investigation of the benefits of energy saving device

8/6/2019 Investigation of the Benefits of Energy Saving Device

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EE-372 FUNDAMENTALS OF POWER SYSTEMS

Project-1 Investigation of the Benefits of Energy Saving Device

B.Sc. Students Involved in This Project:

Mehmet Serdar Teke ID: 260701050

hsan Demir ID: 260701003

Submitted to: Prof. Canbolat Uak

Submission Date: 19.02.2010

Department of Electrical and Electronics Engineering

Engineering and Architecture Faculty

Yeditepe University

2010


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i

TABLE OF CONTENTS

TABLE OF CONTENTS.i

LIST OF FIGURESii

1. The Goal of the Project.......1

2. Introduction.....1

3. Calculations.........3

a. Without Energy Saving Device......4

i. Money paid when machines are working continuously for a month.....5

ii. Money paid for losses in the house for a month...5

iii. Money paid by the utility for the losses between transformer and wattmeter for a

month..5

b. With Energy Saving Device..........5

i. Money paid when machines are working continuously for a

month..............6

ii. Money paid for losses in the house for a

month......6


month..6

4. Comparison of Losses With and Without Energy Saving Device......7

5. Conclusion......7

6. How We Shared the Work as a Group....8

7. Your Opinion About This Type of Projects....8

8. References...........8


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ii

LIST OF FIGURES

Figure 1. Figure 1.An Electricity Bill.....1

Figure 2. Figure 2.The Nameplate of the Refrigerator...

.2

Figure 3. Figure 3.The nameplate of the Washing Machine...2

Figure 4. The Circuit model for the Power System without Energy Saving Device......3

Figure 5. Circuit model for the Power System with Energy Saving Device......4


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1. The Goal of the Project

In this project, it is aimed to discuss whether energy saving device is useful to decrease

the electricity bill by decreasing energy consumption or not. Also, it is targeted to prove if this

device is working or not by making the necessary calculations and using circuit models whichdescribe machines in a house connected to the distribution line.

2. Introduction

In order to achieve the goal mentioned in the previous part, a set of calculations has to

be made. In these calculations, an electricity bill must be used since there exists the price of 1

kWh electricity in this bill. We have to use this price to calculate the money paid whenrefrigerator and washing machine are working continuously for a month after the power

consumption of these machines is calculated by using a circuit model for the power system. In

figure 1, there is shown an electricity bill of August 2009 when the price of 1kWh electrical

energy was 0.162684 TL. In this figure, there also exists the price of 1Wh electrical energy

loss between wattmeter and distribution transformer which was equal to 0.001703 TL inAugust 2009.

Figure 1.An Electricity Bill

Also, the nameplates of a refrigerator and a washing machine are required to determine

the power consumption of each since the nameplates have this value on themselves. These

values are to calculate the total power consumed in a house during one hour.


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Figure 2.The Nameplate of the Refrigerator

In figure 2, there is a nameplate of a refrigerator on which its energy consumption per

year is written. It is needed to be converted to power which means the energy consumption

per hour.

325kWh/year x 1 year/8760hour = 0.037kW/h

Thus, the energy consumed by this refrigerator in one hour is 0.037kW which describes

the power consumption of this refrigerator.

In figure 3, the nameplate of a washing machine is shown. In this nameplate, it is seen

its energy consumption per hour which means its power consumption. Therefore, there is no

need to convert it into power unit unlike refrigerators nameplate. As it is seen in this figure,its power consumption is 0.95kW per cycle.


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Figure 3.The nameplate of the Washing Machine

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After having said these, there is a need to discuss about active and reactive power. In

reference [1], it is said that the active power depends on the cosine function and the reactivepower on the sine function where the angle is taken between the voltage and the current which

means the phase difference between current and voltage. Also in reference [1], it is told about

that when the average in time is calculated, these two types of power behave differently; the

average active power is well defined and not equal to zero whereas the average reactive power

is zero, no matter of the network or state of the power system. In this statement, it might beproven that the work is done by active power since it has some value and not zero and that no

work is done by reactive power. Thats why reactive power is also called the imaginary

power. The article in reference [1] also says that if there exists a net flow from one point of

the network to another for active power, for the reactive power there is a continuously flow

back and forth, but the net flow is zero for a complete cycle, as the amount of energy flowing

in one direction for half a cycle is equal to the amount of energy flowing in the opposite

direction in the next half of the cycle. By this statement, since reactive power flows back and

forth in the transmission line none of this type power is consumed by the machines as the netpower flowing back and forth is equal to zero. Depending on the statements claimed in

reference [1], the main difference between active and reactive power is that active power

executes the work while reactive power does not execute the work at all, it just cycles in the

power system.

While coming to the end of introduction, some words need to be said about energysaving device. The companies who produce this device claim that when this device is

connected somewhere near the wattmeter in the house, it reduces the electricity bill by 40%. It

should not be forgotten that the active power is measured by the wattmeter and the money ispaid for this type power. Therefore, the calculations are required to be made depending on the

active power to investigate if it reduces the electricity bill by 40% or not. Also, it must be said

that this device consists of only one capacitor.

3. Calculations

Before making the calculations, a model for the power system might be useful for thecalculations so that it is tried to create a model for this system. This model is shown in figure

4 and 5.


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Figure 4.The Circuit model for the Power System without Energy Saving Device4

Figure 5.Circuit model for the Power System with Energy Saving Device

In these circuits, Pmeasured is the power which is measured by the wattmeter, R is theresistance of copper wire coming to the house, Rin is the resistance of copper wire inside thehouse. Refrigerator and washing machine are modeled with one coil and one resistor simply

since they have a motor inside themselves and their internal resistances. Depending on this

model, the calculations are made in the following of the report.

Resistivity of copper: Cu=1.77 x 10-9.m

Length of copper wire outside the house: l1=300mLength of copper wire inside the house: l2=20m

Cross-sectional area of copper wire outside the house: A1=50mm2

=50 x 10-6

m2

Cross-sectional area of copper wire inside the house: A1=2.5mm2=2.5 x 10-6m2

R = (Cu x l1) / A1 = (1.77 x 10-9.m x 300m) / 50 x 10-6m2 = 10.62 x 10-3

Rin = (Cu x l2) / A2 = (1.77 x 10-9.m x 20m) / 2.5 x 10-6m2 = 14.16 x 10-3

a. Without Energy Saving DevicePower factor = 0.7 => cos = 0.7 => = 45.57o

To calculate the current through main line, Tellegens Theorem is used, which says the

total active power in the circuit is equal to zero.


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-220 x Irms x cos + 2 x (R + Rin) x I2rms x cos + Prefrigerator + Pwashing machine = 0

Prefrigerator= 37W (written on its nameplate)

Pwashing machine = 950W (written on its nameplate)

-154 x Irms + 2 x 24.78 x 10-3 x I2rms x 0.7 + 37 + 950 = 0

34.692 x 10-3 x I2rms 154 x Irms + 987 = 0

Here, there exists a 2nd degree equation that needs to be solved. By solving this

equation;

Irms1 = 4432.6 A => this cannot be the solution since it is extremely high.

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Irms2 = 6.418 A => this is the solution to the current flowing in the main line.

i. Money paid when machines are working continuously for a month

Pmeasured = 2 x Rin x I2rms x cos + Prefrigerator + Pwashing machine

Pmeasured = 987.816 W

Total energy consumed in one month:

Ehouse = 987.816 W/h x 24h/day x 30day/month = 711228Wh = 711.228kWh

For one kWh, 0.162684 TL is paid according to the bill. Using this:

Money paid for the consumption in the house = 711.228 x 0.162684 = 115.7 TL

Now, money paid for the consumption in the transmission line needs to be found.

PR+ PR= 2 x PR= 2 x (6.418)2 x 0.01062 x 0.7 = 0.612W

In one month;

0.612 x 720 = 440.64Wh

By multiplying this with 0.001703 TL, money paid by the utility for the losses between

transformer and wattmeter is obtained;

Money paid for the losses between transformer and wattmeter = 0.75 TL

Total of them results in money paid when machines are working continuously for a month:

Money paid when machines are working continuously for a month = 115.7 + 0.75 =

116.45 TL

ii. Money paid for losses in the house for a month

Ploss = 2 x Rin x I2rms x cos = 0.81656W = 0.00081656kW

Money paid for losses in house = 0.00081656 x 24 x 30 x 0.162684 = 0.0956 TL


month

This was done in i to find the total money paid for a month: Money paid for the losses between transformer and wattmeter = 0.75 TL


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b. With Energy Saving DevicePower factor = 1 => cos = 1 => = 0o

To calculate the current through main line, Tellegens Theorem is used, which says thetotal active power in the circuit is equal to zero.

-220 x Irms + 2 x (R + Rin) x I2rms + Prefrigerator + Pwashing machine = 0

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Prefrigerator= 37W (written on its nameplate)

Pwashing machine = 950W (written on its nameplate)

-220 x Irms + 2 x 24.78 x 10-3 x I2rms + 37 + 950 = 049.56 x 10-3 x I2rms 220 x Irms + 987 = 0

Here, there exists a 2nd degree equation that needs to be solved. By solving this

equation;

Irms1 = 4434.6 A => this cannot be the solution since it is extremely high.

Irms2 = 4.491 A => this is the solution to the current flowing line.

i. Money paid when machines are working continuously for a month

Pmeasured = 2 x Rin x I2

rms + Prefrigerator + Pwashing machinePmeasured = 987.571 W

Total energy consumed in one month:

Ehouse = 987.571 W/h x 24h/day x 30day/month = 711051Wh = 711.051kWh

For one kWh, 0.162684 TL is paid according to the bill. Using this:

Money paid for the consumption in the house = 711.051 x 0.162684 = 115.67 TL

Now, money paid for the consumption in the transmission line needs to be found.

PR+ PR= 2 x PR= 2 x (4.491)

2

x 0.01062 = 0.428WIn one month;

0.428 x 720 = 308.16Wh

By multiplying this with 0.001703 TL, money paid by the utility for the losses between

transformer and wattmeter is obtained;


Total of them results in money paid when machines are working continuously for a month:

Money paid when machines are working continuously for a month = 115.67 + 0.525 =

116.195 TL


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ii. Money paid for losses in the house for a month

Ploss = 2 x Rin x I2rms = 0.5712W = 0.0005712kW

Money paid for losses in house = 0.000 5712 x 24 x 30 x 0.162684 = 0.0669 TL


month

This was done in i to find the total money paid for a month:


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4. Comparison of Losses With and Without Energy Saving Device

When we take a look at the money paid for the losses, it is less with energy savingdevice. However, very little money paid on the losses due to copper wire. This is because

copper is a very good conductor and therefore has a very little resistance.

Ploss without energy saving device = 0.81656W + 0.612W = 1.42856W

Ploss with energy saving device = 0.5712W + 0.428W = 0.9992W

Money paid for losses without energy saving device = 0.0956 + 0.75 = 0.8456 TL

Money paid for losses with energy saving device = 0.0669 + 0.525 = 0.5919 TL

As seen above, money paid for the losses is nothing when compared to the total money

on the bill. Losses with energy saving device is less because the current through copper wire

decreases from 6.418 A to 4.491 A when energy saving device is connected near wattmeter.

There is a remarkable current decrease, but it does not reflect on the bill.

Decrease in current by percent = [(6.418 4.491) / 6.418] x 100 = 30.02%

5. Conclusion

The companies selling energy saving device claim that this device reduces electricitybill by 40%. In this report, it is proven that it does not reduce the electricity bill that much,indeed it is not even close to this number. It reduces the electricity bill by:

[(116.45 116.195) / 116.45] x 100 = 0.219%. (data from i parts of calculations)

The decrease in the electricity bill is almost zero as it is calculated above. There is a high

decrease in current, but this decrease in current only reduces the money paid for loss, which is

very little and it does not affect the total money paid for electricity. This device saves only

0.8456 0.5919 = 0.2537 TL per month although they are selling this device with a much

higher price, about 50-100 TL. With a basic math, about 200 months must pass for the

consumers to overcome this value.


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In this report, energy saving device is considered with 100% efficiency as power factor

with energy saving device is taken 1 which means no reactive power is consumed. This is

because when the power factor is 1 the angle between current and voltage is equal to 0o which

means they are in the same phase and since sin0o is equal to zero, no reactive power is

consumed. In practical, 100% efficiency is impossible but this device seems like it can

decrease reactive power. However, reactive power is not measured at homes. Therefore, it is

not useful in houses again. The reactive power is measured in factories, so they can sell thisdevice to them but most of the factories are making this compensation themselves. Therefore,

they do not have a market place in factories either.

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6. How We Shared the Work as a Group

The electricity bill needed to learn how much money is taken for 1kWh was obtainedby hsan Demir. The photos of the nameplates of the refrigerator and the washing machine

were taken in a shopping center by Mehmet Serdar Teke. The circuit models are considered to

make calculations easier by both members of the group. These circuit models were drawn on

the computer by Mehmet Serdar Teke. The knowledge about this topic is gained from the

internet, notebook of EE372, and people who are interested in this topic such as a student who

has a friend working in a company producing energy saving devices by both members of thegroup in harmony. The calculations are executed by both members of the group. After making

calculations, comparison of losses with and without energy saving device and conclusion

parts are considered carefully by both members of the group. After making all of these, thereport is written and edited by Mehmet Serdar Teke.

7. Your Opinion About This Type of Projects

This project enables us to learn what is going on in the field of power system

technologies and reminds us not to allow sellers to cheat consumers who do not know the

reality. The reality must be shown them by electrical engineers who have a big responsibility

to show the right way to the public. Also, this project helps us understand the concept of

active and reactive powers. Therefore, in some way it has prepared us to the exam.

8. References

[1] http://www.el.angstrom.uu.se/kurser/water05/Reactive_Power.pdf

investigation of the benefits of energy saving device

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