Answered: The synchronous machine of Examples 5.1 and 5.2 is to be operated as a synchronous generator. For operation at 60 Hz with a terminal voltage of 460 V…

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Question: The synchronous machine of Examples 5.1 and 5.2 is to be operated as a synchronous generator.
For operation at 60 Hz with a terminal voltage of 460 V line-to-line, calculate the field current
required to supply a load of 85 kW, 0.95 power-factor leading.

given Solution:
46.3 A

Can someone explain how to reach this answer? I've tried to figure it out in several ways but the closest answer I got was 54.6 A.

Thank you!

A 60-Hz, three-phase synchronous motor is observed to have a terminal voltage of 460 V
(line-line) and a terminal current of 120 A at a power factor of 0.95 lagging. The field-current
under this operating condition is 47 A. The machine synchronous reactance is equal to 1.68 2
(0.794 per unit on a 460-V, 100-kVA, three-phase base). Assume the armature resistance to be
negligible.
a. Using the motor reference direction for the current and neglecting the armature resistance,
the generated voltage can be found from the equivalent circuit of Fig. 5.3a or Eq. 5.22 as
Ê af = V₁ - jX₂Î₂
We will choose the terminal voltage as our phase reference. Because this is a line-to-
neutral equivalent, the terminal voltage V₂ must be expressed as a line-to-neutral voltage
460
265.6 V, line-to-neutral
V₂ =
A lagging power factor of 0.95 corresponds to a power factor angle = −cos-¹
(0.95) = -18.2°. Thus, the phase-a current is
Î₁ = 120 e-/18.2° A
Thus
Êt = 265.6-j1.68 (120 e-j¹8.2°)
af
= 278.8 e-43.40 V, line-to-neutral
and hence, the generated voltage Eaf is equal to 278.8 V rms, line-to-neutral.
b. The field-to-armature mutual inductance can be found from Eq. 5.21. With ₂ = 120,
√2 Eat
Lat
√2 x 279
120 x 47
22.3 mH
w.le
c. The three-phase power input to the motor Pin can be found as three times the power input
to phase a. Hence,
Pin = 3V₁I₁ x (power factor) = 3 x 265.6 x 120 x 0.95
= 90.8 kW = 122 hp
EXAMPLE 5.1

EXAMPLE 5.2
Assuming the input power and terminal voltage for the motor of Example 5.1 remain constant,
calculate (a) the phase angle & of the generated voltage and (b) the field current required to
achieve unity power factor at the motor terminals.
■ Solution
a. For unity power factor at the motor terminals, the phase-a terminal current will be in
phase with the phase-a line-to-neutral voltage V₁. Thus
From Eq. 5.22,
I₁ =
Pin
1₁ = =
3V₂
Êat = V₁-jX,Î.
= 265.6-j1.68 x 114 = 328 e
Thus, Euf = 328 V line-to-neutral and 8 = -35.8°.
√E
W.L.
90.6 kW
3 x 265.6 V
=
= 114 A
5.2 Synchronous-Machine Inductances; Equivalent Circuits
b. Having found L₁ in Example 5.1, we can find the required field current from Eq. 5.21.
√2 x 328
377 x 0.0223
-135.8
= 55.2 A
V₂, line-to-neutral
273

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Similar questions

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Subject:machines topic: synchronous machines
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