energy conservation


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energy conservation

  2. 2. 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.
  3. 3. 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
  4. 4. 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
  5. 5. 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 nations 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.
  6. 6. 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)
  7. 7. 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.
  8. 8. 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
  9. 9. CONSERVATION OF ENERGY IN LIGHTING SYSTEM Selection of Lighting in Buildings Two methods are available:  Building area method  Space by space method
  10. 10. 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)
  11. 11. 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
  12. 12. 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.
  13. 13. TABLE 2: INTERIOR LIGHTING POWER FOR SPACE BY SPACE METHODSpace Function LPD (W/m2) Space Function LPD (W/m2)Office-enclosed 11.8 Library 11.8Office-open plan 11.8 Card File & Cataloging 11.8Conference/Meeting/Multipurpose 14.0 Stacks 18.3Classroom/Lecture/Training 15.1 Reading Area 12.9Lobby 14.0 HospitalFor Hotel 11.8 Emergency 29.1For Performing Arts Theater 35.5 Recovery 8.6For Motion Picture Theater 11.8 Nurse Station 10.8Audience/Seating Area 9.7 Exam Treatment 16.1For Gymnasium 4.3 Pharmacy 12.9 Patient Room 7.5For Convention Center 7.5 Operating Room 23.7For Religious Buildings 18.3 Nursery 6.5For Sports Arena 4.3 Medical Supply 15.1For Performing Arts Theater 28.0 Physical Therapy 9.7For Motion Picture Theater 12.9 Radiology 4.3For Transportation 5.4 Laundry – Washing 6.5Atrium-first three floors 6.5 Automotive – Service Repair 7.5Atrium-each additional floor 2.2 ManufacturingLounge/Recreation 12.9 Low Bay (<8m ceiling) 12.9 Contd. …………………..
  14. 14. 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
  15. 15. 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.
  16. 16. 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
  17. 17. CASE STUDY ON FINANCIAL EVALUATION 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/-
  18. 18. CASE STUDY ON FINANCIAL EVALUATION 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
  19. 19. CASE STUDY ON FINANCIAL EVALUATION 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/-
  20. 20. 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
  21. 21. 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.
  22. 22. TIPS FOR ENERGY CONSERVATION IN 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
  23. 23. TIPS FOR ENERGY CONSERVATION IN DOMESTIC SECTOR (contd…) 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.
  24. 24. TIPS FOR ENERGY CONSERVATION IN DOMESTIC SECTOR (contd…) 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
  25. 25. 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.
  26. 26. 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/-
  27. 27. 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/-
  28. 28. TIPS FOR ENERGY CONSERVATION IN 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.
  29. 29. TIPS FOR ENERGY CONSERVATION IN 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.
  30. 30. 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.
  31. 31. 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
  32. 32. 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.
  33. 33. 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
  34. 34. 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.
  35. 35. ENERGY INDEXEnergy index is the ratio of energy consumed by the pumping system per unitof work done. This index is the inverse of the pumping system efficiency andis 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
  36. 36. 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