1. ENERGY PERFORMANCE ASSESSMENT
OF BOILERS
1.1
Introduction
Performance of the boiler, like efficiency and evaporation ratio reduces with time, due to
poor combustion, heat transfer fouling and poor operation and maintenance. Deterioration
of fuel quality and water quality also leads to poor performance of boiler. Efficiency testing helps us to find out how far the boiler efficiency drifts away from the best efficiency.
Any observed abnormal deviations could therefore be investigated to pinpoint the problem
area for necessary corrective action. Hence it is necessary to find out the current level of
efficiency for performance evaluation, which is a pre requisite for energy conservation
action in industry.
1.2
•
•
Purpose of the Performance Test
To find out the efficiency of the boiler
To find out the Evaporation ratio
The purpose of the performance test is to determine actual performance and efficiency of the boiler and compare it with design values or norms. It is an indicator for tracking
day-to-day and season-to-season variations in boiler efficiency and energy efficiency
improvements
1.3
Performance Terms and Definitions
1.4
Scope
The procedure describes routine test for both oil fired and solid fuel fired boilers using coal,
agro residues etc. Only those observations and measurements need to be made which can be
readily applied and is necessary to attain the purpose of the test.
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1. Energy Performance Assessment of Boilers
1.5
Reference Standards
British standards, BS845: 1987
The British Standard BS845: 1987 describes the methods and conditions under which a boiler should be tested to determine its efficiency. For the testing to be done, the boiler should
be operated under steady load conditions (generally full load) for a period of one hour after
which readings would be taken during the next hour of steady operation to enable the efficiency to be calculated.
The efficiency of a boiler is quoted as the % of useful heat available, expressed as a percentage of the total energy potentially available by burning the fuel. This is expressed on the
basis of gross calorific value (GCV).
This deals with the complete heat balance and it has two parts:
•
•
Part One deals with standard boilers, where the indirect method is specified
Part Two deals with complex plant where there are many channels of heat flow. In this
case, both the direct and indirect methods are applicable, in whole or in part.
ASME Standard: PTC-4-1 Power Test Code for Steam Generating Units
This consists of
•
•
Part One: Direct method (also called as Input -output method)
Part Two: Indirect method (also called as Heat loss method)
IS 8753: Indian Standard for Boiler Efficiency Testing
Most standards for computation of boiler efficiency, including IS 8753 and BS845 are designed
for spot measurement of boiler efficiency. Invariably, all these standards do not include blow
down as a loss in the efficiency determination process.
Basically Boiler efficiency can be tested by the following methods:
1) The Direct Method: Where the energy gain of the working fluid (water and steam) is
compared with the energy content of the boiler fuel.
2) The Indirect Method: Where the efficiency is the difference between the losses and the
energy input.
1.6
The Direct Method Testing
1.6.1 Description
This is also known as 'input-output method' due to the fact that it needs only the useful output
(steam) and the heat input (i.e. fuel) for evaluating the efficiency. This efficiency can be evaluated using the formula:
x 100
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1. Energy Performance Assessment of Boilers
1.6.2 Measurements Required for Direct Method Testing
Heat input
Both heat input and heat output must be measured. The measurement of heat input requires
knowledge of the calorific value of the fuel and its flow rate in terms of mass or volume, according to the nature of the fuel.
For gaseous fuel: A gas meter of the approved type can be used and the measured volume should
be corrected for temperature and pressure. A sample of gas can be collected for calorific value
determination, but it is usually acceptable to use the calorific value declared by the gas suppliers.
For liquid fuel: Heavy fuel oil is very viscous, and this property varies sharply with temperature. The meter, which is usually installed on the combustion appliance, should be
regarded as a rough indicator only and, for test purposes, a meter calibrated for the particular oil is to be used and over a realistic range of temperature should be installed. Even
better is the use of an accurately calibrated day tank.
For solid fuel: The accurate measurement of the flow of coal or other solid fuel is very
difficult. The measurement must be based on mass, which means that bulky apparatus
must be set up on the boiler-house floor. Samples must be taken and bagged throughout
the test, the bags sealed and sent to a laboratory for analysis and calorific value determination. In some more recent boiler houses, the problem has been alleviated by mounting
the hoppers over the boilers on calibrated load cells, but these are yet uncommon.
Heat output
There are several methods, which can be used for measuring heat output. With steam boilers,
an installed steam meter can be used to measure flow rate, but this must be corrected for temperature and pressure. In earlier years, this approach was not favoured due to the change in
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1. Energy Performance Assessment of Boilers
accuracy of orifice or venturi meters with flow rate. It is now more viable with modern flow
meters of the variable-orifice or vortex-shedding types.
The alternative with small boilers is to measure feed water, and this can be done by previously calibrating the feed tank and noting down the levels of water during the beginning and
end of the trial. Care should be taken not to pump water during this period. Heat addition for
conversion of feed water at inlet temperature to steam, is considered for heat output.
In case of boilers with intermittent blowdown, blowdown should be avoided during the trial
period. In case of boilers with continuous blowdown, the heat loss due to blowdown should be
calculated and added to the heat in steam.
1.6.3 Boiler Efficiency by Direct Method: Calculation and Example
Test Data and Calculation
Water consumption and coal consumption were measured in a coal-fired boiler at hourly intervals. Weighed quantities of coal were fed to the boiler during the trial period. Simultaneously
water level difference was noted to calculate steam generation during the trial period. Blow
down was avoided during the test. The measured data is given below.
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1. Energy Performance Assessment of Boilers
1.6.4 Merits and Demerits of Direct Method
Merits
• Plant people can evaluate quickly the efficiency of boilers
• Requires few parameters for computation
• Needs few instruments for monitoring
Demerits
• Does not give clues to the operator as to why efficiency of system is lower
• Does not calculate various losses accountable for various efficiency levels
• Evaporation ratio and efficiency may mislead, if the steam is highly wet due to water carryover
1.7
The Indirect Method Testing
1.7.1 Description
The efficiency can be measured easily by measuring all the losses occurring in the boilers using
the principles to be described. The disadvantages of the direct method can be overcome by this
method, which calculates the various heat losses associated with boiler. The efficiency can be
arrived at, by subtracting the heat loss fractions from 100.An important advantage of this
method is that the errors in measurement do not make significant change in efficiency.
Thus if boiler efficiency is 90% , an error of 1% in direct method will result in significant
change in efficiency. i.e. 90 ± 0.9 = 89.1 to 90.9. In indirect method, 1% error in measurement
of losses will result in
Efficiency = 100 – (10 ± 0.1) = 90 ± 0.1 = 89.9 to 90.1
The various heat losses occurring in the boiler are:
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1. Energy Performance Assessment of Boilers
The following losses are applicable to liquid, gas and solid fired boiler
L1– Loss due to dry flue gas (sensible heat)
L2– Loss due to hydrogen in fuel (H2)
L3– Loss due to moisture in fuel (H2O)
L4– Loss due to moisture in air (H2O)
L5– Loss due to carbon monoxide (CO)
L6– Loss due to surface radiation, convection and other unaccounted*.
*Losses which are insignificant and are difficult to measure.
The following losses are applicable to solid fuel fired boiler in addition to above
L7– Unburnt losses in fly ash (Carbon)
L8– Unburnt losses in bottom ash (Carbon)
Boiler Efficiency by indirect method = 100 – (L1 + L2 + L3 + L4 + L5 + L6 + L7 + L8)
1.7.2 Measurements Required for Performance Assessment Testing
The following parameters need to be measured, as applicable for the computation of boiler efficiency and performance.
a)
Flue gas analysis
1. Percentage of CO2 or O2 in flue gas
2. Percentage of CO in flue gas
3. Temperature of flue gas
b)
Flow meter measurements for
1. Fuel
2. Steam
3. Feed water
4. Condensate water
5. Combustion air
c)
Temperature measurements for
1. Flue gas
2. Steam
3. Makeup water
4. Condensate return
5. Combustion air
6. Fuel
7. Boiler feed water
d)
Pressure measurements for
1. Steam
2. Fuel
3. Combustion air, both primary and secondary
4. Draft
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1. Energy Performance Assessment of Boilers
e)
Water condition
1. Total dissolved solids (TDS)
2. pH
3. Blow down rate and quantity
The various parameters that were discussed above can be measured with the instruments
that are given in Table 1.1.
TABLE 1.1 TYPICAL INSTRUMENTS USED FOR BOILER PERFORMANCE
ASSESSMENT.
Instrument
Type
Measurements
Flue gas analyzer
Portable or fixed
% CO2 , O2 and CO
Temperature indicator
Thermocouple, liquid in
glass
Fuel temperature, flue gas
temperature, combustion air
temperature, boiler surface
temperature, steam temperature
Draft gauge
Manometer, differential
pressure
Amount of draft used
or available
TDS meter
Conductivity
Boiler water TDS, feed water TDS,
make-up water TDS.
Flow meter
As applicable
Steam flow, water flow, fuel flow,
air flow
1.7.3 Test Conditions and Precautions for Indirect Method Testing
A) The efficiency test does not account for:
•
•
•
•
Standby losses. Efficiency test is to be carried out, when the boiler is operating under a
steady load. Therefore, the combustion efficiency test does not reveal standby losses,
which occur between firing intervals
Blow down loss. The amount of energy wasted by blow down varies over a wide range.
Soot blower steam. The amount of steam used by soot blowers is variable that depends on
the type of fuel.
Auxiliary equipment energy consumption. The combustion efficiency test does not
account for the energy usage by auxiliary equipments, such as burners, fans, and pumps.
B) Preparations and pre conditions for testing
•
•
•
•
•
•
•
•
•
Burn the specified fuel(s) at the required rate.
Do the tests while the boiler is under steady load. Avoid testing during warming up of boilers from a cold condition
Obtain the charts /tables for the additional data.
Determination of general method of operation
Sampling and analysis of fuel and ash.
Ensure the accuracy of fuel and ash analysis in the laboratory.
Check the type of blow down and method of measurement
Ensure proper operation of all instruments.
Check for any air infiltration in the combustion zone.
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1. Energy Performance Assessment of Boilers
C) Flue gas sampling location
It is suggested that the exit duct of the boiler be probed and traversed to find the location of the
zone of maximum temperature. This is likely to coincide with the zone of maximum gas flow
and is therefore a good sampling point for both temperature and gas analysis.
D) Options of flue gas analysis
Check the Oxygen Test with the Carbon Dioxide Test
If continuous-reading oxygen test equipment is installed in boiler plant, use oxygen reading.
Occasionally use portable test equipment that checks for both oxygen and carbon dioxide. If the carbon dioxide test does not give the same results as the oxygen test, something is wrong. One (or both)
of the tests could be erroneous, perhaps because of stale chemicals or drifting instrument calibration.
Another possibility is that outside air is being picked up along with the flue gas. This occurs if the
combustion gas area operates under negative pressure and there are leaks in the boiler casing.
Carbon Monoxide Test
The carbon monoxide content of flue gas is a good indicator of incomplete combustion with all
types of fuels, as long as they contain carbon. Carbon monoxide in the flue gas is minimal with
ordinary amounts of excess air, but it rises abruptly as soon as fuel combustion starts to be incomplete.
E) Planning for the testing
• The testing is to be conducted for a duration of 4 to 8 hours in a normal production day.
• Advanced planning is essential for the resource arrangement of manpower, fuel, water and
instrument check etc and the same to be communicated to the boiler Supervisor and
Production Department.
• Sufficient quantity of fuel stock and water storage required for the test duration should be
arranged so that a test is not disrupted due to non-availability of fuel and water.
• Necessary sampling point and instruments are to be made available with working condition.
• Lab Analysis should be carried out for fuel, flue gas and water in coordination with lab personnel.
• The steam table, psychometric chart, calculator are to be arranged for computation of boiler efficiency.
1.7.4 Boiler Efficiency by Indirect Method: Calculation Procedure and Formula
In order to calculate the boiler efficiency by indirect method, all the losses that occur in the
boiler must be established. These losses are conveniently related to the amount of fuel burnt.
In this way it is easy to compare the performance of various boilers with different ratings.
Conversion formula for proximate analysis to ultimate analysis
%C
= 0.97C + 0.7 (VM + 0.1A) – M(0.6 – 0.01M)
%H2 = 0.036C + 0.086 (VM – 0.1xA) – 0.0035M2 (1 – 0.02M)
%N2 = 2.10 – 0.020 VM
where C
A
VM
M
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=
=
=
=
% of fixed carbon
% of ash
% of volatile matter
% of moisture
8
1. Energy Performance Assessment of Boilers
However it is suggested to get a ultimate analysis of the fuel fired periodically from a
reputed laboratory.
Theoretical (stoichiometric) air fuel ratio and excess air supplied are to be determined first
for computing the boiler losses. The formula is given below for the same.
The various losses associated with the operation of a boiler are discussed below with
required formula.
1. Heat loss due to dry flue gas
This is the greatest boiler loss and can be calculated with the following formula:
m x Cp x (Tf - Ta )
L1
=
x 100
GCV of fuel
Where,
L1
m
=
=
=
% Heat loss due to dry flue gas
Mass of dry flue gas in kg/kg of fuel
Combustion products from fuel: CO2 + SO2 + Nitrogen in fuel +
Nitrogen in the actual mass of air supplied + O2 in flue gas.
(H2O/Water vapour in the flue gas should not be considered)
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1. Energy Performance Assessment of Boilers
Cp
Tf
Ta
=
=
=
Specific heat of flue gas in kCal/kg°C
Flue gas temperature in °C
Ambient temperature in °C
Note–1:
For Quick and simple calculation of boiler efficiency use the following.
A: Simple method can be used for determining the dry flue gas loss as given below.
m x Cp x (Tf – Ta) x 100
a) Percentage heat loss due to dry flue gas =
GCV of fuel
Total mass of flue gas (m)/kg of fuel = mass of actual air supplied/kg of fuel + 1 kg of fuel
Note-2: Water vapour is produced from Hydrogen in fuel, moisture present in fuel and air during the combustion. The losses due to these components have not been included in the dry flue
gas loss since they are separately calculated as a wet flue gas loss.
2. Heat loss due to evaporation of water formed due to H2 in fuel (%)
The combustion of hydrogen causes a heat loss because the product of combustion is water.
This water is converted to steam and this carries away heat in the form of its latent heat.
9 x H2 x {584 + Cp (Tf – Ta )}
L2
=
x 100
GCV of fuel
Where
H2
Cp
Tf
Ta
584
=
=
=
=
=
kg of hydrogen present in fuel on 1 kg basis
Specific heat of superheated steam in kCal/kg°C
Flue gas temperature in °C
Ambient temperature in °C
Latent heat corresponding to partial pressure of water vapour
3. Heat loss due to moisture present in fuel
Moisture entering the boiler with the fuel leaves as a superheated vapour. This moisture loss is
made up of the sensible heat to bring the moisture to boiling point, the latent heat of evaporation of the moisture, and the superheat required to bring this steam to the temperature of the
exhaust gas. This loss can be calculated with the following formula
M x {584 + Cp (Tf – Ta)}
L3
=
X 100
GCV of fuel
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1. Energy Performance Assessment of Boilers
where
M
Cp
Tf
Ta
584
=
=
=
=
=
kg moisture in fuel on 1 kg basis
Specific heat of superheated steam in kCal/kg°C
Flue gas temperature in °C
Ambient temperature in °C
Latent heat corresponding to partial pressure of water vapour
4. Heat loss due to moisture present in air
Vapour in the form of humidity in the incoming air, is superheated as it passes through the boiler. Since this heat passes up the stack, it must be included as a boiler loss.
To relate this loss to the mass of coal burned, the moisture content of the combustion air and
the amount of air supplied per unit mass of coal burned must be known.
The mass of vapour that air contains can be obtained from psychrometric charts and typical
values are included below:
Dry-Bulb
Wet Bulb
Relative Humidity
Temp °C
Temp °C
(%)
Kilogram water
per Kilogram dry
air (Humidity Factor)
20
20
100
0.016
20
14
50
0.008
30
22
50
0.014
40
30
50
0.024
AAS x humidity factor x Cp x (Tf – Ta ) x 100
L4
=
GCV of fuel
where
AAS
Humidity factor
Cp
Tf
Ta
=
=
=
=
=
Actual mass of air supplied per kg of fuel
kg of water/kg of dry air
Specific heat of superheated steam in kCal/kg°C
Flue gas temperature in °C
Ambient temperature in °C (dry bulb)
5. Heat loss due to incomplete combustion:
Products formed by incomplete combustion could be mixed with oxygen and burned again with
a further release of energy. Such products include CO, H2, and various hydrocarbons and are
generally found in the flue gas of the boilers. Carbon monoxide is the only gas whose concentration can be determined conveniently in a boiler plant test.
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1. Energy Performance Assessment of Boilers
%CO x C
=
L5
% CO + % CO2
L5
CO
CO2
C
5744
x
x
100
GCV of fuel
=
=
=
=
% Heat loss due to partial conversion of C to CO
Volume of CO in flue gas leaving economizer (%)
Actual Volume of CO2 in flue gas (%)
Carbon content kg / kg of fuel
or
When CO is obtained in ppm during the flue gas analysis
CO (in ppm) x 10–6 x Mf x 28
CO formation (Mco) =
=
Fuel consumption in kg/hr
Mf
=
Mco x 5744*
L5
* Heat loss due to partial combustion of carbon.
6. Heat loss due to radiation and convection:
The other heat losses from a boiler consist of the loss of heat by radiation and convection from
the boiler casting into the surrounding boiler house.
Normally surface loss and other unaccounted losses is assumed based on the type and size
of the boiler as given below
For industrial fire tube / packaged boiler = 1.5 to 2.5%
For industrial watertube boiler = 2 to 3%
For power station boiler = 0.4 to 1%
However it can be calculated if the surface area of boiler and its surface temperature are
known as given below :
L6
=
0.548 x [ (Ts / 55.55)4 – (Ta / 55.55)4] + 1.957 x (Ts – Ta)1.25 x sq.rt of
[(196.85 Vm + 68.9) / 68.9]
L6
Vm
Ts
Ta
=
=
=
=
Radiation loss in W/m2
Wind velocity in m/s
Surface temperature (K)
Ambient temperature (K)
where
Heat loss due to unburned carbon in fly ash and bottom ash:
Small amounts of carbon will be left in the ash and this constitutes a loss of potential heat in
the fuel. To assess these heat losses, samples of ash must be analyzed for carbon content. The
quantity of ash produced per unit of fuel must also be known.
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1. Energy Performance Assessment of Boilers
7.
Heat loss due to unburnt in fly ash (%).
Total ash collected / kg of fuel burnt x G.C.V of fly ash x 100
L7 =
GCV of fuel
8.
Heat loss due to unburnt in bottom ash (%)
Total ash collected per kg of fuel burnt x G.C.V of bottom ash x 100
L8 =
GCV of fuel
Heat Balance:
Having established the magnitude of all the losses mentioned above, a simple heat balance
would give the efficiency of the boiler. The efficiency is the difference between the energy
input to the boiler and the heat losses calculated.
Boiler Heat Balance:
Input/Output Parameter
kCal / kg of fuel
Heat Input in fuel
=
%
100
Various Heat losses in boiler
1. Dry flue gas loss
=
2. Loss due to hydrogen in fuel
3. Loss due to moisture in fuel
=
4. Loss due to moisture in air
=
5. Partial combustion of C to CO
=
6. Surface heat losses
=
7. Loss due to Unburnt in fly ash
=
8. Loss due to Unburnt in bottom ash
=
Total Losses
=
Boiler efficiency = 100 – (1+2+3+4+5+6+7+8)
1.8
Example: Boiler Efficiency Calculation
1.8.1 For Coal fired Boiler
The following are the data collected for a boiler using coal as the fuel. Find out the boiler efficiency by indirect method.
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1. Energy Performance Assessment of Boilers
Fuel firing rate
Steam generation rate
Steam pressure
Steam temperature
Feed water temperature
%CO2 in Flue gas
%CO in flue gas
Average flue gas temperature
Ambient temperature
Humidity in ambient air
Surface temperature of boiler
Wind velocity around the boiler
Total surface area of boiler
GCV of Bottom ash
GCV of fly ash
Ratio of bottom ash to fly ash
Fuel Analysis (in %)
Ash content in fuel
Moisture in coal
Carbon content
Hydrogen content
Nitrogen content
Oxygen content
GCV of Coal
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
=
5599.17 kg/hr
21937.5 kg/hr
43 kg/cm2(g)
377 °C
96 °C
14
0.55
190 °C
31 °C
0.0204 kg / kg dry air
70 °C
3.5 m/s
90 m2
800 kCal/kg
452.5 kCal/kg
90:10
=
=
=
=
=
=
=
8.63
31.6
41.65
2.0413
1.6
14.48
3501 kCal/kg
Boiler efficiency by indirect method
Step – 1 Find theoretical air requirement
Theoretical air required for
complete combustion
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=
[(11.6 x C) + {34.8 x (H2 – O2/8)} + (4.35 x S)] /100
kg/kg of coal
=
[(11.6 x 41.65) + {34.8 x (2.0413 – 14.48/8)} +
(4.35 x 0)] / 100
=
4.91 kg / kg of coal
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1. Energy Performance Assessment of Boilers
Step – 3 To find Excess air supplied
Actual CO2 measured in flue gas
=
14.0%
7900 x [ ( CO2%)t – (CO2%)a]
% Excess air supplied (EA)
=
(CO2%)a x [100 – (CO2%)t ]
7900 x [20.37 – 14 ]
=
14a x [100 – 20.37]
=
45.17 %
Step – 4 to find actual mass of air supplied
Actual mass of air supplied
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=
{1 + EA/100} x theoretical air
=
{1 + 45.17/100} x 4.91
=
7.13 kg/kg of coal
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1. Energy Performance Assessment of Boilers
Step – 5 to find actual mass of dry flue gas
Mass of dry flue gas
Mass of CO2 + Mass of N2 content in the fuel +
Mass of N2 in the combustion air supplied + Mass of
oxygen in flue gas
=
0.4165 x 44
Mass of dry flue gas
=
12
=
7.13 x 77 (7.13–4.91) x 23
+ 0.016 +
+
100
100
7.54 kg / kg of coal
Step – 6 to find all losses
m x Cp x (Tf – Ta)
1. % Heat loss in dry flue gas (L1)
=
x 100
GCV of fuel
7.54 x 0.23 x (190 – 31)
=
x 100
3501
L1
2. % Heat loss due to formation
of water from H2 in fuel (L2)
=
=
7.88 %
9 x H2 x {584 + Cp (Tf – Ta)}
x 100
GCV of fuel
9 x .02041 x {584 + 0.45(190 – 31)}
=
x 100
3501
L2
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=
3.44 %
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1. Energy Performance Assessment of Boilers
M x {584 + Cp ( Tf – Ta )}
3. % Heat loss due to moisture in
fuel (L3)
=
X 100
GCV of fuel
0.316 x {584 + 0.45 ( 190 - 31) }
=
x 100
3501
=
L3
5.91 %
AAS x humidity x Cp x (Tf – Ta ) x 100
4. % Heat loss due to moisture
in air (L4)
=
GCV of fuel
7.13 x 0.0204 x 0.45 x (190 – 31) x 100
=
3501
=
L4
0.29 %
%CO x C
5. % Heat loss due to partial
conversion of C to CO (L5)
5744
=
x
% CO + (% CO2)a
0.55 x 0.4165
=
=
6. Heat loss due to radiation and
convection (L6)
Bureau of Energy Efficiency
100
5744
x
0.55 + 14
L5
x
GCV of fuel
x
100
3501
2.58 %
=
0.548 x [ (343/55.55)4 – (304/55.55)4] + 1.957 x
(343 – 304)1.25 x sq.rt of [(196.85 x 3.5 + 68.9) / 68.9]
=
=
=
633.3 w/m2
633.3 x 0.86
544.64 kCal / m2
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1. Energy Performance Assessment of Boilers
Total radiation and convection
loss per hour
=
=
% radiation and convection loss
=
544.64 x 90
49017.6 kCal
49017.6 x 100
3501 x 5599.17
L6
=
0.25 %
7. % Heat loss due to unburnt in fly ash
% Ash in coal
=
Ratio of bottom ash to fly ash
=
GCV of fly ash
=
Amount of fly ash in 1 kg of coal =
=
Heat loss in fly ash
=
=
% heat loss in fly ash
=
L7 =
8.63
90:10
452.5 kCal/kg
0.1 x 0.0863
0.00863 kg
0.00863 x 452.5
3.905 kCal / kg of coal
3.905 x 100 / 3501
0.11 %
8. % Heat loss due to unburnt in bottom ash
GCV of bottom ash
= 800 kCal/kg
Amount of bottom ash in 1 kg of
= 0.9 x 0.0863
coal
= 0.077 kg
Heat loss in bottom ash
= 0.077 x 800
= 62.136 kCal/kg of coal
% Heat loss in bottom ash
= 62.136 x 100 / 3501
L8 = 1.77 %
Boiler efficiency by indirect
method
=
100 – (L1 + L2 + L3 + L4 + L5 + L6 + L7 + L8)
=
100 – (7.88 + 3.44 + 5.91 + 0.29 + 2.58 + 0.25
+ 0.11 + 1.77)
100 – 22.23
77.77 %
=
=
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1. Energy Performance Assessment of Boilers
SUMMARY OF HEAT BALANCE FOR COAL FIRED BOILER
Input/Output Parameter
kCal / kg of
coal
% loss
=
3501
100
1. Dry flue gas, L1
=
275.88
7.88
2. Loss due to hydrogen in fuel, L2
=
120.43
3.44
3. Loss due to moisture in fuel, L3
=
206.91
5.91
4. Loss due to moisture in air, L4
=
10.15
0.29
5. Partial combustion of C to CO, L5
=
90.32
2.58
6. Surface heat losses, L6
=
8.75
0.25
7. Loss due to Unburnt in fly ash, L7
=
3.85
0.11
8. Loss due to Unburnt in bottom ash, L8
=
61.97
1.77
Heat Input
Losses in boiler
Boiler Efficiency = 100 – (L1 + L2 + L3 + L4 + L5 + L6 + L7 + L8) = 77.77 %
1.8.2 Efficiency for an oil fired boiler
The following are the data collected for a boiler using furnace oil as the fuel. Find out the boiler efficiency by indirect method.
Ultimate analysis (%)
Carbon
Hydrogen
Nitrogen
Oxygen
Sulphur
Moisture
GCV of fuel
Fuel firing rate
Surface Temperature of boiler
Surface area of boiler
Humidity
Wind speed
=
=
=
=
=
=
=
=
=
=
=
=
84
12
0.5
1.5
1.5
0.5
10000 kCal/kg
2648.125 kg/hr
80 °C
90 m2
0.025 kg/kg of dry air
3.8 m/s
Flue gas analysis (%)
Flue gas temperature
Ambient temperature
Co2% in flue gas by volume
O2% in flue gas by volume
=
=
=
=
190°C
30°C
10.8
7.4
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1. Energy Performance Assessment of Boilers
a) Theoretical air required
b) Excess Air supplied (EA)
c) Actual mass of air supplied/ kg
of fuel (AAS)
Mass of dry flue gas
=
[(11.6 x C) + [{34.8 x (H2 – O2/8)} + (4.35 x S)] /100
kg/kg of fuel. [from fuel analysis]
=
[(11.6 x 84) + [{34.8 x (12 – 1.5/8)}
+ (4.35 x 1.5)] / 100
=
13.92 kg/kg of oil
=
(O2 x 100) / (21 – O2)
=
(7.4 x 100) / (21 – 7.4)
=
54.4 %
=
{1 + EA/100} x theoretical air
=
{1 + 54.4/100} x 13.92
=
21.49 kg / kg of fuel
=
Mass of (CO2 + SO2 + N2 + O2) in flue gas + N2
in air we are supplying
=
=
0.84 x 44
12
+
0.015 x 64
32
+ 0.005 +
7.4x23
100
21.36 kg / kg of oil
m x Cp x (Tf – Ta)
% Heat loss in dry flue gas
=
x 100
GCV of fuel
21.36 x 0.23 x (190 – 30)
=
x 100
10000
L1
=
7.86 %
9 x H2 x{584 + Cp (Tf – Ta )}
Heat loss due to evaporation of
water due to H2 in fuel (%)
=
x 100
GCV of fuel
9 x 0.12 x {584 + 0.45 (190 – 30)}
=
x 100
10000
L2
Bureau of Energy Efficiency
=
7.08 %
20
+
21.49 x 77
100
1. Energy Performance Assessment of Boilers
M x {584 + Cp ( Tf - Ta )}
% Heat loss due to moisture
in fuel
=
X 100
GCV of fuel
0.005 x {584 + 0.45 (190 – 30)}
=
x 100
10000
L3
=
0.033%
AAS x humidity x Cp x (Tf – Ta ) x 100
% Heat loss due to moisture in air =
GCV of fuel
21.36 x 0.025 x 0.45 x (190 – 30) x 100
=
10000
L4
=
0.38 %
Radiation and convection loss =
(L6)
0.548 x [ (Ts / 55.55)4 – (Ta / 55.55)4] + 1.957
x (Ts – Ta)1.25 x sq.rt of [(196.85 Vm + 68.9) / 68.9]
=
0.548 x [ (353 / 55.55)4 – (303 / 55.55)4] + 1.957
x (353 – 303)1.25 x sq.rt of [(196.85 x 3.8 + 68.9)/ 68.9]
=
1303 W/m2
=
1303 x 0.86
=
1120.58 kCal / m2
Total radiation and convection
loss per hour
=
=
1120 .58 x 90 m2
100852.2 kCal
% Radiation and convection loss
=
100852.2 x 100
10000 x 2648.125
L6
Boiler efficiency by indirect
method
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=
0.38 %
Normally it is assumed as 0.5 to 1 % for simplicity
=
=
=
=
100 – (L1 + L2 + L3 + L4 + L6)
100 – (7.86 + 7.08 + 0.033 + 0.38 + 0.38)
100 – 15.73
84.27 %
21
1. Energy Performance Assessment of Boilers
Summary of Heat Balance for the Boiler Using Furnace Oil
Input/Output Parameter
kCal / kg of
furnace oil
%Loss
=
10000
100
1. Dry flue gas, L1
=
786
7.86
2. Loss due to hydrogen in fuel, L2
=
708
7.08
3. Loss due to Moisture in fuel, L3
=
3.3
0.033
4. Loss due to Moisture in air, L4
=
38
0.38
5. Partial combustion of C to CO, L5
=
0
0
6. Surface heat losses, L6
=
38
0.38
Heat Input
Losses in boiler :
Boiler Efficiency = 100 – (L1 + L2 + L3 + L4 + L6) = 84.27 %
Note:
For quick and simple calculation of boiler efficiency use the following.
A: Simple method can be used for determining the dry flue gas loss as given below.
m x Cp x (Tf – Ta ) x 100
a) Percentage heat loss due to dry flue gas =
GCV of fuel
Total mass of flue gas (m) = mass of actual air supplied (ASS)+ mass of fuel supplied
= 21.49 + 1=22.49
%Dry flue gas loss = 22.49 x 0.23 x (190-30) x 100 = 8.27%
10000
1.9
Factors Affecting Boiler Performance
The various factors affecting the boiler performance are listed below:
•
•
•
•
•
•
•
•
•
Periodical cleaning of boilers
Periodical soot blowing
Proper water treatment programme and blow down control
Draft control
Excess air control
Percentage loading of boiler
Steam generation pressure and temperature
Boiler insulation
Quality of fuel
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1. Energy Performance Assessment of Boilers
All these factors individually/combined, contribute to the performance of the boiler and
reflected either in boiler efficiency or evaporation ratio. Based on the results obtained from the
testing further improvements have to be carried out for maximizing the performance. The test
can be repeated after modification or rectification of the problems and compared with standard
norms. Energy auditor should carry out this test as a routine manner once in six months and
report to the management for necessary action.
1.10
Data Collection Format for Boiler Performance Assessment
Sheet 1 – Technical specification of boiler
1
Boiler ID code and Make
2
Year of Make
3
Boiler capacity rating
4
Type of Boiler
5
Type of fuel used
6
Maximum fuel flow rate
7
Efficiency by GCV
8
Steam generation pressure &superheat temperature
9
Heat transfer area in m2
10
Is there any waste heat recovery device installed
11
Type of draft
12
Chimney height in metre
Sheet 2 – Fuel analysis details
Fuel Fired
GCV of fuel
Specific gravity of fuel (Liquid)
Bulk density of fuel (Solid)
Proximate Analysis
1
2
3
4
Fixed carbon
Volatile matter
Ash
Moisture
Ultimate Analysis
1
2
3
Date of Test:
%
%
%
%
Date of Test:
Carbon
Hydrogen
Sulphur
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%
%
%
23
1. Energy Performance Assessment of Boilers
4
5
6
7
Nitrogen
Ash
Moisture
Oxygen
Water Analysis
1
2
3
4
1.11
Date of Test:
Feed water TDS
Blow down TDS
PH of feed water
PH of blow down
Flue gas Analysis
1
2
3
4
%
%
%
%
ppm
ppm
Date of Test:
%
%
%
°C
CO2
O2
CO
Flue gas temperature
Boiler Terminology Danh pháp lò hơi/ Thuật ngữ lò hơi
MCR: Steam boilers rated output is also usually defined as MCR (Maximum Continuous
Rating). This is the maximum evaporation rate that can be sustained for 24 hours and may be
less than a shorter duration maximum rating
Boiler Rating
Conventionally, boilers are specified by their capacity to hold water and the steam generation
rate. Often, the capacity to generate steam is specified in terms of equivalent evaporation (kg
of steam / hour at 100°C). Equivalent evaporation- "from and at" 100°C. The equivalent of the
evaporation of 1 kg of water at 100°C to steam at 100°C.
Efficiency : In the boiler industry there are four common definitions of efficiency:
a. Combustion efficiency
Combustion efficiency is the effectiveness of the burner only and relates to its ability to completely burn
the fuel. The boiler has little bearing on combustion efficiency. A well-designed burner will operate
with as little as 15 to 20% excess air, while converting all combustibles in the fuel to useful energy.
b. Thermal efficiency
Thermal efficiency is the effectiveness of the heat transfer in a boiler. It does not take into account
boiler radiation and convection losses - for example from the boiler shell water column piping etc.
c. Boiler efficiency
The term boiler efficiency is often substituted for combustion or thermal efficiency. True boiler efficiency is the measure of fuel to steam efficiency.
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1. Energy Performance Assessment of Boilers
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1. Energy Performance Assessment of Boilers
d. Fuel to steam efficiency
Fuel to steam efficiency is calculated using either of the two methods as prescribed by the
ASME (American Society for Mechanical Engineers) power test code, PTC 4.1. The first
method is input output method. The second method is heat loss method.
Boiler turndown
Boiler turndown is the ratio between full boiler output and the boiler output when operating at
low fire. Typical boiler turndown is 4:1. The ability of the boiler to turndown reduces frequent
on and off cycling. Fully modulating burners are typically designed to operate down to 25% of
rated capacity. At a load that is 20% of the load capacity, the boiler will turn off and cycle frequently.
A boiler operating at low load conditions can cycle as frequently as 12 times per hour or 288
times per day. With each cycle, pre and post purge airflow removes heat from the boiler and
sends it out the stack. Keeping the boiler on at low firing rates can eliminate the energy loss.
Every time the boiler cycles off, it must go through a specific start-up sequence for safety assurance. It requires about a minute or two to place the boiler back on line. And if there is a sudden load demand the start up sequence cannot be accelerated. Keeping the boiler on line assures
the quickest response to load changes. Frequent cycling also accelerates wear of boiler components. Maintenance increases and more importantly, the chance of component failure
increases.
Boiler(s) capacity requirement is determined by many different type of load variations in
the system. Boiler over sizing occurs when future expansion and safety factors are added to
assure that the boiler is large enough for the application. If the boiler is oversized the ability of
the boiler to handle minimum loads without cycling is reduced. Therefore capacity and turndown should be considered together for proper boiler selection to meet overall system load
requirements.
Primary air: That part of the air supply to a combustion system which the fuel first encounters.
Secondary air: The second stage of admission of air to a combustion system, generally to
complete combustion initiated by the primary air. It can be injected into the furnace of a boiler under relatively high pressure when firing solid fuels in order to create turbulence above the
burning fuel to ensure good mixing with the gases produced in the combustion process and
thereby complete combustion
Tertiary air: A third stage of admission of air to a combustion system, the reactions of which
have largely been completed by secondary air. Tertiary air is rarely needed.
Stoichiometric: In combustion technology, stoichiometric air is that quantity of air, and no
more, which is theoretically needed to burn completely a unit quantity of fuel. 'Sub-stoichiometric' refers to the partial combustion of fuel in a deficiency of air
Balanced draught: The condition achieved when the pressure of the gas in a furnace is the
same as or slightly below that of the atmosphere in the enclosure or building housing it.
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1. Energy Performance Assessment of Boilers
Gross calorific value (GCV): The amount of heat liberated by the complete combustion,
under specified conditions, by a unit volume of a gas or of a unit mass of a solid or liquid fuel,
in the determination of which the water produced by combustion of the fuel is assumed to be
completely condensed and its latent and sensible heat made available.
Net calorific value (NCV): The amount of heat generated by the complete combustion, under
specified conditions, by a unit volume of a gas or of a unit mass of a solid or liquid fuel, in the
determination of which the water produced by the combustion of the fuel is assumed to remain
as vapour.
Absolute pressure The sum of the gauge and the atmospheric pressure. For instance, if the
steam gauge on the boiler shows 9 kg/cm2g the absolute pressure of the steam is 10 kg/cm2(a).
Atmospheric pressure The pressure due to the weight of the atmosphere. It is expressed in
pounds per sq. in. or inches of mercury column or kg/cm2. Atmospheric pressure at sea level is
14.7 lbs./ sq. inch. or 30 inch mercury column or 760mm of mercury (mm Hg) or 101.325 kilo
Pascal (kPa).
Carbon monoxide (CO): Produced from any source that burns fuel with incomplete combustion, causes chest pain in heart patients, headaches and reduced mental alertness.
Blow down: The removal of some quantity of water from the boiler in order to achieve an
acceptable concentration of dissolved and suspended solids in the boiler water.
Complete combustion: The complete oxidation of the fuel, regardless of whether it is
accomplished with an excess amount of oxygen or air, or just the theoretical amount required
for perfect combustion.
Perfect combustion: The complete oxidation of the fuel, with the exact theoretical (stoichiometric) amount of oxygen (air) required.
Saturated steam: It is the steam, whose temperature is equal to the boiling point corresponding to that pressure.
Wet Steam
Saturated steam which contains moisture
Dry Steam
Either saturated or superheated steam containing no moisture.
Superheated Steam Steam heated to a temperature above the boiling point or saturation temperature corresponding to its pressure
Oxygen trim sensor measures flue gas oxygen and a closed loop controller compares the
actual oxygen level to the desired oxygen level. The air (or fuel) flow is trimmed by
the controller until the oxygen level is corrected. The desired oxygen level for each firing
rate must be entered into a characterized set point curve generator. Oxygen Trim maintains
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1. Energy Performance Assessment of Boilers
the lowest possible burner excess air level from low to high fire. Burners that don't have
Oxygen Trim must run with Extra Excess Air to allow safe operation during variations in
weather, fuel, and linkage.
Heat transfer mediums
There are different types of heat transfer medium e.g. steam, hot water and thermal oil. Steam
and Hot water are most common and it will be valuable to briefly examine these common heat
transfer mediums and associated properties.
Thermic Fluid
Thermic Fluid is used as a heat transfer mechanism in some industrial process and heating
applications. Thermic Fluid may be a vegetable or mineral based oil and the oil may be raised
to a high temperature without the need for any pressurization. The relatively high flow and
return temperatures may limit the potential for flue gas heat recovery unless some other system
can absorb this heat usefully. Careful design and selection is required to achieve best energy
efficiency.
Hot water
Water is a fluid with medium density, high specific heat capacity, low viscosity and relatively
low thermal conductivity. At relatively low temperature e.g. 70°C – 90°C, hot water is useful
for smaller heating installations.
Steam
When water is heated its temperature will rise. The heat added is called sensible heat and the
heat content of the water is termed its enthalpy. The usual datum point used to calculate
enthalpy is 0°C.
When the water reaches its boiling point, any further heat input will result in some proportion of the water changing from the liquid to the vapour state, i.e. changing to steam. The heat
required for this change of state is termed the 'latent heat of evaporation' and is expressed in
terms of a fixed mass of water. Where no change in temperature occurs during the change of
state, the steam will exist in equilibrium with the water. This equilibrium state is termed 'saturation conditions'. Saturation conditions can occur at any pressure, although at each pressure
there is only one discrete temperature at which saturation can occur.
If further heat is applied to the saturated steam the temperature will rise and the steam will
become 'superheated'. Any increase in temperature above saturated conditions will be accompanied by a further rise in enthalpy.
Steam is useful heat transfer medium because, as a gas, it is compressible. At high pressure
and consequently density, steam can carry large quantities of heat with relatively small volume.
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1. Energy Performance Assessment of Boilers
QUESTIONS
1)
Define boiler efficiency.
2)
Why boiler efficiency by indirect method is more useful than direct method?
3)
What instruments are required for indirect efficiency testing?
4)
What is the difference between dry flue gas loss and wet flue gas loss?
5)
Which is the best location for sampling flue gas analysis?
6)
Find out the efficiency by direct method from the data given below.
An oil fired package boiler was tested for 2 hours duration at steady state condition.
The fuel and water consumption were 250 litres and 3500 litres respectively. The
specific gravity of oil is 0.92. The saturated steam generation pressure is
7 kg/cm2(g). The boiler feed water temperature is 30°C. Determine the boiler efficiency and evaporation ratio.
7)
What is excess air? How to determine excess air if oxygen / carbon dioxide percentage is measured in the flue gas?
8)
As a means of performance evaluation, explain the difference between efficiency and
evaporation ratio.
9)
Testing coal-fired boiler is more difficult than oil-fired boiler. Give reasons.
10)
What is controllable and uncontrollable losses in a boiler?
REFERENCES
1.
2.
3.
Energy audit Reports of National Productivity Council
Energy Hand book, Second edition, Von Nostrand Reinhold Company - Robert
L.Loftness
Industrial boilers, Longman Scientific Technical 1999
www.boiler.com
www.eng-tips.com
www.worldenergy.org
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2.
ENERGY PERFORMANCE ASSESSMENT OF FURNACES
2.1
Industrial Heating Furnaces
Furnace is by definition a device for heating materials and therefore a user of energy. Heating
furnaces can be divided into batch-type (Job at stationary position) and continuous type (large
volume of work output at regular intervals). The types of batch furnace include box, bogie,
cover, etc. For mass production, continuous furnaces are used in general. The types of continuous furnaces include pusher-type furnace (Figure 2.1), walking hearth-type furnace, rotary
hearth and walking beam-type furnace.(Figure 2.2)
The primary energy required for reheating / heat treatment (say annealing) furnaces are in
the form of Furnace oil, LSHS, LDO or electricity
Figure 2.1: Pusher-Type 3-Zone Reheating Furnace
2.2
Figure 2.2: Walking Beam-Type Reheating Furnace
Purpose of the Performance Test
To find out the efficiency of the furnace
To find out the Specific energy consumption
The purpose of the performance test is to determine efficiency of the furnace and specific
energy consumption for comparing with design values or best practice norms. There are many
factors affecting furnace performance such as capacity utilization of furnaces, excess air ratio,
final heating temperature etc. It is the key for assessing current level of performances and finding the scope for improvements and productivity.
Heat Balance of a Furnace
Heat balance helps us to numerically understand the present heat loss and efficiency and
improve the furnace operation using these data. Thus, preparation of heat balance is a
pre-requirement for assessing energy conservation potential.
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2. Energy Performance Assessment of Furnaces
2.3
Performance Terms and Definitions
1. Furnace Efficiency, η
2. Specific Energy Consumption
2.4
=
Heat output
x 100
Heat Input
=
Heat in stock (material) (kCals)
x 100
Heat in Fuel /electricity (kCals)
=
Quantity of fuel or energy consumed
Quantity of material processed.
Reference Standards
In addition to conventional methods, Japanese Industrial Standard (JIS) GO702 "Method of
heat balance for continuous furnaces for steel" is used for the purpose of establishing the heat
losses and efficiency of reheating furnaces.
2.5
Furnace Efficiency Testing Method
The energy required to increase the temperature of a material is the product of the mass, the
change in temperature and the specific heat. i.e. Energy = Mass x Specific Heat x rise in
temperature. The specific heat of the material can be obtained from a reference manual and
describes the amount of energy required by different materials to raise a unit of weight through
one degree of temperature.
If the process requires a change in state, from solid to liquid, or liquid to gas, then an
additional quantity of energy is required called the latent heat of fusion or latent heat of
evaporation and this quantity of energy needs to be added to the total energy requirement.
However in this section melting furnaces are not considered.
The total heat input is provided in the form of fuel or power. The desired output is the heat
supplied for heating the material or process. Other heat outputs in the furnaces are undesirable
heat losses.
The various losses that occur in the fuel fired furnace (Figure 2.3) are listed below.
1. Heat lost through exhaust gases either as sensible heat, latent heat or as incomplete
combustion
2. Heat loss through furnace walls and hearth
3. Heat loss to the surroundings by radiation and convection from the outer surface of the
walls
4. Heat loss through gases leaking through cracks, openings and doors.
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2. Energy Performance Assessment of Furnaces
Furnace Efficiency
The efficiency of a furnace is
the ratio of useful output to heat
input. The furnace efficiency can
be determined by both direct and
indirect method.
2.5.1 Direct Method Testing
The efficiency of the furnace can
be computed by measuring the
amount of fuel consumed per unit
weight of material produced from
the furnace.
Heat in the stock
Thermal efficiency of the furnace =
Heat in the fuel consumed
The quantity of heat to be imparted (Q) to the stock can be found from the formula
Q
=
m x Cp (t2 – t1)
Where
Q
=
Quantity of heat in kCal
m
=
Weight of the material in kg
Mean specific heat, kCal/kg°C
Cp =
=
Final temperature desired, °C
t2
=
Initial temperature of the charge before it enters the furnace, °C
t1
2.5.2 Indirect Method Testing
Similar to the method of evaluating boiler efficiency by indirect method, furnace efficiency can
also be calculated by indirect method. Furnace efficiency is calculated after subtracting sensible heat loss in flue gas, loss due to moisture in flue gas, heat loss due to openings in furnace,
heat loss through furnace skin and other unaccounted losses from the input to the furnace.
In order to find out furnace efficiency using indirect method, various parameters that are
required are hourly furnace oil consumption, material output, excess air quantity, temperature
of flue gas, temperature of furnace at various zones, skin temperature and hot combustion air
temperature. Efficiency is determined by subtracting all the heat losses from 100.
Measurement Parameters
The following measurements are to be made for doing the energy balance in oil fired reheating
furnaces (e.g. Heating Furnace)
i)
ii)
iii)
iv)
v)
Weight of stock / Number of billets heated
Temperature of furnace walls, roof etc
Flue gas temperature
Flue gas analysis
Fuel Oil consumption
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2. Energy Performance Assessment of Furnaces
Instruments like infrared thermometer, fuel consumption monitor, surface thermocouple
and other measuring devices are required to measure the above parameters. Reference manual
should be referred for data like specific heat, humidity etc.
Example: Energy Efficiency by Indirect Method
An oil-fired reheating furnace has an operating temperature of around 1340°C. Average fuel consumption is 400 litres/hour. The flue gas exit temperature after air preheater is 750°C. Air is preheated from ambient temperature of 40°C to 190°C through an air pre-heater. The furnace has 460
mm thick wall (x) on the billet extraction outlet side, which is 1 m high (D) and 1 m wide. The
other data are as given below. Find out the efficiency of the furnace by both indirect and direct
method.
Flue gas temperature after air preheater
Ambient temperature
Preheated air temperature
Specific gravity of oil
Average fuel oil consumption
= 750°C
= 40°C
= 190°C
= 0.92
= 400 Litres / hr
= 400 x 0.92 =368 kg/hr
Calorific value of oil
= 10000 kCal/kg
= 12%
Average O2 percentage in flue gas
Weight of stock
= 6000 kg/hr
Specific heat of Billet
= 0.12 kCal/kg/°C
Surface temperature of roof and side walls = 122 °C
Surface temperature other than heating and soaking zone = 85 °C
Solution
1. Sensible Heat Loss in Flue Gas:
O2%
= ————
21–O2%
Excess air
× 100
(Where O2 is the % of oxygen in flue gas = 12% )
=
=
=
=
=
=
=
=
=
Theoretical air required to burn 1 kg of oil
Total air supplied
Total air supplied
Sensible heat loss
m
=
=
=
=
Cp
∆T
Bureau of Energy Efficiency
34
12 x 100 / (21 - 12)
133% excess air
14 kg (Typical value for all fuel oil)
Theoretical air x (1 + excess air/100)
14 x 2.33 kg / kg of oil
32.62 kg / kg of oil
m x Cp x ∆T
Weight of flue gas
Actual mass of air supplied / kg of
fuel + mass of fuel (1kg)
32.62 + 1.0 = 33.62 kg / kg of oil.
Specific heat of flue gas
0.24 kCal/kg/°C
Temperature difference
2. Energy Performance Assessment of Furnaces
Heat loss = m x Cp x ∆T
%
= 33.62 x 0.24 x (750- 40)
= 5729 kCal / kg of oil
5729 x 100
= —————— = 57.29%
10000
Heat loss in flue gas
2.
Loss Due to Evaporation of Moisture Present in Fuel
M {584 + 0.45 (Tfg–Tamb)}
% Loss
= ——————————— × 100
GCV of Fuel
Where,
M
Tfg
Tamb
GCV
-
% Loss
0.15 {584 +0.45 (750-40)}
= -------------------------------- x 100
10000
=
3.
kg of Moisture in 1 kg of fuel oil (0.15 kg/kg of fuel oil)
Flue Gas Temperature
Ambient temperature
Gross Calorific Value of Fuel
1.36 %
Loss Due to Evaporation of Water Formed due to Hydrogen in Fuel
% Loss
9 x H2 {584 + 0.45 (Tfg-Tamb)}
= --------------------------------------- x 100
GCV of Fuel
Where, H2 – kg of H2 in 1 kg of fuel oil (0.1123 kg/kg of fuel oil)
4.
=
9 x 0.1123 {584 + 0.45 (750-40)}
------------------------------------------ x 100
10000
=
9.13 %
Heat Loss due to Openings:
If a furnace body has an opening on it, the heat in the furnace escapes to the outside as
radiant heat. Heat loss due to openings can be calculated by computing black body radiation at
furnace temperature, and multiplying these values with emissivity (usually 0.8 for furnace brick
work), and the factor of radiation through openings. Factor for radiation through openings can
be determined with the help of graph as shown in figure 2.4. The black body radiation losses
can be directly computed from the curves as given in the figure 2.5 below.
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2. Energy Performance Assessment of Furnaces
Factor for Determining the Equivalent of Heat Release from Openings to the Quality of Heat
Release from Perfect Black Body
TOTAL BLACK BODY RADIATION (kCal/cm2/hr)
Figure 2.4
Temperature (°C)
Figure 2.5
Graph for Determining Black Body Radiation at a Particular Temperature
The reheating furnace in example has 460mm thick wall (X) on the billet extraction outlet
side, which is 1m high (D) and 1m wide. With furnace temperature of 1340°C, the quantity (Q)
of radiation heat loss from the opening is calculated as follows:
The shape of the opening is square and D/X
The factor of radiation (Refer Figure 2.4)
Black body radiation corresponding to 1340°C
(Refer Figure 2.5 On black body radiation)
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= 1/0.46 = 2.17
= 0.71
= 36.00 kCal/cm2/hr
2. Energy Performance Assessment of Furnaces
Area of opening
= 100 cm x 100 cm
= 10000 cm2
= 0.8
Emissivity
Total heat loss = Black body radiation x area of opening x factor of radiation x emissivity
=
=
36 x 10000 x 0.71 x 0.8
204480 kCal/hr
Equivalent Oil loss
= 204480/10,000
= 20.45 kg/hr
% of heat loss
=
=
20.45 /368 x 100
5.56 %
5. Heat Loss through Skin:
Method 1: Radiation Heat Loss from Surface of Furnace
The quantity of heat loss from surface of furnace body is the sum of natural convection and
thermal radiation. This quantity can be calculated from surface temperatures of furnace. The
temperatures on furnace surface should be measured at as many points as possible, and their
average should be used. If the number of measuring points is too small, the error becomes
large.
The quantity (Q) of heat release from a reheating furnace is calculated with the following
formula:
where
Q
a
tl
t2
E
: Quantity of heat release in kCal / W / m2
: factor regarding direction of the surface of natural convection ceiling = 2.8,
side walls = 2.2, hearth = 1.5
: temperature of external wall surface of the furnace (°C)
: temperature of air around the furnace (°C)
: emissivity of external wall surface of the furnace
The first term of the formula above represents the quantity of heat release by natural convection, and the second term represents the quantity of heat release by radiation.
Method 2 : Radiation Heat Loss from Surface of Furnace
The following Figure 2.6 shows the relation between the temperature of external wall surface
and the quantity of heat release calculated with this formula.
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2. Energy Performance Assessment of Furnaces
Figure 2.6 Quantity of Heat Release at Various
Temperatures
From the Figure 2.6, the quantities of heat release from ceiling, sidewalls and hearth per unit
area can be found.
5a). Heat loss through roof and sidewalls:
Total average surface temperature
Heat loss at 122 °C
Total area of heating + soaking zone
Heat loss
Equivalent oil loss (a)
5b). Total average surface temperature of
area other than heating and soaking zone
Heat loss at 85°C
Total area
Heat loss
Equivalent oil loss (b)
Total loss of fuel oil
Total percentage loss
=
=
=
=
=
=
122°C
1252 kCal / m2 / hr
70.18 m2
1252 kCal / m2 / hr x 70.18 m2
87865 kCal/hr
8.78 kg / hr
=
=
=
=
=
=
=
=
=
85°C
740 kCal / m2 / hr
12.6 m2
740 kCal / m2 / hr x 12.6 m2
9324 kCal/hr
0.93 kg / hr
a + b = 9.71 kg/hr
9.71 / 368
2.64%
6. Unaccounted Loss
These losses comprise of heat storage loss, loss of furnace gases around charging door and
opening, heat loss by incomplete combustion, loss of heat by conduction through hearth, loss
due to formation of scales.
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2. Energy Performance Assessment of Furnaces
Furnace Efficiency (Direct Method)
Fuel input
= 400 litres / hr
= 368 kg/hr
Heat input
Heat output
= 368 x 10000 = 36,80,000 kCal
= m x Cp x ∆T
= 6000 kg x 0.12 x (1340 – 40)
= 936000 kCal
Efficiency
= 936000 x 100 / (368 x 10000)
= 25.43 %
= 25% (app)
Total Losses
=
75% (app)
Furnace Efficiency (Indirect Method)
1. Sensible heat loss in flue gas
2. Loss due to evaporation of moisture in fuel
3. Loss due to evaporation of water
formed from H2 in fuel
4. Heat loss due to openings
5. Heat loss through skin
= 57.29%
= 1.36 %
= 9.13 %
= 5.56 %
= 2.64%
Total losses
=
Furnace Efficiency
= 100 - 75.98
= 24.02 %
Specific Energy Consumption
= 400 litre /hour (fuel consumption)
6 Tonnes/hour
(Wt of stock)
= 66.6 Litre of fuel /tonne of Material (stock)
75.98%
2.5.4 Factors Affecting Furnace Performance
The important factors, which affect the efficiency, are listed below for critical analysis.
Under loading due to poor hearth loading and improper production scheduling
Improper Design
Use of inefficient burner
Insufficient draft/chimney
Absence of Waste heat recovery
Absence of Instruments/Controls
Improper operation/Maintenance
High stack loss
Improper insulation /Refractories
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2. Energy Performance Assessment of Furnaces
2.6
Data Collection Format for Furnace Performance Assessment
The field-testing format for data collection and parameter measurements are shown below
Stock
Charged amount in
furnace
Charging
temperature
Discharging
temperature
Discharge
material
Tons/hr
°C
°C
kg/ton
Fuel Analysis
Fuel
type
Consumption
Kg/hr
Components of heavy oil
Gross
Temperature
C H2 O2 N2 S Water calorific
content
value
% % % % %
%
kCal/kg
°C
Flue gas Analysis
Temperature
°C
Composition of dry exhaust gas
CO2
O2
CO
%
%
%
Cooling water
Amount of Water
kg/ton
Temperature of combustion air
Ambient air temperature
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Inlet temperature
°C
=
=
40
Outlet temperature
°C
2. Energy Performance Assessment of Furnaces
The Table 2.1 can be used to construct a heat balance for a typical heat treatment furnace
TABLE 2.1 HEAT BALANCE TABLE
Heat Input
Item
Heat output
kCal/t
%
Item
Combustion heat of fuel
kCal/t
%
Quantity of heat in steel
Sensible heat in flue gas
Moisture and
hydrogen loss of fuel
Heat loss by
Incomplete combustion
(CO loss)
Heat loss in
cooling water
Sensible heat
of scale
Heat Loss Due To
Openings
Radiation and Other
unaccounted heat loss
Total =
100%
Total =
100%
2.7 Useful Data
Radiation Heat Transfer
Heat transfer by radiation is proportional to the absolute temperature to the power 4.
Consequently the radiation losses increase considerably as temperature increases.
°C1
°C2
K1
(°C1 +273)
K2
(°C2 +273)
(K1/K2)4
Relative
Radiation
700
20
973
293
122
1.0
900
20
1173
293
255
2.1
1100
20
1373
293
482
3.96
1300
20
1573
293
830
6.83
1500
20
1773
293
1340
11.02
1700
20
1973
293
2056
16.91
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2. Energy Performance Assessment of Furnaces
In practical terms this means the radiation losses from an open furnace door at 1500°C are
11 times greater than the same furnace at 700°C. A good incentive for the iron and steel melters
is to keep the furnace lid closed at all times and maintaining a continuous feed of cold charge
onto the molten bath.
Furnace Utilization Factor
Utilization has a critical effect on furnace efficiency and is a factor that is often ignored or
under-estimated. If the furnace is at temperature then standby losses of a furnace occur whether
or not a product is in the furnace.
Standby Losses
Energy is lost from the charge or its enclosure in the way of heat: (a) conduction, (b) convection; or/and (c) radiation
Furnace Draft Control
Furnace pressure control has a major effect on fuel fired furnace efficiency. Running a furnace
at a slight positive pressure reduces air ingress and can increase the efficiency.
Theoretical Heat
Example of melting one tonne of steel from an ambient temperature of 20°C . Specific heat of
steel = 0.186 Wh/kg/°C, latent heat for melting of steel = 40 Wh/kg/°C. Melting point of steel
= 1600°C.
Theoretical Total heat = Sensible heat + Latent heat
Sensible Heat
=
1000 kg x 0.186 Wh /kg °C x (1600-20)°C = 294 kWh/T
Latent heat
=
40 Wh/ kg x 1000 kg
Total Heat
=
294 + 40 = 334 kWh/T
= 40 kWh/T
So the theoretical energy needed to melt one tonne of steel from 20°C = 334 kWh.
Actual Energy used to melt to 1600°C is 700 kWh
Efficiency = 334 kWh x 100 = 48%
700 kwh
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2. Energy Performance Assessment of Furnaces
Typical furnace efficiency for reheating and forging furnaces (As observed in few trials
undertaken by an Energy Auditing Agency on such furnaces)
Pusher Type Billet Reheating Furnace (for rolling mills)
Furnace
Capacity
Specific Fuel
Consumption
Thermal Efficiency
Achieved
Upto 6 T/hr
40-45 Ltrs/tonne
52%
7-8 T / hr
35-40 Ltrs/tonne
58.5%
10-12 T/hr
33-38 Ltrs/tonne
63%
15-20 T/hr
32-34 Ltrs/tonne
66.6%
20 T/hr & above
30-32 Ltrs/tonne
71%
Pusher type forging furnace
Furnace
Capacity
Specific Fuel
Consumption
Thermal Efficiency
Achieved
500-600 kg/hr
80-90 Ltrs/tonne
26%
1.0 T/hr
70-75 Ltrs/tonne
30%
1.5-2.0 T/hr
65-70 Ltrs/tonne
32.5%
2.5-3.0 T/hr
55-60 Ltrs/tonne
38%
The above fuel consumption figures were valid when the furnaces were found to be operating continuously at their rated capacity.
Note: These are the trial figures and cannot be presumed as standards for the furnaces
in question.
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2. Energy Performance Assessment of Furnaces
QUESTIONS
1)
What is a heating Furnace and give two examples?
2)
Define furnace efficiency.
3)
How do you determine the furnace efficiency by direct method?
4)
How do you determine the furnace efficiency by Indirect method?
5)
Between efficiency and specific energy consumption, which is a better mean of comparing furnaces?
6)
List down the various heat losses taking place in oil-fired furnace.
7)
What are the major factors affecting the furnace performance?
8)
Apart from the furnace operating parameters, energy auditor needs certain data from
reference book/manual for assessing furnace. Name few of them
9)
What will be the difference in approach for conducting efficiency testing of batch
and continuous type furnace?
10)
How will you measure the temperature of the stock inside the furnace?
REFERENCES
1.
2.
3.
4.
Handbook of Energy Conservation for Industrial Furnaces, Japan Industrial Furnace
Association.
Energy audit reports of National Productivity Council
Industrial Furnace, Volume 1 and Volume 2, John Wiley & Sons - Trinks
Improving furnace efficiency, Energy Management Journal
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3. ENERGY PERFORMANCE ASSESSMENT OF
COGENERATION AND TURBINES (GAS, STEAM)
3.1
Introduction
Cogeneration systems can be broadly classified as those using steam turbines, Gas turbines and
DG sets. Steam turbine cogeneration systems involve different types of configurations with
respect to mode of power generation such as extraction, back pressure or a combination of backpressure, extraction and condensing.
Gas turbines with heat recovery steam generators is another mode of cogeneration.
Depending on power and steam load variations in the plant the entire system is dynamic. A performance assessment would yield valuable insights into cogeneration system performance and
need for further optimisation.
3.2
Purpose of the Performance Test
The purpose of the cogeneration plant performance test is to determine the power output and
plant heat rate. In certain cases, the efficiency of individual components like steam turbine is
addressed specifically where performance deterioration is suspected. In general, the plant performance will be compared with the base line values arrived at for the plant operating condition rather than the design values. The other purpose of the performance test is to show the
maintenance accomplishment after a major overhaul. In some cases the purpose of evaluation
could even be for a total plant revamp.
3.3
Performance Terms and Definitions
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3. Energy Performance Assessment of Cogeneration and Turbine
kCal/kg
kCal/kg
3.4
Reference standards
Modern power station practices by British electricity International (Pergamon Press) ASME
PTC 22 - Gas turbine performance test.
3.5
Field Testing Procedure
The test procedure for each cogeneration plant will be developed individually taking into consideration the plant configuration, instrumentation and plant operating conditions. A method is
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3. Energy Performance Assessment of Cogeneration and Turbine
outlined in the following section for the measurement of heat rate and efficiency of a
co-generation plant. This part provides performance-testing procedure for a coal fired steam
based co-generation plant, which is common in Indian industries.
3.5.1 Test Duration
The test duration is site specific and in a continuous process industry, 8-hour test data should
give reasonably reliable data. In case of an industry with fluctuating electrical/steam load profile a set 24-hour data sampling for a representative period.
3.5.2 Measurements and Data Collection
The suggested instrumentation (online/ field instruments) for the performance measurement is
as under:
Steam flow measurement
Fuel flow measurements
Air flow / Flue gas flow
Flue gas Analysis
Unburnt Analysis
Temperature
Cooling water flow
Pressure
Power
Condensate
:
:
:
:
:
:
:
Orifice flow meters
Volumetric measurements / Mass flow meters
Venturi / Orifice flow meter / Ion gun / Pitot tubes
Zirconium Probe Oxygen analyser
Gravimetric Analysis
Thermocouple
Orifice flow meter / weir /channel flow/
non-contact flow meters
: Bourdon Pressure Gauges
: Trivector meter / Energy meter
: Orifice flow meter
It is essential to ensure that the data is collected during steady state plant running conditions.
Among others the following are essential details to be collected for cogeneration plant performance evaluation.
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3. Energy Performance Assessment of Cogeneration and Turbine
II. Electrical Energy:
1.
2.
3.
4.
5.
Total power generation for the trial period from individual turbines.
Hourly average power generation
Quantity of power import from utility ( Grid )*
Quantity of power generation from DG sets.*
Auxiliaries power consumption
* Necessary only when overall cogeneration plant adequacy and system optimization / upgradation are the objectives of the study.
3.5.3 Calculations for Steam Turbine Cogeneration System
The process flow diagram for cogeneration plant is shown in figure 3.1. The following calculation procedures have been provided in this section.
•
•
Turbine cylinder efficiency.
Overall plant heat rate
Figure 3.1 Process Flow Diagram for Cogeneration Plant
Step 1 :
Calculate the actual heat extraction in turbine at each stage,
Steam Enthalpy at turbine inlet
Steam Enthalpy at 1st extraction
Steam Enthalpy at 2nd extraction
Steam Enthalpy at Condenser
:
:
:
:
h1 kCal / kg
h2 kCal / kg
h3 kCal / kg
h4* kCal / kg
* Due to wetness of steam in the condensing stage, the enthalpy of steam cannot be considered
as equivalent to saturated steam. Typical dryness value is 0.88 – 0.92. This dryness value can be
used as first approximation to estimate heat drop in the last stage. However it is suggested to calculate the last stage efficiency from the overall turbine efficiency and other stage efficiencies.
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3. Energy Performance Assessment of Cogeneration and Turbine
Heat extraction from inlet
to stage –1 extraction (h5)
:
h1 – h2 kCal / kg
Heat extraction from
1st –2nd extraction (h6)
:
h2 – h3 kCal / kg
Heat extraction from 2nd
Extraction – condenser (h7)
:
h3 – h4 kCal / kg
Step 2:
From Mollier diagram (H-S Diagram) estimate the theoretical heat extraction for the conditions
mentioned in Step 1. Towards this:
a) Plot the turbine inlet condition point in the Mollier chart - corresponding to steam
pressure and temperature.
b) Since expansion in turbine is an adiabatic process, the entropy is constant. Hence draw
a vertical line from inlet point (parallel to y-axis) upto the condensing conditions.
c) Read the enthalpy at points where the extraction and condensing pressure lines meet
the vertical line drawn.
d) Compute the theoretical heat drop for different stages of expansion.
Theoretical Enthalpy after 1st extraction
Theoretical Enthalpy after 2nd extraction
Theoretical Enthalpy at condenser conditions
: H1
: H2
H3
Theoretical heat extraction from inlet to
stage 1 extraction, h8
: h1 – H1
Theoretical heat extraction from
1st – 2nd extraction, h9
: H1 – H2
Theoretical heat extraction from
2nd extraction – condensation, h10
: H2 – H3
Step 3 :
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3. Energy Performance Assessment of Cogeneration and Turbine
Step 4 :
Calculate plant heat rate*
M x (h1 – h11)
Heat rate, kCal / kWh
=
P
M – Mass flow rate of steam in kg/hr
h1 – Enthalpy of inlet steam in kCal/kg
h11 – Enthalpy of feed water in kCal/kg
P – Average Power generated in kW
*Alternatively the following guiding parameter can be utilised
Plant heat consumption = fuel consumed for power generation, kg/hr
Power generated, kW
3.6
Example
3.6.1 Small Cogeneration Plant
A distillery plant having an average production of 40 kilolitres of ethanol is having a cogeneration system with a backpressure turbine. The plant steam and electrical demand are 5.1
Tons/hr and 100 kW. The process flow diagram is shown in figure 3.2.Gross calorific value of
Indian coal is 4000kCal/kg
Figure 3.2 Process Flow Diagram for Small Cogeneration Plant
Calculations :
Step 1 :
Total heat of steam at turbine inlet conditions at 15kg / cm2 and 250°C, h1 =698 kCal/kg
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3. Energy Performance Assessment of Cogeneration and Turbine
Step 2 :
Total heat of steam at turbine outlet conditions at 2 kg/cm2 and 130°C, h2 = 648 kCal/kg
Step 3 :
Heat energy input to turbine per kg of inlet steam (h1– h2)
= (698-648) = 50 kCal/kg
Step 4 :
Total steam flow rate, Q1
Power generation
Equivalent thermal energy
= 5100 kg/hr
= 100 kW
= 100 x 860 = 86,000 kCal /hr
Step 5 :
Energy input to the turbine
= 5100 x 50 = 2,55,000 kCal/hr.
Step 6 :
Power generation efficiency of the turbo alternator =
Energy output
--------------------- x 100
Energy Input
=
Step 7 :
Efficiency of the turbo alternator
Efficiency of Alternator
Efficiency of gear transmission
86,000
------------- x 100 = 34%
2,55,000
= 34%
= 92 %
= 98 %
Step 8 :
Quantity of steam bypassing the turbine
= Nil
Step 9 :
Coal consumption of the boiler
= 1550 kg/hr.
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3. Energy Performance Assessment of Cogeneration and Turbine
Step 10:
Overall plant heat rate, kCal/kWh
= Mass flow rate of steam x ((Enthalpy of steam, kCal/kg – Enthalpy of feed water, kCal/kg)
Power output, kW
= 5100 x (698 – 30)
100
= 34068 kCal/kWh*
*Note: The plant heat rate is in the order of 34000 kCal/kWh because of the use of backpressure turbine. This value will be around 3000 kcal/kWh while operating on fully condensing
mode. However with backpressure turbine, the energy in the steam is not wasted, as it is utilised
in the process.
Overall plant fuel rate including boiler
= 1550/100
= 15.5 kg coal / kW
Analysis of Results:
The efficiency of the turbine generator set is as per manufacturer design specification. There is
no steam bypass indicating that the power generation potential of process steam is fully utilized.
At present the power generation from the process steam completely meets the process electrical demand or in other words, the system is balanced.
Remarks: Similar steps can be followed for the evaluation of performance of gas turbine
based cogeneration system.
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3. Energy Performance Assessment of Cogeneration and Turbine
QUESTIONS
1.
What is meant by plant heat rate? What is its significance?
2.
What is meant by turbine cylinder efficiency? How is it different from turbo-generator efficiency?
3.
What parameters should be monitored for evaluating the efficiency of the turbine?
4.
What is the need for performance assessment of a cogeneration plant?
5.
The parameters for back pressure steam turbine cogeneration plant is given below
T = 310°C, Q = 9000kg/hr
Inlet Steam: P =16 kg/cm2,
Outlet Steam: P = 5.0 kg/cm2, T = 235°C, Q = 9000kg/hr
Find out the turbine cylinder efficiency?
6.
7.
Explain why heat rate for back pressure turbine is greater than condensing turbine.
Explain the methodology of evaluating performance of a gas turbine with a heat
recovery steam generator.
REFERENCES
1.
2.
NPC report on 'Assessing cogeneration potential in Indian Industries'
Energy Cogeneration Handbook, George Polimeros, Industrial Press Inc.
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4. ENERGY PERFORMANCE ASSESSMENT
OF HEAT EXCHANGERS
4.1
Introduction
Heat exchangers are equipment that transfer heat from one medium to another. The proper
design, operation and maintenance of heat exchangers will make the process energy efficient
and minimize energy losses. Heat exchanger performance can deteriorate with time, off
design operations and other interferences such as fouling, scaling etc. It is necessary to
assess periodically the heat exchanger performance in order to maintain them at a high efficiency level. This section comprises certain proven techniques of monitoring the performance of heat exchangers, coolers and condensers from observed operating data of the
equipment.
4.2
Purpose of the Performance Test
To determine the overall heat transfer coefficient for assessing the performance of the heat
exchanger. Any deviation from the design heat transfer coefficient will indicate occurrence of
fouling.
4.3
Performance Terms and Definitions
Overall heat transfer coefficient, U
Heat exchanger performance is normally evaluated by the overall heat transfer coefficient U
that is defined by the equation
When the hot and cold stream flows and inlet temperatures are constant, the heat transfer
coefficient may be evaluated using the above formula. It may be observed that the heat pick up
by the cold fluid starts reducing with time.
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4. Energy Performance Assessment Of Heat Exchangers
Nomenclature
A typical heat exchanger is shown in figure 4.1 with nomenclature.
Heat duty of the exchanger can be calculated either on the hot side fluid or cold side fluid
as given below.
………..Eqn–1,
Heat Duty for Hot fluid, Qh = Wx Cph x (Ti–To)
………...Eqn–2
Heat Duty for Cold fluid, Qc = wx Cpc x ( to–ti)
If the operating heat duty is less than design heat duty, it may be due to heat losses, fouling
in tubes, reduced flow rate (hot or cold) etc. Hence, for simple performance monitoring of
exchanger, efficiency may be considered as factor of performance irrespective of other parameter. However, in industrial practice, fouling factor method is more predominantly used.
4.4
Methodology of Heat Exchanger Performance Assessment
4.4.1 Procedure for determination of Overall heat transfer Coefficient, U at field
This is a fairly rigorous method of monitoring the heat exchanger performance by calculating
the overall heat transfer coefficient periodically. Technical records are to be maintained for all
the exchangers, so that problems associated with reduced efficiency and heat transfer can be
identified easily. The record should basically contain historical heat transfer coefficient data
versus time / date of observation. A plot of heat transfer coefficient versus time permits rational planning of an exchanger-cleaning program.
The heat transfer coefficient is calculated by the equation
U = Q / (A x LMTD)
Where Q is the heat duty, A is the heat transfer area of the exchanger and LMTD is temperature driving force.
The step by step procedure for determination of Overall heat transfer Coefficient are
described below
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4. Energy Performance Assessment Of Heat Exchangers
Density and viscosity can be determined by analysis of the samples taken from the flow
stream at the recorded temperature in the plant laboratory. Thermal conductivity and specific
heat capacity if not determined from the samples can be collected from handbooks.
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4. Energy Performance Assessment Of Heat Exchangers
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4. Energy Performance Assessment Of Heat Exchangers
4.4.2 Examples
a. Liquid - Liquid Exchanger
A shell and tube exchanger of following configuration is considered being used for oil cooler with oil at the shell side and cooling water at the tube side.
Tube Side
• 460 Nos x 25.4mmOD x 2.11mm thick x 7211mm long
• Pitch - 31.75mm 30° triangular
• 2 Pass
Shell Side
• 787 mm ID
• Baffle space - 787 mm
• 1 Pass
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4. Energy Performance Assessment Of Heat Exchangers
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4. Energy Performance Assessment Of Heat Exchangers
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4. Energy Performance Assessment Of Heat Exchangers
Heat Duty: Actual duty differences will be practically negligible as these duty differences
could be because of the specific heat capacity deviation with the temperature. Also, there could
be some heat loss due to radiation from the hot shell side.
Pressure drop: Also, the pressure drop in the shell side of the hot fluid is reported normal
(only slightly less than the design figure). This is attributed with the increased average bulk
temperature of the hot side due to decreased performance of the exchanger.
Temperature range: As seen from the data the deviation in the temperature ranges could be
due to the increased fouling in the tubes (cold stream), since a higher pressure drop is noticed.
Heat Transfer coefficient: The estimated value has decreased due to increased fouling that has
resulted in minimized active area of heat transfer.
Physical properties: If available from the data or Lab analysis can be used for verification
with the design data sheet as a cross check towards design considerations.
Troubleshooting: Fouled exchanger needs cleaning.
b. Surface Condenser
A shell and tube exchanger of following configuration is considered being used for Condensing
turbine exhaust steam with cooling water at the tube side.
Tube Side
20648 Nos x 25.4mmOD x 1.22mm thk x 18300mm long
Pitch - 31.75mm 60° triangular
1 Pass
The monitored parameters are as below:
Parameters
Hot fluid flow, W
Cold fluid flow, w
Hot fluid Temp, T
Cold fluid Temp, t
Hot fluid Pressure, P
Cold fluid Pressure, p
Units
kg/h
kg/h
°C
°C
m Bar g
Bar g
Calculation of Thermal data:
Area = 27871 m2
1. Duty:
Q = qS + qL
Hot fluid, Q = 576990 kW
Cold Fluid, Q = 581825.5 kW
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Inlet
939888
55584000
No data
18
52.3 mbar
4
Outlet
939888
55584000
34.9
27
48.3
3.6
4. Energy Performance Assessment Of Heat Exchangers
2. Hot Fluid Pressure Drop
Pressure Drop = Pi – Po = 52.3 – 48.3 = 4.0 mbar.
3. Cold Fluid Pressure Drop
Pressure Drop = pi – po = 4 – 3.6 = 0.4 bar.
4. Temperature range hot fluid
Temperature Range ∆T = Ti– To = No data
5. Temperature Range Cold Fluid
Temperature Range ∆t = ti – to = 27 – 18 = 9 °C.
6. Capacity Ratio
Capacity ratio, R = Not significant in evaluation here.
7. Effectiveness
Effectiveness, S = (to – ti) / (Ti – ti) = Not significant in evaluation here.
8. LMTD
Calculated considering condensing part only
a). LMTD, Counter Flow = ((34.9 – 18)–(34.9–27))/ ln ((34.9–18)/(34.9–27))
= 11.8 deg C.
b). Correction Factor to account for Cross flow
F = 1.0.
9. Corrected LMTD
MTD = F x LMTD = 1.0 x 11.8 = 11.8 deg C.
10. Heat Transfer Co-efficient
Overall HTC, U = Q/ A ∆T = 576990/ (27871 x 11.8) = 1.75 kW/m2. K
Comparison of Calculated data with Design Data
Parameters
Duty, Q
Hot fluid side pressure drop, ∆Ph
Cold fluid side pressure drop, ∆Pc
Temperature Range hot fluid, ∆T
Temperature Range cold fluid, ∆t
Capacity ratio, R
Effectiveness, S
Corrected LMTD, MTD
Heat Transfer Coefficient, U
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Units
kW
mBar
Bar
°C
°C
--------°C
kW/(m2. K)
63
Test Data
576990
4 mbar
0.4
Design Data
588430
3.7 mbar
(27–18) = 9
(28–19) = 9
11.8
1.75
8.9
2.37
4. Energy Performance Assessment Of Heat Exchangers
Heat Duty: Actual duty differences will be practically negligible as these duty differences
could be because of the specific heat capacity deviation with the temperature. Also, there could
be some heat loss due to radiation from the hot shell side.
Pressure drop: The condensing side operating pressure raised due to the backpressure
caused by the non-condensable. This has resulted in increased pressure drop across the steam
side
Temperature range: With reference to cooling waterside there is no difference in the range
however, the terminal temperature differences has increased indicating lack of proper heat
transfer.
Heat Transfer coefficient: Heat transfer coefficient has decreased due to increased amount of
non-condensable with the steam.
Trouble shooting: Operations may be checked for tightness of the circuit and ensure
proper venting of the system. The vacuum source might be verified for proper
functioning.
C. Vaporizer
A shell and tube exchanger of following configuration is considered being used for vaporizing
chlorine with steam at the shell side.
Tube Side
200 Nos x 25.4mmOD x 1.22mm thick x 6000mm long
Pitch - 31.75mm 30° triangular
2 Pass
Area = 95.7.m2
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64
4. Energy Performance Assessment Of Heat Exchangers
The monitored parameters are as below:
Parameters
Hot fluid flow, W
Cold fluid flow, w
Hot fluid Temp, T
Cold fluid Temp, t
Hot fluid Pressure, P
Cold fluid Pressure, p
Units
kg/h
kg/h
°C
°C
Bar g
Bar g
Inlet
5015
43500
108
30
0.4
9
Outlet
5015
43500
108
34
0.3
8.8
Calculation of Thermal data:
1. Duty:
Q = qS + qL
Hot fluid, Q = 3130 kW
Cold Fluid, Q = qS + qL = 180.3 kW + 2948 kW = 3128.3 kW
2. Hot Fluid Pressure Drop
Pressure Drop = Pi – Po = 0.4 – 0.3 = 0.1 bar
3. Cold Fluid Pressure Drop
Pressure Drop = pi – po = 9 – 8.8 = 0.2 bar.
4. Temperature range hot fluid
Temperature Range ∆T = Ti – To = 0 °C
5. Temperature Range Cold Fluid
Temperature Range ∆t = ti – to = 34 – 30 = 4 °C.
6. Capacity Ratio
Capacity ratio, R = Not significant in evaluation here.
7. Effectiveness
Effectiveness, S = (to – ti) / (Ti – ti) = Not significant in evaluation here.
8. LMTD
Calculated considering condensing part only
a). LMTD, Counter Flow =((108 – 30)–(108–34))/ ln ((108–30)/(108–34)) = 76 °C.
b). Correction Factor to account for Cross flow
F = 1.0.
9. Corrected LMTD
MTD = F x LMTD = 1.0 x 76 = 76 °C.
10. Heat Transfer Co-efficient
Overall HTC, U = Q/ A ∆T = 3130/ (95.7 x 76) = 0.43 kW/m2. K
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65
4. Energy Performance Assessment Of Heat Exchangers
Comparison of Calculated data with Design Data
Parameters
Duty, Q
Hot fluid side pressure drop, ∆Ph
Cold fluid side pressure drop, ∆Pc
Temperature Range hot fluid, ∆T
Temperature Range cold fluid, ∆t
Capacity ratio, R
Effectiveness, S
Corrected LMTD, MTD
Heat Transfer Coefficient, U
Units
kW
Bar
Bar
°C
°C
--------°C
kW/(m2. K)
Test Data
3130
0.1
0.2
Design Data
3130
Neg
4
4
76
0.42
0.44
Heat Duty: There is no difference inferred from the duty as the exchanger is performing as per
the requirement
Pressure drop: The steam side pressure drop has increased in spite of condensation at the
steam side. Indication of non-condensable presence in steam side
Temperature range: No deviations
Heat Transfer coefficient: Even at no deviation in the temperature profile at the chlorine side,
heat transfer coefficient has decreased with an indication of overpressure at the shell side. This
indicates disturbances to the condensation of steam at the shell side. Non-condensable suspected at steam side.
Trouble shooting: Operations may be checked for presence of chlorine at the shell side
through tube leakages. Observing the steam side vent could do this. Alternately condensate pH
could be tested for presence of acidity.
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66
4. Energy Performance Assessment Of Heat Exchangers
d. Air heater
A finned tube exchanger of following configuration is considered being used for heating air
with steam in the tube side.
The monitored parameters are as below:
Parameters
Hot fluid flow, W
Cold fluid flow, w
Hot fluid Temp, T
Cold fluid Temp, t
Hot fluid Pressure, P
Units
kg/h
kg/h
°C
°C
Bar g
Inlet
3000
92300
150
30
Outlet
3000
92300
150
95
Cold fluid Pressure, p
mBar g
200 mbar
180 mbar
Calculation of Thermal data:
Bare tube Area = 42.8 m2; Fined tube area = 856 m2
1.Duty:
Hot fluid, Q = 1748 kW
Cold Fluid, Q = 1726 kW
2. Hot Fluid Pressure Drop
Pressure Drop = Pi – Po = Neg
3. Cold Fluid Pressure Drop
Pressure Drop = pi – po = 200–180 = 20 mbar.
4. Temperature range hot fluid
Temperature Range ∆T = Ti – To = Not required.
5. Temperature Range Cold Fluid
Temperature Range ∆t = ti – to = 95 – 30 = 65 °C.
6. Capacity Ratio
Capacity ratio, R = Not significant in evaluation here.
7. Effectiveness
Effectiveness, S = (to – ti) / (Ti – ti) = Not significant in evaluation here.
8. LMTD
Calculated considering condensing part only
a). LMTD, Counter Flow =((150 – 30)–(150–95)/ ln ((150–30)/(150–95)) = 83.3 °C.
b). Correction Factor to account for cross flow
F = 0.95
9. Corrected LMTD
MTD = F x LMTD = 0.95 x 83.3 = 79 °C.
10. Overall Heat Transfer Co-efficient (HTC)
U = Q/ A ∆T = 1748/ (856 x 79) = 0.026 kW/m2 . K
Bureau of Energy Efficiency
67
4. Energy Performance Assessment Of Heat Exchangers
Comparison of Calculated data with Design Data
Parameters
Duty, Q
Hot fluid side pressure drop, ∆Ph
Cold fluid side pressure drop, ∆Pc
Temperature Range hot fluid, ∆T
Temperature Range cold fluid, ∆t
Capacity ratio, R
Effectiveness, S
Corrected LMTD, MTD
Heat Transfer Coefficient, U
Units
Test Data
Design Data
kW
Bar
Bar
°C
°C
--------°C
kW/(m2. K)
1748
Neg
20
1800
Neg
15
65
65
79
0.026
79
0.03
Heat Duty: The difference inferred from the duty as the exchanger is under performing than
required
Pressure drop: The airside pressure drop has increased in spite of condensation at the steam
side. Indication of choking and dirt blocking at the airside.
Temperature range: No deviations
Heat Transfer coefficient: Decreased because of decreased fin efficiency due to choking on
air side.
Trouble shooting: Operations may be checked to perform pulsejet cleaning with steam / blow
air jet on air side if the facility is available. Mechanical cleaning may have to be planned during any down time in the immediate future.
Bureau of Energy Efficiency
68
4. Energy Performance Assessment Of Heat Exchangers
4.4.3
Instruments for monitoring:
The test and evaluation of the performance of the heat exchanger equipment is carried out by
measurement of operating parameters upstream and downstream of the exchanger. Due care
needs to be taken to ensure the accuracy and correctness of the measured parameter. The instruments used for measurements require calibration and verification prior to measurement.
Parameters
Fluid flow
Units
kg/h
Temperature
Pressure
°C
Bar g
Density
kg/m3
Viscosity
MpaS
Specific heat capacity
J/(kg.K)
Thermal conductivity
W/(m.K)
Composition+
%wt (or) % Vol
Instruments used
Flow can be measured with instruments like
Orifice flow meter, Vortex flow meter, Venturi
meters, Coriollis flow meters, Magnetic flow
meter
as applicable to the fluid service and flow ranges
Thermo gauge for low ranges, RTD, etc.
Liquid manometers, Draft gauge, Pressure
gauges Bourdon and diaphragm type, Absolute
pressure transmitters, etc.
Measured in the Laboratory as per ASTM
standards, hydrometer, etc
Measured in the Laboratory as per ASTM
standards, viscometer, etc.
Measured in the Laboratory as per ASTM
standards
Measured in the Laboratory as per ASTM
standards
Measured in the Laboratory as per ASTM
standards using Chemical analysis, HPLC, GC,
Spectrophotometer, etc.
4.4.4 Terminology used in Heat Exchangers
Terminology
Capacity ratio
Co current flow
exchanger
Counter flow
exchanger
Cross flow
Definition
Ratio of the products of mass flow rate and specific
heat capacity of the cold fluid to that of the hot fluid.
Also computed by the ratio of temperature range of the
hot fluid to that of the cold fluid.
Higher the ratio greater will be size of the exchanger
An exchanger wherein the fluid flow direction of the
cold and hot fluids are same
Exchangers wherein the fluid flow direction of the cold and
hot fluids are opposite. Normally preferred
An exchanger wherein the fluid flow direction of the
cold and hot fluids are in cross
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69
Unit
4. Energy Performance Assessment Of Heat Exchangers
Density
Effectiveness
Fouling
Fouling Factor
Heat Duty
Heat exchanger
Heat Flux
Heat transfer
Heat transfer
surface or heat
Transfer area
Individual
Heat transfer
Coefficient
It is the mass per unit volume of a material
Ratio of the cold fluid temperature range to that of
the inlet temperature difference of the hot and
cold fluid. Higher the ratio lesser will be requirement
of heat transfer surface
The phenomenon of formation and development of
scales and deposits over the heat transfer surface
diminishing the heat flux. The process of fouling will
get indicated by the increase in pressure drop
The reciprocal of heat transfer coefficient of the
dirt formed in the heat exchange process.
Higher the factor lesser will be the overall heat
transfer coefficient.
The capacity of the heat exchanger equipment
expressed in terms of heat transfer rate, viz.
magnitude of energy or heat transferred per time.
It means the exchanger is capable of performing at
this capacity in the given system
Refers to the nomenclature of equipment designed
and constructed to transmit heat content
(enthalpy or energy) of a comparatively high
temperature hot fluid to a lower temperature cold
fluid wherein the temperature of the hot fluid
decreases (or remain constant in case of losing latent
heat of condensation) and the temperature of the cold
fluid increases (or remain constant in case of gaining
latent heat of vaporisation). A heat exchanger will
normally provide indirect contact heating. E.g. A
cooling tower cannot be called a heat exchanger
where water is cooled by direct contact with air
The rate of heat transfer per unit surface of a heat
exchanger
The process of transport of heat energy from a
hot source to the comparatively cold surrounding
Refers to the surface area of the heat exchanger that
provides the indirect contact between the hot and cold
fluid in effecting the heat transfer. Thus the heat transfer
area is defined as the surface having both sides
wetted with one side by the hot fluid and the other
side by the cold fluid providing indirect contact for
heat transfer
The heat flux per unit temperature difference across
boundary layer of the hot / cold fluid film formed
at the heat transfer surface. The magnitude of heat
transfer coefficient indicates the ability of heat
conductivity of the given fluid. It increases with increase
in density, velocity, specific heat, geometry of the film
forming surface
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70
kg/m3
(m2.K)/W
W
W/ m2
m2
W/( m2.K)
4. Energy Performance Assessment Of Heat Exchangers
LMTD
Correction
factor
Logarithmic
Mean
Temperature
difference,
LMTD
Overall Heat
transfer
Coefficient
Pressure drop
Specific
heat capacity
Temperature
Approach
Temperature
Range
Terminal
temperature
Thermal
Conductivity
Viscosity
Calculated considering the Capacity and effectiveness
of a heat exchanging process. When multiplied with
LMTD gives the corrected LMTD thus accounting
for the temperature driving force for the cross flow
pattern as applicable inside the exchanger
The logarithmic average of the terminal temperature
approaches across a heat exchanger
°C
The ratio of heat flux per unit difference in approach
across a heat exchange equipment considering the
individual coefficient and heat exchanger metal surface
conductivity. The magnitude indicates the ability of
heat transfer for a given surface. Higher the coefficient
lesser will be the heat transfer surface requirement
The difference in pressure between the inlet and
outlet of a heat exchanger
The heat content per unit weight of any material per
degree raise/fall in temperature
The difference in the temperature between the hot and
cold fluids at the inlet / outlet of the heat exchanger.
The greater the difference greater will be heat
transfer flux
The difference in the temperature between the inlet
and outlet of a hot/cold fluid in a heat exchanger
The temperatures at the inlet / outlet of the hot / cold
fluid steams across a heat exchanger
The rate of heat transfer by conduction though any
substance across a distance per unit
temperature difference
The force on unit volume of any material that
will cause per velocity
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W/(m2.K)
Bar
J/(kg.K)
°C
°C
°C
W/(m2.K)
Pa
4. Energy Performance Assessment Of Heat Exchangers
QUESTIONS
1.
What is meant by LMTD ?
2.
Distinguish between heat exchanger efficiency and effectiveness.
3.
Explain the terms heat duty and capacity ratio.
4.
What is meant by fouling?
5.
List five heat exchangers used in industrial practice.
6.
What are the parameters, which are to be monitored for the performance assessment
of heat exchangers?
7.
In a heat exchanger the hot stream enters at 70°C and leaves at 55°C. On the other
side the cold stream enters at 30°C and leaves at 55°C. Find out the LMTD of the
heat exchanger.
8.
In a condenser what type of heats are considered in estimating the heat duty?
a) Latent Heat b) Sensible heat c) Specific heat d) Latent heat and sensible heat
9.
What is the need for performance assessment of a heat exchanger?
10.
The unit of overall coefficient of heat transfer is
a) kCal/hr/m2 °C b) kCal/kg °C c) kCal/m2 hr d) kCal/hg m2
REFERENCES
1.
2.
3.
4.
5.
6.
"Process Heat Transfer" by D.Q.Kern, Edn. 1965.
"Modern Power Station Practice" - British Electricity International- Volume - G;
Chapter - 7 - " Plant performance and performance monitoring.
Coulsons & Richardson's CHEMICAL ENGINEERING Volume 3 third edition
Scimod " Scientific Modeling Software", techno software International, India
Ganapathy. V, "Fouling factor estimated quickly", O&G Journal, Aug 1992.
Liberman, Norman P, Trouble shooting Process Operations, Penwell Books, Tulsa,
Oklahoma
Bureau of Energy Efficiency
72
5. ENERGY PERFORMANCE ASSESSMENT OF MOTORS
AND VARIABLE SPEED DRIVES
5.1
Introduction
The two parameters of importance in a motor are efficiency and power factor. The efficiencies of induction
motors remain almost constant between 50% to 100%
loading (Refer figure 5.1). With motors designed to
perform this function efficiently; the opportunity for
savings with motors rests primarily in their selection
and use. When a motor has a higher rating than that
required by the equipment, motor operates at part load.
In this state, the efficiency of the motor is reduced.
Replacement of under loaded motors with smaller
motors will allow a fully loaded smaller motor to operate at a higher efficiency. This arrangement is generally most economical for larger motors, and only when
they are operating at less than one-third to one-half
capacity, depending on their size.
5.2
Figure 5.1
Efficiency vs. Loading
Performance Terms and Definitions
Efficiency :
The efficiency of the motor is given by
Pout
Ploss
η = —— = 1 – ——
Pin
Pin
Where Pout – Output power of the motor
Pin – Input power of the motor
PLoss – Losses occurring in motor
Motor Loading :
Motor Loading % =
Bureau of Energy Efficiency
Actual operating load of the motor
x 100
Rated capacity of the motor
73
5. Energy Performance Assessment of Motors and Variable Speed Drives
5.3 Efficiency Testing
While input power measurements are fairly simple, measurement of output or losses need a laborious exercise with extensive testing facilities. The following are the testing standards widely used.
Europe: IEC 60034-2, and the new IEC 61972
US: IEEE 112 - Method B
Japan: JEC 37
Even between these standards the difference in efficiency value is up to 3%.
For simplicity nameplate efficiency rating may be used for calculations if the motor load is in
the range of 50 -100 %.
Field Tests for Determining Efficiency
(Note: The following section is a repeat of material provided in the chapter-2 on Electrical
Motors in Book-3.)
No Load Test :
The motor is run at rated voltage and frequency without any shaft load. Input power, current,
frequency and voltage are noted. The no load P.F. is quite low and hence low PF watt meters
are required. From the input power, stator I2R losses under no load are subtracted to give the
sum of Friction and Windage (F&W) and core losses. To separate core and F & W losses, test
is repeated at variable voltages. It is worthwhile plotting no-load input kW versus Voltage; the
intercept is F & W kW loss component.
F&W and core losses = No load power (watts) – (No load current)2 x Stator resistance
Stator and Rotor I2R Losses :
The stator winding resistance is directly measured by a bridge or volt amp method. The resistance must be corrected to the operating temperature. For modern motors, the operating temperature is likely to be in the range of 100°C to 120°C and necessary correction should be made.
Correction to 75°C may be inaccurate. The correction factor is given as follows :
R2
—–
R1
=
235 + t2
———– , where, t1 = ambient temperature, °C & t2 = operating temperature, °C.
235 + t1
The rotor resistance can be determined from locked rotor test at reduced frequency, but rotor
I2R losses are measured from measurement of rotor slip.
Rotor I2R losses = Slip x (Stator Input - Stator I2R Losses - Core Loss)
Accurate measurement of slip is possible by stroboscope or non-contact type tachometer. Slip
also must be corrected to operating temperature.
Stray Load Losses :
These losses are difficult to measure with any accuracy. IEEE Standard 112 gives a complicated
method, which is rarely used on shop floor. IS and IEC standards take a fixed value as 0.5 % of
Bureau of Energy Efficiency
74
5. Energy Performance Assessment of Motors and Variable Speed Drives
output. It must be remarked that actual value of stray losses is likely to be more. IEEE - 112 specifies values from 0.9 % to 1.8 %.
Motor Rating
Stray Losses
1 – 125 HP
125 – 500 HP
501 – 2499 HP
2500 and above
1.8 %
1.5 %
1.2 %
0.9 %
Points for Users :
It must be clear that accurate determination of efficiency is very difficult. The same motor tested by different methods and by same methods by different manufacturers can give a difference
of 2 %.
Estimation of efficiency in the field can be summarized as follows:
a) Measure stator resistance and correct to operating temperature. From rated current
value, I2R losses are calculated.
b) From rated speed and output, rotor I2R losses are calculated
c) From no load test, core and F & W losses are determined for stray loss
The method is illustrated by the following example :
Example :
Motor Specifications
Rated power
Voltage
Current
Speed
Insulation class
Frame
Connection
No load test Data
Voltage, V
Current, I
Frequency, F
Stator phase
resistance at 30°C
No load power, Pnl
=
=
=
=
=
=
=
34 kW/45 HP
415 Volt
57 Amps
1475 rpm
F
LD 200 L
Delta
=
=
=
415 Volts
16.1 Amps
50 Hz
=
=
0.264 Ohms
1063.74 Watts
a)
Calculate iron plus friction and windage losses
b)
Calculate stator resistance at 120°C
235 + t2
R2 = R1 x ————
235 + t1
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75
5. Energy Performance Assessment of Motors and Variable Speed Drives
c)
Calculate stator copper losses at operating temperature of resistance at 120°C
d)
Calculate full load slip(s) and rotor input assuming rotor losses are slip times rotor input.
e)
Determine the motor input assuming that stray losses are 0.5 % of the motor rated power
f)
Calculate motor full load efficiency and full load power factor
Solution
a)
Let Iron plus friction and windage loss, Pi + fw
No load power, Pnl = 1063.74 Watts
Stator Copper loss, P st-30°C (Pst.cu)
= 3 x (16.1 / √3)2 x 0.264
= 68.43 Watts
Pi + fw = Pnl - Pst.cu
= 1063.74 – 68.43
= 995.3 W
b)
Stator Resistance at 120°C,
120 + 235
R120°C = 0.264 x —————
30 + 235
= 0.354 ohms per phase
c)
Stator copper losses at full load, Pst.cu 120°C
= 3 x (57 / √3)2 x 0.354
= 1150.1 Watts
d)
Full load slip
S = (1500 – 1475) / 1500
= 0.0167
Rotor input, Pr
= Poutput/ (1-S)
= 34000 / (1-0.0167)
= 34577.4 Watts
e)
Motor full load input power, P input
= Pr + Pst.cu 120°C + (Pi + fw) + Pstray
= 34577.4 + 1150.1 + 995.3 + (0.005* x 34000)
= 36892.8 Watts
*
where, stray losses = 0.5% of rated output (assumed)
f)
Motor efficiency at full load
Poutput
Efficiency
= ——– x 100
Pinput
Bureau of Energy Efficiency
=
34000
——–
36892.8
=
92.2%
76
5. Energy Performance Assessment of Motors and Variable Speed Drives
=
Pinput
= —————–
= √3 x V x Ifl
Full Load PF
=
=
=
36892.8
——————–
√3 x 415 x 57
=
0.90
Comments :
a) The measurement of stray load losses is very difficult and not practical even on test beds.
b) The actual value of stray loss of motors up to 200 HP is likely to be 1 % to 3 % compared
to 0.5 % assumed by standards.
c) The value of full load slip taken from the nameplate data is not accurate. Actual measurement under full load conditions will give better results.
d) The friction and windage losses really are part of the shaft output; however, in the above
calculation, it is not added to the rated shaft output, before calculating the rotor input
power. The error however is minor.
e) When a motor is rewound, there is a fair chance that the resistance per phase would
increase due to winding material quality and the losses would be higher. It would be interesting to assess the effect of a nominal 10 % increase in resistance per phase.
5.4
Determining Motor Loading
1. By Input Power Measurements
•
First measure input power Pi with a hand held or in-line power meter
Pi = Three-phase power in kW
•
Note the rated kW and efficiency from the motor name plate
•
The figures of kW mentioned in the name plate is for output conditions.
So corresponding input power at full-rated load
Nameplate full rated kW
Pir
= ————————————————
ηfl
ηfl = Efficiency at full-rated load
Pir = Input power at full-rated load in kW
•
The percentage loading can now be calculated as follows
Pi
Load = — x 100%
Pir
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77
5. Energy Performance Assessment of Motors and Variable Speed Drives
Example
The nameplate details of a motor are given as power = 15 kW, efficiency η = 0.9. Using a power
meter the actual three phase power drawn is found to be 8 kW. Find out the loading of the motor.
Input power at full-rated power in kW, Pir
Percentage loading
= 15 /0.9
= 16.7 kW
= 8/16.7
= 48 %
2. By Line Current Measurements
The line current load estimation method is used when input power cannot be measured and only
amperage measurements are possible. The amperage draw of a motor varies approximately linearly with respect to load, down to about 75% of full load. Below the 75% load point, power
factor degrades and the amperage curve becomes increasingly non-linear. In the low load
region, current measurements are not a useful indicator of load. However, this method may
be used only as a preliminary method just for the purpose of identification of oversized motors.
% Load
=
Input load current
———————— *100 (Valid up to 75% loading)
Input rated current
3. Slip Method
In the absence of a power meter, the slip method can be used which requires a tachometer. This
method also does not give the exact loading on the motors.
Load
Slip
= —— *100%
Ss–Sr
Where:
Load = Output power as a % of rated power
Slip = Synchronous speed - Measured speed in rpm
Ss = Synchronous speed in rpm at the operating frequency
Sr = Nameplate full-load speed
Example: Slip Load Calculation
Given: Synchronous speed in rpm
(Synchronous speed
Nameplate full load speed
Measured speed in rpm
Nameplate rated power
= 1500 at 50 HZ operating frequency.
= 120f/P) f: frequency, P: Number of poles
= 1450
= 1480
= 7.5 kW
Determine actual output power.
1500 – 1480
Load = —————
1500 – 1450
*100% = 40%
From the above equation, actual output power would be 40% x 7.5 kW = 3 kW
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78
5. Energy Performance Assessment of Motors and Variable Speed Drives
The speed/slip method of determining motor part-load is often favored due to its simplicity
and safety advantages. Most motors are constructed such that the shaft is accessible to a
tachometer or a strobe light.
The accuracy of the slip method, however, is limited. The largest uncertainty relates to
the accuracy with which manufacturers report the nameplate full-load speed. Manufacturers
generally round their reported full-load speed values to some multiple of 5 rpm. While 5 rpm is
but a small percent of the full-load speed and may be considered as insignificant, the slip
method relies on the difference between full-load nameplate and synchronous speeds. Given a
40 rpm "correct" slip, a seemingly minor 5 rpm disparity causes a 12% change in calculated
load.
Slip also varies inversely with respect to the motor terminal voltage squared. A voltage correction factor can, also, be inserted into the slip load equation. The voltage compensated load
can be calculated as shown
Slip
Load = ———————– x 100%
(Ss – Sr) x (Vr/V)2
Where:
Load = Output power as a % of rated power
Slip
= Synchronous speed - Measured speed in rpm
Ss
= Synchronous speed in rpm
Sr
= Nameplate full-load speed
V
= RMS voltage, mean line to line of 3 phases
Vr
= Nameplate rated voltage
5.5
Performance Evaluation of Rewound Motors
Ideally, a comparison should be made of the efficiency before and after a rewinding. A
relatively simple procedure for evaluating rewind quality is to keep a log of no-load input
current for each motor in the population. This figure increases with poor quality rewinds. A
review of the rewind shop's procedure should also provide some indication of the quality of
work. When rewinding a motor, if smaller diameter wire is used, the resistance and the I2R
losses will increase.
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79
5. Energy Performance Assessment of Motors and Variable Speed Drives
5.6
Format for Data Collection
The motor loading survey can be performed using the format given below:
Motor Field Measurement Format
Company_________________________ Location_______________________
Date ________
Process________________________
Department_____________________
General Data
Driven Equipment__________________
Motor Operating Profile:
Motor Name Plate Data
Manufacturer ______________________
Model ___________________________
Serial Number _____________________
Type :Squirrel cage/Slp ring__________
Size (hp/kW)______________________
Synchronous Speed (RPM) ___________
Full-Load Speed (RPM) _____________
Voltage Rating _____________________
Full-Load Amperage ________________
Full-Load Power Factor (%) __________
Full-Load Efficiency (%) ____________
Temperature Rise __________________
Insulation Class ____________________
No of hours of operation
I Shift _____________
II Shift _____________
III Shift _____________
Annual Operating Time ______ hours/year
Type of load
1.Load is quite steady, motor "On" during shift
2.Load starts, stops, but is constant when "On"
3.Load starts, stops, and fluctuates when "On"
Stator resistance per phase =
Measured Data
Supply Voltage
By Voltmeter
VRY ________
V avg ______
VYB ________
VBR ________
Input Amps
By Ammeter
A a __________
A b __________ A avg ______
A c __________
Power Factor (PF) _____________________
Input Power (kW) ______________________
Yes ,if yes How many
times rewound ?-- No
Motor Loading %_________________
Motor Operating Speed ____________RPM
At frequency of __________
Driven Equipment Operating Speed
__________RPM
Type of Transmission (Direct/Gear/Fluid coupling)
From Test Certificate
Load
100%
75%
25%
No Load
Current
PF
Efficiency
Rewound
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5. Energy Performance Assessment of Motors and Variable Speed Drives
The monitoring format for rewound motor is given below:
5.7
Application of Variable Speed Drives (VSD)
Although there are many methods of varying the speeds of the driven equipment such as
hydraulic coupling, gear box, variable pulley etc., the most possible method is one of
varying the motor speed itself by varying the frequency and voltage by a variable
frequency drive.
5.7.1 Concept of Variable Frequency Drive
The speed of an induction motor is proportional to the frequency of the AC voltage applied to
it, as well as the number of poles in the motor stator. This is expressed by the equation:
RPM = (f x 120) / p
Where f is the frequency in Hz, and p is the number of poles in any multiple of 2.
Therefore, if the frequency applied to the motor is changed, the motor speed changes in
direct proportion to the frequency change. The control of frequency applied to the motor is the
job given to the VSD.
The VSD's basic principle of operation is to convert the electrical system frequency and voltage to the frequency and voltage required to drive a motor at a speed other than its rated speed.
The two most basic functions of a VSD are to provide power conversion from one frequency to
another, and to enable control of the output frequency.
VSD Power Conversion
As illustrated by Figure 5.1,
there are two basic components,
a rectifier and an inverter, to
accomplish power conversion.
The rectifier receives the
50-Hz AC voltage and converts it to direct current (DC)
voltage. A DC bus inside the
VSD functions as a "parking
lot" for the DC voltage. The
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Figure 5.1 Components of a Variable Speed Drive
81
5. Energy Performance Assessment of Motors and Variable Speed Drives
DC bus energizes the inverter, which converts it back to AC voltage again. The inverter can
be controlled to produce an output frequency of the proper value for the desired motor shaft
speed.
5.7.2 Factors for Successful Implementation of Variable Speed Drives
a) Load Type for Variable Frequency Drives
The main consideration is whether the variable frequency drive application require a variable
torque or constant torque drive. If the equipment being driven is centrifugal, such as a fan or
pump, then a variable torque drive will be more appropriate. Energy savings are usually the primary motivation for installing variable torque drives for centrifugal applications. For example,
a fan needs less torque when running at 50% speed than it does when running at full speed.
Variable torque operation allows the motor to apply only the torque needed, which results in
reduced energy consumption.
Conveyors, positive displacement pumps, punch presses, extruders, and other similar type
applications require constant level of torque at all speeds. In which case, constant torque variable frequency drives would be more appropriate for the job. A constant torque drive should
have an overload current capacity of 150% or more for one minute. Variable torque variable
frequency drives need only an overload current capacity of 120% for one minute since centrifugal applications rarely exceed the rated current.
If tight process control is needed, then you may need to utilize a sensor less vector, or flux
vector variable frequency drive, which allow a high level of accuracy in controlling speed,
torque, and positioning.
b) Motor Information
The following motor information will be needed to select the proper variable frequency drive:
Full Load Amperage Rating. Using a motor's horsepower is an inaccurate way to size variable frequency drives.
Speed Range. Generally, a motor should not be run at any speed less than 20% of its specified
maximum speed allowed. If it is run at a speed less than this without auxiliary motor cooling,
the motor will overheat. Auxiliary motor cooling should be used if the motor must be operated
at very slow speeds.
Multiple Motors. To size a variable frequency drive that will control more than one motor, add
together the full-load amp ratings of each of the motors. All motors controlled by a single drive
must have an equal voltage rating.
c) Efficiency and Power Factor
The variable frequency drive should have an efficiency rating of 95% or better at full load.
Variable frequency drives should also offer a true system power factor of 0.95 or better
across the operational speed range, to save on demand charges, and to protect the equipment
(especially motors).
d) Protection and Power Quality
Motor overload Protection for instantaneous trip and motor over current.
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5. Energy Performance Assessment of Motors and Variable Speed Drives
Additional Protection: Over and under voltage, over temperature, ground fault, control or
microprocessor fault. These protective circuits should provide an orderly shutdown of the VFD,
provide indication of the fault condition, and require a manual reset (except under voltage)
before restart. Under voltage from a power loss shall be set to automatically restart after return
to normal. The history of the previous three faults shall remain in memory for future review.
If a built-up system is required, there should also be externally-operated short circuit protection, door-interlocked fused disconnect and circuit breaker or motor circuit protector (MCP)
To determine if the equipment under consideration is the right choice for a variable speed
drive:
The load patterns should be thoroughly studied before exercising the option of VSD. In effect
the load should be of a varying nature to demand a VSD ( refer figure 5.3 & 5.4).
Figure 5.3
Example of an excellent variable
speed drive candidate
Figure 5.4
Example of a poor variable speed
drive candidate
The first step is to identify the number of operating hours of the equipment at various load
conditions. This can be done by using a Power analyzer with continuous data storage or by a
simple energy meter with periodic reading being taken.
5.7.3 Information needed to Evaluate Energy Savings for Variable Speed Application
1. Method of flow control to which adjustable speed is compared:
o output throttling (pump) or dampers (fan)
o recirculation (pump) or unrestrained flow (fan)
o adjustable-speed coupling (eddy current coupling)
o inlet guide vanes or inlet dampers (fan only)
o two-speed motor.
2.
Pump or fan data:
o head v's flow curve for every different type of liquid (pump) or gas (fan) that is
handled
o Pump efficiency curves.
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5. Energy Performance Assessment of Motors and Variable Speed Drives
3. Process information:
o specific gravity (for pumps) or specific density of products (for fans)
o system resistance head/flow curve
o equipment duty cycle, i.e. flow levels and time duration.
4. Efficiency information on all relevant electrical system apparatus:
o motors, constant and variable speed
o variable speed drives
o gears
o transformers.
If we do not have precise information for all of the above, we can make reasonable assumptions for points 2 and 4.
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5. Energy Performance Assessment of Motors and Variable Speed Drives
QUESTIONS
1)
Define motor efficiency.
2)
Why it is difficult to measure motor efficiency at site?
3)
Describe the various methods by which you calculate motor loading.
4)
If no instrument other than tachometer is available, what method you would suggest
for measuring the motor load?
5)
A 20 kW rated motor is drawing actual measured power of 14 kW. If the rated efficiency is 92%, determine the motor loading?
6)
What are the limitations of slip method in determining motor loading?
7)
A 4 pole motor is operating at a frequency of 50 Hz. Find the RPM of the motor?
8)
What are the two factors influencing the speed of induction motor?
9)
A fan's operating hours and loading are given below:
15 hours at 100% load
8 hours at 95% load
1 hour at 40% load
Is the application suitable candidate for application of VSD?
11)
The losses in a variable speed drive is
a) 12%
b) 8%
c) <5% d) no losses at all
REFERENCES
1.
2.
Motor challenge: Office of Industrial Technologies, Department of Energy, USA
Energy audit Reports of National Productivity Council
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6. ENERGY PERFORMANCE ASSESSMENT OF
FANS AND BLOWERS
6.1
Introduction
This section describes the method of testing a fan installed on site in order to determine the
performance of the fan in conjunction with the system to which it is connected.
6.2
Purpose of the Performance Test
The purposes of such a test are to determine, under actual operating conditions, the volume flow
rate, the power input and the total pressure rise across the fan.
These test results will provide actual value for the flow resistance of the air duct system,
which can be compared with the value specified by supplier.
6.3
Performance Terms and Definitions
Static Pressure:
The absolute pressure at a point minus the reference atmospheric pressure.
Dynamic Pressure: The rise in static pressure which occurs when air moving with specified
velocity at a point is bought to rest without loss of mechanical energy. It is also known as velocity pressure.
Total Pressure:
The sum of static pressures and dynamic pressures at a point.
Fan Shaft Power:
The mechanical power supplied to the fan shaft
Motor Input Power: The electrical power supplied to the terminals of an electric motor
drive.
6.4
Scope
The procedure describes field testing of centrifugal fans and blowers for assessing performance
and efficiency.
6.5
Reference Standards
British Standard, BS 848 - Fans for general purposes Part 1, Methods of testing performance
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6. Energy Performance Assessment of Fans and Blowers
6.6
Field Testing
6.6.1 Instruction for Site Testing
Before site tests are carried out, it should be ensured that:
•
•
Fan and its associated equipment are functioning properly, and at the rated speed
Operations are at stable conditions, e.g. steady temperatures, densities, system resistance
etc.
6.6.2
Location of Measurement Planes
General: The flow measurement plane shall be located in any suitable straight length,
(preferably on the inlet side of the fan) where the airflow conditions are substantially axial,
symmetrical and free from turbulence. Leakage of air from or into the air duct shall be negligible between the flow measuring plane and the fan. Bends and obstructions in an air duct can
disturb the airflow for a considerable distance downstream, and should be avoided for the purposes of the test.
Test length: That part of the duct in which the flow measurement plane is located, is termed
the 'test length' and shall be straight, of uniform cross section and free from any obstructions
which may modify the airflow. It shall have a length equal to not less than twice the equivalent diameter of the air duct (i.e. 2De). For rectangular duct, equivalent diameter, De is given by
2 LW/(L + W) where L, W is the length and width of the duct. For circular ducts De is the same
as diameter of the duct.
Inlet side of the fan: Where the 'test length' is on the inlet side of the fan, its downstream end
shall be at a distance from the fan inlet equal to atleast 0.75De. See figure 6.1. In the case of a
fan having an inlet box , the downstream end of the test length shall be at a distance from the
nearest part of the inlet cone of the fan equal to at least 0.75De.
Outlet side of the fan: Where the 'test length' is on the outlet side of the fan, the upstream
end of the 'test length' shall be at a distance from the fan outlet of at least 3De. See figure 6.2.
For this purpose, the fan outlet shall be considered as being the outlet of any expander on the
outlet side of the fan.
Location of the Flow Measurement Plane within the 'Test Length': The flow measurement plane shall be located within the 'test length' at a distance from the downstream end of the
'test length' equal to at least 1.25De.
Location of Pressure Measurement Plane: For the purpose of determining the pressure rise
produced by the fan, the static pressure shall be measured at planes on the inlet and/or the outlet side of the fan sufficiently close to it to ensure that the pressure losses between the measuring planes and the fan are calculable in accordance with available friction factor data without
adding excessively to the uncertainty of fan pressure determination.
If conveniently close to the fan, the 'test length' selected for air flow measurement should
also be used to pressure measurement. Other planes used for pressure measurement should be
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6. Energy Performance Assessment of Fans and Blowers
2
no closer than 0.25De from the fan inlet and no closer than 4De from the fan outlet. The plane
of pressure measurement should be selected at least 4De downstream of any bend, expander or
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6. Energy Performance Assessment of Fans and Blowers
obstruction which are likely to cause separated flow or otherwise interfere with uniformity of
pressure distribution.
6.6.3 Measurement of Air Velocity on Site
Velocity shall be measured by either pitot tube or a rotating vane anemometer. When in use, the
pitot tube shall be connected by means of airtight tubes to a pressure measuring instrument. The
anemometer shall be calibrated before the test.
Pitot Tube: In Figure 6.4, note that separate static connections (A) and total pressure connections (B) can be connected simultaneously across a manometer (C). Since the static pressure
is applied to both sides of the manometer, its effect is canceled out and the manometer indicates
only the velocity pressure.
In practice this type of measurement is usually made with a Pitot tube which incorporates both
static and total pressure sensors in a single unit. Essentially, a Pitot tube consists of an impact tube
(which receives total pressure input) fastened concentrically inside a second tube of slightly larger
diameter which receives static pressure input from radial sensing holes around the tip. The air space
between inner and outer tubes permits transfer of pressure from the sensing holes to the static pressure connection at the opposite end of the Pitot and then, through connecting tubing, to the low or
negative pressure side of a manometer. When the total pressure tube is connected to the high pressure side of the manometer, velocity pressure is indicated directly. See Figure 6.5.
To ensure accurate velocity pressure readings, the Pitot tube tip must be pointed directly into
(parallel with) the air stream. As the Pitot tube tip is parallel with the static pressure outlet tube,
the latter can be used as a pointer to align the tip properly. When the Pitot tube is correctly
aligned, the pressure indication will be maximum.
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6. Energy Performance Assessment of Fans and Blowers
Figure 6.4
Types of Pressure Measurement
Figure 6.5 Pitot tube senses total and static pressure. Manometer measures
velocity pressure (Difference between total and static pressures)
Traverse readings: In practical situations, the velocity of the air stream is not uniform across
the cross section of a duct. Friction slows the air moving close to the walls, so the velocity is
greater in the center of the duct.
To obtain the average total velocity in ducts of 100 mm diameter or larger, a series of
velocity pressure readings must be taken at points of equal area. A formal pattern of sensing
points across the duct cross section is recommended. These are known as traverse readings.
Figure 6.6 shows recommended Pitot tube locations for traversing round and rectangular
ducts.
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6. Energy Performance Assessment of Fans and Blowers
Figure 6.6
Traverse on Round and Square Duct Areas
In round ducts, velocity pressure readings should be taken at centers of equal concentric areas.
At least 20 readings should be taken along two diameters. In rectangular ducts, a minimum of 16
and a maximum of 64 readings are taken at centers of equal rectangular areas. Actual velocities for
each area are calculated from individual velocity pressure readings. This allows the readings and
velocities to be inspected for errors or inconsistencies. The velocities are then averaged.
By taking Pitot tube readings with extreme care, air velocity can be determined within an
accuracy of ± 2%. For maximum accuracy, the following precautions should be observed:
Example-Traverse point determination for round duct
Round duct: Let us calculate various traverse points for a duct of 1 m diameter. From Figure
6.4, for round duct of 1 m diameter (D). The radius, R is 0.5 m. The various points from the
port holes are given below:
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0.5 – 0.949 x 0.5
0.0255
0.5 – 0.837 x 0.5
0.0815
0.5 – 0.707 x 0.5
0.1465
0.5 – 0.548 x 0.5
0.226
0.5 – 0.316 x 0.5
0.342
0.5 + 0.316 x 0.5
0.658
0.5 + 0.548 x 0.5
0.774
0.5 + 0.707 x 0.5
0.8535
0.5 + 0.837 x 0.5
0.9185
0.5 + 0.949 x 0.5
0.9745
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6. Energy Performance Assessment of Fans and Blowers
Example-Traverse point determination for rectangular duct
Rectangular duct: For 1.4 m x 0.8 m rectangular duct, let us calculate the traverse points. 16
points are to be measured.
Dividing the area 1.4 x 0.8 = 1.12 m2 into 16 equal areas, each area is 0.07 m2. Taking
dimensions of 0.35 m x 0.20 m per area, we can now mark the various points in the rectangular duct as follows:
In small ducts or where traverse operations are otherwise impossible, an accuracy of ± 5%
can frequently be achieved by placing Pitot in center of duct.
Calculation of Velocity: After taking velocity pressures readings, at various traverse points,
the velocity corresponding to each point is calculated using the following expression.
Anemometer: The indicated velocity shall be measured at each traverse point in the cross
section by holding the anemometer stationary at each point for a period of time of not less than
1 minute. Each reading shall be converted to velocity in m/s and individually corrected in accordance with the anemometer calibration. The arithmetic mean of the corrected point velocities
gives the average velocity in the air duct and the volume flow rate is obtained by multiplying
the area of the air duct by the average velocity.
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6. Energy Performance Assessment of Fans and Blowers
6.6.4 Determination of Flow
Once the cross-sectional area of the duct is measured, the flow can be calculated as follows:
Flow, (m3/s) = Area (m2) x Velocity (m/s)
6.6.5 Determination of Fan Pressure
General: Precautions shall be taken so that the measurements of the static pressure on the
inlet and outlet sides of the fan are taken relative to the atmosphere pressure.
Measurement of Static Pressure: This shall be done by using a manometer in conjunction
with the static pressure connection of a pitot tube or a U tube manometer.
When using a pitot tube it is necessary to carry out a traverse in the pressure measurement
plane taking individual point pressure readings in a manner similar to that for determining flow
rate. In general, a smaller number of readings will be found adequate where individual readings
do not vary by more than 2% from each other. The average of all the individual readings shall
be taken as the static pressure of that section.
6.6.6 Determination of Power Input
Power Measurement: The power measurements can be done using a suitable clamp- on
power meter. Alternatively by measuring the amps, voltage and assuming a power factor of 0.9
the power can be calculated as below:
Transmission Systems: The interposition of a transmission system may be unavoidable
introducing additional uncertainties. The following values shall be used as a basis for transmission efficiency in the case of drives rated at 20 kW and above unless other reliable information
is available:
Properly lubricated precision spur gears
Flat belt drive
V-belt drive
98% for each step
97%
95%
Other Prime Movers: When the fan forms one unit with a non-electric prime mover it is recommended that the fuel consumption (oil, steam, compressed air etc.) should be specified and
determined in place of the overall power.
input to fan shaft in kW
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6. Energy Performance Assessment of Fans and Blowers
6.7
Example: Performance Test Report on Cooling Air Fan
The following is a typical report on measurements taken and calculations made for a
double inlet fan in a palletizing plant.
A. Design Parameters:
Volume
Static Pressure
=
=
292 m3/sec.
609.6 mmwc
=
=
32°C
740 RPM
B. Measurements:
Temperature
Speed
Inlet
Damper
Position
%
80%
Suction
Pressure
(-) mmwc
Outlet
Pressure
(+) mmwc
Measured
Velocity
∆ p),
Pressure (∆
mmwc
ONE SIDE
25, 22, 20
Average=22.33
455, 462,
480,478
Avg.=468.75
Average = 70
ANOTHER
SIDE
15, 18, 23, 21
Average=19.25
459, 464, 473
479, 480, 470
Avg.=470.83
Instruments used
a) Suction pressure, outlet pressure
b) For differential pressure
c) For temperature
d) Fan speed
e) Line current
C. Performance calculations:
a) Gas Density
=
(Corrected to NTP)
=
=
=
=
=
Volume
m3/Sec.
Amps
Power
(I)
Consumption
(kW)
166.6
220
Average = 70
'U' tube manometer
Inclined tube manometer
Mercury in glass thermometer
Tachometer
Tong tester
273 x 1.293
273 + T°C (at site condition)
=
273 x 1.293
273 + 32°C (at site condition)
=
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1.15 kg/m3
95
2127 KW
6. Energy Performance Assessment of Fans and Blowers
d) P = Power input to the fan shaft
=
Power input to the motor (kW) x Efficiency of motor (%) at the
operating load x transmission efficiency
Motor efficiency = 0.94
P
=
2263 x 0.94 x 1 (as motor was direct coupled)
=
2127 kW
Volume in m3 / Sec x total pressure in mmwc
e) Fan Efficiency %
=
102 x Power input to the shaft in (kW)
Where 102 is a conversion constant
For double inlet fan,
The total Volume of
air, m3 / Sec
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=
166.6 x 2 = 333.2
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6. Energy Performance Assessment of Fans and Blowers
Total static pressure,
mmwc (∆ pStatic,
across the fan)
=
Fan Efficiency =
Static Fan Efficiency
6.8
•
•
•
•
•
•
=
468.75 – (–22.33) = 491
333.2 x 491
102 x 2127
x
100
75%
Factors that Could Affect Performance
Leakage, re-circulation or other defects in the system;
Inaccurate estimation of flow resistance;
Erroneous application of the standardized test data;
Excessive loss in a system component located too close to the fan outlet;
Disturbance of the fan performance due to a bend or other system component located too
close to the fan inlet;
Error in site measurement
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6. Energy Performance Assessment of Fans and Blowers
QUESTIONS
1)
What is the relationship between static pressure, dynamic pressure and total pressure?
2)
For determining fan efficiency, why static pressure readings should be taken as close
to fan as possible?
3)
What is the significance of having traverse points in velocity measurement?
4)
What is fan efficiency?
5)
Determine various traverse points for a round duct of 0.5 m diameter.
6)
Why flow should not be measured very close to inlet and outlet of fan?
7)
Calculate the flow rate for the following data:
Diameter of duct: 0.5 m, differential pressure: 100mmWC,
Density of air at 0°C: 1.293, Temperature of air in the duct: 100°C, pitot coefficient:
0.85
8)
How many traverse points you would propose for a rectangular duct of 1 m x 1 m
dimensions?
9)
What are the various ways of measuring the flow?
10)
What are the various factors, which can affect fan performance?
REFERENCES
1.
2.
British Standard: BS 848 : Part 1 : 1980
Energy and Environmental Audit Reports of National Productivity Council
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7. ENERGY PERFORMANCE ASSESSMENT
OF WATER PUMPS
7.1
Introduction
Pumping is the process of addition of kinetic and potential energy to a liquid for the
purpose of moving it from one point to another. This energy will cause the liquid to do work
such as flow through a pipe or rise to a higher level. A centrifugal pump transforms mechanical energy from a rotating impeller into a kinetic and potential energy required by the
system.
The most critical aspect of energy efficiency in a pumping system is matching of pumps to
loads. Hence even if an efficient pump is selected, but if it is a mismatch to the system then the
pump will operate at very poor efficiencies. In addition efficiency drop can also be expected
over time due to deposits in the impellers. Performance assessment of pumps would reveal the
existing operating efficiencies in order to take corrective action.
7.2
•
•
7.3
Purpose of the Performance Test
Determination of the pump efficiency during the operating condition
Determination of system resistance and the operating duty point of the pump and compare
the same with design.
Performance Terms and Definitions
Pump Capacity, Q = Volume of liquid delivered by pump per unit time,m3/hr or m3/sec
Q is proportional to N, where N- rotational speed of the pump
Total developed head, H = The difference of discharge and suction pressure
The pump head represents the net work done on unit weights of a liquid in passing from
inlet of the pump to the discharge of the pump.
There are three heads in common use in pumps namely
(i) Static head
(ii) Velocity head
(iii) Friction head.
The frictional head in a system of pipes, valves and fittings varies as a function (roughly as
the square) of the capacity flow through the system.
System resistance: The sum of frictional head in resistance & total static head.
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7. Energy Performance Assessment of Water Pumps
Pump Efficiency: Fluid power and useful work done by the pump divided by the power input
in the pump shaft.
7.4
Field Testing for Determination of Pump Efficiency
To determine the pump efficiency, three key parameters are required: Flow, Head and Power.
Of these, flow measurement is the most crucial parameter as normally online flow meters are
hardly available, in a majority of pumping system. The following methods outlined below can
be adopted to measure the flow depending on the availability and site conditions.
7.4.1 Flow Measurement, Q
The following are the methods for flow measurements:
•
•
•
•
Tracer method BS5857
Ultrasonic flow measurement
Tank filling method
Installation of an on-line flowmeter
Tracer Method
The Tracer method is particularly suitable for cooling water flow measurement because of their
sensitivity and accuracy.
This method is based on injecting a tracer into the cooling water for a few minutes at an
accurately measured constant rate. A series of samples is extracted from the system at a point
where the tracer has become completely mixed with the cooling water. The mass flow rate is
calculated from:
qcw
where qcw
q1
C1
= q1 x C1/C2
= cooling water mass flow rate, kg/s
= mass flow rate of injected tracer, kg/s
= concentration of injected tracer, kg/kg
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7. Energy Performance Assessment of Water Pumps
C2
= concentration of tracer at downstream position during the 'plateau' period
of constant concentration, kg/kg
The tracer normally used is sodium chloride.
Ultrasonic Flow meter
Operating under Doppler effect principle these meters are non-invasive, meaning measurements
can be taken without disturbing the system. Scales and rust in the pipes are likely to impact the
accuracy.
•
•
•
Ensure measurements are taken in a sufficiently long length of pipe free from flow disturbance due to bends, tees and other fittings.
The pipe section where measurement is to be taken should be hammered gently to enable
scales and rusts to fall out.
For better accuracy, a section of the pipe can be replaced with new pipe for flow measurements.
Tank filing method
In open flow systems such as water getting pumped to an overhead tank or a sump, the flow
can be measured by noting the difference in tank levels for a specified period during which
the outlet flow from the tank is stopped. The internal tank dimensions should be preferable
taken from the design drawings, in the absence of which direct measurements may be
resorted to.
Installation of an on-line flowmeter
If the application to be measured is going to be critical and periodic then the best option would
be to install an on-line flowmeter which can get rid of the major problems encountered with
other types.
7.4.3 Determination of total head, H
Suction head (hs)
This is taken from the pump inlet pressure gauge readings and the value to be converted in to
meters (1kg/cm2 = 10. m). If not the level difference between sump water level to the centerline of the pump is to be measured. This gives the suction head in meters.
Discharge head (hd)
This is taken from the pump discharge side pressure gauge. Installation of the pressure gauge
in the discharge side is a must, if not already available.
7.4.4 Determination of hydraulic power (Liquid horse power),
Hydraulic power Ph(kW) = Q x (hd – hs) x ρ x g / 1000
Q = Volume flow rate (m3/s), ρ = density of the fluid (kg/m3), g = acceleration due to gravity
(m/s2), (hd - hs) = Total head in metres
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7. Energy Performance Assessment of Water Pumps
7.4.5 Measurement of motor input power
The motor input power Pm can be measured by using a portable power analyser.
7.4.6 Pump shaft power
The pump shaft power Ps is calculated by multiplying the motor input power by motor efficiency at the existing loading.
Ps = Pm x ηMotor
7.4.7 Pump efficiency
This is arrived at by dividing the hydraulic power by pump shaft power
ηPump =
Ph
Ps
Example of pump efficiency calculation
Illustration of calculation method outlined
A chemical plant operates a cooling water pump for process cooling and refrigeration
applications. During the performance testing the following operating parameters were
measured;
Measured Data
0.40 m3/ s
325 kW
+1 M
55 M
5M
88 %
Direct coupled
996 kg/ m3
Pump flow, Q
Power absorbed, P
Suction head (Tower basin level), h1
Delivery head, h2
Height of cooling tower
Motor efficiency
Type of drive
Density of water
Pump efficiency
0.40 m3/s
54 M
0.40 x 54 x 996 x 9.81/1000 = 211 kW
325 kW
(211 x 100) / 325 = 65 %
65/0.88 = 74 %
Flow delivered by the pump
Total head, h2 –(+h1)
Hydraulic power
Actual power consumption
Overall system efficiency
Pump efficiency
7.5
Determining the System resistance and Duty point
Determination of the system resistance curve and imposing the pump curve over it will give an
idea of the operating efficiency of the pump and also the drop in efficiencies when the system
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7. Energy Performance Assessment of Water Pumps
curve changes from normal / design. The example following from the earlier example outlines
the method of constructing a system curve.
Example:
Location of equipments
The Refrigeration plant is located at +0.00 level and the Process plant condensers are located at
+15 M level. One cooler having a design pressure drop of 1.9 kg/cm2 is located at the 0.00 level
(ground level). Other relevant data can be inferred from the earlier section. See schematic in
Figure 7.1.
The step-by-step approach for determining system resistance curve is given
below.
Step-1 Divide system resistance into Static and dynamic head
Find static head;
Static head (Condenser floor height) ; 15M
Find dynamic head;
Dynamic Head
Dynamic head
Bureau of Energy Efficiency
= Total Head – Static Head
= (54–15) = 39 M
103
7. Energy Performance Assessment of Water Pumps
Step-2 Check the maximum resistance circuit
Resistance in the different circuits is as under
S.no
System
Condenser loop
resistance, M
Reactor loop
resistance, M
Cooler loop
resistance, M
1.
Supply line from
pump
15
10
15
2.
Static head
15
5
Nil (cooler at
ground level)
3.
Equipment
5
5
19
4.
Return line from
equipment to CT
15
10
15
5.
Tower head
-
-
5
6.
Total
50
30
54
It can be noted that at full load the condenser and cooler circuits offer the maximum resistance to flow.
Step 3; Draw system resistance curve
Choose the condenser loop as it offers maximum resistance and is also having a static head
component
Static head: 15 M
Dynamic head at full load; 39 M
Compute system resistance at different flow rates
S.No. Flow (%)
Dynamic head
= 39x (%flow)2
Static head M
Total head M
1.
100
39
15
54
2.
75
21.9
15
36.9
3.
50
9.75
15
24.75
4.
25
2.44
15
17.44
Step 4 - Plot the system resistance against flow in the pump efficiency curves (see Figure 7.2)
provided by the vendor and compare actual operating duty point and see whether it operates at
maximum efficiency. In the example provided it is found that the pump system efficiency is
lower by 4 % due to change in operating conditions.
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7. Energy Performance Assessment of Water Pumps
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7. Energy Performance Assessment of Water Pumps
QUESTIONS
1)
How would you measure the flow by using tracer method?
2)
What are the various ways of measuring flow?
3)
A pump motor draws 75 A current. The voltage is 415 V. Assuming a power factor of
0.9. Calculate the power drawn?
4)
The suction head is 1m below the pump centerline. The discharge pressure shows 3
kg/cm2. The flow is calculated to be 100 m3/hr. Find out the pump efficiency.
5)
The pump efficiency is 70%. The hydraulic power is calculated to be 22 kW. Find
out the motor power required to drive the pump.
REFERENCES
1.
2.
Pump handbook by Karassik
Energy Audit Reports of National Productivity Council
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8. ENERGY PERFORMANCE ASSESSMENT
OF COMPRESSORS
8.1
Introduction
The compressed air system is not only an energy intensive utility but also one of the least energy efficient. Over a period of time, both performance of compressors and compressed air system reduces drastically. The causes are many such as poor maintenance, wear and tear etc. All
these lead to additional compressors installations leading to more inefficiencies. A periodic performance assessment is essential to minimize the cost of compressed air.
8.2
Purpose of the Performance Test
To find out:
•
•
•
•
Actual Free Air Delivery (FAD) of the compressor
Isothermal power required
Volumetric efficiency
Specific power requirement
The actual performance of the plant is to be compared with design / standard values for
assessing the plant energy efficiency.
8.3
Performance Terms and Definitions
8.4
Field Testing
8.4.1 Measurement of Free Air Delivery (FAD) by Nozzle method
Principle: If specially shaped nozzle discharge air to the atmosphere from a receiver getting
its supply from a compressor, sonic flow conditions sets in at the nozzle throat for a particular
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8. Energy Performance Assessment of Compressors
ratio of upstream pressure (receiver) to the downstream pressure (atmospheric) i.e. Mach number equals one.
When the pressure in the receiver is kept constant for a reasonable intervals of time, the airflow output of the compressor is equal to that of the nozzle and can be calculated from the
known characteristic of the nozzle.
8.4.2 Arrangement of test equipment
The arrangement of test equipment and measuring device shall confirm to Figure 8.1.
8.4.3 Nozzle Sizes
The following sizes of nozzles are recommended for the range of capacities indicated below:
Flow Nozzle: Flow nozzle with profile as desired in IS 10431:1994 and dimensions
Capacity (m3/hr)
Nozzle size (mm)
6
10
16
22
33
50
80
125
165
3–9
9 – 30
27 – 90
60 – 170
130 – 375
300 – 450
750 – 2000
1800 – 5500
3500 – 10000
8.4.4 Measurements and duration of the test.
The compressor is started with the air from the receiver discharging to the atmosphere through
the flow nozzle. It should be ensured that the pressure drop through the throttle valve should be
equal to or twice the pressure beyond the throttle. After the system is stabilized the following
measurements are carried out:
•
•
•
•
•
Receiver pressure
Pressure and temperature before the nozzle
Pressure drop across the nozzle
Speed of the compressor
kW, kWh and amps drawn by the compressor
The above readings are taken for the 40%, 60%, 100% and 110% of discharge pressure values.
Measuring instruments required for test
•
•
•
•
Thermometers or Thermocouple
Pressure gauges or Manometers
Differential pressure gauges or Manometers
Standard Nozzle
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108
8. Energy Performance Assessment of Compressors
•
•
•
8.5
Psychrometer
Tachometer/stroboscope
Electrical demand analyser
Calculation Procedure for Nozzle Method
k
d
T1
P1
P3
T3
Ra
P3–P4
:
:
:
:
:
:
:
:
Flow coefficient – as per IS
Nozzle diameter M
Absolute inlet temperature °K
Absolute inlet pressure kg/cm2
Absolute Pressure before nozzle kg/cm2
Absolute temperature before nozzle °K
Gas constant for air 287.1 J/kg k
Differential pressure across the nozzle kg/cm2
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8. Energy Performance Assessment of Compressors
II.
Isothermal Efficiency
=
Isothermal power/Input power
Isothermal power(kW)
=
P1 x Qf x loger
36.7
P1
Qf
r
III. Specific power consumption
=
=
=
Absolute intake pressure kg/ cm2
Free air delivered m3/hr.
Pressure ratio P2/P1
=
Power consumption ,kW
Free Air Delivered, m3/hr
at rated discharge pressure
IV. Volumetric efficiency
Free air delivered m3/min x 100
=
Compressor displacement, m3/min
Compressor Displacement
=
π
4
8.6
D
L
S
χ
=
=
=
=
n
=
x D2 x L x S x χ x n
Cylinder bore, metre
Cylinder stroke, metre
Compressor speed rpm
1 for single acting and
2 for double acting cylinders
No. of cylinders
Example
Calculation of Isothermal Efficiency for a Reciprocating Air Compressor.
Step – 1 : Calculate Volumetric Flow Rate
k
d
P2
P1
T1
P3
T3
P3 – P4
Ra
:
:
:
:
:
:
:
:
:
Flow coefficient (Assumed as 1)
Nozzle diameter : 0.08 metre
Receiver Pressure – 3.5 kg / cm2 (a)
Inlet Pressure – 1.04 kg / cm2(a)
Inlet air temperature 30°C or 303°K
Pressure before nozzle – 1.08 kg / cm2
Temperature before the nozzle 40°C or 313°K
Pressure drop across the nozzle = 0.036 kg / cm2
Gas constant : 287 Joules / kg K
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8. Energy Performance Assessment of Compressors
Step – 2 : Calculate Isothermal Power Requirement
Isothermal Power (kW)
=
P1 x Qf x loger
36.7
P1 - Absolute intake pressure
=
1.04 kg / cm2 (a)
Qf - Free Air Delivered
=
1407.6 m3 / h.
Compression ratio
r =
3.51
= 3.36
1.04
Isothermal Power
=
1.04 x 1407.6 x loge3.36
36.7
=
48.34 kW
Step – 3 : Calculate Isothermal Efficiency
8.7
Motor input power
Motor and drive efficiency
Compressor input power
=
=
=
100 kW
86 %
86 kW
Isothermal efficiency
=
Isothermal Power x 100
Compressor input Power
=
48.34 x 100
86.0
=
56%
Assessment of Specific Power requirement
Specific power consumption
=
Actual power consumed by the compressor
Measured Free Air Delivery
In the above example the measured flow is 1407.6 m3/hr and actual power consumption is
100 kW.
Specific power requirement =
100
1407.6
= 0.071 kW/m3/hr
8.8
Measurement of FAD by Pump Up Method
(Note: The following section is a repeat of material provided in the chapter-3 on Compressed
Air System in Book-3.)
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8. Energy Performance Assessment of Compressors
Another way of determining the Free Air Delivery of the compressor is by Pump Up
Method - also known as receiver filling method. Although this is less accurate, this can be
adopted where the elaborate nozzle method is difficult to be deployed.
Simple method of Capacity Assessment in Shop floor
•
•
•
•
•
Isolate the compressor along with its individual receiver being taken for test from main
compressed air system by tightly closing the isolation valve or blanking it, thus closing
the receiver outlet.
Open water drain valve and drain out water fully and empty the receiver and the pipeline.
Make sure that water trap line is tightly closed once again to start the test.
Start the compressor and activate the stopwatch.
Note the time taken to attain the normal operational pressure P2 (in the receiver) from initial pressure P1.
Calculate the capacity as per the formulae given below:
Actual Free air discharge
P2 – P1
Q
=
V
Nm3/Minute
X
P0
T
Where
P2
P1
P0
V
=
=
=
=
T
=
Final pressure after filling (kg/cm2 a)
Initial pressure (kg/cm2a) after bleeding
Atmospheric Pressure (kg/cm2 a)
Storage volume in m3 which includes receiver,
after cooler, and delivery piping
Time take to build up pressure to P2 in minutes
The above equation is relevant where the compressed air temperature is same as the ambient air temperature, i.e., perfect isothermal compression. In case the actual compressed air temperature at discharge, say t2°C is higher than ambient air temperature say t1°C (as is usual case),
the FAD is to be corrected by a factor (273 + t1) / (273 + t2).
EXAMPLE
An instrument air compressor capacity test gave the following results (assume the final compressed air temperature is same as the ambient temperature) - Comment?
Piston displacement
Theoretical compressor capacity
Compressor rated rpm 750
Receiver Volume
Additional hold up volume,
i.e., pipe / water cooler, etc., is
Total volume
Bureau of Energy Efficiency
:
:
:
:
16.88 m3/minute
14.75 m3/minute @ 7 kg/cm2
Motor rated rpm : 1445
7.79 m3
:
:
0.4974 m3
8.322 m3
112
8. Energy Performance Assessment of Compressors
Initial pressure P1
Final pressure P2
Atmospheric pressure P0
Time taken to buildup pressure
from P1 to P2
:
:
:
0.5 kg/cm2
7.03 kg/cm2
1.026 kg/cm2,a
:
4.021 minutes
(P2 – P1) × Total Volume
Compressor output m3/minute
:
:
Atm. Pressure × Pumpup time
(7.03 – 0.5) × 8.322
= 13.17 m3/minute
1.026 × 4.021
Capacity shortfall with respect to 14.75 m3/minute rating is 1.577 m3/minute i.e.,
10.69 %, which indicates compressor performance needs to be investigated further.
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8. Energy Performance Assessment of Compressors
QUESTIONS
1)
What is meant by Free Air Delivery?
2)
Describe the method of estimating flow by nozzle method.
3)
Describe the method of estimating flow by pump up method.
4)
Define the term isothermal efficiency and explain its significance.
5)
Define the term volumetric efficiency and explain its significance.
6)
How is specific power requirement calculated?
REFERENCES
1.
2.
3.
IS 10431:1994: Measurement of airflow of compressors and exhausters by nozzles.
IS 5456:1985 code of practice for testing of positive displacement type air compressors
and exhausters
Compressor performance – Aerodynamics for the user by M Theodore GreshButterworth Heinemann.
Bureau of Energy Efficiency
114
9. ENERGY PERFORMANCE ASSESSMENT
OF HVAC SYSTEMS
9.1
Introduction
Air conditioning and refrigeration consume significant amount of energy in buildings and in
process industries. The energy consumed in air conditioning and refrigeration systems is sensitive to load changes, seasonal variations, operation and maintenance, ambient conditions etc.
Hence the performance evaluation will have to take into account to the extent possible all these
factors.
9.2
Purpose of the Performance Test
The purpose of performance assessment is to verify the performance of a refrigeration system
by using field measurements. The test will measure net cooling capacity (tons of refrigeration)
and energy requirements, at the actual operating conditions. The objective of the test is to estimate the energy consumption at actual load vis-à-vis design conditions.
9.3
Performance Terms and Definitions
Tons of refrigeration (TR): One ton of refrigeration is the amount of cooling obtained by one
ton of ice melting in one day: 3024 kCal/h, 12,000 Btu/h or 3.516 thermal kW.
Net Refrigerating Capacity. A quantity defined as the mass flow rate of the evaporator water
multiplied by the difference in enthalpy of water entering and leaving the cooler, expressed in
kCal/h, tons of Refrigeration.
kW/ton rating: Commonly referred to as efficiency, but actually power input to compressor
motor divided by tons of cooling produced, or kilowatts per ton (kW/ton). Lower kW/ton indicates higher efficiency.
Coefficient of Performance (COP): Chiller efficiency measured in Btu output (cooling) divided by Btu input (electric power).
Energy Efficiency Ratio (EER): Performance of smaller chillers and rooftop units is frequently measured in EER rather than kW/ton. EER is calculated by dividing a chiller's cooling
capacity (in Btu/h) by its power input (in watts) at full-load conditions. The higher the EER, the
more efficient the unit.
9.4
Preparatory for Measurements
After establishing that steady-state conditions, three sets of data shall be taken, at a minimum
of five-minute intervals. To minimize the effects of transient conditions, test readings should be
taken as nearly simultaneously.
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9. Energy Performance Assessment of HVAC Systems
9.5 Procedure
9.5.1 To determine the net refrigeration capacity
The test shall include a measurement of the net heat removed from the water as it passes
through the evaporator by determination of the following:
a. Water flow rate
b. Temperature difference between entering and leaving water
The heat removed from the chilled water is equal to the product of the chilled water flow
rate, the water temperature difference, and the specific heat of the water is defined as
follows
The net refrigeration capacity in tons shall be obtained by the following equation:
The accurate temperature measurement is very vital in refrigeration and air conditioning and
least count should be at least one decimal.
Methods of measuring the flow
In the absence of an on-line flow meter the chilled water flow can be measured by the following methods
•
•
•
In case where hot well and cold well are available, the flow can be measured from the tank
level dip or rise by switching off the secondary pump.
Non invasive method would require a well calibrated ultrasonic flow meter using which
the flow can be measured without disturbing the system
If the waterside pressure drops are close to the design values, it can be assumed that the
water flow of pump is same as the design rated flow.
9.5.2 Measurement of compressor power
The compressor power can be measured by a portable power analyser which would give reading directly in kW.
If not, the ampere has to be measured by the available on-line ammeter or by using a tong
tester. The power can then be calculated by assuming a power factor of 0.9
Power (kW) = √3 x V x I x cosφ
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9. Energy Performance Assessment of HVAC Systems
9.5.3 Performance calculations
The energy efficiency of a chiller is commonly expressed in one of the three following
ratios:
First calculate the kW/ton rating from the measured parameters.
a) kW/ton rating = Measured compressor power, kW
Net refrigeration Capacity (TR)
Use this data to calculate other energy efficiency parameters with the following
relations
COP
= 0.293 EER
EER = 3.413 COP
kW/Ton = 12 / EER
EER = 12 / (kW/Ton)
kW/Ton = 3.516 / COP
COP = 3.516 / (kW/Ton)
* Source : American Refrigeration Institute
b) Coefficient of performance (COP)
=
3.516
kW/ton rating
c) Energy Efficiency Ratio (EER) =
12
kW/ton rating
9.5.4 Performance evaluation of air conditioning systems
For centralized air conditioning systems the air flow at the air handling unit (AHU) can be measured with an anemometer. The dry bulb and wet bulb temperatures can be measured at the
AHU inlet and outlet. The data can be used along with a psychrometric chart (Figure 9.1) to
determine the enthalpy (heat content of air at the AHU inelt and outlet)
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9. Energy Performance Assessment of HVAC Systems
Heat load (TR) = m x (hin – hout)
4.18 x 3024
m – mass flow rate of air, kg/hr
hin – enthalpy of inlet air at AHU, kJ/kg
hout – enthalpy of outlet air at AHU, kJ/kg
Heat load can also be calculated theoretically by estimating the various heat loads, both sensible and latent, in the air-conditioned room (refer standard air conditioning handbooks). The
difference between these two indicates the losses by way of leakages, unwanted loads, heat
ingress etc.
9.6
Measurements to be Recorded During the Test
All instruments, including gauges and thermometers shall be calibrated over the range of test
readings for the measurement of following parameters.
a.
b.
c.
d.
Evaporator
Temperature of water entering evaporator
Temperature of water leaving evaporator
Chilled water flow rates
Evaporator water pressure drop (inlet to outlet)
Compressor
e. Power input to compressor electrical power, kW
9.7
Example
In a brewery chilling system, ethylene glycol is used a secondary refrigerant. The designed
capacity is 40 TR. A test was conducted to find out the operating capacity and energy performance ratios. The flow was measured by switching off the secondary pump and measuring the
tank level difference in hot well.
Measurements data:
Temperature of ethylene glycol entering evaporator
Temperature of ethylene glycol leaving evaporator
Ethylene glycol flow rates
Evaporator ethylene glycol pressure drop (inlet to outlet)
Power input to compressor electrical power, kW
Specific heat capacity of ethylene glycol
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118
=
=
=
=
=
=
(-) 1°C
(-) 4°C
13200 kg/hr
0.7 kg/cm2
39.5 kW
2.34 kCal/kg°C
9. Energy Performance Assessment of HVAC Systems
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9. Energy Performance Assessment of HVAC Systems
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120
9. Energy Performance Assessment of HVAC Systems
QUESTIONS
1)
What is meant by a ton of refrigeration?
2)
Define the terms net refrigeration capacity, COP, energy efficiency ratio.
3)
What is the relation between COP and kW/ton of refrigeration?
4)
How would you calculate the heat load for a room to be air-conditioned?
5)
If the power consumed by a refrigerating unit / ton of refrigeration is 2 kW then find
energy efficiency ratio?
REFERENCES
1.
2.
Refrigeration and Air Conditioning by Richard C.Jordan & Gayle B.Priester - Prentice
Hall of India pvt.ltd.
Modern Air Conditioning Practice by Norman C.Harris - McGraw-Hill International
Edition.
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10. ENERGY PERFORMANCE ASSESSMENT OF LIGHTING
SYSTEMS
10.1
Introduction
Lighting is provided in industries, commercial buildings, indoor and outdoor for
providing comfortable working environment. The primary objective is to provide the
required lighting effect for the lowest installed load i.e highest lighting at lowest power
consumption.
10.2
Purpose of the Performance Test
Most interior lighting requirements are for meeting average illuminance on a horizontal plane,
either throughout the interior, or in specific areas within the interior combined with general
lighting of lower value.
The purpose of performance test is to calculate the installed efficacy in terms of
lux/watt/m² (existing or design) for general lighting installation. The calculated value can
be compared with the norms for specific types of interior installations for assessing
improvement options.
The installed load efficacy of an existing (or design) lighting installation can be assessed
by carrying out a survey as indicated in the following pages.
10.3
Performance Terms and Definitions
Lumen is a unit of light flow or luminous flux. The lumen
rating of a lamp is a measure of the total light output of the
lamp. The most common measurement of light output (or
luminous flux) is the lumen. Light sources are labeled with
an output rating in lumens.
Lux is the metric unit of measure for illuminance of a surface. One lux is equal to one lumen per square meter.
Circuit Watts is the total power drawn by lamps and ballasts in a lighting circuit under
assessment.
Installed Load Efficacy is the average maintained illuminance provided on a horizontal working plane per circuit watt with general lighting of an interior. Unit: lux per watt per square
metre (lux/W/m²)
Lamp Circuit Efficacy is the amount of light (lumens) emitted by a lamp for each watt of
power consumed by the lamp circuit, i.e. including control gear losses. This is a more
meaningful measure for those lamps that require control gear. Unit: lumens per circuit
watt (lm/W)
Installed Power Density. The installed power density per 100 lux is the power needed per
square metre of floor area to achieve 100 lux of average maintained illuminance on a horizonBureau of Energy Efficiency
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10. Energy Performance Assessment of Lighting Systems
tal working plane with general lighting of an interior. Unit: watts per square metre per 100 lux
(W/m²/100 lux)
100
Installed power density (W/m²/100 lux) = —————————————–
Installed load efficacy (lux/W/m²)
Installed Load Efficacy Ratio (ILER)
= Actual Lux/W/m²
——————— or
Target Lux/W/m²
Target W/m²/100lux
————————
Actual W/m²/100lux
Average maintained illuminance is the average of lux levels measured at various points in a
defined area.
Color Rendering Index (CRI) is a measure of the effect of light on the perceived color of objects.
To determine the CRI of a lamp, the color appearances of a set of standard color chips are measured
with special equipment under a reference light source with the same correlated color temperature as
the lamp being evaluated. If the lamp renders the color of the chips identical to the reference light
source, its CRI is 100. If the color rendering differs from the reference light source, the CRI is less
than 100. A low CRI indicates that some colors may appear unnatural when illuminated by the lamp.
10.4
Preparation (before Measurements)
Before starting the measurements, the following care should be taken:
•
•
All lamps should be operating and no luminaires should be dirty or stained.
•
Accuracies of readings should be ensured by
– Using accurate illuminance meters for measurements
– Sufficient number and arrangement of measurement points within the interior
– Proper positioning of illuminance meter
– Ensuring that no obstructions /reflections from surfaces affect measurement.
•
Other precautions
– If the illuminance meter is relatively old and has not been checked recently, it
should be compared with one that has been checked over a range of illuminances,
e.g. 100 to 600 lux, to establish if a correction factor should be applied.
– that the number and arrangement of measurement points are sufficient and
suitable to obtain a reasonably accurate assessment of the average illuminance
throughout an interior. The procedure recommended in the CIBSE Code for
such site measurements is as follows:
There should be no significant obstructions to the flow of light throughout the interior,
especially at the measuring points.
The interior is divided into a number of equal areas, which should be as square as possible.
The illuminance at the centre of each area is measured and the mean value calculated. This
gives an estimate of the average illuminance on the horizontal working plane.
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10. Energy Performance Assessment of Lighting Systems
10.5
Procedure for Assessment of Lighting Systems
10.5.1 To Determine the Minimum Number and Positions of Measurement Points
Calculate the Room Index: RI =
LxW
————–
Hm(L + W)
Where L = length of interior; W = width of interior; Hm = the mounting height, which
is the height of the lighting fittings above the horizontal working plane. The working plane
is usually assumed to be 0.75m above the floor in offices and at 0.85m above floor level in
manufacturing areas.
It does not matter whether these dimensions are in metres, yards or feet as long as the
same unit is used throughout. Ascertain the minimum number of measurement points from
Table10.1.
TABLE 10.1 DETERMINATION OF
MEASUREMENT POINTS
Room Index
Minimum number of
measurement points
Below 1
9
1 and below 2
16
2 and below 3
25
3 and above
36
To obtain an approximately "square array", i.e. the spacing between the points on
each axis to be approximately the same, it may be necessary to increase the number of
points.
For example, the dimensions of an interior are:
Length = 9m, Width = 5m, Height of luminaires above working plane (Hm) = 2m
Calculate RI =
9 x 5 = 1.607
2(9 + 5)
From Table 10.1 the minimum number of measurement points is 16
As it is not possible to approximate a "square array" of 16 points within such a rectangle it is necessary to increase the number of points to say 18, i.e. 6 x 3. These should be
spaced as shown below:
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10. Energy Performance Assessment of Lighting Systems
Therefore in this example the spacing between points along rows along the length of the
interior = 9 ÷ 6 = 1.5m and the distance of the 'end' points from the wall = 1.5 ÷ 2 = 0.75m.
Similarly the distance between points across the width of the interior = 5 ÷ 3 = 1.67m with
half this value, 0.83m, between the 'end' points and the walls.
If the grid of the measurement points coincides with that of the lighting fittings, large errors
are possible and the number of measurement points should be increased to avoid such an
occurrence.
10.5.2 Calculation of the Installed Load Efficacy and Installed Load Efficacy Ratio of a
General Lighting Installation in an Interior
STEP 1
Measure the floor area of the interior:
Area = -------------------- m²
STEP 2
Calculate the Room Index
RI
STEP 3
Determine the total circuit watts of the installation by a power
meter if a separate feeder for lighting is available. If the actual
value is not known a reasonable approximation can be obtained by
totaling up the lamp wattages including the ballasts:
Total circuit watts = --------
STEP 4
Calculate Watts per square metre, Value of step 3 ÷ value of step 1
W/m² =
STEP 5
Ascertain the average maintained illuminance by
using lux meter, Eav. Maintained
Eav.maint. = ----------------
STEP 6
Divide 5 by 4 to calculate lux per watt per square Metre
Lux/W/m² =
STEP 7
Obtain target Lux/W/m² lux for type of the type of
interior/application and RI (2):
Target Lux/W/m² =
Calculate Installed Load Efficacy Ratio ( 6 ÷ 7 ).
ILER =
STEP 8
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= -----------------------
----------------------
---------------
10. Energy Performance Assessment of Lighting Systems
TABLE 10.2 Target lux/W/m² (W/m²/100lux) values for
maintained illuminance on horizontal
plane for all room indices and applications:
Ra : Colour rendering index
The principal difference between the targets for Commercial and Industrial Ra: 40-85
(Cols.2 & 3) of Table 10.2 is the provision for a slightly lower maintenance factor for the latter. The targets for very clean industrial applications, with Ra: of 40 -85, are as column 2.
10.5.3 ILER Assessment
Compare the calculated ILER with the information in Table 10.3.
TABLE 10.3 INDICATORS OF
PERFORMANCE
ILER
Assessment
0.75 or over
Satisfactory to Good
0.51 – 0.74
Review suggested
0.5 or less
Urgent action required
ILER Ratios of 0.75 or more may be considered to be satisfactory. Existing installations
with ratios of 0.51 - 0.74 certainly merit investigation to see if improvements are possible. Of
course there can be good reasons for a low ratio, such as having to use lower efficacy lamps or
less efficient luminaires in order to achieve the required lighting result -but it is essential to
check whether there is a scope for a more efficient alternative. Existing installations with an
ILER of 0.5 or less certainly justify close inspection to identify options for converting the
installation to use more efficient lighting equipment.
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10. Energy Performance Assessment of Lighting Systems
Having derived the ILER for an existing lighting installation, then the difference between
the actual ILER and the best possible (1.0) can be used to estimate the energy wastage. For a
given installation:
Annual energy wastage (in kWh)
= (1.0 - ILER) x Total load (kW) x annual operating hours (h)
This process of comparing the installed load efficacy (ILE) with the target value for the
Room Index and type of application can also be used to assess the efficiency of designs for new
or replacement general lighting installations. If, when doing so, the calculated ILE (lux/W/m²)
is less than the target value then it is advisable to ascertain the reasons. It may be that the
requirements dictate a type of luminaire that is not as efficient as the best, or the surface
reflectances are less than the normal maxima, or the environment is dirty, etc., Whatever the
reasons, they should be checked to see if a more efficient solution is possible.
10.6
Example of ILER Calculation (for the room as mentioned in
paragraph 10.5.1)
STEP 1
Measure the floor area of the interior:
Area = 45 m²
STEP 2
Calculate the Room Index
RI
STEP 3
Determine the total circuit watts of the installation by a power
meter if a separate feeder for lighting is available. If the actual
value is not known a reasonable approximation can be obtained by
totaling up the lamp wattages including the ballasts:
= 1.93
Total circuit watts = 990 W
STEP 4
Calculate Watts per square metre, 3 ÷1 :
W/m² =
STEP 5
Ascertain the average maintained illuminance, Eav. Maintained
(average lux levels measured at 18 points)
Eav.maint. = 700
STEP 6
Divide 5 by 4 to calculate the actual lux per watt per square Metre
Lux/W/m² =
STEP 7
Obtain target Lux/W/m² lux for type of the type of interior/
application and RI (2):(Refer Table 10.2)
Target Lux/W/m² = 46
Calculate Installed Load Efficacy Ratio ( 6 ÷ 7 ).
ILER = 0.7
STEP 8
22
31.8
Referring to table 3, ILER of 0.7 means that there is scope for review of the lighting system.
Annual energy wastage = (1 - ILER) x watts x no. of operating hours
= (1 - 0.7) x 990 x 8 hrs/day x 300 days
= 712 kWh/annum
10.7
•
•
•
•
Areas for Improvement
Look for natural lighting opportunities through windows and other openings
In the case of industrial lighting, explore the scope for introducing translucent sheets
Assess scope for more energy efficient lamps and luminaries
Assess the scope for rearrangement of lighting fixtures
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10. Energy Performance Assessment of Lighting Systems
10.8
Other Useful Information
10.8.1 IES - Recommendations
The Illuminating Engineering Society (IES) has published illuminance recommendations for
various activities. These tables cover both generic tasks (reading, writing etc), and 100's of
very specific tasks and activities (such as drafting, parking, milking cows, blowing glass and
baking bread).
All tasks fall into 1 of 9 illuminance categories, covering from 20 to 20,000 lux, (2 to 2000
foot candles). The categories are known as A - I, and each provide a range of 3 iluminance values (low, mid and high). See Table 10.4.
TABLE 10.4 IES ILLUMINANCE CATEGORIES AND VALUES - FOR GENERIC INDOOR
ACTIVITIES
ACTIVITY
CATEGORY
LUX
FOOTCANDLES
Public spaces with dark surroundings
A
20-30-50
2-3-5
Simple orientation for short temporary visits
B
50-75-100
5-7.5-10
Working spaces where visual tasks are only
occasionally performed
C
100-150-200
10-15-20
Performance of visual tasks of high contrast
or large size
D
200-300-500
20-30-50
Performance of visual tasks of medium
contrast or small size
E
500-750-1000
50-75-100
Performance of visual tasks of low contrast
or very small size
F
1000-1500-2000
100-150-200
Performance of visual tasks of low contrast
or very small size over a prolonged period
G
2000-3000-5000
200-300-500
Performance of very prolonged and exacting
visual tasks
H
5000-7500-10000
500-750-1000
Performance of very special visual tasks of
extremely low contrast
I
10000-15000-20000
1000-1500-2000
A-C for illuminances over a large area (i.e. lobby space)
D-F for localized tasks
G-I for extremely difficult visual tasks
10.8.2 Example Using IES Recommendations
Let us determine the appropriate light level for a card file area in a library.
Step 1: The visual task is reading card files in a library. A number of tasks are accomplished in
the room. In such a cases, a category is chosen based on the generic descriptions in the IES
Illuminance Category and Illuminance table discussed in step 3. For example, offices will usually require Category E: 500-750-1000 lux.
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10. Energy Performance Assessment of Lighting Systems
Step 2: More detailed task descriptions are given in the recommended illuminance level tables
in the IES Handbook. (For an intensive lighting survey) Under the task category "Libraries,"
subheading "Card files," the illuminance category is E.
Step 3: From the IES Illuminance Category and Ranges table, find category E and choose 500-7501000 lux for the range of illuminance recommended. The first column in the table is illuminance
values in units of lux, the metric version of footcandle. Notice that categories A through C are for
general illumination throughout the area, but D through I are for illuminance on the task. Categories
G through I would require a combination of general lighting and task lighting.
Step 4: Use the weighting factors to decide which of the values in the illuminance range to use.
Since libraries are public facilities, there may be many individuals over 55 years of age so select
the category 'Over 55' for a weighting factor of +1.
Next, decide whether the demand for speed and accuracy is not important, important or critical. Filing of cards correctly is not a critical activity, so the weighting factor of zero (0) is
selected. An example of critical might be drafting work. The task background reflectance for
black type on a white page is 85%. So choose "greater than 70 percent" for a weighting factor
of -1. The total weighting factor is 0. So use the middle recommended illuminance, or 750 lux.
For more detailed information on this the IES handbook may be referred.
10.8.3 Characteristics of Different Types of Lamps
Type
of
Lamp
Lamp Lumens
Wattage
(Watts)
Lamp Efficiency
(Lumens/Watt)
Choke Life of Capacitor
Color
Rating Lamp
Rating
Rendering
(Watts) (Hours) Required
Index
(Micro
farads)
13
15000 - 0.2 - 0.39
12
20000
20
15000 - 0.2 - 0.39
20
20000
20
15000 - 0.2 - 0.39
32
20000
40
15000 - 0.2 - 0.39
45
20000
---------
HPSV
70
5600
80
HPSV
150
14000
93
HPSV
250
25000
100
HPSV
400
47000
118
HPSV
Super
HPSV
Super
HPSV
Super
HPSV
Super
HPSV
70
---
---
100
9500
95
18
150
15500
103
20
250
30000
120
25
400
54000
129
40
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15000 20000
15000 20000
15000 20000
15000 -
0.2 - 0.39
---
0.2 - 0.39
---
0.2 - 0.39
---
0.2 - 0.39
---
10. Energy Performance Assessment of Lighting Systems
Super
HPSV
Super
HPMV
20000
---
600
---
---
---
80
3400
43
9
HPMV
125
6300
50
12
HPMV
250
13000
52
16
HPMV
400
22000
55
25
Metal
Halide
Metal
Halide
Metal
Halide
Metal
Halide
Metal
Halide
FTL
FTL
Super
70
4200
84
26
4000 5000
4000 5000
4000 5000
4000 5000
10000
150
10500
70
20
250
19000
76
400
31000
1000
40
36
---
---
0.6 - 0.69
8
0.6 - 0.69
10
0.6 - 0.69
18
0.6 - 0.69
18
0.9 - 0.93
---
10000
0.9 - 0.93
---
25
10000
0.9 - 0.93
---
76
60
10000
0.9 - 0.93
---
80000
80
65
10000
0.9 - 0.93
---
2400
3250
60
90
15
5
4400
14000
0.8 - 0.89
0.8 - 0.89
3.2 - 3.8
3.2 - 3.8
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10. Energy Performance Assessment of Lighting Systems
QUESTIONS
1)
What is circuit watts?
2)
Define ILER and its significance.
3)
Distinguish between lux and lumens.
4)
What do you understand by the term colour rendering index?
5)
Define room index?
6)
For a room of length 10 m and width 20 m, calculate room index?
7)
For a room of 9 x 6 m, determine the appropriate number of measuring points for lux
levels?
8)
What possible improvement measures you would look for in a general lighting system?
9)
Which of the following lamps has the maximum lamp efficiency?
(lumes/Watt) a) Metal Hallide b) Fluorescent c) Incandescent d) HPSV
REFERENCES
1.
2.
Illumination engineering for energy efficient luminous environments by Ronald N.
Helms, Prentice-Hall, Inc.
The 'LIGHTSWITCH' programme, Energy Saving Trust, UK
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11. PERFORMING FINANCIAL ANALYSIS
11.1
Introduction
When planning an energy efficiency or energy management project, the costs involved
should always be considered. Therefore, as with any other type of investment, energy management proposals should show the likely return on any capital that is invested. Consider the
case of an energy auditor who advises the senior management of an organisation that capital should be invested in new boiler plant. Inevitably, the management of the organisation
would ask:
•
•
How much will the proposal cost?
How much money will be saved by the proposal?
These are, of course, not unreasonable questions, since within any organisation there are
many worthy causes, each of which requires funding and it is the job of senior management to
invest in capital where it is going to obtain the greatest return. In order to make a decision about
any course of action, management needs to be able to appraise all the costs involved in a project and determine the potential returns.
This however, is not quite as simple as it might first appear. The capital value of plant
or equipment usually decreases with time and it often requires more maintenance as it gets
older. If money is borrowed from a bank to finance a project, then interest will have to be
paid on the loan. Inflation too will influence the value of any future energy savings that
might be achieved. It is therefore important that the cost appraisal process allows for all
these factors, with the aim of determining which investments should be undertaken, and of
optimising the benefits achieved. To this end a number of accounting and financial appraisal
techniques have been developed which help energy managers and auditors make correct and
objective decisions.
The financial issues associated with capital investment in energy saving projects are investigated in this chapter. In particular, the discounted cash flow techniques of net present value
and internal rate of return are discussed in detail.
11.2
Fixed and Variable Costs
When appraising the potential costs involved in a project it is important to understand the difference between fixed and variable costs. Variable costs are those which vary directly with the
output of a particular plant or production process, such as fuel costs. Fixed costs are those costs,
which are not dependent on plant or process output, such as site-rent and insurance. The total
cost of any project is therefore the sum of the fixed and variable costs. Example 1 illustrates
how both fixed and variable costs combine to make the total operating cost.
Example 1
The capital cost of the DG set is Rs.9,00,000, the annual output is 219 MWh, and the maintenance cost is Rs.30,000 per annum. The cost of producing each unit of electricity is 3.50
Rs./kWh. The total cost of a diesel generator operating over a 5-year period, taking into consideration both fixed and variable cost is:
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11. Performing Financial Analysis
Item
Type of cost
Calculation
Cost
Capital cost of generator
Fixed
-
9,00,000
Annual maintenance
Fixed
30,000 x 5 (years)
1,50,000
Fuel cost
Variable
219,000 x 3.50 x 5
3,83,2500
Total cost
4,88,2500
From Example 1, it can be seen that the fixed costs represent 21.5% of the total cost. In fact,
the annual electricity output of 219 MWh assumes that the plant is operating with an average
output of 50 kW. If this output were increased to an average of 70 kW, then the fuel cost would
become Rs. 53,65,500, with the result that the fixed costs would drop to 16.37% of the total.
Thus the average unit cost of production decreases as output increases.
The concept of fixed and variable costs can be used to determine the break-even point
for a proposed project. The break-even point can be determined by using the following
equation.
Example 2
If the electricity bought from a utility company costs an average of Rs.4.5/kWh, the breakeven point for the generator described in Example 1, when the average output is 50 kW is
given by:
4.5 x 50 x n =
n =
(9,00,000 + 150000) + (3.5 x 50 x n)
21000 hours
If the average output is 70 kW, the break-even point is given by:
4.5 x 70 x n =
n =
(9,00,000 + 150000) + (3.50 x 70 x n)
15000 hours
Thus, increasing the average output of the generator significantly reduces the break-even
time for the project. This is because the capital investment (i.e. the generator) is being better
utilised.
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11. Performing Financial Analysis
11.3
Interest Charges
In order to finance projects, organizations often borrow money from banks or other leading
organizations. Projects financed in this way cost more than similar projects financed from
organisation's own funds, because interest charges must be paid on the loan. It is therefore
important to understand how interest charges are calculated. Interest charges can be calculated
by lending organization in two different ways: simple interest and compound interest.
(i) Simple interest: If simple interest is applied, then charges are calculated as a fixed percentage of the capital that is borrowed. A fixed interest percentage is applied to each year of
the loan and repayments are calculated using the equation.
(ii) Compound interest: Compound interest is usually calculated annually (although this is
not necessarily the case). The interest charged is calculated as a percentage of the outstanding
loan at the end of each time period. It is termed 'compound' because the outstanding loan is
the sum of the unpaid capital and the interest charges up to that point. The value of the total
repayment can be calculated using the equation.
Example 3
A company borrows Rs.3,00,00,00 to finance a new boiler installation. If the interest rate is
10% per annum and the repayment period is 5 years, let us calculate the value of the total
repayment and the monthly repayment value, assuming (i) simple interest and (ii) compound
interest.
(i) Assuming simple interest:
Total repayment
Monthly repayment
=
=
30,00,000 + (10/100 x 30,00,000 x 5) = Rs.45,00,000
45,00,000 / (5 x 12) = Rs.75,000
(ii) Assuming compound interest
Repayment at end of year 1
=
30,00,000 + (10/100 x 30,00,000) = Rs.33,00,000
Repayment at end of year 2
=
33,00,000 + (10/100 x 33,00,000) = Rs.36,30,000
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11. Performing Financial Analysis
Similarly, the repayments at the end of years 3, 4 and 5 can be calculated:
Repayment at end of year 3
=
Rs. 39,93,000
Repayment at end of year 4
=
Rs. 43,92,300
Repayment at end of year 5
=
Rs. 48,31530
Alternatively, the following equation can be used to determine the compound interest repayment value.
Total repayment value
=
Monthly repayment
=
30,00,000 x (1 + 10 / 100)5
=
Rs.48,31,530
4831530
= Rs.80,525
5 x 12
It can be seen that by using compound interest, the lender recoups an additional Rs.33,1530.
It is not surprisingly lenders usually charge compound interest on loans.
11.4
Simple Payback Period
This is the simplest technique that can be used to appraise a proposal. The simple payback period can be defined as 'the length of time required for the running total of net savings before
depreciation to equal the capital cost of the project'. In theory, once the payback period has
ended, all the project capital costs will have been recouped and any additional cost savings
achieved can be seen as clear 'profit'. Obviously, the shorter the payback period, the more
attractive the project becomes. The length of the maximum permissible payback period generally varies with the business culture concerned. In some companies, payback periods in excess
of 3 years are considered acceptable.
The payback period can be calculated using the equation
The annual net cost saving (AS) is the least savings achieved after all the operational costs have
been met. Simple payback period is illustrated in Example 4.
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11. Performing Financial Analysis
Example 4
A new small cogeneration plant installation is expected to reduce a company's annual energy
bill by Rs.4,86,000. If the capital cost of the new boiler installation is Rs.22,20,000 and the
annual maintenance and operating costs are Rs. 42,000, the expected payback period for the
project can be worked out as.
Solution
PB = 22,20,000 / (4,86,000 – 42,000) = 5.0 years
11. 5
Discounted Cash Flow Methods
The payback method is a simple technique, which can easily be used to provide a quick evaluation of a proposal. However, it has a number of major weaknesses:
•
•
The payback method does not consider savings that are accrued after the payback period
has finished.
The payback method does not consider the fact that money, which is invested, should
accrue interest as time passes. In simple terms there is a 'time value' component to cash
flows. Thus Rs.1000 today is more valuable than Rs.1000 in 10 years' time.
In order to overcome these weaknesses a number of discounted cash flow techniques have
been developed, which are based on the fact that money invested in a bank will accrue annual
interest. The two most commonly used techniques are the 'net present value' and the 'internal
rate of return' methods.
Net Present Value Method
The net present value method considers the fact that a cash saving (often referred to as a
'cash flow') of Rs.1000 in year 10 of a project will be worth less than a cash flow of Rs.1000
in year 2. The net present value method achieves this by quantifying the impact of time on
any particular future cash flow. This is done by equating each future cash flow to its current
value today, in other words determining the present value of any future cash flow. The present value (PV) is determined by using an assumed interest rate, usually referred to as a discount rate. Discounting is the opposite process to compounding. Compounding determines
the future value of present cash flows, where" discounting determines the present value of
future cash flows.
In order to understand the concept of present vale, consider the case described in Example 3.
If instead of installing a new cogeneration system, the company invested Rs.22,20,000 in a
bank at an annual interest rate of 8%, then:
The value of the sum at the end of year 1 = 22,20,000 + (0.08 x 22,20,000) = Rs.23,97,600
The value of the sum at the end of year 2 = 23,97,600 + (0.08 x 23,97,600) = Rs.25,89,408
The value of the investment would grow as compound interest is added, until after n years
the value of the sum would be:
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11. Performing Financial Analysis
Example :
The future value of the investment made at present, after 5 years will be:
FV = 22,20,000 x (1 + 8/100)5 = Rs.32,61,908.4
So in 5 years the initial investment of 22,20,000 will accrue Rs.10,41,908.4 in interest and will
be worth Rs.32,61,908.4. Alternatively, it could equally be said that Rs.32,61908.4 in 5 years
time is worth Rs.22,20,000 now (assuming an annual interest rate of 8%). In other words the
present value of Rs.32,61,908.40 in 5 years time is Rs.22,00,000 now.
The present value of an amount of money at any specified time in the future can be determined by the following equation.
The net present value method calculates the present value of all the yearly cash flows (i.e.
capital costs and net savings) incurred or accrued throughout the life of a project, and summates
them. Costs are represented as a negative value and savings as a positive value. The sum of all
the present values is known as the net present value (NPV). The higher the net present value,
the more attractive the proposed project.
The present value of a future cash flow can be determined using the equation above.
However, it is common practice to use a discount factor (DF) when calculating present value.
The discount factor is based on an assumed discount rate (i.e. interest rate) and can be determined by using equation.
DF = (1 + IR/100)–n
The product of a particular cash flow and the discount factor is the present value.
PV = S x DF
The values of various discount factors computed for a range of discount rates (i.e. interest rates)
are shown in Table 11.1. The Example 5 illustrates the process involved in a net present value
analysis.
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11. Performing Financial Analysis
TABLE 11.1COMPUTED DISCOUNT FACTORS
Discount rate % (or interest rate %)
Year
2
4
6
8
10
12
14
16
0
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1.000
1
0.980
0.962
0.943
0.926
0.909
0.893
0.877
0.862
2
0.961
0.825
0.890
0.857
0.826
0.797
0.769
0.743
3
0.942
0.889
0.840
0.794
0.751
0.712
0.675
0.641
4
0.924
0.855
0.792
0.735
0.683
0.636
0.592
0.552
5
0.906
0.822
0.747
0.681
0.621
0.567
0.519
0.476
6
0.888
0.790
0.705
0.630
0.564
0.507
0.456
0.410
7
0.871
0.760
0.665
0.583
0.513
0.452
0.400
0.354
8
0.853
0.731
0.627
0.540
0.467
0.404
0.351
0.305
9
0.837
0.703
0.592
0.500
0.424
0.361
0.308
0.263
10
0.820
0.676
0.558
0.463
0.386
0.322
0.270
0.227
11
0.804
0.650
0.527
0.429
0.350
0.287
0.237
0.195
12
0.788
0.625
0.497
0.397
0.319
0.257
0.208
0.168
13
0.773
0.601
0.469
0.368
0.290
0.229
0.182
0.145
14
0.758
0.577
0.442
0.340
0.263
0.205
0.160
0.125
15
0.743
0.555
0.417
0.315
0.239
0.183
0.140
0.108
16
0.728
0.534
0.394
0.292
0.218
0.163
0.123
0.093
17
0.714
0.513
0.371
0.270
0.198
0.146
0.108
0.080
18
0.700
0.494
0.350
0.250
0.180
0.130
0.095
0.069
19
0.686
0.475
0.331
0.232
0.164
0.116
0.083
0.060
20
0.673
0.456
0.312
0.215
0.149
0.104
0.073
0.051
Example 5
Using the net present value analysis technique, let us evaluate the financial merits of the proposed
projects shown in the Table below. Assume an annual discount rate of 8% for each project.
Project – 1
Project – 2
Capital cost (Rs.)
30 000.00
30 000.00
Year
Net annual saving (Rs.)
Net annual saving (Rs.)
1
+6 000.00
+6 600.00
2
+6 000.00
+6 600.00
3
+6 000.00
+6 300.00
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11. Performing Financial Analysis
4
+6 000.00
+6 300.00
5
+6 000.00
+6 000.00
6
+6 000.00
+6 000.00
7
+6 000.00
+5 700.00
8
+6 000.00
+5 700.00
9
+6 000.00
+5 400.00
10
+6 000.00
+5 400.00
Total net saving at
end of year 10
+60 000.00
+ 60 000.00
Solution
The annual cash flows should be multiplied by the annual discount factors for a rate of 8% to
determine the annual present values, as shown in the Table below:
Year
Discount
Factor for
8%
Net
savings
Project 1
Present
value (Rs.)
Net
savings
Project 2
Present
value (Rs.)
(a)
(Rs.)
(b)
(a x b)
(Rs.)
(a x c)
(c)
0
1.000
–30 000.00
–30 000.00
–30 000.00
–30 000.00
1
0.926
+6 000.00
+5 556.00
+6 600.00
+6 111.60
2
0.857
+6 000.00
+5 142.00
+6 600.00
+5 656.20
3
0.794
+6 000.00
+4 764.00
+6 300.00
+5 002.20
4
0.735
+6 000.00
+4 410.00
+6 300.00
+4 630.50
5
0.681
+6 000.00
+4 086.00
+6 000.00
+4 086.00
6
0.630
+6 000.00
+3 780.00
+6 000.00
+3 780.00
7
0.583
+6 000.00
+3 498.00
+5 700.00
+3323.10
8
0.540
+6 000.00
+3 240.00
+5 700.00
+3 078.00
9
0.500
+6 000.00
+3 000.00
+5 400.00
+2 700.00
10
0.463
+6 000.00
+2 778.00
+5 400.00
+2 500.20
NPV = +10 254.00
NPV = +10 867.80
It can be seen that over a 10 year life-span the net present value for Project 1 is
Rs.10,254.00, while for Project 2 it is Rs.10,867.80. Therefore Project 2 is the preferential
proposal.
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11. Performing Financial Analysis
The whole credibility of the net present value method depends on a realistic prediction of
future interest rates, which can often be unpredictable. It is prudent therefore to set the discount
rate slightly above the interest rate at which the capital for the project is borrowed. This will
ensure that the overall analysis is slightly pessimistic, thus acting against the inherent uncertain
ties in predicting future savings.
Internal rate of return method
It can be seen from Example 5 that both projects returned a positive net present value over 10
years, at a discount rate of 8%. However, if the discount rate were reduced there would come a
point when the net present value would become zero. It is clear that the discount rate which
must be applied, in order to achieve a net present value of zero, will be higher for Project 2 than
for Project 1. This means that the average rate of return for Project 2 is higher than for Project
1, with the result that Project 2 is the better proposition.
Example 6 illustrates how an internal rate of return analysis is performed.
Example 6
A proposed project requires an initial capital investment of Rs.20 000. The cash flows generated by the project are shown in the table below:
Year
Cash flow (Rs.)
0
–20,000.00
1
+6000.00
2
+5500.00
3
+5000.00
4
+4500.00
5
+4000.00
6
+4000.00
Given the above cash flow data, let us find out the internal rate of return for the project.
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11. Performing Financial Analysis
Solution
Cash
flow
(Rs.)
(Rs.)
8% discount rate
Discount
Present
factor
value
(Rs.)
12% discount rate
Discount
Present
factor
value
16% discount rate
Discount
Present
factor
value
(Rs.)
0
–20000
1.000
–20000
1.000
–20000
1.000
–20000
1
6000
0.926
5556
0.893
5358
0.862
5172
2
5500
0.857
4713.5
0.797
4383.5
0.743
4086.5
3
5000
0.794
3970
0.712
3560
0.641
3205
4
4500
0.735
3307.5
0.636
3862
0.552
2484
5
4000
0.681
2724
0.567
2268
0.476
1904
6
4000
0.630
2520
0.507
2028
0.410
1640
NPV = 2791
NPV = 459.5
NPV = –1508.5
It can clearly be seen that the discount rate which results in the net present value being zero
lies somewhere between 12% and 16%.
For12% discount rate, NPV is positive; for 16% discount rate, NPV is negative. Thus
for some discount rate between 12 and 16 percent, present value benefits are equated to
present value costs. To find the value exactly, one can interpolate between the two rates
as follows:
459.5
Internal rate of return
=
x 100
0.12 + (0.16 – 0.12) x
(459.5 – (–1508.5))
459.5
Internal rate of return
=
x 100 = 12.93%
0.12 + (0.16 – 0.12) x
(459.5 + 1508.5)
Thus the internal rate of return for the project is 12.93 %. At first sight both the net present
value and internal rate of return methods look very similar, and in some respects are. Yet there
is an important difference between the two. The net present value method is essentially a comparison tool, which enables a number of projects to be compared, while the internal rate of
return method is designed to assess whether or not a single project will achieve a target rate of
return.
Profitability index
Another technique, which can be used to evaluate the financial viability of projects, is the profitability index. The profitability index can be defined as:
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11. Performing Financial Analysis
The application of profitability index is illustrated in Example 7.
Example 7
Determine the profitability index for the projects outlined in Example 5
10254
For Project 1: Profitability index
=
= 0.342
30,000
10867
For Project 2: Profitability index
=
= 0.362
30,000
Project 2 is therefore a better proposal than Project 1.
11.6
Factors Affecting Analysis
Although the Examples 5 and 6 illustrate the basic principles associated with the financial
analysis of projects, they do not allow for the following important considerations:
•
•
The capital value of plant and equipment generally depreciates over time
General inflation reduces the value of savings as time progresses. For example, Rs.1000
saved in 1 year's time will be worth more than Rs.1000 saved in 10 years time.
The capital depreciation of an item of equipment can be considered in terms of its salvage
value at the end of the analysis period. The Example 8 illustrates the point.
Example 8
It is proposed to install a heat recovery equipment in a factory. The capital cost of installing the
equipment is Rs.20,000 and after 5 years its salvage value is Rs.1500. If the savings accrued by
the heat recovery device are as shown below, we have to find out the net present value after
5 years. Discount rate is assumed to be 8%.
Data
Year
1
2
3
4
5
7000
6000
6000
5000
5000
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11. Performing Financial Analysis
Solution
Year
Discount
Factor for
8%
(a)
Capital
Investment
(Rs.)
(b)
0
1,000
–20,000.00
1
0.926
+7000.00
+6482.00
2
0.857
+6000.00
+5142.00
3
0.794
+6000.00
+4764.00
4
0.735
+6000.00
+3675.00
5
0.681
+5000.00
+4426.50
+1,500.00
Net
Savings
(Rs.)
(c)
Present
Value
(Rs.)
(a) x (b + c)
–20,000.00
NPV = +4489.50
It is evident that over a 5-year life span the net present value of the project is Rs.4489.50.
Had the salvage value of the equipment not been considered, the net present value of the project would have been only Rs.3468.00.
Real value
Inflation can be defined as the rate of increase in the average price of goods and services. In
some countries, inflation is expressed in terms of the retail price index (RPI), which is determined centrally and reflects average inflation over a range of commodities. Because of inflation, the real value of cash flow decreases with time. The real value of sum of money (S)
realised in n years time can be determined using the equation.
RV = S x (1 + R/100)–n
Where RV is the real value of S realized in n years time. S is the value of cash flow in n
years time and R is the inflation rate (%).
As with the discount factor it is common practice to use an inflation factor when assessing
the impact of inflation on a project. The inflation factor can be determined using the equation.
IF = (1 + R/100)–n
The product of a particular cash flow and inflation factor is the real value of the cash flow.
RV = S x IF
The application of inflation factors is considered in Example 9.
Example 9
Recalculate the net present value of the energy recovery scheme in Example 8, assuming the
discount rate remains at 8% and that the rate of inflation is 5%.
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11. Performing Financial Analysis
Solution
Because of inflation; Real interest rate = Discount rate – Rate of inflation
Therefore Real interest rate = 8 – 5 = 3%
Year
Capital
Investment
(Rs.)
0
–20,000.00
Net real
Savings
(Rs.)
Inflation
Factor
For 5%
Net real
Savings
(Rs.)
Real
Discount
Factor
For 3%
Present
Value
(Rs.)
1.000
–20,000.00
1.000
–20,000.00
1
+7000.00
0.952
+6664.00
0.971
+6470.74
2
+6000.00
0.907
+5442.00
0.943
+5131.81
3
+6000.00
0.864
+5184.00
0.915
+4743.36
4
+5000.00
0.823
+4145.00
0.888
+3654.12
+5000.00
0.784
+5096.00
0.863
+4397.85
5
+1500.00
NPV = +4397.88
The Example 9 shows that when inflation is assumed to be 5%, the net present value of the
project reduces from Rs.4489.50 to Rs.4397.88. This is to be expected, because general inflation will always erode the value of future 'profits' accrued by a project.
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11. Performing Financial Analysis
QUESTIONS
1.
Why fresh investments are needed for energy conservation in industry ?
2.
Cost of an heat exchanger is Rs.1.00 lakhs. Calculate simple pay back period considering annual saving potential of Rs.60,000/- and annual operating cost of
Rs.15,000/-.
3.
What is the main draw back of simple pay back method?
4.
Calculate simple pay back period for a boiler that cost Rs.75.00 lakhs to purchase
and Rs.5 lakhs per year on an average to operate and maintain and is expected to
save Rs.30 lakhs.
5.
What are the advantages of simple pay back method?
6.
What do you understand by the term " present value of money"?
7.
Define ROI.
8.
What is the objective of carrying out sensitivity analysis?
9.
You are investing Rs.100 in a bank. The bank gives 10% interest per year for two
years. What is the present value and what is the future value?
REFERENCES
1.
Energy Management, Supply and Conservation, Dr. Clive Beggs, .Butterworth
Heinemann
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12. APPLICATION OF NON-CONVENTIONAL &
RENEWABLE ENERGY SOURCES
12.1
Concept of Renewable Energy
Renewable energy sources also called non-conventional energy, are sources that are continuously replenished by natural processes. For example, solar energy, wind energy, bio-energy bio-fuels grown sustain ably), hydropower etc., are some of the examples of renewable energy
sources
A renewable energy system converts the energy found in sunlight, wind, falling-water, seawaves, geothermal heat, or biomass into a form, we can use such as heat or electricity. Most of
the renewable energy comes either directly or indirectly from sun and wind and can never be
exhausted, and therefore they are called renewable.
However, most of the world's energy sources are derived from conventional sources-fossil
fuels such as coal, oil, and natural gases. These fuels are often termed non-renewable energy
sources. Although, the available quantity of these fuels are extremely large, they are nevertheless finite and so will in principle 'run out' at some time in the future
Renewable energy sources are essentially flows of energy, whereas the fossil and nuclear
fuels are, in essence, stocks of energy
Various forms of renewable energy
Solar energy
Wind energy
Bio energy
Hydro energy
Geothermal energy
Wave and tidal energy
This chapter focuses on application potential of commercially viable renewable energy
sources such as solar, wind, bio and hydro energy in India.
12.2
Solar Energy
Solar energy is the most readily available and free
source of energy since prehistoric times. It is estimated that solar energy equivalent to over 15,000
times the world's annual commercial energy consumption reaches the earth every year.
India receives solar energy in the region of 5 to
7 kWh/m2 for 300 to 330 days in a year. This energy is sufficient to set up 20 MW solar power plant per square kilometre land area.
Solar energy can be utilised through two different routes, as solar thermal route and solar
electric (solar photovoltaic) routes. Solar thermal route uses the sun's heat to produce hot water
or air, cook food, drying materials etc. Solar photovoltaic uses sun's heat to produce electricity
for lighting home and building, running motors, pumps, electric appliances, and lighting.
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12. Application of Non-Conventional & Renewable Energy Sources
Solar Thermal Energy Application
In solar thermal route, solar energy can be converted into thermal energy with the help of solar
collectors and receivers known as solar thermal devices.
The Solar-Thermal devices can be classified into three categories:
Low-Grade Heating Devices - up to the temperature of 100°C.
Medium-Grade Heating Devices -up to the temperature of 100°-300°C
High-Grade Heating Devices -above temperature of 300°C
Low-grade solar thermal devices are used in solar water heaters, air-heaters, solar cookers
and solar dryers for domestic and industrial applications.
Solar water heaters
Most solar water heating systems have two main parts: a
solar collector and a storage tank. The most common collector is called a flat-plate collector (see Figure 12.1). It
consists of a thin, flat, rectangular box with a transparent
cover that faces the sun, mounted on the roof of building
or home. Small tubes run through the box and carry the
fluid - either water or other fluid, such as an antifreeze
solution – to be heated. The tubes are attached to an
absorber plate, which is painted with special coatings to
absorb the heat. The heat builds up in the collector, which
Figure 12.1 Solar Flat plate collector
is passed to the fluid passing through the tubes.
An insulated storage tank holds the hot water. It is
similar to water heater, but larger is size. In case of systems that use fluids, heat is passed from
hot fluid to the water stored in the tank through a coil of tubes.
Solar water heating systems can be either active or passive systems. The active system,
which are most common, rely on pumps to move the liquid between the collector and
the storage tank. The passive systems rely on gravity and the tendency for water to
naturally circulate as it is heated. A few industrial application of solar water heaters are listed below:
❑
❑
❑
❑
❑
❑
Hotels: Bathing, kitchen, washing, laundry applications
Dairies: Ghee (clarified butter) production, cleaning and sterilizing, pasteurization
Textiles: Bleaching, boiling, printing, dyeing, curing, ageing and finishing
Breweries & Distilleries: Bottle washing, wort preparation, boiler feed heating
Chemical /Bulk drugs units: Fermentation of mixes, boiler feed applications
Electroplating/galvanizing units: Heating of plating baths, cleaning, degreasing applications
❑ Pulp and paper industries: Boiler feed applications, soaking of pulp.
Solar Cooker
Solar cooker is a device, which uses solar energy for cooking, and thus saving fossil fuels, fuel
wood and electrical energy to a large extent. However, it can only supplement the cooking fuel,
and not replace it totally. It is a simple cooking unit, ideal for domestic cooking during most of
the year except during the monsoon season, cloudy days and winter months
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12. Application of Non-Conventional & Renewable Energy Sources
Box type solar cookers: The box type solar cookers with a
single reflecting mirror are the most popular in India.
These cookers have proved immensely popular in rural
areas where women spend considerable time for collecting
firewood. A family size solar cooker is sufficient for 4 to
5 members and saves about 3 to 4 cylinders of LPG every
year. The life of this cooker is upto 15 years. This cooker
costs around Rs.1000 after allowing for subsidy. Solar
cookers.(Figure 12.2) are widely available in the market. Figure 12.2 Box Type Solar Collector
Parabolic concentrating solar cooker:
A parabolic solar concentrator comprises of sturdy Fibre
Reinforced Plastic (FRP) shell lined with Stainless Steel
(SS) reflector foil or aluminised polyester film. It can accommodate a cooking vessel at its focal point. This cooker is
designed to direct the solar heat to a secondary reflector
inside the kitchen, which focuses the heat to the bottom of a
cooking pot. It is also possible to actually fry, bake and roast
food. This system generates 500 kg of steam, which is
enough to cook two meals for 500 people (see Figure 12.3).
This cooker costs upward of Rs.50,000.
Positioning of solar panels or collectors can greatly
influence the system output, efficiency and payback. Tilting
mechanisms provided to the collectors need to be adjusted
according to seasons (summer and winter) to maximise the
collector efficiency.
The period four to five hours in late morning and early
Figure 12.3 Parabolic Collector
afternoon (between 9 am to 3pm) is commonly called the
"Solar Window". During this time, 80% of the total collectable energy for the day falls on
a solar collector. Therefore, the collector should be free from shade during this solar window throughout the year - Shading, may arise from buildings or trees to the south of the
location.
Solar Electricity Generation
Solar Photovoltaic (PV): Photovoltaic is the technical
term for solar electric. Photo means "light" and voltaic
means "electric". PV cells are usually made of silicon,
an element that naturally releases electrons when
exposed to light. Amount of electrons released from
silicon cells depend upon intensity of light incident
on it. The silicon cell is covered with a grid of
metal that directs the electrons to flow in a path
to create an electric current. This current is guided
into a wire that is connected to a battery or DC appliance. Typically, one cell produces about 1.5 watts of
power. Individual cells are connected together to form a
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Figure 12.4 Solar Photovoltaic Array
12. Application of Non-Conventional & Renewable Energy Sources
solar panel or module, capable of producing 3 to 110 Watts power. Panels can be connected
together in series and parallel to make a solar array (see Figure 12.4), which can produce
any amount of Wattage as space will allow. Modules are usually designed to supply
electricity at 12 Volts. PV modules are rated by their peak Watt output at solar noon on a
clear day.
Some applications for PV systems are lighting for commercial buildings, outdoor (street)
lighting (see Figure 12.5),
rural and village lighting
etc. Solar electric power
systems can offer independence from the utility grid
and offer protection during
extended power failures.
Solar PV systems are
found to be economical
Figure 12.5 Photovoltaic Domestic and Streetlights
especially in the hilly and
far flung areas where conventional grid power supply will be expensive to reach.
PV tracking systems is an alternative to the fixed, stationary PV panels. PV tracking systems are mounted and provided with tracking mechanisms to follow the sun as it moves through
the sky. These tracking systems run entirely on their own power and can increase output by
40%.
Back-up systems are necessary since PV systems only generate electricity when the sun is
shining. The two most common methods of backing up solar electric systems are connecting the
system to the utility grid or storing excess electricity in batteries for use at night or on cloudy
days.
Performance
The performance of a solar cell is measured in terms of its efficiency at converting sunlight into
electricity. Only sunlight of certain energy will work efficiently to create electricity, and much
of it is reflected or absorbed by the material that make up the cell. Because of this, a typical
commercial solar cell has an efficiency of 15%—only about one-sixth of the sunlight striking
the cell generates electricity. Low efficiencies mean that larger arrays are needed, and higher
investment costs. It should be noted that the first solar cells, built in the 1950s, had efficiencies
of less than 4%.
Solar Water Pumps
In solar water pumping system, the pump is driven by motor run by solar electricity instead
of conventional electricity drawn from utility grid. A SPV water pumping system consists of
a photovoltaic array mounted on a stand and a motor-pump set compatible with the photovoltaic array. It converts the solar energy into electricity, which is used for running the motor
pump set. The pumping system draws water from the open well, bore well, stream, pond,
canal etc
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12. Application of Non-Conventional & Renewable Energy Sources
Case Example:
Under the Solar Photovolatic
Water Pumping Programme
of the Ministry of Nonconventional Energy Sources
during 2000-01 the Punjab
Energy Development Agency
(PEDA) has completed
installation of 500 solar
pumps in Punjab for agricultural uses.
Under this project, 1800
watt PV array was coupled
with a 2 HP DC motor pump
Figure 12.6 Photovoltaic Water Pumping
set. The system is capable of
delivering about 140,000 litres water every day from a depth of about 6 – 7 metres. This quantity of water is considered adequate for irrigating about 5 – 8 acres land holding for most of the
crops. Refer Figure 12.6.
12.3 Wind Energy
Wind energy is basically harnessing of wind power to
produce electricity. The kinetic energy of the wind is
converted to electrical energy. When solar radiation enters
the earth's atmosphere, different regions of the atmosphere
are heated to different degrees because of earth curvature.
This heating is higher at the equator and lowest at the poles.
Since air tends to flow from warmer to cooler regions,
this causes what we call winds, and it is these airflows that
are harnessed in windmills and wind turbines to produce
power.
Wind power is not a new development as this power, in the
form of traditional windmills -for grinding corn, pumping
water, sailing ships - have been used for centuries. Now wind
power is harnessed to generate electricity in a larger scale with
better technology.
Wind Energy Technology
The basic wind energy conversion device is the wind turbine. Although various designs and
configurations exist, these turbines are generally grouped into two types:
1. Vertical-axis wind turbines, in which the axis of rotation is vertical with respect to the
ground (and roughly perpendicular to the wind stream),
2. Horizontal-axis turbines, in which the axis of rotation is horizontal with respect to the
ground (and roughly parallel to the wind stream.)
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12. Application of Non-Conventional & Renewable Energy Sources
Figure 12.7 Wind Turbine Configuration
The Figure 12.7 illustrates the two types of turbines and typical subsystems for an electricity generation application. The subsystems include a blade or rotor, which converts the energy
in the wind to rotational shaft energy; a drive train, usually including a gearbox and a generator, a tower that supports the rotor and drive train, and other equipment, including controls, electrical cables, ground support equipment, and interconnection equipment.
Wind electric generators (WEG)
Wind electric generator converts kinetic energy available in wind to electrical energy by using
rotor, gear box and generator. There are a large number of manufacturers for wind electric generators in India who have foreign collaboration with different manufacturers of Denmark,
Germany, Netherlands, Belgium, USA, Austria, Sweden, Spain, and U.K. etc. At present,
WEGs of rating ranging from 225 kW to 1000 kW are being installed in our country.
Evaluating Wind Mill Performance
Wind turbines are rated at a certain wind speed and annual energy output
Annual Energy Output = Power x Time
Example: For a 100 kW turbine producing 20 kW at an average wind speed of 25 km/h, the calculation would be:
100 kW x 0.20 (CF) = 20 kW x 8760 hours = 175,200 kWh
The Capacity Factor (CF) is simply the wind turbine's actual energy output for the year
divided by the energy output if the machine operated at its rated power output for the entire
year. A reasonable capacity factor would be 0.25 to 0.30 and a very good capacity factor would
be around 0.40. It is important to select a site with good capacity factor, as economic viability
of wind power projects is extremely sensitive to the capacity factor.
Wind Potential
In order for a wind energy system to be feasible there must be an adequate wind supply. A wind
energy system usually requires an average annual wind speed of at least 15 km/h. The following
table represents a guideline of different wind speeds and their potential in producing electricity.
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12. Application of Non-Conventional & Renewable Energy Sources
Average Wind Speed
km/h (mph)
Suitability
Up to 15 (9.5)
No good
18 (11.25)
Poor
22 (13.75)
Moderate
25 (15.5)
Good
29 (18)
Excellent
A wind generator will produce lesser power in summer than in winter at the same wind
speed as air has lower density in summer than in winter.
Similarly, a wind generator will produce lesser power in higher altitudes - as air pressure as
well as density is lower -than at lower altitudes.
The wind speed is the most important factor influencing the amount of energy a wind turbine can produce. Increasing wind velocity increases the amount of air passing the rotor, which
increases the output of the wind system.
In order for a wind system to be effective, a relatively consistent wind flow is required.
Obstructions such as trees or hills can interfere with the wind supply to the rotors. To avoid
this, rotors are placed on top of towers to take advantage of the strong winds available
high above the ground. The towers are generally placed 100 metres away from the nearest
obstacle. The middle of the rotor is placed 10 metres above any obstacle that is within
100 metres.
Wind Energy in India
India has been rated as one of the most promising countries for wind power development, with
an estimated potential of 20,000 MW. Total installed capacity of wind electric generators in the
world as on Sept. 2001 is 23270 MW. Germany 8100 MW, Spain- 3175 MW, USA 4240 MW,
Denmark 2417 MW, and India - 1426 MW top the list of countries. Thus, India ranks fifth in
the world in Wind power generation.
There are 39 wind potential stations in Tamil Nadu, 36 in Gujarat, 30 in Andhra Pradesh, 27
in Maharashtra, 26 in Karnataka, 16 in Kerala, 8 in Lakshadweep, 8 Rajasthan, 7 in Madhya
Pradesh, 7 in Orissa, 2 in West Bengal, 1 in Andaman Nicobar and 1 in Uttar Pradesh. Out of
208 suitable stations 7 stations have shown wind power density more than 500 Watts/ m2.
Central Govt. Assistance and Incentives
The following financial and technical assistance are provided to promote,
support and accelerate the development of wind energy in India:
Five years tax holiday
100% depreciation in the first year
Facilities by SEB's for grid connection
Energy banking and wheeling and energy buy back
Industry status and capital subsidy
Electricity tax exemption
Sales tax exemption
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12. Application of Non-Conventional & Renewable Energy Sources
Applications
• Utility interconnected wind turbines generate power which is synchronous with the grid and
are used to reduce utility bills by displacing the utility power used in the household and by
selling the excess power back to the electric company.
• Wind turbines for remote homes (off the grid) generate DC current for battery charging.
• Wind turbines for remote water pumping generate 3 phase AC current suitable for driving
an electrical submersible pump directly. Wind turbines suitable for residential or village
scale wind power range from 500 Watts to 50 kilowatts.
12.4
Bio Energy
Biomass is a renewable energy resource derived from the carbonaceous waste of various human and natural activities. It is
derived from numerous sources, including the by-products from
the wood industry, agricultural crops, raw material from the forest, household wastes etc.
Biomass does not add carbon dioxide to the atmosphere as it absorbs the same amount of
carbon in growing as it releases when consumed as a fuel. Its advantage is that it can be used
to generate electricity with the same equipment that is now being used for burning fossil fuels.
Biomass is an important source of energy and the most important fuel worldwide after coal, oil
and natural gas. Bio-energy, in the form of biogas, which is derived from biomass, is expected
to become one of the key energy resources for global sustainable development. Biomass offers
higher energy efficiency through form of Biogas than by direct burning (see chart below).
Application
Biogas Plants
Biogas is a clean and efficient fuel, generated from cow-dung,
human waste or any kind of biological materials derived through
anaerobic fermentation process. The biogas consists of 60%
methane with rest mainly carbon-di-oxide. Biogas is a safe fuel
for cooking and lighting. By-product is usable as high-grade
manure.
A typical biogas plant has the following components: A digester in which the slurry (dung
mixed with water) is fermented, an inlet tank - for mixing the feed and letting it into the
digester, gas holder/dome in which the generated gas is collected, outlet tank to remove the
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12. Application of Non-Conventional & Renewable Energy Sources
spent slurry, distribution pipeline(s) to transport the gas into the kitchen, and a manure pit,
where the spent slurry is stored.
Biomass fuels account for about one-third of the total fuel used in the country. It is the most
important fuel used in over 90% of the rural households and about 15% of the urban households.
Using only local resources, namely cattle waste and other organic wastes, energy and manure are
derived. Thus the biogas plants are the cheap sources of energy in rural areas. The types of biogas
plant designs popular are: floating drum type, fixed dome-type and bag-type portable digester.
Biomass Briquetting
The process of densifying loose agro-waste into a
solidified biomass of high density, which can be
conveniently used as a fuel, is called Biomass
Briquetting (see Figure 12.8). Briquette is also
termed as "Bio-coal". It is pollution free and ecofriendly. Some of the agricultural and forestry
residues can be briquetted after suitable pre-treatment. A list of commonly used biomass materials
that can be briquetted are given below:
CornCob, JuteStick, Sawdust, PineNeedle,
Bagasse, CoffeeSpent, Tamarind, CoffeeHusk,
Figure 12.8 Biomass Briquetting
AlmondShell,
Groundnutshells,
CoirPith,
BagaseePith, Barleystraw, Tobaccodust, RiceHusk, Deoiled Bran
Advantages
Some of advantages of biomass briquetting are high calorific value with low ash content,
absence of polluting gases like sulphur, phosphorus fumes and fly ash- which eliminate the need
for pollution control equipment, complete combustion, ease of handling, transportation & storage - because of uniform size and convenient lengths.
Application
Biomass briquettes can replace almost all conventional fuels like coal, firewood and lignite in
almost all general applications like heating, steam generation etc. It can be used directly as fuel
instead of coal in the traditional chulhas and furnaces or in the gasifier. Gasifier converts solid
fuel into a more convenient-to-use gaseous form of fuel called producer gas.
Biomass Gasifiers
Biomass gasifiers (see Figure 12.9) convert the
solid biomass (basically wood waste, agricultural residues etc.) into a combustible gas mixture normally called as producer gas. The conversion efficiency of the gasification process is
in the range of 60%–70%. The producer gas
consists of mainly carbon-monoxide, hydrogen, nitrogen gas and methane, and has a lower
calorific value (1000–1200 kcal/Nm3).
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Figure 12.9 Biomass Gasifiers
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12. Application of Non-Conventional & Renewable Energy Sources
Gasification of biomass and using it in place of conventional direct burning devices will
result in savings of atleast 50% in fuel consumption. The gas has been found suitable for combustion in the internal combustion engines for the production of power.
Applications:
Water pumping and Electricity generation: Using biomass gas, it possible to operate a diesel
engine on dual fuel mode-part diesel and part biomass gas. Diesel substitution of the order of
75 to 80% can be obtained at nominal loads. The mechanical energy thus derived can be used
either for energizing a water pump set for irrigational purpose or for coupling with an alternator for electrical power generation - 3.5 KW - 10 MW
Heat generation: A few of the devices, to which gasifier could be retrofitted, are dryers- for
drying tea, flower, spices, kilns for baking tiles or potteries, furnaces for melting non-ferrous
metals, boilers for process steam, etc.
Direct combustion of biomass has been recognized as an important route for generation of
power by utilization of vast amounts of agricultural residues, agro-industrial residues and forest wastes. Gasifiers can be used for power generation and available up to a capacity 500 kW.
The Government of India through MNES and IREDA is implementing power-generating system based on biomass combustion as well as biomass gasification
High Efficiency Wood Burning Stoves
These stoves save more than 50% fuel wood consumption. They reduce drudgery of women
saving time in cooking and fuel collection and consequent health hazards. They also help in saving firewood leading to conservation of forests. They also create employment opportunities for
people in the rural areas.
Bio fuels
Unlike other renewable energy sources, biomass can be converted
directly into liquid fuels— biofuels— for our transportation needs
(cars, trucks, buses, airplanes, and trains). The two most common
types of biofuels are ethanol and biodiesel. See Figure 12.10.
Ethanol is an alcohol, similar to that used in beer and wine. It
is made by fermenting any biomass high in carbohydrates (starches, sugars, or celluloses) through a process similar to brewing beer.
Ethanol is mostly used as a fuel additive to cut down a vehicle's
carbon monoxide and other smog-causing emissions. Flexible-fuel
Figure 12.10 Biodiesel
vehicles, which run on mixtures of gasoline and up to 85%
Driven Bus
ethanol, are now available.
Biodiesel, produced by plants such as rapeseed (canola), sunflowers and soybeans, can be
extracted and refined into fuel, which can be burned in diesel engines and buses. Biodiesel can
also made by combining alcohol with vegetable oil, or recycled cooking greases. It can be used
as an additive to reduce vehicle emissions (typically 20%) or in its pure form as a renewable
alternative fuel for diesel engines.
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Biopower
Biopower, or biomass power, is the use of biomass to generate electricity. There are six
major types of biopower systems: direct-fired, cofiring, gasification, anaerobic digestion,
pyrolysis, and small - modular.
Most of the biopower plants in the world use direct-fired systems. They burn bioenergy
feedstocks directly in boiler to produce steam. This steam drives the turbo-generator. In some
industries, the steam is also used in manufacturing processes or to heat buildings. These are
known as combined heat and power facilities. For example, wood waste is often used to produce both electricity and steam at paper mills.
Many coal-fired power plants use cofiring systems to significantly reduce emissions, especially sulfur dioxide emissions. Cofiring involves using bio energy feedstock as a supplementary fuel source in high efficiency boilers.
Gasification systems use high temperatures and an oxygen-starved environment to convert
biomass into a gas (a mixture of hydrogen, carbon monoxide, and methane). The gas fuels a gas
turbine, which runs an electric generator for producing power.
The decay of biomass produces methane gas, which can be used as an energy source.
Methane can be produced from biomass through a process called anaerobic digestion.
Anaerobic digestion involves using bacteria to decompose organic matter in the absence of oxygen. In landfills -scientific waste disposal site - wells can be drilled to release the methane from
the decaying organic matter. The pipes from each well carry the gas to a central point where it
is filtered and cleaned before burning. Methane can be used as an energy source in many ways.
Most facilities burn it in a boiler to produce steam for electricity generation or for industrial
processes. Two new ways include the use of microturbines and fuel cells. Microturbines have
outputs of 25 to 500 kilowatts. About the size of a refrigerator, they can be used where there are
space limitations for power production. Methane can also be used as the "fuel" in a fuel cell.
Fuel cells work much like batteries, but never need recharging, producing electricity as long as
there is fuel.
In addition to gas, liquid fuels can be produced from biomass through a process called
pyrolysis. Pyrolysis occurs when biomass is heated in the absence of oxygen. The biomass then
turns into liquid called pyrolysis oil, which can be burned like petroleum to generate electricity. A biopower system that uses pyrolysis oil is being commercialized.
Several biopower technologies can be used in small, modular systems. A small, modular
system generates electricity at a capacity of 5 megawatts or less. This system is designed
for use at the small town level or even at the consumer level. For example, some farmers
use the waste from their livestock to provide their farms with electricity. Not only do
these systems provide renewable energy, they also help farmers meet environmental
regulations.
Biomass Cogeneration
Cogeneration improves viability and profitability of sugar industries. Indian sugar mills are
rapidly turning to bagasse, the leftover of cane after it is crushed and its juice extracted, to generate electricity. This is mainly being done to clean up the environment, cut down power costs
and earn additional revenue. According to current estimates, about 3500 MW of power can be
generated from bagasse in the existing 430 sugar mills in the country. Around 270 MW of
power has already been commissioned and more is under construction.
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12.5
Hydro Energy
The potential energy of falling
water, captured and converted to
mechanical energy by waterwheels, powered the start of the
industrial revolution.
Wherever sufficient head, or
change in elevation, could be
found, rivers and streams were
dammed and mills were built.
Water under pressure flows
through a turbine causing it to
spin. The Turbine is connected to
a generator, which produces
electricity (see Figure 12.11). In
order to produce enough electricity, a hydroelectric system
requires a location with the following features:
Figure 12.11 Hydro Power Plant
Change in elevation or head: 20 feet @ 100 gal/min = 200 Watts.
100 feet head @ 20 gal/min gives the same output.
In India the potential of small hydro power is estimated about 10,000 MW. A total of 183.45
MW small Hydro project have been installed in India by the end of March 1999. Small Hydro
Power projects of 3 MW capacity have been also installed individually and 148 MW project is
under construction.
Small Hydro
Small Hydro Power is a reliable, mature and proven technology. It
is non-polluting, and does not involve setting up of large dams or
problems of deforestation, submergence and rehabilitation. India
has an estimated potential of 10,000 MW
Micro Hydel
Hilly regions of India, particularly the Himalayan belts, are endowed with
rich hydel resources with tremendous potential. The MNES has launched
a promotional scheme for portable micro hydel sets for these areas. These
sets are small, compact and light weight. They have almost zero maintenance cost and can provide electricity/power to small cluster of villages.
They are ideal substitutes for diesel sets run in those areas at high generation cost.
Micro (upto 100kW) mini hydro (101-1000 kW) schemes can
provide power for farms, hotels, schools and rural communities, and help create local
industry.
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12.6
Tidal and Ocean Energy
Tidal Energy
Tidal electricity generation involves the construction of a barrage across an estuary to block the incoming and outgoing tide.
The head of water is then used to drive turbines to generate
electricity from the elevated water in the basin as in hydroelectric dams.
Barrages can be designed to generate electricity on the ebb
side, or flood side, or both. Tidal range may vary over a wide
range (4.5-12.4 m) from site to site. A tidal range of at least 7
m is required for economical operation and for sufficient head
of water for the turbines.
Ocean Energy
Oceans cover more than 70% of Earth's surface, making them the world's largest solar collectors. Ocean energy draws on the energy of ocean waves, tides, or on the thermal energy (heat)
stored in the ocean. The sun warms the surface water a lot more than the deep ocean water, and
this temperature difference stores thermal energy.
The ocean contains two types of energy: thermal energy from the sun's heat, and mechanical energy from the tides and waves.
Ocean thermal energy is used for many applications, including electricity generation. There
are three types of electricity conversion systems: closed-cycle, open cycle, and hybrid. Closed
cycle systems use the ocean's warm surface water to vaporize a working fluid, which has a low
boiling point, such as ammonia. The vapour expands and turns a turbine. The turbine then activates a generator to produce electricity. Open-cycle systems actually boil the seawater by operating at low pressures. This produces steam that passes through a turbine / generator. The hybrid
systems combine both closed-cycle and open-cycle systems.
Ocean mechanical energy is quite different from ocean thermal energy. Even though
the sun affects all ocean activity, tides are driven primarily by the gravitational pull of the
moon, and waves are driven primarily by the winds. A barrage (dam) is typically used
to convert tidal energy into electricity by forcing the water through turbines, activating a
generator.
India has the World's largest programmes for renewable energy. Several renewable energy
technologies have been developed and deployed in villages and cities of India. A Ministry of
Non-Conventional Energy Sources (MNES) created in 1992 for all matters relating to NonConventional / Renewable Energy. Government of India also created Renewable Energy
Development Agency Limited (IREDA) to assist and provide financial assistance in the form
of subsidy and low interest loan for renewable energy projects.
IREDA covers a wide spectrum of financing activities including those that are connected to
energy conservation and energy efficiency. At present, IREDA's lending is mainly in the following areas: Bureau of Energy Efficiency
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12. Application of Non-Conventional & Renewable Energy Sources
•
•
•
•
•
•
Solar energy technologies, utilization of solar thermal and solar photo voltaic systems
Wind energy setting up grid connected Wind farm projects
Small hydro setting up small, mini and micro hydel projects
Bio-energy technologies, biomass based co-generation projects, biomass gasification, energy from waste and briquetting projects
Hybrid systems
Energy efficiency and conservation
The estimated potential of various Renewable Energy technologies in India by IREDA are
given below.
Energy source estimated potential
Solar Energy
Wind Energy
Small Hydro
Ocean Thermal Power
Sea Wave Power
Tidal Power
Bio energy
Draught Animal Power
Energy from MSW
Biogas Plants
Improved Wood Burning Stoves
Bagasse-based cogeneration
20 MW / sq. km
20,000 MW
10,000 MW
50,000 MW
20,000 MW
10,000 MW
17,000 MW
30,000 MW
1,000 MW
12 Million Plants
120 Million Stoves
3500 MW
Cumulative achievements in renewable energy sector (As on 31.03.2000)
Sources / Technologies
Wind Power
Small Hydro
Biomass Power & Co-generation
Solar PV Power
Urban & MSW
Solar Heater
Solar Cookers
Biogas Plants
Biomas Gasifier
Improved Chulhas
Bureau of Energy Efficiency
Unit Upto31.03.2000
MW 1167
MW 217
MW 222
MW / Sq. km 42
MW 15.21
m2. Area 480000
No. 481112
Nos. in Million 2.95
MW 34
Nos. in Million 31.9
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12. Application of Non-Conventional & Renewable Energy Sources
QUESTIONS
1
What do you mean by renewable energy
2
Why is solar energy potential high in India?
3.
Explain working of solar water heater?
4.
List few applications of low temperature water heaters in domestic and industrial use
5.
What are the two methods by which energy can be recovered from solar radiation
6.
How can the performance of solar collectors be improved?
7.
Explain any two applications of concentrated solar energy?
8.
What do you mean by photovoltaic?
9.
Explain the terms cell, module and array as applicable to photovoltaic.
10.
What are the typical applications of photovoltaic power?
11.
Name the few states with high wind energy potential in India.
12.
What are the criteria for selection of wind mill installation?
13.
What ere the incentives available for wind mill installation?
14.
Explain the bio-energy potential in India and its applications.
15.
What are the various methods by which power can be generated from biomass?
16.
What is the role of IREDA in renewable energy sector
17.
India has recorded good growth in wind energy sector. Do you agree? What are the
factors responsible for such a high growth?
REFERENCES
1.
2.
3.
4.
Alternate Energy Sources by T H Taylor.Adam Hilger Ltd, Bristol
Renewable Energy Sources for rural areas in Asia and Pacific, APO, Tokyo, 2000
www.ireda.org
www.windenergy.com
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13. WASTE MINIMISATION AND RESOURCE
CONSERVATION
13.1
Introduction
Traditionally, waste is viewed as an unnecessary element arising from the activities of any
industry. In reality, waste is a misplaced resource, existing at a wrong place at a wrong time.
Waste is also the inefficient use of utilities such as electricity, water, and fuel, which are
often considered unavoidable overheads. The costs of these wastes are generally underestimated by managers. It is important to realise that the cost of waste is not only the cost of waste disposal, but also other costs such as:
Disposal cost
Inefficient energy use cost
Purchase cost of wasted raw material
Production cost for the waste material
Management time spent on waste material
Lost revenue for what could have been a product instead of waste
Potential liabilities due to waste.
What is waste minimisation?
Waste minimisation can be defined as "systematically reducing waste at source". It means:
•
•
•
•
•
Prevention and/or reduction of waste generated
Efficient use of raw materials and packaging
Efficient use of fuel, electricity and water
Improving the quality of waste generated to facilitate recycling and/or reduce hazard
Encouraging re-use, recycling and recovery.
Waste minimisation is also known by other terms such as waste reduction, pollution prevention, source reduction and cleaner technology. It makes use of managerial and/or technical interventions to make industrial operations inherently pollution free
It should be also clearly understood that waste minimization, however attractive, is not a
panacea for all environmental problems and may have to be supported by conventional treatment/disposal solutions.
Waste minimization is best practiced by reducing the generation of waste at the source itself.
After exhausting the source reduction opportunities, attempts should be made to recycle the
TABLE 13.1 WASTES AND POSSIBLE RESOURCES
Wastes
Resources
Fly ash from power plant
Raw material for cement or brick
manufacture
Bagasse wastes from sugar manufacture
Fuel for boiler
CO2 release from ammonia plant
Raw material for Urea manufacture
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13. Waste Minimisation and Resource Conservation
waste within the unit. Finally, modification or reformulation of products so as to manufacture
it with least waste generation should be considered. Few wastes and possible resources are indicated in the Table 13.1
13.2
Classification of Waste Minimization (WM) Techniques
The waste minimization is based on different techniques. These techniques are classified as
hereunder.
Source Reduction
Under this category, four techniques of WM are briefly discussed below:
a) Good Housekeeping- Systems to prevent leakages & spillages through preventive maintenance schedules and routine equipment inspections. Also, well-written working instructions,
supervision, awareness and regular training of workforce would facilitate good housekeeping.
b)
Process Change:
(i)
Under this head, four CP techniques are covered:
Input Material Change - Substitution of input materials by eco-friendly (nontoxic or less toxic than existing and renewable) material preferably having longer
service time.
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(ii)
Better Process Control - Modifications of the working procedures, machine-operating instructions and process record keeping in order to run the processes at higher
efficiency and with lower waste generation and emissions.
(iii) Equipment Modification - Modification of existing production equipment and
utilities, for instance, by the addition of measuring and controlling devices, in order
to run the processes at higher efficiency and lower waste and emission generation
rates.
(iv)
c)
d)
Technology change - Replacement of the technology, processing sequence
and/or synthesis route, in order to minimise waste and emission generation during
production.
Recycling
i)
On-site Recovery and Reuse - Reuse of wasted materials in the same process or
for another useful application within the industry.
ii)
Production of Useful by-product - Modification of the waste generation process
in order to transform the wasted material into a material that can be reused or recycled for another application within or outside the company.
Product Modification
Characteristics of the product can be modified to minimise the environmental impacts
of its production or those of the product itself during or after its use (disposal).
13.3
Waste Minimization Methodology
For an effective Waste Minimization programme, it is essential to bring together various
groups in the industry to ensure implementation. How formalised the programme would
be depends upon the size and composition of the industry and its waste and emission problems. The programme should be flexible enough so that it can adapt itself to changing circumstances. A methodical step-by-step procedure ensures exploitation of maximum waste
minimization opportunities. The steps in a typical waste minimization progamme are
illustrated below:
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Step 1: Getting Started
Form a Waste Minimization Team
Waste Minimization (WM) concept encompasses all departments and sections in an industry.
Activities involved in WM audits require assistance and cooperation not only from the personnel belonging to concerned department, but also from other related departments. Hence, making
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an inter-disciplinary and inter-departmental team is a prerequisite for successful conduct of a
WM audit. In special cases, it would be advantageous to have external experts in order to have
an unbiased, professional approach.
List Process Steps / Unit Operations
The WM team should familiarize itself with the manufacturing processes including utilities,
waste treatment and disposal facilities, and list all the process steps. Preparation of sketches of
process layout drainage system, vents and material-loss areas would be useful. This helps in
establishing cause-effect relationships and ensuring that important areas are not overlooked.
Periodic, intermittent and continuous discharge streams should be appropriately labeled.
Identify and Select Wasteful Process Steps
In multi-process type industries, it may be difficult to start detailed Waste minimization activities covering the complete unit. In such cases, it is advisable to start with fewer process steps
to begin with. The selected step(s) could be the most wasteful and / or one with very high waste
minimization potential.
This activity could also be considered a preliminary prioritization activity. All the various
wasteful steps identified in 1.2 should be broadly assessed in terms of volume of waste, severity of impact on the environment, Waste minimization opportunities, estimated benefits (specially cost savings), cost of implementation etc. Such assessment would help in focusing on the
process steps / areas for detailed analysis.
Step 2: Analysing Process Steps
Prepare Process Flow Charts
This activity follows the activity described at 1.2. Flow charts are diagrammatic / schematic representation of production, with the purpose of identifying process steps and the
source of waste streams and emissions. A flow chart should list, and characterize the input
and output streams, as well as recycle streams. Even the so called free or less costly inputs
like water, air, sand, etc should be taken into account as these often end up in being the
major cause of wastes. Wherever required, the process flow diagram should be supplemented with chemical equations to facilitate understanding of the process. Also the materials which are used occasionally and / or which do not appear in output streams (for example catalysts, coolant oil) should be specified. The periodic / batch / continuous steps
should also be appropriately highlighted. Preparation of a detailed and correct process flow
diagram is a key step in the entire analysis and forms the basis for compilation of materials and energy balance.
Make material and Energy Balances
Material and Energy balances are important for any Waste minimization programme since they
make it possible to identify and quantify, previously unknown losses or emissions. These balances are also useful for monitoring the progress achieved in a prevention programme, and
evaluating the costs and benefits. Typical components of a material balance and energy balance
are given below (see Figures 13.1 & 13.2):
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It is not advisable to spent more time and money to make a perfect material balance. Even
a rough / preliminary material balance throws open Waste Minimization opportunities which
can be profitably exploited.
On the other hand, the precision of analytical data and flow measurements is important as
it is not possible to obtain a reliable estimate of the waste stream by subtracting the materials
in the product from those in the raw materials. In such cases, a direct monitoring and analysis
of waste streams should be carried out.
Assign Costs To Waste Stream
In order to assess the profit potential of waste streams, a basic requirement would be to assign
costs to them. This cost essentially reflects the monetary loss arising from waste. Apparently, a
waste stream does not appear to have any quantifiable cost attached to it, except where direct
raw material / product loss is associated with it. However, a deeper analysis would show several direct and indirect cost components associated with waste streams such as:
Cost of raw materials in waste.
Manufacturing cost of material in waste
Cost of product in waste
Cost of treatment of waste to comply with regulatory requirements
Cost of waste disposal
Cost of waste transportation
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Cost of maintaining required work environment
Cost due to waste cess.
Based on this, for each waste stream, total cost per unit of waste (Rs/KL or Rs/Kg)
should be worked out. This figure would be useful in working out the feasibility of the waste
minimization measures. The result can also be used to categorize the waste streams for
priority action.
Review of Process
Through the material and energy balances, it is possible to carry out a cause analysis to locate
and pinpoint the causes of waste generation. These causes would subsequently become the tools
for evolving Waste Minisation measures. There could be a wide variety of causes for waste
generation ranging from simple lapses of housekeeping to complex technological reasons as
indicated below.
Typical Causes Of Waste
Poor housekeeping ;
Leaking taps / valves / flanges
Spillages
Overflowing tanks
Worn out material transfer belts
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Operational and maintenance negligence
Unchecked water / air consumption
Unnecessary running of equipment
Sub optimal loading
Lack of preventive maintenance
Sub-optimal maintenance of process condition
Ritualistic operation
Poor raw material quality
Use of substandard cheap raw material
Lack of quality specification
Improper purchase management system
Improper storage
Poor process / equipment design
Mismatched capacity of equipment
Wrong material selection
Maintenance prone design
Adoption of avoidable process steps
Lack of information / design capability
Poor layout
Unplanned / adhoc expansion
Poor space utilization plan
Bad material movement plan
Bad technology
Continuation of obsolete technology
Despite product / raw material change
High cost of better technology
Lack of availability of trained manpower
Small plant size
Lack of information
Inadequately trained personnel
Increased dependence on casual / contract labour
Lack of formalized training system
Lack of training facilities
Job insecurity
Fear of losing trade secrets
Lack of availability of personnel
Understaffing hence work over pressure
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Employee Demotivation
Lack of recognition
Absence of reward
Emphasis only on production, not on people
Lack of commitment and attention by top management
Step 3: Generating Waste Minimization Opportunities
Develop Waste Minimization Opportunities
Once the origin and causes of waste and emissions are known, the process enters the creative
phase. The WM team should now start looking for possible opportunities for reducing waste.
Finding potential options depends on the knowledge and creativity of your team members. The
potential sources of help in finding Waste Minimization Opportunities are:
Other personnel from the same or similar plant elsewhere
Trade associations
Consultancy organizations
Equipment suppliers
Consultants
The process of finding Waste Minimization opportunities should take place in an
environment, which stimulates creativity and independent thinking. It would be beneficial to move away from the routine working environment for better results. Various
analysis tools and techniques like "brainstorming", "group discussions" etc would be useful in this step.
Select Workable Opportunities
The Waste Minimization opportunities developed should be screened and those, which are
impractical, should be discarded.
The discarding process should be simple, and straightforward and may often be only
qualitative. There should be no ambiguity or bias. Unnecessary effort in doing detailed feasibility analysis of opportunities, which are impractical or non-feasible, should be avoided.
The remaining Waste Minimization opportunities are then subjected to a more detailed
feasibility analysis.
Step 4: Selecting Waste Minimization Solutions
The selection of a Waste Minimization solution for implementation requires that it should not
only be techno-economically viable, but also environmentally desirable.
Assess Technical Feasibility
The technical evaluation determines whether a proposed Waste Minimization option will work
for the specific application. The impact of the proposed measure on product production rate
should be evaluated first. In case of significant deviation from the present process practices, laboratory testing trial runs might be required to assess the technical feasibility.
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A typical checklist for technical evaluation could be as follows:
Availability of equipment
Availability of operating skills
Space availability
Effect on production
Effect on product quality
Safety aspects
Maintenance requirements
Effect on operational flexibility
Shut down requirements for implementation
Assess Economic Viability
Economic viability would often be the key parameter to promote or discourage Waste
Minimization. For a smooth take-off, it is therefore essential that the first few Waste
Minimization measures should be economically very attractive. Such a strategy helps in creating more interest and sustains commitment.
Options requiring little investment, but involving more of procedural changes (housekeeping measures, measures, operational improvements) do not need an intensive economic analysis and simple methods like pay back period could be used. However as Waste Minimization
measures become more involved and capital intensive, other profitability analysis methods viz.
Return on Investment (IRR) or Net Present Value (NPV) are recommended to get the total
picture.
While doing the economic investment, the costs may include fixed capital i.e. cost of equipment, working capital cost, shutdown cost, O & M costs etc. The savings could be direct savings of input materials / energy, increased production levels and hence lower specific input cost,
lower O & M cost, value of by products, reduction in environmental cost i.e. waste treatment
transportation and disposal cost etc.
Evaluate Environmental Aspects
The options for Waste Minimization with respect to their impact on the environment
should be assessed. In many cases the environmental advantage will be obvious if there
is a net reduction in the toxicity and / or quantity of waste. Other impacts could be
changes in treatment of the wastes. In the initial stages, environmental aspects may not
appear to be as compelling as economic aspects. In future as in developed countries,
environmental aspects would become the most important criteria irrespective of the economic viability.
Select Solutions for Implementation
After technical, economic and environmental assessment, Waste Minimization measures should
be selected for implementation. Understandably, the most attractive ones would be those having best financial benefits, provided technical feasibility is favourable. However, in a growing
number of cases especially when active pressure groups are present, environmental priorities
may become the final deciding factor.
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The work done so far should be documented. Apart from becoming a reference document
for seeking approvals in implementation, the document would also be useful in obtaining
finances from external finance institutions, reporting status to other agencies, and establishing
base levels for performance evaluation and review.
Step 5: Implementing Waste Minimization Solutions
Prepare for Implementation
The selected solutions could be taken for implementation. Apart from simple housekeeping
measures several others would require a systematic plan of implementation.
The Waste Minimization team should be well prepared to take up the job of implementation. The preparation would include arranging finances, establishing linkages in case of multidepartment solutions, technical preparations, etc.
Implement Solutions
The task comprises layout and drawing preparation equipment fabrication / procurement, transportation to site, installation and commissioning. Whenever required, simultaneous training of
manpower should be taken up as many excellent measures have failed miserably because of
non-availability of adequately trained people.
Monitor and Evaluate Results
The WM solutions should be monitored for performance. The results obtained should be
matched with those estimated / worked out during technical evaluation to establish causes for
deviation, if any. The implementation job is considered to be over, only after successful commissioning and sustained stable performance over a reasonable length of time.
Step 6: Sustain Waste Minimization
The biggest challenges in Waste Minimization lies in sustaining Waste Minimization. The
enthusiasm of the Waste Minimization team wanes off with time. Such tragic ends should be
avoided. Backing out from commitment, predominance of production at any cost, absence of
rewards and appreciation, and shifting of priorities are some of the commonly encountered reasons, which one should check and avoid.
Also monitoring and review of the implemented measures should be communicated to all
employees in the industry so that it fans the desires for minimizing wastes. Involvement of as
large a number of employees as possible and rewarding the deserving ones, will help long term
sustenance of Waste Minimization.
Having implemented Waste Minimization solutions in the area under study, the Waste
Minimization team should go back to Step-2 i.e. analysing the process steps and identifying
and selecting the next wasteful area. In this way, the cycle continues, till all the steps are
exhausted.
In a nutshell, a philosophy of minimizing waste must be developed within the company. This means that Waste Minimization should become an integral part of company's
activities. All successful Waste Minimization programmes, till date, have been founded on
this philosophy.
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13.4
Case Study
Maximising Cullet Recovery Reduces Batch Costs
At a Lead Crystal Glass Works, glass was produced by melting a charge of blended raw materials together with process cullet. However, only about 30% of the cullet produced at the glassworks was of a size appropriate for remelting. Concern about the significant amounts of valuable raw materials lost in this cullet and being sent for waste disposal, led to the installation of
a crushing unit to reduce the cullet to an optimum size for recovery. Operation of the crushing
unit was subsequently enhanced by the addition of a vibrating screen and cullet washing
system.
The ideal size for cullet pieces, to produce a high quality melt of uniform composition and
avoid faults in the blown glass, is 12 – 20 mm. Most of the heavy cullet at the company was
present in large pieces that cannot be easily broken up manually to the optimum size. Lighter
pieces such as those from wine glasses, were also difficult to recycle because they have to be
broken up manually to obtain a satisfactory charge weight. This generates a lot of fine material, which was unsuitable because it tends to result in air bubbles being trapped in the glass gathered from the furnace pot by the glass blower.
Before waste minimization programme, about 560 tonnes of cullet were disposed for waste
disposal each year, costing the company considerably in terms of lost raw materials. The company therefore decided to install a crushing plant capable of producing a consistent output of a
size suitable for remelting and with minimum generation of fine material. Such a plant would
allow more cullet to be recycled, leading to a reduction in the cost of both primary raw materials and cullet disposal.
Following discussions with suppliers of crushing plant, the company installed a rotary hammer mill. This resulted in recycling of 74% of process cullet as against 30% previously. Also
alternative uses avoiding waste disposal have been found for the crusher fines and other forms
of waste glass. Crushing has also increased the bulk density of the cullet by a factor of three and
improved its size distribution.
The benefits of maximising inhouse cullet recovery include:
•
•
•
•
Cost savings
Reduction of 37% in the purchase of primary raw materials
Improved yield of first quality glass
Payback period of three weeks
Associated Waste Minimization Measures
In addition to installing the cullet crusher, the company had initiated a number of other associated
waste minimization measures such as segregation by source of inhouse cullet, segregating stones
from cullet, lead recovery from reject cullet, crusher fines and waste glass prior to disposal.
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13. Waste Minimisation and Resource Conservation
QUESTIONS
1.
Explain the concept of waste minimization with suitable examples.
2.
"Waste is a misplaced resource" Explain.
3.
What are the 3R's in waste minimization techniques?
4.
Which would you prefer between recycling and source reduction? Justify.
5.
List down few housekeeping measures by which wastes can be reduced.
6.
Explain how modifying a product can help minimize the wastes with few examples.
7.
For a coal-fired boiler, draw a block diagram and indicate various material and energy inputs, outputs and wastes.
8.
Can employee be a factor in reducing wastes? Explain.
REFERENCES
1.
2.
From Waste to Profits, Guidelines for Waste Minimization by National Productivity
Council, New Delhi
Waste Minimization Manual for Textile Processing by National Productivity Council,
Chennai.
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