energy conservation

WHAT IS ENERGY CONSERVATION?
 Energy conservation is the practice of decreasing
the quantity of energy used. It may be achieved
through efficient energy use, in which case energy
use is decreased while achieving a similar outcome,
or by reduced consumption of energy services.
 Individuals and organizations that are direct
consumers of energy may want to conserve energy
in order to reduce energy costs and promote
economic security.
 Industrial and commercial users may want to
increase efficiency and thus maximize profit.

NECESSITY OF ENERGY CONSERVATION
 Limited fuels available on earth (most of fuels
will be exhausted and we will have to switch
over to alternative sources of energy)
 Ever increasing demand of energy
 Cost of energy has increased substantially
 The product cost has a bearing of energy cost
 Energy efficient technology options available
 Reduction in green house gases emissions
 Over all environment friendly option

ENERGY CONSERVATION IN INDIA
 Indian government has passed Energy conservation
act, 2001 for better utilization of energy and its
conservation
 25,000 MW capacity creation through energy
efficiency in electricity sector alone has been
estimated in India
 23% energy conservation potential
 Industrial and agriculture sectors have the
maximum potential
 Assessment of potential savings in different sectors
 Bureau of Energy Efficiency (BEE) created

BUREAU OF ENERGY EFFICIENCY (BEE)
 BEE is an agency of the Government of India,
under the Ministry of Power set up in March
2002 under the provisions of the nation's
2001 Energy Conservation Act.
 The function of agency is to develop
programs which promotes the conservation
and efficient use of energy in India.
 The government has proposed to make it
mandatory for all appliances in India to have
ratings by the BEE starting in January 2010.

ENERGY CONSERVATION BUILDING CODE
(ECBC)
 Mandatory scope covers commercial buildings
 Applies to new construction only
 Building components included
 Building Envelope (Walls, Roofs, Windows)
 Lighting (Indoor and Outdoor)
 Heating Ventilation and Air Conditioning
(HVAC) System
 Solar Water Heating and Pumping
 Electrical Systems (Power Factor,
Transformers)

ENERGY CONSERVATION IN BUILDINGS
 Current best practices in building design and
construction result in homes that are
profoundly more energy conserving than
average new homes like
 Passive house
 Super-insulation
 Smart ways to construct homes, such that
minimal resources are used for cooling and
heating the house in summer and winter
respectively, can significantly reduce energy
costs.

ENERGY EFFICIENT BUILDING
CONSTRUCTION
 Very good insulation of walls, roofs
and Basement.
 Windows with high quality double
or triple glazing.
 Air-tight construction to be
checked by blower door test
 Avoid cooling demand
 Sun shading in summer
 Natural cooling sources

CONSERVATION OF ENERGY IN LIGHTING
SYSTEM

 Selection of Lighting in Buildings

 Two methods are available:

 Building area method

 Space by space method

BUILDING AREA METHOD
 Interior lighting power allowance (W) by the building
area method is determined in accordance with the
following:
a) Determine the allowed lighting power density (LPD)
from Table 1 for each appropriate building area type.
b) Calculate the gross lighted floor area (GLFA) for each
building area type.
c) The interior lighting power allowance (ILPA) is the
sum of the products of the gross lighted floor area of
each building area times the allowed lighting power
density for that building area types i.e.
ILPA = Σ(GLFA x LPD)

TABLE 1: INTERIOR LIGHTING POWER FOR BUILDING
AREA METHOD
Building Area Type LPD (W/m2) Building Area Type LPD (W/m2)
Automotive Facility 9.7 Multifamily Residential 7.5
Convention Center 12.9 Museum 11.8
Dining: Bar Lounge/ Leisure 14.0 Office 10.8
Dining: Cafeteria/ Fast Food 15.1 Parking Garage 3.2
Dining: Family 17.2 Performing Arts Theater 17.2
Dormitory/Hostel 10.8 Police/Fire Station 10.8
Gymnasium 11.8 Post Office/ Town Hall 11.8
Healthcare-Clinic 10.8 Religious Building 14.0
Hospital/ Health Care 12.9 Retail/ Mall 16.1
Hotel 10.8 School/ University 12.9
Library 14.0 Sports Arena 11.8
Manufacturing Facility 14.0 Transportation 10.8
Motel 10.8 Warehouse 8.6
Motion Picture Theater 12.9 Workshop 15.1

SPACE BY SPACE METHOD
 Determination of interior lighting power allowance (Watts)
by the space by space method shall be in accordance with
the following:
a) Determine the appropriate building type from Table 2
and the allowed lighting power density.
b) For each space enclosed by partitions 80% or greater of
ceiling height, determine the gross interior floor area by
measuring to the center of the partition wall.
c) The interior lighting power allowance is the sum of the
lighting power allowances for all spaces. The lighting
power allowance for a space is the product of the gross
lighted floor area of the space times the allowed lighting
power density for that space.

TABLE 2: INTERIOR LIGHTING POWER FOR SPACE BY SPACE METHOD
Space Function LPD (W/m2) Space Function LPD (W/m2)
Office-enclosed 11.8 Library 11.8
Office-open plan 11.8 Card File & Cataloging 11.8
Conference/Meeting/Multipurpose 14.0 Stacks 18.3
Classroom/Lecture/Training 15.1 Reading Area 12.9
Lobby 14.0 Hospital
For Hotel 11.8 Emergency 29.1
For Performing Arts Theater 35.5 Recovery 8.6
For Motion Picture Theater 11.8 Nurse Station 10.8
Audience/Seating Area 9.7 Exam Treatment 16.1
For Gymnasium 4.3 Pharmacy 12.9
Patient Room 7.5
For Convention Center 7.5 Operating Room 23.7
For Religious Buildings 18.3 Nursery 6.5
For Sports Arena 4.3 Medical Supply 15.1
For Performing Arts Theater 28.0 Physical Therapy 9.7
For Motion Picture Theater 12.9 Radiology 4.3
For Transportation 5.4 Laundry – Washing 6.5
Atrium-first three floors 6.5 Automotive – Service Repair 7.5
Atrium-each additional floor 2.2 Manufacturing
Lounge/Recreation 12.9 Low Bay (<8m ceiling) 12.9
Contd. …………………..

TABLE 2: INTERIOR LIGHTING POWER FOR SPACE BY SPACE METHOD
Space Function LPD (W/m2) Space Function LPD (W/m2)
For Hospital 8.6 High Bay (>8m ceiling) 18.3
Dining Area 9.7 Detailed Manufacturing 22.6
For Hotel 14.0 Equipment Room 12.9
For Motel 12.9 Control Room 5.4
For Bar Lounge/Leisure Dining 15.1 Hotel/Motel Guest Rooms 11.8
For Family Dining 22.6 Dormitory – Living Quarters 11.8
Food Preparation 12.9
Laboratory 15.1 General Exhibition 10.8
Restrooms 9.7 Restoration 18.3
Dressing/Locker/Fitting Room 6.5 Bank Office – Banking Activity Area 16.1
Corridor/Transition 5.4
For Hospital 10.8 Sales Area 18.3
For Manufacturing Facility 5.4 Mall Concourse 18.3
Stairs-active 6.5 Sports Arena
Active Storage 8.6 Ring Sports Area 29.1
For Hospital 9.7 Court Sports Area 24.8
Inactive Storage 3.2 Indoor Field Area 15.1
For Museum 8.6
Electrical/Mechanical 16.1 Fine Material Storage 15.1
Workshop 20.5 Medium/Bulky Material Storage 9.7
Sleeping Quarters 3.2 Parking Garage – Garage Area 2.2
Convention Center – Exhibit Space 14.0
Airport – Concourse 6.5
Air/Train/Bus – Baggage Area 10.8
Ticket Counter 16.1

ENERGY CONSERVATION IN INDUSTRIES
THROUGH MOTOR
 Electric motors converts electrical energy into
mechanical energy.
 Energy can be saved by using Energy Efficient
Motor (EEM) in place of standard motor.
 An EEM produces the same shaft output
power (HP) but uses less input power (kW)
than a standard motor.

CASE STUDY ON FINANCIAL EVALUATION
OF EEM
 A 60 HP standard AC motor operating at 75% load.
At this load efficiency of motor is 82%. The motor
power factor is 0.8. Motor operates 20 hrs a day and
300 days in a year. Per unit energy charge is Rs. 3.10
and per month per kVA demand charge is Rs. 175.
Cost of 60 HP standard AC motor is Rs. 1,00,000/-
(Take 1 HP = 750 W)
 Rating of standard AC motor
= 60 x 0.75= 45 kW
 Input power to motor
= 45 x 0.75 / 0.82 = 41.16 kW

OF EEM (contd…)
kVA demand
= 41.16 / 0.8 = 51.45 kVA

 kVA charges/year
= 51.45 x 175 x 12 = Rs. 1,08,045/-

 Energy (kWh) charges/year
= 41.16 x 3.1 x 20 x 300
= Rs. 7,65,576/-

 Total (Demand + Energy) Cost/year
= 1,08,045 + 7,65,576 = Rs. 8,73,621/-

OF EEM (contd…)
 A 60 HP Energy Efficient Motor is operating a
75% load. At this load, efficiency of the motor
is 87% and the power factor is 0.83. Cost of
60 HP EEM is Rs. 1,20,000/-

 Rating of standard AC motor
= 60 x 0.75= 45 kW

 Input power to motor
= 45 x 0.75 / 0.87 = 38.79 kW

OF EEM (contd…)
 kVA demand
= 38.79 / 0.83 = 46.73 kVA

 kVA charges/year
= 46.73 x 175 x 12 = Rs. 98,133/-

 Energy (kWh) charges/year
= 38.79 x 3.10 x 20 x 300 = Rs. 7,21,494/-

 Total (Energy + Demand) charges/year
= 7,21,494 98,133 = Rs. 8,19,627/-

FINANCIAL EVALUATION
 Energy savings achieved by using E.E.M. over
standard motor
= 8,73,621– 8,19,627 = Rs. 53,994/-

 Payback period for replacement of existing standard
motor with E.E.M.
= 1,20,000 / 53,994 = 2.3 years

 Payback period for purchase of E.E.M. for new
installation
= 20,000 / 53,994 = 0.37 yr = 4.5 months

TIPS FOR ENERGY CONSERVATION IN
DOMESTIC SECTOR
 Organized cooking activity can save about 20%
Energy.
 Use right quantity of water required for cooking and
reduce gas/kerosene usage by 65%.
 Cook on low flame as far as possible and save 6 to
10% energy.
 The pressure cooker should be loaded 2/3rd of the
foodstuff is solid & hard and ½ if loaded with liquid.
Properly used pressure cookers can save upto 50 to
75% of energy as well as time.

DOMESTIC SECTOR (contd…)
 Cook the food in solar cookers. Cook anything
except roti and save cost of 2 LPG Cylinders
annually.
 Allow refrigerated foodstuff to come to room
temperature before heating and allow heated
foodstuff to cool down to normal temperature
before placing it in the refrigerator. Avoid
frequent opening and closing of refrigerator &
air-conditioned rooms.
 Electricity is an expensive as cooking fuel – avoid
it

 Use proper lighting & efficient lighting devices
 A tube light (36/40 W) gives more light than a 60 or
100 W bulb and consumes 40 to 60% less power.
 Tube light with electronic choke is even more energy
efficient means of lighting
 Use Daylight as far as Possible
 Lighting devices like bulb, tube light, etc. consume
energy according to their capacities. Use
appropriate lighting according to your requirement.
A so-called zero bulb uses 12 to 15 W per hour.
 Compact Fluorescent Lamps (CFL) are available in 5,
7, 9, 11 W capacities and give more light output.

 Use your refrigerator & air conditioners efficiently
 Place the refrigerator about 6 inches away from the
wall to allow air circulation
 Air-conditioned room must be leak proof
 Instant geysers are considered to be more efficient
than storage type geysers. solar water heaters
operate on solar energy which is available free of
cost.
 Switch off lights & fans when not required
 Stop wastage – we cannot afford it
 conserve energy
 Plant trees

ENERGY CONSERVATION IN STREET LIGHTS

Normally a tube light of 40W rating with a choke
of 20 W is being utilized for street lights with a
total power of 60 W. Alternately the use of CFLs
of 18 W rating which has an equivalent
luminosity would lead to a power saving to the
extent of 70% ie. 42 W. Also the life of the CFLs
are much longer than that of the tube lights
with a cumulative savings on life and as well as
the energy consumption for the entire life.

CASE STUDY ON ENERGY CONSERVATION
IN STREET LIGHTS
The experience of such a reported transition
from tube light to CFL in Thiruvallur District is as
follows:

340 tube lights originally deployed were
replaced with 18 W CFL lamps, Consequently,
the annual savings in the cost of energy
reported is of Rs.2,12,704/-

CASE STUDY ON ENERGY CONSERVATION
IN STREET LIGHTS (contd…)
 Parameter
 40W Tube Light with a 20 W choke
 18 W Compact Fluorescent Lamp
 Savings
 Annual Consumption in units (with an average of 12 hours
per day for 365 days)
With Tube light: 60 x 12 x 365 = 263 kWh
With CFL: 18 x1 2 x 365 = 79 kWh
 Energy saving = 263 - 79 = 184 kWh
 Annual saving per light point (with cost of energy as
Rs.3.40 per unit)
= 184 x 3.40 = Rs. 625.60
 Total annual saving =Rs.625.60 x 340 =Rs. 2,12,704/-

AGRICULTURAL PUMP
 Selection of proper capacity of pumps according to the
irrigation requirement.
 Matching of pump set with source of water-canal or
well.
 Matching of motor with appropriate size pump.
 Proper installation of the pump system-shaft alignment,
coupling between motor and pump.
 Use of efficient transmission system.
 Maintain right tension and alignment of transmission
belts.
 Use of low friction rigid PVC pipes and foot valves.
 Avoid use unnecessary bends and throttle valves.
 Use bends in place of elbows.

AGRICULTURAL PUMP (contd…)
 The suction depth of 6 m is recommended as optimum
for centrifugal pumps. The delivery line should be kept
at minimum required height according to requirement.
 Periodically check pump system and carry out corrective
measures - like lubrication, alignment, tuning of engines
and replacement of worn-out parts.
 Over irrigation can harm the crops and waste vital water
resource. Irrigate according to established norms for
different crops.
 Use drip irrigation for specific crops like vegetable,
fruits, tobacco, etc. Drip systems can conserve upto 80%
water and reduce pumping energy requirement.

ENERGY EFFICIENCY IN AGRICULTURAL
PUMPING SYSTEM

 In centrifugal pumps, the fluid is fed to the
centre of a rotating impeller and is thrown
outward by centrifugal action.
 The high speed of rotation of the impeller
imparts high kinetic energy to the fluid. This
kinetic energy when converted into pressure
energy results in pressure difference between
the suction and delivery of the pump.

GENERAL PERFORMANCE CHARACTERISTICS
 The hydraulic performance of a centrifugal pump is based on
operating characteristics like:
 Capacity, Q: expressed in units of volume per unit of time,
such as, m3/s or lps
 Head, H : expressed in units of height of liquid column, to
which the liquid is pumped, such as, ft or m
 Power, P: expressed in units of energy, kW or HP
 Efficiency, η: expressed in %
 The variables influencing the performance of pump are:
 Speed, N : expressed speed at which pump runs, in RPM
 Diameter, D: expressed as diameter of impeller or wheel,
generally, in mm

PUMP POWER OUTPUT
 The power output of a pump is the product of the total
dynamic head and the mass of liquid pumped in a given time.
The power output is given by:
Pout = (Q x H x ) / 102
Where,
Pout = Pump power output (fluid power), in kW
Q = Capacity, in m3/s
H = Total dynamic head, in m of liquid column (LC)
= Density of liquid, in kg/m3

 The power required for driving the pump is the fluid power
(output) divided by the pump efficiency.

 The power output of a pump is less than the power input.
This is because of internal losses resulting from friction,
leakage etc.

PUMP EFFICIENCY
 The pump efficiency is expressed as:
ηpump = (Q x H x ) / (102 x kW x motor)
Where
Q = Capacity, in m3/s
H = Dynamic head, in m of LC
= Density of liquid, in kg/m3
kW = Motor input power
motor = Efficiency of motor

 The pump efficiency ( pump) is the product of three efficiencies:
pump = m v h
Where
m = Mechanical efficiency

v = Volumetric efficiency

h = Hydraulic efficiency

ENERGY CONSUMPTION
 The energy consumption is a factor, which is
affected by three aspects of hydrogeology, namely,
the geology of the area, depth of ground water and
the current level of ground water extraction.
 For example, the pumps operating in alluvial areas
with low pump density will consume less energy,
while the pumps operating in hard rock areas with
high pump density, will consume more energy.
 The energy consumption in the agricultural sector is
influenced by the cropping pattern also.

ENERGY INDEX
Energy index is the ratio of energy consumed by the pumping system per unit
of work done. This index is the inverse of the pumping system efficiency and
is given by:
(i) ELECTRIC PUMP SET
Energy index = ( 3 VI cos x 102) / (Q x H)
where V = Supply voltage, in V
I = Current measured, in A
cos = Power factor (0.70 or 0.80)
Q = Discharge rate, in lps
H = Static head, in m
(ii) DIESEL PUMP SET
Energy index = (27.78 x D) / (Q x H)
where D = Diesel consumed, in lph
Q = Discharge rate, in lps
H = Static head, in m

REDUCING FRICTION ACROSS SUCTION
PIPING, Hfs
 The reduction of Hfs can result in the following
benefits:
 Reduction of total system head (H)
 Cavitations-free operation
 Energy efficient performance
 The friction loss across pipes is given by:
Hfs = (4 x f x L x v2) / (2 x g x D)
Where,
f = coefficient of friction of pipe
L = equivalent length of pipe, in m
v = velocity of flow, in m/s
D = diameter of pipe, in m
g = acceleration due to gravity, in m/s2

energy conservation

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