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 69P

A certain store contains three balanced three-phase loads. The three loads are:

Load 1: 16kVA at 0.85 pf lagging

Load 2: 12 kVA at 0.6 pf lagging

Load 3: 8 kW at unity pf

The line voltage at the load is 208 V rms at 60 Hz, and the line impedance is 0.4 + j 0.8 Ω . Determine the line current and the complex power delivered to the loads.

Expert Solution & Answer
Check Mark
To determine

Find the line currents and the complex power delivered to the loads.

Answer to Problem 69P

The line currents Ia, Ib and Ic are 94.3262.05°A_, 94.32177.95°A_, and 94.3257.95°A_  respectively.

The complex power delivered to the loads is 28.8+j18.03kVA_.

Explanation of Solution

Given data:

The given three balanced three-phase loads are,

The reactive power of the Load 1 is 16kVA and the power factor is 0.85 (lagging).

The reactive power of the Load 2 is 12 kVA and the power factor is 0.6 (lagging).

The real power of the Load 3 is 8kW and the power factor is unity.

The line voltage at the load is 208V(rms) at 60 Hz.

The line impedance is 0.4+j0.8Ω.

Formula used:

Write the expression to find the complex power S1 for 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 average power P1.

P1=S1cosθ1 (2)

Here,

S1 is the apparent power of the Load 1.

cosθ1 is the power factor of the Load 1.

θ1 is the power factor angle of the Load 1.

Write the expression to find the reactive power Q1.

Q1=S1sinθ1 (3)

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 of the Load 2.

P2=S2cosθ2 (5)

Here,

S2 is the apparent power of the Load 2.

cosθ2 is the power factor of the Load 2.

θ2 is the power factor angle of the load 2.

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

Q2=S2sinθ2 (6)

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

S3=P3+jQ3 (7)

Here,

P3 is the real power of the Load 3, and

Q3 is the reactive power of the Load 3.

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

P3=S3cosθ3 (8)

Here,

S3 is the apparent power of the Load 3.

cosθ3 is the power factor of the Load 3.

θ3 is the power factor angle of the load 3.

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

Q3=S3sinθ3 (9)

Write the expression to find the total complex power.

S=S1+S2+S3 (10)

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.

Write the expression to find the phase voltage.

Vp=VL3 (11)

Here,

VL is the line voltage at the load.

Write the expression to find the line to neutral voltage (Van).

Van=Vp30° (12)

Here,

Vp is the phase voltage.

Write the expression to find the complex power S.

S=3VI* (13)

Here,

V is the voltage, and

I* is the conjugate of the current Ia.

Write the expression to find the line current Ib.

Ib=Ia+240° (14)

Here,

Ia is the line current.

Write the expression to find the line current Ic.

Ic=Ia+120° (15)

Calculation:

The given lagging power factor of the Load 1 is,

cosθ1=0.85

Rewrite the above equation to find the angle θ1.

θ1=cos1(0.85)=31.788°θ131.79°

Substitute 0.85 for cosθ1, and 16kVA for S1 in equation (2) to find P1.

P1=(16kVA)(0.85)=13.6kVA{1W=1VA}P1=13.6kW

Substitute 31.79° for θ1, and 16kVA for S1 in equation (3) to find Q1.

Q1=(16kVA)sin(31.79°)=(16kVA)(0.5268)=8.428kVARQ18.43kVAR

Substitute 13.6kW for P1, and 8.43kVAR for Q1 in equation (1) to find S1.

S1=13.6kW+j8.43kVAR=(13.6+j8.43)kVA{1W=1VA}

The given lagging power factor of the Load 2 is,

cosθ2=0.6

Rearrange the above equation to find the angle θ2.

θ2=cos1(0.6)=53.130°

Substitute 0.6 for cosθ2, and 12kVA for S2 in equation (5) to find P2.

P2=(12kVA)(0.6)=7.2kW

Substitute 53.130° for θ2, and 12kVA for S2 in equation (6) to find Q2.

Q2=(12kVA)sin(53.130°)=(12kVA)(0.8)Q2=9.6kVAR

Substitute 9.6kVAR for Q2, and 7.2kW for P2 in equation (4) to find S2.

S2=7.2kW+j9.6kVAR=(7.2+j9.6)kVA{1W=1VA}

The given unity power factor of the Load 3 is,

cosθ3=1

Rewrite the above equation to find the angle θ3.

θ3=cos1(1)=0

Substitute 1 for cosθ3, and 8kW for P3 in equation (8).

8kW=S3(1)

Rewrite the above equation to find S3.

S3=8kW=8kVA{1W=1VA}

Substitute 8kVA for S3, and 0 for θ3 in equation (9) to find Q3.

Q3=(8kVA)sin(0)=0kVAR

Substitute 8kW for P3, and 0kVAR for Q3 in equation (7) to find S3.

S3=8kW+j0kVAR=8kVA{1W=1VA}

Substitute 13.6+j8.43kVA for S1, 7.2+j9.6kVA for S2, and 8kVA for S3 in equation (10) to find S.

S=[13.6+j8.43kVA]+[7.2+j9.6kVA]+[8kVA]=28.8+j18.03kVA

Substitute 208V for VL in equation (11) to find Vp.

Vp=208V3=120.09V

Substitute 120.09V for Vp in equation (12) to find Van.

Van=120.0930°V

Substitute 120.0930°V for V, and 28.8+j18.03kVA for S in equation (13).

28.8+j18.03kVA=3(120.0930°V)I*

Re-write the above equation to find the current I*.

I*=28.8+j18.03kVA3(120.0930°V)=33.9832.05°kVA360.2730°V=33.98360.27(32.05°+30°)kAI*=0.09431862.05°kA

The complex current I* is approximated as follows.

I*=94.3262.05°kA

The line current Ia is the complex conjugate of the current I*.

Ia=94.3262.05°A

Substitute 94.3262.05°A for Ia in equation (14) to find the line current Ib.

Ib=(94.3262.05°A)(+240°)=94.32(62.05°+240°)AIb=94.32177.95°A

Substitute 94.3262.05°A for Ia in equation (15) to find the line current Ic.

Ic=(94.3262.05°A)(120°)=94.32(62.05°+120°)AIc=94.3257.95°A

Conclusion:

Thus,

The line currents Ia, Ib and Ic are 94.3262.05°A_, 94.32177.95°A_, and 94.3257.95°A_ respectively.

The complex power delivered to the loads is 28.8+j18.03kVA_.

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Chapter 12 Solutions

Fundamentals of Electric Circuits

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