This document discusses various topics related to energy conservation. It begins by defining energy conservation as decreasing the quantity of energy used through efficient energy use or reduced consumption. Individuals and organizations want to conserve energy to reduce costs and promote economic security. It then discusses the necessity of energy conservation due to factors like increased energy costs, the impact of energy costs on product costs, available efficient technology options, reductions in greenhouse gas emissions, and overall environmental benefits. The document also covers energy conservation efforts and policies in India, including the Energy Conservation Act of 2001, targets for capacity creation through efficiency, identified conservation potential, and the role of the Bureau of Energy Efficiency. Specific sectors like buildings, industries, and domestic use are examined

1. ENERGY
CONSERVATION

2. 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.

3. NECESSITY OF ENERGY CONSERVATION
 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

4. ENERGY CONSERVATION IN INDIA
 Energy conservation act, 2001
 25,000 MW capacity creation through energy efficiency
 23% energy conservation potential
 Industrial and agriculture sectors have the maximum
potential
 Bureau of Energy Efficiency (BEE) created
 Set up in 2002 Under Ministry of power
 Assessment of potential savings in different
sectors

5. 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 to cooling and heating the house in summer and winter
respectively can significantly reduce energy costs.

6. Energy efficient building
construction
• Very good insulation of walls,
roofs and Basement.
• Windows with high quality double
or triple glazing.
• Air-tight construction
• check by blower door test
• Avoiding cooling demand
- sun shading in summer
- natural cooling sources

7. •
a)
b)
c)
Conservation of Energy in Lighting System
- Selection of Lighting in Buildings
- Building Area Method
Determination of interior lighting power allowance
(watts) by the building area method shall be in
accordance with the following:
Determine the allowed lighting power density from
Table 1 for each appropriate building area type.
Calculate the gross lighted floor area for each building
area type.
The interior lighting power allowance 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.

8. Table 1 Interior Lighting Power – 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

9. Space Function Method
Determination of interior lighting power allowance (watts)
by the space function method shall be in accordance with
the following:
a)
b)
c)
Determine the appropriate building type from Table 2 and
the allowed lighting power density.
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.
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.

10. Table 2 Interior Lighting Power – Space Function 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. …………………..

11. 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

12. •
•
Energy Conservation in Industry through
Motor:
Electric motors converts electrical energy
into mechanical energy.
By using Energy Efficient Motor (EEM) in
place of Standard Motor we save the energy.
An Energy Efficient Motor produces the same
shaft output power (HP) but uses less input
power (kW) than a standard efficiency
motor.

13. Case study on financial evaluation of EEM
The following case can be evaluated financially using energy saving
criteria.
A 60 HP standard AC motor operating at 75% load. At this load efficiency
of motor is 82%. Motor operates 20 hrs a day and 300 days a year. Per
unit energy charge is Rs. 3.10 and per month per kVA demand charge is
Rs. 175. (Take 1 HP = 750 W)
Input power to motor
= 45 kW x 0.75 / 0.82 = 41.15 kW
Power factor of motor = 0.80
kVA
= kW/Power Factor
kVA demand
= 41.15 / 0.80 = 51.43 kVA
kVA charges/year
= 51 x Rs. 175/kVA x 12 months
= Rs. 1,07,100/Energy (kWh) charges/year
= 41.15 kW x Rs. 3.1/kWh x 20 hrs x
300 days
= Rs. 7,65,390/Total (Demand + Energy) Cost/year = 1,07,100 + 7,65,390
= Rs. 8,72,490/-

14. A 60 HP Energy Efficient Motor operating a 75% load.
Efficiency of the motor is 87%.
Input power = 45 kW x 0.75 / 0.87 = 38.79 kW = 38.8 kw
P.F. of E.E.M. = 0.83
kVA demand = 38.8 / 0.83 = 46.75 kVA = 47 kVA
Demand charges/year = 47 kVA x Rs. 175/kVA x 12 months =
Rs. 98,700/Energy Charges/year = 38.8 kW x Rs. 3.10/kWh x 20 hrs/day x
300 days = Rs. 7,21,680/Total (Energy + Demand) charges/year = Rs. 7,21,680 + Rs.
98,700 = Rs. 8,20,380/-

15. Financial Evaluation
Cost of 60 HP standard AC motor = Rs. 1,00,000/Cost of 60 HP EEM
= Rs. 1,20,000/-
Energy savings achieved by
E.E.M. over standard motor
= Rs. 8,72,490 – Rs. 8,20,380
= Rs. 52,000 (approx.)

16. Payback period for replacement of = 1,20,000 / 52,000
existing standard motor with E.E.M.= 2.3 years
Payback period for purchase of
E.E.M. for new installation
= 20,000 / 52,000
= 0.381 yr = 4.6 months

17. Tips for Energy Conservation in Domestic Sector
•Organised 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.
•COOK YOUR FOOD IN SOLAR COOKERS Cook anything except ROTI AND save cost of 2 LPG
Cylinders annually.
USE YOUR REFRIGERATOR & AIRCONDITIONERS EFFICIENTLY
•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 & closing of
refrigerator & air-conditioned rooms.
PLANT TREES
•Place the refrigerator about 6 inches away from the wall to allow air circulation.
•Air-conditioned room must be leakpfoof.
USE PROPER LIGHTING & EFFICIENT LIGHTING DEVICES
•A tubelight (36/40 watt) gives more light than a 60 or 100 watt bulb and will consume 40 to 60% less
power. Tubelight with electronic choke is even more energy efficient means of lighting USE DAYLIGHT
AS FAR AS POSSIBLE.
•Lighting devices like bulb, tubelight, etc. consume energy according to their capacities. Use appropriate
lighting according to your requirement. A so-called zero bulb uses 12 to 15 watt per hour. Compact
Flouroscent Lamps (CEL) are available in 5.7.9.11 watts capacities and they give more light output.
•Instant geysers are considered to be more efficient than storage type geysers. SOLAR WATER
HEATERS OPERATE ON SOLAR ENERGY WHICH IS FREE OF COST.
ELECTRICITY IS AN EXPENSIVE AS COOKING FUEL – AVOID IT SWITCH OFF LIGHTS & FANS
WHEN NOT REQUIRED STOP WASTAGE – WE CANNOT AFFORD IT – CONSERVE ENERGY.

18. Energy conservation in Street Lights:
Normally a tube light of 40W rating with a choke of 20W are being utilized for
street lights with a total power of 60W. Alternately the use CFL bulbs of 18W
rating which has an equivalent luminosity would lead to a power saving to the
extent of 70% ie. 42W. Also the life of the CFL bulbs 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.
The experience of such a reported transition from tube light to CFL in Thiruvallur
District is as follows:
' 340 Nos of tube lights originally deployed were replaced with 18 W CFL lamps
' Consequent to the switching over to 18 W CFL lamps the annual savings in the
cost of energy reported to have been generated is to the tune of Rs.2,12,704/- (~
625.60x340)

19. This is based on the Savings derived from a single street light which is as follows:
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)
60x12x365= 263 kWh
18x12x365 =79 kWh
Energy Saving = 263-79 =184
kWh
Cost of Annual consumption @ Rs.3.40 per unit
= 184x3.40 = Rs.625.60
Total annual saving=Rs.625.60x340 =Rs212704

20. Tips for Energy Conservation in Agricultural Pump
•
•
•
•
•
•
•
•
•
•
•
•
Selection of right 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 of 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.
The suction depth of 6 metres is recommended as optimum for centrifugal pumps.
The delivery line should be kept at minimum require height according to
requirement.
Periodically check pump system and carryout 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 crop.
Use drip irrigation for specific crops like vegetable, fruits, tobacco, etc. Drip
systems can conserve upto 80% water and reduce pumping energy requirement.

21. Energy Efficiency in Agricultural
Pumping System
Operating Principles of Pump
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.

22. 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/h 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, h : expressed as %
The variables that influence these are:
Speed, N
: expressed speed at which pump runs, in RPM
Diameter, D : expressed as diameter of impeller or wheel, generally,
in mm

23. 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 ρ) / 102
Where, Pout
Q
H
ρ
= Pump power output (fluid power), in kW
= Capacity, in m3/s
= Total dynamic head, in m of liquid column (LC)
= Density of liquid, in kg/m3
Pump Power Input
The power required for driving the pump is the water horsepower 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.

24. Pump Efficiency
The pump efficiency is the ratio of fluid power to the total power input. This is
expressed as:
hP
= (Q x H x ρ) / 102 x kW x ηmotor
Where
Q
H
ρ
kW
ηmotor
= Capacity, in m3/s
= Dynamic head, in m of LC
= Density of liquid, in kg/m3
= Motor input power
= Efficiency of motor
The efficiency (η) is the product of three efficiencies:
η
Where
= ηm ηv ηh
ηm
= Mechanical efficiency
ηv
= Volumetric efficiency
ηh
= Hydraulic efficiency

25. 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.

26. 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)
(ii)
Electric pumpset
Energy index
where
I
Cos ϕ
Q
H
= (√3 VI Cos ϕ x 102) / (Q x H)
= Current measured, in Amperes
= Power factor (0.70 or 0.80)
= Discharge rate, in lps
= Static head, in m
Diesel pumpset
Energy index
where
D
Q
H
= (27.78 x D) / (Q x H)
= Diesel consumed, in lph
= Discharge rate, in lps
= Static head, in m

27. Reducing Friction Across Suction Piping, Hfs
The reduction of Hfs can result in the following benefits:
 Reduction of total system head (H)
 Cavitation-free operation
 Energy efficient performance
The friction loss across pipes is given by:
Hf
Where,f
L
v
D
g
= (4 x f x L x v2) / (2 x g x D)
= coefficient of friction of pipe
= equivalent length of pipe, in m
= velocity of flow, in m/s
= diameter of pipe, in m
= acceleration due to gravity, in m/s2