Boise State University Department of Electrical and Computer Engineering ECE 212L Circuit Analysis and Design Lab

Transcription

1 Objectives Boise State University Department of Electrical and Computer Engineering ECE 22L Circuit Analysis and Design Lab Experiment #4: Power Factor Correction The objectives of this laboratory experiment are: to calculate and measure complex power, observe lagging power factor, and correct lagging power factor by adding capacitive compensation. 2 Theory The apparent power or voltamperes (VA) in a given circuit is the product of the rms voltage and rms current magnitudes. The real power (also referred to as average power or active power) is the apparent power times the cosine of the angle between the voltage and the current waveforms or phasors. This cosine term is known as the power factor, and it is desirable to operate at or near unity power factor. This is the case since equipment costs are largely proportional to conductor size and insulation, which are determined by the voltamperes required. Also, low power factor operation implies high current with resulting small useful work. This is clear since a purely inductive load may draw a large current and thus large voltamperes demand. The real power lost in the transmission or distribution line feeding this load could be substantial. These Ĩ 2 R losses are costly to utilities and large energy consumers. It is common to add capacitive reactance to an inductive circuit in order to bring the voltage and current in phase (and thus bring the power factor to unity). This practice is known as power factor correction. Clearly, power factor correction can reduce the cost of electric power system operation and can permit generators to produce more active power at rated kva. In order to maintain a higher voltage at the load, and reduce line losses, the correction should always be made at the load rather than the source. P S jq S Z = R jx jq C R 2 P L jq L V jx C V 2 jx 2 Figure : Capacitive Compensation at Load End

2 Without the shunt capacitor in dotted lines and in parallel with the load, the source must supply both real power P L and reactive power Q L to the load, thus resulting in high current, high voltage drop and high losses in the transmission line. When the capacitor is added as shown, the capacitor supplies reactive power to the load and thus relieves the source and line of this unnecessary load. As a load, a capacitor has zero power factor leading, thus making its reactive power consumption negative (i.e., it generates reactive power). If the capacitors were shunted at point instead of point 2, the compensation would still relieve the source of the unnecessary reactive power Q L of the load, but the line would still carry the full P L jq L, thus still causing substantial voltage drop and line losses. The convention used for leading and lagging power factor is standard. When computing the power factor of a load (assuming load notation with the current entering the terminal), the power factor is lagging if o < φ < 8 o (i.e. current lags voltage), and leading if 8 o < φ < o (i.e. current leads voltage). Note that for passive RLC loads, 9 o φ 9 o. 3 Equipment LTspice IV 4 Procedure I R jx I 2 R 2 V jx C V 2 jx 2 Figure 2: AC Study Circuit. Build the AC circuit shown in Figure 2 using a 6-Hz, -Vrms source (peak voltage of 2 or 69.7 volts) delivering real and reactive power to a series RL-load with values R 2 = Ω and L 2 = 26 mh through a transmission line with values R = Ω and L = 26. mh. Assume initially that there is no shunt capacitor. (In the LTspice circuit, use a value C =. µf to represent a nearly perfect open circuit.) 2

3 2. Simulate your circuit with the following parameters: SINE Voltage Source (right click on voltage icon): DC offset[v]: Amplitude[V]: 69.7 Frequency[Hz]: 6 Simulation Command (under Simulate menu): Stop Time: ms Time to Start Saving Data: ms Maximum Timestep:. ms Choosing a small time step will give you smoother sine waveforms. Place two voltage markers: One at the top of the voltage source and the other at the top of the R 2 -L 2 load. Run your simulation. In the output plot, right-click on the top label of a waveform of interest and activate two sets of cursors by choosing the option Attached Cursor: st & 2nd. Measure the following: peak-to-peak source voltage V,pp (V). peak-to-peak load voltage V 2,pp (V). 3. Remove the voltage markers and place two current markers: One through R or L and the other through R 2 or L 2. Measure the following: peak-to-peak source current I,pp (A). peak-to-peak load current I 2,pp (A). 4. Repeat Steps 2-3 with increasing values of C from µf to µf in steps of µf. Tabulate your results as shown below. C (µf) V,pp (V) V 2,pp (V) I,pp (A) I 2,pp (A)

4 Data Analysis and Interpretation. Fill out the table below with the following quantities using the following equations: (a) The rms source voltage magnitude V = V,pp /2 2. (b) The rms source current magnitude I = I,pp /2 2. (c) The apparent power S = V I delivered by the source. (d) The rms load voltage magnitude V 2 = V 2,pp /2 2. (e) The rms load current magnitude I 2 = I 2,pp /2 2 in the RL part of the load. (f) The apparent power S 2 = V 2 I absorbed by the RLC load. C (µf) V (V) I (A) S (VA) V 2 (V) I 2 (A) S 2 (VA) Fill out the table below with the following quantities using the following equations: (a) The real power absorbed by the RLC load P 2 = R 2 I 2 2. (b) The reactive power absorbed by the load Q 2 = X 2 I 2 2 V 2 2 /X C = ωli 2 2 ωcv 2 2. (c) The power factor pf 2 = cos φ 2 = R 2 I 2 2 /V 2I of the RLC load and specify whether it is lagging or leading using the sign of Q 2. (d) The power factor angle φ 2 = cos (pf 2 ) of the RLC load. (e) The real power P = R I 2 P 2 delivered by the source. (f) The reactive power Q = X I 2 Q 2 = ωl I 2 Q 2 delivered by the source. C (µf) P 2 (W) Q 2 (VAr) pf 2 (lag/lead) φ 2 (deg) P (W) Q (VAr)

5 3. Comment on the effect of the capacitive compensation on the source apparent power S (VA), the line current magnitude I (A), the load voltage magnitude V 2 (V), and the load power factor pf Plot V 2 (Vrms) as a function of C (µf). By interpolation, find the value of C that makes V 2 = Vrms (same as the source voltage). Assuming that V < < V 2 are the two points closest to V = (V), use the following interpolation formula V 2 C 2 C = V 2 V C 2 C and solve for C (µf).. Fill out the table below with the following quantities using the following equations: (a) The rms voltage drop V line = I R 2 X2 = I R 2 (ωl ) 2 across the line. (b) The real power losses P line = R I 2 in the transmission line. (c) The reactive power losses Q line = X I 2 = ωl I 2 in the transmission line. C (µf) V line (V) P line (W) Q line (VAr) Comment of the effect of the capacitive compensation on the line voltage drop V line, the real power losses P line in the line, and the reactive power losses Q line in the line. 7. Plot P line (W) as a function of C (µf). Find the optimum value C min that yields the lowest losses in the line. To do this, fit a parabola P line = αc 2 βc γ through the three points closest to the minimum, solve for α, β, and γ, and find C min = β/(2α).

6 Boise State University Department of Electrical and Computer Engineering ECE 22L Circuit Analysis and Design Lab Experiment #4: Power Factor Correction Date: Recorded by: Data Sheet C (µf) V,pp (V) V 2,pp (V) I,pp (A) I 2,pp (A) 2 3 4