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A 3 phase, 50 Hz, 434-V, supply is connected to a, Y-connected induction motor and synchronous motor. Impedance of each phase of induction motor is (1.25 +j2.17). The 3-phase synchronous motor draws a current of 120 A at 0.87 leading p.f. Two-wattmeter method is used to measure the total power. If the phase sequence is positive. Calculate a) reading on each wattmeter b) combined power factor.

A 3 phase, 50 Hz, 434-V, supply is connected to a, Y-connected induction motor and synchronous motor. Impedance of each phase of induction motor is (1.25 +j2.17). The 3-phase synchronous motor draws a current of 120 A at 0.87 leading p.f. Two-wattmeter method is used to measure the total power. If the phase sequence is positive. Calculate a) reading on each wattmeter b) combined power factor.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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A 3 phase, 50 Hz, 434-V, supply is connected to a, Y-connected induction motor and synchronous motor. Impedance of each phase of induction motor is (1.25 +j2.17) 2. The 3-phase synchronous motor draws a current of 120 A at 0.87 leading p.f. Two-wattmeter method is used to measure the total power. If the phase sequence is positive. Calculate a) reading on each wattmeter b) combined power factor. You can use the sample problem below as guide answering the problem, please use 3 decimal points for the final answer! Sample Problem solution for guide (the given from this problem is different from the problem above) Solution. It will be assumed that the synchronous motor is Y-connected. Since it is over-excited it has a leading p.f. The wattmeter connections and phasor diagrams are as shown in Fig. 19.63. Z₁ = 1.25 +2.17 = 25/60° 12-120 /29.5° voor R$ 1-1002-60° 434V 29.5° B ww Yo- Govor ly Induction Motor Synchronous Motor VRB VYB (b) Fig. 19.63 Phase voltage in each case = 434/√3 = 250 V 1₁=250/2.5=100 A lagging the reference vector V by 60°. Current 1₂ = 120 A and leads V by an angle = cos¹ (0.87) = 29.5° 1₁ = 100 60 50 86.6; 1₂ 120 29.5 104.6 j59 1-1, +1₂=154.6-27.6= 156.82-10.1° (a) As shown in Fig. 19.63 (b), 1 lags V by 10.1°. Similarly, I, lags V, by 10.1⁰. R As seen from Fig. 19.63 (a), current through W, is I, and voltage across it is VRB = VR-VB- As seen, VRB-434 V lagging by 30°. Phase difference between VR and I is = 30-10.1 = 19.9⁰. .. reading of W, = 434 x 156.8 x cos 19.9° = 63,970 W Current I, is also (like I) the vector sum of the line currents drawn by the two motors. It is equal to 156.8 Á and lags behind its respective phase voltage V, by 10.1°. Current through W₂ is I, and voltage across it is V, V. As seen, Vy=434 V. Phase difference between V and YB ly = 30° +10.1° = 40.1° (lag). .. reading of W₂ = 434 x 156.8 x cos 40.1° = 52,050 W (b) Combined p.f.= cos 10.1°= 0.9845 (lag) W₂ www 10.1° ܕܐ IR
Transcribed Image Text:A 3 phase, 50 Hz, 434-V, supply is connected to a, Y-connected induction motor and synchronous motor. Impedance of each phase of induction motor is (1.25 +j2.17) 2. The 3-phase synchronous motor draws a current of 120 A at 0.87 leading p.f. Two-wattmeter method is used to measure the total power. If the phase sequence is positive. Calculate a) reading on each wattmeter b) combined power factor. You can use the sample problem below as guide answering the problem, please use 3 decimal points for the final answer! Sample Problem solution for guide (the given from this problem is different from the problem above) Solution. It will be assumed that the synchronous motor is Y-connected. Since it is over-excited it has a leading p.f. The wattmeter connections and phasor diagrams are as shown in Fig. 19.63. Z₁ = 1.25 +2.17 = 25/60° 12-120 /29.5° voor R$ 1-1002-60° 434V 29.5° B ww Yo- Govor ly Induction Motor Synchronous Motor VRB VYB (b) Fig. 19.63 Phase voltage in each case = 434/√3 = 250 V 1₁=250/2.5=100 A lagging the reference vector V by 60°. Current 1₂ = 120 A and leads V by an angle = cos¹ (0.87) = 29.5° 1₁ = 100 60 50 86.6; 1₂ 120 29.5 104.6 j59 1-1, +1₂=154.6-27.6= 156.82-10.1° (a) As shown in Fig. 19.63 (b), 1 lags V by 10.1°. Similarly, I, lags V, by 10.1⁰. R As seen from Fig. 19.63 (a), current through W, is I, and voltage across it is VRB = VR-VB- As seen, VRB-434 V lagging by 30°. Phase difference between VR and I is = 30-10.1 = 19.9⁰. .. reading of W, = 434 x 156.8 x cos 19.9° = 63,970 W Current I, is also (like I) the vector sum of the line currents drawn by the two motors. It is equal to 156.8 Á and lags behind its respective phase voltage V, by 10.1°. Current through W₂ is I, and voltage across it is V, V. As seen, Vy=434 V. Phase difference between V and YB ly = 30° +10.1° = 40.1° (lag). .. reading of W₂ = 434 x 156.8 x cos 40.1° = 52,050 W (b) Combined p.f.= cos 10.1°= 0.9845 (lag) W₂ www 10.1° ܕܐ IR
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