Please answer Question 5-4 ONLY! **Problem 5-4: Analysis of Generator Performance at Rated Conditions**Assume that the field current of the generator mentioned in Problem 5-2 is adjusted to maintain a rated voltage of 2300 V under full load conditions. Analyze the performance of this generator by addressing the following questions:**(a)** What is the efficiency of the generator at rated load?**(b)** Determine the voltage regulation of the generator when it is loaded to rated kilovolt-amperes (kVA) with a power factor of 0.8 lagging.**(c)** Determine the voltage regulation of the generator when it is loaded to rated kilovolt-amperes (kVA) with a power factor of 0.8 leading.**(d)** Determine the voltage regulation of the generator when it is loaded to rated kilovolt-amperes (kVA) with a unity power factor.---No graphs or diagrams are included with this question set. **5-2.**A 2300-V 1000-kVA 0.8-PF-lagging 60-Hz two-pole Y-connected synchronous generator has a synchronous reactance of 1.1 Ω and an armature resistance of 0.15 Ω. At 60 Hz, its friction and windage losses are 24 kW, and its core losses are 18 kW. The field circuit has a dc voltage of 200 V, and the maximum \( I_f \) is 10 A. The resistance of the field circuit is adjustable over the range from 20 to 200 Ω. The OCC of this generator is shown in Figure P5-1.(a) How much field current is required to make \( V_T \) equal to 2300 V when the generator is running at no load?(b) What is the internal generated voltage of this machine at rated conditions?(c) How much field current is required to make \( V_T \) equal to 2300 V when the generator is running at rated conditions?(d) How much power and torque must the generator’s prime mover be capable of supplying?**Graph Description:**The graph provided is labeled, "Open-circuit terminal voltage, V" on the vertical axis and "Field current, A" on the horizontal axis. The voltage ranges from 0 to 3000 V and the current ranges from 0 to 10 A.- At approximately 0 A, the voltage is approximately 0 V.- As the field current increases, the open-circuit terminal voltage increases non-linearly, forming a curve that gradually flattens out as it approaches 10 A.- The curve starts steeply, indicating that small increases in field current result in relatively large increases in voltage at low currents.- As the field current reaches higher values, the increase in voltage per increase in current diminishes. This curve is typical for an open-circuit characteristic (OCC) of a synchronous generator, demonstrating the relationship between the field current and the open-circuit generated voltage.

The synchronous generator rotates at the synchronous speed. So, the speed of the rotation of the synchronous generator is expressed as : NS=120 fP where, f=frequency P= No. of Poles So, N=120×60 Hz2N=3600 rpm Since the synchronous generator is connected in star connection, thus the per-phase current is equal to the line current.The per-phase terminal voltage across the generator is determined by the following expression : E∠δ=V^+I^ (jXL) where, V^=Per phase voltageI^=Line current of the generatorXL=Per-phase inductive reactive of the generator V^ph=V^L3V^ph=23003 V The per-phase terminal voltage for loaded with the rated current at 0.8 PF lagging. Thus, E∠δ=23003∠0°+10 ∠-cos-10.8Aj1.1ΩE∠δ=1328∠0° V+10 ∠-36.86°Aj1.1ΩE∠δ=1328+j0 V+8-j6 A j1.1ΩE∠δ=1336+6.6+j8.8E∠δ=1334.6+j8.8E∠δ=1334.63∠0.38° V The terminal line voltage of the synchronous generator is given as : EL=1334.63V×3EL=2311.64 V We know that the voltage regulation is given by the expression : % Voltage regulation =EL-VV×100 %=2311.64-23002300×100 %=0.50 %(ii) The per-phase terminal voltage for loaded with the rated current at 0.8 PF leading. Thus, E∠δ=23003∠0°+10 ∠cos-10.8Aj1.1ΩE∠δ=1328∠0° V+10 ∠36.86°Aj1.1ΩE∠δ=1328+j0 V+8+j6 A j1.1ΩE∠δ=1336+-6.6+j8.8E∠δ=1321.4+j8.8E∠δ=1321.43∠0.38° V The terminal line voltage of the synchronous generator is given as : EL=1321.63V×3EL=2289.13 V We know that the voltage regulation is given by the expression : % Voltage regulation =EL-VV×100 %=2289.13-23002300×100 %=-0.47 % (iii) The per-phase terminal voltage for loaded with the rated current at 1 PF (unity). Thus, E∠δ=23003∠0°+10 ∠cos-11Aj1.1ΩE∠δ=1328∠0° V+10 ∠0°Aj1.1ΩE∠δ=1328+j0 V+10+j0 A j1.1ΩE∠δ=1328+j11E∠δ=1328.05∠0.47° V EL=1328.05V×3EL=2300.25 V % Voltage regulation =EL-VV×100 %=2300.25-23002300×100 %=0.0108 %The efficiency of the synchronous generator is expressed as : %η=outputoutput+losses So, the efficiency of the synchronous generator at rated load of 2300 V will be : %η=3×2311.64×10 A0.83×(2311.64)×(10 A)(0.8)+24 kW+18 kW×100 %%η=32031.023432031.0234 kW+(24 kW+18 kW)×100 %%η=99.86 %