EE  314  –  Power  Lab  Experiment  1  –  Power,  Phasors,  Impedance  in  AC  Circuits  

 By:  Sabrina  Sandoval  

Partners:  Eduardo  Antunez,  Cruz  Olivares  Date:  Tuesday,  January  20,  2015  

     

   Power  in  AC  Circuits    Objective  To  discover  how  to  improve  the  power  factor  and  find  the  active,  reactive,  and  apparent  power  in  an  inductive  load  through  basic  current  and  voltage  measurements.      Procedure  –  Power  in  AC  Circuits    

1. In  addition  to  the  power  supply  we  installed  the  following  necessary  modules:  resistive,  inductive,  and  capacitive    

2. We  connected  the  power  to  a  three-­‐phase  wall  receptacle  (i.e-­‐  the  voltmeter  was  connected  across  node  4  and  neutral)  

3. We  then  connected  the  following  circuit.  The  resistive  and  inductive  load  modules  were  connected  in  parallel        

                         

4. We then connected the main power supply to our computer station and ran configuration ES15-1

5. The voltage was set to 120V. The resistance and inductance were both set to 100Ω.

6. After the power supply was turned on, our metering window yielded the following values: E = 114.8 V, I= 1.695 A, P= 143.1 W

7. After turning off the power supply, our team then modified the circuit by adding capacitive reactance in parallel with our load as shown in the figure below. All switches were initially set to the open position.

7. We turned the power back on, and begin adding capacitance by closing each

switch on the module one at a time and recording the values upon each addition. 8. We then recorded our data for the RLC Circuit. See next section for

data/calculations. 9. Then, we adjusted our circuit so the capacitive load was 100Ω, which yielded the

minimum line current. 10. See the next section for data/calculations concerning minimum line current.

Calculations/Data – RL Circuit The following equations and calculations were performed using our team’s measured values E = 114.8 V, I= 1.695 A, P= 143.1 W: Apparent Power, S S = E x I S = 114.8V(1.695A) = 194.586 VA Power Factor, PF PF = cos θ = P/S PF = 0.735 ≈ 0.74 Reactive Power, Q

Q = √S2 – P2 Q = √(194.586)2 – (143.1)2 = 131.856 Vars Do the values calculated above demonstrate a low power factor and a notable amount of reactive power for the simulated motor load? Yes. Calculations/Data – Capacitive reactance added (RLC Circuit)

E (Volts) I ( Amps) P (Watts) 0 116.2 1.72 146.51 1 116.53 1.66 147.65 2 116.4 1.55 147.38 3 116.39 1.38 147.57 4 116.3 1.35 147.9 5 116.34 1.32 147.49 6 116.34 1.34 147.05 7 116.46 1.36 147.49 8 116.35 1.44 147.51 9 116.31 1.64 147.53

As can be seen from the graph, the current begins decreasing, then ceases to decrease and begins to increase as more capacitance is added to the circuit. Calculations/Data – Minimum Line Current (RLC)

E = 115.7 V, I = 1.303A, P = 145.1W Xc= 1 / 2πfC = 1/(2π*60Hz*100Ω) = 2.56x10-5 Ω

Apparent Power, S S = E x I S = 115.7V(1.303A) = 150.757 VA Power Factor, PF PF = cos θ = P/S PF = (145.1)/(150.757) = 0.962 ≈ 0.96 Reactive Power, Q Q = √S2 – P2 Q = √(150.757)2 – (145.1)2 = 40.910 Vars Has the reactive power consumed by the circuit decreased between step 9 (RL Circuit) and 18 (RLC – Circuit)? The reactive power consumed by the circuit has significantly decreased from the previous RL-Circuit calculations. Has the line current been reduced by a significant amount with the addition of capacitance? No. Is the active power consumed by the RL Load approximately the same with and without capacitance? Yes, but it is not exactly the same, but approximate. Conclusion We determined the active, reactive, and apparent power of an inductive load, then observed the effects of adding capacitance to the circuit. In adding capacitance to the load the power factor increased from 0.74 to 0.96. This means that adding capacitance improved the power factor since, ideally, we’d like to reach unity power factor. Adding capacitance also did not significantly change the amount of active power consumed by the load, which aids in a more efficient power supply system. Review Questions

1. An electromagnet draws 3kW of active power and 4 kvars of inductive reactive power. What s the apparent power? 5 kVA

2. What is the power factor cos Φ for the electromagnet in Question 1? 0.60 3. A capacitor drawing 4 kvars of reactive power is plced in parallel with the

electromagnet in Question 1. How does this effect the apparent power and the power factor? Apparent power remains the same and the cos Φ decreases.

4. What is the formula used to determine reactive power Q? Q = √S2-P2 5. A capacitor drawing 8 kvars is placed in parallel with an electromagnet that draws

3 kW of active power and 4 kvars of reactive power Q provided by the ac power source and the power factor cos Φ ? Q goes from +4 to -4 kvars and cos Φ is less than unity.

Impedance Objective To determine and demonstrate the impedance of ac circuits as outline by our lab manual.

Procedure

1. In  addition  to  the  power  supply  we  installed  the  following  necessary  modules:  resistive,  inductive,  and  capacitive    

2. We  connected  the  power  to  a  three-­‐phase  wall  receptacle  (i.e-­‐  the  voltmeter  was  connected  across  node  4  and  neutral)  

3. We  then  constructed  the  circuit  below  with  the  resistive  and  inductive  components  connected  in  parallel. We set the resistance to 80Ω, the inductance to 60Ω, and the voltage to 120V.

Circuit for figure 5-19  4. We then connected the main power supply to our computer station and ran

configuration ES15-4, resulting in the following values given in the next section. 5. After completing our measurements we replaced the inductive reactance with

capacitive reactance. Both the capacitor and resistor was set to 60 Ω. 6. After recording values for the previous step, we constructed an RLC series circuit

shown below. The values of the resistance, inductance, and capacitance were set to 80, 60 and 120 Ω respectively.

Is  =  1.0  A  R  =  80  Ω  XL  =  60  Ω  Line  voltage  =  120  V  

7. After recording values for the above circuit, we constructed the RL parallel circuit shown below. The values of the resistance, inductance, and source voltage were set to 80Ω, 60Ω and 120V respectively.

7. After recording values for the above circuit, we substituted the inductor for a

capacitor, creating an RC parallel circuit with the same values given previously. 8. Finally, we set up the below RLC parallel circuit with the resistance, inductance,

and capacitance values to 80, 80, and 60 Ω respectively.

Calculations/Data – Figure 5-19 Circuit

Impedance, Z Φ Z = √R2+XL = 101.88Ω Φ = arctan XL/R = 61.55 / 81.19 = 0.758 Circuit Voltages, Es, ER, EL Es-IsZ = 0 à ES=ISZ à ES=(1.0A)(101.88Ω) = 101.88 V ER-ISR = 0 à ER= ISR à ER= (1.0A)(80Ω) = 80 V EL-ISXL=0 à EL= ISXL à EL= (1.0A)(60Ω) = 60V Compare the calculated and measured values of Z and those for the different voltages. Are they approximately the same? Yes, approximately. Calculated Values vs. Measured Values Z = 101.88 Ω Z = 109. 0 Ω ES = 101.88 V ES = 115.4 V ER= 80 V ER= 86.14 V EL= 60 V EL= 66.02 V Are the measured values of R and XL approximately the same as the values set on the load modules? Yes. We then used the Phasor Analyzer to observe the phase angle between the voltage and the current.

Measured  Values  Es=  115.4  V     Z  =  109.0Ω  ER=  86.14  V     R  =  81.19Ω  EL=  66.02     XL  =  61.55Ω  

Is the measured value of Φ approximately equal to the calculated value? Yes. Data/Calculations – Capacitive Reactance

Calculated Z = √R2+XL= 86.245 Ω, we should get a phase angle of -45 degrees. Es-IsZ = 0 à ES=ISZ à ES=(1.34A)(86.245Ω) = 115.56 V ER-ISR = 0 à ER= ISR à ER= (1.34A)(60Ω) = 80.4 V EL-ISXC=0 à EL= ISXC à EC= (1.34A)(60Ω) = 80.4V After comparing the measured and calculated values for Z, we noticed that the voltages are approximately the same. The measured and calculated values for R and XC are not the same. The phase angle between the voltage and current is as expected. See diagram below.

Data/Calculations – RLC Series Circuit Measured values for Z, R, and XEQ :

Compare the calculated and measured values of Z. Are they approximately the same? Yes, they are: 107.6 ≈108.3 Are the measured values of R and XEQ approximately equal to those set in the circuit. Yes they are close, but not exact. Observing the graph below, is the measured value approximate to the calculated value? Yes, it’s close. -35.1≈-34.21

Z  =  √R2  +  (XL  -­‐  XC)2  =  107.618Ω      Φ  =  arctan  (XEQ/R)  =  arctan(-­‐60.81/89.34)  =  -­‐34.24  degrees  

Data/Calculations – Parallel Capacitive Circuit

The measured and calculated values of Z are very close. The same can be said for the R and XC values and the phase angle.

Z  =  RXC/√R2+XC2    =  80*60/√802+602  =  48  Ω    Φ=  arctan  80/-­‐60)  =  53.1  degrees    IS=  ES/Z  =  120/48  =  2.5A  IR=  ER/R=114.7/80=  1.4  A  IC=EC/XC=115.3/60=1.9  A  

Data/Calculations – RLC Parallel Impedance

Compare the calculated and measured values of Z. Are the approximately the same? They should be, but they are not. Are the measured values of R and XC approximately the same as those set in the circuit? The R values are similar, but the XEQ measured and calculated values are not similar. After observing the measured phase angle and comparing it to the calculated values, the results are similar. Conclusion We calculated and compared the impedance for RL and RLC circuits in series and in parallel. We proved that impedance is equal to the source voltage multiplied by the current. We were then able to use the voltage analyzer to double check that our angle calculations were correct. Review Questions

1. The total opposition to current flow in an ac circuit is called impedance. 2. Circuit impedance can be determined from a.) Z = E/I 3. What is the impedance of an RLC parallel circuit when R, XL and XC are all equal,

ES= 120V and IS=2A? 120/2 =60, 60Ω/3 = 20Ω ea. 4. The combined reactance in a series ac circuit results in the impedance having a

positive phase angle. Does the circuit lead or lag the source voltage? The current leads the voltage because the capacitance is reactive.

5. The impedance of an RLC parallel circuit can be determined from both a and c.

XEQ=  XCXL  /Xc  –  XL  =  60*80/60-­‐80  =  -­‐240  Z  =  RXEQ/√R2*XEQ  =  15.59  Ω  Φ=  arctan  80/-­‐240)  =  -­‐18.43  degrees