pleasee provide me ur complete solution! A 3 phase, 50 Hz, 434-V, supply is connected to a, Y-connected induction motor and synchronousmotor. Impedance of each phase of induction motor is (1.25 +j2.17) 2. The 3-phase synchronousmotor draws a current of 120 A at 0.87 leading p.f. Two-wattmeter method is used to measure thetotal power. If the phase sequence is positive. Calculate a) reading on each wattmeter b) combinedpower factor.You can use the sample problem below as guide answering the problem, please use 3 decimal pointsfor 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-excitedit 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°voorR$1-1002-60°434V29.5°BwwYo-Govorly Induction MotorSynchronous MotorVRBVYB (b)Fig. 19.63Phase voltage in each case = 434/√3 = 250 V1₁=250/2.5=100 A lagging the reference vector V by 60°. Current 1₂ = 120 A and leads Vby an angle = cos¹ (0.87) = 29.5°1₁ = 100 60 50 86.6; 1₂ 120 29.5 104.6 j591-1, +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.RAs 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 WCurrent I, is also (like I) the vector sum of the line currents drawn by the two motors. It isequal 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 andYBly= 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)W₂www10.1°ܕܐIR

Assume synchronous motor is Y connected As I2=120∠cos-10.87I1=VphZVph=VL3=4343Vph=250…

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.

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter3: Power Transformers

Section: Chapter Questions

Problem 3.25P: Consider a single-phase electric system shown in Figure 3.33. Transformers are rated as follows:...

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