Question

7.19 A 1000 kW, 3-phase. 6.6kV, 50 Hz, 6 pole, delta connected unity power factor synchronous motor has following parameters: X= 40 92, R = 0. rated field current = 5 A. Machine is controlled by variable frequency control at a constant (V/f) ratio. Calculate (i) torque and field current for rated armature current, 500 rpm and unity power factor. (ii) armature current and power factor for regenerative braking torque equal to the rated motor torque. 500 rpm and the rated field current.

Answer 1

Motor Specifications ! Rated power = 100 kw (S) Terminal Vollage. Vt) = 6.6kV Rated field current (ff) = 54 Power factor (coso) = 1 Rated Armature Current (La) 3 vecoso S ) こ 1000xio √3x6.68103 x1 Ia 87.47 ZA 4 Rf radi di/sec. Synchronous Speed (No) = P 4TX 50 6 104-12 rad./see - Rated Power Rated Torque synchronous Speed 1000 x 10 3 104.72 Rated Tosque - 9. 549 KN-m circuit Per Phase equivalent Ra=0 0 + la Xs auor 4 6600kv VE ExL-o .: Efik-s= Ut - j Ia Xs 6600 - 87.47 x 40 v3 -( 3810,5 s 349 8.8) v IEHT = V13810.5) +(3498.8) 2 51 73.16 V As Torque Rotation speed for constant armature current, 500rbon Torque at 1500rpm x 1500 Tosque at 50 9-549 x 3 KN-in 1 Torque 28.647 kN-m No= 1200 rpm at 5oor Dorpm IP frequency NsP fa 120 5 od оо хь G 120 f 2542 for operation : vi fi 6.610 50Hz V 2 Tz V2 257 V2 = 3.3kV Per Phase circuit at 25 Hz rated m t Ia L 2 x sijzor ve 3300 kW vtz N Ef LS a X= (at 25nz ) 2 jzon 330 Era = Tea- j Ia X3 j x 87.47 x 20 3 1905.256 -j 1749.4 (Efal (1905.25) ²+Q749.422 Velt/phase 25 8 6.57 Machine - We know for synchronous field current If & Ep Ifi Eti Efz I fa 5173.16 5 2586.57 Ifr I fa = 2.497 A T yaled from 9. 549 kN m ( part ci) cui Rated Motor Torque 3 to Ian g. 549X103 هو لا Ef lan g. 5498103 x 3 x 120x25 3 6 Ns = 12of P 3 Ella 1591.5 x 10 (X) As Ef Vt - jta L-p.X Where to power factory angle, we can Wrize 2 2 و LEA [Ut sind - Ia Xs)+(ut Gosol) LE+ 1² = ut sin & +22² x ² - 2 Vt kaya sing the cost of VE 2 Vt I a Xs sind 2 2 2 ارت | + Ia xo V4 SERTA -(1905-254) ? + ta’x1209*- 2 x3300 20*** 3300 kV at 2542, 23 3300 x 20 x Kx 3 Ia sind (ii) testt from part i 2586.57 v As Rated field Current 76210. 235 Ia sind = you la ... 3060366.8 Eftai 159). 5x103 equation (x) i so Ef As given that rated field current, = 5173, 16V at rated field current as calculated in part is Hence Ital 1597.5*163 5) 73.16 1 Ial - 307.645 A 2 putting & & ras in eq ci). 51 73.16) ? (1905.25) + Yoox (307.645) ² 4 x 3300 307.145 sine sind = -0.628 & -38.91 حمل مع العدو ... cast Cos (-38-91) 0.778 & legging Power factor