calculations

Capacity of plant = 100,000 m/day
Raw material used = Gray Fabrics, Dye Colors, Chemicals
Type of fuel used = Reliance Peat Coke, Lignite / Barmeri Coal
BOILER CALCULATION
HEAT OUTPUT DATA
Quantity of steam generated = 2 TPH
Steam pressure = 8 Kg/cm2
Enthalpy of steam at 8 Kg/cm2 = 659 kcal/kg
Feed water temperature = 65°C
Enthalpy of feed water = 65 kcal/kg
Fuel used = Lignite Coal
HEAT INPUT DATA
Quantity of coal consumed = 450 kg/hr
Gross calorific value of coal = 3500 kcal /kg
Boiler Efficiency (η) = (Q*(H-h))/(q*GCV) * 100
Where
Q= Quantity of steam generated per hour (kg/hr)
q = Quantity of fuel used per hour (kg/hr)
GCV= Gross calorific value of fuel used (kcal/kg)
H= Enthalpy of steam (kcal/kg)
h = Enthalpy of feed water (kcal/kg)
Boiler Efficiency (η) = (2000 ×(659-65))/(450×3500)*100
= 75.4 %
Evaporation Ratio = (RATE OF STEAM GENERATION )/(RATE OF FUEL CONSUMPTION )
=(2000 (kg )/hr OF STEAM)/(450 kg/hr OF COAL )
= 4.44 kg OF STEAM PRODUCED / kg OF COAL

BOILER EFFICEINCY BY INDIRECT METHOD
Fuel Analysis (in %)
Ash Content in Fuel = 8.63, Moisture in Coal = 31.65, Carbon Content = 41.65, Hydrogen Content= 2.0413, Nitrogen Content = 1.6, Oxygen Content = 14.48, GCV of Coal = 3500 Kcal/Kg Flue Gas Temperature = 180°C, Ambient Temperature = 30°C, Percentage of O2 and Co2 in Flue Gas = 11 & 14, Theoretical Air Requirement.
THE CHEMICAL REACTION ARE
C + O2 ------------------------ CO2
2H2 + O2 -------------------2H2O
S + O2 -------------------- SO2
C + O2 ---------------------CO2
12 + 32 ---------------------- 44
12 kg of carbon requires 32 kg of oxygen to form 44 kg of carbon dioxide. Therefore 1 kg of carbon requires 32/ 12 kg i,e 2.67 kg of oxygen
41.65 kg C + (41.65 X 2.67) kg O2 ---------------------------- 152.85 kg CO2
2H2 + O2 ------------------------------- 2H2O
4 + 32 ------------------------------- 36
4 kg of hydrogen requires 32 kg of oxygen to form 36 kg of water. Therefore 1 kg of hydrogen requires 32/ 4 kg i,e 8 kg of oxygen

2.0413 kg H2 + (2.0413 X 8) kg O2 ------------------------ 18.37 kg H2O
S + O2 ------------------------------------ SO2
32 + 32 ------------------------------------ 64
32 kg of sulphur requires 32 kg of oxygen to form 64 kg of sulphur dioxide. Therefore 1 kg of sulphur require 32/32 kg i,e, 1 kg of oxygen .
0 kg S + (0 X 1) kg O2 ------------------------------------------ 0 kg OF SO2
Total Oxygen Required
(111.20 + 16.33 + 0) = 127.53 kg
Oxygen already present in 100 kg fuel = 14.28 kg (from ultimate analysis)
Additional oxygen required = 127.53 – 14.28 = 113.25kg
Therefore quantity of dry air required = 113.25/0.23 = 492.39 kg OF AIR
(Air Contains 23 % Oxygen By Wt.)
Theoretical air required = 492.39/100
= 4.9239 kg OF AIR /kg OF FUEL

Theoretical CO2 % present in exhaust gases
% CO2 at Theoretical Condition = MOLES OF C
MOLES OF N2 + MOLES OF C


Moles of N2 = WT. OF N2 IN THEORITICAL AIR + WT. OF N2 IN FUEL
MOL. WT. OF N2 MOL. WT. OF N2

Moles of N2 = (4.91 ×7.7)/28 +0.016/28 = 0.1356

Where Moles of C = (WT.OF C IN FUEL )/(MOL.WT OF C)
= 0.4165/12
= 0.0347
(CO2)t = 0.0347/(0.1332+0.0347)
(CO2)t = 20.37%

Excess air present in exhaust gases
Actual Co2 measured in flue gas = 14%
% Excess air supplied (EA) = 7900 X [( CO2%)t – (CO2%)a]
(CO2%)a X[100 – (CO2%)t]

= (7900 ×[20.37-14])/(14 ×[100-20.37])
= 45.17%

Actual mass of air supplied for combustion
Actual mass of air supplied = [1+ EA/100] X theoretical air
= [1+ 45.17/100] X 4.91
= 7.13 kg of air/kg OF COAL
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
Mass of dry flue gas = (0.4165 ×44)/12 +0.016+ (7.13 ×77)/100+((7.13-4.9))/100X23
= 7.54 kg/kg OF COAL

Flue Gas Temperature (Tf) = 180°C
Ambient Temperature (Ta) = 30°C
% Heat Loss in Dry Flue Gas (L1) = (m ×Cp×(Tf-Ta))/(GCV OF FUEL ) X 100
= (7.54 ×0.23 ×(180-30))/3500 X100
L1 = 7.43%
% Heat loss due to formation
of water from h2 in fuel (L2) = (9 ×H2×{584+Cp (Tf-Ta})/(GCV OF FUEL ) X 100
=(9 ×0.02041 ×{584+0.45 (180-30))/3500 X 100
L2 = 3.41 %

% Heat Loss Due To Moisture in Fuel (L3) = (M ×{584+Cp (Tf-Ta)})/(GCV OF FUEL ) X 100
= (0.316 ×{ 584+0.45 (180-30)})/3500 X100
L3 = 5.8 %
% Heat Loss Due To Moisture in Air = (AAS ×HUMIDITY ×Cp ×(Tf-Ta))/(GCV OF FUEL ) X 100
= (7.13 ×0.0204 ×0.45 ×(180-30))/3500 X 100
L4 = 0.28%
% Heat Loss Due To Partial Conversion of C to Co
= %CO X C X 5744
%CO + CO2% GCV OF FUEL

= (0.55 ×0.4165)/(0.55+14) X 5744/3500 X 100
L5 = 2.58%
Heat Loss Due To Radiation and Convection
= 0.548 X [(Ts/55.55)4-(Ta/55.55)4] + 1.957 X (Ts-Ta) 1.25 X
SQRT OF [(196.85 Vm + 68.9)/68.9]
Where
Vm = wind velocity in m/s
Ts = Surface Temperature (K)
Ta = Ambient Temperature (K)
Surface area of boiler = 32.31m2
= O.548 [(353/55.55)4- (303/55.55)4] + 1.957 X (353-303)1.25√((196.85×3.5+68.9))/68.9)
=1271.48 w/m2
=1271.48 X 0.86
=1093.47 Kcal /m2
TOTAL RADIATION AND CONVECTION LOSS PER HOUR
= 1093.47 X 32.31.
= 35330.01 Kcal
% RADIATION AND CONVECTION LOSS = (35330.01 ×100)/(3500 ×450)
L6= 2.24%


Heat loss due to unburnt in fly ash
% ash in coal = 8.63
ratio of bottom ash to fly ash = 90:10
Gcv of fly ash = 452.5 kcal/kg
Amount of fly ash in 1 kg of coal = fly ash % x amount of ash in 1 kg of coal
Total ash %
0.1 x 0.0863
= 0.00863 kg
Heat loss in fly ash = 0.00863 X 452.5
= 3.905 kcal/kg OF COAL
% Heat loss in fly ash = heat loss in fly ash / GCV of coal = 3.905 X 100/3500
L7= 0.11%
% Heat loss due to unburnt in bottom ash
GCV of bottom ash = 800 kcal / kg
Amount of bottom ash in 1 kg of coal = 0.9X 0.0863
= 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 / 3500
L7 =1.77%
BOILER EFFICIENCY BY INDIRECT METHOD
=100-(L1 + L2 + L3 +L4 +L5 +L6 +L7 +L8)
=100-(7.43+3.41+5.8+0.28+2.58+2.24+0.11+1.77)
=76.39%





BOILER FEED PUMP
Rated Data
Motor HP= 5 HP + 5HP
Motor RPM = 2860 RPM
Full Load Current = 7.1 amp
Efficiency = 65%
Measured Data
Measured Current = 3.5 amp AND 3.4 amp
Power Factor = 0.50
Operating Hr = 5 Hr/day
Annual Working Day = 330 days
Power Consumption = (HP ×0.746 ×POWER FACTOR )/(MOTOR EFFICIENCY)
= (5 ×2 ×0.746 ×0.50)/0.65
= 5.73 kWh
Daily electric power consumed = actual power consumption x time of usage/day
= 5.73 X 5
= 28.69kWh/day
Annual electric power consumed = daily consumption x no. of days /year
= 28.69 X 330
= 9468 kWh/yr
Cost of electricity
=9468 kWh/yr. x 4.41 Rs/kWh
=41756Rs/yr.

PROPOSED PUMP
The motor replace with 5 hp ksb kds 527++ pump
Proposed pump data
Flow rate = 14.3 ltr/sec.
Head = 16m
Efficiency = 65%
Specific gravity of water = 1
Power required (P) = (H ×Q×K)/(102×η)
P= Brake power at pump shaft in kW
η = Pump efficiency
H =Total heat in meters
Q= Discharge required in LPS
K= Specific gravity of liquid
102 = Constant
P=(16 ×14.3 ×1)/(102 ×0.65)
P= 3.45 kW
Daily consumption (kWh) = Actual power consumed (kW) x time of usage (hrs/ day)
= 3.45 X 5
= 17.25kWh/day
Annual consumption = Daily consumption (kWh/day) x no. of use days/yr
= 17.25 X 330
= 5692.25kWh/yr.
Cost of electricity
= 5692.5 kWh/yr. X 4.41 Rs/kWh
= 25103.925 Rs/yr.
Economics of proposed pump
The cost of 5 hp ksb kds 527++ pump is of 12000/- Rs
Net annual saving = 41756 – 25103.925
= 16652 Rs.
Simple payback period = (NET COST)/(ANNUAL SAVING ) X 12
= 12000/16652 X 12
= 8.6 MONTHS
BOILER BLOW DOWN PROCESS:
Boiler pressure = 8 kgf/cm2
Boiler rating = 2000kg/hr
Boiler blow down TDS (total dissolved solid) = 4500PPM
Feed water TDS = 500 PPM
% Blow down = (FEED WATER TDS)/(BLOWDOWN TDS-FEED WATER TDS)
= 500/(4500-500)
= 12.5%

If boiler evaporation rate is 2000 kg/hr then required blow down rate is
= (2000 X 12.5 )/100
=250kg/hr
Calculated blow down rate = 250kg/hr
To obtain the energy flow in kW
The blow down rate in kg/s = (250 Kg/hr)/3600
= 0.0694 kg/s
The amount of energy in each kg of blow down sludge, hf AT 8 kg/cm2 = 720kJ/kg
Rate of energy wasted in blow down = 0.0694 X 720
= 49.9kW
Assuming the blow down water is released to a flash steam system operating at 0.5 BAR g. Steam tables may be used to quantify this energy excess.
Specific enthalpy of water at 0.5 BAR g(1.013 bar + 0.5 bar= 1.51 bar abs) = 468 kJ/kg
Excess energy = energy at 8 bar discharge – energy at 1.51 bar = 720kJ/kg – 468 kJ/kg=252 kJ/kg
The specific enthalpy of evaporation ( hfg ) at 1.5 bar from steam tables is 2226 kJ/kg
% Flash steam = (hf HIGH PRESSURE-hf LOW PRESSURE )/(hfg LOW PRESSURE ) X 100
% Flash steam = (720-468)/2226 X 100
= 11.3%
Rate of flash steam generation = 250 kg/hr X 11.3 %
= 28.25 kg/hr (0.0078 kg/s)
Total energy per kg of steam = 2694 kJ/kg (hg AT 1.5 BAR)
Energy flow rate in flash steam = 0.0078 kg/s X 2694kJ/kg
=21.14kJ/s
Compare this to the 49.9 kW rate of energy blow down from the boiler
It may be possible to use this flash steam in this it represent almost 43% of the energy flow rate in the blow down and 11.3% of the water blow down

THERMOPACK (BOILER NO.2)
Capacity = 20, 000, 00 kcal/hr.
Fuel= Reliance peat coke
Avg. coal consumption = 16 TONNE / day.
Pump detail
Rated data
Capacity flow rate (Q) = 160 m3/hr.
Head = 90 m
Pump motor HP =40 HP
Pump motor RPM = 2940 RPM
Full load current = 52 AMP.
Efficiency = 85%
Measured data
Measured current = 53.56 AMP.
Load factor = 1.03
Measured discharge capacity
Brake power = (Q ×S×G×H )/η
Where
Q= Flow rate m3/hr.
S= Specific gravity
G= Gravity acc. =9.8 m/sec
H= Head in m
η = Efficiency
Power = (HP ×0.746×LOAD FACTOR)/(EFFICIENCY )

=(40 ×0.746×1.03×3600)/0.85 = (Q ×0.72 ×9.81 ×90)/0.85
Q= 174 m3/hr.
Heat output
= m Cp ∆T
∆T = AVERAGE ∆T (DIFFERENCE OF SUPPLY & RETURN TEMPERATURE)
= 174 X 1000 X 0.48 X (220 -200)
=1670400 kcal.
Heat input
Total coal consumption = 666 kg/ hr.
GCV of coal = 3500 kcal /kg
Heat input = 666 X 3500
= 2333333kcal /hr.
Efficiency = (HEAT OUTPUT )/(HEAT INPUT )

=1670400/2333333
= 71 %
THERMOPACK EFFICIENCY BY INDIRECT METHOD
FUEL ANALYSIS (IN %)
ASH CONTENT IN FUEL = 8.63
MOISTURE IN COAL = 31.65
CARBON CONTENT = 41.65
HYDROGEN CONTENT= 2.0413
NITROGEN CONTENT = 1.6
OXYGEN CONTENT = 14.48
GCV OF COAL = 3500 kcal/kg
FLUE GAS TEMPERATURE = 220°C
AMBIENT TEMPERATURE = 30°C
PERCENTAGE OF O2 AND CO2 IN FLUE GAS = 11.1 & 8.5
THEORITICAL AIR REQUIREMENT
THE CHEMICAL REACTION ARE
C + O2 ------------------------ CO2
2H2 + O2 -------------------2H2O
S + O2 -------------------- SO2
C + O2 ---------------------CO2
12 + 32 ---------------------- 44
12 KG OF CARBON REQUIRES 32 KG OF OXYGEN TO FORM 44 KG OF CARBON DIOXIDE. THEREFORE 1 KG OF CARBON REQUIRES 32/ 12 kg I,e 2.67 kg OF OXYGEN
41.65 C + (41.65 X 2.67) kg O2 ---------------------------- 152.85 kg CO2
2H2 + O2 ------------------------------- 2H2O
4 + 32 ------------------------------- 36
4 kg OF HYDRIGEN REQUIRES 32 kg OF OXYGEN TO FORM 36 kg OF WATER. THEREFORE 1 kg OF HYDROGEN REQUIRES 32/ 4 kg I,E, 8 kg OF OXYGEN

2.0413 kg H2 + (2.0413 X 8) kg O2 ------------------------ 18.37 kg H2O
S + O2 ------------------------------------ SO2
32 + 32 ------------------------------------ 64
32 kg OF SULPHUR REQUIRES 32 kg OF OXYGEN TO FORM 64 kg OF SULPHUR DIOXIDE. THEREFORE 1 kg OF SULPHUR REQUIRE 32/32 kg ie, 1 kg OF OXYGEN.
0 kg S + (0 X 1) kg O2 ------------------------------------------ 0 kg OF SO2
TOTAL OXYGEN REQUIRED
(111.20 + 16.33 + 0) = 127.53
OXYGEN ALREADY PRESENT IN 100 kg FUEL = 14.28 kg (FROM ULTIMATE ANALYSIS)
ADDITIONAL OXYGEN REQUIRED = 127.53 – 14.28 = 113.25kg
THEREFORE QUANTITY OF DRY AIR REQUIRED = 113.25/0.23 = 492.39 kg OF AIR
(AIR CONTAINS 23 % OXYGEN BY WT.)
THEORITICAL AIR REQUIRED = 492.39/100
= 4.9239 kg OF AIR /kg OF FUEL

THEORITICAL CO2 % PRESENT IN EXHAUST GASES
% CO2 AT THEORITICAL CONDITION = MOLES OF C
MOLES OF N2 + MOLES OF C

MOLES OF N2 = WT. OF N2 IN THEORITICAL AIR + WT. OF N2 IN FUEL
MOL. WT. OF N2 MOL. WT. OF N2

MOLES OF N2 = (4.91 ×7.7)/28 +0.016/28 = 0.1356

WHERE MOLES OF C = (WT.OF C IN FUEL )/(MOL.WT OF C)
= 0.4165/12
= 0.0347
(CO2)t = 0.0347/(0.1332+0.0347)
(CO2)t = 20.37%

EXCESS AIR PRESENT IN EXHUAST GASES
ACTUAL CO2 MEASURED IN FLUE GAS = 8.5%
% EXCESS AIR SUPPLIED (EA) =7900 X [(CO2%)t – (CO2%)a]
(CO2%)a X[100 – (CO2%)t]

= (7900 ×[20.37-8.5])/(8.5 ×[100-20.37])
= 138.54%

ACTUAL MASS OF AIR SUPPLIED FOR COMBUSTION
ACTUAL MASS OF AIR SUPPLIED = [1+ EA/100] X THEORITICAL AIR
= [1+ 138.54/100] X 4.91
= 11.71 kg/kg OF COAL
ACTUAL MASS OF DRY FLUE GAS
MASS O 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
MASS OF DRY FLUE GAS = (0.4165 ×44)/12 +0.016+ (7.13 ×77)/100+((7.13-4.9))/100X23
= 7.54 kg/kg OF COAL

FLUE GAS TEMPERATURE (Tf) = 220°C
AMBIENT TEMPERATURE (Ta) = 30°C
% HEAT LOSS IN DRY FLUE GAS (L1) = (m ×Cp×(Tf-Ta))/(GCV OF FUEL ) X 100
= (7.54 ×0.23 ×(220-30))/3500 X100
L1 = 9.41%
% HEAT LOSS DUE TO FORMATION
OF WATER FROM H2 IN FUEL (L2) = (9 ×H2×{584+Cp (Tf-Ta})/(GCV OF FUEL ) X 100
=(9 ×002041 ×{584+0.45 (220-30))/3500 X 100
L2 = 3.51 %

% HEAT LOSS DUE TO MOISTURE IN FUEL (L3) = (M ×{584+Cp (Tf-Ta)})/(GCV OF FUEL ) X 100
= (0.316 ×{ 584+0.45 (220-30)})/3500 X100
L3 = 5.71 %
% HEAT LOSS DUE TO MOISTURE IN AIR = (AAS ×HUMIDITY ×Cp ×(Tf-Ta))/(GCV OF FUEL ) X 100
= (11.71 ×0.0204 ×0.45 ×(220-30))/3500 X 100
L4 = 0.58%
% HEAT LOSS DUE TO PARTIAL CONVERSION OF C TO CO
= (% CO ×C)/(% CO+% CO2 ) X 5744/(GCV OF FUEL ) X 100
= (0.55 ×0.4165)/(0.55+8.5) X 5744/3500 X 100
L5 = 4.15%
HEAT LOSS DUE TO RADIATION AND CONVECTION
= 0.548 X [(Ts/55.55)4-(Ta/55.55)4] + 1.957 X (Ts-Ta)1.25 X
= SQRT OF [(196.85 Vm + 68.9)/68.9]
WHERE
Vm = WIND VELOCITY IN m/s
Ts = SURFACE TEMPERATURE (K)
Ta =AMBIENT TEMPERATURE (K)
= O.548 [(353/55.55)4- (303/55.55)4] + 1.957 X (353-303)1.25√((196.85×3.5+68.9))/68.9)
=1271.48W/m2
=1271.48 X 0.86
=1093.47 Kcal /m2
TOTAL RADIATION AND CONVECTION LOSS PER HOUR
= 1093.47 X 90
= 98412.3 Kcal
% RADIATION AND CONVECTION LOSS = (98412.32 ×100)/(3500 ×450)
L6 = 4.22%



HEAT LOSS DUE TO UNBURNT FUEL IN FLY ASH
% ASH IN COAL = 8.63
RATIO OF BOTTOM ASH TO FLY ASH = 90.10
GCV OF FLY ASH = 452.5 kcal/kg
AMOUNT OF FLY ASH IN 1 kg OF COAL = 0.1 X 0.0863
= 0.00863 kg
HEAT LOSS IN FLY ASH = 0.00863 X 452.5
= 3.905 kcal/kg OF COAL
HEAT LOSS IN FLY ASH = 3.905 X 100/3500
L7 = 0.11%
% HEAT LOSS DUE TO UNBURNT IN BOTTOM ASH
GCV OF BOTTOM ASH = 800 kcal / kg
AMOUNT OF BOTTOM ASH IN 1 kg OF COAL = 0.9X 0.0863
= 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 / 3500
L8 =1.77%
BOILER EFFICIENCY BY INDIRECT METHOD
100-(L1 + L2 + L3 +L4 +L5 +L6 +L7 +L8)
100-(9.41+3.51+5.71+0.58+4.15+4.22+0.11+1.77)
=76.25%

COMPRESSOR DETAIL
RATED DATA COMPRESSOR 1 COMPRESSOR 2
Motor HP 10 HP 10HP
Motor Rpm 1430 RPM 1430 RPM
Full load current 11 AMP 11 AMP
Efficiency 7.3 AMP 7.5 AMP
Measured Data
Measured current 7.3 AMP 7.5 AMP
Power factor 66% 67%
Avg. running hrs/day 4 hrs 6hrs
Annual operating days/hr 330days 330 days

COMPRESSOR 1
Power consumption rating of motor = 10 HP
= 10 X 0.746 KW
= 7.46 KW
Actual power consumed = (POWER CONSUMPTION ×POWER FACTOR )/(EFFICIENCY OF MOTOR)
= (7.4 ×0.66)/0.85
= 5.74KW
Daily electric power consumed by the motor = Actual power consumed x time of usage
= 5.74 X 4
=22.98 kWh
Annual electric power used by motor = Daily consumption x no. of days use
= 22.98 X 330
= 7584.56kWh/yr.
Cost of electricity used by motor/yr. = 7584.56 kWh/yr X 4.41 Rs/kWh
= 33447.9 Rs/yr.

COMPRESSOR 2
Power consumption rating of motor = 10 HP
= 10 X 0.746 KW
= 7.46KW
Actual power consumed = (POWER CONSUMPTION ×POWER FACTOR )/(EFFICIENCY OF MOTOR )
= (7.46 ×0.67)/0.85
= 5.81 kW
Daily electric power consumed by the motor= Actual power consumed x time of usage
= 5.81 X 6 kWh
= 34.87 kWh
Annual electric power used by motor = Daily consumption x no. of days /yr
= 34.87kWh X 330
= 11507kWh/yr.
Cost of electricity used by month / yr. = 11507 kWh/yr. X 4.41 Rs/yr
= 50748Rs/yr.
PERFORMANCE TEST AND MEASUREMENT
P1= Initial pressure = 5 kg/cm2
P2=Final pressure = 7kg/cm2
Pa= Atmospheric pressure (ABSOLUTE)
Vr= Volume of air receiver and interconnected pipes = 300 ltr = 0.300 m3
t =Time in seconds for pressure to rise from P1 to P2 = 56 SEC.
TC= Temperature correction factor
TR=Air temperature in receiver K = 31°C= 304K
Ta=Ambient temperature K= 30°C= 303K
TC=(273+Ta)/(273+Tr) = (273+303 )/(273+304 ) = 0.99
Q= (P2-P1)/1.026 X Vr X 60/t XTCX (35.32 FOR m3 TO CUB. FEET)
= (7-5)/1.026 X 0.300 X 60/56 X 0.99 X 35.52
= 21.90 CFM
Specific power consumption
Actual output = 21.90 CFM
Actual consumption = 5.74 kW
= Actual output / actual consumption
=21.90/5.74
=3.81 CFM/kW
%Leakage in the compressor air system
= T/(T+t ) X100
T= 64 SEC.
T= 172 SEC.
= 64/(64+172 ) X 100
= 27%
The quantity of compressed air leakage
Leakage quantity =Percentage leakage x actual output
= 0.27 X 21.90
= 5.91 CFM
Calculate the power lost due to leakage
= Leakage quantity / specific energy
= 5.91 CFM / 3.81 CFM/kW
= 1.549 kW
Daily electric power lost by motor
= Actual power lost x time of usage
=1.549 X 4
6.196 kWh
Annual electric power lost by motor
= 6.196 X 330
= 2044.68 kWh / yr.
Cost of electricity lost by motor /yr.
= 1548.624 kWh/ yr. X 4.41 Rs/kWh
=9017.11 Rs/yr.
PERFORMANCE TEST AND MEASUREMENT OD COMPRESSOR 2
P1= Intial pressure = 4 kg/cm2
P2= Final pressure= 6 kg/cm2
Pa= Atmospheric pressure (ABSOLUTE)
Vr= Volume of air receiver and interconnected pipes= 300 LTR= 0.300 m3
t = Time in second for pressure to rise from P1 to P2= 58 SEC
TC= Temperature correction factor
Tr = Air temperature in receiver K = 31 °C= 304K
Ta= Ambient temperature K= 30°C= 303K
TC=(273+Ta)/(273+Tr ) = (273+303 )/(273+304 ) = 0.99
Q= (P2-P1)/1.026 X Vr X 60/t X TC X (35.32 FOR m3 TO CU. FEET)
=(6-4)/1.026 X 0.300 X 60/58 X 0.99 X 35.32
Q= 21.53 CFM
Specific power consumption
=Actual output = 21.53 CFM
=Actual consumption = 5.81 kW
=Actual output / actual consumption
=21.53/5.81
3.70 CFM / kW
% Leakage in the compressed air system
% leakage in the system
= T/(T+t) X 100
T= 52 sec
T = 66 sec
=52/(52+66) X 100
= 44 %
The quantity of compressed air leakage
Leakage quantity= Percentage leakage x actual output
= 0.44 X 21.153
= 9.30 CFM
Calculate the power lost due to leakage
=Leakage quantity/ specific energy
= 9.30 CFM / 5.81 CFM/kW
=1.60kW
Daily electric power lost by motor
= Actual power lost x time of usage
= 1.60 X 6
= 9.60 kWh
Annual electric power lost by motor
= 9.60 X 330
= 3168kWh/yr
Cost of electricity lost by motor / yr
= 3168kWh/yr X 4.41 Rs/kWh
= 13970Rs/yr
LIGHNING SYSTEM
MAKE: Crompton greaves/ Aluminum choke
Assumption
1 Daytime use = 12 Hrs
2 Night time use = 12 Hrs
Energy consumption of each fluorescent tube
1 Tube light = 40W
2 Choke = 14W
3 Total = 54 W
Sr. LOCATION NO. OF TUBE USED DAYTIME NO. OF TUBE USED NIGHT TIME
1 Main Gate --- 4
2 D.G.Sets --- 7
3 Boiler house --- 9
4 Zero-Zero M/C 1 2
5 Felt M/C 3 4
6 Fitter room 2 2
7 Stenter 1 2 16
8 Stenter2 2 14
9 Front fitter room 2 7
10 Jet Dyeing M/C - 7
11 Store 4 15
12 Laboratory 4 4
13 F.B. Printing M/C 10 10
14 F.B. Printing M/C2 11 11
15 F.B. Printing M/C3 12 12
16 F.B. Printing M/C4 11 11
17 F.B. Printing M/C5 11 11
18 F.B. Printing M/C6 10 10
19 F.B. Printing M/C7 8 8
20 F.B. Printing M/C8 9 9
21 Miscellaneous - 15
TOTAL 102 185
Daily electric power consumed by tube light in daytime
= Actual power consumed x no. of tube light x time of usage
= (54 W ×102 ×12 hr)/1000
=66kWh/daytime
Annual electric power consumed by tube light
= Daily consumption x no. of days use/yr.
=66kWh X 330
=21780 kWh/yr.
Cost of electricity consumed by tube light
= 21780kWh/yr. X 4.41 Rs/kWh
= 96049.8 Rs/yr.
Daily electric power consumed by tube light in nighttime
=Actual power consumed x no. of tube light x time of usage
= (54 W ×185 ×12)/1000
=119.88kWh/nighttime
Annual electric power consumed by tube light
=Daily consumption x no. of days use /yr.
=119.88 X 330
=39560.4kWh/yr.
Cost of electricity used by tube light
= 39560.4kWh/yr X 4.41 Rs/kWh
=174461.36 Rs/yr.
Case 1
Interlock F.B. Printing M/C tube light
Avg. tube light on one F.B. Printing M/C= 10 NOS.
Total M/C = 8 NOS.
Total no. of tube light = 8 X 10
= 80 NOS.
Approx. hrs saving = 5 hr/day
Annual working days = 330
Unit cost = 4.41/ unit
Annual cost saving= (ACTUAL POWER CONSUMPTION ×NO.OF TUBELIGHT ×TIME OF USAGE× NO.OF DAYS×UNIT COST)/1000
=(54 ×80 ×5 ×330 ×4.41)/1000
=31434
=31500Rs/yr
Case 2 retrofit the tube light with CFL of 15w
No. of tube light = 56
Daily power consumption of tube light before retrofitting
= Actual power consumed x no. of tube light x time of usage
= (54 ×56 ×24)/1000
=72.57kWh/day
Annual electric power used
= Daily consumption x no. of days use / yr
= 72.57 kWh X 330
= 23950kWh/yr
Cost of electricity
= 23950 kWh/yr. X 4.41 Rs/kWh
= 105619.8 Rs/yr.
Daily power consumption of tube light after retrofit with CFL of 15 W
= Actual power consumed x no. of tube light x time of usage
=(15 ×56 ×24)/1000
=20.16kWh/DAY
Annual electric power used
= Daily consumption x no. of days use /yr.
= 20.16 X 330
=6652.8kWh/yr
Cost of electricity
= 6652.8 kWh/yr. x 4.41Rs/kWh
29338.84 Rs/yr
Net annual saving = 105619.8- 29338.84
= 76280.9 Rs/yr.
Payback period
=(TOTAL COST)/(ANNUAL SAVING ) X 12
Total cost of CFL = 85 Rs X 56
= 4760 Rs
= 4760/76280.9 X 12
= 0.748 Yr
Case 3 retrofit the tube light with energy efficient tube light of 28 W
No. of tube light =47
Energy consumed by tube = 27W
Energy consumed by choke = 1 W
Total = 28 W
Daily power consumption before retrofit
Daily power consumption of tube light before retrofitting
= Actual power consumed x no. of tube light x time of usage
= (54 ×47 ×24)/1000
=60.92kWh/day
Annual electric power used
= Daily consumption x no. of days use / yr
= 60.92 kWh X 330
= 20100kWh/yr
Cost of electricity
= 20100 kWh/yr. X 4.41 Rs/kWh
= 88645Rs/yr.
Daily power consumption after retrofit
=(ACTUAL POWER CONSUMED ×NO.OF TUBELIGHT ×TIME OF USAGE )/1000
=(28 ×47 ×24)/1000
=31.48kWh/day
Annual electric power consumption
= Daily consumption x no. of days use /Yr.
= 31.58 X 330
= 10421.4kWh/Yr.
Cost of electricity
= 10421.4kWh/yr. X 4.41 Rs/kWh
= 45958.37 Rs/yr.
Net annual saving = 88645-45958
= 42686Rs/yr
Payback period
=(TOTAL COST)/(ANNUAL SAVING ) X 12
Total cost of energy efficient lightning = 500 Rs X 47
= 23500 Rs
= 23500/42686 X 12
= 6.6 MONTH

DETERMINATION OF ELECTRIC MOTOR LOAD
Jet Dyeing machine 1,2
Measured data
V= 414 VOLT
I= 11 AMP
P.F=95%
P=7.4 kW
Rated data
Hp= 10 Hp
Efficiency = 75%
Case1
PI= (V ×I×PF×√3)/1000
Where
PI=Three phase power in kW
V= RMS VOLTAGE
I= RMS CURRENT
P.F= Power factor
PI=(414 ×11×0.95×1.732)/1000
=7.49 kW
PIR= (HP ×0.7457)/η
Where
Pir= Input power at full rated
Hp=Name plate rated horse power
η = Efficiency at full rated load
=(10 ×0.7457)/(0.75 )
= 9.94 kW
Load = Pi/Pir X100
Load = Output power as a % of rated power
Pi = Measured three phase power in kW
Pir= Input power at full rated load in kW
Load = 7.49/9.94 X 100
= 74.4%
Case 2
Motor efficiency saving
Replace the standard efficiency (S.E) motor with energy efficient (E.E) motor whose efficiency = 86%
Then
kW saved = 0.746 X [ HP1/(EFFICIENCY )–HP2/(EFFICIECNY )]
HP1 =Existing motor load
HP2 = Motor load for new motor
kW saved = 0.746[10/0.75-10/0.86]
= 1.269 KW
Now
Input power at full rated load with 86% efficiency
Pir= (HP X 0.7457)/(EFFICIENCY )
= (10 ×0.7457)/0.86
= 8.67 kW
Load = 7.49/8.6 X 100
= 86 %
Comment
Most electric motors are designed to runs at 50% to 100% of rated load. In case 1 motor runs at 74.4% which is quiet good. But in case 2 when we replace S.E motor with E.E motor it runs at 86% load which is better
Jet Dyeing machine 3, 4
Measured data
V= 414 volt
I = 13 amp
PF = 77 %
P= 6.4 kW
Rated Data
HP = 10 HP
Efficiency = 75%
PI=(V ×I×PF×√3)/1000
=(414 ×13×0.77×1.732)/1000
=7.17 kW
Pir=(HP ×0.7457)/η
= (10 ×0.7457)/0.75
= 9.94 kW
Load =PI/PIR X 100
= 7.17/9.94 X 100
= 72.1%
Case 1
Motor efficiency saving
Replace the standard efficiency motor (S.E) with energy efficient motor whose efficiency = 86%
KW saved = 0.746X [HP1/EFFICIENCY-HP2/EFFICIENCY]
= 0.746X [10/0.75-10/0.86]
= 1.269 kW
Now input power at full rated load with 86% efficiency
Pir= (HP ×0.7457)/EFFICIENCY
= (10 ×0.7457)/0.86
= 8.69 kW
Load = Pi/Pir X 100
= 7.17/8.67 X 100
= 82%
Case3
Replace the 10 Hp S.E. motor with 8 Hp (E.E) motor then
Motor efficiency saved
kW saved = 0.746 X [HP1/EFFICIENCY-HP2/(EFFICIENCY )]
= 0.746 X [10/0.75-8/0.86]
= 3.004 kW
Now input power at full rated load with 8 HP motor
Pir =HP X 0.7457/(EFFICIENCY )
= 8 X 0.7457/0.86
= 6.93 kW
Load = Pi/Pir X 100
= 7.17/6.93 X 100
= 103 %
Comment
In case1 motor runs at 72.1% which is quiet good .in case 2 when we replace S.E. motor with E.E motor it runs at 82% load which is better but in case3 when motor is replaced with 8 HP E.E motor.Motor runs at 103 % load which is not feasible
Stenter machine 1
Measured data
V= 431 volts
I= 58.1 amp
PF= 85.1%
P= 42.69 kW
Rated data
HP = 75 HP
Efficiency = 0.75
Pi = (V ×I×PF× √3)/1000
= (431 ×58.1×0.851×1.732)/1000
= 36.90 kW
Pir =HP X 0.7457/EFFICIENCY
= 75 X 0.7457/0.75
= 74.57 kW
Load = Pi/Pir X 100
= 36.90/7457 X 100
= 49.48%
Case 2
Replace the S.E motor with E.E motor when efficiency is 86%
Motor energy saved
kW saved = 0.746 X [(HP1 )/(EFFICIENCY )–(HP2 )/(EFFICIENCY )]
= 0.746 X [75/0.75-75/0.86]
= 9.54 kW
Now input power at full rated load with 86% efficiency
Pir =HP X 0.7457/EFFICIENCY
= 75 X (0.7457 )/0.86
= 65.03 kW
Load = Pi/Pir X 100
= 36.90/65.03 X 100
= 56%
Case 3
Replace the 75 HP S.E. motor with 50 HP E.E motor
Motor efficiency saved
kW saved = 0.746 X [HP1/EFFICIENCY-HP2/(EFFICEINCY )]
= 0.746 X [75/0.75-50/0.86]
= 31.22 kW saved
Pir =HP X 0.7457/EFFICIENCY
= 50 X 0.7457/0.86
= 43.35 kW
Load = Pi/Pir X 100
= 36.90/43.35 X 100
= 85.1%
Comment
In case 1 motor runs at 49.48% loading which is not good. in case 2 motor runs at 56% loading but in case 3 motor runs at 85% loading which is better than case 1 and case 2

STENTER NO. 2
MEASURED DATA
V= 424 VOLTS
I= 76 AMP
PF= 80%
P=51.73 kW
RATED DATA
HP = 75 HP
EFFICIENCY = 75%
Pi = (V ×I ×PF ×√3)/1000
= (424 ×76 ×0.80 ×1.732)/1000
= 44.64 kW
Pir = HP X 0.7457/EFFICIENCY
= 75 X 0.7457/0.75
= 74.57 kW
LOAD = Pi/Pir X 100
= 44.64/75.57 X 100
= 59.86%
CASE 2
REPLACE THE S.E MOTOR WITH E.E MOTOR WHOSE EFFICEINCY IS 86%
MOTOR ENERGY SAVED
kW SAVED = 0.746 X [(HP1 )/(EFFICIENCY )–(HP2 )/(EFFICIENCY )]
= 0.746 X [75/0.75-75/0.86]
= 9.54 kW
NOW INPUT POWER AT FULL RATED LOAD WITH 86% EFFICIENCY
Pir =HP X 0.7457/EFFICIENCY
= 75 X (0.7457 )/0.86
= 65.03 kW
LOAD = Pi/Pir X 100
= 44.64/65.03 X 100
= 68.6%
CASE 3
REPLACE 75 HP S.E MOTOR WITH 50 HP E.E MOTOR
MOTOR EFFICIENCY SAVED
kW saved = 0.746 X [HP1/EFFICIENCY-HP2/(EFFICEINCY )]
= 0.746 X [75/0.75-50/0.86]
= 31.22 kW saved
Pir =HP X 0.7457/EFFICIENCY
= 50 X 0.7457/0.86
= 43.35 kW
LOAD = Pi/Pir X 100
= 44.64/43.35 X 100
= 103%
COMMENT
IN CASE 1 MOTOR RUNS AT 59.86% WHICH IS QUIET GOOD . IN CASE 2 MOTOR RUNS AT 68.6% LOAD WHICH IS BETTER BUT IN CASE 3 MOTOR RUNS AT 103% LOAD WHICH IS NOT FEASIBLE
LOOP MACHINE
MEASURED DATA
V= 423 VOLT
I= 15.47 AMP
PF=95 %
P= 6.35 kW
RATED DATA
HP = 15 HP
EFFIECIENY = 86%

Pi = (V ×I ×PF ×√3)/1000
= (423 ×15.47 ×0.95 ×1.732)/1000
= 10.76 kW
Pir = HP X 0.7457/EFFICIENCY
= 15 X 0.7457/0.86
= 13 kW
LOAD = Pi/Pir X 100
= 10.76/13 X 100
= 82.7%
COMMENT
IN THIS CASE MOTOR RUNS AT 82.7% LOADING WITH 86% EFFICIENCY OF MOTOR SO NEED TO REPLACE THE MOTOR

THRMOPACK
MEASURED DATA
V= 423 VOLT
I= 56.1 AMP
P.F= 86.2%
P= 44.92 kW
RATED DATA
HP = 65 HP
EFFICIENCY = 86%
Pi = (V ×I ×PF ×√3)/1000
= (423 ×56.1 ×0.862 ×1.732)/1000
= 35.42 kW
Pir = HP X 0.7457/EFFICIENCY
= 65 X 0.7457/0.75
= 56.36 kW
LOAD = Pi/Pir X 100
= 35.42/56.36 X 100
= 62.8%
CASE 2
REPLACE THE 65 HP S.E MOTOR WITH 50 HP E.E MOTOR
MOTOR EFFICIENCY SAVED
kW saved = 0.746 X [HP1/EFFICIENCY-HP2/(EFFICEINCY )]
= 0.746 X [65/0.86-50/0.86]
= 13.01 kW saved
Pir =HP X 0.7457/EFFICIENCY
= 50 X 0.7457/0.86
= 43.35 kW
LOAD = Pi/Pir X 100
= 35.42/43.35 X 100
= 81%
COMMENT
IN CASE 1 MOTOR RUNS AT 62.8% LOADING WHICH IS GOOD AND IN CASE 2 MOTOR RUNS AT 81% LOADING WHICH IS BETTER
COMPRESSOR
MEAUSRED DATA
V= 423 VOLT
I= 10 AMP
P.F= 64.5%
P=5.23 kW
RATED DATA
HP = 10 HP
EFFICIENCY = 75%
Pi = (V ×I ×PF ×√3)/1000
= (423 ×10 ×0.645 ×1.732)/1000
= 4.72 kW
Pir = HP X 0.7457/EFFICIENCY
= 10 X 0.7457/0.86
= 8.67 kW
LOAD = Pi/Pir X 100
= 4.72/8.67 X 100
= 54.4%
CASE 2
REPLACE 10 HP S.E MOTOR WITH 8 HP E.E. MOTOR THEN
MOTOR EFFICIENCY SAVED
kW saved = 0.746 X [HP1/EFFICIENCY-HP2/(EFFICEINCY )]
= 0.746 X [10/0.86-8/0.86]
= 3.04 kW saved
Pir =HP X 0.7457/EFFICIENCY
= 8 X 0.7457/0.86
= 6.9 kW
LOAD = Pi/Pir X 100
= 4.72/6.93 X 100
= 68%
COMMENT
IN CASE 1 MOTOR RUNS AT 54.4% LOADING WHICH IS GOOD BUT IN CASE2 MOTOR RUNS AT 68% LOADING WHICH IS BETTER

INTRODUCTION ABOUT TEXTILE INDUSTRY

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1 INTRODUCTION

The textile industry occupies a vital place in the Indian economy and contributes substantially to its exports earnings. Textile exports represents nearly 30 percent of the country‘s total exports. It has a high weight age of over 20 percent in the national production. India is the world’s second largest producer of textiles after China. It is the world’s third largest producer of cotton – after China and the USA – and the second largest cotton consumer after china. The textile industry in India is one of the oldest manufacturing sectors in the country and is currently it’s largest.

The textile industry occupies a unique place in our country. One of the earliest to come into existence in India , it accounts for 14 percent of the total industrial production, contributes to nearly 30 percent to the total export and is the second largest employment generator after agriculture. The Indian textile industry is one of the largest in the world with a massive raw material and textile – manufacturing base. Indian economy is largely dependent on the textile manufacturing and trade in addition to other major industries about 27 percent of the exchange earning are on account of export of textile and clothing alone. The textiles and clothing sector contributes about 14 percent to the industrial production and 3 percent to the gross domestic product of the country. Around eight percent of the total excise revenue collection is contributed by the textile industry. So much so, the textile industry accounts for as large as 21 percent total employment generated in the economy. Around 35 million people are directly employed in the textile manufacturing activities.

1.1 HISTORY OF TEXTILE INDUSTRY IN INDIA

India has been well known for her textile goods since very ancient times. The traditional textile industry of India was virtually decayed during the colonial regime. However, the modern textile industry took birth in India in the early nineteenth century when the first textile mill in the country was established at fort gloster near Calcutta in 1818. The cotton textile industry, however, made its real beginning in Bombay, in 1850s. The first cotton textile mill of Bombay was established in 1854 by a Parsi cotton merchant then engaged in overseas and internal trade. Indeed, the vast majority of the early mills were the handiwork of Parsi merchants engaged in yarn and cloth trade at home and Chinese and African markets. The first cotton mill in Ahmadabad, which was eventually to emerge as a rival centre to Bombay, was established in 1861. The spread of the textile industry to Ahmadabad was largely due to the Gujarat trading class. The cotton textile industry made rapid progress in the second half of the nineteenth century and by the end of the century there were 178 cotton textile mills; but during the year 1900 the cotton textile industry was in bad state due to the great famine and a number of mills of Bombay and Ahmadabad were to be closed down for long periods.

The partition of the country at the time of independence affected the cotton textile industry also. The Indian union got 409 out of the 423 textiles mills of the undivided India. 14 mills and 22 per cent of the land under cotton cultivation went to Pakistan. Some mills were closed down for some time. For a number of years since independence, Indian mills had to import cotton from Pakistan and other countries. After independence, the cotton textile industry made rapid strides under the plans. Between 1951 and 1982 the total number of spindles doubled from 11 million to
22 million. It increased further to well over 26 million by 1989-90.

1.2 GENERAL DESCRIPTION ABOUT THE ENERGY CONSERATION PATTERN IN TEXTILE INDUSTRY

Textile industry is an energy intensive industry and more so is textile wet processing. The break – up of energy consumption during the various operation of spinning, weaving and wet processing of textile report for UK and the Indian textile industry is as follows

UK India

Spinning 25% 24%

Weaving 23% 28%

Wet processing 52% 48%

From the above figures it is evident that wet processing operation consumes almost 50% of the total energy requirement of a composite textile mill. This is an attributable to the fact that wet processing operation involves heating of large quantities of water, drying of wet fabric, high temperature such as heat setting, high temperature, dyeing and curing operation etc.

Steam plays a vital role in the wet processing of fabrics not only due to the fact that steam cost is more than 30% of the total processing cost but also if steam quality is good, cloth processing is also good.

The consumption of utilities for mill processing cotton will be as under:

Steam 12.3 kg/kg

Power 0.73 kwh/kg

Water 13.3 lit/m

In spite of such an important role of steam for cloth, processing, generally the aspects of steam generation, distribution and the Utilization are neglected in most of the textile mills.

For a textile unit, processing around 40,000 meters of cloth per day the fuel consumption varies around 10,000 (Rs.) which is 10% of total sales

Due to depleting nature of fossil fuels the cost and avail ability of coal/oil both will be problem in near future. So the role of steam will be further enhanced and this is going to be a major deciding factor in the survival of composite and processing units.

1.3 TEXTILE CLUSTER OF PALI DISTRICT

In textile cluster of Pali district (Rajasthan) mostly wet processing is carried out. In wet processing, dyeing & finishing of fabric take place. Dyeing & finishing process are the major energy consuming areas. In these processes mainly thermal & electrical energy are used. It is known that thermal energy in textile mill is largely consumed in two operations, in heating of water & drying of water. Fuel consumption in textile mills is mostly directly proportional to amount of water consumed. Hence if consumption of water can be reduced, it will also save energy. Conservation of energy can be affected through process and machine modification, proper chemical recipes and new technologies. In Pali the annual turnover is shown in fig. 1 where the colors of the bars represent the rupees in lakh and no. On the bars represent the no. of textile units. The annual production is shown in fig 2 where the color bars represent the no. of production in meters and the no. On the bars represents the no. of textile unit.

Fig. 1.1 Annual turnover of cluster of textile unit located in Pali district

Fig.1.2 Annual production of cluster of textile unit located in Pali district

1.4 PHYSICAL LOCATION OF RESEARCH ACTIVITY

The small scale textile industries of Pali city are located at Mandia road industrial phase 1 of Pali city and medium scale textile industries are located at industrial phase 2 and phase 3 of Pali city

Fig 1.3 Physical location of research activity

Fig 1.4 Physical location of research activity

1.5 HORIZON OF PROBLEM AND OBJECTIVE OF PRESENT STUDIES

A comprehensive survey of the issues associated with the current performance level of the textile industries of Pali cluster was carried out. The objective of the survey was to identify the barriers that are responsible for the poor energy efficiency levels and technological obsolescence of the industry as compared to the developed countries. The textile industries of Pali cluster mainly comprises of Small and Medium Enterprises (SMEs). Besides these the survey unveils the real facts associated with the Textile industries of Pali Cluster. Ten Textile industries were surveyed. In this assessment the industries of different types and capacities have been thoroughly analyzed.

Through this sectoral analysis of Textile industry an attempt has been made to identify the energy conservation opportunities in the Textile industries sector and to find out the various ways to improve the overall performance of these industries. Further the effect of implementation and financial analysis of Energy Efficient (EE) technologies in the sector has been examined.

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