Topic 3 Industrial Energy Management

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BEF 44903 IndustrialIndustrial

Power Systems

Topic 3

BEF44903

Topic 3Topic 3Topic 3Topic 3

INDUSTRIAL ENERGY MANAGEMENT

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BEF 44903 Industrial

Outline

Power Losses in Industrial Power SystemsIndustrial

Power Systems

Topic 3 Loss Reduction Methods Energy Efficiency in Industrial Power

Systems Battery Application in Industrial Power

SystemsSystems Application of Uninterruptible Power Supplies

(UPS)

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Power Losses in IPS

• Losses occurred in the electrical

• Losses due to theft of energyIndustrial

Power Systems

Topic 3

in the electrical elements mainly comprises of ohmic and iron losses

• Mostly depends on electrical

theft of energy and unrecovered billed amount

• Also called as commercial Losses

loading

Technical losses

Non-technical losses

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Power Losses in IPS

TECHNICALIndustrial Power Systems

Topic 3

TECHNICAL LOSSES

Current dependent

Voltage dependent

Series Losses

Shunt Lossesdependent

Copper Losses

dependent

Iron Losses

Dielectric Losses

Corona Losses

Losses Losses

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Power Losses in IPS

TX

I2R

FD

Industrial Power Systems

Topic 3

I2R

I2R I2R I2R

Power Transformer

Power CableFeeder

I2R

Power Cable

I2R I2R I2R

Lighting L Motor Loads M Heating Elements H5


Power Losses in IPS

Line Losses are a result of passing current through an imperfect conductor such as


Topic 3

copper. The line losses can be calculated based on

the measured current load as:Ploss = I * (I * r/l * L) = I2RWhere,,I is currentr/l is resistance / KilometreL is length of cable in Kilometres

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Power Losses in IPS

For a 3-phase system, the losses for each phase are calculated separately according to h dIndustrial

Power Systems

Topic 3

the measured current as:Ploss total = Ploss-a + Ploss-b + Ploss-c

Assumptions made: All load currents maintain a constant ratio to

ccbbaa RIRIRI 222

the total current The voltage at every source bus remain

constant in magnitude Power factor remains constant

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Power Losses in IPS

It is not possible to measure the distribution losses incurred by individual customers due

d l f hIndustrial Power Systems

Topic 3

to many unexpected losses factors such as network utilisation, network configuration, the shape of the load profile and the level of reactive power support.

Thus, an estimated losses calculation app oach sing Load Facto (LF) and Loadapproach using Load Factor (LF) and Load Loss Factor (LLF) is introduced.

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Power Losses in IPS

Load Factor – Ratio of average demand over maximum demand.


Topic 3

Load Loss Factor – Ratio of average power loss over power loss at maximum demand

demandMaximum

kW)(in demandAverage LF

periodduring(kWh)lossActualLLF

Relationship between LF and LLF:LLF = k * LF + (1 - k) * LF2

(kWh)current maximumat Loss

pg( ) LLF

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Power Losses in IPS

where k = a constant, typically 0.1, 0.2 or 0.3


Topic 3

Typically, k = 0.3 for subtransmission systems k = 0.2 for medium voltage feeders and

distribution substations k = Sample sections of the network can be

analysed to produce an estimate of the k factoranalysed to produce an estimate of the k factor applicable for the rest of the system

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Power Losses in IPS

Loss in a feeder for a given period (which is one month in this case) is given by:


Topic 3

Technical loss in MU = I2 × R × L × LLF ×24 × 30 × 10−9

Where,I = Load in ampereR = Resistance of the conductor in ohms/ kilometreL = Length of the feeder in kilometresLLF = Load loss factor

Yearly technical loss for 3- feeder in MU,= 3 x I2 x R x L x LLF x 8760 / 10911


Power Losses in IPS

Example 3.1 Calculate LF, LLF,


Topic 3

and k. Calculate technical

losses per month. Consider following parameters: 300mm2 = 0.1173/km 240mm2 = 0.1267 /km 120mm2 = 0.2176 /km L1 = 1.2 km, L2 =

0.5km, L3 = 2.4km12

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Power Losses in IPS

Technical loss of Transformer in MU for year = {No load loss + [(% loading in p.u.)2

* d C l * ]} * 2 * 36 / 06Industrial Power Systems

Topic 3

* rated Cu. loss * LLF]} * 24 * 365 / 106

Example 3.2: What is the total MU loss?

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Loss Reduction Methods

Proper Sizing Regular MaintenanceIndustrial

Power Systems

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p g Maintenance

Convert LV to HV Lines

MinimiseOverload

Improve Power Factor

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Energy Efficiency in Industrial Power Systems

Reduce iIndustrial

Power Systems

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Energy efficiency

consumption

Reduce CO2emission

Optimisecost

Improve reliability

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Transport – 27%Industrial

Power Systems

Topic 3

Residential – 16%

Commercial – 8%

Industrial – 49%

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Save energy

• Proper production planning to avoid peak hour charges

• Use automation systemIndustrial Power Systems

Topic 3

Reduce energy

use

energy cost

Energy reliability

Use automation system in the plant

U How to increase energy

efficiency?

• Avoid energy losses• Proper maintenance• Treat PQ properly

• Use energy efficient devices/ equipment

• Proper control and monitoring

• Use variable speed drives17


Energy Efficiency in Industrial Power Systems Some of energy efficiency classes for LV

motors:Industrial

Power Systems

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Energy Efficiency in Industrial Power Systems Energy savings based on reducing fan speed

by half:Industrial

Power Systems

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Energy Efficiency in Industrial Power Systems Load management strategy:


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Energy Efficiency in Industrial Power Systems Monitoring architecture for a small site:


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Battery Application in Industrial Power Systems

UPSIndustrial

Power Systems

Topic 3

Solar power systemVoltage support

Battery Applications

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Tele-communication

Emergency Control/lighting

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Battery Application in Industrial Power Systems

Lead acid (Mostly used in

• Most economical• Not much sensitive to

temperature• 5-7 Years lifespanIndustrial

Power Systems

Topic 3

IPS)p

• Needs special battery room

Nickel Cadmium

• Most expensive • 20 years lifespan• Large size as

compared to same Ah

Sealed Mainte-nanceFree

• No maintenance• User friendly• Very sensitive to

temperature• Lowest lifespan23


Battery Application in Industrial Power Systems Why battery sizing?

Ensure the loads being supplied or the powerIndustrial Power Systems

Topic 3

Ensure the loads being supplied or the power system being supported are adequately provided by the battery for the period of time (i.e. autonomy) for which it is designed.

Improper battery sizing can lead to poor autonomy times, permanent damage to battery cells from over-discharge low load voltages etccells from over discharge, low load voltages, etc.

Methodology – Normal industry practice and technical standards: IEEE Std 485 (Lead-Acid) and IEEE Std 1115 (Nickel-Cadmium), based on the ampere-hour method.

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Battery Application in Industrial Power Systems Steps to be considered in battery sizing:

Step 1 – Collect load data to be supported byIndustrial Power Systems

Topic 3

Step 1 Collect load data to be supported by the battery

Step 2 – Construct load profile and calculate the design load (VA) and design energy (VAh)

Step 3 – Select battery type and determine the characteristics of the cellSt 4 S l t b f b tt ll t b Step 4 – Select number of battery cells to be connected in series

Step 5 – Calculate required Ampere-hour (Ah) capacity of the battery

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Battery Application in Industrial Power Systems Design load (VA) (Step 2):

Cgpd kkSS 11Industrial

Power Systems

Topic 3 Design energy demand (VAh) (Step 2):

where, Sp is the peak load apparent power (VA)

gp

Cgtd kkEE 11

Et is the total load energy (VAh) kg is a contingency for future load growth (%)

(10%) kc is a design margin (%) (10%)

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Battery Application in Industrial Power Systems Factors to be considered for battery type

selection (Step 3):Industrial

Power Systems

Topic 3 Physical characteristics Expected life of cell Frequency and depth of discharge Ambient temperature Charging characteristicsg g Maintenance requirements Ventilation requirements Seismic factors (vibration and shock)

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Battery Application in Industrial Power Systems Most common number of cells (Step 4):

Rated Voltage Lead-Acid Ni-CdIndustrial Power Systems

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g

12 V 6 9 – 10

24 V 12 18 – 20

48 V 24 36 – 40

125 V 60 92 – 100

250 V 120 184 – 200

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Battery Application in Industrial Power Systems Number of cells calculation (Step 4):

lDC VVN max,1 lDC VV

N min,1Industrial

Power Systems

Topic 3

where,

Nmax/ Nmin is the max. or min. of cells VDC is the nominal battery voltage

V i th l d lt t l (%)

c

lC

VN a,

max

eod

lC

VN ,

min

Vl,max is the max. load voltage tolerance (%) Vl,min is the min. load voltage tolerance (%) VC is the cell charging voltage (charge cycle) Veod is the cell end of discharge voltage

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Battery Application in Industrial Power Systems Min. battery capacity required (Step 5):

tcad kkkEC

)1()1(Industrial

Power Systems

Topic 3

where,

Cmin is the min. battery capacity (Ah) Ed is the design energy over autonomy time (VAh) V is the nominal battery voltage

dodDC

tcad

kVC

)()(min

VDC is the nominal battery voltage ka is a battery ageing factor (%) (25%) kc is a capacity rating factor (%) (10%) kt is a temperature correction factor kdod is the max. depth of discharge (%) (80%)30

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Battery Application in Industrial Power Systems IEEE Std 485-1997

reIndustrial Power Systems

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acto

rs fo

r te

mpe

ratu

Cell

corr

ectio

n fa

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Design of Uninterruptible Power Supplies (UPS) The primary function of uninterruptible

power supplies is to ensure continuity of l l dIndustrial

Power Systems

Topic 3

electrical power during power outage or contingencies.

UPS can be used to improve PQ especially voltage quality in IPS.

Static transfer switches can be used to supply power to a load via two independent power sources. In the event of a fault, they automatically transfer the loads from one source to the other.

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Design of Uninterruptible Power Supplies (UPS)


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Line-interactive

OnlineOffline

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Design of Uninterruptible Power Supplies (UPS)Online or Double conversion UPS In normal operation, the load is supplied with power

l hIndustrial Power Systems

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permanently via the inverter. Bypass channel can be used to increase the

availability of the power supply in the event of an overload or if the inverter shuts down following a fault or for maintenance purposes.

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Design of Uninterruptible Power Supplies (UPS)Line-interactive UPS The load is supplied from power supply via power

f h h d ll l h hIndustrial Power Systems

Topic 3

interface in which connected in parallel with the inverter. Can improve voltage quality/ harmonics.

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Design of Uninterruptible Power Supplies (UPS)Offline or passive stand-by UPS In normal operation, the load is supplied with power

d l h lIndustrial Power Systems

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directly via the line. If the line voltage strays outside the permissible

load tolerance limits, the inverter takes place. This configuration is designed for low-sensitivity

loads such as personal computers.

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Design of Uninterruptible Power Supplies (UPS) Steps in UPS sizing:

Step 1 – Prepare load list Consider apparentIndustrial Power Systems

Topic 3

Step 1 Prepare load list. Consider apparent power and autonomy hour:

Step 2 – Construct load profile Step 3 – Calculate peak design load (in kVA) and

cos

PS

design energy demand (kVAH)

cgpd kkSS 11

cgtd kkEE 11

kg – future load growth (10%)kc – design margin (10%)Sp – peak loadEt – total load energy (VAh)

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Step 4 – Battery sizing Step 5 – Rectifier/ Charger sizing. The design

DC load current:Industrial Power Systems

Topic 3

DC load current:

The maximum charging current:DC

DCL V

SI ,

lC t

kCI

C is the selected battery capacity (Ah)kl is the battery recharge efficiency/ loss

S is the selected UPS rating (kVA)VDC is the nominal battery/ DC link voltage

The total min. DC rectifier/ charger current:

Ct factor (pu) (1.1)tc is the min. battery recharge time (hrs)

CDCLDC III ,38

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Step 6 – Inverter/ Static switch sizing.For a three-phase UPS:


Topic 3

For a single-phase UPS:

where

03 V

SIL

0V

SIL

where,S is the selected UPS rating (kVA)V0 is the nominal AC voltage

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Design of Uninterruptible Power Supplies (UPS) Example 3.3

An inverter based UPS is installed in a Industrial

Power Systems

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factory to run 10 Nos of 250 VA ESD cabinets for at least 3 hours during blackout period. The battery bank output voltage is 120 VDC. The future growth contingency and the design margin are both considered as 10%10%.

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Design of Uninterruptible Power Supplies (UPS) Example 3.3a) The cell charging voltage is 2.25 VDC/cell, the end-

f d h l 8 / ll d hIndustrial Power Systems

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of-discharge voltage is 1.8 VDC/cell, and the minimum and maximum load voltage tolerance are 10% and 20%, respectively. Decide the number of cells in series.

b) Given a depth of discharge is 80%, battery ageing factor is 20%, temperature correction factor for

t d ll t 30C i 0 956 d th it tivented cell at 30C is 0.956 and the capacity rating factor is 10%. Compute the minimum battery size (in AH).

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Design of Uninterruptible Power Supplies (UPS)a)

cells 64

252

2.011201 max,max

lDC

V

VVN


Topic 3

The selected n mbe of cells in se ies is 62

25.2acV

cells 60

8.1

1.011201 min,min

eod

lDC

V

VVN

The selected number of cells in series is 62 cells.

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Design of Uninterruptible Power Supplies (UPS)b)

VAHEd 90751.011.01325010 Industrial

Power Systems

Topic 3

The minimum battery size,

d

AHkV

kkkEC

dodDC

tcad 29.1198.0120

956.01.12.19075)1()1(min

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Design of Uninterruptible Power Supplies (UPS) Example 3.4

Table E3.4 shows the critical loads for a Industrial

Power Systems

Topic 3

plastic packaging plant that to be supplied by UPS during power outage period. Figure E3.4 depicts the connection diagram of the proposed UPS and its main configurations.

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Design of Uninterruptible Power Supplies (UPS)a) Construct the load profile (VA versus period

in hour) for the UPS loading Industrial

Power Systems

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b) Compute the peak design load and design energy demand if the future growth contingency and the design margin are both considered as 10%

c) Design the UPS based on the ratings of the battery bank, the rectifier system, the static switch and the inverter system

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Design of Uninterruptible Power Supplies (UPS) Table E3.4

Rating Nos Autonomy Industrial

Power Systems

Topic 3

Load Description (VA) (Unit) Time (Hours)

Distributed Control System Cabinet

350 10 4

Electrostatic Discharge Cabinet

400 10 6

Telecommunications Cabinet

180 3 8

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CabinetComputer Console 120 5 2HMI Units 150 10 4

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Design of Uninterruptible Power Supplies (UPS) Figure E3.4

Static Switch

Bypass AC Input Supply

415V ACIndustrial Power Systems

Topic 3Rectifier Inverter

Battery Bank

Main AC Input Supply

415V AC Output

Output voltage is 120 VDC

Depth of discharge is 85%

f

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Battery ageing factor is 25%

Temperature correction factor for vented cell at 30C is 0.956

Capacity rating factor is 8%

Recharge efficiency factor is 1.1

Minimum recharge time is 3 hours


Design of Uninterruptible Power Supplies (UPS)a) Load profile of the UPS loading:

VAIndustrial

Power Systems

Topic 3

6000

8000

10000

12000

48

2000

4000

4 6 8 Operating hour

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Design of Uninterruptible Power Supplies (UPS)b) The peak design load = 540 + 4000 + 3500

+ 1500 + 600 = 10,140 VAIndustrial

Power Systems

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Total energy demand,= 540(8) + 4000(6) + 3500(4) + 1500(4) + 600(2) = 49,520 VAH

The design energy demand,

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VAHEd 2.919,591.011.01520,49


Design of Uninterruptible Power Supplies (UPS)c) Battery bank sizing,

AHkkkE

C tcad 15758956.008.125.12.919,59)1()1(


Topic 3

A battery bank capacity of 800 AH is selected.

Rectifier sizing

AHkV

CdodDC

15.75885.0120min

Rectifier sizing,Design load = 10140 (1.1)(1.1) = 12,269.4 VAThe 15,000 VA of design load is considered.

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Design of Uninterruptible Power Supplies (UPS)c) Design DC load current,

AVA

I 125000,15


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Maximum battery charging current,

Total minim m DC ectifie and cha ging

AV

I DCL 125120,

AIC 3.293

3

1.1800

Total minimum DC rectifier and charging currents,IDC = IL,DC + IC = 125A + 293.3A = 418.3AA DC rectifier rating of 420A is selected.51



Inverter and static switch sizing,


Topic 3Output AC voltage is 415V,

Design AC load current,

AV

VAI ACL 87.20

4153

000,15,

An inverter and static switch rating of 30A is selected.52

V4153

Topic 3 Industrial Energy Management

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Transcript of Topic 3 Industrial Energy Management