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ENERGY CONSERVATION WHY ? WHAT IS THE NEED? WHO WILL GAIN ?

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Presentation on theme: "ENERGY CONSERVATION WHY ? WHAT IS THE NEED? WHO WILL GAIN ?"— Presentation transcript:

1 ENERGY CONSERVATION WHY ? WHAT IS THE NEED? WHO WILL GAIN ?

2 All India Power Shortages $Existing Supply : 368,046 M Units $Existing Demand : 414,000 M Unit $ Shortage : 11.5 %

3 POWER COST FOR LAST TEN YEARS 3 – Fold Increase in Last Nine Years

4 NATIONAL POWER SCENE $Precarious $Adverse Impact On Industry $New Project Additions very slow $Action Plan Required for Improvement

5 COMPARISION OF POWER COST Comparative Power Cost In Different Countries

6 ENERGY SAVING POTENTIAL IN INDIAN INDUSTRY $Energy Saving Potential Rs.100,000.00 Million/yr $Equivalent 2500 MW $Investment Opportunity Rs.200,000.00 Million

7 Equipment MW Saving Savings in Rs. Million VFD3508750 Soft Starter802000 Auto Star Delta802000 EE Motors1253125 Lighting Volt stabilizers802000 Energy Effi Chokes401000 Auto Lux Level Controller 5125 On/Off temp Control 802000 Total84021000 ENERGY SAVING POTENTIAL In Industry – Electrical Devices Investment Potential Rs.42000 million

8 ENERGY COST AS % OF MANUFACTURING COST $Chlor Alkali Industry 65% $Cement Industry 40% $Paper Industry 25% $Chemical Industry 15% $Foundry 25% $Engineering Industry 10%

9 Energy Conservation : An Excellent opportunity for enhancing profit and improving competitiveness

10 PROBLEMS FACED Lack of Awareness Doubting High Capital Investment Lack Of Attractive Financing Schemes Over Promises Reliability Of Equipment Pricing of Energy need for Policy - Changes

11 POSSIBLE SOLUTIONS Awareness Campaign Encon Mission Training to Industry Energy Summit Energy Norms Demo Projects Award Schemes

12 ENERGY CONSERVATION AT MACRO LEVEL 3-Pronged Approach a. Capacity Utilization b. Fine Tuning c. Technology Up gradation (Target – Reduction in specific energy consumption)

13 MACRO LEVEL METHODOLOGY Energy Input (a) = Unavoidable losses (c) +Theoretical Requirement (b) +Avoidable losses (d)

14 MACRO LEVEL METHODOLOGY FOCUS SHOULD BE a. To concentrate on avoidable losses b. Quantify the losses c. Identify ways and means for reduction d. Implementation

15 ENERGY CONSERVATION IN ELECTRICAL MOTORS A device which converts electrical energy into mechanical energy Major source of energy consumption Major population- Induction motors

16 Motor Efficiency =O/P Power/Input PowerX100 Watt Losses- Stator & rotor losses Iron Losses Friction & Windage losses Stray Load losses

17 Range Of losses In An Induction Motor

18 Motor Losses #Voltage dependent – Iron Losses Magnetization Eddy Current #Current Dependent – copper losses Stator Rotor #Mechanical losses – Friction and windage losses

19 Energy Waste- Causes @Use of less efficient motors @Oversized/undersized motors @Improper supply voltage @Voltage fluctuations @Poor power factor @Less efficient driven equipment @Idle running

20 Motor Efficiency Improvement Motor operation in lightly loaded condition which is common practice in industry – Forced to operate in less efficient zone

21 Voltage Optimisation Impact on motor operating parameters $ Red. in volt. Dependent losses $ Capacity reduces. $ PF Improves $ Load Current drops. $ Load factor improves $ Efficiency improves

22 Optimisation of Lightly Loaded Motors Options – Lightly loaded motors + Delta to permanent star connection +Auto star delta convertor +Soft start cum energy saver +Down Sizing +Overall voltage optimisation

23 Soft Start Cum Energy Saver % of Loading % Saving 10 60 20 38 30 20 40 11 50 6.5 60 4.5 70 2.5 80 1.5 90 1.0

24 Optimise The Plant Operating Voltage Overall - Plant operating voltage plays vital role in energy saving -Suggested to have on line voltage optimization (OLTC) -Magnetization losses vary exponentially with the voltage * Capacity prop V2 *Volt. Opt. will vary capacity *Should be implemented after analyzing the loading pattern of all motors

25 Energy Conservation in Electrical Distribution System * Componenets in electrical distribution a. HT/LT Circuit breakers b. Switches and Fuses c. Transformers d.Busbars/Cables (HT/LT)

26 Measures in Minimizing Distribution Losses * HT/LT Circuit Breakers Maintain the contact surface uniformity, through vigorous maintenance Select energy efficient fuses

27 Measures in Minimizing Distribution Losses * Transformers #Select energy efficient #wherever possible run in parallel #Loading should be optimal

28 Measures in Minimizing Distribution Losses * Bus Ducts /Power Cables # Select correct size # Bus duct with minimum joints and bends # Cables with minimum joints #Panel should be placed near to load wherever possible to minimize cable length & its losses # Cable should be terminated with proper crimping sockets

29 Methods and Procedures To Minimize The Distribution Losses * Voltage drop measurement # In a large complex distribution system voltage drops are common # Acceptable limit is 4-5 V/PHASE # More than 5V/phase indicates energy loss in system * Reasons For Voltage Drop # Poor Power Factor # Inadequate Cable size laid # Poor contact surface at Cable termination Cable joints contactors/switches

30 Case Study – Voltage Drop From Engineering Industry Voltage at substation – 415 V Voltage at LT panel – 398 V Load Current – 180 to 200A PF - 0.4 LAG Cable size – 1RX3CX300 mm. sq Relocate 90 KVAR Cap bank from SS to LT panel Reduced 50% of energy losses Annual Saving – Rs. 0.6 lacs

31 Energy Conservation in Transformers Transformer Efficiency – 98-99% Optimum Efficiency Occurs when Iron Losses = Copper Losses (Optimum eff. Occurs between 40% to 60 % of loading ) Selection of Transformer should be based on TOC TOC = Price +(No load loss x loss value) +(load loss x loss value)

32 Three Phase Transformer Typical Loss Chart KVA Iron Loss FL Copper Loss 500 1030 6860 750 1420 9500 1000 1770 11820 1250 1820 12000 *loss is in watts

33 CB 2000KVA 11KV/433V CB 2000KVA 11KV/433V 11KV 415V CASE STUDY

34 CASE STUDY Background *Cap of Xmer = 1600 KVA *Load on Xmer is 80 % *Iron Loss = 2.3 kw *copper loss = 21 kw Suggestion – Operate both transformer in parallel *One Xmer operation loss = 2.3+21(0.8)2= 15.7 kw *Both Xmer in operation loss = [2.3+21x(0.4)2]x2=11.3 kw Annual Saving = Rs.0.78 lacs

35 THANK YOU


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