please i need handwritting step by step In the figure, bus 1 is the reference bus, V1 = 1.06 p.u and the angle is 0°. A load with active power= 100MW and reactive power = 50 Mvar is connected to bus 2. The line impedance is z12 = 0.12 +j0.16 p.u ona base of 100 MVA. Start with an initial estimate of |V2|0) = 1.0 p.u and 8200) = 0°. Find |V2|(¹) (oneiteration).2100 MWHO*12= 0.12 + j0.16+ 50 MvarV₁ = 1.040⁰O a.|V2|(¹) = 0.4 p.u.O b.|V2|(¹) = 0.5 p.u.OC. |V2|(¹) = 0.7 p.u.O d.|V2|(¹) = 0.8 p.u.O e.|V2|(¹) = 0.6 p.u.

Answered: (1)= 0.4 p.u. V₁ = 1.040⁰ #12 = 0.12 +…

Engineering

Electrical Engineering

(1)= 0.4 p.u. V₁ = 1.040⁰ #12 = 0.12 + j0.16 100 MW + 50 Mvar

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter6: Power Flows

Section: Chapter Questions

Problem 6.61P

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Concept explainers

Winding and Constructions

Alternating Current (AC) winding is generally of the 3-phase type due to the inherent advantages of the 3-phase machine. Armature coils must be connected to the yield balance (equivalent in magnitude and 2/3 of the series phase difference of the rod) 3-phase electromotive forces (emf). The armature should be divided into step-bands to start with slots around the perimeter.In the construction of AC machines, the frame is used as an external protective shield to protect it from environmental conditions. The frame also serves as the outer perimeter, allowing the interior to be easily placed. The static state section of the devices is the stator winding attached stator.

Question

please i need handwritting step by step

In the figure, bus 1 is the reference bus, V1 = 1.06 p.u and the angle is 0°. A load with active power= 100
MW and reactive power = 50 Mvar is connected to bus 2. The line impedance is z12 = 0.12 +j0.16 p.u on
a base of 100 MVA. Start with an initial estimate of |V2|0) = 1.0 p.u and 8200) = 0°. Find |V2|(¹) (one
iteration).
2
100 MW
H
O
*12= 0.12 + j0.16
+ 50 Mvar
V₁ = 1.040⁰
O a.
|V2|(¹) = 0.4 p.u.
O b.
|V2|(¹) = 0.5 p.u.
OC. |V2|(¹) = 0.7 p.u.
O d.
|V2|(¹) = 0.8 p.u.
O e.
|V2|(¹) = 0.6 p.u.

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