University of Rochester Department of Electrical & Computer Engineering ECE111 Laboratory #9 Power Factor Correction week of 30 Nov 2008 --------------------------------------------------------------------------------------------------------- Your write-up must provide a detailed description of the exercise, with diagrams of circuits and a description of procedures. Your lab TA must sign and date each page of your lab notebook. Separate from the report, include a typed abstract that meets the requirements for a good technical abstract. Your grade will be based in part upon conciseness, grammar, and spelling. Late work will not be accepted. --------------------------------------------------------------------------------------------------------- 0. Laboratory preparation Read closely sections 11-5 to 11-9 of the ECE111 text, paying particular attention to example #11.15 found on pg. 482. I. Power factor of AC circuits Consider the AC circuit shown in Fig. 1, consisting of a series L-R plus a capacitor C connected in parallel. CRLIV+-Inductiveload Figure 1. AC circuit consisting of inductive load and shunt capacitor C. The series L-R represents a load such as an induction motor. The power factor (without the capacitor) is PF = cos φ (lagging), where φ = -tan-1(ωL/R). In terms of the rms current I, the real and reactive power quantities are, respectively, PR = RI2 and QX = +XI2. The reactive power does not contribute to the time-average real power consumed by the load, but does increase the required current-carrying capacity (ampacity) of the electrical wiring. The higher current flow results in inefficiency because of ohmic losses in the wiring while the need for heavier wiring increases the capital cost of the distribution system. Power engineers, striving to provide electrical service to such loads at minimum cost, try to "correct" the power factor by adding capacitance in parallel with the load. The capacitance is chosen so that its reactive power: QB = -BV2 (where V = rms voltage and B = ωC-2- is capacitive susceptance) at least partially cancels out the reactive power to the inductor: i.e., QB + QX ≈ 0. Power factor correction works because the current flowing through the capacitor is 180° out of phase with respect to the reactive component of the load current. As a result, the net current to the combined network, consisting of the LR load and the parallel capacitor, can be reduced, as long as the capacitance is correctly specified. II. Experimental procedure & lab questions A. Assemble the series R-L circuit shown in Figure 2, using RL = 1.8 kΩ, LL = 30 to 60 mH, and Rm = 100 Ω. The resistor Rm is chosen to be small compared to the load impedance, that is, Rm « |RL + jωLL|. As long as this inequality is maintained, then the resistor serves as a convenient method to measure the load current i(t) = vR(t)/Rm without significantly disturbing the voltage across the load. CRLi(t)+--+v (t)sv (t)RRmshunt capacitor isto be added in part E~ Figure 2. Laboratory setup for part II of lab. B. Set vs(t) = 2cos(2π•104) and then measure the peak value of i(t) and the phase angle φ between vs(t) and i(t). What is the power factor? Is it leading or lagging? C. Ignoring the effect of the resistor Rm on the network, calculate theoretical predicted values for φ and the peak value of i(t) and compare to your measurements. Explain any discrepancies observed between the calculated and measured results.-3- D. Plot the current phasor with respect to the voltage reference. E. Now add shunt capacitance to the network as shown. Use values for C as follows: C ≈ 1000 pF, 2200 pF, 4700 pF, 0.01 µF, 0.022 µF, and 0.047 µF. Measure φ and the peak magnitude of the current for each capacitance value. F. From your experimental data, calculate the power factor PF and tabulate these results, along with all experimental data from E above. Use your data to plot PF versus C. What value of C has given you the minimum current i(t)? G. Now use the equation QB + QX = 0 to calculate the value of C at which the power factor is corrected to unity, that is, PF = 1.0. To do this, you must correctly calculate the current through the L-R load. Compare your result to the experimental results from part F. H. What is the relationship between maximum power transfer and power factor correction? III. 60 Hz power factor correction of a choke coil Warning: In this section of the laboratory, you will be working with potentially lethal voltages and currents. Follow all procedures exactly and do not touch any circuit or re-wire anything unless the Variac supply is unplugged!! watt-meterVA••fuse~120 v60 HzvariacLOAD12 µF12 µF12 µF Figure 3. Setup for 60 Hz power factor determination. The setup for this exercise is shown in Fig. 3. A variable autotransformer (Variac®) provides adjustable 60 Hz AC to a load consisting of a choke (inductor with series resistance). Several shunt-connected capacitors may be switched into the network to adjust PF. This network is instrumented with an AC voltmeter, AC ammeter (both reading rms), and a power meter (reading average real power). With this arrangement, the power factor can be determined directly using the equation: PF = P/VrmsIrms.-4- A. First, characterize the choke by measuring its power factor PF and then calculating Qload with the voltage set to 80 v-rms. Is power factor leading or lagging? Explain how you arrived at your answer. B. Use the results from part III.A to determine the resistance R and reactance X for the load. Specify units for these quantities. * TO AVOID OVERHEATING CHOKE COIL, DO NOT EXCEED 80 WATTS * C. Adjust the Variac so that the wattmeter reads 80 watts. Record voltage and current for all the possible values of shunt capacitance that can be switched in: C = 0, 12 µF, 24 µF, and 36 µF. D. Construct a table containing all data from the above measurements plus all the calculated results shown below. The measured and calculated values for current, Imeas and Icalc, should agree within measurement error. C(µF) Vmeas Imeas Pmeas PFcalc Qcalc Icalc 0 12 24 36 Use the following expressions to fill out the table: PFcalc = Pmeas/VmeasImeas, Qcalc = (VmeasImeas)2 - Pmeas2, Icalc = Pmeas2- (Qload+Qc)2Vmeas, with Qload = Qcalc @ C = 0, and Qc = - !CVmeas2. Note: ω = 2π•60 ≈ 377 rad/sec. E. Write a 150 word