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Chapter: Electrical and electronics - Circuit Theory - Analysing Three Phase Circuits

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Solved Problems: Analysing Three Phase Circuits

Electrical and electronics - Circuit Theory - Analysing Three Phase Circuits - Solved Problems: Analysing Three Phase Circuits


Problems

 

1.     The input power to a 3-phase a.c. motor is measured as 5kW. If the voltage and current to the motor are 400V and 8.6A respectively, determine the power factor of the system?

 

Power      P=5000W,

 

line voltage VL = 400 V,

line current, IL = 8.6A and

power,    =√3 VLIL cos φ

 

Hence

power factor = cos φ = P √3 VLIL

 

= 5000 √3 (400) (8.6)

 

= 0.839

 

2.     Two wattmeters are connected to measure the input power to a balanced 3-phase load by the two-wattmeter method. If the instrument readings are 8kW and 4kW, determine (a) the total power input and (b) the load power factor. 

 

(a)Total input power,

P=P1 +P2 =8+4=12kW

 

(b) tan φ =√3(P1 − P2)/(P1 + P2)

=√3 (8 – 4) / (8 + 4)

=√3 (4/12)

 

=√3(1/3)

= 1/ √3

 

Hence                   φ= tan−1 1 √3       =30◦

Power factor=      cos φ= cos 30◦ =0.866

 

3.     Two wattmeters connected to a 3-phase motor indicate the total power input to be 12kW. The power factor is 0.6. Determine the readings of each wattmeter. 

 

If the two wattmeters indicate P1 and P2 respectively

Then P1 + P2 = 12kW

---(1)

tan φ =√3(P1 − P2)/(P1 + P2)

And   power factor=0.6= cos φ.

 

Angle φ= cos−10.6=53.13◦ and

tan 53.13◦ =1.3333.

 

Hence

1.3333 =√3(P1 − P2)/12

From which,

 

P1 − P2          = 12(1.3333) /√3

i.e.    P1 −P2 =9.237kW          ----(2)

 

Adding Equations (1) and (2) gives:

2P1 = 21.237

i.e     P1 = 21.237/2

 

= 10.62kW Hence wattmeter 1 reads 10.62kW From Equation (1), wattmeter 2 reads

(12−10.62)=1.38kW

 

4.     Three loads, each of resistance 30, are connected in star to a 415 V, 3-phase supply. Determine

(a) the system phase voltage, (b) the phase current and (c) the line current.

 

A ‘415 V, 3-phase supply’ means that 415 V is the line voltage, VL

 

(a) For a star connection, VL =√3Vp Hence phase voltage, Vp = VL/√3

= 415 /√3

 

= 239.6 V or 240 V

correct to 3 significant figures

 

(b) Phase current, Ip = Vp/Rp

= 240/30

 

= 8 A

 

(c) For a star connection, Ip = IL Hence the line current, IL = 8 A

 

5.     Three identical coils, each of resistance 10ohm and inductance 42mH are connected (a) in star and (b) in delta to a 415V, 50 Hz, 3-phase supply. Determine the total power dissipated in each case.

 

(a) Star connection

Inductive reactance,

XL =2πf L =2π (50) (42×10−3) =13.19

Phase impedance,

Zp =√(R2 +XL2)

=√(102 +13.192) =16.55

Line voltage, VL =415 V

And phase voltage,

VP =VL/√3=415/√3=240 V.

 

Phase current,

 

Ip =Vp/Zp =240/16.55=14.50 A. Line current,

IL =Ip =14.50 A.

 

Power factor= cos φ=Rp/Zp =10/16.55 =0.6042 lagging.

 

Power dissipated,

 

=√3 VLIL cos φ =√3 (415) (14.50)(0.6042) = 6.3kW (Alternatively,

 

=3I2R =3(14.50)2(10)=6.3kW)

(b) Delta connection

 

VL = Vp = 415 V,

 

Zp = 16.55_, cos φ = 0.6042 lagging (from above). Phase current,

 

Ip =Vp/Zp =415/16.55=25.08ALine current,

IL =√3Ip =√3(25.08)=43.44A.

 

Power dissipated,

=√3 VLIL cos φ

=√3 (415)(43.44)(0.6042) = 18.87kW

(Alternatively,

=3I2R

=3(25.08)2(10) =18.87 kW)

 

6.     A 415V, 3-phase a.c. motor has a power output of 12.75kW and operates at a power factor of 0.77 lagging and with an efficiency of 85 per cent. If the motor is delta-connected, determine (a) the power input, (b) the line current and (c) the phase current.

 

 

(a)  Efficiency=power output/power input.

Hence

 

(85/100)=12.750 power input from which, Power input = 12. 750 × 10085

= 15 000W or 15Kw

 

(b)  Power, P=√3 VLIL cos φ, hence

 

(c)    line current,

IL = P/ √3 (415) (0.77)

 

= 15 000/ √3 (415) (0.77)

 

= 27.10A

 

(d)  For a delta connection, IL =√3 Ip,

Hence

 

Phase current, Ip = IL/√3

= 27.10 /√3

 

= 15.65A

 

7.     A 400V, 3-phase star connected alternator supplies a delta-connected load, each phase of which has a resistance of 30_ and inductive reactance 40_. Calculate (a) the current supplied by the alternator and (b) the output power and the kVA of the alternator, neglecting losses in the line between the alternator and load.

 

A circuit diagram of the alternator and load is shown in Fig.

 

(a)  Considering the load:

 

Phase current, Ip =Vp/Zp

Vp =VL for a delta connection,

Hence Vp =400V.

 

Phase impedance,

Zp =√ (R2+XL2)

 

=√ (302 +402) =50


Figure

Hence Ip =Vp/Zp =400/50=8A.

For a delta-connection,

 

Line current, IL =√3 Ip =√3 (8) =13.86 A.

Hence 13.86A is the current supplied by the alternator.

 

(b) Alternator output power is equal to the power Dissipated by the load

 

I.e. P =√3 VLIL cos φ, Where cos φ = Rp/Zp = 30/50 = 0.6.

 

Hence P =√3 (400) (13.86) (0.6) = 5.76kW.

Alternator output kVA,

 

S =√3 VLIL =√3 (400) (13.86)

 

9.60 kVA.


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