LOAD TEST ON A THREE PHASE TRANSFORMER

 AIM:

     To conduct a direct load test on the three phase transformer to determine the efficiency and regulation at different load conditions.

INSTRUMENTS AND EQUIPMENTS REQUIRED:

                       

THEORY:

      Large scale generation of electric power is usually 3-phase at generated voltages of 13.2KV or somewhat higher. Transmission is generally accompanied at higher voltages of 110, 132, 275, 400, and 750KV for which purpose 3-phase transformers are necessary to step up the generated voltage to that of the transmission line. Next, at load centers, the transmission voltages are reduced to distributed voltages of 6600, 4600 and 2300 volts. Further, at most of the consumers, the distribution voltages are still reduced utilization voltages of 440, 220 or 110 volts. Year ago, it was a common practice to use suitably interconnected three single-phase transformer. But these days, the latter is gaining popularity because of improvement in design and manufacture but principally because of better acquittance of operating men with three-phase type. As compared to a bank of single-phase transformers, the main advantage of a 3-phase transformer are that it occupies less floor space for equal rating, weight less, cost about 15% less and further, that only one unit is to be handled and connected.

PROCEDURE:

 1. The connections are made as per the circuit diagram.

 2. Keeping the autotransformer and load in its minimum position the main supply is switched ON by closing the TPSTS.

3. By slowly and carefully operating the autotransformer the rated voltage (415V) is applied to the primary side of the transformer.

4. Under this no-load condition one set of reading namely V1, W1, V2, I2 recorded in the tabular column.

5. Now the load is increases in gradual steps and at each step primary voltage is maintained at rated voltage and then all meter readings are noted down in the tabular column.

6. The procedure is continued until the current on the secondary side is equal to its full load value.

7. After the experiment is completed, the load is decreased to its minimum, the auto transformer is brought, back to its original position and then the main supply is switched OFF.

8. Calculate output power, efficiency and regulation.

CIRCUIT DIAGRAM:

                      

MODEL GRAPH:

                

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NUCLEAR POWER PLANT-PART:3 - GAS- COOLED REACTOR

       The first gas-cooled reactors with CARBON-DI-OXIDE (at a pressure of 16 bar) as coolant and GRAPHITE as moderator were developed in Britain during 1956-69. The fuel was a NATURAL URANIUM, clad with an alloy of magnesium called MAGNOX.

                                  

       Several types of gas-cooed reactor have been designed and built, with England developing an ADVANCED GAS-COOLED REACTOR (AGR) SYSTEM, and GERMANY and the USA developing HELIUM-COLED, GRAPHITE-MODERATOR SYSTEM  (HTGH). The AGR uses URANIUM OXIDE as the fuel clad in stainless steel tubes with CARBON-DI-OXIDE gas as coolant and graphite as moderator.

   The graphite moderator HELLIUM-COOLED HTGR is designed to use U-233 as the fissile material and thorium as fertile material. Initially, the system would have to be fueled with U-235, until sufficient U-233 is available for makeup fuel. Because of the very high melting point of graphite, these fuel elements can operate at very high temperatures, and it is possible to generate steam at conditions equivalent to those in modern coal-fired power plant. The basic fuel forms are small spheres of fissile and fertile material as carbide, URANIUM CARBIDE AND THORIUM CARBIDE. The fissile spheres are 0.35 to  0.5 mm in diameter and the fertile spheres are 0.6 to 0.7 mm in diameter. Each spheres is coated with two or three layers of CARBON  and SILICON CARBIDE to prevent fission products from escaping from the particles. HELIUM is a suitable coolant in sense that it is chemically INERT, as good heat transfer characteristic and low neutron absorption. Being an MONOATOMIC gas, it can produce for given temperature in the BRAYTON CYCLE and HIGH EFFICIENCY.

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NUCLEAR POWER PLANT- PART:2 - PRESSURIZED WATER REACTOR (PWR)

    The excellent properties of water as a moderator and coolant make it a natural choice for power reactor, and the PWR has been extensively developed in USA. The most important limitation in PWR is the critical temperature of water, 374* C. This is the maximum possible temperature of the coolant in the reactor, and in practice it is considerably less, possibly at about 300*C . In PWR, the coolant pressure must be greater than the saturation pressure at, say, 300*C (85.93 bar) to suppress boiling. The pressure is maintained at about 155 bar so as to prevent bulk boiling.

                      

   A PWR power plant is composed of two loops in series , the coolant loops called as the PRIMARY LOOP, and WATER- STEAM or WORKING FLUID LOOP. The coolant picks up heat in the reactor and transfers it to the working fluid in the steam generator. The steam is then used in a Rankine type cycle to produce electricity.

    The fuel in PWR is slightly ENRICHED URANIUM in the form of thin rods or plates. The cladding is either of stainless steel or zircaloy. because of very high coolant pressure, the steel pressure vessel containing the core must be about 20 to 25  cm thick. A typical PWR contain  about 200 fuel assemblies , each assembly being an array of rods. In a typical fuel assembly, there are 264 fuel rods and 24 guide tubes for control rods. Grid spacers maintain a separation between the fuel rods to prevent excessive vibration and allow some axial thermal expansion.

   The coolant leave the reactor and enter the steam generator which can be either shell and tube type with U- tube bundles or once-through type, the former being more common. In the U- tube steam generator, the hot coolant enters an inlet channel head at the bottom, flows through the U-tubes, and reverse the direction to an outlet at the bottom. It can produce only saturated steam. In the once-through design, the primary coolant enters at top, flows downward through tubes and exits at the bottom to the main pumps. feedwater is on the shell side. A dry or low degree of superheat stream is possible.

    The first land based PWR for power generation was built at SHIPPINGPORT, USA in 1957. Its thermal output is 231 MW, the pressure in the primary circuit is 141 bar, and the water temperature at outlet from the reactor is 283*C . Dry saturated steam is generated in the heat exchangers at 41 bar, 252*C. For a gross electrical output of 68 MW, the efficiency is 29.4%.

                             

   The shippingport cycle has been modified in the Indian point (USA) PWR by the inclusion of an oil-fired superheater between the main heat exchangers and the turbines. There is also an economizer along with some feedwater heaters. The steam condition improves to 25.5 bar and 538*C at turbine inlet, and so the cycle efficiency increases.

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NUCLEAR POWER PLANT - PART:1 - COMPONENTS

 INTRODUCTION:

      Nuclear power plants are a type of power plant that use the process of nuclear fission in order to generate electricity. They do this by using nuclear reactors in combination with the Rankine cycle, where the heat generated by the reactor converts water into steam, which spins a turbine and a generator.

COMPONENTS OF NUCLEAR REACTOR:

MODERATOR:

 The moderator of a nuclear reactor is a substance that slows neutrons down.  When fast neutrons strike the hydrogen atoms in H₂O, they slow down. eg-  graphite, beryllium, Heavy water, Normal water, Helium gas.

PROPERTIES:

 1. It should have high thermal conductivity.
 2. It should have high corrosion resistance.
 3.It should lighter in weight.
 4. It should has good stability to heat and power radiation.

CONTROL RODS:

  It use to control the incidence of the neutrons and prevent the nuclear reactor from the explosions.

FUNCTION OF CONTROL RODS:

 1. To control the rate of fission.
 2. To start the nuclear chain reaction when the reactor is started from cold.
 3. To shunt down the reactor in the emergency condition.
 4. To prevent the melting of fuel rods.

CONTROL RODS SHOULD POSSES THE FOLLOWING PROPERTIES:

 1. It should have good stability under heat and radiation.
 2. It should have adequate heat transfer properties.
 3. It should have better corrosion resistance.

BORON, CADMIUM, HAFNIUM ARE MOSTLY USED AS  CONTROL RODS.

REFLECTOR:

 1. Reflector materials placed around the core to reflect back some of neutrons that they leak- out from the surface of the core.

 2. The reflected neutrons cause more fission and they improve the neutrons economy of reactor.
 3. It made as same material as moderator.

HEAVY WATER, NORMAL WATER, BERYLLIUM, GRAPHITE, HELIUM GAS.

COOLING SYSTEM:

   The coolant are used to carry away heat produced inside the reactor to the heat exchanger. From heat exchanger , heat is transferred to another working medium for further utilization of power generation.

NORMAL WATER, GASES , LIQUID METALS ARE USED.

PROPERTIES:

 1. It should have low melting point.

 2. It should have low viscosity

 3. It should be Non - Corrosiveness.

 4. It should have high density.

 5. It should have high specific heat.

BIOLOGICAL SHIELDING:

     Shielding is necessary to protect the walls of the reactor vessel from radiation damage and it protect the operation personnel from exposure to radiation.
 1. Thick layer of lead concrete or steel are provided all around the reactor.
 2. These layer absorbs the gamma rays, neutrons, etc.,

A GOOD SHIELDIG MATERIALS SHOULD HAVE THE FOLLOWING PROPERTIES:

 1. It should absorb Alpha, Beta and Gamma radiation.
 2. It should have uniform density.
 3. It should be fire resistance.

HEAT EXCHANGER:

  It is used to convert feed water into stream from heat produced in a nuclear reactor core.

STEAM TURBINE:

  A steam turbine is a device that extracts thermal energy from pressurized stream and uses it to do mechanical work on a rotating output shaft.

GENERATOR:

  A generator is a device that convert mechanical energy of the steam turbine to electrical energy.

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PARALLEL OPERATION OF TRANSFORMERS

          The transformers are connected in parallel when the load on them is more than the rating of the individual transformers.so several smaller units are connect in parallel. 

                                          

          from the above diagram, it can be seen that the primary winding are connected to the supply bus bars while the secondary winding are connected to load bus bars.

          Thus two or more transformers are connected in parallel to carry common load. If the given transformer is insufficient in capacity to deliver a particular  load it may replaced by the larger unit or an additional unit. Then the primary side is connected to the same supply and the secondary side is connected to the same load circuit.

                                  

         From the above diagram , It is the another form of parallel operation. The difference from the previous  diagram is that the two units are first connected together and then connected to the respective supply and load circuit.

ABOVE TWO FORMS ARE APPLICABLE IN PARALLEL OPERATION OF TRANSFORMERS .☝☝

CONDITION FOR SATISFACTORY PARALLEL OPERATION:

 1. The transformers are connected with same polarity. In case of three phase transformers should have same angular displacement and same phase transformers.

 2. The voltage ratio of primary and secondary are should be same.

 3. The  frequency ratio primary and secondary must be same.

 4. Percentage impedance must be equal to magnitude.

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BIPOLAR JUNCTION TRANSISTOR

INTRODUCTION:

      A transistor is a semiconductor device used to amplify or switch the electronic signal and electrical power. Transistors are also called as current controlling device because it is use to control the electrical power. In this section we will see about BIPOLAR JUNCTION TRANSISTORS  (BJT) .

                           

EXPLANATION AND WORKING OF BJT :

  1. BJT is classified as two , they are NPN Transistors and PNP Transistors.

                

  2.We will take the NPN Transistors for more in details

                         

.

  3.There are three terminals in the transistors . They are EMITTER, BASE , COLLECTOR .

  4. In this , EMITTER - BASE is connected in FORWARD BIAS.

                                      

 5. The BASE - COLLECTOR  is connected in REVERSE BIAS .

 6. When the biasing is activated , the electrons from emitter is move to the BASE and and also move to the COLLECTOR.

 7. We will lead to reduce the flow of current in the BASE. so, we increase the reverse bias voltage  and also reduce the thickness of the BASE.

                           

 

 8. By vary the bias voltage to get the desired current.

APPLICATION: 

    

Like AMPLIFIER AND SWITCHING APPAATUS etc.

                       

NODAL ANALYSIS

 

 USES OF NODAL ANALYSIS :

       It is an important technique to finding solution in a network

ABOUT NODE :

 Three or more branches connected at a point is called as node point or node.

STEPS INVOLVES IN NODAL ANALYSIS :

STEP - 1 : Find the all nodal points and choose one point as a reference point.

STEP - 2 : Label the node voltage and mark current.

STEP - 3 : Apply Kirchhoff's current law at node point.

NOTE :

Node voltage is obtain by below pneumatics.

Node voltage = Voltage at arrow tail - Voltage at arrow head

          

APPLICATION OF NODAL ANALYSIS :

Determine the current through 10 ohm resistor.

                

STEP - 1 : There are three nodal point in the above circuit.

STEP - 2 :Mark the nodal point and current.

                

STEP - 3 : Apply Kirchhoff's current law at each node point.

       At Node V1 :

              I1 - I2 - I3 = 0                                                                               -----> 1

      Get the value of current refer notes,

             I1 = 8 - V1/4 , I2 = V1 - 0/8 , I3 = V1 - V2/10

        Substitute the current value in equation 1

           ((8 - V1)/4) - (V1/8) - ((V1 - V2)/10) =0

            0.475 V1 - 0.1 V2 = 2                                                                  ------> 2

       At Node V2 :

                I3 - I4 - I5 =0                                                                              ------> 3

     Get the value of current refer notes,

            I3 = V1 - V2/10 , I4 = V2 - 0/4 , I5 = V2 - 4/2

      Substitute the value of current in the equation 3

            ((V1 - V2)/10) - ((V2 - 0)/4) - ((V2 - 4)/2) = 0

             0.1 V1 - 0.85 V2 = -2                                                               ------> 4

       Solving the equation 2 and 4, we get

               V1 = 4.825 V , V2 = 2.920 V

        For the current at 10 ohm,

           I10 ohm = ((V1-V2)/10) = ((4.825 - 2.920)/10) = 1.905/10 = 0.1905 ohm.

IN THIS WAY WE USE THE NODAL ANALYSIS.

MESH ANALYSIS

 USES OF MESH ANALYSIS :

       It is an important technique to finding solution in a network.

ABOUT MESH ANALYSIS :

1. If the network has a large number of voltage source , the mesh analysis is applicable.

2. If there is a current source then it will convert into voltage source for apply mesh analysis.

3. The mesh analysis s only applicable to the planar circuit .

        Example of planar circuit :-

        Example of Non Planar circuit :- 

STEPS INVOLVES IN MESH ANALYSIS :

  Step - 1 :  To find whether the circuit is planar or Non - Planar.

  Step - 2 :  Select the mesh current.

  Step - 3 : Using the Kirchhoff's voltage law to get the equation in terms of unknown to solve the problem.

NOTE :

No . of Equation = No . of mesh current

APPLICATION OF MESH ANALYSIS :

Write the MESH CURRENT equation in the circuit.

                  

STEP - 1 : The above circuit is planar.

STEP - 2 :  The selected mesh currents are  I1 and I2

STEP - 3 :  Using the Kirchhoff's' voltage law

          There are two loops in the above circuit. so, apply law in both first loop and second loop . The equation is

                      5 I1 + 2( I1 -I2 ) = 10

                      10 I2 + 2 ( I2 -I1 ) + 50 =0                                                   

          The following equation is

                       7 I1 - 2 I2 = 10                                                                     ----> 1 

                        -2 I1 + 12 I2 = -50                                                             -----> 2

        By solving the above equations , we have

          I1 = 0.25 A   ,    I2 =  -4.125 A

ANSWER THE QUESTIONS :

1. If the circuit is planar then which analysis is preferable ?

2. MESH ANALYSIS need which source ?

3.  Number of equation is equal to ?   

ORANIC LIGHT EMITTING DIODE ( OLED )

                           

 PRINCIPLE :

            It is the solid state device made up of thin films of organic molecules that produce light with the application of electricity. An electron moves from the cathode to the emissive layer ( ELECTRON TRANSPORTING LAYER ) and the hole moves from the anode to the conductive layer (HOLE INJECTIVE LAYER ) and they recombine to produce photon ( Light ). 

STRUCTURE : 

1. The OLED consist of a cathode and a anode, in between there are two organic layers present

2. The layers are emissive layer (ELECTRON TRANSPORTING LAYER ) and Conductive layer   (HOLE INJECTION LAYER )

3. All the layers are grown over a transparent substrate through which the light will be emitted.

4. ANODE is connected to the positive terminal and CATHODE is connected to the negative terminal of the battery.

                          

WORKING :

1. Forward bias voltage is applied across the OLED.

2. By apply voltage to the cathode will get negative charge electrons. Then the electrons in the cathode will repel  the electrons in the next layer i.e. (ELECTRON EMITTING LAYER).

3.By apply voltage to the anode will get positive charge holes. Then the holes in the anode will repel the holes in the previous layer i.e. (HOLE INJECTING LAYER ).

4.By the electrostatic force the electron in the ELECTRON EMITTING LAYER attract the holes in the hole injection layer and the recombination take place in the organic emitters .

5. The recombination take place near to the emissive layer of the organic emitters because the movement of the hole is faster in the organic semiconductor.

6.The recombination is result in the production of POTON (Light).The light is pass through the transparent substrate.

                       

OTHER DETAILS :

• OLED is made up of 100 to 500 nm thick.
•  The Organic Light Emitting Diode (OLED) is also called Light Emitting Polymer (LEP) or Organic Electro Luminescence (OEL) and it consist of a film of organic compound.
• eg of ELECTRON TRANSPORTIN LAYER - POLYFLURENE
• eg of HOLE INJECTION LAYER - POLYANILINE 

SOLAR CELL OR PHOTOVOLTAIC CELL

CONSTRUCTION AND FUNCTION OF SOLAR CELL 

 BELOW  QUESTIONS  GIVE THE ANSWER TO THE FUNCTION OF THE SOLAR CELL 

1. How to make a  N - type semiconductor ?

        In the normal silicon material the doping is take place with the pentavalent  atoms like PHOSPHORUS  . In this type of semiconductor  electrons are the majority charge carriers .

               

2. How to make a  P - type semiconductor ?

        In the normal silicon material the doping is take place with the Tetravalent atoms like BORON. In this type of semiconductor Holes are the majority charge carriers.

                    

3. What will happen when the light is fall on the  N - type semiconductor ?

         When the light fall on the  N - type semiconductor, the bond brakes to the extra electron present in the atoms which are loosely  bonded. So the electrons are random in motion.

4. What will happen when both  P - type and  N - type are combines ?

        When we combine both types of semiconductor , the electrons in the  N - type is combine with the holes in the  P - type by the driving force . It produce the depletion  region in the          middle  

                             

5. What is the depletion layer ?

         It is the region , in which there is no free electrons and holes.

                        

6. What happen when the light fall on the depletion layer ?

        When light fall on the depletion region, leads to the generation of ELECTRON - HOLE pair.

                         

CONSTRUCTION :

1. Place a thin heavily doped  N - type material at the top .

2.Place a thick lightly doped  P - type material at the bottom.

3. Both the  P - type and  N - type are packed in a can with glass window on top.

4. A nickel ring is provided around the  P - type is act as positive terminal.

5. A metal is contact with the  N - type is act as negative terminal .

FUNCTION :

      When the light is fall on the depletion region it generate the ELECTRON - HOLE pairs which create the potential difference which produce the DIRECT CURRENT to flow in the load.(refer above questions)