Paper Plant Energy Audit Report

ANSH ENERGY SOLUTIONS PVT. LTD. [email protected]

Task Specific Audit: Steam Generating

Boilers & Electrical distribution netwrok

Client: Sandeep Paper Mills (P) Ltd.

Auditor: Anshul Singh Yadav

Auditor Registration: EA-3267

Page | 0

TASK SPECIFIC ENERGY AUDIT

OBJECT: EFFICIENCY TESTING OF BOILERS & LOSS IDENTIFICATION

IN ELECTRICAL DISTRIBUTION SYSTEM

CLIENT: M/S SANDEEP PAPER MILLS (P) LTD., NOIDA

AUDITORS: ANSH ENERGY SOLUTIONS PVT. LTD., Gayatri Dham, Lower Bazar,

Modinagar – 201204 (UP)

AUDITOR BEE REGISTRATION: EA – 3267 (Anshul Singh Yadav)

& EA-10465 (Anubhav Gupta)

DURATION: 21 JANUARY, 2010 – 20 FEBRUARY, 2010

REPORT NUMBER: 09-10/R/AG/104







pg. 1

Contents

S.No. Description Page No.

1. Executive Summary 2

2. Overview 4

3. Scope of Audit 4

4. Assumptions 5

5. Equipment Verbal Picture 5

6. Energy management Plan 7

7. Maintenance recommendations 8

8. Technical Supplement 8

9. Standard Calculations and Assumptions 17

10. SWOT 19

11. Thermal references 20

12. Appendix – 1 21

13. Appendix – 2 23

14. Electrical bill analysis 2008 24

15. Electrical Bill Analysis 2009 25

16. Electrical Parameters 32

17. Harmonics 41

18. Motor Power Factor 47

19. Scope of Energy Savings in electrical motors 56

20. Comparison of Low efficiency and EED 60

21. Conversion of LV Drives to MV drives 74







pg. 2

EXECUTIVE SUMMARY

The boiler in this report is referred as Boiler No. 1 (IBL). The efficiency was calculated as

below:

Boiler 1 Boiler 2

Average Direct Efficiency 80.20 % NA

The operation has been observed carefully and few areas of carelessness have been

observed. The best achievable efficiencies of 85% + have a path of nearly 5% to be

travelled and continuous monitoring and efforts are required.

The suggested methodology of improvement is diagrammatically represented as below:







pg. 3

The summary of possible savings is tabulated as below:

S.No. Action Tentative Savings Pay Back

Boiler action plan

1. Installation of fixed type oxygen analyzer INR 416114 per month

i.e. INR 49,93,368 Per

annum

19 Days

2. Installation of Automatic Blow Down

Control System

INR 52014 per month

i.e. INR 6,24,168 per

annum

5 months

3. Replacement of fire doors and better

mud coating

INR 1,83,089 per month

i.e. INR 21,97,068 per

annum

7 days

Total Boiler savings INR 78,14,604 per

annum

Electrical action plan

1. Installation of only 90 KVAR capacitors on

transformer LT

INR 4017 per annum Immediate

2. Conversion of V-Belt to Flat belt drives INR 8,39,446 per annum 6 months

3. Conversion of all old motors to new

Energy efficient series motors

INR 29,32,356 per

annum

3 months

4. Conversion of LV panels to MV panels INR 9,67,887 NOT suggested

Total Savings in Electrical system INR 37,75,819 per

annum

Total savings by complete

implementation

INR 1,15,90,423







pg. 4

OVERVIEW

The client M/s Sandeep Paper Mills (P) Ltd. with registered office at A-20/6, Noida is

operating for production of various grades of paper for which utilities used are Steam,

Water, Compressed air and Electricity (Grid power). The top management of company

includes Mr. Jatinder Singh Jolly and Mr. Sandeep Arora (ED) and they are committed to

cost reduction and energy conservation for betterment of company as well as

environment. The Audit has been done specifically for the steam generating unit and

Electricity Load Distribution. Company possesses 6 TPH capacity of steam generation F &

A 100oC with pressure rating of 10 kg/cm2 and a CD of 1250 KVA at 33,000V distributed

through a 1250 KVA 33KV/433V single Core outlet transformer. The target is to achieve

a reduction of recurring cost to improve business viability.

Scope of Audit

The aim and scope of audit is to quantify and justify the fuel consumption as well as

energy consumption and consumption quality, total and specific steam generation,

boiler efficiency monitoring, load balancing, run-ability optimization and achieving best

possible fuel to steam ratio. The audit will thus cover parameter detection of:

1. Feed water inlet flow.

2. Blow Down flow estimation (If possible).

3. Inlet air temperature.

4. Temperature of exhaust to stack.

5. Feed water quality.

6. Cycle of concentration

7. Variance in phase loading of motors ACB’s and Transformer

8. Heat loss in cabling

9. Heat loss due to poor operation of motors

10. Losses due to poor capacitor behavior or installation faults.

11. Losses due to uneven load pattern.

12. Losses due to voltage variations.







pg. 5

The completion of audit will achieve identification of all types of boiler losses and

almost all possible ECOs (Energy Conservation Opportunities). It will highlight the

efficiency improvement possibilities in motors, capacitors and voltage variations.

This inspection report reflects the conditions of the equipment at the time of the inspection only.

Please note that equipment conditions change with time and use and the conditions noted in this

report may change in appearance and severity as time progresses or with mishandling. Hidden or

concealed defects cannot be included in this report. An earnest effort was made on our behalf to

discover all fallacies; however in the event of an oversight no liability is acceptable. No warranty is

either expressed or implied. This report is not an insurance policy, nor a warranty service.

Assumptions (if Any)

The following assumptions were made taking in consideration to have minimal effect on

the results:

1. Nil

Equipments’ Verbal Picture

The Plant is operating two separate boilers one of which operates continuously using

petcoke as fuel, the second boiler is standby and is capable of using multiple fuels. The

feed water supply quality is common and the water treatment facilities include RO

plant. The condensate from main machine hall is pumped and mixed into the feed water

tank which stands common for both the boilers. The description of boilers is as below:

1. Boiler no. 1 is an IBL make boiler with 6 TPH Evaporation capacity F & A 100oC

and is equipped with high temperature fluidized bed furnace beneath it. Boiler

number 1 is a smoke tube boiler. The boiler is operating at loading pressure of 10

kg/cm2 with both forced and induced drafts involved. The furnace is divided into

two separate chambers to ensure maximum burning of high CV fuel. The fuel is

fed into the primary chamber and the ash carryover is collected in secondary

chamber where next feed of air makes it to burn further due to high CV content.

Both furnace chambers on a 40 HP FD fan which is common, and have side fire

doors for cleaning. The blow down arrangement is manual and is being operated

once for 50 seconds in every 12 hours.







pg. 6

2. Boiler no.2 is similar to above boiler in make and fire side arrangement however

it carries a Brownian furnace and is capable of burning baggasse as fuel.

The electrical circuit consists of an incoming state electricity board line of 33KV which is

received by a 1250 KVA 33/0.433KV single transformer. The transformer is installed with

OLTC AVR and is capable of handling voltage range of +7.5% to – 12.5% of the rated

incomer. The State owned meter and VCB precedes transformer. Multiple single core

cables carry energy from transformer to the MCC which is installed with ACB through

which 2 Separate PCC are connected.

The plant operates mostly inductive load along with some inverting electronic load to

govern the use of inductive load. The premises has been observed to be maintaining

good power factor and good voltage conditions however ACBs installation seems to be

very negligent.

A wide range of VFDs have been installed to compensate for major electrical energy

losses and more or less the aim has been successfully achieved.







pg. 7

Energy Management Plan







pg. 8

Maintenance Recommendations

The following maintenance activities need constant attention:

1. The furnace fire doors need to be proper and closed during operation to avoid

major heat loss at temperatures nearly 900oC. The ambient temperature at 2

mtrs of distance from furnace fire doors was found to be 74 degree celcius which

is dramatically high. WE RECOMMEND USE OF CI DOORS WITH INTERNAL LINING

OF FIRE PROOF CEMENT.

2. Record should be kept, monitored and counter checked for APH cleaning in Boiler

1 as it was observed to be carrying fly ash in itself despite of rotary dust remover

at bottom.

3. The feed water quality needs to be monitored very closely as this may lead to

corrosion and slow damage.

The Technical Supplement

1. Calculation of Efficiency for boiler by direct method.

The data collected shows:

S. No. Feed Water

Opening

Reading

Feed Water

Closing

Reading

Fuel Weight

(kg)

Fuel

Moisture

Content (%)

Steam

Produced

(kg)

1. 224.015 m3 336.023 m3 9096 6 112008

2. 230.683 240.382 m3 835 6 9699

Calculation of actual fuel available on air dry basis (M) =

1 !"# $"%&'"(

Thus, )*%#+#" ,-".&/ ), 0 123 4!"#

And, ,-".&/ 55%-"( %- 5"6 0 5"6 ,-5'#7/ 5 10 +. 49 :"67.

Finally, <%#". " %=%"-=/ 0 >? @A B@>







pg. 9

Hence, For Boiler Number 1

S.No. Fuel Calorific

value (kcal/kg)

Available

energy (kcal)

Energy attained

in steam at

10kg/cm2

Overall

Efficiency

1 9340 79859241 65860704 82.47%

2 9340 7330966 5703012 77.79%

2. Various types of losses:

With help of the Ultimate analysis of fuel we can followed the below mentioned

procedure:







pg. 10







pg. 11







pg. 12







pg. 13

Ultimate analysis of fuels:

Content Petcoke as fuel

Carbon % 87.17

Hydrogen % 4.67

Nitrogen % 1.96

Oxygen % Remainder

Ash % 0.64

Volatiles % 1.83

GCV (Kcal/KG) 9340

a) Dry Flue Gas Loss

Boiler 1 Boiler 2

11.24% NA

b) Loss Due to Hydrogen in fuel

Boiler 1 Boiler 2

3.02 % NA

c) Loss due to moisture in fuel

Boiler 1 Boiler 2

0.40 % NA

d) Loss due to moisture in air

Boiler 1 Boiler 2

3.22 % NA

e) Approx. heat loss due to surface radiation and leakages

Boiler 1 Boiler 2

2.76 % NA







pg. 14

f) Heat loss due to unburnt in fly ash

Boiler 1 Boiler 2

0.13 % NA

g) Heat loss due to un burnt in bottom ash

Boiler 1 Boiler 2

0.01 % NA

h) Heat loss due to blow down.

Boiler 1 Boiler 2

1.02 % NA

3. Calculations of tentative and stepwise investment:

The major losses of boilers are clearly evident to be dry flue gas loss, blow down

losses. The suggested tasks to be taken care of shall be:

a) To counteract dry flue gas loss we shall control the excess air (Presently

98.113%) supplied to the furnace by adjusting fan speed balances and by

continuous monitoring of the excess oxygen going into the flue gases. An

estimated 1% reduction in excess oxygen shall bring about 0.9% reduction in

loss. Stack monitoring can be done by either continuous type oxygen scanner

or by handheld type portable oxygen probes.

b) Hydrogen loss is inherent with fuel and is dependent only on fuel type. The

only variation in our control shall be allowance of residence floor time to fuel

so that hydrogen can be slowly converted into water vapors. The process is

slow and may cause not more than 00.25 percent improvement.

c) Again loss due to moisture in air is beyond manual control as it is purely based

on atmospheric conditions. The only precaution that can be taken is to avoid

any spray or source of water on suction side of FD fan.







pg. 15

d) Loss due to furnace leakage can be brought down by nearly 1 % in this case as

the furnace doors and feed hoppers allow lot of heat to escape. This is due to

carelessness of man power involved and due to absence of any provision of

keeping the furnace pressure negative. The flames have been evidently

coming out due to furnace high pressure. The best solution to this problem

shall be to control fan speeds by variable speed drives for which controls are

either with wise operators or alternately a pressure transmitter controls the

flow of FD air.

e) Lastly the blow down losses has been observed to be very high. The water test

report clearly indicates the malpractice being followed. The blow down

should only be made when TDS of drum water reaches 3500ppm. The present

values show operation at very low cycle of concentrations which lead to heat

loss and more significantly costly water loss. The drum pH is also low. The

ideal pH shall be 11. The solution to above mentioned problem is either

vigilant checking of water parameters or installation of an automatic blow

down control system.

f) The list of improvements suggested is not exhaustive and can be revisited.

Further suggested approach is pinch analysis.

4. Approximate costs of equipments and payback periods.

Equipment Tentative Cost Payback period

Oxygen probe

(continuous)

INR 2,70,000/- 19 days

Vigilant Feed water

monitoring

INR ZERO

Automatic blow down

control

INR 2,50,000/- 4.8 months

5. Pay back calculations

The Average Steam used from Boiler No. 1 = 5.333 T/hour







pg. 16

Average operating hours per month = 24 x 26 = 624 hours

Steam produced = 18302 tonnes per month.

Fuel used per tonne of team produced = weight of petcoke/steam produced =

81.2 kg per tonne of steam

Cost of 1 kg of Petcoke = Rs. 7 per kg approx.

Thus cost of 1 tonne of steam = 81.2 x 7 = Rs. 568 (for boiler number 1)

Thus 1% saving in boiler 1 will lead to 568 18302 0 H. 1,04,028being

saved per month.

Now, Oxygen monitoring can reduce excess oxygen by 6% in flue gases. This will

lead to increase in efficiency by 4% in boiler 1.

Thus saving of boiler 1 shall be 4 x Rs. 104028 per month = Rs. 416114

Hence simple payback for Continuous oxygen probe in boiler 1 = KLMNM 30 0

19 (/ and

Alternatively payback for one portable probe shall be = PMNM 30 0 5.76 (/

However the most lucrative option shall be a continuous monitoring probe for

boiler 1 as it will also iron out any possibilities of man power negligence.

The saving on blow down can be nearly 0.5%.

Hence cost saving in Boiler 1 = 0.5 x Rs. 104028 = Rs. 52014 per month

Hence simple payback after installing continuous blow down for Boiler 1

0 KKM 0 4.8 6-5'

The payback is although lucrative but we do not recommend installation of

automatic blow down as this result can be very well achieved by vigilant blow

down water quality monitoring, manually.

There is as strong need to bring down the radiation and surface losses to under 1

% i.e. savings of 1.76 % or saying monetarily = 1.76 R 104028 0 ST. UVWXVY







pg. 17

The tentative investment in fire doors and better mud coating and insulation all

together is estimated to be Rs. 40,000.

Hence a payback period of MPZP[ R 30 (/ 0 6.55 is attainable which is

remarkable.

We strongly recommend immediate change of fire doors and better coating of

boiler and water wall body.

Standard Calculations and Assumptions

1. The calculations along with formulae are mentioned above.

2. Calculations for payback have been done taking in consideration simple payback

only as the payback is extremely fast and IRR calculations are not significant.

3. It is assumed that the equipment suggested shall be allowed to work properly

with adequate staff training or else the payback calculations may vary.

Audit Recommendations

The point wise recommendations are listed below:

1. Installation of Continuous oxygen monitoring on stack of boiler no. 1

2. Adequate staff training and cross checking so as to control blow down frequency

and water’s operating parameters.

3. For maintaining feed water pH we shall suggest installation of a water softener

with capacity of 2 m3/hr. this softener shall feed the RO plant totally and 0.3

m3/hr shall be bypassed directly from RO plant i.e. 0.3 m

3/hr shall flow directly

from softener to feed water tank. Alternatively pH booster chemical should be

used as per demand to maintain the pH.

4. The doors of furnace need to be changed to CI casted fire grade doors with cavity

on back so as to be filled with casteable cement for reduction in heat loss.

Appendix

1. Lab Testing Reports.

2. Data collection tables.







pg. 18

3. SWOT Analysis.

Other possible utilities to undergo audits

The possible utilities to undergo audits can be pumping system.







pg. 19

SWOT Analysis

Situation being analyzed: ____Boiler Efficiency Improvement_________

STRENGTHS WEAKNESSES

1. Sufficient Capacity installed. 2. Availability of down time for

installation of additional equipments as interconnecting pipeline is already installed.

3. Willingness of management for improving existing set-up

1. Unavailability of process instrumentation.

2. Negligence of staff to monitor actual parameters.

OPPORTUNITIES THREATS

1. Vast scope of improvement of efficiencies.

2. Reduction of loss can be achieved by using instrumentation and monitoring.

3. Monitoring instruments are readily available in market so no long lead times.

1. Instrumentation if installed might not work properly due to lack of monitoring or knowledge.

2. Biased mindset regarding possibility of improvements.

3. Lack of equipment maintenance.







pg. 20

References

1. BEE Manual on Energy Efficiency testing (Book 4)

2. www.emerson.com

3. Perry’s Handbook of Chemical Engineers (2003)

4. www.lenntech.com/boiler-feedwater.htm







pg. 21

Appendix 1

Client: M/s Sandeep Paper Mills Pvt. Ltd., A-20/6, Noida

Tests to be done: Blow Down (TDS, pH, T.Alk, P. Alk.), Feed Water (TDS, pH, T.Alk, P. Alk.) RO Water

(TDS, pH, T.Alk, P. Alk.)

Sample collected by: Dr.Shubhangi Gupta (MD, AESPL)

Sample collection date: 20.01.09

Qualitative Analysis

PURPOSE

The purpose of this experiment is to identify the boiler operating parameters by getting to know the

TDS, Alkalinity and ph of blow down and feed water of boiler system. The results will help in better

water management and thus will let us control the proper operation practices to enhance equipment

life and maintain low fuel consumption.

DATA

STEP SUBSTANCE Parameter (Std.

Value)

RESULT INFERENCE

1

Blow down water

Boiler 1

pH (11-12)

9.20

A lot of fresh

treated water is

being inputted or

excessive blow

down is being

operated.

2

Blow down water

Boiler 1

T. Alk as CaCO3

(750 ppm)

140 ppm

Excessive Blow

Down Operation.

3

Blow down water

Boiler 1

Caustic Alk.

(300 or 30% of

above)

60

Excessive Blow

Down Operation.

4

Blow down water

Boiler 1

Total Dissolved

Solids

(2500 ppm)

1824 ppm

Excessive Blow

Down Operation.







pg. 22

5

Boiler feed water

pH

7.16

No presence of

softener unit.

6

Boiler feed Water

Total Hardness

(Strictly <5ppm)

5.8 ppm

No effect of RO

plant on hardness

or too hard fresh

water

Conclusions

Boiler water monitoring at site is poor and careless. Excessive blow down is being operated and lot of

valuable heat is being wasted. The water treatment for feed water is done by complete Reverse Osmosis

Plant. However this plant is usually not suitable to maintain pH and hardness of feed water. The

observed pH is neutral which could cause mild damage to the boiler tubes and drum. Also the Hardness

of water is higher than the set limits.

Recommendations

Installation of a Soft water treatment plant and that soft water should be inputted to RO system. Only

60% of total feed water should go through RO plant and balance 40% should be taken from softener

itself so as to maintain pH value. The operation of such a system after installation is to be monitored

strictly so that the ratio of 60:40 shall be balanced as per site requirements.







pg. 23

APPENDIX - 2

Client: M/s Sandeep Paper Mills Pvt. Ltd., A-20/6, Noida

Tests to be done: Proximate Analysis of Fuel, Ultimate Analysis of Fuel, GCV

Sample collected by: Dr. Shubhangi Gupta (MD, AESPL)

Sample collection date: 20.01.09

Sample: Petcoke (Source Jamnagar Refinery of Reliance Industries Ltd.)

Qualitative Analysis

PURPOSE

The purpose of this experiment is to identify the Fuel Chemical Composition so the reaction and

consequences inside the furnace can be calculated. The results will help in better Furnace operation,

loss detection and will let us control the proper operation practices to enhance equipment life and

maintain low fuel consumption.

DATA Proximate Analysis

S.NO. SUBSTANCE Parameter RESULT INFERENCE

1

Petcoke

Moisture

6.25

2

Petcoke

Ash

0.64

3

Petcoke

Volatile Matter

1.83

4

Petcoke

Fixed Carbon (By

Difference)

91.28

DATA Ultimate Analysis

S.NO. SUBSTANCE Parameter RESULT INFERENCE

1

Petcoke

Ash

0.64







pg. 24

2

Petcoke

Carbon

87.17

3

Petcoke

Hydrogen

4.67

4

Petcoke

Nitrogen

1.96

5

Petcoke

Other

Remainder

FUEL GROSS CALORIFIC VALUE: 9340kcal/kg







Page | 24

NAME OF THE CONSULTANT:- M/s Ansh Energy Solutions Pvt Ltd.

ANALYSIS OF ELECTRICITY BILLS FOR THE YEAR:- 2008 (FROM JAN-2008 TO DEC- 2008)

( ENERGY AUDIT ANALYSIS SHEET)

( 1 ) NAME OF THE INDUSTRIES :- M/S Sandeep paper Mills Pvt. Ltd. (6 ) SUPPLY VOLTAGE :- 33000 VOLTS

(2 ) TYPE OF INDUSTRIES :- PAPER MILL ( 7 ) TARIFF APPLICABLE :- HV - 2

( 3 ) CONTRACT DEMAND :1000KVA ( 8 ) CONTINIOUS PROCESS INDUSTRY:-

( 4 ) CONNECTION NO. :- LP-304 ( 9 ) PASCHIMANCHAL VIDHYUT VITRAN NIGAM LTD

( 5 ) PLACE:- A-20/6 Noida

SR MONTH CONTRACT MINIMUM ACTUAL DEMAND P. F. ENERGY ENERGY ENERGY TOTAL TOTAL TOTAL NET

NO DEMAND BILLING MAX. CHARGES BILLED CONSUMED CONSUMED CONSUMED ENERGY ENERGY BILL UNIT

DEMAND DEMAND RS. FROM FROM FROM CONSUMED CHARGE AMOUNT IN

BILLED 22.00 - 6.00 06.00 - 17.00 17.00 - 22.00

KVAh RS. RS. RS.

1 JAN 1000 750 925 166500 0.98 150500 191280 90480 432260 1483483.00 1648028.00 3.81

2 FEB 1000 750 987 177660 0.99 116380 144940 66260 327580 1122293.00 1306745.00 3.99

3 MAR 1000 750 980 176400 0.97 0 0 0 0 0.00 0.00 0.00

4 APR 1000 750 1023 184140 0.99 223740 307460 138360 669560 2298250.00 2504593.00 3.74

5 MAY 1000 750 1099 219800 0.99 143720 203840 103520 451080 1527067.00 1803058.00 4.00

6 JUN 1000 750 1099 219800 0.99 193080 243020 107500 543600 1826607.00 2110901.00 3.88

7 JUL 1250 937.5 1126 225200 0.99 199120 268700 122300 590120 1988408.00 2202844.00 3.73

8 AUG 1250 937.5 1080 216000 0.99 171180 223620 98460 493260 1658888.00 1837558.00 3.73

9 SEP 1250 937.5 1074 214800 0.99 183340 332100 157400 672840 2287123.00 2533176.00 3.76

10 OCT 1250 937.5 1072 214400 0.99 121660 173300 80500 375460 1267835.00 1505793.00 4.01

11 NOV 1250 937.5 1106 221200 1.00 153600 199920 87720 441240 1483721.00 1699130.00 3.85

12 DEC 1250 937.5 1064 212800 0.99 211540 278300 132400 622240 2730525.00 2366463.00 3.80

13 TOTAL 2448700 11.87 1867860 2566480 1184900 5619240 19674200.00 21518289 42.31

14 AVG 1060 206573 0.99 169805 233316 107718 510840 1788564 1956208 3.85

SD 35671.92 58208.91 27366.23 117480.125

Mean +1 SD 205477.38 291525.27 135084.41 628320.12

Mean-1sd 134133.53 175107.46 80351.95 393359.88

MINIMUM 925 166500 0.98 116380 144940 66260 327580 1122293 1306745 3.72533

MAXIMUM 1126 225200 1.00 223740 332100 157400 672840 2730525 2533176 4.01

LOAD FACTOR = TOTAL ENERGY CONSUMED DURING YEAR X 100 73.7 74%







pg. 25

Name of Consultant : M/s Ansh Energy Solutions Pvt. Ltd.

ANALYSIS OF ELECTRICIY BILLS FOR YEAR 2009 (Jan 2009 – Dec 2009)

1 Name of industry: Sandeep Paper Mills Pvt. Ltd. 6 Supply Voltage: 33 KV

2 Type of Industry: Paper Mill 7 Connection Tarrif: HV-2

3 Contract Demand: 1250 KVA 8 Process type: Continuous

4 Connection No. LP 304 9 Supplier: Paschimanchal Vidhyut Vitran Nigam Ltd.

5 Address: A-20/6, Noida

SR MONTH CONTRACT MINIMUM ACTUAL DEMAND P. F. ENERGY ENERGY ENERGY TOTAL TOTAL TOTAL NET

NO DEMAND BILLING MAX. CHARGES BILLED CONSUMED CONSUMED CONSUMED ENERGY ENERGY BILL UNIT

DEMAND DEMAND RS. FROM FROM FROM CONSUMED CHARGES AMOUNT IN

BILLED 22.00 - 6.00 06.00 - 17.00 17.00 - 22.00 KVAh RS. RS. RS.

1 JAN 1250 937.5 1059 211840 0.993 180280 245780 114060 540120 1821421.00 2042097 3.78

2 FEB 1250 937.5 1059 211840 0.990 167240 215280 102500 485020 1634304.00 1870344 3.86

3 MAR 1250 937.5 1129 203220 0.987 175140 209800 107520 492460 1682465.00 1881232 3.82

4 APR 1250 937.5 1034 186120 0.986 202700 222160 121240 546100 1860189.00 2024631 3.71

5 MAY 1250 937.5 1072 214400 0.987 185280 233300 110940 529520 1783247.00 1990820 3.76

6 JUN 1250 937.5 1085 217000 0.989 198320 256900 118260 573480 1930875.00 2148093 3.75

7 JUL 1250 937.5 1039 207800 0.990 192140 268960 122040 583140 1966608.00 2166587 3.72

8 AUG 1250 937.5 1018 203600 0.996 151520 283750 124370 559640 1899141.00 2031389 3.63

9 SEP 1250 937.5 1092 218400 0.996 80760 268920 121980 471660 1621123.00 1866308 3.96

10 OCT 1250 937.5 1070 214000 0.994 77860 217380 119220 414460 1531041.00 1767243 4.26

11 NOV 1250 937.5 1044 208800 0.995 182340 243900 114280 540520 1822354.00 2079598 3.85

12 DEC 1250 937.5 1056 211200 0.996 198920 274880 122000 595800 2007256.00 2167039 3.64

13 TOTAL 2508220 11.90 1992500 2941010 1398410 6331920 21560024 24035381 45.7

14 AVG 1063 209018.33 0.99 166041.67 245084.17 116534.17 527660.00 1796668.67 2002948 3.81

SD 42990.88 25633.26 6770.89 52669.52

MEAN+1SD 209032.55 270717.43 123305.06 580329.52

MEAN-1SD 123050.79 219450.90 109763.28 474990.48

MINIMUM 1018 186120 0.99 77860 209800 102500 414460 1531041 1767243 3.63

MAXIMUM 1129 218400 1.00 202700 283750 124370 595800 2007256 2167039 4.26







pg. 26

925

987 9801023

1099 10991126

1080 1074 10721106

1064

0

200

400

600

800

1000

1200

JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC

MONTH vs MAX. DEMAND - GRAPH- YEAR 2008







pg. 27

0.98

0.99

0.97

0.99

0.99 0.99 0.99 0.99 0.99 0.99

1.00

0.99

0.955

0.96

0.965

0.97

0.975

0.98

0.985

0.99

0.995

1


MONTH vs POWER FACTOR - GRAPH - 2008

P. F. …







pg. 28

3.81

3.99

3.74

4.00

3.88

3.73 3.73 3.76

4.01

3.853.80

2.00

2.50

3.00

3.50

4.00

4.50


MONTH vs UNIT RATE 2008







pg. 29

1059 1059

1129

1034

1072

1085

1039

1018

1092

1070

1044

1056

960

980

1000

1020

1040

1060

1080

1100

1120

1140


MONTH vs MAX. DEMAND - GRAPH- YEAR 2009







pg. 30

0.993

0.990

0.9870.986

0.987

0.9890.990

0.996 0.996

0.9940.995

0.996

0.975

0.98

0.985

0.99

0.995

1


MONTH vs POWER FACTOR - GRAPH - 2009

P. F. …







pg. 31

3.783.86 3.82

3.713.76 3.75 3.72

3.63

3.96

4.26

3.85

3.64

2.00

2.50

3.00

3.50

4.00

4.50


MONTH vs UNIT RATE







Page | 32

Measurement of Electrical Parameters of Transformer no. 1 (1250 KVA) L. T. side measurement:

Voltage - R - Y = 411 V Current - R phase = 917 Amp

Y - B = 417 V Y phase = 945 Amp

B - R = 412 V B phase = 909 Amp

Average phase to phase Voltage = 413 V Average phase Current = 924 A

Average Measured Power Factor:- 0.996

Power output on secondary side

\ 0 1.73 R 413 R 924 R 0.9961000 0 658.31 ]3) 0 660.95 ]9

Calculation of % loading with respect to full load capacity of transformer:

Rated full load of transformer in KVA. – 1250 KVA

∴ % Loading on Transformer

0 NP.Z ^ K 0 52.66%







pg. 33

Performance Evaluation of Transformer No. 1 - 1250 KVA:

For a Transformer,

Losses at present loading 0 ` a( a" b c dAeA dA fK R a( a"

Where,

No load losses = Power consumed to sustain the magnetic effect in the

transformer’s steel core. Core loss occurs whenever the transformer is energized;

core losses do not vary with the load. Core losses are caused by two factors:

hysteresis and eddy current losses.

Load losses = these losses are associated with full load current in the transformer

windings. Copper loss is power lost in the primary and secondary windings of a

transformer due to the ohmic resistance of the windings and this loss varies with

square of the current.

For the transformer of 1250 KVA

No load losses are - 2000 W

Load losses are - 11880 W

Losses at loading of 658.31 KVA 0 2000 b cNP.ZK fK R 11880 0 5295 9

% ghhijikljm 0 nopqop r UXXnopqop b stTTkT

0 ukvTowkT nopqop r UXXX r UXXukvTowkx nopqop r UXXX b ys b ss r z

Where,

NL = No load Losses in Watts

LL = Load losses in watts at full load at 75 C

TF = Temperature correction factor

PF = Load power factor

The basic D.C. resistance copper losses are assumed to be 90% of the load losses.

Eddy current losses (in conductors) are assumed to be 10% of the load losses.

Basic I2R losses increase with temperature, while eddy losses decreases with

increase in temperature. Thus, 90% of the load losses vary directly with rise in

temperature and 10% of the load losses vary inversely with temperature.







pg. 34

Calculations are done for an assumed temperature rise, and the rise in

temperature is dependent on the total losses to be dissipated.

Operating temperature = Ambient temperature + Temperature rise.

To estimate the variation in resistance with temperature, which in turn depends

on the loading of the transformer, the following relationship is used.

Rt – op = F + T amb + T rise

Rt – ref F + T ref

Where F = 234.5 for Copper

= 225 for Aluminium

= 227 for Alloyed Aluminium

Rt. op = Resistance at operating temperature

T ref = Standard reference temperature, 75 C

Temperature correction factor, T

= Load losses at operating temperature

Load losses at reference temperature

= 0.9 X (Rt – op) + 0.1 X (Rt – ref)

(Rt – ref) (Rt – op)

Here, Rt – op = 234.5 + 31.5 + 53 = 1.024

Rt – ref 234.5 + 75

So, T = 0.9 x 24 + 0.1 x1.024 = 1.024

% Efficiency * = 658 x 1000 x 100

(658 X 1000) + 2000 + [(658/1250) 2 x 11880 x 1.024]

= 658000 x 100

663370.91

= 99.19% say 99 %







pg. 35

The efficiency of transformer no – 1 at the 52 % load is 99%, which is observed to

be very good.

-: Efficiency Curve of A Transformer :-







pg. 36

System unbalance checking with respect to voltage and current

Transformer no – 1 - L. T. side measurement:-

L. T. side measurement:-

Voltage:- R to Y = 422 V Current:- R phase = 1024 Amp

Y to B = 436 V Y phase = 1098 Amp

B to R = 434 V B phase = 1001 amp

Average phase to phase Voltage = 430.66 V, Average phase Current = 1041 A

Measured Average Power Factor = 0.991

Calculation of Voltage unbalance:-

% Voltage unbalance = Maximum voltage on any phase – Average voltage x 100

Average voltage

= 436 – 430.66 x 100 = 1.24 %

430.66

% Voltage unbalance is observed to be 1.24% which is more than 1 % and hence

it is not good.

One percentage System unbalance voltage wills increases System losses by five

percent so you are advised to keep voltage balance in each Phase for Energy

conservation.

We have observed that you already have OLTC voltage control panel but it is

running on Last step of its setting. So we are giving you some other suggestion for

voltage balance in your system.

1. Check all major Electrical Equipment connections and tighten if necessary.

2. Check H.T. Side and Transformer side connection and tighten if necessary.

3. Check all capacitor terminal connection and tighten it if necessary.

4. Try to keep all loads in balance on each phase.







pg. 37

5. Transformer body to earth insulation and Neutral connection checking.

Calculation of Current unbalance:-

% Current unbalance = Maximum Current on any phase – Average Current x 100

Average Current

= 1098 – 1041 x 100 = 5.47 %

1041

Comments:- Since current unbalance is more than 1.5 %, Following suggestions

are given to bring

% Current unbalance will come under 1 %.

Suggestions are made for balancing unbalance currents in the plant:-

a. Lighting loads of the plants are to equally balance on all the three phases.

b. Single-phase motor loads of the plants are to equally balance on all the three

phases

c. All the loose connections in the plants are to be tightened.

d. Request power supply board. to tightened their H.T. line connections for

balancing incoming loads.







pg. 38

Table for Transformer Losses as per Standards: -

KVA Rating No Load Losses in

W

Full Load Copper

losses at 75oC in

W

Efficiency %

100 500 2000 97.50

125 570 2350 97.66

160 670 2840 97.81

200 800 3400 97.90

250 950 4000 98.02

315 1150 4770 98.15

400 1380 5700 98.23

500 1660 6920 98.28

630 1780 8260 98.37

750 1800 9200 98.40

800 1900 9980 98.45

1000 2000 11880 98.54

1500 2100 15500 99.00

2000 2500 22000 99.00

2500 2800 26000 99.00

3200 3240 31050 99.00

Energy saving Opportunities in Transformer:-

Power and distribution transformer which works on the principle of

electromagnetic induction and

∴∴∴∴ These transformers consume reactive power (KVAr) for their own needs for

formation of magnetic flux even when its secondary is not connected to any load

or at very low load.

The power factor will be very low under such a situation which is COSØ = COS 75o

= 0.2588 and this can be confirmed from the below mention vector diagram of no

load transformer.

The low power factor condition of transformer will increase no load losses and

hence for reduction of no load losses and to improve no load power factor a fixed







pg. 39

capacitor or banks of a fixed capacitors are required to be install at the L. T.

(secondary side) side of the transformer.

The table gives the approximate KVAr of capacitor required for improving no load

poor P. F. of the transformer:-

Sr. No KVA rating of Transformer KVAr required for compensation

1 Up to and including 315 KVA 5 % of KVA rating

2 315 KVA – 1000 KVA 6 % of KVA rating

3 1000 KVA – 2000 KVA and

above

7 % of KVA rating

In our case Sr. no 3 will apply.

The KVAr required for compensation is 1250 KVA x 0.07 = 87.5 KVAr say 90 KVAR

The calculation of no load losses and reduction due to improvement of no load

P.F. by installing fixed capacitors of 90 KVAr at the L.T. side of the each

transformer. (COSØ = COS75° = 0.2588 to new COSØ = COS74.5° = 0.2672)

% No load losses reduction in Transformer 0 |1 cA ~ ~f

K R 100

0 |1 c.KPP.KNLKfK R 100

0 1 0.9382 R 100

= 6%

No load vector diagram of transformer: Primary Voltage

Iw

↓ = 75° E1

Iw = Active current component or working or iron loss component

Im = Magnetizing current component

Io = No load Primary input current

= 1.0 % of rated primary current

In our case primary full load current = 21.86 Amp

∴ 1.0 % of primary input current =21.86 x 0.01 = 0.2186 Amp

Im Io







pg. 40

Assuming, transformer works 365 days in a year. The no load losses generated

during the year as under.

6 % of No load losses of the year = 17520 KW H x 0.06 = 1051.20 KWH say 1052

KWH

(Detailed calculation In Final Report)

Cost of 6 % no load losses of the year = 1052 KWh x Rs. 3.81/-

(Average unit rate for year – 2009)

= Rs. 4008.12/- say Rs. 4010/- per year.

The saving per year by installing 90 KVAr capacitor to one Transformer

= Rs. 4010/-

Please note that we will have to reduce the capacitors installed directly on the

transformers as similar calculations are reversed if power factor is leading instead

of lagging.

As the transformers already have a higher installation of capacitors we have to

make no investment. Simple reshuffling of capacitors involving some man hours

will solve the problem and lead us towards better efficiency.







pg. 41

What is Harmonics:-

At the time of the designing any A.C. machine, it is assumed that voltage and

current wave from at the output terminals of A. C. machines is assumed to be

sinusoidal and consists of only one frequency which is called fundamental

frequency or 1st

harmonics and such sinusoidal wave from dose not contain

harmonics of other frequency.

Due to non linear system load such as thyristorised control, variable frequency

drive and D. C. motor, a harmonics are generated at the output side of the A.C.

machines and hence original sinusoidal wave form are disturbed and wave form

becomes complex and non sinusoidal in nature generating 2nd, 3rd, 4th and so on

frequencies of the fundamental frequency.

The above phenomenon is shown in the below given diagram.

These 2nd

, 3rd

, 4th

frequencies are called harmonics of the fundamental frequency.

In short waveform with frequencies other than fundamental frequency is called

harmonics.

2nd

, 4th etc frequencies are called even harmonics and 3rd

, 5th

, 7th

, etc frequencies

are called harmonics.

Harmonics in transformer:-

The non-sinusoidal nature of the magnetizing current necessary to produce a sine

wave of flux produces harmonics in current and voltage wave –forms of the three

phase transformers.

The effects of current harmonics:-

1. Increased heating of winding.

2. Inductive interference with communication circuits.

3. Increased iron losses.

The effects of voltage harmonics:-

a) Increased heating of winding.

b). Capacitive interference with communication circuits.

c). Production of large resonant voltages.







pg. 42

Major Causes of Harmonics

Devices that draw non-sinusoidal currents when a sinusoidal voltage

is applied create harmonics. Some of these devices are listed below:

Electronic Switching Power Converters

1. Computers, UPS, Solid-state rectifiers.

2. Electronic process control equipments

3. Electronic Lightning Ballasts.

4. Reduced voltage motor controllers.

Arcing Devices

1. Discharge lighting.

2. Arc furnaces, welding equipments.

Ferromagnetic devices.

1. Transformers operating near saturation level.

2. Magnetic ballasts.

3. Induction heating equipment chokes.

Appliances

1. TV sets air conditioners, washing machines, and microwave

ovens.

2. Fax machines, photocopiers, and printers.

Higher RMS current and voltage in the system are caused by

harmonic currents, which can result in any of the problems listed

below:

1. Blinking of Incandescent Lights- Transformer Saturation.

2. Capacitor Failure- Harmonic Resonance.

3. Circuit Breakers Tripping- Inductive Heating and Overload.

4. Electronic Equipment Shutting down- Voltage Distortion.

5. Flickering of Fluorescent lights- Transformer Saturation.

6. Fuses Blowing for no apparent reason- Inductive heating and

Overload.







pg. 43

7. Motor Failures (overheating) – Voltage Drop.

8. Conductor Failure- Inductive heating.

9. Neutral conductor and terminal failures – Additive Triplen

currents.

10. Electromagnetic Load Failures – Inductive heating.

11. Overheating of Metal Enclosures- Inductive heating.

12. Power Interference on voice communication- harmonic noise.

13. Transformer failures- Inductive Heating.

Overcoming Harmonics

Tuned Harmonics filters consisting of a capacitor bank and reactor in

series are designed and adopted for suppressing harmonics by

providing low impedance path for harmonic component. The

harmonic filters connected suitably near the equipment generating

harmonics help to reduce THD to acceptable limits. For overcoming

and troubleshooting of some problems in the electrical power system

HARMONICS WAVE FORM







pg. 44

Measurement of Harmonic Level in the Plant:-

We have measured Harmonic Level in the plant and results are mentioned as

under.

Location THD % Harmonics in Voltage %

Harmonics in Current %

V A 3rd 5th 7th 3rd 5th 7th Main ACB incomer 11.2 24.7 0.8 10.3 3.1 4.7 19.6 5.2

Table – Harmonics measurements

Observations-

1. THD Levels in both Voltage and Current are above the permissible limits.

This is because of many VFD Drives in your system.

2. These higher THD Level create problems in system for example ACB tripping

without any fault flag, Burning of capacitors, Motor winding Bursting,

Frequent tripping of machinery on voltage fluctuation, and we advise to

install harmonic Filters in your Electrical network.

Power factor improvement with the use of static capacitors:-

In case of alternating current power supply system current is always lag behind

the voltage. This is due to the fact that the A.C. machines works on the principle

of electromagnetic induction and these A.C. machines consume reactive power

for their own needs for formation of magnetic flux and this phenomenon will

cause current vector to lag behind the voltage vector and this will generates the

P.F. in the system. The above fact is shown in below sine wave diagram.







pg. 45

What is Power Factor:-

The P. F. = CosØ is the ratio of KW = Active Power

KVA Apparent Power

Methods of improving power factor:-

1. With the use of static capacitors.

2. With the help of synchronous condenser.

3. With use of phase advancers.

Here we can discuss the use of static capacitor and there advantages for

improving

How Power factor improves with the use of static capacitors:-







pg. 46

The static capacitor generates reactive current of opposite nature at leading

power factor when connected to the supply mains parallel to inductive load and

compensates reactive current of the inductive load, which is running at lagging

power factor.

∴∴∴∴ When static capacitor is connected parallel to the inductive load, the inductive

load starts receiving reactive power of opposite nature at leading power factor

from the capacitors and thus this reactive power neutralizes the inductive power

requirement of the load and thereby improves the P. F. of the load.

The above Explanations are made simple with the below mentioned Vector

Diagram.







pg. 47

The P. F. = CosØ is the ratio of KW = Active Power KVA Apparent Power

Vector diagram and physical diagram of inductive load with use of capacitor







pg. 48

Effect of Different Power Factor on 100 KW Industrial Motor Working

Load:-

Assume 3 phase, 100 KW rating inductive motor., V = 415, P.F. = Cos ↓↓↓↓ =

Cos 0°°°° = 1,

∴∴∴∴↓↓↓↓ = 0°°°°,

F = 50 HZ, Efficiency = 90 % Sin ↓↓↓↓ = Sin 0°°°° = 0

η Of Motor = Output Input = 1.73 x V x I x Cos ↓ = output x 100

Input η I = 100 x1000 1,73 x 415 x1 x 0.90 I (line) = 155 Amp.

Active Current = I active = I(line) x Cos ↓ = 155 x 1 = 155 Amp.

Reactive Current = I reactive = I(line) x Sin ↓ = 155 x 0 = 0 Amp. The KVA = KW = 100 =100 KVA

Cos ↓ 1







pg. 49

KW = KVA x Cos ↓ = 100x1 = 100

KVAr = KVA x Sin ↓ = 100x0 = 0

Vector Diagram

Active Power – (KW) = 100 Reactive power-(KVAr) = 0 Apparent or resultant power – (KVA) = 100

(B) 100 KW load working at Cos ↓↓↓↓ = P. F. =0.90, ∴∴∴∴↓↓↓↓ = 25°°°°

The KVA = KW = 100 = 111 KVA

Cos ↓↓↓↓ 0.90

Line Current = I (line) at P.F of 0.90 = 155 Amp./ 0.90 = 172 Amp

Active Current = I active = I(line) x Cos ↓ = 172 x 0.90 = 155 Amp

Reactive Current = I reactive = I(line) x Sin ↓ = 155 x 0.422 = 64.4 Amp.

KW = KVA x Cos ↓ = 111x0.90 = 100 KVA = 111

KVAr = KVA x Sin ↓ = 111x0.422 = 47

-: Vector Diagram :-

Active power – (KW) = 100 KW

↓ = 25° Voltage vector - V

Current vector – A . Reactive power–(KVAr) = 47 Apparent or resultant power – (KVA) = 111







pg. 50

(C) 100 KW load working at Cos ↓↓↓↓ = P. F. = 0.80, ∴∴∴∴↓↓↓↓ = 36.8°°°°

Line Current = I (line) at P.F of 0.90 = 155 Amp./ 0.80 = 194 Amp

Active Current = I active = I(line) x Cos ↓ = 194 x 0.80 = 155 Amp

Reactive Current = I reactive = I(line) x Sin ↓ = 194 x 0.599 = 116 Amp The KVA = KW = 100 = 125 KVA

Cos ↓ 0.80

KW = KVA x Cos ↓ = 125x0.80 = 100 KVA = 125

KVAr = KVA x Sin ↓ = 125x0.599 = 75

-: Vector Diagram :- Active power – (KW) = 100 KW

↓ = 36.8° Voltage Vector - V Reactive power–(KVAr) = 75

Current vector - A

Apparent resultant power – (KVA) = 125

From the above vector diagrams and below mentioned calculation it can be seen that at

(A) 100 KW load and P. F. = 1 ∴KVA = 100 KVA

∴∴∴∴ Demand charges = 100 x Rs.200 = Rs.20000/- (Assuming Demand Charges = Rs. 200 /KVA)

(B) 100 KW load and P. F. = 0.90 ∴∴∴∴KVA = 111 KVA

∴∴∴∴ demand charges = 111 x Rs. 200 = Rs.22200/-







pg. 51

Effects of Transmission & Distribution Losses Due Power Factor

Improvement:-

Transmission losses of plant + Distribution losses of plant = 3 % + 7 % = 10 % (On total yearly Consumption) The above Transmission and Distribution losses of plant are called plant losses. Transmission losses of plant are the losses occurring in main

transformers, H.T. Cables, Switch Gear etc. = 3 % (Of total yearly

Consumption)

Distribution losses of plant are the losses occurring in main L.T. Cables, L.T. Switch Gear, L.T. Bus -Bar etc. = 7 % (Of total yearly Consumption)

Now % loss reduction due to improvement of average P.F. from 0.94 to 0.99

Now Plant Copper losses in any A.C. System = I2 xR α 1

Cos ↓2

∴∴∴∴ Plant Copper losses = K

Cos ↓2

Now Plant copper losses at 0.94 P.F. 0 .[M ^ .[M

0 .PPZN

0 ] R 1.1317

Now Plant copper losses at 0.99 P.F . 0 .[[ ^ .[[







pg. 52

0 ] R 1.0203

Thus Plant heat loss reduction by improving PF at consumption point

0 277". a 5 0.94 \4 277". # 5 0.99 \4277". # 5 0.94 \4

0 ] R 1.1317 1.0203] R 1.1317

0 0.11141.1317

0 0.0984 / 10 %

Thus, by improvement in P.F. from 0.94 to 0.99, there will be a plant loss reduction of 10 %. However the exact saving percentage can’t be calculated as all motors are running at various PF varying form 0.7 – 0.9. But it is evident that heat loss reduction is enormous and substantial, as per the example calculation above. * What are Transmission and Distribution losses of plant? Transmission losses of plant are those electrical losses occurring in the Transformers, H.T. Cables, and Switch Gears etc and which is standardized at the rate of 3 % of the yearly energy consumptions of the Plant.

Distribution losses of plant are those electrical losses occurring in the L.T. Cables, Fuse links, bus bars and Meters etc and which is standardized at the rate of 7 % of the yearly energy consumptions of the plant.

∴Transmission losses of plant + Distribution losses of plant = 3 % + 7 % = 10 % (On yearly Consumption)

Power factor improvement with the use of static capacitors:-







pg. 53

In case of alternating current power supply system current is always lag behind the voltage. This is due to the fact that the A.C. machines works on the principle of electromagnetic induction and these A.C. machines consume reactive power for their own needs for formation of magnetic flux and this phenomenon will cause current vector to lag behind the voltage vector and this will generates the P.F. in the system. The above fact is shown in below sine wave diagram. Methods of testing & checking of running capacitors:-

1. With the help of AVO meter - A good capacitor will show dead short

between any two terminals first & then charge up to battery voltage.

2. Megger test- A good capacitor will show infinity resistance between any

terminals & earth.

3. With the help of Ampere meter - A good capacitor will draw rated current at

rated voltage.

We have observed that Most major motors in your plant are working with VFDs

with proper capacitor installation and they are maintaining motor and panel PF

over 0.95 which is excellent in terms of controlling the transmission and

distribution losses. We suggest to check the APFC capacitor current ratings,

every week and replace any faulty capacitors as soon as possible.

Importance of good Power Factor and various benefits thereof: -

1. The KW capacity of the prime movers is better utilized.

2. The KVA capacity of the transformers and cables are increased.

3. The efficiency of every plant is increased.

4. The overall production cost per unit decreased.







pg. 54

5. Heat losses in any electrical machine = k x 1/P.F. and hence high P. F. will

generate less heat.

6. Reduction of plant electrical losses due to improvement of P. F. =

1 A ~ ~

7. KVA reduction =a6 R ~

~ K

Disadvantages of poor power factor: -

1. Losses in any electrical equipment are proportional to i² which means

proportional to 1/P.F.² thus losses at P.F. = Unity = 1 and losses at P.F. =

0.8 are 1/ [0.8] ² = 1.57 times higher than those at unity P. F.

2. Rating of motors and transformers etc. are proportional to current hence to

1/P. F. therefore large motors and transformers are required.

3. Poor P. F. causes a large voltage drop, hence extra regulation equipment is

required to keep voltage drop within prescribed limits.

Indirect benefits of improved P. F. to the industries:- (Example for

understanding)

1. Losses reduction of the plant due to improvement in P. F. from P. F. 1 (0.92)

to P. F. 2 (0.98)

Now we are raising the existing P. F. of 0.92 to new P. F. of 0.98.

Therefore, monthly energy loss reduction in the plant, due to improvement of PF

= 1 ~ ~ K







pg. 55

= 1 .[K.[P

= 1 – 0.8812

= 0.1188

= 11.88 %

When current I amperes flow through any electrical machines having resistances

R ohms for t seconds the electrical energy expended is I² x R x t joules.

∴∴∴∴Heat produced = I² x R x t / 4187 kilocalories.

∴∴∴∴Heat produced at PF 1 (0.92) = k [1/ (PF 1)2] & Heat produced at PF 2 (0.98) = k

[1/(PF 2)2]

∴∴∴∴Reduction in heat generation due to improvement of P. F.

= R ~

~ K = k x [ 1.1814 – 1.0412 ]

= k x 0.1402 Calories

Sample Calculation for input to the motor, % shaft loading on motor and load

factor Table – (A) for Sr. No.1. (Assuming average η of motor = 82 %)

Input Power in KW

0 1.73 R 3 R R 2 1000

0 1.73 R 425 R 155 R 0.871000

0 99.14 ]9 / 99 ]9

η Of Motor = Output x 100

Input

∴Output of motor (Shaft Loading) = ηx Input = 82 x 99 = 81.18 KW.







pg. 56

100 100 say 81 KW

∴% Shaft Loading = Measured output in KW x100 = 81 x100 = 36 %

Rated output power in KW 225

% Load factor of motor = Input Power in KW x 100

Rated Output Power in KW

= 99 x100 = 44 %

225

Similarly the others motor’s shaft loading and Load factor had been summer

breaks but completion.

Scope of Energy Saving In Electrical Motors

(A)

Energy Saving By Replacing Existing V- Belt Drive Into Energy Saving Flat Belt

Drive:-

Engineering for Energy Saving in Energy Saving Flat Belt Drive:-

Energy Saving Flat Belt Drive is made from Nylon (Super Polyamide) and Chrome

leather materials while that of V – belt drive is made from Rubber and hence

Energy Saving in Flat Belt Drive is possible.

Disadvantages of V – belt drive:-

1. Thick rigid sections of V- belt drive absorb a great deal of useful power and

add to running costs. Theses power losses are typically dissipated in the

form of heat. Which in-turn has a deleterious effect on the belt life.







pg. 57

2. The wedging action of V- belt drive involves an irreversible energy loss, as

each V- belt is continuously wedged into the pulley groove and pulled out

again. This loss is aggravated by misalignment and unmatched V- belts in a

set. Further loss of energy occurs in the flexing of solid rubber and in the

shuffling of power between V – belts in a set.

3. Over all Weight of the V – belt drive is higher.(Including Pulleys)

The above three negative points of the V – belt drive can be converted in to

positive points by converting existing V – belt drive in to Energy Saving Flat Belt

Drive.

Advantages of Energy Saving Flat Belt Drive over V – belt drive:-

1. Efficiency of Energy Saving Flat Belt Drive is 98 % while that of V – belt drive

is 86 % and hence 12 % Energy Saving is possible with Flat Belt Drive.

2. Energy Saving Flat Belt Drive is generate very less vibrations while that of V

– belt drive generate more vibrations and hence long life of bearings and

machines with less noise are possible with Flat Belt Drives.

3. Energy Saving Flat Belt Drive is made from Nylon (Super Polyamide) and

Chrome leather while that of V – belt drive is made from Rubber and hence

less weight are possible with Flat Belt Drive which results high load transfer

with

4. High transmission efficiency.

5. Energy Saving Flat Belt Drive is vibrations free, slip free and maintenance

free while in that of V – belt drive above actions are not possible.

6. 5 % Energy Saving in Flat Belt Drive is possible while in that of V – belt drive

is not possible

Calculations for Energy Saving in Flat Belt Drive:-

There is total 14 big and major Motors which are running on V – Belt drive , so

total KW of those motors is 916 KW.







pg. 58

Cost benefits analysis are mentioned as under. Considering 83 % Load Factor

(From Billing Analysis sheet – 2009)

1. The Total Effective Load V – belt drive Motor = 916 KW x 0.83 = 760 KW

1 Energy Saving with flat belt drive = 4 %

Daily load reduction due to replacement of existing V belt drive in to Flat Belt

Drive

= 760 x 0.04 = 30 KW

= 30/0.99 KVA = 31 KVA

Monthly monitory saving due to 31 KVA reduction

= 31 x Rs. 200 /-

= Rs. 6200/--

Annual monitory saving due to 31 KVA reduction

= Rs. 6200 x12 Month

= Rs. 74400/-

So total saving by Replacing V – belt Drive in to Flat belt drive by reduction of KVA

demand is Rs. 74400/-

2. Annual consumptions on 83 % load factor of 12 months working

= 30 KW x 24 x 365 Days x 0.83

= 218124 KWH

= 220327 KVAh

Annual Monitory saving in Rs by above conversion = 220327 KVAh x Rs.

3.81/-

= Rs. 8,39,446/-







pg. 59

4. Cost of converting one no. Of V – belt drive in to Flat Belt Drive with other

Exp.

= Rs32000/- (assumed average)

5. Salvage value of existing Of V – belt drive = Rs.2000/-

6. Net Cost of converting one no. Of V – belt drive in to Flat Belt Drive

= Rs32000/-- - Rs.2000/- = Rs.30000/-

7. Net Cost of converting 14 no. Of V – belt drive in to Flat Belt Drive

= Rs.30000/- x 14

= Rs. 4,20,000/-

8. The payback period for all conversion = Net cost of V - belt drive

conversion

Total yearly saving by conversion

= Rs. 420000/-

Rs. 839446/-

= 0.50 year = 0.50 x12 months =6 months

9. Rate of yearly return = 200 %

Remarks:-

(1) The annual saving in energy bill is 4 % = Rs. 839446/- for all 14 Nos. Motors.

(2) KVA demand saving of Rs. 74400/-

(3) Total saving in Rs. Due to conversion of V belts to flat belt drive. = (1) + (2)

= Rs.

913846/-

(4) The payback period on investment is only Six months

(5) Rate of annual interest on capital invested is 200 % and hence installations of

Flat Belts drive are strongly recommended.







pg. 60

S.No. Motor Details KW

1. Main Motor PM1 110

2. Vacuum Pump PM1 165

3. Turbo Main 75

4. Boiler 4 Ton ID Fan 18

5. Vacuum Pump PM2 200

6. Machine Chest Pump x 2 Nos. 15 KW Each

7. Couch Pit Pump PM1 15

8. TDR Pump x 2 Nos. 37

9. Chest Agitator x 3 Nos. 15 KW Wach

10. Rewinder 22

11. Blower PM2 x 2 Nos. 15 KW Each

12. Pulper 132

(B) Energy saving by replacing Old Low Efficiency Motors by New Energy

Efficient Motors.

We have observed that in your plant there are so many old Low Efficiency motors

are working. Old Low Efficiency motor and rewinded motors consumed more

energy as compared to New Energy efficient Motor.

Motor Efficiency Test (No Load Method)

We have taken measurement on your old rewind 10 HP motor for calculation of

efficiency for this motor.

Motor Specifications

Rated power = 7.5 kW/10 HP

Voltage = 415 Volt

Current = 17 Amps

Speed = 935 RPM

Connection = Delta







pg. 61

No load test Data

Voltage, V = 424 Volts

Current, I = 5.9 Amps

Frequency, F = 50 Hz

Stator phase resistance at 20 °C = 2.5 Ohms

No load power, Pnl = 156 Watts

( a) Let Iron Plus Friction and windage Loss , Pi + fw

No load Power Pnl – 156 Watt

Stator copper Loss, Pst @ 20o

C (Pst.Cu)

= 3 x (5.9/1.73) 2 x 2.5

= 87.23 Watt

Now Pi + fw = Pnl – Pst.cu = 156 – 87.23 = 68.77 Watt

(b) Stator Resistance at 120 C

HK 0 2.5 R KKZK KZ

0 3.48 '6

(c) Stator Copper Losses at Full Load Pst.cu 120 0

C

= 3 x (17/1.73) 2 x 3.48

= 1008.32 Watt

(d) Full Load Slip = 0

[Z

= 0.065

Thus, Input to Rotor =

0 L.N

0 L.[Z







pg. 62

= 8021.40

(e) Total Full Load Input Power

= 8021.40 + 1008.32 + 68.77 + 37.5 ( stray Losses 5 % of rated Output)

= 9135.99 Watt Say 9136 watt

(E) Motor efficiency at Full Load = nopqoplqop r UXX

= 82.09 % say 82 %

Above test clearly shows that Old and many times rewind Motors have very low

efficiency as compared to new Energy efficient Motor. New Energy Efficient

Motors have efficiency up to 95%.

So you are advised to change all possible old motors in one go and install New

Energy efficient Motors.

Cost benefits analysis of New Energy Efficient Motor in place of Old Motor

Sample calculations for 100 KW Motor – (Old) with compare to new 100 KW

Energy efficient Motor.

1. Power Consumption for 100 KW – Old Motor Considering 82 % efficiency On

Full load.

= 100 KW/0.82 = 121.95 Kw Input Power

2. Power Consumption for 100 KW – New Motor Considering 0.95 % On Full load.

= 100 KW/0.95 = 105.26 KW Input Power

Energy Saving by Installing New Energy Efficient Motor = 121.95 - 105.26

= 16.68 KW







pg. 63

From above calculations we can save 16.68 Unit per Hour By replacing Old motor

with new motor

(Considering running motor on Full Load)

Area S.No. Motor

Details

HP VFD

(kw)

Make RPM Rated

current(A)

Voltage

Measured

Measured

current(A)

Efficiency Present

%

loading

of

Motors

Required

current at

90%

efficiency

and 75%

loading

PM-1

1 Vacuum

separator

7.5 5.5 Kirloskar 2800 7.8 410 5 68.41 75.0 4.5

2 Holly Roll 3.5 2.5 ABB 960 3.5 410 4.4 100.0 3.5

3 Nozzle

Pump

1 0.75 CG 2800 1.5 410 2.7 100.0 1.5

PM-2

4 Pope reel 20 15 ABB 1440 28 14.3 55.15 51.1 13.5

5 MG 75 55 ABB 1440 98.5 58.8 58.42 59.7 50

6 Second

Press

75 55 ABB 1440 98.5 53 51.40 53.8 46.5

7 Wire 120 90 ABB 1440 157 72 38.00 45.9 66

8 Vacuum

Pump

270 200 NGEF 1440 340 260 96.64 76.5 142.8

9 Blower 1 20 15 ABB 1440 21 13.5 48.39 63.4 11.5

10 Blower 2 20 15 ABB 1440 21 12.5 47.06 58.7 11

11 Nozzle

Pump

1 0.75 CG 2800 2 410 2.6 100.0 2.00

Pulp

mill

12 Turbo Main 100 75 106.5 136 100.0 106.5

13 Turbo

Pump

50 35 49.7 40 69.40 80.5 36

14 Refiner

Pump

40 30 42.6 30 56.01 70.4 24

15 Back water 15 10 14.2 10 64.88 70.4 8

Boiler

16 Compressor

Boiler

15 10 14.20 15.5 100.00 14.2







pg. 64

KVA required by above machines @

present loading conditions

321.332 KVA

KVA required after replacing above

machines @ present loading conditions

238.5296 KVA

Reduction in KVA demand by replacing

above machines by new energy efficient

machines

82.80242 KVA

Monthly Reduction in KVAh required by

using new energy efficient motors

59617.75 KVAh

Monitory saving @Rs3.81/KVAh/

annum

244363 Rs

Annual Savings 2932356 Rs

The approx. Net cost of conversion of all old Motor Rs. 6,50,000/- (we have

considered a thumb rule of Rs. 2300 per KW) (Salvage value of Old Motor has

been Considered)

The payback Period = Cost Of Conversion = Rs.650000/- = 0.22 Years x 12 Month = 3 Months

Yearly Saving Rs. 2932356/-

Looking to the above benefits and yearly saving of Rs. 2932356/- every year, you

are recommended to change motors in one go with new energy efficient motor.

New motors have energy saving benefits, long life and trouble free operation, less

maintenance cost and with manufacturer guarantee. So we suggest you to

implement this change as early as possible.

-: Power Stages In An Induction Motor:-

ANSH ENERGY SOLUTIONS PVT. [email protected]

Variation of Motor Efficiency /P. F / Stator Current / Torque & Speed

with receipt to Load Demand

Power Loss Due to Under Load Operation of Induction Motor

(% of Power Loss in Motor):

Whenever induction motors rlosses are observed and hence under loading and no load running of the inductions motor are to be avoided.


Task Specific Audit:




Auditor Registration:


with receipt to Load Demand


(% of Power Loss in Motor):-

Whenever induction motors runs in under load conditions, losses are observed and hence under loading and no load running of the inductions motor are to be avoided.

Steam Generating


Sandeep Paper Mills (P) Ltd.

Anshul Singh Yadav


pg. 65



uns in under load conditions, heavy power losses are observed and hence under loading and no load running of the







pg. 66

For our customers knowledge a following chart of power losses is attached. Power Loss Due to Under Load Operation of Induction Motor

(% of Power Loss in Motor) :-

Motor Capacity in H.P.

No Load 25 % Load 50 % Load Full Load

5 50 40 25 18 7.5 45 30 20 17 10 44 26 18 17 15 43 23 17 14 20 42 20 15 14 25 41.5 19 15 13 30 41 18 14 13 40 40 17 15.5 10.5 50 40 16 12.5 10 60 39 15.5 12.5 10 75 38.5 13 12 9

100 38 13 12 9

Motor Rating

(HP)

Capacitor rating (kVAr) for Motor Speed

3000 1500 1000 750 600 500 5 2 2 2 3 3 3 7.5 2 2 3 3 4 4 10 3 3 4 5 5 6 15 3 4 5 7 7 7 20 5 6 7 8 9 10 25 6 7 8 9 9 12 30 7 8 9 10 10 15 40 9 10 12 15 16 20 50 10 12 15 18 20 22 60 12 14 15 20 22 25 75 15 16 20 22 25 30 100 20 22 25 26 32 35 125 25 26 30 32 35 40 150 30 32 35 40 45 50 200 40 45 45 50 55 60







pg. 67

250 45 50 50 60 65 70

Table – Rating of Capacitor required for different rating and

speed.







pg. 68

Performance Evaluation of Motors

Electrical motors accounts for a major part of the total electrical consumption. So

a careful attention should be given to the performance of this utility.

Measurements of the different electrical parameters of the major motors of the

plant are given in Table

The efficiency of the induction motor and loading condition of the motors are

directly proportional to each other. Higher the loading and higher is the efficiency

of the motors. The best efficiency of the motors is achieved at a load very much

near to the rated load.







pg. 69

Moreover at lower loads the power factor is on the lower side increasing the load

current and thereby increasing the copper losses, resulting in lower efficiency, the

rating of the motors should be decided after carefully understanding the process

requirement in the absence of which the motor might come out to be oversized.

Also one should run a motor, which has been rewound more than once as every

time a motor is rewinded it losses 2 – 5% of its actual efficiency.

Motor performance is affected considerably by the quality of input power that is

the actual volts and frequency available at motor terminals, vis-à-vis rated values

as well as voltage and frequency variations and voltage unbalance across the

three phases.

Mostly all the motors are old or rewound at least once. A good saving can be

achieved if higher efficient ones replace them. Though it’s a scheme with higher

initial investment but can be implemented phase wise. Induction motors are

characterized by power factors less than unity, leading to lower overall efficiency

(and higher overall operating cost) associated with a plant’s electrical system.

Capacitors connected in parallel (shunted) with the motor are typically used to

improve the power factor. The impacts of PF correction include reduced KVA

demand (and hence reduced utility demand charges), reduced I2R losses in cables

(leading to improved voltage regulation), and an increase in the overall efficiency

of the plant electrical system.

It should be noted that PF capacitor improves power factor from the point of

installation back to the generating side. It means that, if a PF capacitor is installed

at the starter terminals of the motor, it won’t improve the operating PF of the

motor, but the PF from starter terminals to the power generating side will

improve, i.e., the benefits of PF would be only on upstream side.







pg. 70

The size of capacitor required for a particular motor depends upon the no-load

reactive KVA (KVAR) drawn by the motor, which can be determined only from no-

load testing of the motor. Higher capacitors could result in over-voltages and

motor burnouts. Alternatively, typical power factors of standard motors can

provide the basis for conservative estimates of capacitor ratings to use for

different size motors. The capacitor rating for power connection by direct

connection to induction motors is shown in Table

We are suggesting some Measures to Improve Efficiency of Motors and

Distribution system

1: Electrical Distribution Correction

Measures available to improve power quality and reduce electrical losses are

1. Maintain voltage level close to nameplate level as far as possible, with a

maximum deviation of 5% (at 5% under voltage, copper loss is increased to

10%).

2. Minimize phase imbalance within a tolerance of 1%. As deviation of one

phase voltage from average phase voltage increases, it will result in

increased winding temperature.

3. Maintain high power factor to reduce distribution losses.

4. Avoid excessive harmonic content in the power supply system, as increased

harmonic content in power supply system will increase motor temperature.

5. Use oversize distribution cable in the new installation to reduce copper

losses. This will also help in reducing voltage drop during starting and

running and minimizing the motor losses.

2: Motor Efficiency Improvement

The measures available to improve motor efficiency are:

1. If motor is running at partial load then convert motor from delta to star

connection. This will improve motor efficiency.







pg. 71

2. Replace rewound induction motor (with reduced efficiency) with new

energy efficient motor.

3. If process demands oversized motor then possibility of use of VFD may be

explored to save energy. This is also applicable in case of varying load duty

cycle motor application.

4. Control the motor drive temperature. This will reduce copper losses and

increase motor life.

3: Better System Matching

Measures available are:

1. Use an on/off control system using timer, PLCs, etc to provide motor power

only when required.

2. Size the motor to avoid insufficient low load operation. Motor should run at

65% to 100% of its nameplate rating to get maximum efficiency.

4: Driven Load and Process Optimization

Measures available to optimize the process and its operation are:

1. Change or reconfigure the process or application so that less input power is

required.

2. Downsize the over sized pumps, fans, compressors or other driven loads if

possible.

3. Install more efficient mechanical subsystems. Check that coupling, gearbox

fan or pump must be energy efficient.

Miscellaneous Measures to Improve Motor Efficiency

Maintenance Energy savings of 10 to 15 percent of motor energy consumption

can typically be realized, depending on change from existing maintenance

practices.

These are:







pg. 72

1. Proper lubrication: it will minimize wear on moving parts. Lubrication is best

done on a regular schedule to ensure wear is avoided. Once it occurs, no

lubricant can undo it. It is crucial that the correct lubricant is applied in the

right quantities.

2. Correct shaft alignment: It ensures smooth, efficient transmission of power

from the motor to the load. Incorrect alignment puts strain on bearings and

shafts, shortening their lives and reducing system efficiency. Shafts should

be parallel and directly in line with each other. It is necessary to use

precision instruments to achieve this. Shaft alignment is an important part

of installation and should be checked at regular intervals.

3. Proper alignment: Belts and pulleys must be properly aligned and tensioned

when they are installed, and regularly inspected to ensure alignment and

tension stay within tolerances. Abnormal wear patterns on belts indicate

specific problems that may require correction. Loose bests may squeal and

will slip on the pulleys, generating heat. Correctly tensioned pulleys run

cool. Excess tension strains bearings and shafts, shortening their lives.

4. Painting of motor: Avoid painting motor housing because paint acts as

insulation, increasing operating temperatures and shortening the lives of

motors. One coat of paint has little effect, but paint buildup accumulated

over years may have a significant effect.

Maintenance Schedule for MOTOR for Energy Conservation

i). Daily: -

Clean the motor and starter.

ii). Weekly: -

Clean slip rings with soft brush dipped in white spirit.

iii). Monthly: -

Check earth connections of motor and starter.

Blow through motor and starter with dry compressed air at 2 Kg/Cm .

Check tightness of cable connections.







pg. 73

Check motor for overheating and abnormal noise / sound, sparking and

for proper bedding of brushes.

Tighten belts and pulleys to eliminate excessive losses.

iv). Quarterly: -

Check motor terminal voltage for balanced supply. If more than +1% of

average, then check from transformer onward.

Carry out SPM checks viz. vibrations and sound of bearing. Record reading

and

compare With earlier / other motor readings.

Slip Ring: - Inspect the brushes and make sure that they move freely in the

brush holder clips.

Clean brushes, holder chip and wipe with cloth dipped and in gasoline.

Replace

the brush if they are worn out less than 5 mm in length from brush

holder.

Clean the starter and motor contacts with white spirit.

v). Six Monthly Maintenance: -

Check over load mechanism of starter.

Check alignment of motor with driven equipment.

Check no load current and compare with earlier / original.

Check / change lubrication as per lubrication schedule given on next

pages.

Check the securing foundation nuts for tightness.

Inspect the paint coating and do-touching wherever required.

Check IR(Insulation) Resistance of motor and starter with 500 V megger.

It should be more than 2 MΩ







pg. 74

Projected savings by converting LV drives to MV drives @ 6.6 KV

As it is clearly evident in all cases that the cabling, transmission, core and iron losses etc are

related to current which in turn is inversely proportional to the voltage it is a fact that changing

certain drives from 415 V to 6.6 KV may reduce energy consumption due to reduced losses and

variations.

Below is the list of proposed motors which can be converted to the MV operation.

Saving calculation by replacement of following motors with HT drives

Location S.No. Motor

Details

HP VFD

(kw)

Motor

Make

RPM Rated

current

(Amps)

Measured

current @

415v level

(Amps)

Measured

KW

Current

required

@ 6.6kv

level

(Amps)

PM-1 1 Main Motor 120 110 CG 987 162 138 91.5 10

2 Vacuum

pump

180 132 ABB 960 229 162 120 14.5

PM-2 5 Wire 120 90 ABB 1440 157 72 34.4 10

6 Vacuum

Pump

270 200 NGEF 1440 340 260 192 21

Pulp mill

motors

1 Pulper 180 132 229 197 96 14.4

2 Refiner 150 110 150 131 57.8 9.5

Total 1267 79.4

Tentative savings are calculated below.

KVA required by existing(above) motors 557.48 KVA

KVA required by new HT (above) motors 524.04 KVA

Reduction in KVA required by new HT (above)

motors

33.44 KVA

Monthly Reduction in KVAh required by new HT

(above) motors

24076.8 KVAh

Monitory saving @Rs3.35/KVAh/ annum 967887.4 Rs







pg. 75

As the market price of drives for 6.6 KV is not very standard as of date and

availability of the panels in this segment is poor we have not calculated the

payback or investments for this operation. However the idea can be stored for

future reference as we may sometime achieve better pricing or better availability

in the same. Higher rating motors like 300 KW are still viable for this operation.

Paper Plant Energy Audit Report

Documents