Electrical & Electronics Technology

S.B.A. (CAPE)

Model Power Line

Name: Tarick Edwards

Date: 06 Feb 2009

Lecturer: Mr. A. Clarke

Reg#: 1000101042

Territory: Jamaica

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Table of Content

Acknowledgement ………………………………………………………..2

Purpose of the report…………………………………………………….3

Aim……………………………………………………………………….……….4

Scope& Specification……………………………………………………..5

Theory……………………………………………………………………………6

Methodology……………………………………………………………......8

Design & Construction detail...…………………………………..….9

Summary of Tests and Troubleshooting procedure…….…18

Tests and Troubleshooting Results………………………………..19

Discussion……………………………………………………………………..20

Constraints & Difficulties/limitations………………………....…21

Conclusion/Recommendations…………………………………..…22

1

Acknowledgement

I sincerely Mr. Clarke guide us in construction of the project, and also in the

assembling of the information gathered in the correct format. I also wish to give

thanks to the pass student for their marvellous pass project.

2

Purpose of the report

The purpose of this project is to satisfy the requirements of CAPE Electrical and

Electronic Technology.

3

Aim

The aim of this project is to construct a model AC power line at high voltage in order to investigate the power losses in the transmission lines. The AC model will deliver up to 12 volts at the rated output. Also a DC power line to be compared with the AC lines.This project also gives us a lot of practical experience working with the device we leant by theory.

4

Scope and Specifications

This project is based on the creation of a Power Transmission System. This Power Transmission should entail: Two Transformers, one bulb, Length Constantan wire

Transformers

Step down Transformer – 110V/12V 50Hz, KVA

Step up Transformer – 12V/240 50Hz, 4KVA

Step down Transformer – 240V/12V 50Hz, 4KVA

Wire

Constantan Wires length – 1m each

Daimeter-0.32mm

Resistivity- 49x10-8

Wire gauge- 30 SWG

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Theory

Power lines are overhead wires supported by high towers that transmit electric energy from power supplies or plants. The centre strands of power lines are made of steel to give them strength and the outer strands are made of aluminum because of its lightness and ability to carry current. The wires are insulated from the towers by porcelain insulators to prevent the loss of electric energy. Electric power transmission is a process in the delivery of electricity to consumers and also the bulk transfer of electric power.

D.C. power transmission is the distribution of power using direct current. Direct current can be transmitted at high voltages but to lower it for consumers would be costly.

A.C. power transmission is the distribution of power using alternating current. A.C. voltages has the advantage of increasing and decreasing in value more readily than that of the (D.C).Transmission efficiency is improved by increasing the voltage using a step up transformer which reduces the current in the conductors while keeping the power transmitted nearly equal to the power input. The reduced current flowing through the conductor reduces the losses in the conductor and since, according to Joule's Law, the losses are proportional to the square of the current, halving the current makes the transmission loss one quarter the original value.

In alternating current systems, energy loss across power lines is reduced because transformers make it possible to raise the A.C. voltage to very high values. These high voltages allow the same level of electric power made available at a lower current. This results in less power loss, smaller transmission cables and higher efficiency. In addition to stepping up or raise the voltage for long distance transmission, transformers also step-down or lower the voltage to the requirements of the load. PLoss = I2R

The transformer is based on two principles: firstly that an electric current can produce a magnetic field (electromagnetism) and secondly that a changing magnetic field within a coil of wire induces a voltage across the ends of the coil (electromagnetic induction). By changing the current in the primary coil, it changes the strength of its magnetic field; since the changing magnetic field extends into the secondary coil, a voltage is induced across the secondary.

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Diagram showing a simplified transformer design

A current passing through the primary coil creates a magnetic field. The primary and secondary coils are wrapped around a core of very high magnetic permeability, such as iron; this ensures that most of the magnetic field lines produced by the primary current are within the iron and pass through the secondary coil as well as the primary coil.

The voltage induced across the secondary coil may be calculated from Faraday's law of induction, which states that:

where VS is the instantaneous voltage, NS is the number of turns in the secondary coil and Φ equals the magnetic flux through one turn of the coil. If the turns of the coil are oriented perpendicular to the magnetic field lines, the flux is the product of the magnetic field strength B and the area A through which it cuts. The area is constant, being equal to the cross-sectional area of the transformer core, whereas the magnetic field varies with time according to the excitation of the primary.

Methodology and Approach

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1) The information on Power Transmission Systems was gathered2) The value of various circuit components was calculated (e.g. resistance of wire).3) The equipment and components needed were then gathered and assembled.4) The current and voltage readings of both circuits were measured (for the 2V D.C. and

12V A.C.)5) The current and voltage readings of the lamps and the voltage across the transmission

line were recorded.6) The power losses were then calculated.

Design and Construction details

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These Materials were gathered: 12-240 step up Transformer, 240-12 step down Transformer, 110-12 step down transformer, switch (SPDT), 12 lamp, 2 Piece 1m constantan wire, multi-meter, 12 DC supply

12V D.C.

The lamp was connected across the 12V D.C. Supply as shown in fig.1

Source12 V

Lamp12V, 50W

R4

4.99

12

R1

4.87

12V A.C.

1) The lamp was connected to the secondary winding of the (110V-12V) transformer.2) The primary winding of the transformer is connected to a 12V A.C. Supply. 3) (step down transformer, 110V/12V) as shown in fig.2

Source120 V 50 Hz 0Deg

Lamp12V, 50W

Transformer

NLT_PQ_4_120

1

5R2

4.99

46

R1

4.87

3 2

240V A.C.

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1) The 110V/12V transformer was connected to one of the bulbs

2) The bulb was then connected across a 12V/240V transformer

3) The transmission line was then connected to the transformer

4) The other ends of the transmission line was then connected to a 240V/12V transformer

5) The transformer was then connected to a 12V AC power supply.

(Shown in fig.3)

Source12 V 50 Hz 0Deg

Lamp12V, 50W

R2

4.99

Transformer1

NLT_PQ_4_12

1

4 6

Transformer2

NLT_PQ_4_12

2

38

R1

4.87

57

Calculations

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12 Volt AC Supply

Expected Values

Expected Resistance of Transmission lines

R= ρl/A

R = 49 x 10 -8 x 2m

8.042 x 10-8m2

R = 12.19Ω

The Area of constantan wire

A = Π r2

A = Π x (0.16 x 10-3)2

A = Π x 2.56 x 10-8

A = 8.042 x 10-8m2

Expected Current

I = V/ RT

I = 12 / 15.07

I = 0.80A

RT = 12.19 + 2.88 = 15.07 Ω

Expected voltage Drop in Lines

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Side Notes / Key

ρ –Resistivity of constantan- 4.9x10-7 Ωm

Diameter of wire used - 0.32 x 10-3 m

Radius of wire – 0.16 x 10-3m

Expected Bulb Resistance – 2.88 Ω

Expected Vdrop in transmission Lines(9.75v) Expected

V supply -12v

Expected wire resistance (R Line) - 12.19Ω

Actual Bulb Resistance - 0.6Ω

Actual Current Supplied - 1.03A

Actual Resistance of Lines - 9Ω

Actual voltage supplied - 11.28v

Actual voltage across lamp - 2v

Power delivered = Output Power -2.06W

Power Supplied = Input Power- 11.62W

V = I x R line

V = 0.8A x 12.19 Ω

V = 9.75v

P Loss Expected =( V drop )2 / R Line

P Loss = (9.75)2 / 12.19

P Loss = 7.80W

Expected Voltage at Load

V Lamp =V supply - V drop in Lines

V Lamp = 12 – 9.75

V Lamp = 2.25v

Measured Values

Power loss measured in Transmission Lines

P Loss measured = I2R

P Loss measured = 1.032 x 9Ω

P Loss measured = 9.55W

Power Measured at Lamp/Load.

P Delivered to Lamp= I Supplied x V Delivered to lamp

P Delivered to Lamp = 1.03A x 2V

P Delivered to Lamp = 2.06W

Measured Input Power

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P Generated = I Supplied x V supplied

P Generated = 1.03A x 11.28V

P Generated = 11.62W

%η = P Delivered x 100

P Generated

%η = 2.06 x 100

11.62

%η = 17.7

Where η is the Efficiency.

12 Volt Dc Supply

Expected Values

Expected Current

I = V/ RT

I = 12 / 15.07 = 0.80A

RT = 12.19 + 2.88 = 15.07 Ω

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Side Notes for calculations

Total Resistance - 15.07Ω

Voltage Supplied - 12v

Current in circuit 1.12A

Actual voltage across lamp 2.23v

Expected wire resistance Power -2.24W

Expected voltage Drop in Transmission Lines

V = I x R Line

V = 0.8A x 12.19 Ω

V = 9.75v

P Loss Expected =( V drop )2 / R

P Loss = (9.75)2 / 12.19

P Loss = 7.80W

Expected Voltage at Load

V Lamp =V supply - V drop in Lines

V Lamp = 12 – 9.75

V Lamp = 2.25v

Measured Values

Power measured in Transmission lines

P Loss measured = I2R

P Loss measured = 1.122 x 9Ω

P Loss measured = 11.28W

Power Measured at the Load

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P Delivered to Lamp = I Supplied x V Delivered to lamp

P Delivered = 1.12A x 2V

P Delivered = 2.24W

Input Power

P Generated = I Supplied x V supplied

P Generated = 1.12A x 12V

P Generated = 13.44W

%η = P Delivered x 100

P Generated

%η = 2.24 x 100

13.44

%η = 18.15

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High Voltage Transmission (110V AC)

Expected Values

Expected power loss

P loss = (Vdrop) 2 /R line

P loss = 2.072/12.19

= 0.352w

Expected current in transmission line : V p / Vs = Is /I p

9/110 = Is/2

Is = 0.082 x 2

Is = 0.164A

* Is is also the voltage transmitted to the Transmission lines

Expected voltage drop

Vdrop = I x R line

= 0.164 x 12.19

= 2V

– Expected = 107.93V

Expected V After T line = Expected V bef. T line – Vdrop in line =110- 2.07

= 107.93V

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Side Notes for Calculations

V p – Primary voltage in Step up Transformer 9V

I p – Primary Current in Step up Transformer 2A

Vs – Voltage in secondary of Step up Transformer 110V

Is – Current in Secondary of Step up Transformer = 0.164A

Expected Current in Lines = Is =0.164A

Measured Values

Measured V bef. T line = 45.8V

Measured VAfter T line = 43.8

Measured Voltage drop in Transmission lines

V drop = V before T lines – V after T Lines

V drop = 45.8 – 43.1

V drop = 2.7v

Measured current in lines – 0.17A

Power Loss Measured

Ploss = I2R

Ploss = 0.172 x 9Ω

Ploss = 0.260W

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Testing & Troubleshooting Procedures 

1) The Multimeter was used to test the Resistance of the Length of Constantan. 2) It was then used to test the current and voltage through/across each bulb for

the D.C. Supply.3) The digital multi-meter was also used to measure the voltage and current

across/through the bulb when connected as shown in the schematics for the A.C. Power Transmissions.

4)  The digital multi-meter was again used to measure the voltage across the power line in fig.3.

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Testing & Troubleshooting Results

12 Volt AC supply

Parameters Value Expected Value Measured

Error

Current 0.80A 1.03A 0.23Supply Voltage 12V 11.28v 0.72Voltage Drop 9.75V 9.76V 0.01Power Loss 7.80W 9.55W 1.75Resistance of Lines 12.19Ω 9 Ω 3.19

12 Volt Dc Supply

Parameters Value Expected

Value Measured

Error

Current 0.8A 1.12A 0.32Supply Voltage 12V 12V -Voltage Drop 9.75V 9.55V 0.20Power Loss 7.80W 11.28W 3.48Resistance of Lines 12.19 Ω 9 Ω 3.19

240v ac Supply with Mounted Transformer at Both Ends

Parameters Value Measured

Value Expected

Error

Transformer 1Primary Voltage 9.27v 9V 0.27Secondary Voltage 45.8v 110V 64.2Primary Current 2.14A 2.0A 0.14Secondary Current 0.17A 0.164A 0.006Power Loss in lines 0.26W 0.35W 0.09Voltage Delivered to user 1.85V 1.53V 0.32Current Delivered to user 1.58A 2A 0.42

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Discussion

The circuits were constructed using the required components; hence, our requirements were achieved. The resistances of both lengths of constantan wires were both taken and recorded. The resistance of the lamp was also measured and recorded.

The 12 AC circuit was constructed using a step down transformer (120V-12V). The primary windings of the transformer are connected to the power supply, and the secondary windings connected to two 1 meter length of constantan wire. The constantan wires are then connected to the lamp. Ideally, the reading on the secondary windings would be 12v, but it was measured to be 11.28V. The voltage drops across the constantan wires were measured. The current in the circuit was 1.03A; this is constant since it is a series circuit. The lamp had small illumination due to the fact that most of the voltage drops across the constantan wires. But it would definitely show better in a dark area. After the circuit was powered up for a while, the temperature of the constantan wires began to increase.

After completing the test of the 12V AC Circuit, the circuit was disassembled and reconstructed into a high voltage transmission line. With the use of two transformers refer to diagram below.

Source12 V 50 Hz 0Deg

Lamp12V, 50W

R2

4.99

Transformer1

NLT_PQ_4_12

1

4 6

Transformer2

NLT_PQ_4_12

2

38

R1

4.87

57

The voltage across the secondary winding of the transformer was measured. A voltage 43.1V was measured at primary end of transformer2. The voltage at the load was found to be1.81V. The voltage drop across the transmission line R1 and R2 were both measured; both result in a value of 0.6V. The current was then measured and a value of 0.17A was obtained. The circuit has the least power lost of all. The lamp light could be seen clearly. That is why Power Company transmits electricity at high Voltage and low current. It is also less expensive to transmit at low current (smaller wires can be used).

After completing the circuit illustrated of the circuit above. A 12V DC circuit was assembled. The 12V DC supply was connected across the 12V lamp. The voltage across the transmission line was measured and recorded to be 9.55V, it was assumed to be 9.75V. The circuit was open and a multi-meter used to test the current in the circuit, which was found to be 1.12A. This circuit has the highest power loss. Hence, the lamp did not light. That is why this is a poor method of transmitting electricity. It is expensive and inefficient to distribute due to high temperature, overheat in the wire. As experienced in the experiment. It is also difficult to convert from Dc to Ac.

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Constraints & Difficulties/ Limitations

1) It was difficult to source the constantan wire.2) The temperature of the room was constantly changing thus affecting the results. (Drawn

from ohms law).3) It was difficult to get a 12 volt AC supply. (An 110V to 12V step-down transformer was used).

4) The AC voltage fluctuated thus causing a variety of voltages and currents readings on the multi-meter.

5) The some of voltages, currents, resistance, and reading on the multi-meter never kept still and hence were estimated.

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Conclusion

 

The Power loss in a Power Transmission line is I2R, where I is the current through the power line and R is the resistance of the line. It can also be concluded that the larger the voltage being transmitted, the smaller the power loss. That is why power companies transmit electricity at high voltages. AC and also recommended over DC, because, AC transmission is less expensive transmit as I can be stepped up and down easily by a transformer. AC is also preferable because it is easily converted.

Recommendations

1) It is recommended that A.C. is used over D.C. for transmission as A.C. is less expensive to transmit as it can be stepped up by the use of a transformer.

2) It is also recommended that when transmitting electrical power, it should be transmitted at high voltages so as to limit the power loss.

 

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