Lab 3 Transformers short and open circuit test
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Lab 3 Transformers short and open circuit test Transformers Short Circuit and Open Circuit Tests Introduction One of the universal electrical machines is the transformer, reciveing power at one voltage and delivering it to another. Making use of the laws involving electromagnetism. Working on the basis of Faradays law of induction stating that “ When the magnetic flux linking a circuit changes, and electromotive force (EMF) is induced in the circuit proportional to the rate of change of flux linkage”. This applies to the primary coils having current pushed through it causing for a magnetic field to be produced, this in turn interacting with the magnetic field of the secondary winding inducing an EMF and therefore a current. Furthermore the the operation of these machines aids the effiecent long distance transmission of electrical power from the generating stations. Since power lines incur signinficant πΌ 2 π losses, it is important to minimize these losses by the use if high voltages. The same power can be delievered by high voltage circuits at a fraction of the current required for low voltage circuits” reference hughes book In this laboratory two transfomers were analysed during practical sessions where a number of methods were undergone to enable for the extraction of values from the loaded transformers. Results were recorded from both an open circuit test and a short circuit test on each tranfomer ultimately allowing for calculations to be made in order to gain values for the paramters of the equivalent circuit show in figure(). These results were then used to calculate the effeincncy of each transformer. π π = πππ ππ π‘ππ£π πππ π ππ ππ πππππππ¦ π€πππππππ ππ = πππππππ πππ’π₯ ππ πππππππ¦ π€ππππππ π π = ππππ πππ π ππ’πππππ‘ ππππππ πππ‘ππ ππ¦ πππ ππ π‘ππππ ππ = ππππππ‘ππ ππ‘πππ ππ ππ’πππππ‘ πππ π ππ ππππππ πππ‘ππ ππ πππππ‘ππππ The nominal values of the two transfomers parameters are shown in table () Transformer Design frequency(Hz) 1 50Hz 2 50HZ Nominal VA 500 700 Nominal Vin 240 240 Noimnal Iin 2.083 2.917 Nominal Vout 240 35 Nominal Iout 2.083 20.0 Transformer 1 results The first test was the open circuit (oc) the circuit of which can be seen in figure () The transformer was connected as seen above with the supply as the rated nominal voltage and frequency as shown in table (). The ratio of the voltmeters V1/V2 gives the ratio of the number of the turns. The ammeter labelled A provides the no-load current reading, allowing for the values of the magnetic elements to be calculated. As the current across π π and ππ are negliable in comparison to the magnetic elements π π and ππ the winding elements can be disregarded. Therefore, the voltage and current values across and through the magnetic elements are that of ππ and πΌπ . Thus finding values for ππ and πΌπ aids in the caluclations for the values of the magnetic elements. The results obtained from the practical can be seen in table () Transformer Frequency π1ππ πππ 1 50 Hz 240.5Vrms 26.2W 2 50.02 Hz 239.8Vrms 38.11W ππ΄ππ ππππ 67.61VA 0.387 109.2VA 0.349 πΌ1ππ 0.281 0.456 π20π 248.7Vrms 35.92Vrms The second test was the short circuit (sc) the circuit of which can be seen in figure() The short circuit test works on the basis that the electrons will always flow in the shortest route and therefore by shorting the circuit current will flow around the circuit not passing through the magnetic elements of the transformer. Therefore across the two winding no voltage is dropped; only the winding elemetns are effected. Thus finding ππ and πΌπ aids in calculating the values of the winding elements π π and ππ . Table () shows the results obtained from this practical. Transformer Frequency π1π π ππ π 1 49.3 Hz 8.43Vrms 18.1W 2 50.06 Hz 17.78Vrms 52.42W ππ΄π π ππ π 18.45VA 0.982 53VA 0.989 πΌ1π π 2.209 2.970 π2π π 2.127Vrms 19.55Vrms Equations ()()() are used to calculate the power factor value in tables () and (). ππ΄ = πππππππ¦ ππππ‘πππ(π) × πππππππ¦ πΆπ’πππππ‘ (π΄) π πππ πππ€ππ (π) ππ΄ The efficiency and voltage regulation can be calculated using the readings obtained from the short and open circuit tests using equations ()() πππ€ππ ππππ‘ππ (ππ) = Transformer 1 : πππ = ππ π’ππππ¦ + ππππ π ππ 500π + (26.2π + 18.1π) = 544.3π π0π’π‘ × 100 = 91.8% πππ assuming that the transformer is 100% efficient allows for the parameters to be calculated as follows : πΈπππππππ£π¦ (ο¨) = π π = πΌππ = ππ2 240.5π 2 = = 2207.64ο ππ 26.2π πππ 240.5π = = 108.94ππ΄ π ππ 2207.64ο πΌπ = √0.2812 − 0.108942 = 259.02ππ΄ ππ (ο) = πΏπ = π1ππ 240.5 = = π928.5ο πΌπ 0.25902 ππ 928.5ο = = 2.96π» 2ππ 2π × 50π»π§ These calculations show that the equivalent circuit during the open circuit test can can be simplified down to the magnetic elements connected to the primary voltage. Below shows the calculations required for the short circuit test to obtain the values of the winding components connected to the supply voltage. π π (ο) = ππ π 18.1π = = 3.71ο 2 πΌπ π 2.209π΄2 πππ = πΌππ × π ππ = 2.209π΄ × 3.71ο = 8.2π ππ₯ = √π2π π − π2ππ = √8.432 − 8.22 = 1.96π ππ = πΏπ (π») = ππ₯ πΌ2π π = 1.96π 2.209π΄ = 0.89ο 0.89ο = 2.84ππ» 2 × π × 49.3π»π§ Now that the parameters of the circuit have been calculated the overall efficiency can be computed this is done as follows: 500 πππππππππ¦ ππ’πππππ‘(π΄) = = 2.083π΄ 240 240 π ππππ = = 115.2ο 2.083 1 2 π πππ’πππ£ππππ‘ = 115.2ο × ( ) = 107.75ο 1.034 π2 = π2 = π π //ππ 2207.64ο × 2π × 50π»π§ × π2.96ππ» 2207.64ο + (2π × 50π»π§ × π2.96ππ») = 331.28 + π788.9 π3 = π π //π2 π3 = 107.75ο × (331.28 + π788.91) = 101.5 + π11.24 107.75ο + (331.28 + π788.91) Utlizing the voltage divider rule the calculations below could be made: π3 101.5 + π11.24 ×π = π₯240 = 231.41 − π1.025 π1 + π3 101.5 + π11.24 + 3.713 + π0.87 Written is polar form as : 231.41∠ − 0.254° 1.034 πππ = ( ) × 231.41 = 239.28π 1 π 2 231.412 ππ = = = 497.055π π π 115.2ο Thus to find the effeiceny the power loss in both the magnetic elements and the windings must be calculated as follows: ππππ π = 231.412 = 24.23π 2207.64 240.5 − 231.41 = 8.59π ππππ π π€ = 8.592 3.713 = 19.87π πππ‘ππ πππ π = 24.23 + 19.87 = 44.01π 497.055 π% = × 100 = 91.37% 500 + 44.01 Following the same method and procedure the parameters and efficiency of transformer 2 could be calculated, the results are shown below: Efficiency from results obtained: 700π + (38.11π + 52.42π) = 790.53π πΈπππππππ£π¦ (ο¨) = π0π’π‘ × 100 = 88.5% πππ Parameters: π π = 1388.087β¦ πΌπ = 0.173π΄ πΌππ = 0.4219π΄ ππ = π568.381β¦ πΏπ = 1.809π» π π = 3.71β¦ ππ π = 8.2π πππ = 1.965 ππ = 0.89β¦ πΏπ = 2.84ππ» πππππππππ¦ ππ’πππππ‘, πΌ2 = 2.083π΄ (VA and secondary voltage taken from table 4.6 and plugged into formula from the first part of equation 4.15). π πΏπππ = 77.886β¦ ππ’πππ πππ‘ππ π = 0.1498 π πΏ = 107.962β¦ π3 = 74.8891 + π9.6342ο ππ πΏ = 221.93π ππ πΏ = 33.2432 ππ = 631.48π ππ π = 54.16 ππ π = 32.64 π = 87.92% Discussion and evaluation Calculating the parameters of each transformers required for the assumption that the transformers were 100% efficient to be made or in other words the transformers were ideal. In reality however transfomers are highly effieicent but not completely, these losses that occur can be divided into two groups; the losses that occur in the primary and secondary windings and the core losses due to hysteresis and eddy currents. The two methods were used to calculate the effienecy of the transformers and it serves that the second method (that using th parameters ) is the mores accurate of the two. None the less the two methods and results mirror each other. Conclusion The oc and sc test permitted the characteristics of the transformers through influencing the parameters in order to obtain the power losses within the winding and the core. This in turn allowed for the efficiency of the transformers to be calculated. These values were then double checked through mathematical calculations, these can be seen from equations () through to (). As the values calculated were similar to that from the practical work it helped confirm and validate that the calculation was done correctly.
