A 3phase synchronous motor of 10 KW at 1.3 kV has a synchronous reactance of 10 Ω/phase. The efficiency of the machine is 75%. Find input power for the motor.
Given that, output to the motor (P_{out}) = 10 kW
Efficiency (η) = 0.75
Motor line current is minimum at cos ϕ = 1
Input to motor
= 13.33 kW
A synchronous motor with negligible armature resistance runs at a load angle of 20° at the rated frequency. If supply frequency is increased by 10%, keeping the other parameters constant, the new load angle will be
Power delivered by the synchronous motor
From data, other parameters such as P, E and V are constant.
The per phase circuit equivalent of a synchronous generator feeding a synchronous motor through a transmission line is shown in fig. Each machine is rated 10 kVA, 400V, 50 Hz. The motor is driving a load of 8 kW. The field current of the two machine are so adjusted that the motor terminal voltage is 400V and its pf is 0.8 leading. Find the minimum value of E_{m} (in P.U) for the machine to remain in synchronism.
For maximum E_{m}, δ_{m} = 90° limit of stability
The rotor of a 50 Hz, 4 poles, 3ϕ synchronous motor is directly coupled to the rotor of a 50 Hz, 6 poles induction motor. If the stators of both machines are given balanced 3ϕ, 50 Hz supplies of rated voltage values. The possible induced frequencies in the rotor of induction motor are
We know that, N = 120f/P
Speed of synchronous motor
Speed of induction motor
Given that both the rotors are directly coupled.
If rotating field is in the opposite direction to that of rotor speed.
Speed = 1500 + 1000 = 2500 rpm
If rotating field is in the same direction to that of rotor speed,
Speed = 1500 – 1000 = 500 rpm
A cylindrical rotor synchronous motor is used to improve the power factor of a load 5 MW at a power factor of 0.6 lag. The motor is required to meet an additional demand of 0.2 MW and raise the overall power factor to 0.8 (lag). If the efficiency of the motor is 75%. The power factor of the motor is – (up to two decimal places)
Rating of load is –
P_{1} = 5 MW, cos ϕ_{1} = 0.6
Q_{1} = P_{1} tan ϕ
= 5 × tan (cos^{1} (0.6)) = 6.67 MVAR
Required motor output = 0.2 MW
Efficiency = 0.75
Total load = 5 + 0.267 = 5.267 MW
New power factor cos ϕ = 0.8 lag.
Reactive power, Q = P tan ϕ
= 5.267 × tan (cos^{1} (0.8))
= 3.95 MVAR
MVA supplied by motor, Q_{2} = Q_{1} – Q = 6.67 – 3.95 = 2.72 MVAR
Q_{2} = P_{2} tan ϕ_{2}
⇒ 2.72 = 0.267 tanϕ_{2}
⇒ ϕ_{2} = 84.39°
Cosϕ_{2} = 0.098
A 60 MVA, 11 kV, 50 Hz, Y connected, 3 phase salient pole synchronous motor has direct axis synchronous reactance X_{d} = 1.8 Ω and quadrature axis synchronous reactance X_{q} = 1.2 Ω. If the armature resistance is negligible and full load power factor is 0.8 log. The power angle is
For logging load in salient pole motor,
ψ = 0.338°
δ + ψ = 3.338°
⇒ δ = 0.338°  36.86° = 36.52°
A 3phase synchronous motor of 10 KW at 1.3 kV has synchronous reactance of 10 Ω/phase. The efficiency of the machine is 75%. The induced emf (per phase) corresponding to minimum current for full load condition is ____ (in V). Neglect armature resistance.
Given that, output to motor (P_{out}) = 10 kW
Efficiency (η) = 0.75
Motor line current is minimum at cos ϕ = 1
= 13.33 kW
Impedance drop = I × synchronous impedance
= 5.92 × 10 = 59.20 V
= 752.89 V
A 400 V, 3ϕ delta connected synchronous motor runs at rated voltage and with an excitation emf of 500 V. Its winding and iron losses are 1800 W and synchronous impedance per phase is (0.7 + 3j) Ω. Calculate the shaft power output in kW
Z_{s} = (0.7 + 3j) = 3.08 ∠76.87
V_{t} = 400 V, E_{f} = 500 V
Maximum out put power
Shaft power = [3 × 46.49 – 1.8]kW = 137.66 kW
The maximum power delivered by 1500 kW, three phase star connected 4 kV, 48 pole 50 Hz synchronous motor, with synchronous reactance of 4 Ω per phase and unity power factor is___(in MW)
A 2300 V, 400 h.p, 60 Hz, 8 pole Y connected synchronous motor has a rated power factor of 0.85 leading. At full load, the efficiency 85%. The armature resistance is 0.4, and the synchronous reactance is 4.4 Ω. Find the E_{a} for this machine when it is operating at full load.
The input power is
The armature current
= 1328 ∠0°  0.4 × 104∠31.8°  j × 4.4 × 104∠31.8°
E_{a} = 1588 ∠15° V
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