Tr faculty polytech s. verma

Training Workshop
on Energy Conservation

2
Presenter's Profile
Sandeep Verma
BE (Electricals) Honours, Gold Medalist
University of Roorkee
Accredited Energy Auditor
Lead Auditor (EnMS: ISO 50001)
ECBC Master Trainer
Area of Expertise: Energy Management

Various Forms of Energy
 Potential Energy
 Chemical
 Nuclear
 Mechanical stored energy
 Gravitational
 Kinetic Energy
 Radiant
 Thermal
 Motion
 Sound
 Electrical

Electrical Energy Basics
 Work
Work done on a body, in Joules W = Fs
Where F is the force in Newtons and s is the distance in meters moved by the body in the
direction of the force.
 Power
power in watts, P = W/t
Where W is the work done or energy transferred in Joules and t is the time in seconds.
 Direct current:
Non-varying unidirectional current. Current produced by batteries
 Alternating Current
Reverses in regular recurring intervals with alternate +ve and –ve values at specified
number of times per second. In 50 cycle AC, current reverses direction 100 times a second
(twice in one cycle)
 Amps or Ampere (A)
Current is the rate of flow of charge. Ampere is the basic unit of electric current.
 Voltage or Volts (V)
It is a measure of electric potential or electromotive force. A potential of one Volt (V) appears
across a resistance of one Ohm when a current of one Ampere flows through the
resistance.

 Resistance and Conductance
R = Volts /Amp = V/I
The reciprocal of resistance is called conductance and is measured in siemens (S). Thus,
conductance, in mho or Siemens G = 1/R, where R is the resistance in ohms.
 Frequency (Hertz)
The supply frequency is the number of cycles at which alternating current changes. The unit of
frequency is cycles / second or Hz.
 Electrical Energy
When a direct current of I amperes is flowing in an electric circuit and the voltage across the circuit
is V volts, then,
Power, in watts P = VI
Electrical energy = Power x time
= V x I x t Joules
Although the unit of energy is the Joule, when dealing with large amounts of energy, the unit used
is the kilowatt hour (kWh ).
 Power Factor kW = kVA Cos 
For a balanced three-phase load,
Power, Watts = 3 VL x IL Cos
For a balanced single-phase load,
Power, Watts = VL x IL Cos

Electrical Tariff
 Billing Demand is highest KVA recorded during a demand
interval(15/ 30 minutes) in one month -measured by digital tri-
vector meter
 Maximum Demand is maximum KVA or KW over one billing cycle
 Contract Demand is amount of electric power (KVA or KW)
demanded by consumer from power utility
Connected Load, Demand Factor & Load Factor
 Connected Load is name plate readings (in KVA or KW) of all
equipment in a industry
 Demand Factor = Maximum Demand/
Connected Load
 Load factor = Average Load/ Maximum Load

Time of Day (TOD)
Utilities prefer flat demand curve
Encourage users to draw power during off-peak
hours (i.e. Night time)
–Utilities have TOD offer or incentives or
disincentives
–Separate energy meters to record peak and off-
peak power consumption
–Opportunity for user as off-peak power tariff is
usually lower

Thermal Energy Basics
 Temperature is the degree of hotness or coldness measured on a
definite scale and is denoted in Fahrenheit (F) or Celsius (C)
 Calorie is unit of heat: 1 kilocalorie can raise 1 kg of water by 1oC
 Specific heat is heat required to raise 1 kg of water by 1oC
 Heat quantity = mass x specific heat x rise in temperature
 Latent heat of fusion is amount of heat applied to solid to change its
state to liquid
 Latent heat of vaporization is amount of heat applied to liquid to
change its state to vapour
 Superheat is heating of vapour to a temperature higher than its boiling
point at existing pressure
 Dew point is temperature at which condensation of water vapour from
the air begins as the temperature of the air-water vapour mixture falls

 Moisture content in air is humidity
 Specific Humidity is Actual weight of water vapour mixed in 1 kg of dry air
 Humidity Factor is kg of water vapour per kg of dry air
 Relative Humidity is degree of saturation of air at any dry bulb temperature
(given as % of actual water content of air divided by moisture content of air at
exiting temperature)
 RH = Percentage (%) at temperature (oC)
 Absolute Pressure (Ps) is total or true pressure. It is measured relative to
the absolute zero pressure - the pressure that would occur at absolute vacuum.
 Gauge Pressure (Pg) is the pressure indicated by a gage. All gages are
calibrated to read zero at atmospheric pressure (Pa).
• Pg = Ps - Pa
 Heat Transfer: The rate of energy transfer, more commonly called heat
transfer, is measured in Watts (J/s). Heat is transferred by three primary modes:
Conduction (Energy transfer in a solid)
Convection (Energy transfer in a fluid)
Radiation (doesn't need a material to travel through)

 Absolute Pressure (Ps) is total or true pressure. It is measured relative to
the absolute zero pressure - the pressure that would occur at absolute vacuum.
 Gauge Pressure (Pg) is the pressure indicated by a gage. All gages are
calibrated to read zero at atmospheric pressure (Pa).
Pg = Ps - Pa
 Heat Transfer: The rate of energy transfer, more commonly called heat
transfer, is measured in Watts (J/s)
Heat is transferred by three primary modes:
Conduction (Energy transfer in a solid)
Convection (Energy transfer in a fluid)
Radiation (doesn't need a material to travel through)

 Dry Bulb Temperature is an indication of the sensible heat
content of air-water vapour mixtures
 Wet Bulb Temperature is a measure of total heat content or
enthalpy. It is the temperature approached by the dry bulb and
the dew point as saturation occurs
 Specific Gravity is density of fuel relative to water
 Viscosity is measure of internal resistance to flow
 Calorific Value is measure of heat content in organic matter
– Gross or high calorific value and Net or low calorific value

 1 Joule (J) = 0.2390057 Calorie (Cal) = 9.478172×10-4 British
Thermal Unit (Btu)
 1 Cal = 4.184 J = 3.965667×10-3 Btu
 1 Btu = 1055.056 J = 252.1644 cal = 2.930711×10-4 kilowatt-hour (kWh)
 1 kWh = 3.6 Megajoule (MJ) = 0.8604207 Mcal = 3412.142 Btu
Power (Energy Rate) Equivalents
1 kilowatt (kW) 1 kilo joule /second (kJ/s)
1 kilowatt (kW) 3413 BTU/hour (Btu/hr.)
1 horsepower (hp) 746 watts (0.746 kW)
1 Ton of refrigeration 12000 Btu/hr.
Units of Energy

 Energy is one of the major inputs for the economic
development of any country.
 The global need for energy is increasing on an average by
about 1.5% every year.
 Criteria for Classification of Energy
 Primary Energy and Secondary Energy
 Commercial and Non-Commercial Energy
 Renewable and Non-Renewable Energy
Introduction

 Primary energy refers to all types of energy extracted or captured directly
from natural resources.
 Primary Energy Content of all the fuel is expressed as TOE (Tonne of Oil
Equivalent) ==>> 1 TOE = 10 x 106 kCal = 11630 kWh = 41870 MJ
 Further Classified into
 Renewable (solar, wind, geothermal, tidal, biomass etc.)
 Non-Renewable (fossil fuels: crude oil and its products, coal, natural gas,
nuclear etc.)
 Secondary Energy Sources are derivatives from Primary Energy Sources.
(ex: electricity, steam etc.)
Primary & Secondary Energy

Major Primary and Secondary
Energy Sources

Commercial Energy:
Energy that is available in the market for a definite price
Eg.: electricity, coal, refined petroleum products and natural gas
Non-Commercial Energy:
Energy which is sourced within a community and its surrounding area, and
which is not normally traded in the commercial market
Eg: Firewood and agro waste in rural areas, solar energy for water heating,
electricity generation, and for drying grain, fish and fruits; animal power for
transport, threshing, lifting water for irrigation, crushing sugarcane etc.; wind
energy for lifting water and electricity generation.

Renewable energy is the energy
obtained from natural sources
which are essentially
inexhaustible.
A non-renewable resource is a natural
resource which cannot be produced,
grown, replenished, or used on a scale
which can sustain its consumption rate.
Renewable & Non-Renewable Energy

Global Primary Energy Reserves &
Commercial Energy Production
Coal

Natural Gas
 Natural gas is a gaseous fossil fuel consisting primarily of methane but
also includes small quantities of ethane, propane, butane and pentane.

Global Primary Energy Consumption

Natural Gas Sector
 Termed as the fuel of the 21st century.
 Natural gas is cooled to -161oC, it is transformed into Liquefied Natural Gas (LNG).
 Compressed Natural Gas (CNG) is made by compressing natural gas (which is
mainly composed of methane [CH4]) to less than 1% of the volume it occupies
at standard atmospheric pressure and is stored and distributed in hard containers
of cylindrical or spherical shapes, at a pressure of 200–248 kg/cm2.
 Bulk of onshore production comes from Gujarat, Assam and Nagaland.
 The disadvantages with the use of natural gas are unpredictability in its price and
uncertainty in its availability

Reserves/Production (R/P): If the reserves remaining at the end of the year are
divided by the production in that year, the result is the length of time that the
remaining reserves would last if production were to continue at that level.

Energy Conservation and its Importance
Energy conservation is achieved when
growth of energy consumption is reduced
in physical terms
Energy efficiency is achieved when energy
intensity in a specific product, process or
area of production or consumption is
reduced without affecting output,
consumption or comfort levels.
Promotion of energy efficiency will contribute
to energy conservation and is therefore an
integral part of energy conservation
promotional policies.

ENERGY CONSERVATION ACT- 2001
AND RELATED POLICIES

Definition of Energy Management
“The strategy of adjusting and optimizing energy,
using systems and procedures so as to reduce
energy requirements per unit of output while
holding constant or reducing total costs of
producing the output from these systems”
“The judicious and effective use of energy to
maximize profits (minimize costs) and enhance
competitive positions”

As per the Energy Conservation Act, 2001, Energy
Audit is defined as:
“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 “
Definition of Energy Audit

Three top operating expenses are energy (both
electrical and thermal), labour and materials.
Energy would emerge as a top ranker for cost reduction
primary objective of Energy Audit is to determine ways
to reduce energy consumption per unit of product
output or to lower operating costs
Energy Audit provides a “ bench-mark” (Reference
point) for managing energy in the organization
Need for Energy Audit

Preliminary energy audit
Detailed energy audit
Type of energy audit chosen depends on
Function and type of industry
Depth to which final audit is needed
Potential and magnitude of cost
reduction desired
Types of Energy Audit

 Establish energy consumption in the organization (sources: energy bills and
invoices)
 Obtain related data such as production for relating with energy consumption
 Estimate the scope for energy savings
 Identify the most likely and the easiest areas for attention (e.g. unnecessary
lighting, higher temperature settings, leakage etc.)
 Identify immediate (especially no-/low-cost) improvements/ savings
 Set up a baseline or reference point for energy consumption
 Identify areas for more detailed study/measurement
Preliminary Energy Audit
Methodology

Targeted Energy Audits
 Targeted energy audits often results from preliminary audits.
 They provide data and detailed analysis on specified target projects.
 For example, an organization may target its lighting system or boiler
system or steam system or compressed air system with a view of
effecting energy savings.
 Targeted audits therefore involve detailed surveys of the target subjects
and analysis of the energy flows and cost associated with the targets.
 Final outcome is the recommendations regarding actions to be taken.

Evaluates all energy using system,
equipment and include detailed energy
savings and costs
Carried out in 3 phases:
Pre-audit Phase
Audit Phase
Post-Audit
Detailed Energy Audit

Ten Steps Methodology for
Detailed Audit
Step No PLAN OF ACTION PURPOSE / RESULTS
PHASE I –PRE AUDIT PHASE
Step 1  Plan and organise
 Walk through Audit
 Informal Interview with
Energy Manager, Production
/ Plant Manager
 Establish/organize a Energy audit team
 Organize Instruments & time frame
 Macro data collection (suitable to type of
industry.)
 Familiarization with process/plant activities
 First hand observation & Assessment of current
level of operation and practices
Step 2  Introductory meeting with
all divisional heads and
persons concerned with
energy management (1-2
hrs.)
 To built up cooperation and rapport
 Orientation, awareness creation
 Issue questionnaire tailored for each
department

PHASE II –AUDIT PHASE
Step 3  Primary data gathering,
Process Flow Diagram, &
Energy Utility Diagram
 Historic data collection and analysis for setting up
Baseline energy consumption
 All service utilities system diagram (e.g. Single line
power distribution diagram, water, and compressed
air & steam distribution).
 Prepare process flow charts
 Design, operating data and schedule of operation
 Annual Energy Bill and energy consumption
pattern
(refer manual, logbook, name plate etc.)
Step 4  Conduct survey and
monitoring
 Measurements :
Motor survey, Insulation, lighting survey etc. with
portable instruments for operating data. Confirm and
compare operating data with design data.
Step 5  Conduct of detailed trials
/tests for selected major energy
equipment
 Trials/ Tests
- 24 hours power monitoring (MD, PF, KWH etc.).
- Load variations trends in pumps, fan compressors
etc.
- Boiler/Efficiency trials for (4 – 8 hours)
- Furnace Efficiency trials
- Equipments Performance tests etc

Step 6  Analysis of energy use  Energy and Material balance
 Energy loss/waste analysis
Step 7  Identification and
development of Energy
Conservation (ENCON)
opportunities
 Conceive, develop and refine ideas
 Review ideas suggested by unit personal
 Review ideas suggested in previous energy audit
report if any
 Use brainstorming and value analysis techniques
 Contact vendors for new/efficient technology
Step 8  Cost benefit analysis  Assess technical feasibility, economic viability and
prioritization of ENCON options for implementation
 Select the most promising projects
 Prioritise by low, medium, long term measures
Step 9  Reporting & Presentation to
the Top Management
 Documentation, draft Report Presentation to the top
Management.
 Final report preparation on feedback from unit
PHASE II –POST AUDIT PHASE
Step 10  Implementation and Follow-
up
Implementation of ENCON recommendation
measures and Monitor the performance
Action plan, schedule for implementation
Monitoring and periodic review

Energy audit team should ensure that the following baseline data are collected:
 Quantity and type of raw materials
 Technology, process used and equipment used
 Capacity utilization
 Efficiencies / yield
 Percentage rejection / reprocessing
 Quantity and types of wastes
 Consumption of fuel, water, steam, electricity, compressed air, cooling water,
chilled water

Identification of
Energy Conservation
Opportunities
Energy generation
Energy distribution
Energy usage by processes
Fuel substitution

Summary Of Energy Saving Recommendations
S.No.
Energy Saving
Recommendations
Annual Energy
(Fuel & Electricity)
Savings (kWh/MT
or KL/MT)
Annual
Savings
Rs. Lakhs
Capital
Investment
(Rs. Lakhs)
Simple
payback
period
1
2
3
4
Total

Types and Priority of Energy Saving Measures
Type of Energy
Saving Options
Annual
Electricity/Fuel
savings
Annual
Savings
Priority
kWh/MT or
kJ/MT
(Rs. Lakhs)
A
No Investment (Immediate)
Operational Improvement
Housekeeping
B
Low Investment (Short to Medium
Term)
Controls
Equipment Modification
Process change
C
High Investment (Long Term)
Energy efficient Devices
Product modification
Technology Change

Sources of Waste Heat
 Boilers
 Kilns
 Ovens
 Furnaces
 DG Sets
 Thermic Fluid Heaters
 Low Temperature Process Streams

High Temperature Heat Recovery
Types of Device Temperature, oC
Nickel refining furnace 1370 –1650
Aluminium refining furnace 650-760
Zinc refining furnace 760-1100
Copper reverberatory furnace 900-1100
Cement kiln (Dry process) 620- 730
Glass melting furnace 1000-1550

Types of Device Temperature, oC
Steam boiler exhausts 230-480
Gas turbine exhausts 370-540
Reciprocating engine exhausts 315-600
Heat treating furnaces 425 – 650
Catalytic crackers 425 – 650
Annealing furnace cooling
systems
425 - 650
Medium Temperature
Heat Recovery

Source Temperature, oC
Process steam condensate 55-88
Cooling water from:
Furnace doors 32-55
Annealing furnaces 66-230
Air compressors 27-50
Air conditioning and
refrigeration condensers
32–43
Low Temperature Heat Recovery

Benefits of WHR
 Direct Benefits
 Indirect Benefits
Reduction in Pollution
Reduction in
Equipment Sizes
Reduction in auxiliary
Energy consumption

 Recuperator
 Regenerator
 Heat Wheel
 Heat Pipe
 Economiser
 Shell and tube heat exchanger
 Waste heat boiler
 Heat Pump
 Thermo-compressor
 Direct contact heat exchanger
Commercial WHR Devices

Recuperator
In a recuperator, heat exchange
takes place between the flue
gases and the air through
metallic or ceramic walls. Duct
or tubes carry the air for
combustion to be pre-heated,
the other side contains the
waste heat stream.

Metallic Radiation Recuperator
The radiation recuperator
gets its name from the fact
that a substantial portion of
the heat transfer from the
hot gases to the surface of
the inner tube takes place by
radiative heat transfer
Figure 8.2 Metallic Radiation Recuperator

Convective Recuperator
The hot gases are carried
through a number of
parallel small diameter
tubes, while the incoming
air to be heated enters a
shell surrounding the
tubes and passes over the
hot tubes one or more
times in a direction
normal to their axes.

Convective Radiative Recuperator
For maximum effectiveness of
heat transfer, combinations of
radiation and convective
designs are used, with the
high-temperature radiation
recuperator being first followed
by convection type.

Ceramic Recuperator
In order to overcome the temperature
limitation of metallic recuperators which is
about 1000oC on the gas side, ceramic tube
recuperators have been developed.
The materials of ceramic recuperators allow
operation on the gas side up to 1300 oC and
on the preheated air side up to 850 oC.

Regenerator
The Regeneration which is
preferable for large capacities
has been very widely used in
glass and steel melting
furnaces.
Important relations exist
between the size of the
regenerator, time between
reversals, thickness of brick,
conductivity of brick and heat
storage ratio of the brick.

Heat Wheels
Widely used in low to medium
temperature waste heat
recovery systems.
A disk rotates between two side-
by-side ducts: one a cold gas
duct, the other a hot gas duct.
As the disk slowly rotates,
sensible heat (moisture that
contains latent heat) is
transferred to the disk by the
hot air and, as the disk rotates,
from the disk to the cold air.

Heat Pipe
Heat pipe is a thermal energy absorbing and
transferring system and have no moving parts
and hence require minimum maintenance.
wall

Economiser
Widely used in a
boiler system
1. To pre-heat the
boiler feed water.
2. To pre-heat the
combustion air.
by utilizing the heat
in the flue gas.

Shell and Tube Heat Exchanger
When the medium containing waste heat is a liquid or a
vapor which heats another liquid, then the shell and tube
heat exchanger must be used since both paths must be
sealed to contain the pressures of their respective fluids.

Plate Heat Exchanger
The cost of heat exchange surfaces is a major factor when the temperature
differences are not large. In such case this is used. Hot liquid passing through a
bottom port in the head is permitted to pass upwards between every second
plate while cold liquid at the top of the head is permitted to pass downwards
between the odd plates. When the directions of hot & cold fluids are
opposite, the arrangement is described as counter current
Figure 8.10 Plate Heat Exchanger

Waste Heat Boilers
Waste heat boilers are ordinarily
water tube boilers in which the
hot exhaust gases from gas
turbines, incinerators, etc., pass
over a number of parallel tubes
containing water. The water is
vaporized in the tubes and
collected in a steam drum from
which it is drawn off for use as
heating or processing steam

Thermo-compressor
SUCTION
STEAM L.P.
Discharge
Steam
M.P
Motive
Steam
H.P
The thermo-compressor is a simple equipment with a
nozzle where HP steam is accelerated into a high velocity
fluid. This entrains the LP steam by momentum transfer
and then recompresses in a divergent venturi.

Transmission and Distribution
 Sub transmission at 132 kV,110 kV, 66 kV ,33kV,22kV
 Distribution at 11kV, 6.6 kV, 3.3 kV
The power plants produce 50 cycle/second (Hertz), (AC)
electricity with voltages between 11kV and 33kV.
At the power plant site, the 3-phase voltage is stepped up to
a higher voltage for transmission on cables strung on cross-
country towers.
High voltage (HV) and extra high voltage (EHV) transmission
is the next stage from power plant to transport A.C. power
over long distances at voltages like; 220 kV & 400 kV.
Where transmission is over 1000 kM, high voltage direct
current transmission is also favoured to minimize the losses.

Generation
Efficiency 1
Efficiency ranges 28-35 % with respect to size of thermal plant,
age of plant and capacity utilisation
Step-up Station
2
Step-up to 400 / 800 kV to enable EHV transmission.
Envisaged max. losses 0.5 % or efficiency of 99.5 %
EHV
Transmission &
3 Station
EHV transmission and substations at 400 kV / 800 kV.
Envisaged maximum losses 1.0 % or efficiency of 99 %
HV
Transmission &
Station 4
HV transmission & Substations for 220 / 400 kV.
Envisaged maximum losses 2.5 % or efficiency of 97.5 %
Sub-
transmission 5
Sub-transmission at 66 / 132 kV
Envisaged maximum losses 4 % or efficiency of 96 %
Distribution
Station 6
Step-down to a level of 11 / 33 kV.
Envisaged losses 0.5 % or efficiency of 99.5 %
Primary
Distribution 7
Distribution is final link to end user at 11 / 33 kV.
Envisaged losses maximum 5 % of efficiency of 95 %
End user
Premises
Cascade efficiency from Generation to end user
= 1  2  3  4  5  6  7
The cascade efficiency in the T&D system from output of the power plant to the end use is
87% (i.e. 0.995 x 0.99 x 0.975 x 0.96 x 0.99.5 x 0.95 = 87%)

Why high voltage is preferred in
T&D?
• V=IR
– Higher the voltage lesser the Voltage drop
– Voltage drop proportional to the ratio of voltages
• Power loss = I2R
– Higher the voltage, lesser the current and lesser the power
loss
• For ex. If voltage is raised from 11 kV to 33 kV the
voltage drop would be lowered by a figure of 1/3 and
line loss would be lowered by (1/3)2
• Higher voltage can also bring down the conductor
sizes on account of lower currents handled

MECHANICALELECTRICAL
Electrical Distribution System –Single Line Diagram
FANS & BLOWERS PUMPS R & AC COMPRESSORS LIGHTING, HEATING LOAD
M M M M C
M
M
DG SET
EB POWER SUPPLY
Metering
Transformer
.Effi.98-99.5%
Feeders,
Capacitors
Distribution
Panels
Motors
Effi.85-96%
TRIVECTOR METER
11 KV/440V
Transformer
C
C
Dist. loss
1 -6%
Sub-station
Industrial End User
ONE Unit Saved = TWO Units Generated

What are the components of
Electricity Billing?
For Industry: Two Part Tariff for HT Consumers
The consumer pays for two components.
- Energy Charges for kWh consumed
- Maximum demand Charges (kVA) registered
- Plus
• PF penalty or PF incentives
• Fuel Cost adjustments
• Electricity Duty Charges
• Meter rentals
• Lighting and fan power consumption
• TOD, (peak and non-peak )
• Penalty for exceeding contract demand
• Surcharge if metering is at LT side in some of the utilities

1. Load Curve Generation
2. Rescheduling of Loads
3. Storage of Products/in process material/ process
utilities like refrigeration
4. Shedding of Non-Essential Loads
5. Operation of Captive Generation and Diesel
Generation Sets
6. Reactive Power Compensation
What are the Load Management
Strategies

Inductive loads require two kinds of power:
1. Active (or working) power to perform the work (motion) and
2. Reactive power to create and maintain electro-magnetic fields.
The vector sum of the active power and reactive power make up the
total (or apparent) power used. This is the power generated by the
utility for the user to perform a given amount of work.

Transformer loading – 1160/1500 = 78 %
Capacitors totaling 410 kVAr installed in
each of the 13 large motors
Transformer loading – 913/1500 = 61 %
Example

Example:
Method-1
The utility bill shows an average power factor of .72 with an average KW of 627. How much KVAR is required
to improve the power factor to .95 ?
STEPS:
Cos F1 = 0.72 , Tan F1 = 0.963
Cos F2 = 0.95 , Tan F2 = 0.329
Kvar required = P ( Tanf1 - Tanf2 )
= 627 (0.964 – 0.329)
= 398 kVAr
Method-2
1. Locate 0.72 (original power factor) in column (1).Refer table.
2. Read across desired power factor to 0.95 column. We find .635 multiplier
3. Multiply 627 (average KW) by .635 = 398 KVAR.
4. Install 400 KVAR to improve power factor to 95%.
Now that we have determined that capacitors totaling 400 KVAR must be installed, we must decide where to
locate them.
How to determine the Rating of capacitors
required?

Where to Locate Capacitors?
For motors of 50 hp and above, it is best to install power factor correction capacitors at the
motor terminals since distribution circuit loading is reduced. (1,2)
The second arrangement shows capacitor banks connected at the bus for each motor
control centre. This compromise to Method 1 will reduce installation costs. (3)
The least expensive method shows capacitor banks connected at the service entrance.
However, the disadvantage is that higher feeder currents still flow from the service
entrance to the end of line equipment. (4,5,6)
Power Distribution Diagram Illustrating Capacitor Locations

Reduction in Distribution Loss
As current flows through conductors, the conductorsheat. This heating is
power loss
Power loss is proportional to current squared (P Loss =I2R)
Current is proportional to P. F.:
Conductor loss can account for as much as 2- 5% of total load
Capacitors can reduce losses by 1- 2% of the total load
x 100 ) 2
21 PF/PF-1% Loss Reduction =

What are the Types
of Transformers
•Power transformers : Used in transmission network of higher
voltages, deployed for step-up and step down transformer application
(400 kV, 200 kV, 110 kV, 66 kV, 33kV,22kW)
•Distribution transformers: Used for lower voltage distribution
networks as a means to end user connectivity.
(11kV, 6.6 kV, 3.3 kV, 440V, 230V)

How to calculate Transformer losses ?
1. Load loss (or copper loss)
2. No load loss (or iron loss)
• The total transformer loss, PTOTAL, at any load level can then be
calculated from:
PTOTAL = PNO-LOAD+ (% Load)2 x PLOAD
• Where transformer loading is known, the actual transformers
loss at given load can be computed as:
)(
2
lossloadfullx
kVARated
loadkVA
lossloadNo










=

Case Example:
For a load of 1500 KVA the plant has installed three numbers of 1000 KVA transformers. The
No load loss is 2.8 KW and the full load loss 11.88 KW. Estimate the total loss with 3
transformers in operation and 2 transformers in operation.
a) 2 transformers in operation :
No load loss = 2 x 2.8 = 5.6
Load loss = 2 x (750)2 x 11.88
(1000)
= 13.36 kW
Total Loss = 5.6 + 13.36 = 18.96
b) 3 transformers in operation :
No load loss = 3 x 2.8 = 8.4 KW
Load loss = 3 x (500)2 x 11.88 = 8.91 KW
(1000)
Total loss = 17.31 KW
Savings by loading all the 3 transformers = 13200 kWh.

Energy Efficient Transformers
• Most energy loss in dry-type transformers
occurs through heat or vibration from the
core. The new high-efficiency transformers
minimise these losses.
• The iron loss of any transformer depends on
the type of core used in the transformer.
However the latest technology is to use
amorphous material – a metallic glass alloy
for the core the expected reduction in
energy loss over conventional (Si Fe core)
transformers is roughly around 70%, which is
quite significant.
• By using an amorphous core– with unique
physical and magnetic properties- these new
types of transformers have increased
efficiency even at low loads - 98.5%
efficiency at 35% load.
1600 kVA AmorphousCore Transformer

What are Harmonics ?
• Harmonics are multiples of the fundamental frequency of an
electrical power system.
• If, for example, the fundamental frequency is 50 Hz, then the 5th
harmonic is five times that frequency, or 250 Hz.
• Likewise, the 7th harmonic is seven times the fundamental or 350
Hz, and so on for higher order harmonics.
Some of the Harmonic problems are
1. Blinking of Incandescent Lights
2. Capacitor Failure
3. Conductor Failure
4. Flickering of Fluorescent Lights
5. Motor Failures (overheating)
6. Transformer Failures

Total Harmonic Distortion
Harmonics can be discussed in terms of current or voltage.
Total Harmonic Distortion (THD) expresses the amount of harmonics.
The following is the formula for calculating the THD for current:
n
1
I
I
2
100
2
currentTHD
n n
n
 
=  
 
=

=
Mostly, 3rd,5th and 7th harmonics causes distortion.
When expressed as a percentage of fundamental voltage THD is given
by,
THDvoltage = 100
2
2
1
n
V
V

=
=





nn
n

Harmonics Limits
Maximum Harmonic Current Distortion in % of IL
Isc/IL <11 11 ≤ h < 17 17 ≤ h < 23 23 ≤ h < 35 TDD
< 20 4.0 2.0 1.5 0.6 5.0
20 < 50 7.0 3.5 2.5 1.0 8.0
50 < 100 10.0 4.0 4.0 1.5 12.0
100 < 1000 12.0 5.5 5.0 2.0 15.0
1000 15.0 7.0 6.0 2.5 20.0
1000
Total Harmonic Distribution for Different Voltage Levels in %
Bus Voltage at PCC Individual Voltage
Distortion (%)
Total Voltage Distortion
THD (%)
69 kV and below 3.0 5.0
69.001 kV Thru 161 kV 1.5 2.5
161 kV and above 1.0 1.5

Harmonic filters
• Tuned Harmonic filters consisting of a capacitor bank and
reactor in series are designed and adopted for suppressing
harmonics, by providing low impedance path for harmonic
component. The Harmonic filters connected suitably near the
equipment generating harmonics help to reduce THD to
acceptable limits.
• Benefits of Harmonic Filters
Improve power factor ratings through harmonic recombination
Isolate harmonic currents
Minimize equipment problems through THD cancellation
Reduce wasted energy
Minimize current carried in plant cables

Electrical Motors
•Motors convert electrical energy into mechanical energy
•90 % of industrial motors are induction motors

Types of Motors
• Induction motors
• Direct current motors
• Synchronous motors
All motor types have the same
four operating components:
1. Stator (stationary windings)
2. Rotor (rotating windings)
3. Bearings
4. and Frame (enclosure)

Motor Characteristics
The actual speed, which the motor operates, will be less
than the synchronous speed. The difference between
synchronous and full load speed is called slip and is
measured in percent. It is calculated using this equation:

Motor Efficiency
A typical EFFICIENCY value for an 11 kW standard motor is around 90
per cent and, for 100 kW, up to 94 per cent.
% Load vs. Power factor, Efficiency Speed vs. Power factor

What are the Motor losses?
The % losses indicated are for 3000 rpm motors
Core Loss : approx 18% of total loss at full load
Stator and Rotor Resistance (I2R) Loss: app 42% of total loss at full Load
Friction and Windage Loss approx 30% of total loss at full load
Stray Load Loss : approx 10%of total loss at full load
• fixed losses - independent of motor load
• Variable losses - dependent on load.
• Fixed losses =Magnetic core losses +Friction &Windage losses. Magnetic core losses
(Iron losses) consist of eddy current and hysteresis losses in the stator.
• Variable losses =Resistance losses in the stator and in the rotor and miscellaneous
stray losses.

Motor Selection
Few of salient selection issues are given below:
• In the selection process, the power drawn at 75 % of loading can be a
meaningful indicator of energy efficiency.
• Reactive power drawn (kVAr) by the motor.
• Allows 15 % tolerance on efficiency for motors upto 50 kW rating and 10 % for
motors over 50 kW rating.
• Stray losses are assumed as 0.5 % of input power.
• Procure motors based on test certificates rather than labeled values.
• kW savings = kW output  [ 1/old - 1/ new ] where old and new are the
existing and proposed motor efficiency values.
• The cost benefits can be worked out on the basis of premium required for high
efficiency vs. worth of annual savings.

Energy Efficient Motors
Power Loss Area Efficiency Improvement
1. Iron Use of thinner gauge, lower loss core steel reduces eddy current losses.
Longer core adds more steel to the design, which reduces losses due to
lower operating flux densities.
2. Stator I 2 R Use of more copper and larger conductors increases cross sectional
area of stator windings. This lowers resistance (R) of the windings and
reduces losses due to current flow (I).
3. Rotor I 2 R
Use of larger rotor conductor bars increases size of cross section,
lowering conductor resistance (R) and losses due to current flow (I).
4. Friction &
Windage
Use of low loss fan design reduces losses due to air movement.
5. Stray Load Loss Use of optimized design and strict quality control procedures minimizes
stray load losses.

Factors Affecting Energy Efficiency &
Minimising Motor Losses in Operation
• Motor Loading
or
• Reducing Under-loading
• Improving the Motor Loading by Operating in Star Mode
• Sizing to Variable Load
• Power Factor Correction
• Proper Maintenance
100
/
)(
% x
efficiencymotorloadfullplatenameratingkWloadfullplateName
loadexistingatkWmotorthebydrawnpowerInput
Loading =
100
3
)(
% x
CosxIxkVx
loadexistingatkWmotorthebydrawnpowerInput
Loading
f
=

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
• ŋfl = Efficiency at full-rated load
• Pir = Input power at full-rated load in kW
• The percentage loading can now be calculated as follows

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.
100*%
currentratedInput
currentloadInput
Load = (Valid up to 75% loading)

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.
*100%
s r
Slip
Load
S S
=
-
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 = 1500 at 50 HZ operating frequency.
( Synchronous speed = 120f/P) f: frequency, P: Number of poles
Nameplate full load speed = 1450
Measured speed in rpm = 1480
Nameplate rated power = 7.5 kW
Determine actual output power.
%40%100*
14501500
14801500
=
-
-
=Load
From the above equation, actual output power would be 40% x 7.5 kW = 3 kW
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.

Rewinding Effects on Energy Efficiency
• Rewinding can reduce motor efficiency and reliability.
• Majority of the users would wish to rewind the motor.
• Rewind-versus-replace decision
• Population of rewound motors may exceed 50 %
• Compare of no load current and stator resistance per
phase of a rewound motor with the original no-load
current and stator resistance at the same voltage to
assess the efficacy of rewinding.

Speed Control of AC Induction Motors
 Direct Current Drives (DC)
 Multi-speed motors
 Slip Ring Induction Motors
 Slip Power Recovery Systems
 Application of Variable Frequency AC Drives

Principle of VFDs
• The basic function of the VFD is to
act as a variable frequency
generator in order to vary speed of
the motor as per the user setting.
• The rectifier and the filter convert
the AC input to DC with negligible
ripple.
Components of a Variable Speed Drive

Other methods of speed control
of motors
• Eddy Current Drives
• Fluid Coupling
• Soft Starter
Eddy Current Drive
Fluid Coupling
Soft Starter
Soft Starter: Starting current, Stress profile during starting

Energy Saving Recommendations in motors
The margins in motor efficiency may be less than 5% of consumption often, but the load
survey would help to bring out savings in driven machines / systems, which can give 30 – 40
% energy savings.
•Identify motors with low voltage / power factor /
voltage imbalance for needed improvement
measures
•Identify the motors with <35, 35-50 %, 50 -75 % ,
75-100 % loading, over 100 % loading.
• Identify motors with machine side losses /
inefficiencies like idle operations, throttling / damper
operations for avenues like automatic controls /
interlocks, variable speed drives

Consider on/off controls
– Use controls to turn off idling motors.(Timer)
– Use of soft starters prove helpful in increasing the
number of starts and stops per hour.

Operate motor in Star instead of Delta
if load is very low….
• Operating in star mode leads to a voltage reduction by a factor
of '√3'.
• Full-load operation in star mode gives higher efficiency and
power factor than partial load operation in the delta mode
• However, motor operation in the star mode is possible only for
applications where the torque-to-speed requirement is lower at
reduced load.
• If you operate at light loads, there will be no problem, however
if you operate at high loads, the slip of the motor will be
increased dramatically and it may stall.
• Use automatic star delta change over devise if load varies
between 35 to 50

Optimize Transmission Efficiency
Transmission equipment including shafts, belts, chains,
and gears should be properly installed and maintained.
• When possible, use flat belts in place of V-belts.
• Helical gears are more efficient than worm gears; use
worm gears only with motors under 10 hp.
• It is better to have a direct drive for avoiding losses in
transmission system.

Air Is Free !!!
Compressed Air System

 Compressed Air System Efficiency:
60 to 80% of the power of the prime mover is converted into an unusable
form of energy (HEAT)
And to a lesser extent, into friction, misuse and noise
Approximately
10% gets to
the point of
use!!
Introduction

Axial Compressor
Axial Compressor

Compressor Efficiency Definitions
Isothermal Efficiency
Isothermal Efficiency =
Actual measured input power
IsothermalPower
Isothermal power(kW) = P1 x Q1 x loger/36.7
P1 = Absolute intake pressure kg/ cm2
Q1 = Free air delivered m3
/hr.
r = Pressure ratio P2/P1

Compressor Efficiency Definitions
Volumetric Efficiency
3
Free air delivered m /min
Volumetric efficiency =
Compressor displacement
Compressor Displacement =  x D2
x L x S x  x n
4
D = Cylinder bore, metre
L = Cylinder stroke, metre
S = Compressor speed rpm
 = 1 for single acting and
2 for double acting cylinders
n = No. of cylinders

Energy Conservation in Compressors
Blowers in place of Compressed Air System
Using a blower that is designed for lower pressure operation will cost only a fraction of
compressed air generation energy and cost.
Capacity Control of Compressors
In many installations, the use of air is intermittent. Therefore, some means of controlling the
output flow from the compressor is necessary. The type of capacity control chosen has a
direct impact on the compressor power consumption. Some control schemes commonly used
are discussed below:
• Automatic On / Off Control:
• Load and Unload
• Multi-step Control
• Throttling Control
Avoiding Misuse of Compressed Air
Avoiding Air Leaks and Energy Wastage

Air Dryers
• Most commonly used dryers in the industry are:
1. Refrigerant type and
2. Adsorption type - can be of the following type:
(a) Blower reactivated type
(b) Heatless purge type
(c) Heat of compression (HOC) type

Typical Dew Point and Power Consumption Data for Dryers
Typical Dew Point and Power Consumption Data for Dryers
Type of Dryer
Dew Point
0C
First
Cost
Operating
Cost
Power
Cons. For
1000 m3/hr
Refrigeration -20 Low Low 2.9 kW
Desiccant regenerative
(blower reactivated type)
-40 Medium Medium 18.0 kW
(heatless purge type)
-40 Low High 20.7 kW
(heat of compression type)
-40 High Very low 0.8 kW

Steps in simple shop-floor method
for leak quantification
• Shut off compressed air operated equipments (or conduct test when no
equipment is using compressed air).
• Run the compressor to charge the system to set pressure of operation
• Note the sub-sequent time taken for ‘on load’ and ‘off load’ cycles of the
compressors. For accuracy, take ON & OFF times for 8 – 10 cycles
continuously. Then calculate total ‘ON’ Time (T) and Total ‘OFF’ time (t).
• The system leakage is calculated as
• System leakage (cmm) = Q  T / (T + t)
• Q = Actual free air being supplied during trial, in cubic meters
per minute
• T = Time on load in minutes
• t = Time unload in minutes

Leak test: example
• Compressor capacity (CMM) = 35
• Cut in pressure kg/SQCMG = 6.8
• Cut out pressure kg/SQCMG = 7.5
• On load kW drawn = 188 kW
• Unload kW drawn = 54 kW
• Average ‘On-load’ time = 1.5 minutes
• Average ‘Unload’ time = 10.5 minutes
Comment on leakage quantity and avoidable loss of power due to air
leakages.
a) Leakage quantity (CMM) =
= 4.375 CMM
b) Leakage per day = 6300 CM/day
c) Specific power for compressed air generation=
=0.0895 kwh/m3
d) Power lost due to leakages/day = 563.85 kWh
 )
 )  )
35
5.105.1
1.5


 )CMH6035
kWh188


Capacity Assessment in Shop-floor
• Isolate the compressor along with its individual receiver being taken for test from main compressed
air system by tightly closing the isolation valve or blanking it, thus closing the receiver outlet.
• Open water drain valve and drain out water fully and empty the receiver and the pipe line. Make
sure that water trap line is tightly closed once again to start the test.
• Start the compressor and activate the stop watch.
• Note the time taken to attain the normal operational pressure P2 (in the receiver) from initial
pressure P1.
• Calculate the capacity as per the formulae given below :
Actual Free air discharge Min./NM
T
V
P
PP
Q 3
0
12

-
=
Where
P2 = Final pressure after filling (kg/cm2
a)
P1 = Initial pressure (kg/cm2
a) after bleeding
P0 = Atmospheric Pressure (kg/cm2
a)
V = Storage volume in m3
which includes receiver,
after cooler, and delivery piping
T = Time take to build up pressure to P2 in minutes
The above equation is relevant perfect isothermal compression. For actual compressed
air temperature at discharge, say t20C is higher than ambient air temperature say t1 0C ),
the FAD is to be corrected by a factor (273 + t1) / (273 + t2).

Example
• Piston displacement : 16.88 CMM
• Theoretical compressor capacity : 14.75 CMM @ 7 kg/SQCMG
• Compressor rated rpm 750 : Motor rated rpm : 1445
• Receiver Volume : 7.79 CM
• Additional hold up volume,
i.e., pipe / water cooler, etc., is : 0.4974 CM
• Total volume : 8.322 CM
• Initial pressure P1 : 0.5 Kgf / SQCMG
• Final pressure P2 : 7.03 Kgf / SQCMG
• Atmospheric pressure P0 : 1.026 Kgf/cm2A
• Compressor output CMM :  )
timePumpupPressureAtm.
VolumeTotalPP 12

-
 )
4.021.0261
8.3225.003.7

-
= 13.17 CMM

Pumps and Pumping Systems
Centrifugal Pumps

Hydraulic power, pump shaft power and
electrical input power
• Hydraulic power Ph = Q (m3/s) x Total head, hd - hs (m) x  (kg/m3) x g (m2/s)
1000
Where hd - discharge head, hs – suction head,  - density of the fluid, g –
acceleration due to gravity
• Pump shaft power Ps = Hydraulic power, Ph
pump efficiency, Pump
• Electrical input power = Pump shaft power P
Motor

System Characteristics
Static Head Static Head vs. Flow

Dynamic (Friction) Head
Friction Head vs. Flow

Typical pump characteristic curves

Head
Meters
System Curve
Flow (m
3
/hr)
Selecting a pump

Selecting a pump
Head
Meters
82%
Pump Curve at
Const. Speed
System Curve
Flow (m3
/hr)
Operating Point
500 m3/hr

Selecting a pump
Head,
m
82%
Pump Curve at
Const. Speed
System Curve
Flow (m3
/hr)
Operating Point
500300
50

Selecting a pump
Pump Efficiency
77%
82%
Pump Curve at
Const. Speed
Partially
closed valve
Full open valve
System Curves
Flow (m3
/hr)
500300
Head,
m
50
70

Selecting a pump
Head
Meters
Pump Efficiency 77%
82%
Pump Curve at
Const. Speed
Partially
closed valve
Full open valve
System Curves
Flow (m3
/hr)
Operating Points
A
B
500 m3
/hr300 m3
/hr
50 m
70 m
Static
Head
C
42 m

Efficiency Curves
28.6 kW
14.8 kW

If we select E, then the pump
efficiency is 60%
• Hydraulic Power = Q (m3/s) x Total head, hd - hs (m) x  (kg/m3) x g (m2/s)
1000
= (68/3600) x 47 x 1000 x 9.81
1000
= 8.7 kW
• Shaft Power - 8.7 / 0.60 = 14.5 Kw
• Motor Power - 14.8 / 0.9 = 16.1Kw
(considering a motor efficiency of 90%)

If we select A, then the pump
efficiency is 50%
• Hydraulic Power = Q (m3/s) x Total head, hd - hs (m) x  (kg/m3) x g (m2/s)
1000
(68/3600) x 76 x 1000 x 9.81
1000
= 14 kW
Shaft Power - 14 / 0.50 = 28 Kw
Motor Power - 28 / 0.9 = 31 Kw (considering a
motor efficiency of 90%)

Flow Vs Speed
If the speed of the impeller is increased from N1
to N2 rpm, the flow rate will increase from Q1 to
Q2 as per the given formula:
Flow:
Q1 / Q2 = N1 / N2
Example: 100 / Q2 = 1750/3500
Q2 = 200 m3/hr
The affinity law for a
centrifugal pump with
the impeller diameter
held constant and the
speed changed:

Head Vs speed
The head developed(H) will be
proportional to the square of the
quantity discharged, so that
Head:
H1/H2 = (N12) / (N22)
Example: 100 /H2 = 1750 2 / 3500
H2 = 400 m

Power Vs Speed
The power consumed(W) will be
the product of H and Q, and,
therefore
Power(kW):
kW1 / kW2 = (N13) / (N23)
Example: 5/kW2 = 17503 / 35003
kW2 = 40

Flow:
Q1 / Q2 = D1 / D2
Example: 100 / Q2 = 8/6
Q2 = 75 m3/hr
Head:
H1/H2 = (D1) x (D1) / (D2) x (D2)
Example: 100 /H2 = 8 x 8 / 6 x 6
H2 = 56.25 m
Horsepower(BHP):
kW1 / kW2 = (D1) x (D1) x (D1) / (D2) x (D2) x (D2)
Example: 5/kW2 = 8 x 8 x 8 / 6 x 6 x 6
kW2 = 2.1 kW
The affinity law for a centrifugal pump with the
speed held constant and the impeller diameter
changed

Impeller Diameter Reduction on Centrifugal
Pump Performance

Pump suction performance (NPSH)
• Net Positive Suction Head Available – (NPSHA)
• NPSH Required – (NPSHR)
• Cavitation
• NPSHR increases as the flow through the pump
increases
• as flow increases in the suction pipework, friction
losses also increase, giving a lower NPSHA at the
pump suction, both of which give a greater chance
that cavitation will occur

Pump control by varying speed:
Pure friction head
• Reducing speed in the
friction loss system
moves the
intersection point on
the system curve
along a line of
constant efficiency
• The affinity laws are
obeyed

Pump control by varying speed:
Static + friction head
• Operating point for the pump
moves relative to the lines of
constant pump efficiency when the
speed is changed
• The reduction in flow is no longer
proportional to speed
• A small turn down in speed could
give a big reduction in flow rate
and pump efficiency
• At the lowest speed illustrated,
(1184 rpm), the pump does not
generate sufficient head to pump
any liquid into the system

Pumps in parallel switched to meet demand

Pumps in parallel with system curve

Control of Pump Flow by Changing System Resistance Using a
Valve

Fixed Flow reduction
Impeller trimming
Before Impeller trimming After Impeller Trimming
1BHP
3
D1
D2
2BHP
1H
2
1D
2D
2H
1Q
1D
2D
2Q




=









=
=

Meeting variable flow reduction
Variable Speed Drives (VSDs)
Effect of VFD

Energy Conservation Opportunities in
Pumping Systems
1. Ensure adequate NPSH at site of installation
2. Operate pumps near best efficiency point.
3. Modify pumping system/pumps losses to minimize
throttling.
4. Adapt to wide load variation with variable speed drives
5. Stop running multiple pumps - add an auto-start for an on-
line spare or add a booster pump in the problem area.
6. Conduct water balance to minimise water consumption
7. Replace old pumps by energy efficient pumps

What is a Ton of refrigeration?
1 ton of refrigeration = 3024 kCal/hr heat rejected.
The cooling effect produced is quantified as tons of refrigeration.
HVAC and Refrigeration System
Introduction
Refrigeration deals with the transfer of heat from a low temperature level at the heat
source to a high temperature level at the heat sink by using a low boiling refrigerant.

Conceptual view of a chilled-water
air-conditioning system

Psychrometrics and Air-Conditioning Processes
• Psychrometrics is the science of moist air properties and
processes, which is used to illustrate and analyze air-
conditioning cycles.
• It translates the knowledge of heating or cooling loads (which
are in kW or tons) into volume flow rates (in m3/s or cfm) for the
air to be circulated into the duct system.
• The most commonly used psychrometric quantities include the
dry and wet bulb temperatures, dew point, specific humidity,
relative humidity.

Comfort Zone
• One of the major applications of the Psychrometric Chart is in air
conditioning, and we find that most humans feel comfortable when the
temperature is between 22°C and 27°C, and the relative humidity between
40% and 60%.

How do the chillers work ?
1. Boiling point of the water is a function of pressure. At atmospheric pressure water boils at 100
deg. C. When maintained at high vacuum, water will boil and subcool itself. The boiling point
of the water at 6 mmHg (abs) is 3.7 deg. C.
Vapour Absorption Refrigeration

2. Lithium Bromide (LiBr) has the property to absorb water due to its chemical affinity. At higher
concentration and lower temperature LiBr absorbs water vapour (refrigerant vapour) very effectively.

3. As Lithium Bromide becomes dilute it loses its capacity to absorb water vapour. It thus needs to be
reconcentrated using a heat source. Heat source may be Steam or Flue gases or even Hot water.

Evaporative Cooling
• Humidity up to 50 % for human comfort or for
process,
• Cheaper and less energy intensive
• Comfort cooling in dry regions

Performance Assessment: Refrigeration
The specific power consumption kW/TR is a useful indicator of the
performance of refrigeration system. By messing refrigeration duty
performed in TR and the Kilo Watt inputs measured, kW/TR is used
as a reference energy performance indicator.
The refrigeration TR is assessed as TR = Q Cp  (Ti – To) / 3024
Where
TR- is cooling TR duty
Q - is mass flow rate of coolant in kg/hr
Cp - is coolant specific heat in kCal /kg / 0C
Ti - is inlet. Temperature of coolant to evaporator (chiller) in 0C.
To- is outlet temperature of coolant from evaporator (chiller) in 0C.

Calculating the operating load of a chiller plant
Refrigeration
plant
Hot well
12OC
Cold well
8OC
Process
Chilled water flow – 100 m3/hr
Refrigeration TR - 100,000 kg/hr x 1 x 4
3000
- 133.33 TR
Efficiency -
Power drawn by compressor, kW
TR
m Cp
120
133.33
- = 0.9
DT

Overall energy consumption
• Compressor kW
• Chilled water pump kW
• Condenser water pump kW
• Cooling tower fan kW
Overall kW/TR = sum of all above kW/ TR

COP
The theoretical Coefficient of Performance (Carnot), COPCarnont - a standard measure of
refrigeration efficiency of an ideal refrigeration system- depends on two key system
temperatures, namely, evaporator temperature Te and condenser temperature Tc with COP
being given as:
COPCarnot = Te / Tc - Te
This expression also indicates that higher COPCarnot is achieved with higher evaporator
temperature and lower condenser temperature.
But COPCarnot is only a ratio of temperatures, and hence does not take into account the type of
compressor. Hence the COP normally used in the industry is given by
Cooling effect (kW)
COP =
Power input to compressor (kW)

Performance Assessment: Air conditioning
In case of air conditioning units, the airflow at the Fan Coil Units (FCU) or the Air Handling
Units (AHU) can be measured with an anemometer. Dry bulb and wet bulb temperatures are
measured at the inlet and outlet of AHU or the FCU and the refrigeration load in TR is
assessed as ;
 )
3024
hhρQ
TR outin -
=
Where, Q is the air flow in m3
/h
 is density of air kg/m3
h in is enthalpy of inlet air kCal/kg
h out is enthalpy of outlet air kCal/kg

Effect of Variation in Evaporator Temperature on
Compressor Power Consumption
Evaporator
Temperature
(0
C)
Refrigeration
Capacity
(tons)
Specific
Power
Consumption
Increase in
kW/ton (%)
5.0 67.58 0.81 -
0.0 56.07 0.94 16.0
-5.0 45.98 1.08 33.0
-10.0 37.20 1.25 54.0
-20.0 23.12 1.67 106.0
A 10C raise in evaporator temperature can help to save almost 3 % on power
consumption.
Factors affecting Performance

Effect of Variation in Condenser Temperature on
Compressor Power Consumption
Condensing
Temperature
(0
C)
Refrigeration
Capacity
(tons)
Specific
Power
Consumption
Increase in
kW/TR
(%)
26.7 31.5 1.17 -
35.0 21.4 1.27 8.5
40.0 20.0 1.41 20.5

Effect of Poor Maintenance on Compressor
Power Consumption
Condition
Evap.
Temp
(0
C)
Cond.
Temp
(0
C)
Refrigeration
Capacity
(tons)
Specific
Power
Consumption
(kW/ton)
Increase
in
kW/Ton
(%)
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

Performance Assessment of window, split and
package air conditioning units
• Energy Efficiency Ratio (EER): EER is calculated by dividing a chiller's cooling
capacity (in watts) by its power input (in watts) at full-load conditions. This
definition of EER has been adopted in BEE star labeling programme.
• The energy efficiency ratio (EER) = Refrigeration effect in watts/ Power input in
watts
• Based on the condition of the air, the air properties such as specific volume
and enthalpy at both inlet and out let conditions can be obtained from
psychrometric charts. From these parameters the capacity delivered by the air
conditioner can be evaluated, which when compared with power drawn would
reveal its performance in terms of kW/TR and EER.

Performance Assessment of Package Air Conditioner – Example
• Package air conditioner capacity - 10 TR
• Average Air Velocity - 2.27 m/s (across suction side
filter)
• Cross Sectional Area - 0.58 m2
• Air Flow Rate - 1.32 m3/sec - 4751 m3/hr
Inlet Air Condition
• DBT - 20 oC
• WBT - 14 oC
• Sp.Vol - 0.8405 m3/kg
• Enthalpy - 9.37 kCal/kg
Outlet Air Condition
• DBT - 12.7 oC
• WBT - 11.3 oC
• Enthalpy - 7.45 kCal/kg
• Cooling Effect Delivered - 3.6 TR = 12.7 kW

Performance Assessment of Package Air Conditioner –
Example (Contd..)
Power Drawn
• Compressor - 4.71 kW
- 4.3 kW (shaft power@ 90%
motor efficiency)
• Pump - 2.14 kW
• C.T Fan - 0.384 kW
Specific Energy Consumption
• Compressor - 1.31 kW/TR
• Overall - 2 kW/TR
• EER - 12700/4300
- 2.95 W/W

Cold Storage Systems
• A Refrigerated storage which includes cold storage and frozen food storage is the best known
method of preservation of food to retain its value and flavor.
• The refrigeration system in a cold storage is usually a vapour compression system comprising
the compressor, condenser, receiver, air cooling units and associate piping and controls.
• In smaller cold rooms and walk-ins the practice is to use air cooled condensing units with
sealed, semi-sealed or open type compressors. In the light of the CFC phase out the trend
now is to use HCFC-22, HFC-134a or other substitute refrigerants. In the medium and large
sized units the practice is to use a central plant with ammonia as the refrigerant.
• In some present day medium and large sized units with pre-fabricated (insulated) panel
construction the trend is to use modular HCFC-22/HFC units which are compact, light weight
and easy to maintain.

Energy Saving Opportunities in Cold Storage Systems
• Cold Store Building Design: Proper orientation, compact arrangement of chambers, shading
of exposed walls, adequate insulation etc. are some of the important factors.
• Refrigeration System: The system must be designed for optimum operating conditions like
evaporating and condensing temperatures, as these conditions have a direct bearing on
energy consumption.
• Compressor capacity control system helps in energy savings during partial load operation.
• Control System: The proper control systems for refrigerant level, room temperature,
compressor capacity etc., are required to further optimize energy consumption.
• Air Curtain or Strip Curtain: The use of air curtains and strip curtains is a common feature in
present day cold stores as they help reduce air infiltration due to frequent and some times
long door openings. Fan operated air curtains are expensive and work on electrical power
whereas strip curtains are cheaper and need no energy for operation.
• Heat Recovery System: In processing plant cold stores, a heat reclaim system can be installed
to recover a part of the heat rejected by the refrigeration system. This can be gainfully utilised
in generating hot water free of cost.

Heat Pumps and Their Applications
• Heat Pump Technology
• A heat pump is same as an air conditioner except that the heat rejected in an air
conditioner becomes the useful heat output. Heat flows naturally from a higher to
a lower temperature. Heat pumps, however, are able to force the heat flow in the
other direction, using a relatively small amount of high quality drive energy
(electricity, fuel, or high-temperature waste heat).
Heat Pump

Heat Pump Applications
Industrial heat pumps are mainly used for:
• Space heating:
• Heating of process streams;
• Water heating for washing, sanitation and cleaning;
• Steam production;
• Drying/dehumidification;
• Evaporation;
• Distillation;
• Concentration.

Difference between fans, blowers
and compressors
Equipment Specific Ratio Pressure rise (mmWc)
Fans Up to 1.11 1136
Blowers 1.11 to 1.20 1136 – 2066
Compressors more than 1.20 -

Fan types
Centrifugal fan Axial fan

Centrifugal Fan: Types
Paddle Blade
(Radial blade)
Forward Curved
(Multi vane)
Backward Curved

Axial Flow Fan: Types
Tube Axial Vane Axial Propeller

Pressure Drop Across Various Equipment

System characteristics and Fan curves

Fan Laws
Flow ? Speed Pressure ? (Speed)2
Power ? (Speed)3
1 1
2 2
Q N
Q N
=
2
1 1
2 2
SP N
SP N
 
=  
 
3
1 1
2 2
kW N
kW N
 
=  
 
Varying the RPM by 10%
decreases or increases air
delivery by 10%.
decreases or increases the
static pressure by 19%.
decreases or increases the
power requirement by
27%.
Where Q – flow, SP – Static Pressure, kW – Power and N – speed (RPM)

Fan System Curve
ID Fan Characteristic
Curve
System Resistance
Curve
Operating Point
Totalpressuredrop
240
mmWc
30,000 M3/hr
Flow rate of Flue gases

Flow Control Strategies
• Normally, fan is designed for operating at constant speed
• Practically, there may be need for increase in flow or
decrease in flow. Various strategies are
– Damper controls
– Pulley change
– Inlet guide vanes,
– Variable speed drives
– Series and parallel operation

Pulley Change
8”
10”
6” 10”
1470 RPM
1470 RPM
1181 RPM
882RPM
26 kW
14 kW
Fan speed can be increased or decreased with
change in drive pulley or driven pulley

Series and Parallel operation
• Parallel operation
– Two fans operating in parallel ideally will result in doubling of flow
at free delivery
– Higher the system resistance, less increase in flow with parallel
fan operation
• Series operation
– Two fans operating in series ideally will result in doubling of static
pressure at given flow
– Higher the system resistance, better the result

Figure 5.13 Series and Parallel Operation

Fans Performance Assessment
• Static pressure
– Potential energy put into the system by the fan
• Velocity pressure
– Pressure arising from air flowing through the duct. This is used to
calculate velocity
• Total pressure
– Static pressure + velocity pressure
– Total pressure remains constant unlike static and velocity pressure

Traverse points for velocity measurement
Figure 5.17 Traverse Points for Circular Duct

Ct
x
DensityGas o

=
273
293.1273
)(
pC x 2 9.81
Velocity v, m/s =
p 

 D 
100
*102
)(*sec/
%),(
3
x
kWinshaftfantoinputpower
mmWCinpressuretotalpminVolume
EfficiencyMechanicalFan mechanical
D
=
100
*102
)(*sec/
%),(
3
x
kWinshaftfantoinputpower
mmWCinpressurestaticpminVolume
EfficiencyStaticFan static
D
=
Volumetric flow (Q), m3/s = Velocity, m/s x Area(m2/s)

Energy Saving Opportunities
• Avoid unnecessary demand- excess air reduction, idle running,
arresting leaks
• Match fan capacity to demand – downsizing, pulley change, VSD,
impeller de-rating
• Reduce pressure drops – remove redundant drops, modify ducts
with minimum bends
• Drive system- direct drive, V belt by Flat belt, two speed motors
• Replace with energy efficient fan, impeller
• Change to hallow FRP impeller
• Inlet guide vane in place of discharge damper control

Tr faculty polytech s. verma

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