Question

Q3) A 400-V, 3-phase, 50-HZ, 4-pole Y-connected synchronous machine is operated as a motor and draws 27A at 0.77 PF leading. The motor armature resistance is 0.50/Phase and synchronous reactance is j12Ω /phase. The rotational loss is 200 watts. The field winding has a resistance of 0.4Ω and is supplied by 50V DC sources. Determine the following: a) Real and reactive power drawn by the motor b) The power developed c) The torque angle d) The shaft torque e) Efficiency 17 watts, 11,927 vars, b) 13,292 watts, c) 31.2° d)83.3 N-M, e) 90.8%

Answer 1

Ans)

Given

Vi = 400 V. f = 50 H: P = 4 poles. Ia-27 A. Pf = cosp = 0.77 lead

a)

Phase voltage is given as

V400 =-=-= 230.942001 3 V3

ph = 230.9420。1

The phase current with phase is

1,-27 Lcos-pf= 27 Lcos-10.77=27し39.65° A

$I_{a}=27\angle 39.65^{o}\: \: A$

Now complex power taken by motor is

$S=3V_{ph}I_{a}^{*}=3*230.94\angle 0^{o}*27\angle -39.65^{o}=14417-j11927\: \: VA$

S= P+ jQ= 14417-j11927 VA

Now real power and reactive power is

$\mathbf{P=14417\: \: W,\: \: Q=11927\: \: var}$

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b)

The power developed is

Pa-P-31 R,-14417-3 * 272 * 0.5 = 13323.5 W

$\mathbf{P_{d}=13323.5\: \: W}$

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c)

The back emf per phase is

$E_{a}=V_{ph}-I_{a}(R_{a}+jX_{s})=230.94\angle 0^{o}-27\angle 39.65^{o}*(0.5+j12)$

$E_{a}=499.2\angle -31.2^{o}\: \: V$

torque angle is phase of back emf i.e

$\delta =\angle E_{a}= -31.2^{o}$

$\mathbf{\delta = -31.2^{o}}$

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d)

The shaft power is

Speed of machine is

$N=\frac{120f}{P}=\frac{120*50}{4}=1500\: \: rpm$

Now shaft torque is

$T_{shaft}=\frac{P_{shaft}}{2\pi \frac{N}{60}}=\frac{13123}{2\pi *\frac{1500}{60}}=83.3\: N-m$

$\mathbf{T_{shaft}=83.3\: N-m}$

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e)

Efficiency is

$\eta =\frac{P_{shaft}}{P}=\frac{13123}{14417}*100=91.02\: \: o/o$

$\mathbf{\eta =91.02\: \: o/o}$