Energy Auditing Final Phase

Tuesday, May 2, 2023

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ENERGY AUDITING IN THE COLLEGE CAMPUS

GROUP MEMBERS:

BINOY P B

SRUTHI C.K

RADHIKA A

SANJUSH S

SUJINI M

GUIDED BY :

SREEJITH KAsst Prof Dept of EEE


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ABSTRACT An energy audit is an inspection, survey and analysis of energy flows for

energy conservation in a building, process or system to reduce the amount of energy input into the system without negatively affecting the output(s).

By identifying the sources of energy use in the college campus such as wastage of energy that can be found out and can be reduced or neglected.

Also by replacing the old inefficient items such as( fans, fluorescent lamp, motors etc...) by efficient ones.

By auditing we can save the wastage of energy resources and also we can reduce consumptions also.

Effective and concrete method to achieve rapid improvement in energy efficiency in buildings and industrial process.

First step in identifying opportunities to reduce energy expense.

Also called as Energy labeling , Energy assessment , Energy survey etc..

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INTRODUCTION

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It is a systematic procedure that

Obtain adequate knowledge of the existing energy consumption profile of the site.

Identifies the factors that affect on energy consumption.

Identifies and scales the cost effective energy saving opportunities.

Subject of energy consumption analysis are energy resources consumed in the industrial area, i.e.:

Electricity

Coal

Wood, wood chips/pellets

Fuels (petrol, oil, propane-butane).....

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AREAS OF ENERGY AUDITING

• Gas or other energy resources consuming appliances (furnaces, cleaning devices,..)

• Indoor/outdoor lighting,

• Measurement and regulation.

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1. ADMINISTRATOR

2. OPERATING AGENT

3. ENERGY AUDITOR

4. AUDIT CLIENT

KEY PLAYERS OF AN ENERGY AUDITING PROGRAMME

Responsible body for the energy auditing program.

Sets the energy audit program goals.

Promotes energy audit on a general level.

Provide subsidies for energy audit.

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ADMINISTRATOR

Assists the administrator in developing auditing program.

Responsible for actual running of audit program.

Undertakes the monitoring of results , provides advice and training to auditors.

Take care of information transfer between administrator and auditor.

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OPERATING AGENT

The technical expert undertaking actual auditing work in the clients.

Responsible for actual marketing of energy audits to clients.

Receives finance from the administrator.

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ENERGY AUDITOR

The body , person or organization responsible for the energy cost of a building or an industrial site.

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AUDIT CLIENT

1. Start up meeting

2. Collecting the basic data

3. Field work

4. Analysis of collected and measured data

5. Reporting

6. Implementing some or all of the proposed energy saving measures02/05/2023 12

AUDIT PROCEDURE

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ENERGY CONSUMPTION FOR LAST YEAR

jan feb mar apr may june july aug sep oct nov dec0

100020003000400050006000

3955436845534107

344534223636377137734338

51305162

Consumption

Months

KWh


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Above graph shows the previous year (2015) electricity bills.

We have supply voltage of 11KV.

We have average power factor of 0.92.

Energy charges as follows:- Normal time: 6.20 per KWh₹ Peak time : 9.30 per KWh₹ Off peak : 4.650 per KWh₹


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LOAD CALCULATION

We have 7 blocks in college Main block Seminar hall block Canteen Fluid lab Civil workshop Mechanical lab Pump room


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IN MAIN BLOCK WE HAVE....

Total no of lights: 440 each 36 watt Total no of fans: 406 each 60 watt Projectors: 3 each 165watt Computers : 25 each 150 watt Exhaust : 8 each 50watt Photostat machine: 2 each 900 watt Power sockets : 21 max output 1k watt Ordinary sockets : 242 max output 100 watt Water cooler: 2 each 200watt


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Total lamp load= 440*36= 15.84kw Total fan load= 406*60= 24.36kw Projector load= 3*165= .495kw Total computer load= 150*25= 3.75kw Total exhaust fan load= 8*50= .24kw Photostat machine= 2*900= 1.8kw Water cooler= 2*200= .4kw

Total load= 46.88KW


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IN SEMINAR HALL BLOCK

Total no of lights: 183 each 36w Total no fans: 145 each 60w Total no of sockets: 106 max 100w Total no of computers: 150 each 150w Projectors: 1 each 165w

Total Load = 48.53kw


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Machine load in EEE lab= 64.4kw Machine load in Mech lab= 23.57kw

Total load = 136.7kw


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IN CANTEEN

Total no of lights: 25 each 36w Total no of fans: 20 each 60w Refrigerator: 2 each approx 400w Freezer 1 having 900w Grinder: 1 having 1.5KW Mixer: 1 having .75KW

TOTAL LOAD= 6.05KW


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IN CIVIL LAB.....

No of tube lights: 48 each 36watt No of fans: 14 each 60 watt No of sockets: 23 each max 100watt No of power sockets: 6 each max 1kw Total Machine Load: 8.541KW

Total load= 19.4089KW


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IN FLUID LAB....

No of tube lights: 35 each 36watt No of fans: 11 each 60watts No of sockets: 31 each max100watts Total Machine load: 14.73KW

Total Load= 19.75KW


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IN MECHANICAL LAB...

No of tube lights: 19 each 36watts No of fans: 6 each 60watts No of sockets: 15 each max 100watts Total Machine load: 5.34KW

TOTAL LOAD= 7.88KW


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Water pump having 5HP ie 3.7285KW From the observations and calculations total

connected load to our campus is 240.39 KilloWatts


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SANKEY DIAGRAM


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SELECTION OF BETTER LAMPS..!!!

In the campus we are using fluorescent lamps.

As compared with LED the lux produced by fluorescent lamp are low for same wattage.

Exterior lightining of our campus is done using LED’s.


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https://www.bijlibachao.com/wp-content/uploads/2013/11/Infographic12.jpg?0bc077



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ADVANTAGES OF LED OVER FLUORESCENT TUBES



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CALCULATIONS OF ILLUMINATIONTo find number of light fittings ‘N’

N= (E*A)/(O*CU*MF)

N: No of fittings neededE: Required illumination (lux)A: Working Area (sq.m)O: Luminous flux produced per lamp( lumens)CU: Coefficient of utilizationMF: Maintenance factor



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LUX CALCULATION

When lights are off and windows are closed

20 lux

When all lights are on and windows are closed

90 lux

When all lights are on and windows are opened

150 lux

When all lights are off and windows are opened

95 lux


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Considering the seminar hall...

We have 16 tube lights each having 42 watts and lumens of 2100

Length of seminar hall= 26.6mBreadth of seminar hall= 9.8 mArea of seminar hall= 260.68 sq.m

a) In case of Tube lightsCost : 400per tube set & 50Rs per tubeWattage : 42 wattlumens : 2100life : 4000hrs

So, required no:of tube lights (42W) for seminar hall having area of 260.68sq.m isN=(150*260.68)/(2100*0.6*1)= 30 nos.



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Required no of fittings: 30nos

Seminar hall is used 100hrs/month. (5Hrs per day for 5 days in a weak)

Total light load =1260Watts.

Consumption = 126KWhr per month.

For 126KWhr monthly charge will be Rs 780.

Annual Energy consumption charge= Rs 9360

For every 3 year we have to replace two tube lights from an tube set.So we have to replace 60 tubes in an year from seminar hall.Replacement cost = Rs 3000 per year



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Annual energy cost = Rs 9360Installation cost = 30*400= Rs 12000

So for 1st year Approx Rs 21,360 while using tube lights.

For 2nd year Rs 21360 + Rs9360= Rs 30720 (ie previous year cost plus this year consumption)

For 3rd year Rs 30720 + Rs 9360 + Rs 3000 = Rs 43080 (i.e. Previous year cost + this year consumption + replacement cost)



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b) In case of LED

Cost : 500Rs per led bulbWattage : 18wattlumens : 1800life : 50000hrs

So, required no:of led bulb (18W) for seminar hall having area of 260.68sq.m is

N=(150*260.68)/(1800*0.6*1)= 34 nos.

Required no of fittings: 34nos

Seminar hall is used 100hrs/month.



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Total light load = 612Watts

Consumption = 61.2KWhr per month

For 126KWh monthly charge will be Rs 380aprox

Annual consumption charge will be Rs 4560

Installation cost 500*34= Rs 17000

So for 1st year approx Rs 21,560

For 2nd year total cost = Rs 21560+Rs 4560 = Rs 26120



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For 3rd year total cost = Rs 26120 + Rs 4560 = Rs 30680 As compared with tube lights initial cost of LED’s are high.

So by replacing tube lights with LED’s we can reduce the monthly consumption by half.

We will get back profits after 1 years & 1 months.



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REPLACING WHOLE TUBES BY LED’STotal no:of tube lights: 750

Wattage: 42wattCost : 400 per Tube set & 50 per an tube lightLife : 4000hrs (approx 6 months)Usage : 5hrs a day for 5 days in a week

ie 100hrs per month

Total load : 42*750= 31.5KW

Monthly consumption= 3150KWh



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SEB’s charge Rs 6.20 (approx) per Unit

Charge for monthly consumption is Rs 19530

For an year, energy charges will be Rs 234360

For every 3 year we have to replace all tube lights by once because of its life so its cost Rs 37500

Energy charge + Replacement cost = Rs 233760 per year

Installation cost = 750*400= Rs 300,000

So for 1st year total cost = Rs 534360

So for 2nd year total cost= Rs 534360 + Rs 234360 = Rs 768720 ( ie previous year cost plus this years cost)



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Using LED’sWattage : 18wattsCost : 500per led bulbLife : 50000hrs Usage : 100hrs per month

Total load : 18*750= 13.5KW

Monthly consumption= 1350KWh

Charge for monthly consumption= Rs 8370(6.20per unit)

Annual consumption charge= Rs 100440



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Installation cost : 500*750= Rs 375,000

For the 1st year total cost= Rs 475,440

For 2nd year total cost= Rs 475,440 + Rs 100440 = Rs 575,880

So by replacing tube lights with LED’s we can reduce the monthly consumption by half.

We will get back profits after approx 8 months.



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ENERGY SAVINGS BY AUTOMATIC CONTROLLED FANS

1) Energy savings by controlled fans and conventionally controlled fans= 60w x operating hours (per day per fan) =60W x 10 hours/day =600Wh/day/fan

Energy cost per day per fan =0.6 x 6.20= Rs3.72/- Cost of Energy consumption for 602 fans per day

=Rs3.72x602 = Rs22400/-Annual Cost of Energy consumption by regulator controlled fans

=Rs22400x288days =Rs 645120 -----(1)



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2) Automatic Controlled Fans Can Be Operating Based On User Requirement May Reduce The Operating Time

Let us operate the fans on need basis as remote control is available, it will reduce the operating hours (Assume that the wattage is same) =60W x 7 hr

= 420 Wh/day =0.42 kWh/day = 0.42 x 6.20 = RS 2.6/day

Annual Energy consumption by automatic controlled fan = 2.6x602x288= Rs. 450777-------------(2)



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Cost saving (1)-(2) = Rs 194343

Total cost of additional unit automatic operating switch = Rs.400 x602

=Rs240800

Payback period = 240800/194343=Approx 1 year 2 months



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ENERGY SAVING IN XEROX MACHINEWhich Consume The Energy As Follows Power Consumption of Xerox M/C in Non Operating

Mode = 1x100W =100W (sleep mode)

Energy Saving for Approximate Sleepy Mode Hours For 2hours In A Day =100W x 2hr/day

= 200Wh/Day



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Energy in kWh = 200/1000 = 0.2kWh/day

Energy for a Month = 0.2kWh x 24days =4.8kWh

Units Monthly Energy Cost =4.8x 6.20 =Rs29.76/-

Annual Energy Cost Saving = Rs29.76x12 =Rs 357.12/-

So avoid sleep mode on xerox machines to save Rs 357.12/- per year



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In the campus we are using fluorescent lamps.

As compared with LED the lux produced by fluorescent lamp are low for same wattage.

Exterior lightining of our campus is done using LED’s which is fine.



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POWER FACTOR CALCULATION

We have calculated the pf of pump set with the help of an pf meter.

A 3phase 400V delta connected motor having .79 lag power factor.



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DISADVANTAGES OF LOW PF Large Line Losses (Copper Losses)

Large kVA rating and Size of Electrical Equipments

Greater Conductor Size and Cost

Poor Voltage Regulation and Large Voltage Drop

Low Efficiency

Penalty from Electric Power Supply Company on Low Power factor



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METHODS FOR POWER FACTOR IMPROVEMENT

The following devices and equipment are used for Power Factor Improvement.

Static CapacitorSynchronous CondenserPhase Advancer

http://electricaltechnology.org/?p=117



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1. Static Capacitor Most of the power system loads are inductive that take

lagging current which decrease the system power factor. For Power factor improvement purpose, Static capacitors are connected in parallel with those devices which work on low power factor.

These static capacitors provides leading current which neutralize (totally or approximately) the lagging inductive component of load current (i.e. leading component neutralize or eliminate the lagging component of load current) thus power factor of the load circuit is improved



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2. Synchronous Condenser When a Synchronous motor operates at No-Load and

over-exited then it’s called a synchronous Condenser. Whenever a Synchronous motor is over-exited then it provides leading current and works like a capacitor.

When a synchronous condenser is connected across supply voltage (in parallel) then it draws leading current and partially eliminates the re-active component and this way, power factor is improved. Generally, synchronous condenser is used to improve the power factor in large industries.



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3. Phase Advancer Phase advancer is a simple AC exciter which is connected on

the main shaft of the motor and operates with the motor’s rotor circuit for power factor improvement. Phase advancer is used to improve the power factor of induction motor in industries.

As the stator windings of induction motor takes lagging current 90° out of phase with Voltage, therefore the power factor of induction motor is low. If the exciting ampere-turns are excited by external AC source, then there would be no effect of exciting current on stator windings. Therefore the power factor of induction motor will be improved. This process is done by Phase advancer



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ENERGY STAR


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ENERGY STAR products are independently certified to save energy without sacrificing features or functionality. Saving energy helps prevent climate change. Look for the ENERGY STAR label to save money on your energy bills and help protect our environment.

The ENERGY STAR label has undergone a process of inspections, testing, and verification to meet strict requirements set by the BEE. ENERGY STAR certified homes use 15-30% less energy than typical new homes while delivering better comfort, quality, and durability.

Improve financial performance and reduce carbon emissions with EPA's strategic energy management approach.



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STAR RATING OF BUILDING

The Star Rating Program for buildings is based on actual performance of the building in terms of specific energy usage (kWh/sq.m/year).

In our campus yearly consumption is approx 49660KWh, having a build up area of 15000sq.m.

EPI= 49660KWh/15000sq.m/1year = 3.31.


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As per ECBC standards shown in the table given above our campus is labelled as 5 star building.

EPI (KWh/sq.m/year) Star Label

80-70 1 Star

70-60 2 Star

60-50 3 Star

50-40 4 Star

Below 40 5 Star


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ENERGY CONSERVATIONSElectricity Distribution System

Optimise the tariff structure with utility supplier Schedule your operations to maintain a high load factor Shift loads to off-peak times if possible. Minimise maximum demand by tripping loads through a demand

controller Use standby electric generation equipment for on-peak high load

periods. Correct power factor to at least 0.90 under rated load conditions.


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Consider on-site electric generation or cogeneration.

Export power to grid if you have any surplus in

your captive generation.

Check utility electric meter with your own meter. Shut off unnecessary computers, printers, and

copiers at night.


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Motors Properly size to the load for optimum efficiency.

(High efficiency motors offer of 4-5% higher efficiency than standard motors)

Use energy-efficient motors where economical. Use synchronous motors to improve power factor. Check alignment. Provide proper ventilation (For every 10ºC increase

in motor operating temperature over recommended peak, the motor life is estimated to be halved)


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Check for under-voltage and over-voltage conditions.

Balance the three-phase power supply. (An imbalanced voltage can reduce 3 - 5% in motor input power)

Demand efficiency restoration after motor rewinding. (If rewinding is not done properly, the efficiency can be reduced by 5 - 8%).


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Drives Use variable-speed drives for large variable loads. Use high-efficiency gear sets.

Check belt tension regularly.

Use flat belts as alternatives to v-belts. Use synthetic lubricants for large gearboxes. Shut them off when not needed.


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Pumps Operate pumping near best efficiency point.

Modify pumping to minimize throttling. Use booster pumps for small loads requiring higher

pressures. Repair seals and packing to minimize water waste. Balance the system to minimize flows and reduce pump

power requirements.


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RECOMMENDATIONS & SUGGESTIONS

Replacing Fluorescents lamps with Led’s.

Turn off lights and fans when not in use.

Static capacitor banks are recommended to place parallel with the pump to increase the power factor.

Reschedule the time table to reduce the maximum demand.

Good light ventilation and Air ventilation to classrooms may solve the problem of Energy Consumption.


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Walls should be painted with bright coloured paints (enamel) to get more brightness

Outside lightening of the campus should be placed bit more higher.

Use pumps on the off peak time so that we can reduce the consumption cost. If the securities are available. Fill the tank when once pumps are started.


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Shutdown the computers instead of putting sleep mode when not in use.

Turnoff the Photostat machine instead of putting in sleep mode when not in use.

Fans running without capacitor or under rated capacitor will draw more current therefore use of correct rated capacitor will reduce the power consumption.


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Recommended to use solar water cooler in place of conventional one

All major equipments should run with good power factor and the integration of Instrument to read the P. F online should be made mandatory. Therefore immediate care can be taken to improve the power factor.


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CONCLUSION The Proposed project gives strong warning to the consumer not

only in terms of the energy bills also the energy crisis in the near future to all sectors of people.

This project’s recommendations reduces the around 15-20% of the energy and 25-30% of cost reduction excluding some issues takes more payback period.

There is a scope of improvement to include the advanced lighting scheme to reduce further 10% of the cost


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As per the ECBC building ratings 5 Star labeled buildings are most efficient and we found that our campus can labelled as 5 star as per ECBC criterion.


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THANK U........

Engineering

Energy Auditing Final Phase