Question

A 3-phase round rotor synchronous motor has a per phase terminal voltage of 480 V. This machine has a synchronous resistance and reactance of 0.25 ohm and 3.8 ohms, respectively. The excitation and field voltage of this machine is equal to 457 V and 150 V, respectively. The machine is operating at a torque angle of -8° and its field current equal to 3 A. With total rotational losses equal to 700 W, determine the following: a. Input Real and Reactive power b. Power converted C. Shaft output power d. Efficiency of the motor.

Answer 1

Given 3-6 round rotor synchronous motor Per phase terminal voltage VE 480 synchronous resistance Ra 0.25n Synchronous rendance Xg = 3.8 v The exitation voltage Ef 457 V field voltage Voc Isov field current Ige : ЗА Torque angle 8 rotational losses = 700 W Ia 0.25 j 3-8 1000 jxs Ra Field = 480V moo2 65=457 V AIDC = 3A - Voc 150 V → VE 8og Rejºnese 8=-8 V Ia Et = 457 V so Es = 457 E8 V CS Scanned with CamScanner

Ta Vt - Ed LS ZS - VE - EA ES (Ratjxs) 4800 - 457/-89 (0.25 + ; 3.8) 480 - C 452.55 - j 63.602) (0.25 + 33.8) (27.45 + j 63.602) (0.25 + 13.8) - 69.2727 / 66-655° 3.808 | 86.2369 Ia 18.19 | -19.580 A 0 = 419.58 So input power s = petj Qe Ve Lo Ia So Pe tj Qe (480 L8 ) (18.19 [-19.58) 11 873)-2 -19.58 8226.314 - 3 2926.023 Pe ů a e 2 CS Scanned with CamScanner

Hore - Reactive Power Qe negative means, that is delivered by a source. So input Real Power 8226.314 W input Reactive power = 2926.023 VAR Per phase Powers." Power converted Control Solution (a) so 3-Phase input Power 3 times of porphose Power SO input Real Power 3x 8226.314 24678.942 W -3-4" total input Reactive Power 3x 29 26.023 8778.089 VAR 3-0 total Power converted. Solutions Power converted = Ef Infos (EJ, Ia) (Es, la (6-8) ( + from Phelor diagram I ding star (19.58-8) 11.58° so (Power converted = 1 Esilial cos LEG, la "Per phase" 3 CS Scanned with CamScanner

Power converted (457) (18.19) Cos (11:58) Power Converted 8143.63 W Porphase" So Three Phase Power converted Power converted = 3x8143.63 Power converted 24430.89 W "3-8"total (C) Shaft output Power solution Intsinal mechanical Power developed converted Power So Mechanical Pouch developed a 24430.89 W. given rotational losses a Too W 4 CS Scanned with CamScanner

Enitation losses IDC Petiolet) Rfield Resistance Voc Joc Iso 3 Refield) Resistance = son So Enitatian losses Is a efield) 3 (50) 450 W SO shaft output Power (Mechanical Power developed)- ( Rotational losses + Exitation losses) 244 30-89 (700 +450 Shaft out put power 23280.89 W (d) Efficiency of the motor : Total input Power = 2u678.942 W output Power 23280-89 W so efficiency outout input X100 844 23280.89 X100 2u678.942 %n efficiency 94.335 51 CS Scanned with CamScanner