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AC Power: average power v(t ) Vp cost i (t ) I p cost P (t ) i ( t )v (t ) I pV p cost cost I p Vp cos I rmsVrms cos 2 2 ~ ~ I V cos ~ ~ ~ ~ V I Z (notice : V I Z , phasor) Z R jX R X R 2 X 2 j Z Z Z cos R Z tan X ; use this to determine the sign R Example 7.2, P333 The power factor: cos(f •The average ac power (Pav) is the power dissipated on the load resistor. • 0cos1, dependent on the complex load. • ideal power factor: cos =1, Z=R, pure resistive load v(t ) Vp cost i (t ) I p cost ~~ P(t ) I V cos ~ I 2 Z cos ~ I 2R Concept Check: thinking with phasors The power factor in this series circuit is less than ideal. Increasing the capacitance is found to improve the power factor. Is the impedance in the box inductive or capacitive? Answer: inductive Complex Power ~ ~ V V V ~ ~ I I I ~~ ~ ~ S VI * I 2 Z V 2 / Z * Pav jQ • real power Pav: power absorbed by the load resistance. • Q (volt-amperes reactive, VAR): exchange of energy between the source and the reactive part of the load. No net power is gained or lost during the process. • S : compute by measuring the rms load voltage and currents without regard for the phase angle. • if Q<0, the load is capacitive, Q>0, the load is inductive Power factor correction • It is very common that load (motors, power lines, etc) is inductive, i.e. Q>0. • Common practice to bring the power factor to unity is to connect a parallel capacitor across the load, such that QC=Q. (the method is easy and need not break the circuit). Example 7.8, P345: power factor correction ~ VS 1170V ; 2 60( Hz ) 377rad / sec Z L 50 j86.7 1001.047 ~ 1170 ~ VL IL 1.17( 1.047)( A) Z 1001.047 ~~ S VL I L * 1170 1.171.047 68.4 j118.5(W ) ~ 2 VL 1172 XC 115 QC 118.5 C 1 1 23.1F X C 377 115 In theory, the power factor can be corrected to unity by putting a capacitor in series with the load. However, it is found (ex. 7.9) that this configuration will result in significant increase in the current, and thus increase the power required by the generator.