Energy Efficiency Assessment Book

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. Bureau of Energy Efficiency

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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 Bureau of Energy Efficiency

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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 Bureau of Energy Efficiency

= = = =

% 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. Bureau of Energy Efficiency

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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. Bureau of Energy Efficiency

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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. Bureau of Energy Efficiency

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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

Bureau of Energy Efficiency

=

[(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

Bureau of Energy Efficiency

=

{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

Bureau of Energy Efficiency

=

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 %

= = Bureau of Energy Efficiency

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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

Bureau of Energy Efficiency

=

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

Bureau of Energy Efficiency

% % % 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

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. Bureau of Energy Efficiency

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Bureau of Energy Efficiency

25

1. Energy Performance Assessment of Boilers

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. Bureau of Energy Efficiency

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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 Bureau of Energy Efficiency

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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.

Bureau of Energy Efficiency

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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

Bureau of Energy Efficiency

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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. Bureau of Energy Efficiency

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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. Bureau of Energy Efficiency

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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.

Bureau of Energy Efficiency

35

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) Bureau of Energy Efficiency

36

= 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.

Bureau of Energy Efficiency

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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. Bureau of Energy Efficiency

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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 Bureau of Energy Efficiency

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 Bureau of Energy Efficiency

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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. Bureau of Energy Efficiency

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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 Bureau of Energy Efficiency

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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. Bureau of Energy Efficiency

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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 Bureau of Energy Efficiency

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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

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 Bureau of Energy Efficiency

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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 Bureau of Energy Efficiency

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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 Bureau of Energy Efficiency

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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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 Bureau of Energy Efficiency

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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. Bureau of Energy Efficiency

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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

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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

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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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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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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

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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 % =

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Actual operating load of the motor x 100 Rated capacity of the motor

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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

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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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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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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 Bureau of Energy Efficiency

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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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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 Bureau of Energy Efficiency

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. Bureau of Energy Efficiency

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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)