The document is a research paper on energy efficiency presented by Pratap Jung Rai. It discusses various industrial utilities used in energy generation like boilers, furnaces, electric motors, pumps, compressors and HVAC systems. For boilers and furnaces, it explains the components, methodology to assess performance, calculate efficiency and opportunities to improve energy efficiency. For electric motors, it discusses factors affecting efficiency, methods to measure efficiency and load, and opportunities like using efficient motors and avoiding under-loading.
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Energy efficiency of industrial utilities
1. Research Paper
on
Presented by:
Pratap Jung Rai
(065/BIE/047)
Symbol No: 39792
Thapathali Campus, Nepal
1 11/5/2014
2. What is Energy Efficiency?
An Energy Efficiency/Audit is an inspection, survey and analysis of energy for energy conservation in
an industry, building, process or system to reduce the amount of energy input to the system without
negatively affect the output.
Objectives of Energy Efficiency of Industrial Utilities
To minimize energy waste/costs.
To achieve and maintain optimum energy procurement and
utilization.
Enhance environmental performance and minimize GHG
emissions.
Improve reputation with costumer, public and government
Energy Generation
Industrial Utilities:
1) Boiler
2) Furnace
3) Electric Motor
4) Pump
5) Compressor
6) HVAC System
Pratap Jung Rai 2 11/5/2014
3. Types of Energy Audit
1) Preliminary Energy Audit
Shortly, called Walk-Through Audit. Its name implies, is a
tour of the facility to visually inspect each of the energy using
systems.
2) Targeted Energy Audit
It often results from preliminary audits. They provide data and
detailed analysis on specified target projects. For example,
industries may target its lighting system or boiler system.
3) Detailed Energy Audit
It is a comprehensive audit and results in a detailed energy
project implementation plan for a facility, since it accounts for
the energy use of all major equipment. Detailed energy
auditing is carried out in three phases
a) Pre-audit Phase
b) Audit Phase
c) Post-Audit
Methodology of Energy Efficiency
Pratap Jung Rai 4 Source: UNEP
4. 1)Boiler
What is a Boiler?
• Enclosed vessel that heats water to become hot water or
steam
• At atmospheric pressure water volume increases 1,600
times
• Hot water or steam used to transfer heat to a process
BURN
ER
WATER
SOURCE
Brine
SOFTENERS
CHEMICAL
FEED
FUEL
VENT
BLOW DOWN
SEPARATOR
VENT
EXHAUST
STEAM TO GAS
PROCESS
STACK
PUMPS
BOILER
ECO-NOMI-ZER
Causes of poor boiler performance
-Poor combustion
-Heat transfer surface fouling
-Poor operation and maintenance
-Deteriorating fuel and water quality
Pratap Jung Rai 5 11/5/2014
5. Heat Balance
Balancing total energy entering a boiler against the energy
that leaves the boiler in different forms
Heat loss due to moisture in air
Heat in Steam
BOILER
Heat loss due to dry flue gas
Heat loss due to steam in fuel
gas
Heat loss due to moisture in
fuel
Heat loss due to unburnts in
residue
Heat loss due to radiation &
other unaccounted loss
12.7 %
8.1 %
1.7 %
0.3 %
2.4 %
1.0 %
100.0 %
Fuel
73.8 %
Heat Balance
How energy is transformed from fuel into useful energy,
heat and losses
Avoidable losses include: Stoichiometric
Excess Air
Un burnt
FUEL INPUT
STEAM OUTPUT
Stack Gas
Ash and Un-burnt
parts of Fuel in Ash
Blow
Down
Convection &
Radiation
Energy flow diagram
Pratap Jung Rai 6 11/5/2014
6. Steam Output
Boiler Flue gas
Boiler Efficiency
Fuel Input, 100%
Efficiency = 100 – (i + ii + iii + iv + v + vi + vii)
Air
i. Dry Flue gas loss
ii. H2 loss
iii. Moisture in fuel
iv. Moisture in air
v. Fly ash loss
vii. Surface loss
vi. Unborn fuel loss
a) Direct Method (Input output Method)
Boiler efficiency () =
Heat Input
Heat Output
x 100
Q x (hg – hf)
= x 100
Q x GCV
Where,
Q- Quantity of steam generated kg/hr
hg - Enthalpy of saturated steam in kcal/kg of steam
hf - Enthalpy of feed water in kcal/kg of water
GCV- Gross calorific value kcal/kg
b) Indirect Method
Advantages
• Complete mass and energy balance for each individual
stream
• Makes it easier to identify options to improve boiler
efficiency
Disadvantages
• Time consuming
• Requires lab facilities for analysis
Advantages
• Quick evaluation
• Few parameters for computation
• Few monitoring instruments
Disadvantages
• No explanation of low efficiency
• Various losses not calculated
Efficiency of boiler () = 100 – (i+ii+iii+iv+v+vi+vii)
Pratap Jung Rai 7
7. Energy Efficiency Opportunities
1.Stack (flue) temperature control
• Keep as low as possible
• If >200°C then recover waste heat
2. Feed water preheating using economizers
• Proper economizer can reduce 15-20% fuel
consumption
3. Combustion air pre-heating
• If combustion air raised by 20°C = 1% improve
thermal efficiency
4. Incomplete combustion minimization
• Air shortage, fuel surplus, poor fuel distribution
• Poor mixing of fuel and air
5. Excess air control
• 1% excess air reduction = 0.6% efficiency rise
6. Avoid radiation and convection heat loss
• Fixed heat loss from boiler shell, regardless of
boiler output
• Repairing insulation can reduce loss
7. Automatic blow down control
• Sense and respond to boiler water conductivity
and pH
8. Reduction of scaling and soot losses
• Every 22oC increase in stack temperature = 1%
efficiency loss
• 3 mm of soot = 2.5% fuel increase
Pratap Jung Rai 8 11/5/2014
8. 2) Furnace
What is a Furnace?
Equipment to melt metals
• Casting
• Change shape
• Change properties
Low efficiencies due to
• High operating temperature
• Emission of hot exhaust gases
Furnace Components
Furnace chamber:
constructed of
insulating materials
Hearth: support
or carry the steel.
Consists of
refractory
materials
Burners: raise or
maintain chamber
temperature
Chimney: remove
combustion gases
Charging & discharging doors for
loading & unloading stock
What are Refractories:
Materials that
• Withstand high temperatures and sudden changes
• Withstand action of molten slag, glass, hot gases
• Withstand load at service conditions
• Withstand abrasive forces
• Conserve heat
• Have low coefficient of thermal expansion
• Will not contaminate the load
Pratap Jung Rai 9 11/5/2014
9. Assessment of Furnaces
Fuel Input 100%
Useful heat in
stock
(30-50%)
Wall loss (3 -10 %)
Flue loss
(20-50 %)
Opening loss (1-2 %)
Cooling loss (5-10 %)
Stored heat (2 -5 %)
Other loss
Recycled heat (10-30%)
Furnace
Instruments to Assess Furnace Performance
Parameters
to be
measur
ed
Location of
measurement
Instrument
required
Required
Value
Furnace soaking
zone temperature
(reheating furnaces)
Soaking zone and
side wall
Pt/Pt-Rh thermocouple
with indicator and
recorder
1200-1300oC
Flue gas
temperature
In duct near the
discharge end, and
entry to recuperate
Chromel Alummel
Thermocouple with
indicator
700oC max.
Flue gas
temperature
After recuperate Hg in steel thermometer 300oC (max)
Furnace hearth
pressure in
the heating
zone
Near charging end
and side wall over the
hearth
Low pressure ring gauge +0.1 mm of
Wc
Oxygen in flue gas In duct near the
discharge end
Fuel efficiency monitor for
oxygen and temperature
5% O2
Billet temperature Portable Infrared pyrometer or
optical pyrometer
-
Energy Losses Areas
Pratap Jung Rai 10 11/5/2014
10. Furnace Efficiency
a) Direct Method
Thermal efficiency of furnace
= Heat in the stock / Heat in fuel consumed
for heating the stock
Heat in the stock Q:
Q = m x Cp (t1 – t2)
Where,
Q = Quantity of heat of stock in kCal
m = Weight of the stock in kg
Cp= Mean specific heat of stock in kCal/kg ℃
t1 = Final temperature of stock in ℃
t2 = Initial temperature of the stock before it enters the
furnace in ℃
b) Indirect Method
It is similar to the Boiler indirect efficiency
Method
Example:
Heat losses
a) Flue gas loss = 57.29 %
b) Loss due to moisture in fuel = 1.36 %
c) Loss due to H2 in fuel = 9.13 %
d) Loss due to openings in furnace = 5.56 %
e) Loss through furnace skin = 2.64 %
Total losses = 75.98 %
Furnace efficiency =Heat supply minus total
heat loss
Furnace Efficiency = 100% – 76% = 24%
Pratap Jung Rai 11 11/5/2014
11. Energy Efficiency Opportunities
1. Complete combustion with minimum excess air
2. Proper heat distribution
3. Operation at the optimum furnace temperature
4. Reducing heat losses from furnace openings
5. Maintaining correct amount of furnace draft
6. Optimum capacity utilization
7. Waste heat recovery from the flue gases
8. Minimize furnace skin losses
9. Use of ceramic coatings
10. Selecting the right refractories
Pratap Jung Rai 12 11/5/2014
12. 3) Electric Motor
What is an Electric Motor?
• Electromechanical device that converts
electrical energy to mechanical energy
• Mechanical energy used to e.g.
• Rotate pump impeller, fan, blower
• Drive compressors
• Lift materials
• Motors in industry: 70% of electrical load
Motors loose energy when serving a load
Factors that influence efficiency
Age
Temperature
Load
Rewinding
Capacity
Speed
Type
Efficiency of Electric Motors
Efficiency
Pratap Jung Rai 13 Load 11/5/2014
13. Assessment of electric motors
Efficiency of Electric Motors
Motor load is indicator of efficiency
Input power measurement
Ratio input power and rate power at 100% loading
• Three steps for three-phase motors
Step 1. Determine the input power:
Pi = Three Phase power in kW
V = RMS Voltage, mean line to
line of 3 Phases
I = RMS Current, mean of 3
phases
PF = Power factor as Decimal
V x I x PF x 3
1000
Pi
Step 2. Determine the rated power:
• Compare slip at operation with slip at full load
0.7457
r
r P hp x
Where,
Pr = Input Power at Full Rated load
hp = Name plate Rated Horse Power
r = Efficiency at Full Rated Load
Step 3. Determine the percentage load:
• Compare measured amperage with rated
amperage
x 100%
Pi
P
Load
r
Where,
Load = Output Power as a % of Rated Power
Pi = Measured Three Phase power in kW
Pr = Input Power at Full Rated load in kW
Pratap Jung Rai 14 11/5/2014
14. Energy Efficiency Opportunities
1. Use energy efficient motors
Efficiency 3-7% higher
Wide range of ratings
More expensive but rapid payback
2. Reduce under-loading (and avoid
over-sized motors)
If motor operates at <50%
Not if motor operates at 60-70%
3. Improve power quality
too high fluctuations in voltage and
frequency
4. Rewinding
sometimes 50% of motors
5. Power factor correction by capacitors
Benefits of improved PF
•Reduced kVA
•Improved voltage regulation
Capacitor size not >90% of no-load kVAR of motor
6. Improve maintenance
Inspect motors regularly for wear, dirt/dust
Checking motor loads for over/under loading
Lubricate appropriately
Check alignment of motor and equipment
Provide adequate ventilation
7. Speed control of induction motor
Variable speed drives (VSDs)
•Reduce electricity by >50% in fans and pumps
•Convert 50Hz incoming power to variable
frequency and voltage: change speed
Pratap Jung Rai 15 11/5/2014
15. Assessment of compressors and compressed air systems
Simple Capacity Assessment Method
Where,
P2 = Final pressure after filling (kg/cm2a)
P1 = Initial pressure (kg/cm2a) after bleeding)
P0 = Atmospheric pressure (kg/cm2a)
V = Storage volume in m3 which includes receiver, after cooler and
delivery piping
T = Time take to build up pressure to P2 in minutes
Compressor Efficiency
Isothermal efficiency
Isothermal efficiency =
Actual measured input power / Isothermal power
Isothermal power (kW) = P1 x Q1 x loge r / 36.7
Where,
P1 = Absolute intake pressure kg / cm2
Q1 = Free air delivered m3 / hr
r = Pressure ratio P2/P1
Volumetric efficiency
Volumetric efficiency
= Free air delivered m3/min / Compressor displacement
Compressor displacement = Π x D2/4 x L x S x χ x n
D = Cylinder bore, meter
L = Cylinder stroke, meter
S = Compressor speed rpm
χ = 1 for single acting and 2 for double acting cylinders
n = No. of cylinders
Pratap Jung Rai 16 11/5/2014
16. 1. Location
• Significant influence on energy use
2. Elevation
• Higher altitude = lower volumetric efficiency
3. Air Intake
• Keep intake air temperature low
• Every 4 oC rise in inlet air temperature = 1%
higher energy consumption
4. Pressure Drops in Air Filter
• Install filter in cool location or draw air from cool
location
• Keep pressure drop across intake air filter to a
minimum
Every 250 mm WC pressure drop = 2% higher
energy consumption
5. Use Inter and After Coolers
• Inter coolers: heat exchangers that remove heat
between stages
• After coolers: reduce air temperature after final
stage
6. Pressure Settings
a) Reducing delivery pressure
Operating a compressor at 120 PSIG instead of 100
PSIG: 10% less energy and reduced leakage rate
7. Minimizing Leakage
• Tighten joints and connections
One pinpoint of compressed air = 60000 IC
8. Condensate Removal
• Condensate formed as after-cooler reduces
discharge air temperature
• Install condensate separator trap to remove
condensate
Energy Efficiency Opportunities
Pratap Jung Rai 17 11/5/2014
17. 5) HAVC System
High Temperature
Reservoir
Heat Rejected
R Work Input
Heat Absorbed
Low Temperature
Reservoir
Condenser
Evaporator
High
Pressure
Side
Low
Pressure
Side
Compressor
Expansion
Device
1 2
3
4
Choice of compressor, design of condenser,
evaporator determined by
•Refrigerant
•Required cooling
•Load
•Ease of maintenance
•Physical space requirements
•Availability of utilities (water, power) COP increases with
rising evaporator
temperature (Te)
COP increases with
decreasing condensing
Pratap Jung Rai 18 temperature (Tc) 11/5/2014
18. Assessment of Refrigeration and AC
Assessment of Refrigeration
TR = Q xCp x (Ti – To) / 3024
Q = mass flow rate of coolant in kg/hr
Cp = is coolant specific heat in kCal /kg deg C
Ti = inlet, temperature of coolant to evaporator (chiller) in 0C
To = outlet temperature of coolant from evaporator (chiller) in 0C
Coefficient of Performance (COPCarnot)
•Standard measure of refrigeration efficiency
•Depends on evaporator temperature Te and condensing
temperature Tc:
COPCarnot = Te / (Tc - Te)
•COP in industry calculated for type of compressor:
Cooling effect (kW)
COP =
Power input to compressor (kW)
Assessment of Air Conditioning
Measure
• Airflow Q (m3/s) at Fan Coil Units (FCU) or Air
Handling Units (AHU): anemometer
• Air density (kg/m3)
• Dry bulb and wet bulb temperature:
• Enthalpy (kCal/kg) of inlet air (hin) and outlet air
(Hout): psychometric charts
3024
psychrometer
Calculate TR:
Q ρ h
h
TR
in out Pratap Jung Rai 19 11/5/2014
19. Energy Efficiency Opportunities
1.Optimize process heat exchange
• 1oC raise in Te = 3% power savings
2. Multi-staging systems
• 0.55◦C reduction in returning water from
cooling tower = 3.0 % reduced power
Condition
Te
(0C)
Tc
(0C)
Refrigeration
Capacity* (TR)
Specific
Power
Consumption
(kW/TR)
Increase
kW/TR (%)
Normal 7.2 40.5 17.0 0.69 -
Dirty condenser 7.2 46.1 15.6 0.84 20.4
Dirty evaporator 1.7 40.5 13.8 0.82 18.3
Dirty condenser
and evaporator
1.7 46.1 12.7 0.96 38.7
3. Matching capacity to system load
4. Capacity control of compressors
• continuous modulation through vane control
5. Multi-level refrigeration for plant needs
• Monitor cooling and chiller load: 1 chiller full
load more efficient than 2 chillers at part-load
6. Chilled water storage
Economical because
• Chillers operate during low peak demand
hours: reduced peak demand charges
• Chillers operate at nighttime: reduced
tariffs and improved COP
7. System design features
• FRP impellers, film fills, PVC drift eliminators
• Softened water for condensers
• Economic insulation thickness
Pratap Jung Rai 20 11/5/2014
“Energy Audit means the verification, monitoring and analysis of use of energy including submission of technical report containing recommendations for improving energy efficiency with cost benefit analysis and an action plan to reduce energy consumption.”
A boiler is an enclosed vessel that provides a means for combustion heat to be transferred to water until it becomes heated water or steam.
When water at atmospheric pressure is boiled into steam its volume increases about 1,600 times, producing a force that is almost as explosive as gunpowder. This causes the boiler to be an equipment that must be treated with utmost care
The hot water or steam under pressure is then usable for transferring the heat to a process.
Stack (flue) temperature control
Keep as low as possible
If >200°C then recover waste heat
2. Feed water preheating using economizers
Proper economizer can reduce 15-20% fuel consumption
3. Combustion air pre-heating
If combustion air raised by 20°C = 1% improve thermal efficiency
4. Incomplete combustion minimization
Air shortage, fuel surplus, poor fuel distribution
Poor mixing of fuel and air
5. Excess air control
1% excess air reduction = 0.6% efficiency rise
6. Avoid radiation and convection heat loss
Fixed heat loss from boiler shell, regardless of boiler output
Repairing insulation can reduce loss
7. Automatic blow down control
Sense and respond to boiler water conductivity and pH
8. Reduction of scaling and soot losses
Every 22oC increase in stack temperature = 1% efficiency loss
3 mm of soot = 2.5% fuel increase
A furnace’s efficiency increases when the percentage of heat that is transferred to the stock or load inside the furnace increases. The efficiency of the furnace can be calculated in two ways, similar to that of the boiler: direct method and indirect method.
Direct method
The efficiency of a furnace can be determined by measuring the amount heat absorbed by the stock and dividing this by the total amount of fuel consumed.
Thermal efficiency of the furnace =Heat in the stock / Heat in the fuel consumed for heating the stock
The quantity of heat (Q) that will be transferred to stock can be calculated with this equation:
Q = m x Cp (t1 – t2)
Where,
Q = Quantity of heat of stock in kCal
m = Weight of the stock in kg
Cp= Mean specific heat of stock in kCal/kg oC
t1 = Final temperature of stock in oC
t2 = Initial temperature of the stock before it enters the furnace in oC
The furnace efficiency can also be determined through the indirect method, similar to the evaluation of boiler efficiency. The principle is simple: the heat losses are subtracted from the heat supplied to the furnace. (Note that a detailed methodology to calculate each individual heat loss is provided in the chapter)
Adding the losses a to f up gives the total losses:
Flue gas loss = 57.29 %
Loss due to moisture in fuel = 1.36 %
Loss due to H2 in fuel = 9.13 %
Loss due to openings in furnace = 5.56 %
Loss through furnace skin = 2.64 %
Total losses = 75.98 %
(Click once) The furnace efficiency calculated through the indirect method = 100 – 75.98 = 24.02%
Motor performance is affected considerably by the quality of input power, which is determined by the actual volts and frequency compared to rated values. Fluctuation in voltage and frequency much larger than the accepted values has detrimental impacts on motor performance.
Voltage unbalance can be even more detrimental to motor performance and occurs when the voltages in the three phases of a three-phase motor are not equal.
Several factors can affect voltage balance: single-phase loads on any one phase, different cable sizing, or faulty circuits. An unbalanced system increases distribution system losses and reduces motor efficiency.
An example of the effect of voltage unbalance on motor performance is shown in the table. Give one example: small voltage unbalances are acceptable but, for example, a voltage unbalance of 5.4% results in a temperature increase of 40oC!!!
The refrigeration cycle is shown in the Figure and can be broken down into the following stages (note to the trainer: the next slides will explain what is happening between 1 to 4)