energy audit

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By: umair060 (32 month(s) ago)

kindly send this presentation to my email address umair060_tuf@yahoo.com I need it. Thanks

By: GreyP (32 month(s) ago)

Go to http://www.healthyhomediagnostics.com if you need help getting a good home energy audit in Raleigh, NC.

By: wasimdesai (36 month(s) ago)

this presentationis damn good -----------wasim desai

By: gnbgv015 (55 month(s) ago)

Thanks this one is very good PPT.. keep up the good work..

By: thakare.hitesh (57 month(s) ago)

This presentation is damn good.. one of the best i have ever seen ,,, Thank you very much dragontronix.. . I have a query about the capacitors.. how do they generate reactive power ?? & How can you calculate the savings by installing the capacitor, in terms of money & payback period??? Please tell me that ... I will be very much thankful.. Warm Regards, Hitesh (India).

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Slide 1: 

Energy Audit

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Energy loss in any industrial process or plant is inevitable; it is a foregone conclusion. But its economic and environmental impacts are not to be taken lightly, thus explaining the growing need for industrial energy efficiency. Put simply, the level of energy efficiency a plant or process can achieve is inversely proportionate to the energy loss that occurs; the higher the loss, the lower the efficiency. Energy Losses

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Where and how do most of the losses occur, how much energy is actually lost and are they controllable or recoverable? The answers to these questions remain well concealed in a black box where once energy is input, we do not know what really happens to it inside and how much the losses are. It is only when we look into the black box and extract these details that we are able to ascertain the performance of the overall or process levels and respond more effectively to the weaknesses in energy management.

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Overall energy losses in a plant can result from losses due to designs that do not incorporate energy efficient specifications such as heat recovery option; operations that run on inefficient methods; and poor or non-energy efficiency-conscious maintenance programme. Reducing these losses will substantially increase the plant's efficiency, but we need data to identify and quantify the losses and subsequently suggest suitable techno-economic solutions to minimize the losses. This data can be acquired through energy audits.

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Energy audit: Definition and Types   Energy audit is a systematic study or survey to identify how energy is being used in a building or plant, and identifies energy savings opportunities. Using proper audit methods and equipment, an energy audit provides the energy manager with essential information on how much, where and how energy is used within an organization (factory or building).

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This will indicate the performance at the overall plant or process level. The energy manager can compare these performances against past and future levels for a proper energy management. The main part of the energy audit report is energy savings proposals comprising of technical and economic analysis of projects. Looking at the final output, an energy audit can also be defined as a systematic search for energy conservation opportunities.

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This information can be transformed into energy savings projects. It will facilitate the energy manager to draw up an action plan listing the projects in order of priority. He will then present it to the organization's management for approval. Providing tangible data enables the management to be at a better position to appreciate and decide on energy efficiency projects. Adopting this activity as a routine or part of the organization's culture gives life to energy management, and controlling the energy use by energy audit is what we refer to as Energy Management by Facts

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Energy audit stages Energy audit can be categorized into two types, namely walk-through or preliminary and detail audit. Walk-through or preliminary audit   Walk-through or preliminary audit comprises one day or half-day visit to a plant and the output is a simple report based on observation and historical data provided during the visit. The findings will be a general comment based on rule-of-thumbs, energy best practices or the manufacturer's data.

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Preliminary Energy Survey Quick overview of energy use patterns Provides guidance for energy accounting system Provides personnel with perspectives of processes and equipment Identify energy – intensive processes and equipment Identify energy inefficiency ,if any Set the stage for detailed energy survey

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Detail audit   Detail audit is carried out for the energy savings proposal recommended in walk-through or preliminary audit. It will provide technical solution options and economic analysis for the factory management to decide project implementation or priority. A feasibility study will be required to determine the viability of each option.

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Detailed Energy Survey Detailed evaluation of energy use pattern By processes and equipment Measurement of energy use parameters Review of equipment operating characteristics Evaluation of efficiencies Identify DSM options and measures Recommendation for implementation

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Inputs and Outputs of Energy audit

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ELECTRICAL ENERGY AUDIT                                                             Areas covered under Electrical audit are 1. Electrical System : • Electrical Distribution system (substation & feeders study) • PF Improvement study • Capacitor performance • Transformer optimization • Cable sizing & loss reduction • Motor loading survey • Lighting system • Electrical heating & melting furnaces • Electric ovens

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2. Mechanical System :                                        • Fans & Blowers • Exhaust & ventilation System • Pumps and pumping System • Compressed air System • Air Conditioning & Refrigeration System • Cooling Tower System

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THERMAL ENERGY AUDIT Areas covered under Electrical audit are Steam Generation Boilers • Steam Audit and Conversation • Steam Trap Survey • Condensate Recovery System • Insulation Survey • Energy and Material Balance for Unit operation • Heat Exchanger • Waste Heat Recovery System

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WATER AUDIT & CONSERVATION Industry has recognized 'Water Audit' as a important tool for water resource management     Water Audit study is a qualitative and quantitative analysis of water consumption to identify means of reuse and recycling of water. This study includes segregation of effluent streams and schemes for effectively treating them to enable byproduct recovery. Water Audits encourage social responsibility by identifying wasteful use, enables estimation of the saving potential they not only promote water conservation but also deliver cost savings, but also companies to safeguard public health and property, improve external relations and reduce legal liability.

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Benefits of energy audit   There is a lot of potential for energy savings from energy audits.. Technical solutions proposed in the energy audits show massive potential for energy savings in every sub-sector with an average of almost ten percent of the energy usage. However, this can only materialize through replication at other factories within the respective sub-sector.

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Benefits of an industrial energy audit include:   Energy savings Avoiding power factor penalties and environmental compliance costs Quality improvements Productivity improvements Reduced maintenance Fewer breakdowns Better safety and protection A process for repeatable improvements

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Process VS. Equipment Equipment efficiency improvement : Max. 5% Process efficiency improvement : 15% to 30% Conclusion Focus on Processes

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Energy conservation opportunities Electrical Reduce demand by load management Electrical / thermal Reduce consumption by improving energy use efficiency (reducing losses ,utilizing waste heat ,etc..)

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Examples of conservation opportunities - processes 1.Electric furnace - Automation of energy supply control - Substitute electricity by thermal energy ( mazout/solar/gas ) - Reduce radiation losses

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2. Steam generation system - Combustion efficiency improvement - Waste heat recovery from flue gases - Heat loss reduction from boiler surfaces - Reduce leakage 3. Condensate return system 4. Steam distribution system - Pipe insulation - Steam traps - Steam leaks

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Examples of conservation opportunities –equipment Gas/air compression system pre-cooling the gas/air 2. High efficiency motors 3. High efficiency lamps 4. High efficiency pump/fan 5. Change electric dryer and heater to oil/gas fuel 6. Replace motor/generator set with silicon controlled rectifier (SCR)

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Analyze utility bills

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Basic data of the company 1- General description of company and products. 2- Electrical distribution system. 3- Energy price. 4- Energy factor and environmental pollution. 5- Electrical loads of company. Case study

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1- General description of company and products: Company “ A “ lies in Kilo “ XX” Alexandria desert road. The company operates 24 hours with 3 shifts a day for 300 days/ year. The company produces dry batteries. 2- Electrical distribution system: Electric power network of “ A “ Company is supplied by two transformers 500 KVA, 11/ 0.4 kV. Also, there is two standby generators 400 kVA per each. Figure (1) shows the single line diagram for the electric distribution network.

Slide 44: 

Table (2) Figure (2) shows the monthly development of electric consumption for company loads during 2006 and 2007.

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3- Energy price: According to appendix (2) 4- Energy factor and environmental pollution. 0.2196 kg fuel /kWh 3.082 kg CO2/ kg fuel 0.677 kg CO2 /kWh 5- Electrical loads of company: The following tables represent loads and its rating

Slide 48: 

lighting loads:

Slide 49: 

Power Factor

Equipment and appliances requiring reactive energy : 

Equipment and appliances requiring reactive energy All AC equipment and appliances that include electromagnetic devices, or depend on magnetically-coupled windings, require some degree of reactive current to create magnetic flux. The most common items in this class are transformers and reactors, motors and discharge lamps (with magnetic ballasts) The proportion of reactive power (kvar) with respect to active power (kW) when an item of equipment is fully loaded varies according to the item concerned being: * 65-75% for asynchronous motors * 5-10% for transformers

Slide 53: 

Values of cos ϕ and tan ϕ for commonly-used equipment

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1- Example (1) 380 V 380 V± 5% =361 :399 V

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2- Line Current

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3-

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4-

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5- 220 V± 5% =209 :231 V

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6-

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7-

Example (2) : 

Example (2) 1-

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2-

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3- 0.9

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4-

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5-

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6-

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7-

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8-

Slide 73: 

For 3-phase balanced current For 3-phase unbalanced current, PF at certain time is

معامل القدرة و عقد توريد الطاقةعقد توريد كهرباءعلى الجهد المنخفض حتى 500 كيلو وات : 

معامل القدرة و عقد توريد الطاقةعقد توريد كهرباءعلى الجهد المنخفض حتى 500 كيلو وات البند الخامس: أولا: الكهرباء الموردة بموجب هذا العقد يتم حساب قيمتها بناء على ما يلى: 1- سعر الطاقة الكهربائية: لكل كيلووات ساعة طبقا لتعريفة الطاقة الكهربائية السارية 2- معامل القدرة: للمنتفعين بقدرة 10 كيلو وات فأكثر أسعار الطاقة الكهربائية موضوعة على أساس معامل قدرة متوسط 0.9 فى حالة انخفاض هذا المعامل فى السنة المالية عن 0.9 يزاد سعر الطاقة بمقدار 0.5% لكل 0.01 من أنخفاض المعامل حتى 0.7

Slide 75: 

و فى حالة انخفاض المعامل عن 0.7 يزاد سعر الطاقة بمقدار 1% لكل 0.01 من أنخفاض المعامل عن 0.7 يلتزم المنتفع فى هذه الحالة بتركيب أجهزة تحسين معامل القدرة خلال تسعة شهور من تاريخ إخطاره بكتاب مسجل بعلم الوصول. و فى حالة عدم تركيب الأجهزة خلال تلك المدة يكون للشركة الحق فى قطع التغذية عن المنتفع إلى أن يقوم بتحسين معامل القدرة إلى ما لا يقل عن 0.7 و يظل العقد ساريا حتى يتم تحسين المعامل . و فى حالة زيادة المعامل عن 0.92 يخفض سعر الطاقة بمقدار 0.5% لكل 0.01 من أرتفاع المعامل عن 0.92 و بحد أقصى 0.95

عقد توريد كهرباءعلى الجهد المتوسط بقدرةأكبر من 500 كيلو وات : 

عقد توريد كهرباءعلى الجهد المتوسط بقدرةأكبر من 500 كيلو وات البند الخامس: - - - معامل القدرة: أسعار الطاقة الكهربائية موضوعة على أساس معامل قدرة متوسط 0.9 و فى حالة انخفاض هذا المعامل فى السنة المالية عن 0.9 يزاد سعر الطاقة بمقدار 0.5% لكل 0.01 من أنخفاض المعامل حتى 0.7

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و فى حالة انخفاض المعامل عن 0.7 يزاد سعر الطاقة بمقدار 1% لكل 0.01 من أنخفاض المعامل عن 0.7 يلتزم المنتفع فى هذه الحالة بتركيب أجهزة تحسين معامل القدرة خلال تسعة شهور من تاريخ إخطاره بكتاب مسجل بعلم الوصول. و فى حالة عدم تركيب الأجهزة خلال تلك المدة يكون للشركة الحق فى قطع التغذية عن المنتفع إلى أن يقوم بتحسين معامل القدرة إلى ما لا يقل عن 0.7 و يظل العقد ساريا حتى يتم تحسين المعامل . و فى حالة زيادة المعامل عن 0.92 يخفض سعر الطاقة بمقدار 0.5% لكل 0.01 من أرتفاع المعامل عن 0.92 و بحد أقصى 0.95

Slide 78: 

مثال متوسط معامل القدرة المقاس = 0.46 الاستهلاك السنوى = Kwh63933 احسب مقابل انخفاض معامل القدرة الحل: أ ) انخفاض معامل القدرة من 0.9 وحتى 0.7 أي 0.2 تزيد قيمة الطاقة 0.005 لكل 0.01 من انخفاض المعامل للقيمة 0.2 ب) انخفاض معامل القدرة من 0.7 وحتى 0.46 أي 0.24 تزيد قيمة الطاقة 0.01 لكل 0.01 من انخفاض المعامل للقيمة 0.24

Slide 79: 

متوسط معامل القدرة المقاس = 0.46 أ ) انخفاض معامل القدرة من 0.9 وحتى 0.7 أي 0.2 تزيد قيمة الطاقة 0.005 لكل 0.01 من انخفاض المعامل للقيمة 0.2 ب) انخفاض معامل القدرة من 0.7 وحتى 0.46 أي 0.24 تزيد قيمة الطاقة 0.01 لكل 0.01 من انخفاض المعامل للقيمة 0.24

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أي أن: تزيد قيمة الطاقة بالنسبة : (20×0.005) + (24×0.01) = 0.34 = 34% قيمة الطاقة المقابلة لانخفاض معامل القدرة من 0.9 إلى 0.46 = نسبة الزيادة المفروضة × الطاقة المستهلكة خلال العام × سعر الطاقة الكهربائية = 0.34 × (63933 ك.و.س/السنة) × (0.231 جنيه/ك.و.س) =جنية 5021

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Power Factor Improvement Capacitors

What we will learn: : 

What we will learn: Most Industrial loads require both Real power and Reactive power to produce useful work You pay for BOTH types of power Capacitors can supply the REACTIVE power thus the utility doesn’t need to Capacitors save you money!

Why Apply PFC’s? : 

Why Apply PFC’s? Power Factor Correction Saves Money! » Reduces Power Bills » Reduces I2R losses in conductors » Reduces loading on transformers » Improves voltage drop

Why do we Install Capacitors? : 

Why do we Install Capacitors? Capacitors supply, for free, the reactive energy required by inductive loads. » You only have to pay for the capacitor ! » Since the utility doesn’t supply it (kVAR), you don’t pay for it!

Other Benefits: : 

Other Benefits: Released system capacity: » The effect of PF on current drawn is shown below: Decreasing size of conductors required to carry the same 100kW load at P.F. ranging from 70% to 100%

Other Benefits: : 

Other Benefits: Reduced Power Losses: » As current flows through conductors, the conductors heat. This heating is power loss » Power loss is proportional to current squared (PLoss=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

Low Voltage Capacitor Unit –Low Voltage Capacitor : 

Low Voltage Capacitor Unit –Low Voltage Capacitor Cubicle-type automatic capacitor banks are modular in structure.

Slide 88: 

Series CMR Series RCF Series MFA Series MFHC

High Voltage Capacitor Unit –High Voltage Capacitor Units : 

High Voltage Capacitor Unit –High Voltage Capacitor Units There are two types of fuses used for capacitors; internal and external.

High Voltage Capacitor – High Voltage Capacitors : 

High Voltage Capacitor – High Voltage Capacitors High Voltage Capacitors One-Phase Units have all-film dielectric and are impregnated with dielectric liquid which is environmentally safe.

Power Factor Controller – Power Factor Controllers : 

Power Factor Controller – Power Factor Controllers Power Factor controllers 6 step or 12 step models are manufactured for the control of the automatic capacitor banks.

Power Factor Correction Capacitor PFC : 

Power Factor Correction Capacitor PFC Automatic capacitor banks are used for central power factor correction at main and group distribution boards.

Slide 93: 

Different Locations of capacitor banks

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Determining capacitor requirements

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Examples: A plant with a metered demand of 600 KW is operating at a 75% power factor. What capacitor KVAR is required to correct the present power factor to 95%? Solution From Table 1, Multiplier to improve PF from 75% to 95% is 0.553 Capacitor KVAR = KW × Table 1 Multiplier Capacitor KVAR = 600 × 0.553 = 331.8 say 330

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2. A plant load of 425 KW has a total power requirement of 670 KVA. What size capacitor is required to improve the factor to 90%? Solution Present PF = KW/KVA = 425/670 = 63.4% say 63% From Table 1, Multiplier to improve PF from 63% to 90% is 0.748 Capacitor KVAR = KW × Table 1 Multiplier = 425 × 0.748 = 317.9 say 320 KVAR

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3. A plant operating from a 480 volt system has a metered demand of 258 KW. The line current read by a clip-on ammeter is 420 amperes. What amount of capacitors are required to correct the present power factor to 90%? Solution KVA = 1.73 × KV × I = 1.73 × 0.480 × 420 = 349 KVA Present PF = KW/KVA = 258/349 = 73.9% say 74% From Table 1, Multiplier to improve PF from 74% to 90% is 0.425 Capacitor KVAR = KW × Table 1 Multiplier = 258 × 0.425 = 109.6 say 110 KVAR

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4. Assume an uncorrected 460 KVA demand, 380 V, 3-phase, at 0.87 power factor (normally good). Determine KVAR required to correct to 0.97 power factor..

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Solution: KVA × PF = KW 460 × 0.87 = 400 KW actual demand at PF = 0.97 KVA corrected = 400/0.97 = 412 KVA From Table of multipliers, to raise the PF from 0.87 to 0.97 Required Capacitor Multiplier = 0.316 KW × multiplier = KVAR required KVAR required = 400 × 0.316 = 126 KVA = 140 KVAR (use)

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Transformer serving the loads

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As the triangle relationships demonstrate, KVA decreases as power factor increase. At 70% power factor , it required 142 KVA to produce 100 KW. At 95% power factor, it requires only 105 KVA to produce 100 KW. Another way to look at it is that at 70% power factor, it takes 35% more current to do the same work.