Given: Power Machines N5
= 2 460 kg/h
= 760 mm Hg = 0,88
= 31ºC = 688,744 mm Hg
= 4,2 kJ/kg.K =
=
=
= 9,5 kPa
=
=
= 20,62476 kg/kg steam
= 20,62476 X2 460
= 50 736,9096 kg/hour
2.5.6 Throttling calorimeter
An instrument utilizing the principle of constant enthalpy expansion for the
measurement of the moisture content of steam; steam drawn from a steam
pipe through sampling nozzles enters the calorimeter through a throttling
orifice and moves into a well-insulated expansion chamber in which its
temperature is measured. Also known as steam calorimeter see Figure 2.12
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Figure 2.12 A throttling calorimeter
Worked Example 2.4
A dryness fraction test was done using a combined separating and throttling
calorimeter. The pressure in the mains was 580 kPa and 0,12 kg water was
collected in the separating calorimeter. The mass of steam condensed at the
exit from the throttling calorimeter was 11,2 kg. The manometer on the
throttling calorimeter Indicated a pressure of 120 kPa and the temperature
shown on the thermometer was 115ºC. Cp for superheated steam is 2 kJ/kg.
ºC.
Calculate the following with the aid of steam tables:
1. The dryness fraction of the steam in the separating calorimeter
2. The dryness fraction of the steam in the throttling calorimeter
3. The dryness fraction of the steam in the mains
Solution: =
1. =
=
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2. =
=
3. = =
= = 2 038,4
= = 0,976
Activity 2.1
Use steam tables and find the specific liquid enthalpy, the specific enthalpy
of evaporation and the specific enthalpy of dry saturated steam at 0.5 MPa.
[640.1; 2107.4; 2747.5]
Activity 2.2
Find the specific volume of water at saturation temperature for a pressure of
4 MPa.
[250.3; 0.001252]
Activity 2.3
The volume of steam at a pressure of 1 100 kPa is 0,1945 m3/kg. The specific
heat capacity of superheated steam is 2,1 kJ/kg.K.
Calculate the following with the aid of steam tables:
1. The quality of the steam
2. The enthalpy of the steam
3. The temperature of the steam
[1.096; superheated; 2867.19; 225.62]
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Activity 2.4
11,5 kg of water, at a pressure of 880 kPa and a temperature of 48ºC is
heated to boiling point, then dry to saturated steam and then to
superheated steam, with a temperature of 210 ºC. The specific heat
capacity of superheated steam is 2,1 kJ/kg.ºC and that of water is 4,187
kJ/kg. ºC.
Calculate the following with the aid of steam tables:
1. The enthalpy of the sensible heat in MJ
2. The enthalpy of the latent heat in MJ
3. The enthalpy of the superheat in kJ
4. The enthalpy of the total heat required for superheating the water in MJ
5. The saturation temperature of the steam
[6.187; 23.38; 859.74; 30.4267; 174.4]
Activity 2.5
The data provided below refers to steam. Use steam tables and/or by
calculation, determine the condition of the steam, the volume of the steam
and where the steam Is wet, calculate the dryness fraction also.
1. P = 800 kPa t = 200°C h = 2 796 kJ/kg
2. Vg = 0,009308 m3/kg t = 347,3 °C
3. P = 2,35 MPa h = 2 777 kJ/kg
4. P = 16 MPa h = 2777 kJ/kg
5. P = 1 500 kPa t = 200 °C
[70; 9.31; 116]
Activity 2.6
With the aid of steam tables, calculate the following:
1. The enthalpy found in 2 kg of steam which has a pressure of 480 kPa and
is 7,3% wet.
2. The enthalpy of a kilogram of steam which has a pressure of 1,5 MPa, a
temperature of 396 ºC and a specific heat capacity of 2,1 kJ/kg.K.
3. The dryness fraction of wet steam which has an enthalpy of 2 650 kJ/kg at
a pressure of 1,3 MPa.
4. The heat required to change 2,3 kg of water from a temperature of 31 ºC
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to dry saturated steam at a pressure of 1 000 kPa.
[5183.648; 3205.17; 0.19; 6088.1]
Activity 2.7
1.5 kg of steam originally at a pressure of 1 MPa and temperature of 225
degrees Celsius is expanded until the pressure becomes 0.28 MPa.
The dryness fraction of the steam is then 0.9. Find the change in internal
energy that occurs.
[2666; 2504.49; 0.5814; 2341.69; -324.31; 486.47]
Activity 2.8
Steam at 1.4 MPa and dryness fraction 0.7 is throttled to 0.11 MPa. Find the
dryness fraction of the steam after throttle.
[0.787]
Activity 2.9
Steam with a pressure of 5 MPa and dryness fraction of 0.9 has a mass of 1 kg.
Find the enthalpy, volume and density of the mass of steam.
[2630.2; 0.0355; 28.2]
Activity 2.10
The equivalent evaporation from and at 100 ºC of a boiler plant is 9,552. The
pressure In the plant Is 2 000 kPa and the temperature of the steam leaving
the plant Is 250 ºC.
The temperatures at the inlet and outlet of the economiser are 43,8 ºC and
81,4 ºC respectively. The calorific value of the fuel Is 32 MJ/kg and the dryness
factor of the steam entering the superheater is 0,948.
Calculate the following:
1. The mass of steam generated per kilogram of fuel
2. The thermal efficiency of the plant
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3. The heat to the economiser In kJ/kg fuel
4. The heat to the evaporator in kJ/kg fuel
[7.923; 67.37; 158; 2698.772; 2357.772]
Self-Check
I am able to: Yes No
Describe the main types of boilers and their specific application
Describe the different methods of firing a boiler
Calculate the enthalpy, volume and internal energy of wet
steam
Calculate the enthalpy, volume and internal energy of dry
steam
Calculate the enthalpy, volume and internal energy of
superheated steam
Calculate the dryness fraction using the calorimeter
If you have answered ‘no’ to any of the outcomes listed above, then speak to
your facilitator for guidance and further development.
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Learning Outcomes
On the completion of this module the student must be able to:
Describe the construction and operation of the surface condenser
Describe the construction and operation of the jet condenser
Calculate the condensing water required
Calculating cooling surface required
Describe the construction and operation of the air pump
Calculate the condensing and vacuum efficiency
3.1 Introduction
The steam turbine itself is a device to convert the heat in steam to
mechanical power. The difference between the heat of steam per
unit mass at the inlet to the turbine and the heat of steam per unit
mass at the outlet from the turbine represents the heat which is
converted to mechanical power.
Note:
There are two main condenser types in use, namely the surface
condenser and the jet condenser.
The steam turbine itself is a device to convert the heat in steam to
mechanical power. The difference between the heat of steam per unit mass
at the inlet to the turbine and the heat of steam per unit mass at the outlet
from the turbine represents the heat which is converted to mechanical power.
Therefore, the more the conversion of heat per kilogram of steam to
mechanical power in the turbine, the better is its efficiency.
By condensing the exhaust steam of a turbine at a pressure below
atmospheric pressure, the steam pressure drop between the inlet and exhaust
of the turbine is increased, which increases the amount of heat available for
conversion to mechanical power.
Most of the heat liberated due to condensation of the exhaust steam is carried
away by the cooling medium (water or air) used by the condenser.
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3.2 Types of condensers
3.2.1 Surface condenser
A surface condenser is a commonly used term for a water-cooled shell and
tube heat exchanger installed on the exhaust steam from a steam
turbine in power stations. These condensers are heat exchangers which
convert steam from its gaseous to its liquid state at a pressure
below atmospheric pressure.
Figure 3.1 shows a surface condenser that returns the cooling water for
another pass. This is termed a two-pass condenser. Single pass condensers are
also available where the cooling water runs through the condenser once and
the discharges out.
Figure 3.1 The two-pass surface condenser
3.2.1.1 The shell
The shell is the condenser's outermost body and contains the heat exchanger
tubes. The shell is fabricated from carbon steel plates and is stiffened as
needed to provide rigidity for the shell.
When required by the selected design, intermediate plates are installed to
serve as baffle plates that provide the desired flow path of the condensing
steam. The plates also provide support that help prevent sagging of long tube
lengths.
3.2.1.2 The vacuum system
For water-cooled surface condensers, the shell's internal vacuum is most
commonly supplied by and maintained by an external steam jet
ejector system. Such an ejector system uses steam as the motive fluid to
remove any non-condensable gases that may be present in the surface
condenser.
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The venture effect, which is a particular case of Bernoulli’s principal, applies to
the operation of steam jet ejectors.
Motor driven mechanical vacuum pumps, such as the liquid ring type, are also
popular for this service.
3.2.1.3 Tube sheets
At each end of the shell, a sheet of sufficient thickness usually made of stainless
steel is provided, with holes for the tubes to be inserted and rolled. The inlet
end of each tube is also bell mouthed for streamlined entry of water. This is to
avoid eddies at the inlet of each tube giving rise to erosion, and to reduce
flow friction.
To take care of length wise expansion of tubes some designs have expansion
joint between the shell and the tube sheet allowing the latter to move
longitudinally.
Note:
In smaller units some sag is given to the tubes to take care of tube
expansion with both end water boxes fixed rigidly to the shell.
3.2.1.4 Tubes
Generally the tubes are made of stainless steel, copper alloys such as brass or
bronze, cupronickel, or titanium depending on several selection criteria. The
use of Titanium condenser tubes are usually the best technical choice,
however the use of titanium condenser tubes has been virtually eliminated by
the sharp increases in the costs for this material.
The tube lengths range to about 26 m for modern power plants, depending on
the size of the condenser. The size chosen is based on transportability from the
manufacturers’ site and ease of erection at the installation site. The outer
diameter of condenser tubes typically ranges from 3/4 inch to 1-1/4 inch,
based on condenser cooling water friction considerations and overall
condenser size
3.2.1.5 Water boxes
The tube sheet at each end with tube ends rolled, for each end of the
condenser is closed by a fabricated box cover known as a waterbox, with
flanged connection to the tube sheet or condenser shell. The waterbox is
usually provided with man holes on hinged covers to allow inspection and
cleaning.
Note:
In thermal power plants, the purpose of a surface condenser is
to condense the exhaust steam from a steam turbine to obtain
maximum efficiency, and also to convert the turbine exhaust steam
into pure water (referred to as steam condensate) so that it may be
reused in the boiler as boiler feed water.
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3.2.1.6 Evaporative condenser
Where cooling water is in short supply, an air-cooled condenser is often used.
Figure 3.2 shows an evaporative condenser.
An air-cooled condenser is however, significantly more expensive and cannot
achieve as low a steam turbine exhaust pressure (and temperature) as a
water-cooled surface condenser.
Figure 3.2 The evaporative condenser
3.2.3 Jet condenser
This type of condenser is suitable where conditions permit condensation of
exhaust steam by direct contact with the cooling water. It can maintain a
pressure of less than 0,07 bar and can condense over 12000 kg/h of
steam. Figure 3.3.
The vacuum is created in the chamber by an air ejector. The cooling water is
sprayed into the chamber and the fine spray contacts the steam. The steam
condenses and falls to the bottom of the condenser chamber with the
injection water.
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The condensed steam and injection water is withdrawn using a centrifugal
extraction pump. The jet condenser is generally provided with safety features
to guard against flooding.
There are mainly three types of jet condensers:
Low level condenser
High level condenser
Ejector condenser
Figure 3.3 The Jet condenser
3.2.3.1 Low level condenser
Here a condenser chamber is placed at low elevation and overall height of
the unit is low enough so that the condenser may be directly placed beneath
the steam turbine, pump or pumps and are required to extract the cooling
water condensate and air from the condenser.
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3.2.3.2 High level condenser
If a long pipe over 10 m, is closed at top end, filled with water, open at bottom
and bottom is immersed in water, then atmospheric pressure would hold the
water up in the pipe to a height of 10 m at sea level.
On the basis of this principle, high level or Barometric jet condenser is
designed. Figure 3.3 shows a high level jet condenser.
3.2.3.3 Ejector condenser
In this type of condenser, the momentum of falling water is utilized to extract or
ejects air from condensates. The condenser chamber consists of a central
vertical tube in which there is a string of many cones or converging nozzles. The
exhaust steam enters from side way of the cylindrical condenser chamber. The
central tube is provided with number of wholes or steam ports.
3.3 Finding the amount of cooling water required
The principle used is:
Loss of heat by steam = heat gained by the cooling water
The type of steam entering the condenser is wet steam and so we use the
analogy:
Where:
= Sensible enthalpy of steam at constant pressure
= Dryness fraction of steam entering the condenser
= Enthalpy of evaporation of the steam at the condenser
= Enthalpy of condensate
= Mass of cooling water required
Worked Example 3.1
2 460 kg of steam, with a dryness fraction of 0,88, is condensed in a surface
condenser every hour. The barometer reading is 760 mm Hg and the vacuum
meter reading is 688,744 mm Hg. The temperature of the condensate is 31 ºC
and the temperature of the cooling water rises by 25 ºC. The specific heat
capacity of water is 4,2 kJ/kg.K.
Calculate the mass of cooling water required per hour.
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Solution:
=
=
= 9,5 kPa
=
=
=
=
=
Worked Example 3.2
Exhaust steam with a dryness fraction of 0,89 enters a surface condenser,
which has a vacuum of 699,5 mm mercury, and is condensed to water at a
temperature of 34,6 ºC.
The barometer reading Is 760 mm mercury. The suction temperature of the
pump is 34,6 ºC and R for air is 0,287 kJ/kg.K.
Calculate the following:
1. The combined pressure of the steam and air in the condenser
2. The partial pressure of the air
3. The mass of air extracted for every kilogram of steam whilst considering
that air and steam occupy the same volume
Solution:
1. =
=
= 8 kPa
2. =
=
=
=
= 2,5 kPa
3. =
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=
=
=
= 35,31248 m3/kg
=
=
=
=
3.4 Condenser pumps
The air pump is much larger than the feed pump, even though the water
through-put is the same overall. There are several reasons for this. One reason is
the air pump may be required to empty the condenser at a reasonable rate. If
the condenser is full, the air pump may need to move 20 or more litres in a
short time.
Note:
The purpose of the Air Pump is to remove the water from the
condenser and push it into the Hotwell. In doing so a vacuum is
created in the Condenser, often approaching 736 mm Hg (near
perfect Vacuum). There are many air pump designs.
Also the principle of the Air Pumps is to operate as a scavenging pump. That is
the pump is designed to pump more (around 10x) the amount of water as the
feed pump, and so on a stable system the air pump will only be 10% full of
water at each pump stroke. The rest would either be a void (vacuum) or
steam, because water will boil at about 25ºC at a near vacuum.
Also note that the outlet valve is at the top. This allows any air which was
dissolved in the water and has since escaped to be pushed out first thing. This
allows only water to remain in the pump area, allowing the best vacuum to
form.
As a scavenging pump, the air pump has to grab what it can. The best way to
do this is to ensure the pump is as empty as possible before it draws in
condensate. The only way to achieve this is to design the pump with minimal
volume left in the cylinder when the piston is fully in. Valves should also be
designed to leave little space inside the cylinder area.
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3.4.1 Edwards air pump
The Edwards Air pump design, Figure 3.4, is one of the more successful designs.
It is a simple design with only two moving parts, it makes use of the fact that
water contains momentum and at the same time compressing the water to
prevent it from flashing into steam (remember at 736 mm Hg water will boil at
about 25C).
The piston moves down, forming a vacuum. As it approaches the bottom it
opens up ports around the cylinder. This allows the water to fill the void
(vacuum) in the cylinder.
By doing this, water vapour and steam is generated because of the rapid
expansion of the water, taking up valuable space in the cylinder which could
otherwise be filled with water.
To help fill the cylinder with water, when the piston has reached the bottom of
its travel it forces a jet of water from the very bottom of the air pump into the
cylinder. By compressing the water into the cylinder instead of "sucking" the
water in, no steam or water vapour is created, and most which was created
before is forced back to water.
Figure 3.4 Edwards air pump
The piston is then raised shutting off the inlet ports and forcing all air and water
through the top valve. This top valve, often a simple flat ring over a series of
holes, is held shut only due to the vacuum within the cylinder.
The only downside to an Edward’s air pump design is they don't operate to
their design at very low speeds, and they have a tendency to clank. These are
often small factors compared to the benefits of having a decent vacuum.
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3.4.2 Rotary air pump
Water is forced outward by centrifugal force and as it rotates about the inner
rim of the pump, it draws air with it and discharges the water/air mixture at
high velocity.
3.4.3 Air ejector system
An air ejector or steam ejector is a device which uses the motion of moving
fluid (Motive Fluid) to transport another fluid (Suction fluid). It is has a wide
range of application in steam ejector in boiler condenser, fresh water
generator and in priming the centrifugal pump.
It works on the principle of convergent /divergent nozzle as it provides the
venturi effect at the point of diffusion as the tube gets narrows at the throat the
velocity of the fluid increases and because of the venturi affect it pressure
decreases, vacuum will occur in the diffuser throat where the suction line will
be provided.
An air ejector which uses the high pressure motive fluid such as air or steam to
flow through the convergent nozzle the function of the convergent nozzle is to
convert the pressure energy of the motive fluid into the velocity energy. Figure
3.5. As in convergent nozzle the following effect takes place,
Figure 3.5 Air ejector system
As the pressure energy before entering the convergent nozzle is greater and
the velocity is less for the fluid. At the point of discharge, the pressure energy is
converted into the velocity so the velocity will be greater and the pressure will
be less during the discharge.
Divergent nozzle the opposite effect takes place velocity energy is convert into
pressure energy, at the point of diffusion there is a low pressure or vacuum is
created which is used to suck the other fluid for the motion.
It is one of the types of air ejector which is used in the steam like near the
condenser to remove the non-condensable gases and some vapour entering
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into main condenser by an air ejector and it is cooled by the main condensate
and released in the ejector condenser.
3.4.4 Condenser efficiency
Condenser efficiency is influenced by the following:
The difference between the inlet and outlet temperatures of the cooling
water.
The difference between the temperature of the steam in the condenser
and the inlet cooling water temperature.
The efficiency is the ratio of the above two conditions.
Where:
= Cooling water inlet temperature
= Cooling water outlet temperature
= Steam temperature
3.4.5 Vacuum efficiency
This is a ratio of the partial pressure of the steam inside the condenser and the
condenser pressure.
The vacuum efficiency depends on the amount of air there is present inside
the condenser. If there is no air in the condenser then the efficiency will be
100%.
3.4.6 Air leakage into condensers
3.4.6.1 Method of measuring air leakage
This scheme measures the total pressure and temperature in air suction
pipelines of steam condensers with air added in calibrated steps, for two
different combinations of pumps.
With the aid of steam tables, air partial pressures are plotted against added air
leakages. This allows the leakage of the set to be found, and a temperature
obtained from the plots shows whether the water vapour is saturated or
superheated.
The method is useful in routine checks, for showing how much of the leakage is
into particular parts, so that remedial action can be taken, and for revealing if
water temperatures are affecting pump displacements and augmenting air
pressures within condensers, in which case fuel may be saved.
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3.4.6.2 The effects of air leakage
The effect of the air leakage is to introduce non-condensable (air) into the
steam and cause the vacuum that the condenser can produce to rise. That
leads to less horsepower development by the turbine due to the higher
exhaust pressure.
3.4.6.3 Good condenser design
The way to prevent air from the atmosphere from entering the condenser is by
good design and then by careful fabrication and installation.
3.4.7 Daltons law of partial pressures
The problem of an air - steam mixture is solved with this law. The law states that
the pressure exerted by a mixture of gasses is equal to the sum of the pressures
of those individual gasses that make up the mixture.
Each individual gas acts independently of the mixture and the pressure of
each gas taken alone is known as its partial pressure.
Worked Example 3.3
A vacuum gauge on a condenser reads 660 mm Hg and the barometer
height is 765 mm Hg. Steam enters the condenser with a dryness fraction of
0.8 and has a temperature of 41.5 degrees Celsius.
Find the partial pressures of the air and steam in the condenser if the steam is
condensed at a rate of 1500 kg/h.
Solution:
From steam tables at 41.5
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Activity 3.1
A surface condenser handles 4 520 kg of wet steam, with a dryness fraction
of 0,81, every hour. The barometer and vacuum meter readings are 760 mm
Hg and 662 mm Hg respectively. Air leaks In at a rate of 0,9 kg for every 1 000
kg of steam entering the condenser.
The temperature of the condensate and the air pump suction is 39 °C.
Cooling water enters the condenser at 20 °C and leaves at 35,75 °C. The gas
constant for air is 0,288 kJ/kg.K and the specific heat capacity of water is
4,187 kJ/kg.K.
Calculate the following:
1. The quantity of cooling water required In kg/min
2. The capacity of the air pump In m3/mln
3. The capacity of the new air pump in m3/min if the air was cooled to 36,2
°C before being pumped from the condenser
[13; 180; 25.664; 47.7]
Activity 3.2
Every hour a surface condenser processes 6 500 kg exhaust steam, which has
a dryness traction of 0,83. Air leakage Into the condenser Is at a rate of 1 ,2
kg/1 000 kg of steam. The air pump suction pipe and the condensate have a
temperature of 31 ºC each.
The barometer and vacuum gauge reading is 760 mm Hg and 662 mm Hg
respectively. The temperature of the cooling water is increased by 21 ºC after
it passes through the condenser. The specific heat capacity of the water is 4
187 kJ/kg.K and R tor air is 0,88 kJ/kg.K.
Calculate the following:
1. The mass of cooling water required by the condenser every minute
2. The capacity of the air pump in m3/min
[13; 2537.351; 8.5; 304; 13; 1.339]
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Activity 3.3
A container is filled with a mixture of air and wet steam at a temperature of
39 degrees Celsius and a pressure of 100 kPa. The temperature is then raised
to 120.2 degrees.
If the steam remains wet, find:
1. The initial partial pressures of the steam and air
2. The final partial pressures of the steam and air
3. The total pressure in the container after heating.
[7; 97; 122.24; 322.24]
Activity 3.4
A cylinder contains a mixture of air and wet steam at a pressure of 130 kPa
and a temperature of 75.9 degrees Celsius. The dryness fraction of the steam
is 0.92. The air – steam mixture is then compressed to one – fifth of its original
volume with a final temperature of 120.2 degrees.
Find:
1. The final pressure in the cylinder.
2. The final dryness fraction of the steam.
[40; 90; 3.67; 507.1; 200; 707.1; 0.734; 0.83]
Activity 3.5
1. Name 4 auxiliaries in a steam plant and state the function of each.
2. Name two types of steam boilers.
3. Name two types of steam condensers.
Activity 3.6
1. State Daltons law regarding partial pressures.
2. Explain with the aid of a sketch what is meant by the triple point.
3. State three advantages of a condenser.
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Activity 3.7
A surface condenser has the following conditions:
Air that leaks in with the steam = 5 kg/h
Vacuum in the condenser = 86.4 kPa
Barometer reading = 760 mm Hg
Temperature at the steam – air draw-off = 24.1 degrees
Take R = 0.288 for air and find the volume of air extracted per hour and the
mass of steam in this air.
[15; 12; 35.65; 0.781]
Self-Check
I am able to: Yes No
Describe the construction and operation of the surface
condenser
Describe the construction and operation of the jet condenser
Calculate the condensing water required
Calculating cooling surface required
Describe the construction and operation of the air pump
Calculate the condensing and vacuum efficiency
If you have answered ‘no’ to any of the outcomes listed above, then speak to
your facilitator for guidance and further development.
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Learning Outcomes
On the completion of this module the student must be able to:
Describe solid, liquid and gaseous fuels
Describe the higher and lower calorific values of fuel
Calculate the minimum air required for complete combustion
Calculate the products of combustion
Describe the bomb calorimeter and the Orsat apparatus
4.1 Introduction
In thermodynamics, the term exothermic process describes a
process or reaction that releases energy from the system to its
surroundings, usually in the form of heat.
4.1.1 Combustion
Combustion is a high-temperature exothermic chemical reaction between
a fuel and oxygen that releases heat and light.
It is usually atmospheric oxygen that produces oxidized, often gaseous
products, in a mixture termed as smoke.
4.2 Fuels
Most fuels are made up of carbon or hydrogen or a combination of the two
termed hydrocarbons.
4.2.1 Solid fuels
Solid fuels, such as coal, first undergo endothermic pyrolysis to produce
gaseous fuels whose combustion then supplies the heat required to produce
more of them.
Definition:
The term endothermic process describes a process or reaction in
which the system absorbs energy from its surroundings, usually, but
not always, in the form of heat.
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Pyrolysis is a thermochemical decomposition of organic material at elevated
temperatures in the absence of oxygen.
Combustion of an organic fuel in air is always exothermic because the double
bond in O2 is much weaker than other double bonds or pairs of single bonds,
and therefore the formation of the stronger bonds in the combustion products
CO2 and H2O results in the release of energy.
The act of combustion consists of three relatively distinct but overlapping
phases:
Preheating phase, when the unburned fuel is heated up to its flash point and
then fire point. Flammable gases start being evolved in a process similar to dry
distillation.
Distillation phase or gaseous phase, when the mix of evolved flammable gases
with oxygen is ignited. Energy is produced in the form of heat and
light. Flames are often visible. Heat transfer from the combustion to the solid
maintains the evolution of flammable vapours.
Charcoal phase or solid phase, when the output of flammable gases from the
material is too low for persistent presence of flame and the charred fuel does
not burn rapidly and just glows and later only smoulders.
Coal is the solid hydrocarbon fuel that is mostly used in industry.
4.2.2 Liquid fuels
Combustion of a liquid fuel in an oxidizing atmosphere actually happens in the
gas phase. It is the vapour that burns, not the liquid. Therefore, a liquid will
normally catch fire only above a certain temperature: its flash point.
Note:
The flash point of a liquid fuel is the lowest temperature at which it
can form an ignitable mix with air. It is the minimum temperature at
which there is enough evaporated fuel in the air to start
combustion.
Examples of hydrocarbon liquid fuels used to run boilers and furnaces are
diesel and fuel bunker oil. Jet paraffin is used to run gas turbines and there are
many varieties of petrol and other hydrocarbons used to power internal
combustion engines.
In boilers and furnaces, the liquid fuel is atomized through nozzles to create
more burn area so the liquid fuel acts more of a vapour.
Pulverized coal is so fine, that is acts like a liquid when it is sprayed out at high
velocity through nozzles.
4.2.3 Gaseous fuels
Fuel gas is any one of a number of fuels that under ordinary conditions
are gaseous. Many fuel gases are composed of hydrocarbons (such
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as methane or propane) or hydrogen or mixtures thereof. Such gases are
sources of potential heat energy and is readily transmitted and distributed
through pipes from the point of origin directly to the place of consumption.
Fuel gas is contrasted with liquid fuels and from solid fuels, though some fuel
gases are liquefied for storage or transport.
While their gaseous nature has advantages, avoiding the difficulty of
transporting solid fuel and the dangers of spillage inherent in liquid fuels, it also
has limitation. It is possible for a fuel gas to be undetected and collect in
certain areas, leading to the risk of a gas explosion.
For this reason, odorizes are added to most fuel gases so that they may be
detected by a distinct smell.
4.3 Combustion
Definition:
A reagent is a substance or compound added to a system to cause
a chemical reaction, or added to see if a reaction
occurs. a reactant is more specifically a substance consumed in
the course of a chemical reaction.
4.3.1 Calorific value
The heating value (or energy value or calorific value) of a fuel, is the amount
of heat released during the combustion of a specified amount of it. The energy
value is a characteristic for each substance.
It is measured in units of energy per unit of the substance; usually mass, heating
value is commonly determined by use of a bomb calorimeter.
4.3.2 Measuring the calorific value
The higher heating value is experimentally determined in a bomb calorimeter.
The combustion of a stoichiometric mixture of fuel and oxidizer (e.g. two moles
of hydrogen and one mole of oxygen) in a steel container at 25 °C is initiated
by an ignition device and the reactions allowed to complete.
When hydrogen and oxygen react during combustion, water vapour is
produced. The vessel and its contents are then cooled to the original 25 °C
and the higher heating value is determined as the heat released between
identical initial and final temperatures.
When the lower heating value (LHV) is determined, cooling is stopped at
150 °C and the reaction heat is only partially recovered. The limit of 150 °C is
an arbitrary choice.
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Note:
Higher heating value (HHV) is calculated with the product of water
being in liquid form while lower heating value (LHV) is calculated
with the product of water being in vapour form.
A bomb calorimeter is a type of constant-volume calorimeter used in
measuring the heat of combustion of a particular reaction.
Figure 4.1 The bomb calorimeter
Figure 4.1 shows a bomb calorimeter. Bomb calorimeters have to withstand
the large pressure within the calorimeter as the reaction is being measured.
Electrical energy is used to ignite the fuel; as the fuel is burning, it will heat up
the surrounding air, which expands and escapes through a tube that leads the
air out of the calorimeter.
When the air is escaping through the copper tube it will also heat up the water
outside the tube. The change in temperature of the water allows for
calculating calorie content of the fuel.
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The calorific value is obtained by:
In using the results obtained during the experiment, there is a cooling
correction factor to take into account.
This is because of the cooling that takes place during the experiment. This
correction will be incorporated in the question given.
4.3.2.1 The water equivalent
For any given substance the equivalent mass of water W having the same
specific heat as the given substance is called its Water Equivalent.
Worked Example 4.1
A bomb calorimeter was used to test the heat value of coal.
The following readings were:
Mass of coal burned = 0.546 kg
Mass of water in bomb = 2.8 kg
The water equivalent of the calorimeter = 1.05 kg
Corrected temperature rise = 1.056 degrees
Find the calorific value of the coal.
Solution:
Known: The specific heat capacity of water = 4.187 kJ/kg.K
The gas calorimeter determines the calorific value of a fuel. Figure 4.2 shows a
central funnel with a burner, B. A cooling coil around the outside of the funnel
and a condensate trap at the bottom. The outer casing is lagged for
insulation.
Thermometers measure the cooling water inlet and outlet temperatures and
the gas exhaust passes out at the bottom where a thermometer measures the
exhaust temperature.
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Note:
Most gasses contain H2 and this helps to form H2O on the coils that
drip off into the condensate trap.
At the top is an arrangement that gives a constant head of cooling water. The
gas pressure is kept constant by a gas governor.
The calorific value of the gas can be found by the knowledge that the energy
gained by the cooling water is equal to the energy liberated by burning the
gas.
Figure 4.2 The gas calorimeter
An Orsat gas analyser is used to analyse a gas sample (hydrocarbon) for
its oxygen, carbon monoxide and carbon dioxide content. Figure 4.3.
Although largely replaced by instrumental techniques, the Orsat remains a
reliable method of measurement and is relatively simple to use.
The apparatus consists essentially of a calibrated water-jacketed
gas burette connected by glass capillary tubing to two or three absorption
pipettes containing chemical solutions that absorb the gasses it is required to
measure. For safety and portability, the apparatus is usually encased in a
wooden box.
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The absorbents are:
Potassium hydroxide (Caustic Potash)
Alkaline pyrogallol
Ammoniacal Cuprous chloride
Figure 4.3 The Orsat apparatus
The base of the gas burette is connected to a levelling bottle to enable
readings to be taken at constant pressure and to transfer the gas to and from
the absorption media. The burette contains slightly acidulated water with a
trace of chemical indicator (typically menthyl orange) for coloration.
By means of a rubber tubing arrangement, the gas to be analysed is drawn
into the burette and flushed through several times. Typically, 100ml is
withdrawn for ease of calculation.
Using the stopcocks that isolate the absorption burettes, the level of gas in the
levelling bottle and the burette is adjusted to the zero point of the burette.
The gas is then passed into the caustic potash burette, left to stand for about
two minutes and then withdrawn, isolating the remaining gas via the stopcock
arrangements. The process is repeated to ensure full absorption.
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After levelling the liquid in the bottle and burette, the remaining volume of gas
in the burette indicates the percentage of carbon dioxide absorbed.
The same technique is repeated for oxygen, using the pyrogallol, and carbon
monoxide using the ammoniacal cuprous chloride.
4.3.2.2 Analysis by volume
Avogadro’s hypothesis states that equal volumes of different gases at the
same pressure and temperature contain the same number of molecules.
This means that the proportion of volumes is valid so if one gas has twice the
number of molecules than another, then it has twice the volume.
Worked Example 4.2
A gas was calibrated with the following values:
Volume of gas used = 0.0085 m3
Temperature of the gas = 18 degrees C
Gas manometer height = 100 mm Hg
Barometer height = 755 mm Hg
Mass of cooling water used during gas burn = 3.76 kg = M
Mass of condensate in same time = 0.0054 kg = m
Temperature of cooling water in = 14 degrees C = t1
Temperature of cooling water out = 24 degrees C = t2
Solution:
The volume of gas used must be converted to the volume as it would exist at
25 degrees C and 760 mm Hg.
Let this volume = V. then
Energy released by gas = energy gained by water
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4.3.3 Higher heating value
The quantity known as higher heating value (HHV) (or gross energy or upper
heating value or gross calorific value (GCV) or higher calorific value (HCV)) is
determined by bringing all the products of combustion back to the original
pre-combustion temperature, and in particular condensing any vapour
produced.
Note:
Such measurements often use a standard temperature of 15 °C. This
is the same as the thermodynamic heat of combustion since
the enthalpy change for the reaction assumes a common
temperature of the compounds before and after combustion, in
which case the water produced by combustion is liquid.
The higher heating value takes into account the latent heat of
vaporization of water in the combustion products, and is useful in calculating
heating values for fuels where condensation of the reaction products is
practical (e.g., in a gas-fired boiler used for space heat).
In other words, HHV assumes all the water component is in liquid state at the
end of combustion (in product of combustion) and that heat below 150 °C
(302 °F) can be put to use.
4.3.4 Lower heating value
The quantity known as lower heating value (LHV) (net calorific value (NCV)
or lower calorific value (LCV)) is determined by subtracting the heat of
vaporization of the water vapour from the higher heating value. This treats any
H2O formed as a vapour. The energy required to vaporize the water therefore
is not released as heat.
LHV calculations assume that the water component of a combustion process is
in vapour state at the end of combustion, as opposed to the higher heating
value (HHV) (a.k.a. gross calorific value or gross CV) which assumes that all of
the water in a combustion process is in a liquid state after a combustion
process.
The LHV assumes that the latent heat of vaporization of water in the fuel and
the reaction products is not recovered. It is useful in comparing fuels where
condensation of the combustion products is impractical, or heat at a
temperature below 150 °C (302 °F) cannot be put to use.
Another definition, used by Gas Processors Suppliers Association (GPSA) and
originally used by API (data collected for API research project 44), is
the enthalpy of all combustion products minus the enthalpy of the fuel at the
reference temperature (API research project 44 used 25 °C. GPSA), minus the
enthalpy of the stoichiometric oxygen (O2) at the reference temperature,
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minus the heat of vaporization of the vapour content of the combustion
products.
The distinction between the two is that this second definition assumes that the
combustion products are all returned to the reference temperature and the
heat content from the condensing vapour is considered not to be useful. This is
more easily calculated from the higher heating value than when using the
preceding definition and will in fact give a slightly different answer.
Definition: Stoichiometric
A stoichiometric amount or stoichiometric ratio of a reagent is the
optimum amount or ratio where, assuming that the reaction
proceeds to completion:
All of the reagent is consumed
There is no deficiency of the reagent
There is no excess of the reagent.
4.3.5 Stoichiometry (chemically correct air fuel ratio)
Stoichiometry is founded on the law of conservation of mass where the total
mass of the reactants equals the total mass of the products leading to the
insight that the relations among quantities of reactants and products typically
form a ratio of positive integers.
This means that if the amounts of the separate reactants are known, then the
amount of the product can be calculated. Conversely, if one reactant has a
known quantity and the quantity of product can be empirically determined,
then the amount of the other reactants can also be calculated.
Gravimetric analysis of the combining masses of fuel and air before
combustion and the products of combustion.
Air contains about 78 % nitrogen which does not add to the combustion
process and the amount of nitrogen before combustion is the same as the
amount after combustion.
Worked Example 4.3
Complete the combustion equation of hydrogen plus oxygen to form water.
Solution:
The relative atomic mass of hydrogen = 1
The relative atomic mass of oxygen = 16
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Atomic mass of the constituents:
Thus, it appears that:
Simplify by dividing through by 4:
Note that this equation must always be balanced.
23.2 % oxygen is present in air by mass. Hence, the 8 kg of oxygen above is
23.2 % of the mass of air necessary to complete combustion of the H2.
Find the air needed:
The nitrogen present:
In conclusion:
1 kg H2 is burned in 34.5 kg of air and produces the products of combustion of
9 kg H2O with 26.5 kg N2
4.4 Air fuel ratio
Take the combustion of H2 and O2 to form water H2O. This time the equation
has not been balanced:
Notice that with the constituents (on the left) there are two hydrogens and two
oxygens. But with the product (on the right), there are two hydrogens and only
one oxygen.
Balance this equation by:
When the 2 is placed in front of the water molecule, we get four hydrogen and
two oxygen. The 2 placed in front as shown above affects the hydrogen and
oxygen.
Looking to the constituents, a 2 needs to be placed before the hydrogen
molecule to balance the equation:
The equation is now balanced.
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Worked Example 4.4
Hexane, C6H14 a type of paraffin, is combusted with oxygen. Write down the
basic equation and then balance it.
Solution:
This shows that the carbon will form CO2 and the hydrogen will form H2O.
Note the following steps:
Now balance the oxygen:
Total oxygen is = 9.5 O2
The equation id balanced but it is impossible to have half an atom as in 9.5
O2
This equation is now correct.
If the analysis of air is given by mass, then proceed as follows:
Find the mass of oxygen required for each constituent.
From this find the total mass of oxygen required.
Subtract any oxygen that might be in the fuel.
Stoichiometric mass of air = O2 required / 0.232
Worked Example 4.5
The chemical analysis of 1 kg of coal is as follows:
C = 86%, H2 = 3%, S = 0,5% and the rest is non-combustible. 16,5 kg of air is
provided for the complete combustion of 1 kg of the coal.
Calculate the following:
1. The theoretical mass of air required for the complete combustion of 1 kg
of the coal
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2. The percentage of excess air supplied Power Machines N5
3. The masses of the products of combustion
4. Express the mass's in 3. as percentages ]
]
Solution: =[
1. =[ ]
2. =[
= 3,667 x 0,86 =
4. = 9 x 0,03 =( )
= 2 x 0,05 = 50%
= 2,53862 x = 3,15362 kg
= 16,5 x = 0,27 kg
TOTAL = 0,1 kg
= = 2,53862 kg
= = 12,705 kg
= = 18,767 kg
= 16,804%
=
= 1,438%
=
TOTAL = 0,533%
= 13,526%
= 67,699%
= 100%
Worked Example 4.6
The chemical analysis by mass of a fuel is 78% carbon, 14% hydrogen and 8%
oxygen. 35% of excess air is required for complete combustion and it is
supplied at 101 kPa and 26 ºC. The gas constant R for air Is 0,287 kJikg.K.
Calculate the volume of air required for complete combustion of the fuel.
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Solution:
=( ) , -
-{
= ( ) *, } +
= 18,313 kg
=
=
= 15,55934 m3
Worked Example 4.7
A certain sample of fuel contains the following elements by mass:
- Oxygen 3,7%
- Hydrogen 9,2%
- Carbon 84,6%
- Sulphur 2,5%
The calorific value of sulphur, carbon and hydrogen is 9,5 MJ/kg, 35 MJ/kg
and 145 MJ/kg respectively. The partial pressure of the steam Is 6,5 kPa in the
exhaust gases.
Calculate the following:
1. The higher calorific value of the fuel
2. The lower calorific value of the fuel
Solution: = ,( )-
1.1
= ,( )-
= 0,2375 + 29,61 + 12,6694
= 42,5169 MJ/kg
1.2 =
=
= 40,5189 MJ/kg
( )
2.1 =
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=
= 54 334,928 kJ / kg
= 54,335 MJ/kg
2.2 =
=
= 50,981 MJ/kg
Activity 4.1
A bomb calorimeter was used to determine the calorific value of a sample of
coal. The following readings were obtained during the test.
Specific heat capacity of water 4,187 kJ/kg. ºC
Water equivalent of the calorimeter 0,6 kg
Maximum temperature of the water 24,6 ºC
Temperature of water before ignition 17,8 ºC
Mass of water used 2,2kg
Mass of coal used 1,36 grams
Calculate the calorific value of the coal In MJ/kg.
[58.618 MJ/kg]
Activity 4.2
A fuel was tested, using a bomb calorimeter, and the following data was
recorded:
- Mass of fuel used 0,73 g
- Water equivalent of the bomb 50 g
- Mass of the water in the calorimeter 2,5 kg
- Minimum temperature of the water 24,767 ºC
- Maximum temperature of the water 28,482 ºC
- The quantity of hydrogen in the fuel 15,4%
- The partial pressure of the steam In the exhaust gases 5,5 kPa
- The specific heat capacity of water 4,187 kJ/kg. ºC
Calculate the following:
1. The higher calorific value of the fuel In MJ
2. The lower calorific value of the fuel in MJ
[54.335; 50.981]
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Activity 4.3
A fuel consists of 72 % carbon, 20 % hydrogen and 8 % oxygen by mass. Find
the mass of air required to completely burn 1 kg of the fuel.
[3.44; 14.8]
Activity 4.4
A sample of coal has the following analysis by mass: carbon 87%; hydrogen
5%; oxygen 6%; sulphur 1% and ash 1%. The composition of air is 23% oxygen
by mass and the atomic masses are: carbon 12; oxygen 16; sulphur 32 and
hydrogen 1. 30% excess air is supplied for complete combustion of the coal.
Calculate the following:
1. The mass of oxygen required for complete combustion of 1 kg of coal,
from first principles
2. The theoretical mass of air required for the complete combustion of 1 kg
of coal
3. The actual mass of air supplied for complete combustion of 3,5 kg of coal
[3.19; 0.45; 0.02; 2.67; 11.609; 52.821/3.5]
Activity 4.5
A furnace gas made of coal has the following volumetric analysis: 28%
hydrogen; 21% carbon monoxide; 2% oxygen and the remainder are
nitrogen. 25% excess air is supplied and only the hydrogen and carbon
monoxide burn. 3% carbon dioxide.
The atomic masses are:
Hydrogen 1
Carbon 12
Nitrogen 14
Oxygen 16
Calculate the following from first principles:
1. The volume of oxygen required to burn 1 m3 of fuel
2. The theoretical volume of air required to burn 1 m3 of fuel
3. The actual volume of air used to burn 1 m3 of fuel
4. The actual air/fuel ratio
[0.225; 1.0714; 1.339]
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Activity 4.6
A certain fuel has the following composition:
77,7% Carbon
6,8% Hydrogen
1,2% Nitrogen
2,2% Sulphur
8,8% Oxygen
3,3% Ash
The atomic masses are:
Hydrogen 1
Carbon 12
Nitrogen 14
Oxygen 16
Sulphur 32
Calculate the following first principles:
1. The mass of oxygen required for the combustion of carbon
2. The mass of oxygen required for the combustion of hydrogen
3. The mass of oxygen required for the combustion of sulphur
4. The total mass of oxygen required
5. The mass of air required
6. The mass of nitrogen
[2.072; 0.544; 0.022; 2.55; 11.087; 8.549]
Activity 4.7
The volumetric composition of a gas is:
Hydrogen 6,5%
Carbon monoxide 16%
Methane (CH4) 4,5%
Carbon dioxide 13%
Oxygen 5%
Nitrogen 55%
The atomic masses are:
Hydrogen 1
Carbon 12
Nitrogen 14
Oxygen 16
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Draw a table using the following headings: Symbol; % by volume; molecular
mass; % volume x molecular mass, % by mass and determine the composition
of the gas by mass.
[0.463; 15.971; 2.567; 20.392; 5.704; 54.902]
Activity 4.8
A sample of coal has the following analysis by mass:
Carbon 77,9%
Hydrogen 7,9%
Oxygen 5,8%
Sulphur 7,2%
Ash 1,2%
The atomic masses are as follows:
Sulphur 32
Oxygen 16
Carbon 12
Hydrogen 1
Air is supplied at 27,8% excess.
Calculate the following:
1. The theoretical mass of oxygen required for complete combustion of 1,32
kg of the coal from first principles
2. The theoretical mass of air for complete combustion of 3,12 kg of the coal
3. The actual mass of air required for complete combustion of 2,13 kg of the
coal
[15.939; 2.32; 0.56; 2.82; 10.3; 30.9; 118.1682]
Activity 4.9
18,04 kg of air is used for the complete combustion of a kilogram of fuel.
The fuel has the following composition:
Carbon = 87%
Hydrogen = 2,5%
Sulphur = 1%
The balance of the fuel is non-combustible
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Calculate the percentages of the products of combustion.
[11; 7.04; 64; 3.19; 0.225; 1.619; 0.02; 13.89; 18.944]
Activity 4.10
1 kg of fuel oil which has the formula C12 H4 completely burnt with the aid of
40% e3xcess air. The atomic mass of carbon, oxygen and hydrogen is 12,16
and 1, respectively.
Calculate the following:
1. The theoretical mass of air required to burn the fuel
2. The actual mass of air used to burn the fuel
3. The mass of each product of combustion
4. The percentage analysis, by mass, of the combustion products
[12.22; 17.11; 3.568; 0.243; 11244; 13.175; 19.7; 1.34; 6.21; 73]
Self-Check
I am able to: Yes No
Describe solid, liquid and gaseous fuels
Describe the higher and lower calorific values of fuel
Calculate the minimum air required for complete combustion
Calculate the products of combustion
Describe the bomb calorimeter and the Orsat apparatus
If you have answered ‘no’ to any of the outcomes listed above, then speak to
your facilitator for guidance and further development.
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Learning Outcomes
On the completion of this module the student must be able to:
Describe solid, liquid and gaseous fuels
Describe the higher and lower calorific values of fuel
Calculate the minimum air required for complete combustion
Calculate the products of combustion
Describe the bomb calorimeter and the Orsat apparatus
5.1 Introduction
In this course we are going to deal with single stage reciprocating
air compressors only. Figure 5.1 shows a reciprocating air
compressor with the compressor at the top cooled by a fan and
driven by an electric motor, this is mounted on the receiver tank.
Definition:
An air compressor is a device that converts power (using an electric
motor, diesel or gasoline engine, etc.) into potential energy stored
in pressurized air (i.e., compressed air).
Figure 5.1 Reciprocating air compressor
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5.2 Construction
A reciprocating compressor is a positive displacement compressor that
uses pistons driven by a crankshaft to deliver air at high pressure.
The intake air enters the suction manifold, then flows into the compression
cylinder where it gets compressed by a piston driven in a reciprocating motion
via a crankshaft, and is then discharged.
There is an inlet or suction valve and a discharge valve. Both are one way
valves and both are spring loaded so it takes a certain amount of force to
open them. The inlet valve will only let air come into the cylinder and the
discharge valve will only let air out.
As the piston moves downward through the cylinder, air is sucked in the inlet
valve. When the piston moves up through the cylinder the inlet valve closes
and the discharge won't open until a certain amount of force is applied.
This causes the air to be trapped inside while the volume is reduced, increasing
the pressure. When the pressure is enough to open the discharge valve, then
the air goes out at the higher pressure.
5.2.1 Single-acting cylinder
A single-acting cylinder in a reciprocating compressor is a cylinder in which
the working air acts on one side of the piston only. For one revolution of the
crank, one swept volume of air is delivered.
Note:
Single-acting reciprocating compressors only have valves on top of
the cylinder, so there is one compression cycle for every turn of the
crankshaft.
5.2.1.1 The cylinder
This is a metal tube-shaped casing (or body), which is generally fitted with a
metal lining called a 'cylinder liner'. The liner is replaceable when it becomes
worn and inefficient.
The cylinder is also fitted with suction and discharge ports which contain
special spring loaded valves to allow air to flow in one direction only.
5.2.1.2 The piston
The piston consists of a metal drive rod connected to the piston head which is
located inside the cylinder. The piston head is fitted with piston rings to give a
seal against the cylinder lining and minimize internal leakage.
The other end of the drive rod extends to the outside of the cylinder and is
connected to the driver. Modern industry generally used high power electric
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motors and gearing to convert the rotating motion into a reciprocating action.
In a single acting compressor, the backward stroke of the piston causes a
suction which pulls in air through the inlet valve. (The same suction action
keeps the discharge valve closed).
On the forward stroke, the positive pressure generated by the piston, closes the
inlet valve and opens the discharge valve. The air is displaced into the
discharge system.
Safety Warning!
Because the action is positive displacement, a piston compressor
can generate very high pressure and therefore MUST NEVER be
operated against a closed discharge system valve unless it is fitted
with a safety relief system in order to prevent damage to the
compressor and/or the driver and/or other downstream equipment.
Figure 5.2 Single acting air compressor
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Figure 5.3 View of single acting air compressor showing the crank
5.2.2 Double-acting cylinder
A double-acting cylinder, Figure 5.4 is a cylinder in which the working air acts
alternately on both sides of the piston. In order to connect the piston in a
double-acting cylinder to an external mechanism, such as a crankshaft, a hole
must be provided in one end of the cylinder for the piston rod and this is fitted
with a gland or stuffing box to prevent escape of the working air.
Double-acting reciprocating compressor is similar to single-acting, but they
have inlet and discharge valves on both sides of the cylinder. This gives you
two compression cycles for every turn of the crankshaft.
Did you know?
Double-acting reciprocating are some of the most efficient
compressors ever built, but they're not as common as they used to
be. They're still around, but they're extremely large.
You rarely see one under 100 hp. Back in the days, they used to be called
earth-movers, because being around one felt like you were in an earthquake.
If the compressor didn't have the right foundation and vibration isolation, it
could tear down the building it was in. These used to be the workhorses in
manufacturing plants, but in the last 20-30 years that role was taken over by
rotary screw compressors.
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Figure 5.4 Working principle of the double acting air compressor
5.3 Work done by a reciprocating compressor
The function of a compressor is to take a sufficient amount of air and to
increase its pressure. Compressors may be classified into different categories.
Some of them are the reciprocating compressor, rotary compressor, etc.,
Each has their own advantages and disadvantages. As we know, the
reciprocating compressors may be used for high pressure applications.
Now let us consider the following for a single stage single acting reciprocating
compressor without clearance volume.
Let P1, V1, T1 be the Pressure, volume and Temperature of the fluid before
compression, P2, V2, T2 be the pressure, volume and temperature of the fluid
after compression, and r be the compression ratio = P2/P1
Now let us discuss the PV diagram of a single acting, single stage reciprocating
compressor without clearance volume, Figure 5.5.
In this diagram, the path 1-2 represents the polytropic compression.
From the above diagram it is clear that the air is drawn in by the cylinder
during the suction stroke (1-2). The air is compressed during the compression
stroke (2-3) at constant temperature (T1=T2).
Now, the pressure rises from P1 to P2 from 2-3. The air will be delivered when
sufficient pressure has been reached so that the valves can be opened. The
delivery of air takes place from 2-3.
In brief:
4-1 : Suction stroke (air is drawn into the cylinder) Volume of air, V1, at
P1 and T1
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1-2 : compression stroke (air gets compressed inside the cylinder)
According to the law PVn = C from pressure Pi to P2. The
volume decreases from V1 to V2 and the temperature increases
from T1 to T2.
2-3 : air is delivered under pressure at V2, P2 and T2
3-4 : delivery closes and the suction valve opens for admitting fresh
air inside the cylinder.
Figure 5.5 A theoretical PV diagram neglecting clearance
Work done is the are under the graph
The energy equation:
Where:
H = Enthalpy
Also:
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Power Machines N5
()
Since:
()
5.3.1 The clearance volume
This is the volume that remains in the cylinder after the piston has reached the
end of the compression stroke. Figure 5.6.
From point 2 to 3 compressed air is delivered from the cylinder but at 3, the
piston has reached the end of its stroke and so delivery of air stops at 3. V3 is
the clearance volume and is filled with compressed air.
As the piston starts its intake stroke, the residual compressed air expands
according to the law PVn = C and it is not until the pressure is reduced to intake
pressure at 4 that the inlet valve will begin to open. This fresh charge of air (V1 –
V4) is called the effective swept volume.
Figure 5.6 A theoretical PV diagram with clearance
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Power Machines N5
The volumetric efficiency:
The clearance ratio:
Figure 5.7 Actual PV diagram with clearance
The actual PV diagram Figure 5.7 shows the effects at inlet and delivery of gas
inertia and turbulence. The two shaded areas (negative work areas) are in
addition to the theoretical area.
Worked Example 5.1
A single stage, double-acting air compressor takes In air at 15,965 m3/min.
The air intake conditions are 101 kPa and 19 ºC. The delivery pressure is 750
kPa and the speed of the compressor Is 325 r/min. R for air is 0,287 kJ/kg.K, the
mechanical efficiency of the compressor Is 85% and the index n = 1,33. The
stroke length of the piston rod Is 1,25 times the diameter of the cylinder and
there Is no clearance volume.
Calculate the following:
1. The effective stroke volume
2. The diameter of the piston In mm
3. The stroke length of the piston rod In mm
4. The delivery temperature in Kelvin
5. The power required to drive the compressor
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Solution: = Power Machines N5
1. = 1
= 0,02456 m3 / stroke
2. =
=
0,02456 =
D =√
= 0,29247 M
= 292,5 mm
3. =
=
=
4. = ( )
= ()
= ()
= 480,21 K
5. = ( ) 0( ) 1
=( ) 0( )
=
= 82,07 kW
Worked Example 5.2
The clearance volume of a single-stage, single acting, reciprocating
compressor, is 6,5% of the stroke volume. The compressor takes 1n air 33 ºC
and 96 kPa and delivers it to a receiver at 650 kPa. The polytropic index of
compression is 1,35. 'R' for air is 0 ,287 kJ/kg.K and the free air delivery
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Power Machines N5
conditions are 101,3 kPa and 18 ºC.
Calculate the following:
1. The volumetric efficiency at intake conditions
2. The volumetric efficiency referred to free air delivery conditions
3. The effective volume per kilogram of air
4. The work done per kilogram of air
Solution: = ()
1.
= ()
=
= 79,696%
2. =
=
=
= 71,824 %
3. =
=
=
4. = ( ) 0( ) 1
= ( ) 0( ) 1
= 217,44424 kJ / kg
Activity 5.1
0,04 m3/s of air Is drawn into a single-acting, single-cylinder, reciprocating air
compressor, at a pressure of 105 kPa. The air is then delivered to an air
receiver at a pressure of 625 kPa according to the law PV132 = C. The
compressor has a mechanical efficiency of 82% and a motor transmission
efficiency of 88%. The stroke to bore ratio is 1,4 to 1 and the compressor runs
at 8 r/s.
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