Fundamentals of Electric Circuits
Fundamentals of Electric Circuits
6th Edition
ISBN: 9780078028229
Author: Charles K Alexander, Matthew Sadiku
Publisher: McGraw-Hill Education
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Chapter 12, Problem 81CP

A professional center is supplied by a balanced three-phase source. The center has four balance three-phase loads as follows:

Load 1: 150 kVA at 0.8 pf leading

Load 2: 100 kW at unity pf

Load 3: 200 kVA at 0.6 pf lagging

Load 4: 80 kW and 95 kVAR (inductive)

If the line impedance is 0.02 + j0.05 Ω per phase and the line voltage at the loads is 480 V, find the magnitude of the line voltage at the source.

Expert Solution & Answer
Check Mark
To determine

Find the magnitude of the line voltage at the source.

Answer to Problem 81CP

The magnitude of the line voltage at the source is 516V_.

Explanation of Solution

Given data:

A balanced three-phase source connected to four balanced three-phase loads, Those are,

Load 1:

The apparent power of the Load 1 (S1) is 150kVA.

The power factor of the Load 1 is 0.8(leading).

Load 2:

The real power of the Load 2 (P2) is 100kW.

The power factor of the Load 2 is unity.

Load 3:

The apparent power of the Load 3 (S3) is 200kVA.

The power factor of the Load 3 is 0.6 (lagging).

Load 4:

The reactive power of the Load 4 (S4) is 95kVAR (inductive).

The real power of the Load 4 (P4) is 80kW.

The line impedance (Zl) is 0.02+j0.05Ω.

The line voltage at the loads (VL) is 480V.

Formula used:

Write the expression to find the complex power of the Load 1.

S1=P1+jQ1        (1)

Here,

P1 is the real power of the Load 1.

Q1 is the reactive power of the Load 1.

Write the expression to find the real power of the Load 1.

P1=S1cosθ1        (2)

Here,

S1 is the apparent power of the Load 1.

cosθ1 is the power factor of the Load 1.

Write the expression to find the reactive power of the Load 1.

Q1=S1sinθ1        (3)

Here,

θ1 is the power factor angle of the Load 1.

Write the expression to find the complex power of the Load 2.

S2=P2+jQ2        (4)

Here,

P2 is the real power of the load 2.

Q2 is the reactive power of the Load 2.

Write the expression to find the real power (P2).

P2=S2cosθ2        (5)

Here,

S2 is the apparent power of the Load 2.

cosθ2 is the power factor of the Load 2.

Rearrange the equation (5) to find the apparent power (S2).

S2=P2cosθ2        (6)

Write the expression to find the reactive power (Q2).

Q2=S2sinθ2        (7)

Here,

θ2 is the power factor angle of the Load 2.

Write the expression to find the complex power of the Load 3.

S3=P3+jQ3        (8)

Here,

P3 is the real power of the Load 3.

Q3 is the reactive power of the Load 3.

Write the expression to find the real power of the Load 3.

P3=S3cosθ3        (9)

Here,

S3 is the apparent power of the Load 3.

cosθ3 is the power factor of the Load 3.

Write the expression to find the reactive power of the Load 3.

Q3=S3sinθ3        (10)

Here,

θ3 is the power factor angle of the Load 3.

Write the expression to find the complex power of the Load 4.

S4=P4+jQ4        (11)

Here,

P4 is the real power of the Load 4.

Q4 is the real power of the Load 4.

Write the expression to find the total complex power (S) absorbed by the load.

S=S1+S2+S3+S4        (12)

Here,

S1 is the complex power of the Load 1.

S2 is the complex power of the Load 2.

S3 is the complex power of the Load 3.

S4 is the complex power of the Load 4.

Write the expression to find the apparent power (S).

S=3VLIL        (13)

Here,

VL is the line voltage.

IL is the line current.

Write the expression to find the complex power absorbed by the line.

Sl=3IL2ZL        (14)

Here,

ZL is the line impedance.

Write the expression to find the total complex power at the source.

(ST)source=ST+Sl        (15)

Here,

ST is the total complex power absorbed by the load.

Sl is the complex power absorbed by the line.

Write the expression for the apparent power (ST) of the source.

ST=3VTIL        (16)

Here

VT is the line voltage at the source.

IL is the line current.

Calculation:

Load 1:

The given leading power factor of the Load 1 is ,

cosθ1=0.8

Re-write the equation to find the angle θ1.

θ1=cos1(0.8)θ1=36.86°

Substitute 36.86° for θ1, and 150kVA for S1 in equation (3) to find the reactive power of the Load 1.

Q1=(150kVA)sin(36.86°){1k=103}=(150×103VA)(0.6)=90kVAR

Substitute 150kVA for S1, and 0.8 for cosθ1 in equation (2) to find the real power (P1).

P1=(150kVA)(0.8){1k=103}=(150×103VA)(0.8){1W=1VA}=120kW

Substitute 120kW for P1, and 90kVAR for Q1 in equation (1) to find the complex power S1.

S1=120kWj90kVAR=120kVAj90kVAR{1W=1VA}=120j90kVA

Load 2:

Substitute 100kW for P2, and 1 for cosθ2 in equation (6) to find the apparent power (S2).

S2=100kW1=100kVA{1W=1VA}

The given unity power factor of the Load 2,

cosθ2=1

Rewrite the equation to find the angle θ2.

θ2=cos1(1)θ2=0

Substitute 0 for θ2, and 100kVA for S2 in equation (7) to find the reactive power (Q2).

Q2=100kVAsin(0)=0kVAR

Substitute 100kW for P2, and 0kVAR for Q2 in equation (4) to find the complex power (S2).

S2=100kW+j0kVAR=100kVA{1W=1VA}

Load 3:

The given lagging power factor of the Load 3 is,

cosθ3=0.6

Rewrite the equation to find the angle θ3.

θ3=cos1(0.6)θ3=53.13°

Substitute 200kVA for S3, and 0.6 for cosθ3 in equation (9) to find the real power (P3).

P3=(200kVA)(0.6)=120kW{1W=1VA}

Substitute 200kVA for S3, and 53.13° for θ3 in equation (10) to find the reactive power (Q3).

Q3=(200kVA)sin(53.13°)=160kVAR

Substitute 120kW for P3, and 160kVAR for Q3 in equation (8) to find the complex power (S3) of the Load 3.

S3=120kW+j160kVAR=120kVA+j160kVAR{1W=1VA}=120+j160kVA

Load 4:

Substitute 80kW for P4, and 95kVAR for Q4 in equation (11) to find the complex power (S4) of the Load 4.

S4=80kW+j95kVAR=80kVA+j95kVAR{1W=1VA}=80+j95kVA

Substitute 120j90kVA for S1, 100kVA for S2, 120+j160kVA for S3, and 80+j95kVA for S4 in equation (12) to find the total complex power (S).

S=(120j90kVA)+(100kVA)+(120+j160kVA)+(80+j95kVA)=420+j165kVA=451.221.45°kVA

Here, the apparent power is S=451.2kVA.

Substitute 451.2kVA for S, and 480V for VL in equation (13) to find the apparent power (S).

451.2kVA=3(480V)IL

Re-write the above equation to find the line current (IL).

IL=451.2kVA3(480V)=451.2×103VA831.38V{1k=103}=542.7A

Substitute 0.02+j0.05Ω for Zl, and 542.7A for IL in equation (14) to find the complex power absorbed by the line (Sl).

Sl=3(542.7A)2(0.02+j0.05Ω)=17671+j44178A2Ω=17.671+j44.178VA                      {V=AΩ}=17.67+j44.18kVA

Substitute 17.67+j44.18kVA for Sl, and 451.221.45°kVA for ST in equation (15) to find (ST)source.

(ST)source=(17.67+j44.18kVA)+(451.221.45°kVA)=(17.67+j44.18kVA)+(420+j165kVA)=437.67+j209.18kVA(ST)source437.7+j209.2kVA

Re-write the above value as below,

(ST)source=485.125.55° kVA

Substitute  485.1 kVA for ST, 542.7A for IL in equation (16) to find (VT).

485.1 kVA=3(542.7A)VT

Rewrite the above equation to find (VT).

VT=485.1 kVA3(542.7A)=485.1×103VA940A{1k=103}=516.07VVT516V

Conclusion:

Thus, the magnitude of the line voltage at the source is 516V.

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Chapter 12 Solutions

Fundamentals of Electric Circuits

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