Green Energy: A Study  

 

M. Vivek* and G. Rajkumar

Dr. M.G.R. University, Maduravyal, Chennai, Tamilnadu

*Corresponding Author E-mail: tmilselvi@gmail.com

 

ABSTRACT:

The places where the electrical energy sources can be taken naturally, that is from renewable sources, which are called as green energy. Energy is one of the most important in our life, without electrical energy we cannot do any work.  Now a day, most of the energy sources are consumed by us.  At the same time the fossil fuel is being consumed in large quantity. After 20 or 30 years the fuels like coal, petrol, may become extinct from this world.  So now we are forced to develop new energy sources to minimize our energy consumption.    

 

This paper deals about energy generation from the waste heat created, from the ash room in a rice mill boiler. The usage of boiler became one of the most important in industrial applications, like power plant, Dying unit, Rice mill.  

There are different ways for waste heat utilization, at previous systems concentrated only on flu gas.  Fly ash was ignored.  After research, the fly ash was found to have some amount of heat.  Then it was used for boiling water.  But the remaining heat was not considered.  After boiling water, the fly ash was cooled and disposed.  

 


INTRODUCTION:

1. OPERATION

The boilers generate steam, then that steam is used to boil paddy and to dry it.  These type of industries the main fuel for the boiler is husk or wood. To the fuel feeded through the fuel pipe it goes to the furnace.  The fuels have to be burned to convert the water into steam, the steam is used for our applications. The remaining ash will come out from another end of the boiler.  It is to be stored in a ash room.

 

2 .PROCESS IN ASH ROOM5.

The ash room figure is shown in figure – 1.

 

ASH ROOM DIAGRAM:

Figure-1

 

Ash room is used to collect burned and unburned materials from the boiler. Usually, the wastes have some temperature, that heat energy can be converted into electrical energy. For that, at the bottom of ash room we have to arrange parallel plates made of silver to form a solar panel, the plates receives heat energy from the ash and transmit to solar panel.  The solar panel converts heat energy into electrical energy.  It is to be stored in battery or any other sources. The system assembling figure is shown below,

 

Figure  - 2

3. HEAT TRANSFER 1,2,5.

Figure – 3

Now we use HEAT CONDUCTION OF FOURIER LAW

 

     Q =   KA  (   T1 – T2  )  /L                                         

         =   T1 – T2 / L / KA

Where;

L / K/ A -------------------R

R------Thermal resistance

K------Thermal conductivity

Q------Heat conduction

A------Surface area

L-------Wall thickness

 

Rate of heat transfer =        Temperature driving force

_____________________ 

Resistance

 

 

3.1. By practical the data's are collected from M/s. Bharat Modern Rice mill, Kaaranodai, Chennai.

Analysis data sheet 1

shift

Time

Off

On

Room temperature in ºC

Fly ash temperature in  ºC

Shift 1

1 – 2

2 – 3

3 – 4

4 – 5

5 – 6

6 – 7

7 - 8

-

-

-

-

-

 

 

 

 

 

 

on

on

 

 

 

 

 

65

53

 

 

 

 

 

210

180

Shift 2

8 – 9

9 – 10

10 -11

11- 12

12 -13

13 -14

14 – 15

15 - 16

-

-

-

-

-

-

 

 

 

 

 

 

 

on

on

 

 

 

 

 

 

60

55

 

 

 

 

 

 

220

160

Shift 3

16 -17

17 -18

18 – 19

19 – 20

20 -21

21 -22

22 – 23

23 - 24

-

-

-

-

-

-

 

 

 

 

 

 

 

on

on

 

 

 

 

 

 

55

48

 

 

 

 

 

 

180

140

Let us consider the

Ash room temperature 60 ºC - 80 ºC

Waste ash temperature 120 ºC

Consider over all temperature = 60 ºC

                                         T1   = 120 ºC

Assume that

Plate thickness                                    L1 =0.05m

Air medium gap                                                 L2= 0.05M

Thermal conductivity of silver plate           K1 = 419 w/m ºC

                                                                  (Refer data book)

Thermal conductivity of air                    K2 = 0.025w/m k                                                                                 (Refer data book)

Area A                                                     =0.5 m2    

Resistance R                                    =Ra + R1 + R2 + R3 + Rb

Conductivity heat transfer ha, Ra, ha, Rb and the resistant r3 we do know the value so that ignore it, because r3 is solar panel then

The resistance is R = R1 + R2

                          R1 =L1/ K1 A1         

                                     0.05

                                = _________        =0.0002398 m² ºC /w

                                  417*0.5

                          R2 =L2 / K2 A2    

                                    0.05

                               = ______                     = 4

                                0.025*2

Therefore the resistance is

R                             =4 + 0.0002398 m2 ºC /w

Then the total heat flow rate, Q is given by

Q                             = T1 – T2 /R

                                = 120 – 60/ 4

                                = 15 w/m²

Then the overall heat/ temperature is T1 = Q * R1

T1-T2                                                       =15 * 0.0002398

T2                                                             =119.99 ºC

T2 – T3                                                     = Q * R2

119.9 – T3                                                 = 60

 T3                                                             = 60 ºC

Then the solar panel will receive around 60 ºC

The temperature vs time curve is shown below

Time duration vs. Temperature curve

Figure - 4

 

 4. Solar panel Operation: 4,6,7.

Direct conversion of solar energy into electrical energy takes placed through solar cells using semiconductor materials.

 

4.1 DESIGN PROCUDER:

1. Preparation of TiO2 – coated glass

a. Obtain two glass plates and clean it with ethanol.  Do not touch the faces of the plates once they are cleaned

 

Figure - 5

b. Determine which side of each glass plate is conducting with a multimeter.

Put the glass plates side by side with one conducting side up and one conducting side down

c. Cover 1 mm of each long edge of the plates with scotch tape shown in figure 5

Cover 4-5 mm of the short edge of the conductive side up with scotch tape shown in figure 6.

 

Figure - 6

d. Add two drops of the white TiO2 solution on the conductive side up glass

e. Quickly spread the white TiO2 solution evenly with a glass pipette, sweeping first from the another slide, then sweeping the extra TiO2 coated glass on the hot plate, keeping track of where your plate is you will need it

2. Staining of the TiO2 coated glace face down in a Petri dish containing raspberry juice.  And preparing the Carbon coated glass plate.

a. Place the Tio2 coated glace face down in a Petri dish containing raspberry juice.  Soak for about 10 minutes shown in fig (f).  While it is soaking wash the other glass plate with ethanol.  Use the multimeter to find out which side is conducting.         

    b. Use a N'2 pencil to apply a thin carbon layer/carbon coated on the conductive side. of the glass plate is shown in figure 7

 

Figure - 7

 

4.2. ASSMBLY OF THE SOLAR CELL:

a. Remove the first glass plate from the juice, and abound it with deionized water, then with ethanol and dry with a tissue paper.

b. Place the carbon coated glass plate faces down on the TiO2 coated glass plate shown in figure

two glass plates must be slightly offset (5 mm) shown in figure – 8

figure 9

Hold the plates together with binder clips on each side of the longer edges

d. Add two drops of the iodide solution on an offset side and allow it to soak through. 

e. Alternately open and close each side of the solar cell by releasing.  Make sure that all of the stained area is contacted by the iodide solution.

 

4.3. MEASURING THE ELECTRICAL OUT PUT:

a. Fasten alligator clips to the two exposed sides of the solar cell to make and electrical contact shown in fig (k).  Attach the black (-) wire of the millimeter to the Tio2 coated glass plate.

b. Attach the red (+) wire of the multimeter to the carbon coated glass plate.

Place the solar cell on top of an overhead projector measure the current before and after the overhead projector has been turned on.  Measure the voltage before and after the overhead projector has been turned on.

c. To design the suitable panel for the above procedure for our applications.   

 

Now the panel will received around 60 celcicus.

Figure – 10  Time vs. Power

 

4.4. Solar array design:

We utilized power from the plant equipment then the output and current of pv array will be 120 watts load needed for 8 hours at 24V at the plant consumption.  Take mean horizontal insolation from standard map at height region level will take, because we do not consider the full temperature 60˚c.  We consider normal level of the temperature.  The overall heat received time is 6 hours. The size of the array requires data of mean daily insolation at the place of installation.

Mean horizontal daily insolation in kwh/m2=Number of peak sun shine hours. Consider

Here, the peak sun shine hours is the panel received peak temperature (Hpt). Take mean horizontal insolation in normal temperature to take maximum amount of the region 5.59 kwh/m².

 

 Hpt=Yearly insolation of wh/m²        

        ________________________

                          365

Consider the system losses to 20%, the system output can be computed as

 

System out put = Daily load in watt hours + 20% system loss

                          ___________________________________

                                  Panel received peak temperature

System current = System output

                           _____________

                            Voltage output

The photo voltaic cell is generally generate about 0.45 v at normal temperature and the current in full sunlight taken to be 270 amps/sq m.

Htp =Mean horizontal insolation* Number of days in a year

       =5.59 * 365             =2040.35 kwh/m² Yr

Number of peak temperature in hours

 hpt =2040.35 / 365       = 5.59

Load in watt hours/day =Number of watts * Number of hours utilization

                                      =120 * 8 = 960 watts hours

There fore the system output = 960+ 960 * 0.2  

                                              ____________      = 206 watts

                                                        5.59      

For 24 volt output then the system current is = 206

                                                                       ___  =8.58 amps

                                                                          24

Therefore the ampere range of the array is 8.58 amps

Array output considering the efficiency:

Taking lead acid batteries as energy storage medium is

Nbc  =0.9, Nsb =0.9, Nv =0.95

Array output =system output watt hours

                        _____________________

                              Nbc * Nsb * Nv

                            =           206

                               _____________                          =267.88

                                 0.9 * 0.9 * 0.95                         

At 24 volt the system current is         = 267.88  =11.16 amps

                                                              ______

                                                                  24

Number of solar cell is designed as        = Specified voltage

                                                                   ______________

                                                                       Current

                                                                 =24 / 0.45

                                                                 =53.3

Now we reach the desired value to design around 54 solar cells.

Temperature verses power out put analysis is shown in figure 11

Figure 11

5. CONCLUSION:

In this paper, concluded that, the system will reduce the earth heating. By use of waste heat from the Rice mill boiler and to generate new energy sources. Now it’s a added advantage to our power sector and energy managements units. To develop the system by reducing the electricity bill and increasing the fuel efficiency.  

 

6. ACKNOWLEDGMENT:

This work was supported by M/s Bharat Modern Rice Mill, Kaaranodai, Chennai.

 

7. REFERENCE:

1.       Yildiz bayazitoglu, M. Necatiozisik, ”Elements of heat transfer”. International edition 1988, Tata McGraw hill publishers.

2.        Dr.S.Ramachandran,”Heat and Mass transfer, 2004 edition, Airwak publication.

3.       Arora Domkundwar,”A course in power plant engineering with introduction to green house effect”, Danapathi Rai and co publishers.

4.       G.D.Rai,”Non conventional energy sources”, reprint 2001, Kannan publishers.

5.       P.K.Nag,”Power plant engineering steam and nuclear”, 1999 edition, Tata McGraw hill publication.

6.       G.D.Rai,”Solar energy utilization “Fifth edition, Kannan publishers.

7.       M.P.Agwarwal “Solar energy” reprinter 1989, Schand publication.

 

 

 

Received on 02.12.2010        Accepted on 10.01.2011        

©A&V Publications all right reserved

Research J. Engineering and Tech. 2(1): Jan.-Mar. 2011 page 26-30