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    1. DESCRIPTION OF THE COMPANY

    1.1 Name of the Company: Er-Bakir Electrolytic Copper Products Inc 

    1.2 Location of the Company: Denizli / Turkey

    1.3. History of the Company

    Er-Bakir has been in operation since 1981 and manufacturing in the field of

    Electrolytic copper products field. It is one of the leading companies in its field in Turkey. Its

    success is worth full because of its founder, Ahmet Nuri Erikoğlu who has started his working

    life at the age of thirteen at Bakırcılar Carsışı in 1930.His career increased step by step and his

    industry life has been started with foundation of Ergur Kablo in 1970.Erbakır was found in

    order to produce the raw material of Ergur Kablo.

    The construction of the plant had begun in 1983 and it first went into production in

    June 1985 with a 9200m2 closed area,118 workers and 3,809,000 $ production. Today it has a

    32000 m2 closed area composed of modern production units,398 workers with 70 % of highly

    educated and 134,427,000 $ production with 40% import ratio which makes a value of nearly

    50 million $ import sales to the western including USA ,England, Spain, Italy, Greece ,Iraq,

    Germany, etc. Today Er-Bakir is owned by two local families; Abalıoğlu  and Erikoğlu

    families.

    Table 1. Erikoğlu and Abalıoğlu holdings’ companies

    Ahmet Nuri Erikoğlu Holding A.Ş.  Cafer SadıkAbalıoğlu Holding A.Ş. 

    ER-BAKIR ER-BAKIR

    ERIKOĞLU EMAYE  ABALIOĞLU TEKSTIL 

    ERIKOĞLU TEKSTİL  ABALIOĞLU CIRCIR  

    ERIKOĞLU ELEKTRIK   ABALIOĞLU OTOMOTIV 

    ERIKOĞLU PETROL  DENTAS KAĞIT 

    CN WIRE CO CN WIRE CO

    CNR EXIM ROMANIA CNR EXIM ROMANIA

    DENTAS AMBALAJ

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    The factory’s main production units are anode casting, electrolysis, continuous casting

    and wire drawing. The flow chart of the process is given in Appendix 1.In the departments

    copper is processed with high technology and offered for the industrial uses.

    During the production procedure and after production, Er-Bakir is very interested in

    the quality assurance. Quality Assurance Department is responsible for the controlling of

    working conditions of the plant.

    Er-Bakır,  who makes great investments for Quality management systems, owns

    QS9000, ISO9002, ISO 14001 and OHSAS 18001 certifications and is known its world-wide

    relaibility, product quality and fast services.

    Er-Bakir gives opportunity to work in this company for 487 people. 103 of them is

    white collar and 384 of them are blue collar. The engineers and their duties in the company

    are shown in Appendix 2.

    1.4. Organizational Structure of the Company

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    2. INTRODUCTION 

    I have performed my second summer practice in Er-Bakir Electrolytic Copper Products

    Inc. and the scope of this summer practice is production process units of copper wire systems.

    The main objective of the report is to give information about the production of the

    copper wire, its types and its process steps. The explanations of the production process are

    supported with some figures and flowcharts for clear understanding. Also, mass and energy

     balances are shown after producing the production process. The calculations are done for

    some parameters and the variety of the parameters is listed in a table.

    In Er-Bakir, the Blister copper and copper scraps bought from other companies as a raw

    material. They are melted and then they poured into a mold, and then are cooled. Thus, the

    anode is formed. These anodes are used in order to produce cathode. And then, these cathodes

    are sent to the contirod part to be melt, liquid copper. At the end of the contirod, the filmasin

    is obtained. These filmasins are sent to the wire drawing part. At that part, the copper wires

     become thinner or twisted according to the desired contidions. At the end of the process, the

    copper wires are loaded to the trucks.

    Er-Bakir is one of the biggest copper company in Turkey and it plans to become a

    world-class company supplying quality products to users of copper conductors and considers

    its customers an inseparable part of the enterprise. By employing latest creative production

    and management techniques, Er-Bakir both cares for its employees and respects the

    community and environment. This summer practice in Er-Bakir is an opportunity to observe

    the production flow of the factory and to become experienced about working life and

    chemical engineering department.

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    3. REPORT

    3.1. PRODUCTION OF COPPER

    3.1.1 Anode Casting

    Anode casting is the initial stage in the production of the copper. The aim of the

     process is production of the anode that is needed for the electrolysis.

    Charge is the amount of the raw materials putting into furnace which is for melting

    and refinement for one time. Poling is during refinement with wire, the metal is oxidized.

    High oxidization is harmful because the copper that is highly oxidized causes cracking of

    anode, wavy surface and dissolving large amount of copper at electrolysis stage. For

    decreasing the level of the oxygen (250-3500 ppm) the poplar trees are immersed into the

    melt. This process is called polling.

    Scrap copper wires or wire piles are used as raw material. These materials are

    imported from Iran. They are pressed by a machine and 15-20 tons of raw materials are des in

    a charge. The temperature in the furnace is about 1200 0C.When metal becomes incandescent

     burner must rotate. The aim of this action occur homogeneous heating faster reaction and

    oxidization.

    The situation in the furnace is determined by looking inside. At this time a sample are

    taken and looked interface. If it’s clear and color of the interface is tile it’s the point when the

    oxidization complete otherwise process still continues.

    After oxidization completed the ratio of the oxygen is about 6000-10000 ppm and for

    decreasing this ratio “poling” is the best way. Poling 250-3500 ppm is very normal and

    standard. But while poling occurs; a disadvantages must recognize: Heat loosing. But it is

    solved by putting 4-5 shelves of oak coal. At first time; the color of the flame is green but at

    last it turns orange. At this time 1poling1 stops. The metal is heated for about 45 minutes. At

    the end of this process casting starts. But we should heat casting mould by means of heating

     burner.

    A little quartz powder for preventing sticking metal to the mould and silica powder for

    collecting slag on the face of the metal are added.

    The anode are cooled by water spray and immersed in water tanks to prevent surface

    oxidation. After cooling the anodes are prepared for use in the tank house.

    Plates that are produced are sending to the electrolysis department by railway car.

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    3.1.2. Electrolysis (Cathode Casting)

    It can not mention about purify of copper is 100%.There are many elements that stick

    on copper plates come from anode casting.

      Cathode: The negative electrode in electrolysis, where negative electrons are created

    and positive ions are discharged, it is the object that is going to be plated.

      Anode: Positive electrode in the electrolysis, where negative ions are discharge and

     positive ions are created, it is of the same material as the plating metal.

      Electrolyte: Conducting medium that where the flow of current is with the movement

    of matter. Most of the time, done in aqueous solutions such as acids, bases and salts.

      Direct Current: is the electricity that passes from the anode to the cathode.

    As the direct current passes from the anode

    through the electrolyte, it brings positive ions of

    the plating metal to the cathode. It is then joined

    with negative electrons created by the cathode

    and transforms into the metal coating. The

    moral coating bonds to the cathode and thus

    electrolysis process is complete.

    Figure 1. Electrolysis Department

    The chemical reactions which occur in the

    electrolysis department are as follows;

    In Anode: Cu0  Cu +2 + 2e

    In Cathode: Cu +2 + 2e Cu0

    Figure 2. Passing of copper ions from anode to cathode

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    These reactions occur in 2 steps:

    Anode: Cu Cu + + e

    Cu+ Cu +2 + e

    Cathode; Cu +2 + e Cu+

    Cu + + e  Cu

    Electrolysis department is composed of two parts;

      Electrolysis field

      Acid recovery field

    Electrolysis Field: Firstly H2SO4 (96 % purity) comes to heat exchanger and its temperature

    rises to 60 0C. The steam produced by the steam tank. Although company can produce steam

    itself, they buy steam from ARENKO Co., which has a wide field near Er-Bakir. Hot H2SO4 is

     pumped by centrifugal pump. There are two pumps and heat exchanger which one of them is

    spare. Because of the cavitations of the pump, they are changed in a short period. Electrolyte

    is supplied by two tanks which have 10m3 volumes. They are made of concrete.

    Cu2O + H2SO4  CuSO4 +H2O + Cu

    Cu + ½ O2 + H2SO4  CuSO4 + H2O

    CuSO4  Cu+2 + SO4 

    Cu + ½ + H2  Cu +2 + H2O

    The volumetric flow rate of the electrolyte is 0.02 m3/s. 90 pools are ready for

    working. The length of a pool is 4.2 m and the volume is 5 m 3. Anode part is prepared at the

    anode many casting and it includes different elements, even gold (Ni, Pb, As, Bi, Au...).

    Cathode part is prepared before by means of copper coated titanium. Cathode and anode are

    arranged in the pool. Their position is very important. They must not touch to each other.

    Then 16000 DC current is applied. Every pool has 36 anodes and 37 cathodes.

    Cathodes must change every 6 days. Because after increasing of the thickness of the

    cathode; the quality of the electrolysis will decrease and cathode can not collect Cu ions

    easily. The weights of the anodes decrease from the 280-290 kg to 50-60 kg whereas the

    cathode’s weight increases from 10 kg to 70-80 kg. After 18 days cathodes and anodes are

    taken out from the pools. The leaving electrolyte is send to the acid recovery system. The heat

    of the electrolyte is 50 0C.

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    Acid Recovery Field: After electrolysis stage, the churn comes to the filtering part. A sieve is

    used for this process; the remaining part of the mud is taken and put into a special plate to stay

    under sun. The wet section is removes by this process. At the end of the acid recovery field,

    the dry mud is put into sack and sends to the Azerbaijan.

    3.1.3 The Continuous Casting and Rolling Department

    The aim of the department is to produce copper road which is used for wire drawing or

    sold as such is produced by the continuous casting and rolling plant. This department is the

    most important part of the Er-Bakir.

    Contirod: The production capacity of the system is 18 ton/hour. Also this system is supported

     by on-line computer system and quality assurance. Contirod means “CONTInuous ROD”. 

    This department consists of four parts;

      Furnace

      Casting

      Rolling

      Alcohol 

    1) Imposition Crane 11) Copper Bar Align2) Melting Furnace 12) Rolling Machine3) Holding Furnace 13) Cleaner (Alcohol + emulsion)4) Runner 14) Unwinder Pinch-Roll5) Casting Machine 15) Vax6) Slag Vessel 16) Unwinder7) Shaft Controlling Room 17) Wire Distributor8) Pinch-Roll 18) Steelyard9) Scissor 19) Conveyor10) Edge Cutting 20) Package

    Figure 3. Scheme of the Contirod Plant

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    3.1.3.1. Furnace Part The aim is melting of copper in order to prepare for casting at needed conditions.

     Natural Gas is used as fuel for this process. The pressure must be 900 mbar at gas provision

    and 1.8 bars at gas regulator. The copper used as raw material is lifted by means of elecator. It

    will put into the melting furnace which is 7m of height, 2.4 m of diameter.

    Table 2. The properties of the holding furnace

    Inside Outside

    Environment Temperature (oC (input)) 1400 20

    Surface Temperature (oC (calculated)) 1381 114

    Heat Transition Coefficient (W/m K) 120 15.4

    Calculation Formula ASTM680

    Wind Velocity (m/s) 1

    Irradiation Coefficient 0.7

    Solar Irradiation (W/m ) 0

    Diameter (mm) 1270 1980

    Transient Heat (W/m ) 2251 1444

    Transient Heat Tube (W/m tube )

    3.1.3.2 Casting Part

    Producing copper bar from the liquid copper is the aim of this part. There is entrance

    which copper starts to flow and after that there is a gear system which helps transportation.

    The logic is very simple. Cooling the edges of the liquid copper (inner surface not important)

    makes the copper bar, named in Turkish “bara”. Its color is red, and its temperature is 900 oC.

    The metal plates (Dam block)

    help cooling. A special

    solution is used for cooling

    and heat exchanger is used for

    cooling the emulsion. Bara’s 

    length is 7 cm and its width is

    5 cm

    Figure 4. Contirod (copper bar)

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    3.1.3.3 Rolling Part 

    There are many circular steel plates in the rolling part. One of them is at the top

    another one is at the bottom.

    Figure 5. Shape of the rolling part

    The force of the pres and momentum transfer causes deformation of the physical

     properties of the copper bar. It is crushed and turned to the “filmasin”, the raw material of the

    wire drawing. Diameter of filmasin is 8mm. It is cooled at the end of the rolling machine by

    central lubrication system. Before system starts to work, the oil in the tank should be heated.Oil enters to the heater at under 430 C and goes out at 46 0C.

    The capacity of the central lubrication tank is 9 m 3.Two gear types, 435 lit /min

     pumping capacity of a pump is connected to the suction line. One of them is always spare. By

    means of thermostatic valve which is connected to the oil is fixed at 45 0C.A plate type heat

    exchanger is used again. After exchanger oil comes to the fitter and is distributed to gear box.

    Filter composed of two filtering element.

    3.1.3.4. Alcohol Part

    There are two tanks; the capacity of the main tank is 30 m3 and the capacity of the pre-

     preparing tank is 9 m3.Addintionaly,there are two pumps whose capacities are 150 m3.In order

    to cool the system the heat exchanger is used. Its pressure is 5.5 -6 bar. Alcohol line

    comprises of three steps;

      Cleaning the surface of the filmasin

      Decrease the temperature of the filmasin

      Drying the filmasin by means of the air fan ( high pressure air )

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    There are cooling cells and nozzles in which cleaning and cooling occurs. After rolling

    machine, alcohol is not given to the first cell. The reason of this is preventing the leaking

    alcohol into the rolling machine. Alcohol is given to the next 4 cells by means of the filmasin.

    Figure 6. Product at the end of the Contirod (filmasin)

    Table 3. Contents of the filmasin at the end of the contirod plant

    According to ASTM B 49

    Cu (% ) Min 99.90

    Te ( ppm ) Max 2.0

    Se ( ppm ) Max 2.0

    Bi ( ppm ) Max 1.0

    Sb ( ppm ) Max 4.0

    As ( ppm ) Max 5.0

    Sn ( ppm ) Max 5.0

    Pb ( ppm ) Max 5.0

    Fe ( ppm ) Max 10.0

     Ni ( ppm ) Max 10.0

    S ( ppm ) Max 15.0

    Ag ( ppm ) Max 25.0

    O2 ( ppm ) Min 100 –  Max 650

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    3.1.4 Wire Drawing

    The last part of the copper wire production is wire drawing. The scope of this part is

     producing copper and thinned wire at desired qualification in the optimum range.

    The machines are working in the same principles except wire stranding. Firstly copper

    rod comes to this department from contirod plant. Next, they are welded by welding machine.

    Copper rods, whose diameter is 8mm, are dispatched to wire drawing slender. A special

    emulsion is used to prevent heisting of rolling mills and wire. The concentration of the

    emulsion for rolling mills is about 2-4 % and for tempering is 0.5-1.5 %. Cooling towers

    helps to the cooling water and plate type heat exchanger is used as general in the factory. The

    maximum temperature of the wire is 60 0C and the temperature of the emulsion is 35-45 oC

    Tempering is an important factor since it

    decreases the hardness of the wire. A current at low

    ampere is given to the wire; it crushes some

    molecular bonds so copper becomes softer. Wires are

    transported by the rotating heel and rotation speed

    can be adjusted according the products. Products

    have some special codes that explain the size of the

    wire.For instance, instead of “super fine wire”, 

    “M15” is used. 

    Figure 7. Wire Drawing

    There are many types of the wire which are produced according to the customer

    qualifications:

    Table 4. Machines and their ability for producing wire

    MACHINERY SIZE RANGE

    Rod Breakdown 1.15mm - 4.0mm

    Tin Plating 1.15mm -2.6 mm

    Intermediate Wire Drawing 0.20 mm –  1.2 mm

    Fine Wire Drawing 0.05 mm –  0.20 mm

    Multi Wire Drawing 0.05 mm –  0.50 mm

    Bunching / Stranding 0.06 mm  –  16 mm

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    The products that can be produced in the Er-Bakir are listed below;

    Table 5.Electrolytic Copper Cathode

    Conforms ASTM B115

    Size 950 * 950 * 10 mm (37’’ * 37’’ * 0.4’’ ) 

    Weight Each one 65-85 kg (140 –  190 lbs )

    Table 6.Copper Redraw Rod (ETP)

    Conforms ASTM B49, BS EN 1977

    Diameter 8 mm round ( 0.32’’ ) 

    Weight 3500 –  4500 kg ( 7700 –  9900 lbs )

    Package Coil on wooden pallets, fixed by steel straps and wrapped with plastics

    foils.

    Table 7.Bare Copper Wire (Single End)

    Conforms ASTM B1, ASTM B3, DIN 40500 T4, TS EN 13602, TS EN 13601

    Size range 1.15 –  4.00 mm (6 -17 AWG )

    i) Steel or cardboard baskets

    ii)Steel spools of flange 630 mm, 560 mm (25’’ , 22’’ ) 

    iii)Coils on wooden pallets

    Size range 0.20  –  1.20 mm (17-32 AWG) Steel spools of flange. 630 mm, 560 mm,

    400 mm ( 25’’ , 22’’ , 16’’ ) 

    Size range 0.05 –  0.20 mm (32- 44 AWG ) Plastic spools of flange 250 mm ( 10’’ ) 

    Table 8.Tin Plated Copper Wire (Single End)

    Conforms ASTM B33, DIN 40500 TS, BS 4109, EN 16302

    Size range 1.15 –  2.6 MM (10-17 AWG ) Receptacles (baskets) of steel or Card board

    Size range 0.2  –   1.0 mm (18- 32 AWG ) Steel spools of flange 630 mm, 560 mm,

    400mm (25’’ , 22’’ , 16’’ ) 

    Size range 0.05 –  0.2 mm (32- 44 AWG ) Plastic spools of flange 250 mm ( 10’’ ) 

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    Table 9.Bare and Tinplated Copper Wire (Multiword)

    Conforms ASTM B3, ASTM B33, DIN 40500 T4, DIN 40500 T5, B5 4109

    Size range 0.05 –  0.5 mm (24 –  44 AWG ) 20-450 kg (45 –  990 lbs )

    Steel spools of flange 630 mm, 560 mm, 400 mm ( 25’’ , 22’’ , 16’’ ) 

    Plastic spools of flange 250 mm ( 10’’ ) 

    3.2. QUALITY CONTROL

    Quality Assurance which is understood in ER-BAKIR as ensuring that the product we

     produce will fully meet the needs and expectations of the customer is employed at every stage

    of manufacturing from the arrival of the raw materials to shipment of the product and later

    through to after sales service.

    ER-BAKIR always endeavors to meet all our customer requirements, supply

    consistent high quality and provide the best possible service. 

    All training and education at ER-BAKIR is aimed to achieve the continuous

    development of each employee. ER-BAKIR employs quality assurance systems such as ISO

    9001:2000 and ISO/TS 16949 to guarantee the quality of our products and services and uses

    total quality management (TQM) practices to increase the quality and effectiveness of its

    management system as a whole.

    In line with the requirements and expectations of our shareholders; ISO 14001

    "Environmental Management System" and OHSAS 18001 "Occupational Health and Safety

    Systems" are established to protected the environment in all of our activities and to provide

    safe and healthy working conditions for our employees.

    3.3. WASTE WATER TREATMENT

    The waste water comes from production unit to the alkali waste pool and sent to

    reaction tank with the help of pumps. At the reaction tank, they are neutralized after chemical

    reactions and come to Mud Filtrating Pool. The mud is given to the filter press with using

    high pressure pump.

    Then, filtrate is prepared to a second reaction filtrate pool. The mud is put into Mud

    Train and gone away from surroundings. The filtrate in the pool is taken to last refining tank

    where they are precipitated as hydroxides by the metals which are ready for reaction.

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    The same steps are applied at the last refining units and clear water is given to

    environment after testing their pH.

    The emulsion wastes are collected to Pre-Stock Pool and sent to Emulsion Separation

    unit. At this tank, oil and water are separated themselves and filtrates are sent to filtration

     pool.

    There, the acidity of pH last control unit’s material is very important. From this aspect

    the materials which have basic property can be fallen down to 7 levels with HCl. The diagram

     pumps are used for this process and one of them is one stage and another one is two stage

     pumps. One stage pump work with help of emulsion pressure and the other one work with air.

    3.3.1. Waste Water Treatment Department

    3.3.1.1. Acid /Alkali Discontinues Treatment (9 steps)

    1.  The pH setting step ( if pH is greater than 4, pH is decreased to 2.5 )

    2.  The dosage of NaOCl (about 5 min.)

    3.  The reaction step (30 min )

    4.  The dosage of FeCl3 (20 min )

    5.  The dosage of lime ( until pH is 7 )

    6.  The reaction step (90 min )

    7.  The control of the sample (At this section the concentration of Arsenic is checked. It

    must be 0.1 mg/ L )

    8.  The dosage of polyelectrolyte

    9.  The reaction stops

    3.3.1.2 Discontinuous Treatment (5 steps)

    1.  The dosage of lime ( pH is increased to 11-11.5 )

    2.  The reaction time (90 min )

    3.  The control of the sample (Ni concentration is checked. It must be 3 mg/ L )

    4.  The dosage of polyelectrolyte (1.5 min )

    5.  The reaction stops

    By giving NaOCl, the values of the arsenic is increased from +3 to +5, therefore, arsenic

    makes better component. Also, in order to make iron arsenate FeCl3  is given to the system

    since iron arsenate precipitates arsenic. The reason of giving lime (Ca (OH)  2) to the system is just increasing pH.

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    Table 10. S.K.K.Y. (Su Kirliliği Kontrolu Yönetmeliği) Metal Sanayi

    Composite Sample (2 weeks) Composite Sample (24 weeks)

    Chemical Oxygen Requirement 200 mg/ L 160 mg/ L

    Solid Material 150 mg/ L 100 mg/ L

    Oil and Grass 20 mg/ L 10 mg/ L

    Cadmium ( Cd ) 1 mg/ L

    Mercury ( Hg ) 0.05 mg/ L

    Zinc ( Zn ) 5 mg/ L

    Lead ( Pb ) 2 mg/ L

    Copper ( Cu ) 2 mg/ L

    Iron ( Fe ) 10 mg/ L

    Total Chrome (Cr ) 2 mg/ L

    Chrome (Cr + ) 0.5 mg/ L

    Arsenic 0.1 mg/ L

     Nickel 3 mg/ L

    Total Cyanide (Cn- ) 0.1 mg/ L

    Aluminum (Al ) 3 mg/ L 2 mg/ L

     pH 6- 9 6- 9

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    4. ENGINEERING APPLICATIONS

    4.1. Mass Balance

    The mass and energy balances are made in the contirod plant since they are more

    inputs and outputs. In the contirod, 18 ton / hour cathode enter to the melting furnace and it

    melts at 1083 0C.And then it is heated to 1120 0C.Next,it is poured to the runner and goes

    from here to the cooling section. At the end of the process, 18 ton / hour filmasin is obtained.

    Both cathode and filmasin is copper so while doing calculations, these two variables are

    written as copper. In order to melt the copper, 22 (kg / 1 ton copper) natural gas (methane-

    CH4) is used. And air is used to burn the methane. After the reactions O2, N2, NO, NO2, CO,

    CO2 and H2O occur. Their amounts are taken from the report of the melting furnace (in

    Appendix 3).

    Cu, 250C

    m’= 18 ton / h

    Air, 780C

    O2 = 0.21 K 2000C

    N2 = 0.79 K O2 = n1’ (excess O2)

    N2 = n2’ (excess N2)

    NO = n’ NO

    CH4, 770C NO2 = n’NO2 

    m = 22 kg / 1 ton copper CO = n’ CO

    CO2 = n’ CO2 

    Cu, 11200C + H2O = n’ H2O

    m’  = 18 ton / h nT

    Assumptions:

    1.  The dust, SO2, F, Cl2, Bacharach and impurities in the flue gas is neglected since their

    amounts are very small.

    2.  Air content is 21 % O2 and 79 % N2 

    M F

    E U

    L RT N

    I A

    N C

    G E

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    Reactions: 

    CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) (1)

    CH4 (g) + 3/2 O2 (g) CO (g) + 2H2O (g) (2)

     N2 (g) + 2O2 (g) 2NO2 (g) (3)

     N2 (g) + O2 (g) 2NO (g) (4)

    Overall Cu Balance:

    Inlet Cu = Outlet Cu

    18 ton / h Cu * (1 mol Cu / 63.546 kg) = 283.2594 mol / h

    CO Balance:

    From Appendix 3; Emission of CO = 0.876 kg / h

    nCO = m’ / MW = ( 0.876 kg / h ) / (28 kg / kmol ) = 0.0313 kmol / h

    Overall C Atomic Balance:

    Inlet C = Outlet C

    nCO2’ = nCH4 –  nCO = 24.684 –  0.0313 = 24.652 kmol / h (in the outlet stream)

    Overall H Balance:

    Inlet H2 = Outlet H2 

    4 * nCH4 = 2 * nH2O  4* 24.684 kmol /h = 2 * nH2O 

    nH2O’ = 49.367 kmol / h

    NO2 Balance:

    From Appendix 3; Emission of NO2 = 0.120 kg / h

    n NO2’ = (0.120 kg / h) / (46 kg / kmol) = 0.00261 kmol / h

    NO Balance:

    From Appendix 3; Emission of NO = 0.078 kg / h

    n NO’ = ( 0.078 kg / h ) / ( 30 kg / kmol ) = 0.0026 kmol / h

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    Atomic O2 Balance:

    K = kmol of air

    Inlet O2 = Outlet O2 

    0.21 K = n1’ (excess O2) + ½ n NO + n NO2 + ½ nCO + ½ nCO2 

    0.21 K = n1’ + (0.0026 / 2 kmol /h) + (0.00261 kmol / h) + (0.0313 / 2 kmol/ h)

    + (49.367 / 2 kmol / h)

    0.21 K = n1’ + 49.355.......................................... (*)

    N2 Balance:

    Inlet N2 = Outlet N2

    0.79 K = n2’ (excess N2) + ½ n NO + ½ n NO2 

    0.79 K = n2’ + (0.0026 / 2 kmol / h) + (0.00261 / 2 kmol / h)

    0.79 K = n2’ + 0.002605 .................................... (**)

    Required O2 in order to burn all CH4 ( rxn 1 ) = nCH4’ * 2 = 49.368 kmol / h

    Required O2 in order to form NO2 (rxn 2) = n NO2’ = 0.00261 kmol / h

    Required O2 in order to form NO (rxn 3) = ½ n NO’ = 0.0013 kmol / h

    Thus, total required O2 amount = 49.372 kmol / h

      Exact excess air amount is not kwnon, therefore , assume that excess air is %9 then ;

    Inlet O2 = 49.372 * 1.09 kmol / h = 53.815 kmol / h

    Inlet air = 53.815 * 100 / 21 = 256.264 kmol / h

    Inlet N2 = 256.264 –  53.815 = 202.449 kmol / h

    Solve equation (*)

    Inlet O2 = Outlet O2 

    0.21 K = n1 + 49.355 n1= 53.815 –  49.355

    n1’ = 4.46 kmol / h (excess O2)

    Solve equation (**)

    Inlet N2 = Outlet N2 

    0.79 K = n2 + 0.002605n2’ = 202.446 kmol / h (excess N2)

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    As a result; 

    I nl et stream Outlet stream  

    Cu = 283.2594 kmol / h Cu = 283.2594 kmol / h

    CH4 = 24.684 kmol / h

    Flue Gas, 2000C

    Air, 780C O2 = 4.46 kmol / h (excess)

    O2 = 53.815 kmol / h N2 = 202.446 kmol / h (excess)

     N2 = 202.449 kmol / h NO = 0.0026 kmol / h

     NO2 = 0.00261 kmol / h

    CO = 0.0313 kmol / h

    CO2 = 24.652 kmol / h

    H2O = 49.367 kmol / h

    4.2. Energy Balance

    Q in = Q out

    Q CH4 + Q air  + Q reactions = Q loss + Q flue gas + ∆Q copper  

    ∆Q = ∆H + ∆Ek + ∆Ep + ∆Ws

    There is no kinetic or potential energy and no shaft work, so;

    ∆Q = ∆H

    ∆H = m * Cp * ∆T 

    Cp = a + bT + cT2 + dT3 [1] 

    Q in:

    Cp (methane at 25 0C –  77 0C ) = 1.9296 kJ / mol ( calculated )

    Q CH4 = nCH4’ * Cp

    = (24.684 kmol/ h) * (1.9296 * 103 kJ / kmol)

    Q CH4 = 47630.25 kJ / kmol @ 770C

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    Qair  (780C) = Q O2 (inlet) + Q  N2 (inlet)

    Cp O2 (25 –  780C) = 1.543 kJ / mol

    QO2 = nO2’ * Cp = (53.815 kmol / h) * (1543 kJ /kmol)

    QO2 = 83036.54 kJ / h

    Cp  N2 (25 –  780C) = 1.5146 kJ / mol

    Q N2 = (202.449 kmol / h) * (1514.6 kJ / kmol)

    QN2 = 306629.25 kJ / h

    Q air =Q O2 + Q  N2 = 389665.79 kJ / h

    Q rxns = Q rxn 1 + Q rxn 2 + Q rxn 3 + Q rxn 4 

    Q rxn 1 = nCO2’ * ( Hf CO2 + 2 * Hf H2O –  Hf CH4 -2 * Hf O2) 

    [ 2 ]

    = (24.652 kmol / h) * [(- 393.5) + 2 * (-241.83)  –  (-74.85) –  0] * 103 kJ / kmol

    Q rxn 1 = - 19778.5 * 103 kJ / h

    Q rxn 2 = nCO’ * (Hf CO + 2 * Hf H2O –  Hf CH4 -3/2 * Hf O2) 

    = (0.0313 kmol / h) * [(-110.52) + 2 * (-241.83) –  (-74.85) –  0] * 103 kJ / kmol

    Q rxn 2 = - 16255.03 kJ / h

    Q rxn 3 = ½ * n NO2’ * ( 2 * Hf NO2 –  HfN2  –  2* HfO2 )

    = ( 0.00261 / 2 kmol / h ) * ( 2 * 33.8 –  0 –  0 ) *103  kJ / kmol

    Q rxn 3 = 88.218 kJ / h

    Q rxn 4 = ½ * n NO’ * (2 * Hf NO –  HfN2  –  HfO2) 

    = (0.0026 / 2 kmol / h) * (2 * 90.37 –  0 –  0) * 103 kJ / kmol

    Q rxn 4 = 234.218 kJ / h

    Q rxns = -19794.4 * 103 kJ / h

    Q in = (-Qrxns) + QCH4 + Q air  

    = 19794.4 * 103

     kJ / h + 389665.79 kJ / h + 47630.25 kJ / hQ in = 20231.69 * 10

    3 kJ / h

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    Q out; 

    ∆H m (copper) = 13.01 kJ / mol[2] 

    Cp (25 - 1120 0C) = 30.585 kJ /mol

    ∆QCu = n Cu’ * (∆ H m + Cp)

    = (283.2594 kmol / h) * (13.01 + 30.585) * 103 kJ / kmol

    ∆QCu = 12348.69 * 103 kJ / h

    Q flue gas (25 –  200 0C) = QO2 + Q N2 + Q  NO + Q NO2 + QCO2 + QCO + QH2O 

    Cp O2 = 5726.8 kJ / kmol

    Cp  N2 = 5344.4 kJ / kmol

    Cp  NO = 5641.4 kJ / kmol

    Cp  NO2 = 8311.6 kJ / kmol

    Cp CO = 5414 kJ / kmol

    Cp CO2 = 8490 kJ / kmol

    Cp H2O = 6482.9 kJ / kmol

    Q O2 = nO2’ * Cp O2 = 4.46 *5726.8 = 25541.53 kJ / h

    Q N2 = 202.446 * 5344.4 = 1081952.4 kJ / h

    Q NO = 0.0026 * 5641.4 = 14.6676 kJ / h

    Q NO2 = 0.00261 * 8311.6 = 21.693 kJ / h

    Q CO = 0.0313 * 5414 = 169.4582 kJ / h

    QCO2 = 24.652 * 8490 = 209295.48 kJ / h

    QH2O = 49.367 * 6482.9 = 320041.32 kJ / h

    Q flue gas (25 –  200 0C) = 1637.04 * 103 kJ / h

    Q CH4 + Q air  + Q reactions = Q loss + Q flue gas + ∆Q copper  

    19794.4 * 103 + 389665.79 + 47630.25 = Q loss + 12348.69 * 103 + 1637.04 * 103 

    Q loss = 6245.96 * 103 kJ / h

    Efficiency = 1 –  (Q loss / Q in)

    = 1 –  (6245.96 * 103

     / 20231.69 * 103

    )Efficiency = 69 %

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    Results

    1. If all CH 4 is converted to CO2 , then;

    QCH4 = nCH4 * ( Hf CO2 + 2 * Hf H2O –  Hf CH4 -2 * Hf O2) 

    QCH4 = (24.684 kmol / h) * [(-393.5) + 2 * (-241.83) –  (-74.85) –  0] * 103 kJ / kmol

    QCH4 = - 19804.22 * 103 kJ / h

    2. Combustion yield = (Q CO2  –  Q loss ) / QCH4 

    = [(-17322.675 * 103) –  (-6245 .96 * 103)] / (-19804.22 * 103)

    Combustion yield = 56 %

    3. Ration of energy loosing from flue gas to inlet energy

    Q flue gas / Q in = 1637.04 * 103 / 20231.69 * 103 

    Q flue gas / Q in = 8.09 %

    4. Absorbing energy by the copper

    Q Cu / Q in = 12348.69 * 103 / 20231.69 * 103 

    Q Cu / Q in = 61 %

      assume that excess air is % 0 then ;

    Inlet O2 = 49.372 kmol / h

    Inlet air = 49.372 * 100 / 21 = 235.086 kmol / h

    Inlet N2 = 235.086 –  49.372 = 185.718 kmol / h

    Solve equation (*)

    Inlet O2 = Outlet O2 

    0.21 K = n1’ + 49.355

    n1’= 49.372 –  49.355

    n1’ = 0.013 kmol / h (excess O2)

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    Solve equation (**)

    Inlet N2 = Outlet N2 

    0.79 K = n2’ + 0.002605

    n2’ = 185.715 kmol / h (excess N2)

    As a result;

    I nl et stream Outlet stream  

    Cu = 283.2594 kmol / h Cu = 283.2594 kmol / h

    CH4 = 24.684 kmol / h

    Flue Gas, 2000C

    Air, 780C O2 = 0.013 kmol / h (excess)

    O2 = 49.372 kmol / h N2 = 185.715 kmol / h (excess)

     N2 = 185.718 kmol / h NO = 0.0026 kmol / h

     NO2 = 0.00261 kmol / h

    CO = 0.0313 kmol / h

    CO2 = 24.652 kmol / h

    H2O = 49.367 kmol / h

    Energy Balance  

    Q in:

    Cp (methane at 25 0C –  77 0C) = 1.9296 kJ / mol (calculated)

    Q CH4 = nCH4’ * Cp

    = (24.684 kmol/ h) * (1.9296 * 103 kJ / kmol)

    Q CH4 = 47630.25 kJ / kmol @ 770C

    Qair  (780C) = Q O2 (inlet) + Q  N2 (inlet)

    Cp O2 (25 –  780C) = 1.543 kJ / mol

    QO2 = nO2’ * Cp = (49.372 kmol / h) * (1543 kJ /kmol)

    QO2 = 76174.82 kJ / h

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    Cp  N2 (25 –  780C) = 1.5146 kJ / mol

    Q N2 = (185.718 kmol / h) * (1514.6 kJ / kmol)

    QN2 = 281288.48 kJ / h

    Q air =Q O2 + Q  N2 = 357463.3 kJ / h

    Q rxns = Q rxn 1 + Q rxn 2 + Q rxn 3 + Q rxn 4 

    Q rxn 1 = - 19778.5 * 103 kJ / h

    Q rxn 2 = - 16255.03 kJ / h

    Q rxn 3 = 88.218 kJ / h

    Q rxn 4 = 234.218 kJ / h

    Q rxns = -19794.4 * 103 kJ / h

    Q in = (-Qrxns) + QCH4 + Q air  

    = 19794.4 * 103 kJ / h + 389665.79 kJ / h + 357463.3 kJ / h

    Q in = 20199.49 * 103 kJ / h 

    Q out; 

    ∆H m (copper) = 13.01 kJ / mol[2] 

    Cp (25 - 1120 0C) = 30.585 kJ /mol

    ∆QCu = n Cu’ * (∆ H m + Cp)

    = (283.2594 kmol / h) * (13.01 + 30.585) * 103 kJ / kmol

    ∆QCu = 12348.69 * 103 kJ / h

    Q flue gas (25 –  200 0C) = QO2 + Q N2 + Q  NO + Q NO2 + QCO2 + QCO + QH2O 

    Cp O2 = 5726.8 kJ / kmol

    Cp  N2 = 5344.4 kJ / kmol

    Cp  NO = 5641.4 kJ / kmol

    Cp  NO2 = 8311.6 kJ / kmolCp CO = 5414 kJ / kmol

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    Cp CO2 = 8490 kJ / kmol

    Cp H2O = 6482.9 kJ / kmol

    Q O2 = nO2’ * Cp O2 = 0.013 *5726.8 = 74.448 kJ / h

    Q N2 = 185.715 * 5344.4 = 992535.25 kJ / h

    Q NO = 0.0026 * 5641.4 = 14.6676 kJ / h

    Q NO2 = 0.00261 * 8311.6 = 21.693 kJ / h

    Q CO = 0.0313 * 5414 = 169.4582 kJ / h

    QCO2 = 24.652 * 8490 = 209295.48 kJ / h

    QH2O = 49.367 * 6482.9 = 320041.32 kJ / h

    Q flue gas (25 –  200 0C) =1522.15 * 103 kJ / h

    Qin = Q out

    Q CH4 + Q air  + Q reactions = Q loss + Q flue gas + ∆Q copper  

    19794.4 * 103 + 389665.79 + 47630.25 = Q loss + 12348.69 * 103 + 1637.04 * 103 

    Q loss = 6328.65 * 103 kJ / h

    Efficiency = 1 –  (Q loss / Q in)

    = 1 –  (6245.96 * 103 / 20231.69 * 103)

    Efficiency = 68.67 %

    Results

    1. If all CH 4 is converted to CO2 , then;

    QCH4 = nCH4’ * (Hf CO2 + 2 * Hf H2O –  Hf CH4 -2 * Hf O2) 

    QCH4 = (24.684 kmol / h) * [(-393.5) + 2 * (-241.83) –  (-74.85) –  0] * 103 kJ / kmol

    QCH4 = - 19804.22 * 103 kJ / h

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    2. Combustion yield = (Q CO2  –  Q loss ) / QCH4 

    = [(-17322.675 * 103) –  (-6328.65 * 103)] / (-19804.22 * 103)

    Combustion yield = 55.5 %

    3. Ration of energy loosing from flue gas to inlet energy

    Q flue gas / Q in = 1522.15 * 103 / 20199.49 * 103 

    Q flue gas / Q in = 7.5 %

    4. Absorbing energy by the copper

    Q Cu / Q in = 12348.69 * 103 / 20199.49 * 103 

    Q Cu / Q in = 61.1 %

      In order to find out the amount of excess air which causes the energy of flue gas is

    more than energy of loss. After doing calculations, the results are shown in Table 11.

    As it can be seen from the table since the amount of excess air is increased, the

    efficiency increases. Also, the energy of the flue gas increases. The ratio of the energy

    of flue gas to the lost energy is nearly the same when the excess amount is 260 %,

    approximately.

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    Table 11. Results with respect to different excess air amounts (results are in kJ / h)

    amount of excess air 0% 9% 18% 30% 50% 100% 200% 300%

    Q O2 (inlet stream ) 76174.82 83036.54 89885.92 99026.65 114262.24 152349.65 228524.47 304699.3

    Q  N2 (inlet stream ) 281288.48 306629.25 331920 365674.42 421931.21 562575.45 843863.93 1125150.9

    Q air   357463.3 389665.79 421805.97 464701.07 536193.45 714925.101 1072388.4 1429850.2

    Q rxns -19794400 -19794400 -19794400 -19794400 -19794400 -19794400 -19794400 -19794400

    Q CH4 47630.25 47630.25 47630.25 47630.25 47630.25 47630.25 47630.25 47630.25

    Q in  2.02E+07 2.02E+07 2.03E+07 2.03E+07 2.04E+07 2.06E+07 2.09E+07 2.13E+07

    ∆QCu  1.23E+07 1.23E+07 1.23E+07 1.23E+07 1.23E+07 1.23E+07 1.23E+07 1.23E+07

    Q O2 (outlet stream ) 74.448 25541.53 50962.79 84888.36 141434.78 282795.11 565515.77 848236.43

    Q  N2 (outlet stream ) 992535.25 1081952.4 1171193.19 1290298.49 1488805.54 1985097.21 2977632.46 3970140.98

    Q  NO  14.6676 14.6676 14.6676 14.6676 14.6676 14.6676 14.6676 14.6676

    Q  NO2  21.693 21.693 21.693 21.693 21.693 21.693 21.693 21.693

    Q CO  169.4582 169.4582 169.4582 169.4582 169.4582 169.4582 169.4582 169.4582

    QCO2  209295.48 209295.48 209295.48 209295.48 209295.48 209295.48 209295.48 209295.48

    QH2O  320041.32 320041.32 320041.32 320041.32 320041.32 320041.32 320041.32 320041.32

    Q flue gas 1.52E+06 1.64E+06 1.75E+06 1.90E+06 2.16E+06 2.80E+06 4.07E+06 5.35E+06

    Q loss 6.33E+06 6.25E+06 6.16E+06 6.05E+06 5.87E+06 5.41E+06 4.49E+06 3.58E+06

    Efficiency 68.67% 69% 69.60% 70.20% 71.20% 73.70% 78.50% 83.20%

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    0.00

    50.00

    100.00

    150.00

    200.00

    250.00

    300.00

    350.00

    0.00 2.00 4.00 6.00 8.00 10.00

    # of trial

       #

      o   f  e  x  c  e  s  s

      a   i  r

    # of excess air 

    Efficiency

     Figure 8.Amounts of the excess air and efficiency with respect to trails

    Figure 8 indicates the optimum point of the excess air with respect to efficiency. The

    two curves intersect each other when the efficiency is 72% and the amount of excess air, used

    inlet is 75%. Therefore, the optimum value of the excess air for the process is 75%. This

    amount is enough to burn all methane in the inlet stream. Also it is enough for efficiency since

    the optimum value is 72%.

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    5. CONCLUSION

    The summer practice in Er-Bakir Electrolytic Copper Products Inc. was an opportunity

    to observe the flow of a factory and to apply theoretical knowledge on practice. Useful

    information about the administration of the company and the duties of the engineers in the

    factory was learned which is very efficient and useful for carrier as an engineer candidate.

    In this report, the overall process of the production of the copper wire was mentioned.

    Also, the report includes general information about quality control, waste water treatment and

    engineering applications which consist of mass and energy balances.

    Mass and energy balances are done around continuous casting plant. By using

    conservations of mass balance amount of air at the inlet stream is found. If all oxygen is used

    to burn methane (CH4), 235.086 kmol / h air is needed at the beginning. When the energy

     balance is done according to those values, the loss of energy is 6328.65 * 103 kJ / h. Thus,

    the efficiency of using energy is 68.67 %. Additionally, when there is no excess air, the

    energy loss from walls of furnace becomes higher than that of flue gas which is calculated as

    1522.15 * 103 kJ / h. Trial and error method was used by assuming different percentages of

    inlet excess air to find out where the energy loss by flue gas is higher than that of walls. The

    efficiencies were found as 68.67% , 69% , 69.6% ,70 2% , 71.2% ,73.7% , 78.5% and 83.2 %

    at the usage of excess air 0% , 9% ,18% , 30% , 50%, 100 % , 200% and 300%, respectively.

    As a result of the calculations, it was recognized that the efficiency and the energy loss by flue

    gas increase directly proportional to amount of excess air. In addition, while the energy loss

    from walls of furnace increases by using up to 18% excess air, after that point it decreases.

    However, the energy loss by flue gas is higher than that of walls at 260%. After plotting graph

    of the efficiency and excess air, the optimum values are obtained for the process. As a result,

    the efficiency is 72% when the amount of the excess air is 75%.

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    5. NOMENCLATURE

    Cp : specific heat capacity

    : density

    T : temperature

    : mass

    ’ : mass flow rate

    ’ : molar flow rate 

    Q : heat flow rate

    kJ / mol

    kg / m3 

    °C , K  

    kg

    kg / h

    kmol / h

    kJ / h

    REFERENCES

    1.  Sandler S.. Chemical , Biochemical , and Engineering Thermodynamics , Fourth

    Edition.

    2.  Felder R. , Rousseau R. . Elementary Principles of Chemical Processes. Third Edition

    3.  Incorpera F. , Dewitt D. , Bergman T., Lavine A.. Fundamentals of Heat and Mass

    Transfer, Sicth Edition.

    4.  http://www.erbakir.com.tr/indexflash.asp

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    APPENDIX

    A.1 Flow chart of the company

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    A.2. Engineers and their duties

    DUTY EDUCATION

    General Manager Material and Metallurgy EngineeringStrategic Planning Manager Material and Metallurgy Engineering

    Assistant Manager of Technical Mechanical Engineering

    Chief of Continuous Casting Mechanical Engineering

    Assistant Manager of Technical Mechanical Engineering

    Chief of Wire Drawing Mechanical Engineering

    Chief of Wire Drawing Mechanical Engineering

    Chief of Wire Drawing Industrial Engineering

    Chief of Export Industrial Engineering

    Chief of Logistic Industrial Engineering

    Chief of Production Planning Industrial Engineering

    Expert of Human Resources Industrial Engineering

    Production Planning Engineer Industrial Engineering

    Logistic Engineer Industrial Engineering

    Wire Drawing Engineer Material and Metallurgy Engineering

    Wire Drawing Engineer Material and Metallurgy Engineering

    Wire Drawing Engineer Material and Metallurgy Engineering

    Mechanical Maintenance Engineer Mechanical Engineering

    Chief of Mechanical Maintenance Mechanical Engineering

    Quality Assurance Engineer Mechanical Engineering

    Chief of Quality Assurance Chemical Engineering

    Electrical Maintenance Engineer Electrical Engineering

    Electrical Maintenance Engineer Electrical Engineering

    Electrical Maintenance Engineer Electrical EngineeringElectrical Maintenance Engineer Electrical Engineering

    Expert of Maintenance Support Electrical Engineering

    Software Expert Computer Engineering

    Software Expert Computer Engineering

    Computer Systems Engineer Computer Engineering

    Chief of Domestic Sales Environmental Engineering

    Domestic Trade Manager Material and Metallurgy Engineering

    Civil Engineer Civil Engineering

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    A.3. Contirod Ergitme Fırını Bacasında Yapılan Ölçümler  

    Yakıt Tpi  Doğalgaz 

    Yakıt Alt Isıl Değeri (kcal / m3 ) 8250

    Yakıt Sarfiyatı ( m3 / saat ) 466Yakma Isıl Gücü ( kW )  4470

    Cebri Çekiş Uygulaması  Mevcut Değildir  

    Baca Yüksekliği ( Yerden )  22SINIRDEĞER 

    Baca Yüksekliği (Çatıdan )  3 3 1 

    Ölçüm Tarihi  12.08.2008

    Baca Kesiti ( m2 ) 0.817

    1. ÖLÇÜM  2. ÖLÇÜM  3. ÖLÇÜM  ORTALAMA

    Gaz Sıcaklığı ( 0C )

    Gaz Hızı ( m / sn ) 

    İşletme Şartlarındaki Gaz Debisi (m3

     / saat )Normal Şartlardaki Gaz Debisi ( Nm3 / saat )

    İslilik 

    Toz Konsantrasonu (mg / Nm3 )

    Toz Konsantrasonu (kg / saat )

    CO Konsantrasyonu (mg / Nm3 )

    CO Emisyonu ( kg / saat )

    SO2 Konsantrasyonu ( mg / Nm3 )

    SO2 Emisyonu ( kg / saat )

    NO Konsantrasyonu ( mg / Nm3 )

    NO Emisyonu ( kg / saat )NO2 Konsantrasyonu ( mg / Nm

    3 )

    NO2 Emisyonu ( kg / saat )

    F- Konsantrasyonu ( mg / Nm

    3 )

    F- Emisyonu ( kg / saat )

    Cl- Konsantrasyonu ( mg / Nm

    3 )

    Cl-Emisyonu ( kg / Nm3 )

    Cu Konsantrasyonu ( mg/ Nm3 )

    Cu Emisyonu ( kg / saat )

    O2 Konsantrasyonu ( % )

    CO2

     Emisyonu ( % )

    Bu rapor yaknızca ERBAK IR ELEKTRONİK BAKIR MAMULLERİ A.Ş.‘nde 11-13.08.2008 ve 24.11.2008

    tarihlerinde yapılan Emisyon ölçünlşeri için geçerli olup EKOTEST Çevre Danışmanlık Ölçüm Hizmetleri

    Ltd. ‘nin yazılı onayı olmadan kısmen kopyalanıp çoğaltılamaz.İmzasız ve mühürsüz raporlar geçersizdir.Deney

    sonuçları, sadece ölçüm sırasındaki proses koşullarıyla ilgilidir.Akreditasyonumuz sadece Genel Prensipler

    bölümünde verilen deney metotlarınin kapsamı ile sınırlıdır.Bunun dışında verilen görüş ve yorumların

    yeterliliği akreditasyon kapsamında değildir. 

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    A.4. Sketch of the Factory