U of M ECE 4501 - Power Systems Laboratory Manual

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Vac = 120 V rmsVdc = _________ Volts dcVac = _______ Vac100 V rms120 V rmsV ac = 120 V rmsV ac = 120 V rmsS = Vac x Iac =_______ VAV ac = 120 V rmsV ac = 120 V rmsPower Factor, pf = Real Power, P / Apparent Power, SVs = 70.7 V rmsF. Instantaneous Current, Voltage and Power for a Resistive LoadVs = 70.7 V rmsVs = 70.7 V rmsECE 4501 Power Systems Laboratory Manual Rev 1.02.0 AC CIRCUITS2.1 AC VOLTAGE AND CURRENT – CALCULATIONS2.1.1 OBJECTIVETo study sinusoidal voltages and currents in order to understand frequency, period, effective value,instantaneous power and average power.2.1.2 DISCUSSIONAlternating Current (AC) is the world standard for driving motors and other electrical equipment. As the name implies, an alternating current continually and periodically changes direction, going first one way and then reversing. One may consider an alternating current to be a DC current, which is constantly, and periodically changing amplitude and direction. The time necessary for the current to undergo one complete change of amplitude and direction is called a cycle. The number of cycles which occur in one second is called the frequency, which is measured in Hertz (cycles per second). In North America, AC system frequency is standardized at 60 Hertz. Much of the remaining world has chosen a 50 Hz standard.Alternating voltages also reverse polarity in a periodic manner with continually changing amplitude. The shape of the voltage waveform is dependent upon the manner in which it is produced. One may construct a device that produces voltage waveforms which are square waves, triangular waves, etc. However, one type of waveform is most suitable to the use of electric powerin transformers and electric motors, the sine wave. Pure sinusoidal waveforms of a single frequency minimize mechanical and electrical losses in transformers and motors, thus allowing forthe highest efficiency of energy conversion. If one considers the Fourier Transform of a triangularwave or square wave, which consist of a large number of sinusoidal waveforms at many frequencies, it becomes clear that there is increased potential for detrimental effects such as mechanical resonance, eddy currents, sequence currents, etc. A pure sine wave undergoing electrical transformation (either integral or derivative) remains a pure sinusoid, minimizing losses. Another advantage of sine wave voltages is that the resulting current waveforms are also sinusoidal. This is not necessarily true of other waveform shapes. Having consistent wave shapes reduces the burden of power calculations and the analysis of electric systems.2.1.3 INSTRUMENTS AND COMPONENTS(None)2.1.4 PROCEDURE1) Consider the ideal AC generator pictured below. Assume it produces a sinusoidal voltage output waveform between terminals A and B.-1 –ECE 4501 Power Systems Laboratory Manual Rev 1.02) Let the voltage waveform have a peak value of 100 volts, such that the algebraic description is v(t)= 100 sin (t) volts. Calculate the value of v(t) at 15 degree intervals (i.e. when t = 0, 15, 30, …, 330, 360) and plot them on the graph provided (also, write them down, since you willneed these values again). Connect the plotted points with a smooth curve and label it v(t). Remember that the waveform will have a negative value for half its period.-2 –ECE 4501 Power Systems Laboratory Manual Rev 1.03) Read the instantaneous value of v(t) from the graph for each of the following values of t and record it below:t in Degrees v(t) in Volts80 volts160 volts250 volts350 volts4) If a load resistance of 2 Ohms is connected across terminals A and B, a current, i(t), will flow. Knowing the instantaneous value of the voltage from the graph and using Ohm’s Law [ i(t) = v(t)/R ], calculate and record the instantaneous values of i(t) for the following values of t:t inDegreesi(t) in Amps t inDegreesi(t) in Amps30 amps 210 amps60 amps 240 amps90 amps 270 amps120 amps 300 amps150 amps 330 amps180 amps 360 (0) amps5) Plot the instantaneous current values recorded above on the same graph with v(t) and draw a smooth curve through the plotted points. Label this curve i(t).6) Knowing that the instantaneous power, p(t), is the product of the instantaneous voltage and current, calculate p(t) at every 30 degree interval:t in Degreesp(t) = v(t) xi(t)t in Degreesp(t) = v(t) xi(t)30 watts 210 watts60 watts 240 watts90 watts 270 watts120 watts 300 watts150 watts 330 watts180 watts 360 (0) watts-3 –ECE 4501 Power Systems Laboratory Manual Rev 1.07) Plot the instantaneous power values on the same graph with v(t) and i(t), then draw a smooth curve through the plotted points. Label this curve p(t).8) Examine the power curve and determine the maximum, or peak, instantaneous power dissipated by the resistor and the minimum value of instantaneous power. Also make your best estimate as tothe average power dissipated by the resistor.Maximum (peak) p(t) Minimum p(t) Average PowerWatts Watts Watts9) It can be shown mathematically that the average power will equal exactly one-half the peak instantaneous power. The average power is equivalent to that supplied to the two ohm resistor from a DC source. If 2500 Watts were dissipated by a 2 Ohm resistor, supplied by a DC source, what would the DC Voltage of the supply be? (Use the equation, P = V2/R)____________________________________________________________ Volts, DC10) What is the DC current supplied by the source for those conditions? (Use Ohm’s Law)____________________________________________________________ Amps, DCNote that the AC values of 100 Volts, peak and 50 Amps, peak deliver the same average power as DC values of 70.7 Volts, DC and 35.4 Amps, DC. This is a ratio of one over the square root of two (1/2). Thus, the effective value of an AC signal is its peak value divided by the square root of two. This is known mathematically as the root-mean-square or RMS value of the waveform, and leads to the following equations:Vrms = 0.707 VpeakIrms = 0.707 Ipeak2.1.5 CONCLUSIONS1) How long does it take, in seconds, for the voltage to change from 0 to maximum (peak) value on a60 Hz power system? _________________________________________________________________ seconds2) What is the length of time of the positive portion of one complete cycle of a 60 Hz current


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