750kW alternatorsP₁ = 750kW, P₂=750kWare operated in parallel.Speed regulation of one machine is 100% to 103%.Speed regulation of second machine is 100% to104%.Shared load,(P₁)' + (P₂)'arrow_forwardStep 2=(P₂)1000kW....... (A)So the shared load equation for alternator-1 will be,103-100-103-103-(B)'= 103- (Pi)......(₁)The shared ad equation for alternator-2 will be,104 100 104(P₂)750 = 104-1= 104 (P₂)'......(²)Now from equation (1) and (2),103 (P₁) 104- (P₂)'103-104(P)' -(P₂)'(P₁)-(P₂)' = -13(P₁)-4(P₂)-750kW.......On solving equation (A) and (B),(P₁) 464. 28kW(P₂) 535. 714kWHow did he get the value insidethe circle?Can you explain step by step andclear line please?

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750kW alternators are operated in parallel. P₁ = 750kW, P₂ = 750kW Speed regulation of one machine is 100% to 103%. Speed regulation of second machine is 100% to 104%. Shared load, (P₁)' + (P₂)' = 1000kW….…..... (A) arrow_forward Step 2 So the shared load equation for alternator-1 will be, 103-100 103-z 750 (B)' 3 750 103- (P)' x = 103- (P₁)...... (1) The shared load equation for alternator-2 will be, 104 100 104 z 750 (P₂)* 4 750 104- (P₂)' 2 = 104 - (P) .... (2) = Now from equation (1) and (2), 103 (P₁) 104- (P₂) 750 103- 104 = ਰੰਗ (PA)' — ਨ (P2) ' 750 (P₁)-(P₂)' = -1 3(P₁)' — 4(P₂)' = -750kW....... (B) = On solving equation (A) and (B), (P₁)' = 464.28kW (P₂)' = 535.714kW

750kW alternators are operated in parallel. P₁ = 750kW, P₂ = 750kW Speed regulation of one machine is 100% to 103%. Speed regulation of second machine is 100% to 104%. Shared load, (P₁)' + (P₂)' = 1000kW….…..... (A) arrow_forward Step 2 So the shared load equation for alternator-1 will be, 103-100 103-z 750 (B)' 3 750 103- (P)' x = 103- (P₁)...... (1) The shared load equation for alternator-2 will be, 104 100 104 z 750 (P₂)* 4 750 104- (P₂)' 2 = 104 - (P) .... (2) = Now from equation (1) and (2), 103 (P₁) 104- (P₂) 750 103- 104 = ਰੰਗ (PA)' — ਨ (P2) ' 750 (P₁)-(P₂)' = -1 3(P₁)' — 4(P₂)' = -750kW....... (B) = On solving equation (A) and (B), (P₁)' = 464.28kW (P₂)' = 535.714kW

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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8 = 0 Q1: What does happen if we change two currents? Q2: What does happen if we have a three-phase four poles machine? a Q3: What does happen if we have a two-phase two poles machine? Q4: What does happen if we apply a capacitance in phase b and then use same voltage for both phase. Dr. Ali K....p. . --
750kW alternators P₁ = 750kW, P₂=750kW are operated in parallel. Speed regulation of one machine is 100% to 103%. Speed regulation of second machine is 100% to 104%. Shared load, (P₁)' + (P₂)' arrow_forward Step 2 = (P₂) 1000kW....... (A) So the shared load equation for alternator-1 will be, 103-100-103- 103- (B)' = 103- (Pi)...... (₁) The shared ad equation for alternator-2 will be, 104 100 104 (P₂) 750 = 104-1 = 104 (P₂)'...... (²) Now from equation (1) and (2), 103 (P₁) 104- (P₂)' 103-104 (P)' - (P₂)' (P₁)-(P₂)' = -1 3(P₁)-4(P₂)-750kW....... On solving equation (A) and (B), (P₁) 464. 28kW (P₂) 535. 714kW How did he get the value inside the circle? Can you explain step by step and clear line please?
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