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Computer aided Design of Power transformer Presentation Design

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Transformer design principlesLeonardo ENERGY

Transformer design-and-design-parametersMubarak Suliman

Transformer DesignMark Anthony Enoy

Project on Transformer Design | Electrical Machine DesignJikrul Sayeed

Tranformer DesignArnab Nandi

Distribution Transformer Design (Electrical Machine Design)Md. Rayid Hasan Mojumder

- Our group members: Mizanur Rahman khan Reg:09208025 Md. Soriful Islam Rasel Reg:09208008 Md. Toukir Ahmed Reg:09208005 M.K. Masud Hasan Reg:09208031
- Abstract The main objective of this thesis is to design of a Power Transformer using computer program. Six different problems of designing transformer have been solved which are varied from 600 VA to 5 MVA. These problems have been solved by developing a computer program by the authors according to the standard assumed values and principles of the power transformer design.
- Introduction Transformer is an electrical device which electric power in one circuit is transformed into electric power of the same frequency in another circuit. It can rise or lower the voltage in a circuit but with a corresponding decrease or increase in current
- Theory and principles A transformer works on the principle of electromagnetic induction, according to faradey’s law. Whenever a closed conductor is placed into a varying magnetic field, a potential difference is developed across the terminals of the conductor. This phenomenon is known as electromagnetic induction. This is the basic principle of working of any transformer.
- Open circuit test or no-load test From this test, we can determine core loss and no-load current (I0) of the transformer. The high-voltage side is generally kept open because the current in high-voltage winding is less compared to that on low-voltage winding. In low-voltage side, a voltmeter, an ammeter and a wattmeter are connected to measure the input voltage, no-load current and the core loss of the transformer. W=V1I0cosѳ0 IW=I0cosѳ0 Iµ=I0sinѳ0 Fig : Open Circuit Test
- Short circuit or impedance test The main purpose of this test is to compute the following: Equivalent impedance of the transformer referred to primary and secondary. Cu loss of the transformer at any desired load. Total voltage drop of the transformer referred to primary and secondary. Figure : Short Circuit Test
- Voltage regulation • With constant voltage applied in primary, the secondary terminal voltage will decrease due to voltage drop across its internal resistance and leakage reactance. • Let 0V2 and V2 be the secondary terminal voltages at no load and on load respectively. The difference 0V2 – V2 is known as inherent voltage regulation of the transformer. •Voltage regulation down: •Voltage regulation up:
- Voltage regulation The secondary terminal voltage not only depends on load current but also on the power factor of the load. The regulation is said to be at full load provided V2 is determined for full load and at specified power factor condition. V2 drops more and more with increasing load current. For lagging power factor load, V2<E2, the voltage regulation is positive. For, leading power factor load, V2>E2, the voltage regulation is negative.
- Core or Iron Loss From the Nippon Steel Corporation book Core loss for 27-M-4 Grade core. Figure shows core loss vs. flux density. If we include with handling factor and Total weight of core we can found Total core or iron loss. That we have been shown our thesis book
- Copper Loss The loss of power in the form I2R due to the resistances of the primary and secondary windings is known as copper losses. The copper loss also depends on the magnitude of currents flowing through the windings. The total Cu loss is given by Copper losses are determined on the basis of R01 or R02 which is determined from short circuit test. Since the standard operating temperature of electrical machine is taken as 75 °C, it is then corrected to 75 °C.
- No-load current
- Magnetisation Curve From the Nippon Steel Corporation book Magnetising curve for 27-M-4 Grade core. Figure shows flux density (T) vs exciting power (VA/kg)
- Efficiency Due to the losses in a transformer, its output power is less than the input power. Power output = Power input – Total losses Power input = Power output + Total losses = Power output + Pi + PCu The ratio of power output to power input of any device is called its efficiency (η).
- Tank Design The tank dimensions and profile are decided in a layout drawing drawn to scale considering electrical clearances, magnetic clearances, transport size limits and manufacturability. The design of stiffeners is a very important aspect of tank design. An effective stiffening arrangement can reduce the tank plate thickness. The stiffeners are designed in such a way that the tank weight is minimum, and at the same time it should be able to withstand the specified loads.
- Transformer cooling Transformer is non-ideal device that is a practical transformer does not provide 100% efficiency. Transformer windings are made by inductive coils, but a little resistive part is also present in those windings. So, as per the Ohm’s law, that is V=IR, a decent amount of voltage drop will take place across those windings. It is known as the copper loss of the transformer. By the heating action, this amount of energy is directly converted into equivalent amount of heat energy. For better efficiency transformer cooling is required.
- Conservator Design Transformer conservator is considered as one of the important transformer accessories. As the name implies, a transformer conservator conserves some amount of transformer insulating oil. , transformer conservator is an air tight metallic cylindrical drum which is fitted above the transformer main oil tank. A pipeline is provided to create a connection of transformer main tank and conservator and this pipeline usually acts as the circulating path for the insulating oil. This conservator tank is partially filled with insulating oil.
- Parallel operation of Transformer In several power transmission and distribution sector, parallel operations of transformers are effectively done. As the name implies, parallel operation of transformer is achieved by connecting two or more power transformers in parallel with each other. So, in case of parallel operation, the primary windings of the transformers are connected to a common voltage source and the secondary windings are connected to a common load. This is a simple definition of parallel operation of transformers.
- Condition for parallel operation Primary windings should be suitable for the supply system voltage and frequency. The transformers should be properly connected with regard to polarity. The voltage ratings of both primaries and secondaries should be identical. The percentage impedence should be equal in magnetude and have the same X/R ratio in order to avoid calculating currents and operation at different power factor.
- Design procedure of transformer Voltage per turn : Et=k√Q where Q is the rated KVA The Value of ‘K’ for aluminum-wound transformer may b taken between 0.32 to 0.35 For our calculation, let us assume, k=0.33 Therefore, Et=0.33√100=3.3
- Design procedure of transformer value of frequency as : Et= 4.44 × 50 × Bm × Ag × 0.97 × 10-4 For determine the core diameter Ag=K1πd2/4 is the rated frequency which is 50 Hz Bm=is the maximum flux density in tesla Ag=is the gross core area in sq cm 0.97=is the stacking factor The approximate value of K1 may be assume as follows: For steps 6:the approximate value of k1=0.92 For steps 7:the approximate value of k2=0.925 For steps 8:the approximate value of k3=0.93 For steps 9:the approximate value of k4=0.935 For steps 10:the approximate value of k5=0.94 For steps 11:the approximate value of k6=0.945
- According to the advance engineering design computer aided design is the most important part of whole design Problem- 1. 100 KVA, 11000 / 433 V, 50 Hz, 3 Phase Delta / Star connected, Core Type, ON cooled Power Transformer. Output MATLAB coding
- HV Side design values Ph-Voltage in HV winding=11000V Ph-Current in HV winding=3.03A Conductor diameter in LV winding=1.700mm Conductor Area in HV winding=2.270mm Current density in HV winding=1.335A/sq mm For Aluminium wound transformer K=0.33 EMF/Turn = 3.30 Turns in HV winding=3344T 5.00 Percent Turns towards Tapping=167T Total Turns in HV winding=3511T Assume number of coil per phase=4 Turns per c oil=878T Gap between HV to yoke a top & bottom=25.00mm Gap between two tap coil=10.00mm Gap between plain coil=7.00mm Assume no of layer=17 Turns per layer=53T Axial length of HV coil=103.00mm Radial build of HV coil=35.00mm Total axial length of 4 coil=412.00mm Inside diameter of coil in HV winding=187.00mm Outside diameter of coil in HV winding=257.00mm
- LV Side design values Ph-Voltage in LV winding=250V Ph-Current in LV winding=133.34A Conductor diameter in LV winding=10.640mm Gap between to LV coil to yoke=10.00mm Strip size=11.00 X 5.00 X 2 no in parallel A Area of strip=108.280mm Current density in LV winding=1.231A/sq mm Turns in LV winding=76T Assume no of layer=02 Turns per layer=38T Radial build of LV coil=23.00mm Inside diameter of coil in LV winding=121.00mm Outside diameter of coil in LV winding=167.00mmm Length of LV coil with packing=470.00mm
- Core design values Assume 9 step core factor Core diameter=115.000(mm) Core window height=486.00mm Core limb centre=270.000(mm) Step width L=110/100/90/80/70/60/50/40/30/mm Cores tack K=33.541/23.248/14.800/11.024/8.628/6.866/5.454/4.2 58/3.199mm Total core stack K=111.018(mm) Total gross core area=97.832sqcm Net core area=94.897sqcm Grade ofcore:27M-4 Weight of Core=197.984(kg) Flux density=1.561(Tesla)
- Weight of LV Aluminium Mean Diameter of coil=144.000mm Mean Length of turns=453.000mm Total Length of conductor=34428.000mm Specific gravity of aluminium=2.760g/cc Bare weight of conductor for 1 ph=10.289kgs Bare weight of conductor for 3 ph=30.867kgs Weight of the covered conductor=33.000kg (Considering a coverage of 7 percent for LV strip
- Mean Diameter of coil=222.000mm Mean Length of turns=697.434mm Total Length of conductor=2448828.748mm Specific gravity of aluminium=2.760g/cc Bare weight of conductor for 1ph=15.341kgs Bare weight of conductor for 3ph=46.023kgs Weight of the covered conductor=52.000kg (Considering a coverage of 14 percent for HV strip) Weight of HV Aluminium
- Resistance of HV site at 75 Deg-C=35.449ohm Copper loss at 1 Ph of HV site at 75 Deg-C=325.516W Copper loss at 3 Ph of HV site at 75 Deg-C=976.549W Resistance of LV site at 75 Deg-C=0.011ohm Copper loss at 1Ph of LV site at 75 Deg-C=195.023W Copper loss at 3Ph of LV site at 75 Deg-C=585.069W Total Copper loss 3Ph at 75 Deg-C=1561.617W Approximate stray loss for 100.000 KVA=100.000W Load loss as rated load=1662W Load loss(maximum)=1762W Winding Resistance and Load Loss
- No Load Loss From the figure for 1.561 tesla Flax density loss=0.911W/kg Assume Handeling Factor=25percent (ON account of sliting, shearing, knotching,air gap and human error during assemling) Calculated No Load loss=226W
- From the figure for 1.561 tesla Flax Magnetizing power=1.463VA/kg Per phase Magnetizing Current=0.386 A Assume ddd towads 200 percent Handeling Factor (for stress developed duringsliting, shearing, knotching and thikness variation.) Magnetizing Current=1.159A Histeresis and eddy Current=0.301A No load Current=1.197A No load Currentat 100 perct. Voltage as a perct. of Full load current=0.898percent No Load Current at 100 percent voltage
- From the figure for 1.757 tesla Flax Magnetizing power=4.289 VA/kg Per phase Magnetizing Current=1.006A Assume dddtowads200percentHandelingFactor (for stress developed during sliting, shearing, knotching and thikness variation.) Magnetizing Current=3.019A Hysteresis and eddy Current=0.267A No load Current=3.031A No load Current at 112.5 perct. Voltage as a perct. Of Full load current=2.273percent No Load Current at 112.5 percent voltage
- Performance Calculations Efficiency at Unity PF and Full Load: 98.15 percent Efficiency at 0.8 pf: At 100 percent Ld: 97.70 percent, 75 percent Ld:98.10 percent and5 0percent Ld:98.42 percent Max Efficiency=98.79 percent occur at Load =36.84 parcent Mean Daimeter of LV & HV coil=183.0000mm, Average stack Length of LV & HV coil=453.0000mm Reactance= 4.31 percent, Resistance= 1.66 percent, Impedance= 4.62 percent Regulation: at FL and0.8pf=3.94percent, at FL and UPF=1.75percent
- Tank: Length = 847 mm, Width = 337 mm, Height = 901 mm and Volume=0.257m^3 Tank : Surface area for walls side=2.3680 sq.m andTemp-Rise=56is reduced to 40deg-C Unit length of each tube=0.5900m, Surface length of elliptical tube =0.1620 m and No of Tubes =36 Tank Weight: Side sheet ( 3 mm) =52 kg, Bottom sheet ( 5 mm)=11kg, Top sheet( 5 mm)=15kg Base Channel=7kg, Tank Flange=7kg, Stiffeners on all 4s ides=10kg Conservator and its supports=10kg, LV box=3kg, HV Busing adopter=4kg Explosion Vent=4kg, Lifting lugs=2kg, Elliptical tube=39kg Radiators Header=20kg, Miscellaneous items=4kg, Total weight of Tank=187kg Volume of Oil in the tank=186.181litres Volume of Oil in (Radiator + Header)=( 19.0 + 5.0)=24.0litres Volume of Oil in the Conservator (Length= 508.0 mm, Diameter= 235.0 mm)=8.0litres Volume of Oil in LV box and HV box=3.0&4.0litres Total Volume of Oil in the transformer=226.00litresor193.00kg Weight: HV-Wdg( 52.00)+ LV-Wdg( 33.00)+(Core + Yoke)( 197.98 ) Kg Weight(Approx.): Insulation= 15.0 kg, Core Fittings=30.0kg, Other misc.=3.0kg Total weight =735.68kg, Kg/KVA = 7.357 TANK Design and Weights
- Other 5 Problems Problem- 2. 600 VA, 240 / 12 V, 50 Hz, 1 Phase, Delta / Star connected, Core Type, ON cooled Power Transformer. Problem- 3. 1 KVA, 240 / 12 V, 50 Hz, 1 Phase Delta / Star connected, Core Type, ON cooled Power Transformer. Problem- 4. 630 KVA, 11000 / 440 V, 50 Hz, 3 Phase Delta / Star connected, Core Type, ON cooled Power Transformer: Problem- 5. 2 MVA, 33000 / 11000 V, 50 Hz, 3 Phase Delta / Star connected, Core Type, ON cooled Power Transformer: Problem- 6. 5 MVA, 33000 / 11000 V, 50 Hz, 3 Phase Delta / Star connected, Core Type, ON cooled Power Transformer.
- Discussion and Conclusion We have designed transformer according to the proposed specification including primary coil, number of turns, secondary coil, core diameter, step width, core stack, core area, flux density, coil diameter, winding details, loss, no load current, tank, conservator, oil and weights etc. As such, the derivation of various equations involved, explanation for the assumptions made in employing the approximate formula.
- Suggestion for Future work For batter results about transformer design need to internship. If anyone did an internship, they will have the value of specification and assume value those we assumed. They will know about how to design a transformer practically. We did not create the software as a professional one rather we have developed it. It is need to create a software that if give the values of rating and shown the output values.
- Thank you

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