Engineering Science OUTCOME 4 - TUTORIAL 3 CONTENTS. 1. Transformers

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1 Unit : Unit code: QCF Level: 4 Credit value: 5 SYLLABUS Engineering Science L/60/404 OUTCOME 4 - TUTOIAL 3 Be able to apply single phase AC theory to solve electrical and electronic engineering problems AC electrical principles: features of AC sinusoidal wave form for voltages and currents; explanation of how other more complex wave forms are produced from sinusoidal wave forms;, L, C circuits e.g. reactance of, L and C components, equivalent impedance and admittance for -L and -C circuits; high or low pass filters; power factor; true and apparent power; resonance for circuits containing a coil and capacitor connected either in series or parallel; resonant frequency; Q-factor of resonant circuit; transformer fundamentals: construction e.g. double wound; transformation ratio; equivalent circuit; unloaded transformer; resistance (impedance) matching; transformer losses; applications e.g. current transformers, voltage transformers. Transformers. Perfect Transformers. eal Transformers.3 Core Losses.4 Coil Losses.5 Efficiency CONTENTS D.J.Dunn

2 . TANSFOMES Transformers are devices for changing alternating voltages and currents. The types of transformers vary from very large to very small. Very large ones are used for transforming the a.c. power generated at a power station to and from a high voltage grid. Smaller ones are used to transform the mains voltage down to a useable level such as V for use in electronic equipment. These have largely been replaced by modern electronic devices that are much smaller, lighter and cheaper. In other areas of electronics such as radio receivers, small transformers working at high frequencies are common.. PEFECT TANSFOME A perfect transformer has two coils placed in close proximity to each other. An alternating voltage is applied to one coil called the primary winding and this produces an alternating magnetic flux. The flux cuts the turns of the second coil called the secondary winding and generates an e.m.f. at the same frequency. If the secondary voltage is smaller than the primary we have a step down transformer. If the voltage is larger we have a step up transformer. D.J.Dunn

3 In order to make all the flux cut both windings, they are either wound on a core that forms a closed loop or wound very close to each other. In this event, the ratio of the voltages is in direct proportion to the number of turns on each winding such that: V /V = N /N In the ideal transformer, the electric power going in at the primary would be the same as the power coming out of the secondary. In this case: P = V I = P = V I Hence V / V = I / I = N /N Note that if the voltage is stepped up, the current is stepped down and vice versa.. EAL TANSFOMES eal transformers are affected by energy losses. These fall into two main groups - COE LOSS and COIL LOSS.3 COE LOSS Large power transformers have iron cores and these become hot and lose energy because of HYSTEESIS and EDDY CUENTS. The core losses are near constant and are not affected by the current flowing in the coils. EDDY CUENTS The alternating flux generates electricity in the magnetic core material. As this is a short circuit, random currents flow in the material and dissipate energy as heat by its electrical resistance. In order to reduce this, larger transformers have cores made from laminate iron sheets and each layer is insulated from each other. HYSTEESIS You will need to understand magnetism to appreciate the meaning of hysteresis. In simple terms, the molecules of the magnetic core are magnetised in one direction and then the other as the current alternates. If the material has any permanent magnetic properties, current and energy are lost to overcome the permanent magnetism with each reversal..4 COIL LOSSES ESISTANCE The losses in the primary and secondary coils are due to the Ohmic resistance of the copper windings. Energy is lost in the form of heat. These losses are best calculated with the formula I and so they increase as the square of the current. POWE FACTO The inductance in the transformer and more importantly the inductance in the load produces a power factor and so the true power is output is P = I V x PF. OPEN CICUIT TEST If the secondary is left open circuit the current drawn by the primary is called the no load current and this is there to overcome the core losses. A watt meter reading will give the core losses. The output voltage of a transformer falls as current is drawn because the losses increase. The correct voltage should be achieved at the rated current. Modern magnetic materials are produced that minimise loses in the core. Transformers used for weak high frequency signals in radio work need to have a high Q factor and so often do not have a magnetic core and are wound on non magnetic formers. D.J.Dunn 3

4 .5 EFFICIENCY η In simple terms the efficiency of a transformer is the ratio of the Power input to the power output but this is complicated by the power factor of the load. The primary is referred to as and the secondary as. η Power Out P Power In P η (I P P copper loss core I x PF) copper loss core loss loss The copper loss is I + I WOKED EXAMPLE No. A 5/ step down transformer has a full load secondary current of 0 A and is rated at 0 kva. The copper loss at full load is 00 W. The primary winding has a resistance of 0.3 Ω. Find the resistance of the secondary coil and the power loss in the secondary coil. SOLUTION N I 0 5 I 4 A N I 5 I I I 00 - (4 )(0.3) 0.4 I 0 Power Loss in Secondary I 0 x W 70W of power is drawn by the primary when the secondary is open circuit. Calculate the efficiency at the rated load if the power factor is 0.9 SOLUTION V I = 0 kva True Power = x PF = x 0.9 = 9 000Watts P 9000 η 0.98 P copper loss core loss D.J.Dunn 4

5 SELF ASSESSMENT EXECISE. A simple high frequency transformer with negligible losses is required to transform 0.05 V rms to V rms. The primary has 0 turns. How many turns should the secondary have? (00). A 0/ step down transformer has a full load secondary current of 50 A and is rated at 30 kva. The copper loss at full load is 80 W. The primary winding has a resistance of 0.5 Ω. The core losses are 0 W. Find the resistance of the secondary coil and the power loss in the secondary coil. Calculate the efficiency at the rated load when a power factor of 0.9 exists. (0. Ω, 80 W, 98.6%) D.J.Dunn 5