by the equations
P = ∑ Pi for pressure
i
m = ∑ mi for mass
i
A1 • Chapter 4 • Thermodynamics of Perfect Gases
The properties of a mixture can thus be derived from the individual properties of
its constituents, and the resultant mixture can be treated as a perfect gas with the
derived properties.
GIBBS DALTON LAW
Dalton’s law was extended by Gibbs to include the internal energy, enthalpy, and
entropy of a mixture. This extension is known as the Gibbs-Dalton law.
mu = ∑ mi ui for internal energy
i
mh = ∑ mi hi for enthalpy
i
ms = ∑ mi si for entropy
i
For a perfect gas, the specific heat is represented by c
Where:
h = c pT
u = cvT
R = c p − cv
These can be substituted into the previous equations to give the relationships for
specific heat and the universal gas constant.
mi
c pi
i m
m
cv = ∑ i cvi
i m
m
R = ∑ i Ri
i m
cp = ∑
Calculations can now be made for work done and heat flow using the derived
properties and treating the resultant gas as a perfect gas. It should be noted that
this works reasonably well as an approximation at low pressures, but there are
increasing deviations as the pressure increases.
137
138
A1 • First Class • SI Units
Example 9
A mixture of 1 kg of methane and 0.5 kg of carbon dioxide is contained in a
vessel of 10 m3 volume at a pressure of 400 kPa and a temperature of 20°C. The
mixture is expanded to 20 m3. Calculate the work done and heat flow if the
expansion is done at constant pressure.
The individual properties are:
methane: cp = 2.177, cv = 1.675, and R = 1.30
carbon dioxide: cp = 0.825, cv = 0.630, and R = 1.31
Solution
The properties of the mixture are as follows:
cp = ∑
i
mi
c pi
m
1
0.5
c p = ( × 2.177) + ( × 0.825)
1.5
1.5
c p = 1.451 + 0.275
c p = 1.726
cv = ∑
i
mi
cvi
m
1
0.5
cv = ( × 1.675) + ( × 0.630)
1.5
1.5
cv = 1.117 + 0.21
cv = 1.327
R=∑
i
mi
Ri
m
1
0.5
R = ( × 1.3) + ( ×1.31)
1.5
1.5
R = 0.8667 + 0.4367
R = 1.303
A1 • Chapter 4 • Thermodynamics of Perfect Gases
For the constant pressure case, the work done is calculated using equation
work done = P (V2 − V1 ) :
work done = P (V2 − V1 )
work done = 400 kPa(20 − 10) m3
work done = 400 kPa × 10m3
work done = 4000 kN m (Ans.)
The final temperature can be found from Charles’ law, equation
V1 V2
= :
T1 T2
V1 V2
=
T1 T2
T2 = T1
V2
V1
T2 = (273 + 20°C) K ×
20 m3
10 m3
T2 = 293 K × 2
T2 = 586 K
T2 = ( 586 − 273) °C
T2 = 313°C
The heat that has to
heat added = mc p (T2 − T1 ) :
be
supplied
is
obtained
heat added = mc p (T2 − T1 )
heat added = 1.5 kg × 1.726 kJ/kg K × (586 − 293) K
heat added = 1.5 kg × 1.726 kJ/kg K × 293
heat added = 758.58 kJ (Ans.)
using
equation:
139
140
A1 • First Class • SI Units
CHAPTER QUESTIONS
1. State the general energy equation for a non-flow process.
____________________________________________________________
2. Define Joule’s law.
____________________________________________________________
____________________________________________________________
3. One kilogram of methane having an initial temperature of 25°C is heated at
constant volume until the final temperature is 50°C. Using a value for cv of
1.675 kJ/kg K, find the external work and heat required.
____________________________________________________________
____________________________________________________________
4. State the pressure volume relationships for isothermal, adiabatic, and
polytropic expansion or compression and explain the differences.
____________________________________________________________
____________________________________________________________
5. A 5 litre volume of air at a pressure of 100 kPa and a temperature of 20°C is
compressed adiabatically (n =1.40) to a pressure of 250 kPa. Calculate:
a) final volume
b) work done
c) final temperature
6. 0.045 m3 volume of a perfect gas, at a pressure of 100 kPa and a temperature
of 22ºC, is compressed to 3500 kPa. Assuming a polytropic index of
compression of 1.32, values for cv of 718 J/kgK and R, 287 J/kgK, calculate
the following:
a)
b)
c)
d)
final temperature of the gas
work done
change in internal energy
heat transfer
A1 • Chapter 4 • Thermodynamics of Perfect Gases
7. A 50 kW electric motor has an efficiency of 92%. How much heat must be
carried away by the motor ventilation system per hour?
____________________________________________________________
8. A mixture of 2 kg of methane and 1 kg of nitrogen is contained in a vessel of
10 m3 volume at a pressure of 400 kPa and a temperature of 20°C. The
mixture is expanded to 20 m3. Calculate the work done and heat flow if the
expansion is done isothermally.
The individual properties are:
methane: cp = 2.177, cv = 1.675, and R = 1.30
nitrogen: cp = 1.043, cv = 0.745, and R = 1.40
____________________________________________________________
____________________________________________________________
141
Part A1
CHAPTER 5
Expansion and Heat
Transfer
Here is what you will be able to do when you complete each objective:
1. Explain how thermal expansion and contraction is allowed for in boiler
and piping design.
2. Calculate the linear and volumetric expansion of a header or pipe, given
internal temperature conditions.
3. Calculate heat transfer by conduction.
4. Calculate the heat flow through a compound insulated wall.
5. Calculate the thickness of insulation required to maintain a given
temperature gradient.
143
A1 • Chapter 5 • Expansion and Heat Transfer
OBJECTIVE 1
Explain how thermal expansion and contraction is allowed for in
boiler and piping design.
EXPANSION OF SOLIDS
Most solid bodies expand when heated and contract when cooled. This
expansion takes place in all directions, but the increase in length due to rise in
temperature, known as linear expansion, is of particular importance. For
example, pipes that carry hot gases, such as steam, increase in length considerably
when the temperature of the metal is raised by the passage of steam through the
pipes. Provision must be made for this expansion in pipes or in any structure
where long lengths of metal are used.
When designing or erecting any engineering apparatus, neglecting to take into
account the tremendous forces exerted by the expansion and contraction of
metals may have disastrous results. If the members of a structure are rigidly
fixed, expansion will cause compression and contraction will cause tension in the
various members.
Sometimes, alloys with a minimum amount of expansion are used for special
purposes, such as measuring instruments.
Expansion can be considered in one of three ways:
1. Linear expansion
2. Superficial expansion
3. Cubical expansion
145
146
A1 • First Class • SI Units
LINEAR EXPANSION
With objects that are much longer than they are wide, it is sufficient to consider
the increase in length for a certain increase in temperature. The increase in length
depends on three factors:
1. The length of the object.
2. The change in temperature.
3. The coefficient of linear expansion (specific to the type of material in the
object).
The coefficient of linear expansion is denoted by the Greek letter alpha, α .
For a bar of length L, the change in length is:
change in length = original length × coefficient of linear expansion × change in temperature
change in length = Lα (T2 − T1 )
The coefficient of linear expansion is defined as the fraction of its length that
a body expands when heated one degree Celsius. The coefficients of expansion
for various materials have been determined by careful experiment and tabulated.
A few of the most common are given in Table 1. These coefficients are the
amount of expansion per unit length. Thus, a piece of steel 1 m long expands
0.000012 m when heated through 1°C or a piece of steel 1 cm long expands
0.000012 cm when heated through 1°C. If the length is in metres, the expansion
or contraction is given in metres. If the length is in centimetres, the expansion or
contraction is given in centimetres.
Note: the units of measure are per °Celsius.
TABLE 1
Coefficients of Linear
Expansion for
Common Materials
Material
Aluminum
Brass
Bronze
Cast Iron
Copper
Glass
Gold
Ice
Lead
Nickel
Platinum
Silver
Steel
Coefficient of Linear Expansion
(per °C)
23.8 x 10-6
18.4 x 10-6
18.2 x 10-6
10.4 x 10-6
16.5 x 10-6
9.0 x 10-6
14.2 x 10-6
50.4 x 10-6
29.0 x 10-6
13.0 x 10-6
8.6 x 10-6
19.5 x 10-6
12.0 x 10-6
A1 • Chapter 5 • Expansion and Heat Transfer
SUPERFICIAL EXPANSION
Using a similar approach for an increase in area, it can be shown that:
coefficient of superficial expansion = 2 × coefficient of linear expansion
These are very close approximations and are sufficiently accurate for most
engineering purposes. The coefficient of superficial expansion is denoted by the
Greek letter beta, β.
CUBICAL EXPANSION
Cubical, or volumetric, expansion refers to an increase in volume of either a solid
or a liquid. The equation for calculating the increase in volume for a certain
temperature change is similar to the one for linear expansion. The coefficient of
volumetric expansion is denoted by the Greek letter gamma, β .
change in volume = original volume × coefficient of volumetric expansion × change in temperature
change in volume = V β (T2 − T1 )
It is possible to relate the coefficients of linear and volumetric expansion. If one
considers a cube with a unit length of 1 that is heated by 1°C, the new length for
the height, width, and depth will be 1 + α .
original volume = 1× 1× 1
original volume = 1
new volume = (1 + α )3
new volume = 1 + 3α + 3α 2 + α 3
increase in volume = new volume − original volume
increase in volume = 1 + 3α + 3α 2 + α 3 − 1
increase in volume = 3α + 3α 2 + α 3
The second order and third order terms are very small because α is already small,
and therefore, they can be neglected in the equation.
147
148
A1 • First Class • SI Units
Thus:
coefficient of volumetric expansion = 3 × coefficient of linear expansion
β = 3α
The equation for volumetric expansion of a solid can be written as
change in volume = 3V α (T2 − T1 )
EXPANSION OF LIQUIDS
All liquids, with the exception of water, expand when heated and contract when
cooled. Water also obeys this rule, except when near the freezing point. It
contracts in volume as it cools down to 3.9°C. Below this temperature, it begins
to expand. At 0°C it freezes, and the freezing is accompanied by further
expansion. Were it not for the fact that water expands when freezing, streams
and lakes would be frozen solid in winter, the ice sinking to the bottom as fast as
it formed, and no animal or vegetable life could exist in the streams and lakes of
cold countries. This expansion during freezing also accounts for the bursting of
frozen water pipes.
Liquids usually have a greater rate of expansion than solids and likewise exert a
tremendous force when expanding. Hence, if a closed vessel is entirely filled with
a liquid and heated, the resulting pressure due to expansion may be enormous
and may cause serious damage. A hot water heating system is an example of an
apparatus entirely filled with water and subjected to heat. Provision is always
made for expansion in such systems by means of expansion tanks and water
relief valves.
Table 2 shows examples of coefficients of volumetric expansion for liquids
(per °C ).
TABLE 2
Coefficients of
Volumetric Expansion
for Some Liquids
Material
Alcohol (ethyl)
Mercury
Oil
Water (20°C to 100°C)
Coefficient of Volumetric Expansion
(per °C)
0.0011
0.00018
0.0008
0.00037
A1 • Chapter 5 • Expansion and Heat Transfer
EXPANSION OF GASES
The expansion of a gas resulting from temperature change is a more complicated
operation than the thermal expansion of solid or liquid substances. It depends on
the pressure at which the gas is contained and the conditions under which the
expansion was carried out.
The expansion of gases is covered in a different module.
Unequal thermal expansion between components of a boiler or piping system
produces added stresses on the components. Stress and strain calculations are
addressed in a later module. The resulting stress must be allowed for in system
design, especially for systems and components that are routinely heated and
cooled by steam or hot gases. Materials must be selected to withstand the
maximum expected stress. For pressure parts, this is accomplished by designing
systems and selecting materials according to the standards and codes in effect.
Design features, such as expansion joints and piping bends, are also used to
reduce stress caused by thermal expansion.
149
150
A1 • First Class • SI Units
OBJECTIVE 2
Calculate the linear and volumetric expansion of a header or pipe,
given internal temperature conditions.
Example 1
How much expansion should be allowed for in erecting a steel steam main 140 m
long to carry steam which has a temperature of 180°C? Assume that the normal
temperature of the engine room is 22°C.
Solution
The increase in expansion is found using the equation:
change in length = Lα (T2 − T1 ) :
change in length = Lα (T2 − T1 )
change in length = 140 m × 12 × 10−6 / o C × (180 − 22) o C
change in length = 140 m × 0.000012/ o C × 158 o C
change in length = 0.265 m (Ans.)
Example 2
A steel bridge 170 m long is erected in a district where the maximum variation
from summer to winter temperatures is 80°C. What is the maximum change in
length of this bridge?
Solution
change in length = Lα (T2 − T1 )
change in length = 170 m × 12 × 10−6 / o C × 80 o C
change in length = 170 m × 0.000012/ o C × 80 o C
change in length = 0.1632 m (Ans.)
Example 3
A bronze collar is bored out to a diameter of 14.958 cm at an ambient
temperature of 26°C. It will be expanded by heating to a diameter of 15.006 cm
A1 • Chapter 5 • Expansion and Heat Transfer
so that it will slip easily over a hub with a nominal diameter of 15 cm. In order to
fit the hub, to what temperature must the collar be heated?
Solution
Diameter is a linear dimension. Thus, the same approach can be taken as with a
long object such as a bar.
change in length = Lα (T2 − T1 )
15.006 cm − 14.958 cm = 14.958 cm × 18.2 × 10−6 / o C × (T2 − 26) o C
15.006 cm − 14.958 cm = 14.958 cm × 0.0000182/ o C × (T2 − 26) o C
0.048 cm = 0.000272 × (T2 − 26) o C
176.5 o C = (T2 − 26) o C
T2 = 176.5 o C + 26 o C
T2 = 202.5 °C (Ans.)
When the collar cools, it will be in tension but will exert a compressive force on
the shaft. If this collar was made of cast iron, it would probably crack as it cooled
since the tensile strength of cast iron is low.
Example 4
Find the change in volume of an aluminum sphere with a radius of 20 cm when
it is heated from 0°C to 100°C. (α = 0.000 023 8 from Table 1)
Solution
The original volume of the sphere is:
4
V = π r3
3
4
V = π (20 cm)3
3
V = 33510 cm3
Using the equation, change in volume = 3V α (T2 − T1 ) , the increase in volume is:
change in volume = 3V α (T2 − T1 )
change in volume = 3 × 33510 cm3 × ( 23.8 × 10−6 ) / D C × (100 − 0 ) D C
change in volume = 239.26 cm 3 (Ans.)
151
152
A1 • First Class • SI Units
Note: the change in volume of the sphere is the same whether the sphere is solid
or hollow. The same holds true when calculating the thermal expansion of a tank
or a pipeline.
Example 5
2500 litres of oil are heated through 50°C. Find the increase in volume.
Solution
Since there are 1000 litres in a cubic metre, the volume of the oil is
2500L
1000L/m3
= 2.5 m3
volume of oil =
Using equation, change in volume = V β (T2 − T1 ) , the increase in volume is:
change in volume = V β (T2 − T1 )
change in volume = 2.5 m3 × 0.0008 / D C × (50) D C
change in volume = 0.1 m 3 (Ans.)
Example 6
A 20 m long copper pipe with a 10 cm diameter contains a refrigerant at -50°C.
If the ambient temperature is 20°C, find the change in length and volume of the
pipe.
(α = 0000165 from Table 1)
Solution
The change in length is found using equation: change in length = Lα (T2 − T1 ) :
change in length = Lα (T2 − T1 )
change in length = 20 m × 0.0000165 / D C × ⎡⎣ −50 − ( +20 ) ⎤⎦ D C
change in length = 20 m × 0.0000165 / D C × (−70) D C
change in length = -0.0231 m (Ans.)
The minus sign confirms that the pipe contracts, or decreases, in length.
A1 • Chapter 5 • Expansion and Heat Transfer
The original volume of the pipe is:
volume = π r 2 L
volume = π (0.05 m) 2 × 20 m
volume = 0.1571 m3
Using equation, change in volume = 3V α (T2 − T1 ) , the change, or decrease, in
volume is:
change in volume = 3V α (T2 − T1 )
change in volume = 3 × 0.1571 m3 × 0.0000165 / D C × ⎡⎣ −50 − ( +20 ) ⎤⎦ D C
change in volume = 3 × 0.1571 m3 × 0.0000165 / D C × (−70) D C
change in volume = -0.000544 m 3 (Ans.)
153
154
A1 • First Class • SI Units
OBJECTIVE 3
Calculate heat transfer by conduction.
HEAT TRANSFER BY CONDUCTION
Conduction is the transfer of heat by means of a temperature difference through
a body or from one body to another in direct contact. Some materials, such as
most metals, are good conductors. Other materials, such glass wool, cork, and
wood, are poor conductors and are called insulators. Liquids also conduct heat,
although convective currents are often set up as a result, and total heat transfer is
the combination of conduction and convection. The quantity of heat transmitted
through a piece of material, such as a boiler plate or the wall of a house, depends
on the following factors:
1. The temperature difference between its two opposite sides.
2. The nature of the material or its thermal conductivity: steel, for
example, is a much better conductor than boiler scale.
3. The area of the surface: more heat flows through a large surface than a
smaller one.
4. The thickness of the material: less heat passes through a thick material
than through a thin material.
5. The time of heat flow.
Stated as an equation, the quantity of heat transferred becomes:
heat transferred Q =
where Q
λ
A
ΔT
t
d
λ AΔTt
d
= heat transferred, J (joules)
= thermal conductivity or coefficient of heat
transfer, W/m/°C or W/mK
= area, m2
= temperature difference between the surfaces, °C
= time, seconds
= thickness, metres
A1 • Chapter 5 • Expansion and Heat Transfer
155
The units for λ can be derived from the above equation by reorganizing the
λ AΔTt
equation heat transferred Q =
in terms of the Greek letter lambda, λ.
d
Qd
λ=
AΔTt
Jm
λ= 2
m s°C
J
λ=
ms°C
W
λ=
m°C
W
λ=
mΚ
Thermal conductivity (λ) is defined as: the number of joules that can flow
through 1 m2 of a substance 1 m thick with 1°C temperature difference on its
two opposite surfaces in one second.
This quantity may be expressed as “the rate of heat flow of the substance in J per
m2 per m thick, per degree C of temperature difference, per s.”
The thermal conductivity of most substances varies with the temperature. Tables
of conductivity for different materials, through different temperature ranges, are
given in various textbooks. Care should be taken in using these values. For
example, the conductivity of a copper tube at 50°C may be very different from
the conductivity of a copper tube at 500°C. Conductivities of some common
substances are given in Table 3.
Material
Air
Aluminum
Ammonia
Asbestos (air cell)
Brass
Brickwork
Cast iron
Concrete
Copper
Cork
Firebrick
Scale (boiler)
Steel
Water
Wood (pine)
Temperature Range
(°C)
0
0 - 250
0 - 100
100 - 300
0 - 250
20
0 - 250
20
0 - 250
0 - 1300
0 - 100
20
Thermal Conductivity
(W/m/°C)
0.022
140 - 160
0.019 - 0.30
0.014 - 0.210
85 - 31
0.495
73 - 30
0.756 - 0.817
300 - 297
0.05
1.3
2.32
46
0.507 - 0.942
0.15
Note: the conductivity of steel is 20 times that of scale.
TABLE 3
Thermal Conductivity
for Common Materials
156
A1 • First Class • SI Units
Example 7
How much heat will be transmitted per hour through a steel boiler tube 8 cm in
diameter, 3.5 mm thick, and 5 m in length if the water temperature inside is
190°C and the hot gases outside are 690°C? (λ = 46 W/m/ºC from Table 3.)
Solution
The area through which heat is transferred is the surface area of the boiler tube:
A = π dL
A = π × 0.08 m × 5 m
A = 1.257 m 2
Converting from seconds to hours, the heat transmitted is found using the
λ AΔTt
equation: heat transferred Q =
:
d
heat transferred Q =
λ AΔTt
d
46 W/m/ D C × 1.257 m 2 × (690 − 190) D C × 3600 s/h
heat transferred Q =
0.0035 m
D
46 W/m/ C × 1.257 m 2 × 500 D C × 3600 s/h
heat transferred Q =
0.0035 m
6
104.0796 ×10
heat transferred Q =
0.0035 m
heat transferred Q = 29737 × 106 J/h
heat transferred Q = 29 737 MJ/h (Ans.)
A1 • Chapter 5 • Expansion and Heat Transfer
OBJECTIVE 4
Calculate the heat flow through a compound insulated wall.
COMPOUND INSULATION
It is often necessary to construct an insulation partition using various materials.
For example, a furnace enclosure may be constructed of firebrick, sheeting, and
finally building brick. The walls of a cold room may be made up of plaster, wood
lining, rockwool, and brick.
The rate of heat transfer differs across each layer because the thermal
conductivity and wall thickness are different. It is possible to determine a net
thermal conductivity for the compound wall to make calculations easier.
Consider a compound insulation of different materials.
Let:
λ1 λ2 λ3 etc. = coefficient of thermal conductivity for the various
materials.
d1 d2 d3 etc. = thicknesses of the various materials.
Q1 Q2 Q3 etc. = quantities of heat transmitted through the various
materials under unit conditions.
λr = coefficient of thermal conductivity
Qr = quantity of heat transmitted
d r = total thickness
157
158
A1 • First Class • SI Units
From equation, heat transferred Q =
λ AΔTt
d
, the heat passing through a wall is
expressed by:
heat transferred Q =
λ AΔTt
d
Qd
ΔT =
λ At
The total temperature drop ΔTr across the compound wall consists of the
individual temperature drops across each section.
ΔTr = (T1 − T2 ) + (T2 − T3 ) + (T3 − T4 )
Equation ΔT =
Qd
can be substituted into each term as follows:
λ At
ΔTr =
Q1d1 Q2 d 2 Q3 d3
+
+
λ1 At λ2 At λ3 At
Since the heat flow and area are the same, the equation can be reorganized as:
ΔTr =
Q ⎛ d1 d 2 d3 ⎞
⎜ + + ⎟
At ⎝ λ1 λ2 λ3 ⎠
Making the formula more general for any number of partitions, the final equation
for heat transfer across a compound wall is:
Q=
AΔTt
d
∑λ
Example 8
An insulating wall is made up of 12 cm of brick, 20 cm of rockwool, and 2 cm of
wood lining. The wall measures 10 m long by 4 m high. Inside and outside
temperatures are -4°C and 20°C, respectively. Coefficients of thermal
conductivity are: brick = 0.5 W/m/°C, rockwool = 0.04 W/m/°C and wood =
0.1 W/m/°C. Calculate the hourly heat loss through this wall.
Solution
First, calculate the sum of the distances and thermal conductivities. Note that the
distances are expressed in metres since the dimensions for the area are expressed
in metres.
A1 • Chapter 5 • Expansion and Heat Transfer
d
d1
∑λ = λ
+
1
d
d2
λ2
+
d3
λ3
0.12 m
0.2 m
0.02 m
∑ λ = 0.5 W/m/°C + 0.04 W/m/°C + 0.1 W/m/°C
d
∑ λ = 0.24 °C/W + 5 °C/W + 0.2 °C/W
d
∑ λ = 5.44 °C/W
Therefore, the hourly heat loss is:
Q=
Q=
AΔTt
d
∑λ
(10 m × 4 m ) × ( 20 − (−4) ) °C × 3600 s/h
5.44 °C/W
40 m × ( 20 + 4 ) °C × 3600 s/h
Q=
5.44 °C/W
2
40 m × 24°C × 3600 s/h
Q=
5.44 °C/W
2
40 m × 24°C × 3600 s/h
Q=
5.44 °C/W
Q = 635294 J/h
Q = 635 kJ/h (Ans.)
2
159
160
A1 • First Class • SI Units
Example 9
Consider the 8 cm boiler tube in Example 7. Calculate the heat transferred per
hour if the tube has a 1 mm thick skin of scale. Compare the heat transferred
with the clean tube to that with the scale (use thermal conductivities from Table
3).
Solution
First, calculate the sum of the distances and thermal conductivities.
d
d1
∑λ = λ
1
d
+
d2
λ2
0.0035 m
0.001 m
+
D
C 2.32 W/m/ D C
∑ λ = 46 W/m/
d
−5
d
−5 D
∑ λ = ( 7.6 ×10 )
∑ λ = 50.7 ×10
D
C/W + ( 43.1× 10−5 ) D C/W
C/W
The hourly heat loss is:
Q=
AΔTt
d
∑λ
1.257 m 2 × ( 690 − 190 ) D C × 3600 s/h
Q=
50.7 ×10-5 D C/W
1.257 m 2 × 500 D C × 3600 s/h
Q=
50.7 × 10-5 D C/W
Q = 44.63 ×108 J/h
Q = 4463 MJ/h (Ans.)
Comparing the heat transfer rates shows that the ratio of clean to scaled tubes is:
heat rate (clean tube)
heat rate (scaled tube)
29737 MJ/h
ratio of clean to scaled tubes =
4463 MJ/h
ratio of clean to scaled tubes = 6.66 (Ans.)
ratio of clean to scaled tubes =
The clean tube transfers 6.66 times as much heat as the scaled tube.
A1 • Chapter 5 • Expansion and Heat Transfer
OBJECTIVE 5
Calculate the thickness of insulation required to maintain a given
temperature gradient.
INSULATION
It is sometimes necessary or desirable to add more insulation, for example, to a
cold room or to steam pipes to reduce heat loss. It is possible to calculate the
thickness of insulation required to achieve a certain amount of reduction by
using the equations developed in Objective 4.
The process is as follows: first, calculate the amount of heat being transferred.
The desired reduction is used to determine the reduced heat transfer. Then, this
is applied to the heat transfer equation to calculate the thickness of insulation
required.
Example 10
A cold room consists of wood walls that are 15 cm thick. The temperature
maintained in the cold room is -20°C, and the average outside temperature is
22°C. Management wants to reduce the heat loss by 50% in order to increase the
efficiency of the refrigeration system. What thickness of cork needs to be added
to achieve this? (Cork λ = 0.05 from Table 3.)
161
162
A1 • First Class • SI Units
Solution
Taking an area of 1m2, the current heat loss is found from the basic equation for
λ AΔTt
:
heat transfer, heat transferred Q =
d
heat transferred Q =
heat transferred Q =
λ AΔTt
d
0.05 W/m/ D C × 1 m 2 × ⎡⎣ 22 − ( −20 ) ⎤⎦ D C × 3600 s/h
0.15 m
0.05 W/m/ C × 1 m × [ 22 + 20] D C × 3600 s/h
D
heat transferred Q =
2
0.15 m
0.05 W/m/ C × 1 m 2 × 42 D C × 3600 s/h
heat transferred Q =
0.15 m
heat transferred Q = 50400 J/h
D
The desired reduced heat loss is 50% of this amount, or 25 200 J/h.
The equation for heat transfer across a compound wall, Q =
used to find the required thickness of the cork.:
Q=
AΔTt
, can then be
d
∑λ
AΔTt
d
∑λ
1 m 2 × 42 D C × 3600 s/h
d2
0.15 m
+
D
0.15 W/m/ C 0.05 W/m/ D C
42 D C × 3600 s/h
d2
1D C/W +
=
0.05 W/m/ D C
25200 J/h
d2
1D C/W +
= 6 D C/W
D
0.05 W/m/ C
d 2 m/ D C/W
= 5 D C/W
0.05
d 2 = 5 × 0.05 m
25200 J/h =
d 2 = 0.25 m
d 2 = 25 cm (Ans.)
A1 • Chapter 5 • Expansion and Heat Transfer
CHAPTER QUESTIONS
1. Define the coefficient of linear expansion.
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2. A section of steel pipe in a hydraulic system has of a total length of 13.7 m
and an internal diameter of 30 mm. If the coefficient of linear expansion of
pipe is 1.2 × 10-5 per °C, and the coefficient of volumetric expansion of the
oil is 9 × 10-4 per °C, calculate the volumetric allowance in litres to be made
for oil overflow when the temperature rises by 27°C.
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3. What five factors does the quantity of heat transmitted through a piece of
material such as a boiler plate depend on?
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4. A cold storage compartment is 4.5 m long by 4 m wide by 2.5 m high. The
four walls, ceiling, and floor are covered with 150 mm thick insulation that
has a thermal conductivity of 5.8 × 10-2 W/mK. Calculate the amount of heat
leaking from the cold room if the inside temperature is -5°C and the outside
temperature is 15°C.
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163
164
A1 • First Class • SI Units
5. An insulated wall of a cold storage room is 10 m long and 2.75 m high. It
consists of an outer steel plate 22 mm thick and an inner pine wall 24.5 mm
thick. The steel and wooden walls are separated 80 mm apart which forms a
cavity filled with cork. If the difference in temperature across the wall is 22.5
ºC, calculate the following:
a) Total heat transfer per hour through the wall
b) Temperature differential through the cork
The thermal conductivity for steel, cork and pine are 46, 0.05 and 0.15
W/m/ºC
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____________________________________________________________
6. A 10 cm diameter steel pipe, 5 mm thick and 5 m in length, carries steam at
350°C. If insulation with a thickness of 1 cm and a thermal conductivity of
0.015 W/m/ºC is added to the pipe, what is the ratio of heat transfer?
(Assume room temperature is 25°C and the thermal conductivity of steel is
46 W/m/ºC.)
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Part A1
CHAPTER 6
Refrigeration
Calculations
Here is what you will be able to do when you complete each objective:
1. Explain the Carnot Cycle as it applies to refrigeration using
temperature-entropy and pressure-enthalpy diagrams.
2. Calculate the Carnot coefficient of performance of a refrigeration
system and a heat pump system.
3. Calculate the refrigerating effect of a refrigeration system.
4. Calculate the coefficient of performance of a refrigeration system and a
heat pump system.
5. Demonstrate graphically, using temperature-enthalpy diagrams, the
effect on refrigeration capacity of using a throttle valve in place of an
expansion machine, of superheating at the compressor inlet, of
undercooling the condensed refrigerant, and of using a flash chamber.
6. Calculate the mass flow of refrigerant in a system.
7. Calculate the swept volume of a compressor cylinder, given its
volumetric efficiency.
8. Calculate the power requirement of a refrigerant compressor.
165
A1 • Chapter 6 • Refrigeration Calculations
167
OBJECTIVE 1
Explain the Carnot Cycle as it applies to refrigeration using
temperature-entropy and pressure-enthalpy diagrams.
THERMODYNAMICS OF REFRIGERATION
The thermodynamics of refrigeration is very similar to that of engines. In fact,
the refrigeration cycle is the opposite of the heat engine cycle. The heat engine
takes heat from a hot reservoir (usually provided by combustion of a fuel) and
extracts work. Then, the remaining heat is transferred to a cold reservoir. In a
refrigeration and heat pump cycle, heat is removed from a reservoir (that is the
substance to be cooled), work is done on the refrigerant, and heat is expelled to a
hot reservoir. Fig.1 illustrates the general refrigeration cycle.
FIGURE 1
General Refrigeration
Cycle
Refrigeration deals with the transfer of heat from a low temperature level, that of
the location or material to be cooled (sometimes called the heat source), to a
high temperature level, that of the refrigerant condenser (or heat sink).
The basic laws of thermodynamics show that power must be expended in any
continuous refrigeration cycle to induce heat to flow against a temperature
gradient. In a compression cycle, power is supplied through the compressor. The
power input to an absorption cycle is in the form of heat energy.
168
A1 • First Class • SI Units
THE CARNOT CYCLE
The Carnot Cycle can be described as the ideal reversible cycle giving the
maximum possible efficiency for any heat engine.
The Carnot cycle efficiency is:
efficiency =
work done
heat supplied
Which, in this case, reduces to:
T1 − T2
T1
Where:
T1 = lowest temperature in the cycle
T2 = highest temperature in the cycle
The Carnot Cycle, as applied to a heat engine, is shown in the pressure-volume
(P-V) and temperature-entropy (T-S) diagrams in Fig.2.
FIGURE 2 (a) (b)
Carnot Cycle
A1 • Chapter 6 • Refrigeration Calculations
The operations represented are:
1—2: Isothermal heat addition (constant temperature)
2—3: Adiabatic expansion (no heat addition or rejection, constant entropy)
3—4: Isothermal heat rejection
4—1: Adiabatic compression
Applied to a steam power plant, they represent:
1—2: Production of steam in the boiler
2—3: Expansion of steam through the engine or turbine
3—4: Condensing in the condenser
4—1: Pressurization of condensate in the feed pump
The efficiency is given by:
work done
(as in every engine)
heat supplied
Referring to the T-S diagram in Fig 2(b):
area 1256 represents heat supplied
area 4356 represents heat rejected
area 1234 represents work done
For any heat engine:
Work done = heat supplied − heat rejected
Efficiency then is:
area 1234 (T1 − T2 )( S5 − S6 )
=
area 1256
T1 ( S5 − S6 )
Therefore, the ideal Carnot efficiency of a heat engine is:
efficiency =
T1 − T2
T1
169
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A1 • First Class • SI Units
REVERSED CARNOT CYCLE
If the Carnot Cycle is reversed, it appears as in Fig.3 and can be used to
represent the operations performed by a refrigerating machine. Note that this is
the same cycle as drawn in Fig.2, but with a reversed order of operations.
FIGURE 3
Reversed Carnot Cycle
Fig.4 shows a more practical application with the cycle operating between the
liquid and saturation lines for the working fluid (the refrigerant), that is,
operating as a two-phase, liquid to gas, cycle.
FIGURE 4
Refrigeration Cycle
The operations represented are (referring to each of the diagrams):
1—2: Adiabatic expansion of working fluid
2—3: Isothermal heat absorption
3—4: Adiabatic compression
4—1: Isothermal heat rejection
A1 • Chapter 6 • Refrigeration Calculations
171
Applied to a refrigerating plant, they represent:
1—2: The refrigerant passing through the expansion valve (adiabatic
expansion)
2—3: Heat addition to the refrigerant in passing through the evaporator
3—4: Adiabatic compression in the compressor
4—1: Heat rejection at constant temperature in the condenser
Temperature-entropy diagrams are useful in that they provide a convenient
means of understanding and calculating cycle efficiency. The isothermal steps of
the cycle, because they are at constant temperature, are represented by a straight
horizontal line in an ideal cycle, while the adiabatic steps at constant entropy
produce straight vertical lines. Simple mensuration calculations can be used to
derive the area of the resulting graphs, and this provides the amount of work
done. Efficiency is then easily calculated.
Another useful diagram is the pressure-enthalpy diagram, shown in Fig. 5. From
this diagram, the refrigerant enthalpy can be read directly for a given pressure,
which is convenient for heat flow and heat transfer calculations.
FIGURE 5
Pressure - Enthalpy
Diagram
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A1 • First Class • SI Units
OBJECTIVE 2
Calculate the carnot coefficient of performance of a refrigeration
system and a heat pump system.
CARNOT COEFFICIENT OF PERFORMANCE
The coefficient of performance can be determined from Fig.4 (b) as:
coefficient of performance =
heat absorbed
heat equivalent of work done
coefficient of performance =
area 2365
area 1465 − 2365
coefficient of performance =
T2 ( S3 − S 2 )
T1 ( S3 − S 2 ) − T2 ( S3 − S 2 )
Therefore, the Carnot coefficient of performance of a refrigerating machine is:
Carnot coefficient of performance =
T2
T1 − T2
A heat pump is a special form of refrigerating plant whose function is to
transfer heat from one location to another. This is in fact exactly what the
refrigerator does, except that instead of the desired effect being the removal of
heat from the substance being cooled, it is now the rejection of heat to the
condenser.
The output of a heat pump can be considered to be this rejected heat, and the
coefficient of performance becomes as follows:
coefficient of performance =
heat rejected
heat equivalent of work done
A1 • Chapter 6 • Refrigeration Calculations
From Fig.4 (b):
But
COP =
area 1465
area 1465 − area 2365
COP =
T1 ( S 4 − S1 )
T1 ( S4 − S1 ) − T2 ( S3 − S2 )
(S4 - S1) = (S3 - S2)
COP =
T1
T1 − T2
Thus, the Carnot Cycle, when operating in one direction or another, may be used
as a standard of comparison for:
1. A heat engine (transforming heat into mechanical work).
2. A refrigerating machine (using mechanical energy to remove heat).
3. A heat pump (using mechanical energy to transfer heat to a desired
location).
Example 1
A machine operates on the Carnot cycle between the temperature limits of 25ºC
and -5ºC. Calculate the coefficient of performance of the machine when it is
operated as:
a) A refrigerating machine
b) A heat pump
Calculate the efficiency of the machine when it is operated as:
c) A heat engine
Solution
Given:
T2 = −5°C + 273
T2 = 268 K
T1 = 25°C + 273
T1 = 298 K
173
174
A1 • First Class • SI Units
a) Operating as a refrigerating machine, using equation
coefficient of performance =
T2
:
T1 − T2
coefficient of performance =
T2
T1 − T2
coefficient of performance =
268 K
298 K − 268 K
coefficient of performance =
268 K
30 K
coefficient of performance = 8.93 (Ans.)
b) Operating as a heat pump, using equation
coefficient of performance =
T1
:
T1 − T2
coefficient of performance =
T1
T1 − T2
coefficient of performance =
298 K
298 K − 268 K
coefficient of performance =
298 K
30 K
coefficient of performance = 9.93 (Ans.)
c) Operating as a heat engine, using equation efficiency =
efficiency =
T1 − T2
T1
efficiency =
298 K − 268 K
298 K
efficiency =
30 K
298 K
efficiency = 0.101 (Ans.)
T1 − T2
:
T1
A1 • Chapter 6 • Refrigeration Calculations
OBJECTIVE 3
Calculate the refrigerating effect of a refrigeration system.
PRACTICAL REFRIGERATION CYCLES
When applied to a refrigerating machine, the Carnot Cycle suffers from a similar
practical handicap as when applied to the operation of a heat engine.
In order to achieve the compression process, operation 3 – 4 on Fig.4 (b),
compression must begin with a working fluid which is part liquid and part gas.
This brings definite practical difficulties; in fact, the compressor will not be able
to handle such a mixture.
However, if process 2 – 3 in the evaporator was continued until the working
fluid was completely transformed into gas, then point 3 would lie on the gas or
saturation line, the T-S diagram would appear as in Fig.6 (b), and a Rankine cycle
would be used.
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A1 • First Class • SI Units
FIGURE 6 (a) (b) (c)
Vapour-Compression
Refrigeration Cycle
Fig.6 (a) and (b) are the conventional P-V and T-S diagrams. The major
difference between this cycle and the reversed Carnot cycle lies in the
continuation of evaporation (stage 2 – 3) until the refrigerant is a dry gas (point
3). This means that an adiabatic compression 3 – 4 will result in a superheated
discharge from the compressor (point 4).
The actual vapour compression refrigeration machine cycle approaches the
Rankine cycle much more closely than it does the Carnot. The third diagram,
Fig.6(c) pressure-enthalpy, is the one most commonly used in refrigeration
calculations.
A1 • Chapter 6 • Refrigeration Calculations
The operations represented are:
Operation 1—2:
Expansion of the working fluid through the expansion valve
or regulator. This is a throttling operation and will be carried
out at constant enthalpy:
H1 = H 2
Where H1 and H2 are the enthalpies kJ/kg in the refrigerant
under the conditions at points 1 and 2.
Operation 2—3:
Represents passage of the refrigerant through the evaporator
at constant pressure. At point 3 all refrigerant is in gaseous
form.
Heat absorbed during evaporation = ( H 3 − H 2 ) kJ/kg
Operation 3—4:
Represents adiabatic compression. Point 4 shows the gas to
be somewhat superheated.
Operation 4—1:
Represents condensation of the refrigerant gas after discharge
from the compressor. This is a constant pressure operation.
Heat rejected during condensing = ( H 4 − H1 ) kJ/kg
In the ideal operation, with no heat losses to the atmosphere during the cycle:
Heat rejected to condenser = Heat absorbed in evaporator + Work done (W)
( H 4 - H1 ) = ( H 3 - H 2 ) + W
or
W = ( H 4 - H1 ) -
( H3
- H2 )
Work done during compression (W ) = ( H 4 − H1 ) − ( H 3 − H 2 )
But H1 = H 2
W = H 4 − H 3 kJ/kg
The refrigerating effect is the amount of heat absorbed by the refrigerant in its
travel through the evaporator. It could also be called the heat absorbed and is
numerically equal to:
refrigerating effect = (H3 - H 2 ) kJ/kg
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A1 • First Class • SI Units
OBJECTIVE 4
Calculate the coefficient of performance of a refrigeration system
and a heat pump system.
COEFFICIENT OF PERFORMANCE
The coefficient of performance of any refrigeration cycle is given by the ratio
of heat absorbed to work done.
heat absorbed
work done
(H − H2 )
COP = 3
(H4 − H3 )
COP =
Example 2
A Freon-12 refrigerating machine operates on the ideal vapour compression
cycle between the limits of -15ºC and 25ºC. The vapour is dry and saturated at the
end of adiabatic compression, and there is no subcooling of the condensate in
the condenser.
Calculate:
a) the dryness fraction at the suction of the compressor
b) the refrigerating effect per kg of refrigerant
c) the coefficient of performance
A1 • Chapter 6 • Refrigeration Calculations
Solution
From the table for Freon-12 in Table 1
At 25°C,
h f = 59.7 kJ/kg
hg = 197.73 kJ/kg
sg = 0.6869 kJ/kgΚ
At -15°C, h f = 22.33 kJ/kg
hg = 180.97 kJ/kg
h fg = 180.97 − 22.33 = 158.64 kJ/kg
The entropies in the evaporator, at -15ºC, are:
s f = 0.0906 kJ/kgΚ
sg = 0.7051 kJ/kgΚ
sfg = 0.7051 kJ/kgΚ - 0.0906 kJ/kgΚ = 0.6145 kJ/kgΚ
Since the compression is adiabatic, the entropy at compression is 0.6869 kJ/kgK.
a) The dryness fraction at the suction of the compressor is:
sg = s f + qs fg
0.6869 kJ/kgK = 0.0906 kJ/kgK + (q × 0.6145 kJ/kgK)
0.6869 kJ/kgK = 0.0906 kJ/kgK + 0.6145 kJ/kgKq
0.6869 kJ/kgK - 0.0906 kJ/kgK
=q
0.6145 kJ/kgK
0.5963 kJ/kgK
=q
0.6145 kJ/kgK
q = 0.9704 (Ans.)
b) The enthalpy at the exit of the evaporator is
H 3 = h f + qhg
H 3 = 22.33 kJ/kg + (0.9704 ×158.64 kJ/kg)
H 3 = 22.33 kJ/kg + 153.94 kJ/kg
H 3 = 176.27 kJ/kg
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A1 • First Class • SI Units
Refrigerating effect is:
H 3 − H 2 = 176.27 kJ/kg − 59.7 kJ/kg
H 3 − H 2 = 116.57 kJ/kg (Ans.)
c) Work done in compression is:
H 4 − H 3 = 197.73 kJ/kg − 176.27 kJ/kg
H 4 − H 3 = 21.46 kJ/kg
Coefficient of performance is:
COP =
heat absorbed
work done
COP =
(H3 − H2 )
(H 4 − H3 )
COP =
116.57 kJ/kg
21.46 kJ/kg
COP = 5.43 (Ans.)
A1 • Chapter 6 • Refrigeration Calculations
181
TABLE 1
Table of Properties
(Freon-12)
182
A1 • First Class • SI Units
OBJECTIVE 5
Demonstrate graphically, using Temperature-Enthalpy diagrams, the
effect on refrigeration capacity of using a throttle valve in place of an
expansion machine, of superheating at the compressor inlet, of
undercooling the condensed refrigerant, and of using a flash
chamber.
PRACTICAL APPROACH TO THEORETICAL CYCLE
As stated earlier, the actual vapour compression refrigeration machine cycle
approaches the Rankine cycle. There are some practical differences that occur to
the ideal cycle.
The first difference is that expansion usually occurs through the use of a throttle
(or expansion) valve. Throttling results in a drop in pressure without a change in
enthalpy, as shown in Fig.7 (b) from point 1' to point 2.
FIGURE 7
Vapour-Compression
Refrigeration Cycle
with Subcooling and
Superheating
The liquid refrigerant is frequently sub-cooled before it enters the expansion
valve. Point 1 travels to 1' on the diagrams in Fig.7 (a) and 7(b).
Usually, the gas leaving the evaporator is superheated a few degrees before it
enters the compressor (point 3 becomes point 3').
The compression is taken theoretically to be an adiabatic operation (3'—4), a
vertical line on Fig.7 (a), actually, it will be somewhere between adiabatic and
isothermal.
A1 • Chapter 6 • Refrigeration Calculations
183
Finally, the compressor suction and discharge valves are actuated by a pressure
difference, which means that there must be a pressure drop from evaporator to
compressor suction (2—3', Fig.7 (b)).
The diagrams in Fig.7 (a) and 7(b) show only the major differences. The
remaining differences are very small, and a mathematical analysis of a
refrigeration cycle which ignores these differences will still give a very close
approximation to actual working conditions.
If the liquid refrigerant from the receiver is cooled before it passes to the
regulator or expansion valve, the proportion of liquid which flashes off into
vapour as the pressure drops will be reduced. In effect, this also increases the
efficiency of the evaporator and decreases the power consumption of the
compressor unit. Moreover, cooling the liquid reduces the turbulence and assists
the expansion valve to operate smoothly and provide proper control. A flash
tank may be installed between the condenser and evaporator where the flashed
refrigerant is removed and recompressed with a separate flash compressor. The
flash tank, consisting of either a horizontal or vertical vessel (Fig. 8), provides
direct contact cooling and removes superheat from the entering vapour. The heat
of evaporation in the flashed refrigerant is drawn from the remaining liquid
refrigerant, and this provides cooling for the liquid.
FIGURE 8
Flash Tank
In actual refrigerating systems, however, the pressure and temperature of the
liquid refrigerant supplied to the evaporator regulating valve from the highpressure side of the system are usually considerably higher than the pressure and
corresponding temperature in the evaporator. Consequently, part of the liquid
entering the evaporator will flash into vapour due to the pressure drop and the
corresponding excess sensible heat in the liquid.
The amount of liquid that may flash into vapour can be as high as 30%,
depending on the difference between the temperature of the liquid refrigerant
supplied and the evaporator temperature. The larger the difference, the more
liquid will flash.
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A1 • First Class • SI Units
The refrigerant that flashes into vapour will not take part in the actual
refrigerating process. Only the remaining liquid will absorb heat from the
surrounding medium for evaporation. This means that the refrigerating effect per
unit mass of refrigerant is considerably reduced when the liquid refrigerant is
admitted to the evaporator at a temperature higher than the boiling point in the
evaporator. In other words, the refrigerating effect will be lower than the latent
heat of vaporization of the refrigerant.
A1 • Chapter 6 • Refrigeration Calculations
OBJECTIVE 6
Calculate the mass flow of refrigerant in a system.
TONNES OF REFRIGERATION
The standard unit used in the rating of refrigerating machines is the tonne of
refrigeration, defined as the removal of heat at the rate of 13 958 kJ/h, or 233
kJ/min. These units are derived from the amount of heat absorbed by a tonne of
ice when melting from the solid to the liquid phase at 0°C.
latent heat of fusion of 1 kg of ice = 335 kJ
1000 kg = 1 t
heat absorbed/t = 1000 kg/t × 335 kJ/kg
heat absorbed/t = 335 000 kJ/t
If this is done in 24 hours, then:
335 000 kJ
24 h
heat absorbed/h = 13 958 kJ/h
heat absorbed/h =
heat absorbed/m =
13 958 kJ/h
60 min
heat absorbed/m = 233 kJ/min
185
186
A1 • First Class • SI Units
MASS OF REFRIGERANT CIRCULATED
If the refrigerating effect of each kg of refrigerant circulated is known, then the
mass which must be circulated to produce 1 tonne of refrigeration can be
calculated.
refrigerating effect/kg of refrigerant circulated = (H3 – H2) kJ/kg
and:
1 tonne = 233 kJ/min of refrigeration
thus:
mass of refrigerant circulated =
233 kJ/min
kg/min/tonne
( H 3 − H 2 ) kJ/kg
A1 • Chapter 6 • Refrigeration Calculations
OBJECTIVE 7
Calculate the swept volume of a compressor cylinder, given its
volumetric efficiency.
THEORETICAL PISTON DISPLACEMENT
The theoretical dimensions of the compressor required to handle the refrigerant
can be calculated from the foregoing figures in kg/min/t together with the
specific volume of the refrigerant under the pressure and temperature conditions
at exit from the evaporator and inlet to the compressor. At this time, the
refrigerant is in gaseous form.
Thus:
compressor piston displacement m3/min/t = refrigerant circulated
(kg/min/t) × specific volume (m3/kg)
or:
233
× vg 3
(H3 − H 2 )
where vg3 = specific volume of the refrigerant under the
conditions at point 3.
piston displacement m3 /min/t =
VOLUMETRIC EFFICIENCY
In the preceding discussion, it was assumed that at each stroke of the piston the
cylinder would fill completely with vapour at exactly the same pressure and
temperature at which it left the evaporator. This is not true of actual
compressors. The volume, and therefore the mass, of refrigerant that flows into a
cylinder is always less than this theoretical amount for several reasons.
One of the reasons is that the walls of the compressor cylinder are considerably
warmer than the cold vapour leaving the evaporator. The hot cylinder walls raise
the temperature of the vapour that flows into the cylinder. The heated vapour in
the cylinder expands and prevents additional cold vapour from entering. As a
result, the mass of refrigerant that fills the cylinder is less than the mass that the
187
188
A1 • First Class • SI Units
cylinder could hold if the vapour remained at the same temperature at which it
left the evaporator.
Another reason is that the vapour reaching the cylinder must first flow through
the suction valves. In doing so, there is a pressure drop due to the friction of the
vapour flowing through the small valve openings. As a result, the pressure inside
the cylinder is always somewhat lower than the pressure in the evaporator and in
the compressor suction pipe. The vapour inside the cylinder expands because its
pressure is lower than the pressure of the vapour in the suction pipe, and there
is, therefore, a smaller mass of refrigerant in each cubic metre of cylinder space.
Still another reason is that all reciprocating compressors are built with a slight
clearance between the top of the piston and the cylinder head. This clearance
amounts to about 0.4 to 0.8 mm. If the piston could just touch the top of the
cylinder at the end of each stroke, all of the vapour left in the cylinder would be
forced out through the discharge valve. However, since there is clearance space,
a small amount of gas remains in the cylinder after the piston reaches the top of
its travel. As the piston starts on its downward stroke, this trapped vapour
expands. Thus, instead of having an empty cylinder which can fill completely
with vapour from the evaporator, the cylinder is already partially filled with
vapour. Inasmuch as this trapped vapour always remains in the cylinder, it
decreases the mass of vapour that can flow into the cylinder from the evaporator.
For these reasons, the compressor cylinder cannot be filled with a volume of
vapour—at the temperature and pressure in the evaporator—equal to its piston
displacement. Although the effect of clearance volume can be computed, the
total effect of the various factors that decrease the mass of refrigerant vapour
flowing into a cylinder cannot be computed exactly. It is necessary to run tests
on compressors in order to determine how much vapour can actually flow into a
cylinder of a given size. From the results of such tests, it is possible to compute
the mass of vapour that can flow into the cylinder of any compressor.
The effect of all the foregoing factors determines the volumetric efficiency of a
compressor.
Volumetric efficiency is the ratio of the actual mass of refrigerant in a cylinder to
the mass that the cylinder can theoretically hold, or:
volumetric efficiency (%) =
actual mass
× 100
theoretical mass
A1 • Chapter 6 • Refrigeration Calculations
Example 3
The theoretical mass of refrigerant moved in a compressor is 8.28 kg/min. If the
compressor has a volumetric efficiency of 75%, what is the actual mass moved?
Solution
Transposing equation volumetric efficiency (%) =
actual mass
× 100 , the
theoretical mass
actual mass is:
actual mass
×100
theoretical mass
volumetric efficiency × theoretical mass
actual mass =
100
75
actual mass =
× 8.28 kg/min
100
actual mass = 6.21 kg/min (Ans.)
volumetric efficiency (%) =
Thus, the actual refrigerating effect of this system will be considerably less (25%
less) than its theoretical refrigerating effect.
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OBJECTIVE 8
Calculate the power requirement of a refrigerant compressor.
POWER OF A COMPRESSOR
The power of the compressor required per tonne of refrigeration produced can
be calculated from the data collected earlier. This might be a theoretical power if
the effects of mechanical efficiency and volumetric efficiency are not known,
assuming ideal reversible adiabatic compression. If the mechanical and
volumetric efficiencies are available, a more realistic value of power can be
calculated.
work done/tonne of refrigeration = work done/kg of refrigeration × kg of refrigerant circulated/min
work done/tonne = ( H 4 − H 3 ) ×
power/tonne =
233
(H3 − H 2 )
kJ/min/tonne
kJ/min/tonne
60 s/min
Alternatively, the power can be calculated from the work done/cycle in an
adiabatic compression from the formula:
n
( PV
4 4 − PV
3 3 ) kJ/s
n −1
Where P3 and V3 are the pressure and volume before compression, and P4 and
V4 are the values after compression.
C
n = ratio of specific heats p for the refrigerant
Cv
power/tonne =
233
n
1
×
( PV
kW
4 4 − PV
3 3)×
(H3 − H 2 ) n −1
60
A1 • Chapter 6 • Refrigeration Calculations
191
Example 4
Referring to Fig. 9, the following data was obtained during a test on an ammonia
refrigerating machine:
Compressor gauge pressures: suction 180 kPa, discharge 1420 kPa
Temperature of vapour at compressor suction: 5°C
Compressor discharge vapour: 175°C, enthalpy: 1741 kJ/kg
Temperature of liquid entering expansion valve: 31°C
Mass of refrigerant circulated: 2.6 kg/min ammonia
Compressor efficiency: 75%
Barometer: 96.5 kPa
Find:
a)
b)
c)
d)
e)
The power required to drive the compressor
The refrigeration capacity in tonnes
Coefficient of performance
Heat removed/min from condenser
Heat removed/min from compressor
FIGURE 9
Refrigeration Cyde
Solution
atmospheric pressure = 96.5 kPa
absolute pressure at compressor suction = 180 kPa + 96.5 kPa = 276.5
kPa
absolute pressure at compressor discharge = 1420 kPa + 96.5 kPa =
1516.5 kPa
The refrigerating effect is the amount of heat absorbed by the
refrigerant in its travel through the evaporator and is given by:
H3'– H2 kJ/kg refrigerant (see Fig.7)
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A1 • First Class • SI Units
H3' is the enthalpy/ kg of the superheated vapour leaving the evaporator
at 276.5 kPa and 5°C.
From Table 2 or a pressure-enthalpy chart for ammonia:
H3' = 1469.9 kJ/kg
(Use of the table is explained at the end of this module)
H2 is the enthalpy of refrigerant leaving the expansion valve.
From Fig.7 (b)
H2 = H1' enthalpy of liquid entering the expansion valve
From the table
H1' = 327.9 kJ/kg (sensible heat of fluid at 31°C)
Refrigerating effect/kg = 1469.9 kJ − 327.9 kJ = 1142.0 kJ
Work done on refrigerant in compressor during adiabatic compression:
H4 – H3 kJ/kg
H3' = 1469.9 kJ/kg as before (entropy 5.631 kJ/K)
Since entropy remains constant during adiabatic compression (See
Fig.7(a))
Entropy at H4 also = 5.631 kJ/K
From pressure-enthalpy diagram
H4 = 1741 kJ/kg (pressure 1516.5 kPa and entropy 5.631)
ideal work done = 1741 kJ/kg − 1469.9 kJ/kg = 271.1 kJ/kg
271.1 kJ/kg
(since the compressor is 75%
0.75
actual work done = 361.5 kJ/kg
efficient)
actual work done =
A1 • Chapter 6 • Refrigeration Calculations
a) The power required to drive the compressor if 2.6 kg of ammonia are
handled per minute is:
361.5 kJ/kg × 2.6 kg/min
60 s/min
kW = 15.67 kW (Ans.)
kW =
b) The refrigeration capacity in tonnes:
refrigeration effect/kg = 1142.0 kJ/kg
refrigerant circulated = 2.6 kg/min
refrigeration capacity =
1142 kJ/kg × 2.6 kg/min
233 kJ/min
refrigeration capacity = 12.74 t (Ans.)
c) coefficient of performance:
COP =
heat absorbed
work done
COP =
refrigerating effect/kg
actual work done/kg
COP =
1142 kJ/kg
361.5 kJ/kg
COP = 3.16 (Ans.)
d) heat removed/min from condenser
= (H 4 – H1') kJ/kg × mass of refrigerant circulating kg/min
Here, H4 will be the heat content of the ammonia gas entering the
condenser under actual conditions.
H 4 = (1741 kJ/kg − 327.9 kJ/kg) × 2.6 kg/min
H 4 = 1413.1kJ/kg × 2.6 kg/min
H 4 = 3674.06 kJ/min (Ans.)
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e) heat removed/min from compressor:
compression work done/kg = 361.5 kJ
increase in enthalpy in gas passing through compressor/kg
= ( H 4 − H 3 ')
= 1741 kJ/kg − 1469.9 kJ/kg
= 271.1 kJ/kg
heat removed/kg by compressor cooling = 361.5 kJ/kg - 271.1 kJ/kg
heat removed/kg by compressor cooling = 90.4 kJ/kg
heat removed/min = 90.4 kJ/kg × 2.6 kg/min
heat removed/min = 235.04 kJ/min (Ans.)
A1 • Chapter 6 • Refrigeration Calculations
195
TABLE 2
Properties of Ammonia
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A1 • First Class • SI Units
Use of the Table of Properties of Refrigerant 717 (Ammonia)
Use of the refrigerant chart in Table 2 is explained here. We will use the example
of ammonia gas at 276.5 kPa and 5°C, which is the set of conditions of H3' in
Example 4.
Since the saturation temperature at 276.5 kPa is -11.25°C, the amount of
superheat is:
11.25°C + 5°C = 16.25°C
For example, to calculate the temperature, enthalpy, and entropy of ammonia gas
at a pressure of 276.5 kPa with 16.25ºC superheat use the following steps:
1. Search the Pressure column to locate the pressures immediately below and
immediately above 276.5 kPa. Write these pressures and accompanying
properties down as shown in the top and bottom rows of numbers copied
from the table and shown below.
Temp
°C
-12
-11.25
-10
Pressure
kPa
268.0
276.5
290.8
Enthalpy
Hf
hg
126.2
1430.5
129.6
1431.4
135.4
1433.0
Entropy
Sg
5.504
5.493
5.475
50K
h
1548.5
1549.7
1551.7
50K
s
5.919
5.917
5.891
276.5 kPa − 268 kPa
290.8 kPa − 268 kPa
8.5 kPa
using pressure proportion =
22.8 kPa
using pressure proportion = 0.3728
using pressure proportion =
Each interpolated value will be 0.3728 × the difference between the bottom
and top values.
Enthalpy
The interpolated value of enthalpy under 50 K will be:
value of enthalpy = (1551.7 kJ/kg-1548.5 kJ/kg)×0.3728 + 1548.5 kJ/kg
value of enthalpy = 1549.69 kJ/kg
A1 • Chapter 6 • Refrigeration Calculations
Correction for superheat:
at 50°C superheat, difference in enthalpy = 1549.7kJ- 1431.4kJ
at 50°C superheat, difference in enthalpy = 118.3 kJ
16.25°C × 118.3 kJ
at 16.25°C superheat, difference =
50°C
at 16.25°C superheat, difference = 38.45 kJ
total enthalpy at 276.5 kPa and 5° C = 1431.4kJ + 38.45kJ = 1469.9 kJ/kg (Ans.)
Enthalpy
Value of entropy, under 50K, will be:
value of entropy = ( 5.475 kJ/kg − 5.504 kJ/kg ) × 0.3728 + 5.504 kJ/kg
value of entropy = −0.029 × 0.3728 + 5.504 kJ/kg
value of entropy = −0.0108 kJ/kg + 5.504 kJ/kg
value of entropy = 5.4932 kJ/kg
Correct for superheat:
at 50°C superheat, difference in entropy = 5.9173kJ- 5.4932kJ
at 50°C superheat, difference in entropy = 0.4241 kJ
16.25°C × 0.4241 kJ
at 16.25°C superheat, difference =
50°C
6.8916 kJ/°C
at 16.25°C superheat, difference =
50°C
at 16.25°C superheat, difference = 0.1378 kJ
total enthalpy at 276.5 kPa and 5° C = 5.493kJ + 0.1378kJ
total enthalpy at 276.5 kPa and 5° C = 5.631 kJ/kg (Ans.)
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CHAPTER QUESTIONS
1. Sketch the Carnot cycle for a refrigeration system on a temperature-entropy
diagram and describe the four thermodynamic steps of the cycle and how
they apply to a steam power plant.
2. Using the T-S diagram sketched in question 1, derive the Carnot coefficient
of performance. Calculate it for a machine that operates on the Carnot cycle
between the temperature limits of 30oC and -5oC.
____________________________________________________________
3. Sketch the ideal vapour compression cycle on a pressure-enthalpy diagram.
Derive the coefficient of performance using enthalpies.
A1 • Chapter 6 • Refrigeration Calculations
4. A food storage locker requires a refrigeration system of 11 tonne capacity
when the evaporator temperature is at -6°C and the condenser temperature is
30°C. The refrigerant used is ammonia, sub-cooled by 5°C at entry to the
expansion valve and superheated by 5°C at exit from the evaporator. The
compressor used is a two cylinder vertical single-acting, single-stage machine
running at 900 rev/min, each cylinder’s dimension being such that the stroke
is 1 ½ times the bore.
Find the following:
a)
b)
c)
d)
e)
f)
the refrigerating effect/kg
the mass of refrigerant circulating/min
coefficient of performance
compressor drive power, kW
compressor cylinder dimensions
the heat to be removed from the condenser/min
5. The discharge pressure and temperature of an ammonia refrigeration
compressor is 1929 kPa and 85ºC. The refrigerant then leaves the condenser
at this pressure as liquid with no undercooling. The compressor suction
pressure is 246.5 kPa and the compression is isentropic. If the mass flow of
refrigerant is 0.22 kg/s, calculate the following:
a) refrigerating effect
b) compressor work transfer
c) coefficient of performance
199