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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

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

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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Attached are the examples mentioned in the question below. These are examples from a textbook, not a graded exam or assignment so please do not reject.

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
Transcribed Image Text: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
Transcribed Image Text: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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