475 project

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American University of Beirut

Department of Electrical and Computer Engineering

EECE 475 Industrial Electrification

Design Project 6

Fall 2010 2011

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Presented to

Prof. Chahine

Prepared by

Tarik Ibrahim

Jad Karaa

Sabine Khalil

Rand Kassamany

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1.Power Factor Correction1.1Power Factor

Power factor corrections are supposed to reduce electricity consumption of a firm. Othersdispute that and argue it reduces only electricity costs in case the power utility is offering a

tariff where a reactive power demand charge are part of the monthly electricity bill. In

discussing the nature or phenomena of reactive power experts use incomprehensive equations

or explanations.

The vector combination of these two power components (active and reactive) is termed as

Apparent Power (measured in kVA), the value of which varies considerably for the same

active power depending upon the reactive power drawn by the equipment. The ratio of the

active power (kW) of the load to the apparent power (kVA) of the load is known as the

power factor of the load.

Power Factor = __Active Power (kW) __

Apparent Power (kVA)

Thus when the nature of the load is purely resistive the kVAr or the reactive component will

be nil and thus the angle will be equal to 0 degrees and the power factor will be equal to

unity. For a purely inductive load the power factor will be lagging and for a purely capacitive

load the power factor will be leading.

Thus, it is evident from above that, more the power factor departs from unity the more will

be the kVA demand for the same kW load. Since most of the HT tariffs include kVA

Demand charges along with the Energy Charges, the more the kVA demand for the same kW

load the more shall be the electricity bill of the consumer. To say it otherwise, the customers

with a low power factor will pay more for their useful electrical power. (The Billing demand

for the month is generally taken to be the actual maximum kVA demand of the consumer

during the month or a fixed percentage of the contract demand or a fixed kVA value,

whichever is higher, based on the type of the consumer and the tariff structure of the utility.)

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1.2 Disadvantages of low power factor

The disadvantages of a low power factor are:

y The load draws greater current for the same value of the useful power.

Current (I1) = __Power (kW)_

Volts (V) * PF

If the power factor of the motor is increased the current drawn by the motor will decrease

Thus, as the power factor decreases the current required for the same value of active, or

useful, power increases. The result is that the sizes of the equipment, like the switchgear,

cables, transformers, etc., will have to be increased to cater the higher current in the

circuit. All this adds to the cost. Further, the greater current causes increased power loss

or I2

R losses in the circuits. Also due to higher current, the conductor temperature rises

and hence the life of the insulation is reduced.

y Also, with the increased current the voltage drop increases, thereby the voltage at thesupply point is reduced. For different loads it causes voltage drop resulting in :

- Lower output of the illumination system.- Less current is drawn by the heating devices so that the operating temperature drops.

This results in increased consumption for the same rise of temperature.- The induction motors slow down and therefore draw more current to produce a fixed

torque for the loads., and more consumption for the same torque.

In the transmission and distribution of the current itself, from the generating station to the

consumer, heating losses will be greater at low power factor (varying in proportion to the

square of the current) and the voltage drop will be in accordance with relation I*Z (where

Z is the impedance, combination of resistance and reactance) . Since the losses in the

electricity system due to low power factor will incur additional cost, it is evident that

these will have to be reflected to some extent in the charges to the consumers. This is

implemented by metering the maximum demand in kVA or by applying a low power

factor penalty component in the tariffs.

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1.3Advantages of power factor correctionHigher power factors result in:

- Reduced system losses, and the losses in the cables, lines, and feeder circuits andhence lower sizes could be opted.

- Improved system voltages, thus enable maintaining rated voltage to motors, pumpsand other equipment. The voltage drop in supply conductors is a resistive loss, and

wastes power heating the conductors. A 5% drop in voltage means that 5% of your

power is wasted as heat before it even reaches the motor. Improving the power factor,

especially at the motor terminals, can improve your efficiency by reducing the line

current and the line losses.

- Improved voltage regulation.- Increased system capacity, by release of kVA capacity of transformers and cables for

the same kW , thus permitting additional loading without immediate augmentation.

The figure below shows the amount of capacity released for various amounts of

correction. For the example given previously, the dotted lines indicate the system

capacity released is 0.13 times the kilowatt load, which in this case is 10.4 kVA.

- Reduced power costs (depending on the electric utility tariff schedule) due to reducedkVA demand charges and so also by reduced power factor charges.

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1.4Cost saving due to Power Factor improvementTaking the Tariff as below:

1. Demand charges Rs. 144/kVA/month

2. Energy Charges Rs. 4.11 / Unit

3. PF surcharge for each 1% below 90% 1% of (Demand charges + Energy Charges)

As we already know, by improving the power factor there will be a reduction in the kVA

demand of the load. Thus , in this case the kVA MD will drop from 400 kVA (at 79.7%) to

327.78 kVA (at 90%):

Load 1:

SL1 = 100 kVA p.f. = 0.85 = 1

PL1 = SL1 1 = 100 0.85 = 85 kW

Load 2:

SL2 = 300 kVA p.f. = 0.7 = 2

PL2 = SL2 2 = 300 0.7 = 210 kW

PTotal = PL1 + PL2 = 85 + 210 = 295 kW

p.f. = 0.797

S1 = 400 kVA

S2= PTotal /0.9=327.78KVA

Therefore reduction in energy bill due to reduction in maximum demand due to improved power

factor from 0.797 to 0.9 shall be:

Rs. 144.00 * (400-327.78)= Rs. 10399.68 per month

In addition, by increasing the power factor from 79.7% to 90% , there shall be no power factor

penalty/surcharge on account of low power factor. Thus the savings due to avoidance of the PF

surcharge per month would be as below:

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Rs. ((400-327.78)*144*(90-79.7))*1/100= Rs. 1071.16

Thus the total monthly reduction in bill due to P.F improvement from 0.797 to 0.9 would be:

Rs. 10399.68 + 1071.16 = Rs. 11470.84 per month.

Net reduction per annum = 11470.84*12=137650.08 per year

1.5 Cost of investment for Power Factor improvement:

QC = Ptotal[ 1)) comp))]

= Ptotal[ ]

= 295 ( tan 37.16 - tan 25.84)

= 80.73kVAR

QL = S = 400 sin 37.16 = 241.59 kVAR

Qcomp = QL - QC = 241.59 80.73 = 160.86 kVAR

IL =

=

= 607.737 A

IC = I =

=

= 350.877 A

C =

=

= 2.939 mF

Size of capacitor required to improve the PF from 0.797 to 0.9

= kVA1* Sin1 kVA2* Sin2

=400*sin(37.15) 327.78*Sin(25.84) = 98.70 kVAr

If we take the cost of capacitor bank per kVAr as Rs. 200/KVAr,

The cost of the capacitor bank = 98.7*200 = Rs. 19739

Cost of switching and associated equipment = Rs. 1000 and installation, etc.

Total cost = Rs. 20739

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y Static capacitors which are electrical devices without moving parts that have the ability toprovide magnetizing current to the load. Their efficiency is high since losses are less than

one-half of 1 percent of their kVAC (or kVAR) rating.

In the past two other types of equipment were used to supply kVAr to a system: synchronous

condensers and synchronous converters. Both of these machines have been replaced

principally by the use of static capacitors.

Static capacitors are the cheapest and the simplest means for reactive power compensation .

They are installed by power utilities in the transmission and distribution network and also at

the consumers premises on to different loads such as motors, transformers, incoming supply,

etc. In present days automatic switching of the capacitors enables keeping a high power

factor for heavily fluctuating loads as well.

1.7 kVAr Compensation or Power Factor Correction

With the present scenario of power crisis it has become all the more important that whatever

energy is available, it should be used most optimally. For these reasons proper management

of the reactive power is all the more significant. In a power system, it is important that the

power utilities and the consumer both work together for providing the reactive power

compensation. The power utilities have to provide reactive compensation for the transmissionsystems reactive power requirement, reduction of line losses and improved voltage

regulation. The consumers have to compensate for the additional reactive power requirement

by the loads at their installations.

The power utilities take a number of steps for installation of reactive power compensation

equipment. These include:

33 kV series compensation equipment.

220 kV series compensation equipment.

Synchronous condensers.

33 kV shunt capacitors.

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11 kV and LT shunt capacitors.

Static VAR compensation equipment.

Further steps are continuously being taken by the utilities to make their system healthy and

for providing additional compensation based on the system studies carried out.

Based on the reactive power requirement at their installations, the consumers have to provide

for the necessary reactive compensation at their end to achieve the minimum power factor

level prescribed by the utility (0.9). The most economical and reliable method of reactive

compensation is the installation of power capacitors. Lagging power factor can be corrected

by connecting capacitors in parallel with the system. The current in a capacitor produces a

leading power factor. Current flows in the opposite direction to that of the inductive device.

When the two circuits are combined, the effect of capacitance tends to cancel that of the

inductance.

Most customer loads (particularly motors, but many lighting circuits also) are inductive. A

low power factor can generally be corrected by connecting appropriate capacitors. This is not

the case if low power factor is caused by harmonics, in which case the installation of

capacitors will not help, and may cause a serious problem. In high harmonic situations,

expert help should be obtained before attempting to correct power factor problems.

kVAr component before PF correctionResultant kVAr component after correction.

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Note the smaller resultant with the capacitors added, but real power (kW) does not change. A

properly chosen capacitance value will neutralize the inductance and produce unity power

factor. Too little or too much capacitance must be avoided. Too little will not correct to a

sufficiently high power factor to eliminate a surcharge, and too much will result in leading

power factor with possible undesirable effects.

2.Optimum point for Power SupplyFor 400KVA load we need to feed the premise with medium voltage overhead lines with a

distance of 150 m away from the 300KVA Load.

The cable cross section used is 2x (3x240) and thus we will be reducing line losses.

The utility breaker has to be placed according to the electric moment method. Refer to the

figure in Appendix A.

2.1 Overhead Lines

In overhead lines, we have three components:

y Conductors: cross- section and number of conductors (there is usually five conductors-three phases, 1 neutral and one for the municipality lighting). Conductors could be bare

copper or aluminum alloy (mostly alloy because Al doesnt have enough mechanical

strength to withstand the mechanical stresses) or insulated preassembled aerial cable for

MV.

y Support:a. Steel: Various types are present:

i. I- beam: has a lot of weight but takes a very small space for the sameconductors sizes; a higher weight/km is used.

ii. Lattice: Weight/ Km for the same conductors is smaller but larger basearea is needed.

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b. Wood: Special types of wood. Classified according to :i. reputing capacity at the top (how much pull I put on the top of the pole to

break it)

ii. Length of the poleiii. Diameter of the top/ bottom

c. Concrete: It must have special forms (rectangular or square form). Steel barsinside them are sized and then molded into concrete slaps.

On each angle put a wedge, and later install a pole (more stresses).

Partial distance between two wedges must be marked.

The larger the angle the stronger is the support required.

To increase the height of the line above ground, we have to increase either the height of the

supports or increase their number. As the span increases then higher mechanical stresses

(stronger support is needed).

2.2 MV Lines

MV lines have four basic components:

y Conductors :Conductors can be copper as per our case or aluminum alloy.

y Supports:The support can be steel or toughened glass or porcelain.

y Insulator chains and accessories:Insulation could be suspended or rigid

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y Base of supportSupport steel is used in lattice form. We will not use I beam because it is very costly and

we need high Kg of support/ Km.

The connection is considered from M.V. (B.B.) through a M.V. cubicle and the sealing

end (Termination) and it is housed normally in a cubicle.

3.Distribution Sub-StationsAfter the overhead line we need a Distribution Sub-Stations (DSS). We would obviously

save in cable sizes and hence cost, obtain a better voltage regulation, and thus save losses.

Its function is to transfer power from MV to LV. It is composed basically of:

yMV switch gearyLV switch gearyMV/LV transformersyMV busbarsyLV busbarsy

Accessories

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The MV switch gear installed on the nearest accessible steel-pole. The pole supports the

transformer and its accessories, MV feeders, LV feeders & switchgear and panel board.

Anchoring clamp will have to withstand the tension of the MV feeder as it hooked to the

pole. It supports the MV feeder. A grip clamp to elongate the live conductors from the

grounded parts of the DSS. A chain of 3 insulators are used to isolate the anchoring clamp

from the support. Spark gaps are used for protection of the insulators and the transformer

from voltage surges; its insulation is lower than that of the chain so that itt can divert the

surges to ground. Aluminum- Copper joint is used when the line conductors are Aluminum

alloy and transformer windings are made of copper, to avoid voltaic effect. Transformer is

composed of tank, cooling /insulating oil, steel core, MV and LV windings., and MV and LV

brushings (insulators + transversal conductor through it)

The transformer has a built in hook, casted or welded on its tank. This hook is clamped on a

steel bar not less than 3.5cm. in diameter, firmly anchored to the support.

A UPN beam, about 1m long and up to 1ton capacity having a hook on its end to which a

mobile tray could be anchored.

LV feeder goes down from LV winding to the panel board. LV switch gear (a CB housed in a

DSS LV panel board). A minimum of 2 LV feeders to supply the local LV network with 5

conductors. Concrete platform is used for man oeuvre and repair and exploitation. A 25mm2

bare copper conductor or steel equivalent is used for the grounding of the pole.

3.1 Built-in Sub-station

Built in area or where the transformer is larger than 400 KVA (to give the power

required). Most important is the one used in Beirut. The built-in area (has to have housing

i.e needs a room) used to house the DSS is not included in the covered area permitted by

the municipalities of Beirut or the city planning in the region.

3.2Housing the TransformerWe need a minimum of 16 m

2(not included in the covered up area in the city planning) built

in area for a normal DSS. It is composed of:

y Medium voltage switch gear with the incoming and outgoing feeders

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y Transformers with auxiliaryy LV switch gear and feedersy Housing- support

Termination is the junction between the end of the feeder and the switch gear. This ceiling is

applied to medium and low voltages. It must be covered (ambient proof cover) to prevent

entrance of dust, humidity, rats, and to prevent cockroaches to eat the insulation. Joints are

the potential source of faults, so they are used when necessary only.

A minimum of 3 cells or cubicles (on the MV side because if we want to do any

maintenance, it is used to protect personnel) that have the following dimensions

a.

90cm widthb. 3m height

Partitions are made of steel.

Transformer is installed on a pedestal (hollow seat) where the transformer is fixed- it is an

opening with a volume several times that of the transformer oil, then it is put on a railed

beam in order to move/ roll it easily)

Pedestal has a connection to the sewage system (fill the bottom with material that is

inflammable and protect against burning oil). They also allow for ventilation.

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Appendix A:

Figure 1

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Figure 2

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Appendix B:

Meeting Minutes (1) of EECE 475 Design Project 6Date: O

ctober 20, 2010

Location: Professor Chahines Office Bechtel 324A

Attendees:

Prof. Chahine, Tarek Ibrahim, Jad Karaa, Rand Kassamany, Sabine Khalil

Absence:None

Agenda:

1. The meeting was called to order on Wednesday, October 20 at 11:30 am byProfessor Chahine.

2. A primary sketch of the design was drawn and discussed by Prof. Chahine andthe group members.

3.

The following questions were raised:y Do we have to leave the loads at their current consumption?y Should we choose an individual or separate power factor correction?y How can we supply the loads and which is the best alternative?y Where are M.V. and L.V. used?

4. All members agreed upon seeking information from the utility to give us thebasic information about the prices.

5. The meeting was adjourned by Prof. Chahine at 12:05 pm.

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Meeting Minutes (2) of EECE 475 Design Project 6Date: October 27, 2010


Attendees:


Absence:None

Agenda:1. The meeting was called to order on Wednesday, October 27 at 11:40 am by

Professor Chahine.

2. The minute of Wednesday October 20 were approved unanimously.3. The choice of individual or central compensation was discussed.4. A sketch of the main panel along with the loads and the CB were drawn and

discussed.

5. All group members agreed to go to the utility within the coming week to gaininformation about the role of the utility in supplying, whether to use medium

voltage or low voltage, and whether to use overhead or underground live.

6. connected capacitors have to be used, but are they connected locally orcentrally?

7. The meeting was adjourned by Prof. Chahine at 12:05 pm.

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Meeting Minutes (3) of EECE 475 Design Project 6Date: November 10, 2010


Attendees:

Prof. Chahine, Jad Karaa, Rand Kassamany, Sabine Khalil

Absence:Tarek Ibrahim

Agenda:8. The meeting was called to order on Wednesday, November 10 at 11:30 am by

Professor Chahine.

9. The minute of Wednesday October 27 were approved unanimously.10.A discussion of the line losses was held. It has to be taken into consideration

while compensating the power factor.

11.If the power factor has to be taken centrally, where shall the capacitor bankbe placed?

12.The following points are to be discussed in the utility:y Where will the utility give us, centrally or individually?y Where to put the panel board and how to feed the loads?y The types of cables used and how to lay the cables?y The voltage drop?

13.The members agreed that the power factor correction will reduce the powerlosses and the cables used will be of smaller cross-section hence losses are

reduced.

14.The meeting was adjourned by Prof. Chahine at 11:50 am.

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Meeting Minutes (4) of EECE 475 Design Project 6Date: November 10, 2010


Attendees:


Absence:None

Agenda:

15.The meeting was called to order on Monday, November 29 at 11:00 am byProfessor Chahine.

16.The minute of Wednesday November 10 were approved unanimously.17.A discussion of the utility findings was held.18.It depends on us where to place the room of electricity, and what is placed

inside the premises would be on our expense.

19.We have to use a MV source; it has to be overhead and not underground. Thefirst 200m is on the utility.

20.We have to take the rules that govern the utility into consideration and how itapplies on private property with the right of way.

21.The electric moment was done:300x = 100(60-x) 300x = 6000 100x

x = 15m22.A discussion of the cables was held. If we chose underground, the the cost

would roughly be 10 times that of overhead.

23.While calculating the power factor, the losses on the 60m has to beconsidered.

24.The meeting was adjourned by Prof. Chahine at 11:30 am.

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Appendix C:

References:

[1] Grondzik, Kwok, Stein, & Reynolds. Mechanical and Electrical Equipment for Buildings.11

thEdition.

[2] EECE 475 Draft

[3] EDL

Documents

475 project