Lab report 08

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EAST WEST UNIVERSITY DEPARTMENT OF EEE Course code: EEE 201 Course name: Electrical circuit ΙΙ Lab report Experiment no: 08 Experiment name: Study of line and phase relation of voltage and current In a balanced three phase circuit. Student name: B. M. ADNAN Id: 2011-1-80-020 Section: 01 Group no: 01 Group Ids: 2011-1-80-012 2011-1-80-013 2011-1-80-059 Date of performance: 19-07-11 1

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Transcript of Lab report 08

Page 1: Lab report 08

EAST WEST UNIVERSITYDEPARTMENT OF EEE

Course code: EEE 201Course name: Electrical circuit ΙΙ

Lab reportExperiment no: 08

Experiment name: Study of line and phase relation of voltage and currentIn a balanced three phase circuit.

Student name: B. M. ADNANId: 2011-1-80-020

Section: 01Group no: 01

Group Ids: 2011-1-80-0122011-1-80-0132011-1-80-059

Date of performance: 19-07-11Date of submission: 26-07-11

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OBJECTIVE: In this experiment, we verify the relationship between line and phase voltages and currents in balanced three phased circuit.

CIRCUIT DIAGRAM:

Figure 01: Relation between line and phase for Y- connection.

Figure 02: Relation between line and phase for Δ – connection.

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ANSWER TO THE LAB- REPORT QUESTIONS:

ANS: 02

For figure 01:

This circuit is Y – Y connection.The sequence of this circuit is a - b - c.

Line voltage, Vab = 200 <0° VSo, phase voltage, VAB = (200/√3) × <-30° = 115.47 <-30° VAgain,Line voltage, Vbc = 200 <-120° VSo, phase voltage, VBC = (200<-120°/√3) × <-30° = 115.47<-150° V

Again,Line voltage, Vca = 200 <120° VSo, phase voltage, VCA = (200<120°/√3) × <-30° = 115.47 <90° V

KVL l loop n-a-A-N-n:

-115.47<-30° +1200 IaA + 0 = 0 = > IaA = 115.47 <-30° / 1200 = 0.09623 <-30° A

The load is balanced. So, IbB = 0.09623 <-150° A IcC = 0.09623 <90° A

For the balanced load in Y-Y connection the phase current and he line current is same.So, the phase current: IaA = 0.09623 <-30° A IbB = 0.09623 <-150° A IcC = 0.09623 <90° A

For figure 02:

This circuit is Y-∆ connection.

The phase sequence of this circuit is a - b – c.

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We know that for balanced ∆-∆ connection the line and phase voltage are same.

Vab = VAB = 120< 0° V Vbc = VBC = 120 <-120° V Vca = VCA = 120 < 120° V

Then the phase current, IAB = VAB/1200 = 120 < 0° / 1200 = 0.1 < 0° A

Another phase current, IBC = VBC / 1200 = 120 <-120° / 1200 = 0.1 <-120° A

Another phase current, ICA = VCA / 1200 = 120 < 120° /1200 = 0.1 <120° A

So, line current, IaA = √3 × IAB × <-30° = (√3 × 0.1 <0°) × <-30° = 0.173 <-30° A

Another line current, IbB = √3 × IBC × <-30° = (√3 × 0.1 <-120°) × <-30° = 0.173<-150° A

Another line current, IcC = √3 × ICA × <-30° = (√3 ×0.1 <120°) × <-30° = 0.173 <90° A

COMPARISION

Compare between theoretical and experimental values for Y-Y connection.

Elements Theoretical values Experimental valuesLine current, Il 0.09623 A 0.13 A

Phase current, Ip 0.09623 A 0.013 APhase voltage, Vp 115.47 V 110 VLine voltage, Vl 200 V 200 V

Comment: The differences between theoretical and experimental values are very small. So we can ignore them.

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Compare between theoretical and experimental values for Y-∆ connection.

Elements Theoretical values Experimental valuesPhase current, Ip 0.1 A 0.12 ALine current, Il 0.173 A 0.19 A

Phase voltage, Vp 120 V 100 VLine voltage, Vl 120 V 100 V

Comment: The differences between theoretical and experimental values are very small. So we can ignore them.

ANS: 02

For figure 01:

This circuit is Y-Y connection.

From the experimental values,

Line voltage, VAB = 200 <0° VAnother line voltage, VBC = 200<-120° VAnother line voltage, VCA = 200<120° V

So, phase voltage, Van = (Vab/√3) × <-30° = (200<0°/√3) × <-30° = 115.47 <-30° V

Again, phase voltage, Vbn = (Vbc / √3) × <-30° = (200<-120°/ √3) × <-30° = 115.47 <-150° V

Again, phase voltage, Vcn = (Vca / √3) × <-30° = (200<120°/√3) × <-30° = 115.47 <90° A

For figure 02:

This circuit is Y-∆ connection.

Phase current, IAB = 0.1 <30° AAnother phase current, IBC = 0.1 <-90° AAnother phase current, ICA = 0.1<150° A

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Now, the line current, IaA = (√3 × IAB) × <-30° = (√3×0.1<30°) × <-30° = 0.173<0° A

Another line current, IbB = (√3 × IBC) × <-30° = (√3 ×0.1<-90°) × <-30° = 0.173<-120° A

Another line current, IcC = (√3 × ICA) × <-30° = (√3 × 0.1<150°) × <-30° = 0.173<120° A

ANS:03

The current of neutral line is,

InN = - (IaA + IbB +IcC) = - (0.173<0° + 0.173<-120° + 0.173<120°) = 0

So, the theoretical calculation shows that the neutral line of a balanced Y-load is zero.

DISCUSSION: In this experiment we have known that how to construct Y-Y and Y-∆ connection. We also measure line and phase current. From now it is easy for us to construct this type of circuit and calculate it.

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