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10. MEASUREMENT OF REACTIVE VOLT AMPERES

In order to measure reactive power in a single phase circuit a compensated wattmeter is used. In this wattmeter the voltage applied to the pressure coil is 90° out of phase with the actual voltage
and hence it will read VI cos (90° – ɸ) i.e., VI sin ɸ reactive power.

Fig.34

Fig. 34. Measurement of reactive volt-amperes.

In balanced three-phase circuit the reactive power can be determined by using one wattmeter. The necessary connections are shown in Fig. 34. The current coil is inserted in one line and the pressure coil is connected across the other two lines.
The current following through the current coil of the wattmeter
= IY = Iph (say)
The potential difference across the potential coil of wattmeter,
E = ER – EB
= √3 Eph leading vector EY by 90°
= √3 Eph leading the vector of current IY by (90° + ɸ)
∴ Reading of the wattmeter (Fig. 35)
= √3 EphIph cos (90° + ɸ)
= -√3 EphIph sin ɸ = – Wr
Total reactive power of the circuit,
= 3 EphIph sin ɸ = -√3 Wr
Reactive power (as earlier started) can also be determined from two wattmeter readings connected for measurement of power.
W1 – W2 = ELIL cos (30° – ɸ) – ELIL cos (30° + ɸ)
W1 – W2 = ELIL × 2 sin 30° sin ɸ
ELIL sin ɸ = W1 – W2
Reactive power of load circuit.
Wr = √3 ELIL sin ɸ = √3 (W1 – W2).

POWER MEASUREMENT IN 3•PHASE CIRCUITS
Example 18. The power input to a 3-phase induction motor is read by two wattmeters. The readings are 920 W and 300 W. Calculate the power factor of the motor.
Solution. Reading of wattmeter, W1 = 920 W
Reading of wattmeter, W2 = 300 W
Power factor of the motor, cos ɸ:
Using the relation, tan⁡ɸ=(√3 (W_1-W_2))/((W_1+W_2))=(√3 (920-300))/((920+300))=(√3 ×620)/1220=0.88
ɸ = tan-1 0.88 = 41.35°
Power factor of the motor,
cos ɸ = cos 41.35° = 0.75 (lag). (Ans.)
Example 19. While performing a load test on a 3-phase wound-rotor induction motor by two
wattmeters method, the readings obtained on two wattmeters were + 14.2 kW and – 6.1 kW and the line voltage was 440 V. Calculate:
(i) Trite power drawn by the motor
Iii) U/I,(‘ current.

HnluUol1. Itondinlf “rwlli,Lmotor,
1(11,,1111111 (I/’Wlll.l.llIlIl.lll”
1.11111 VIlII,IIIIII.
(ii) Power factor, and
WI. 111.2 kW
W,~ . 0,1 I~W
N” 1.t1O V
(i) True power drawn by the motor
(ii) Power factor, and
(iii) Line current.

Solution. Reading of wattmeter, W1 = 14.2 kW
Reading of wattmeter, W2 = -6.1 kW
Line voltage, EL = 440 V
(i) True power drawn by the motor
= 14.2 – 6.1 = 8.1 kW. (Ans.)
(ii) tan⁡ɸ=(√3(W_1-W_2))/((W_1+W_2))=(√3 [(14.2-(-6.1)])/[14.2+(-6.1)] =(√3×20.3)/8.1=4.34
∴ ɸ = tan-1 4.34 = 77o
∴ Power factor, cos ɸ = cos 77o = 0.2249 (lag). (Ans.)

(iii) Line current, IL:
Using the relation, P =√3 ELIL cos ɸ
8.1 × 1000 =√3 × 440 × IL × 0.2249
∴ I_L=(8.1×1000)/(√3×440×0.2249)=47.26 A
Hence, line current = 47.26 A. (Ans.)

Example 20.A 3-phase, 440 V motor load has a power factor of 0.6. Two wattmeters connected to measure the power show the input to be 25 k W. Find the reading on each instrument.

Solution. Input power = 25 kW
Line voltage, EL = 440 V
Power factor of the motor load, cos ɸ = 0.6

W1 ; W2 :
Using the relation: tan⁡〖ɸ=(√3(W_1-W_2))/((W_1+W_2))〗 …(i)
Now, cos ɸ = 0.6, ∴ ɸ = 53.13o
∴ tan ɸ = tan 53.13o = 1.333
Also W1 + W2 = 25 kW (given) … (ii)
Substituting these values in (i), we get
1.333=(√3(W_1-W_2))/25
(W_1-W_2 )=(1.333×25)/√3=19.24 kW …(iii)
From (ii) and (iii), we get W1 = 22.12 kW
and W2 = 2.88 kW. (Ans.)

Example 21. In a 3-phase circuit two wattmeters used to measure power indicate 1200 W and 600W respectively. Find the power factor of the circuit:
(i) When both wattmeter readings are positive.
(ii)When the latter is obtained by reversing the current coil connections.
(AMIE Summer, 1999)

Solution. (i) When both wattmeter readings are positive:
Reading of wattmeter, W1 = 1200 W
Reading of wattmeter, W2 = 600 W
We know that, tan⁡〖ɸ=(√3(W_1-W_2))/((W_1+W_2))=(√3(1200-600))/((1200+600))〗=0.577
ɸ=〖tan〗^(-1) 0.577=〖30〗^o
Power factor, cos ɸ=〖cos 30〗^o=0.866 (lag.) (Ans.)

(ii) When the reading of wattmeter W2 is obtained by reversing the coil connection:
Reading of wattmeter, W1 = 1200 W
Reading of wattmeter, W2 = – 600 W
We know that, tan⁡〖ɸ=(√3(W_1-W_2))/((W_1+W_2))=(√3 [1200-(600)])/[1200+(-600)] 〗=(√3×1800)/600=5.196
or ɸ = tan-1 5.196 = 79.1o
Hence, power factor = cos ɸ = cos 79.1o = 0.1889. (Ans.)

Example 22. In order to measure the power input and the power factor of an over-excited synchronous motor two wattmeters are used. If the meters indicate (- 3.5 kW) and (+ 8.0 kW) respectively. Calculate:
Power factor of the motor.
Power input to the motor.

Solution. (i) Since an over-excited synchronous motor runs with a leading power factor, we should use the relation,
tan⁡〖ɸ=(√3(W_1-W_2))/((W_1+W_2))〗
Moreover it is W1 that gives negative reading and not W2. Hence W1 = – 3.5 kW.
∴ tan⁡〖ɸ=-(√3(-3.5-8))/((-3.5+8))=(√3×11.5)/5.5=3.62〗
∴ 〖 ɸ〗⁡= 〖tan〗^(-1) 3.62=〖74.6〗^o (lead)

∴ Power factor, 〖cos ɸ〗⁡= cos 〖74.6〗^o=0.2655 (lead) (Ans.)
(ii) Power input = W1 + W2 = – 3.5 + 8 = 4.5. W. (Ans.)

Example 23. Two wattmeters are used to measure power input to a 1.5 kV, 50 Hz, 3-phase motor running on full-load at an efficiency of 85 per cent. Their readings are 250 k W and 80 kW respectively. Calculate:
Input,
Power factor,
Line current, and
Output.

Solution. Since the motor is running at full-load, its power factor must be greater than 0.5. Hence W2 reading is positive
∴ W1 = + 250 kW and W2 = + 80 kW

(i) Input = W1 + W2 = 250 + 80 = 330 kW. (Ans.)
(ii) tan⁡〖ɸ=(√3(W_1-W_2))/((W_1+W_2))=(√3(250-80))/((250+80))=0.892〗
∴ 〖 ɸ〗⁡= 〖tan〗^(-1) 0.892=〖41.74〗^o
and power factor = cos ɸ = cos 41.74o = 0.746 (lag). (Ans.)
(iii)Power, P⁡〖=√3 E_L I_L 〗 cos⁡ɸ
∴ E_L⁡〖=P/(√3 E_L cos⁡Φ )=(330×1000)/(√3×1.5×1000×0.746)=170.27〗
Hence, line current = 170.27 A. (Ans.)
(iv) Output = input × efficiency = 330 × 0.85 = 280.5 kW. (Ans.)

Example 24. Two wattmeters are used to measure power input to a synchronous motor. Each of the indicates 60 kW. If the power factor be changed to 0.866 leading, determine the readings of the two wattmeters, the total input power remaining the same. Draw the vector diagram for the second condition of the load.

Solution. Reading of wattmeter, W1 = 60 kW

Reading of wattmeter, W2 = 60 kW
First case. In the first case, both wattmeters read equal and positive. Hence, motor must be
running at unity power factor.
Second case. Power factor is 0.866 leading.
In this case: W1 =. ELIL cos (30° + ɸ)
W2 = ELIL cos (30° – ɸ)
∴ W1 + W2 = √3 ELIL cos ɸ
W1 – W2 = -ELIL sin ɸ
∴ tan⁡ɸ=-(√3(W_1-W_2))/((W_1+W_2))
Since cos ɸ = 0.866 (given)
∴ 1/√3=-(√3(W_1-W_2))/120
[∵W_1+W_2=120 kW (given)] ∴ W_1-W_2=-120/3=-40 …(i)
But W_1+W_2=120 …(ii)

Fig.36

Fig. 36

From (i) and (ii), we get W1 = 40 kW, W2 = 80 kW. (Ans.)

For connection diagram, please refer to Fig. 29. The vector or phasor diagram is shown in
Fig. 36.

Example 25. Two wattmeters connected to read the total power in a 3-phase system supplying a balanced load read 10.5 kW and - 2.5 kW respectively. Calculate the total power and power factor.

Also, explain the significance of (i) equal wattmeter readings and (ii) a zero reading on one wattmeter.

(AMIE Winter, 1998) 

Solution. Given: W1 = 10.5 kW ; W2 = – 2.5 kW
For two wattmeter method,
total power = W1 + W2 = 10.5 + (- 2.5) = 8.0 kW. (Ans.) \
We know that, tan⁡ɸ (√3 (W_1-W_2))/(W_1+W_2 )=(√3[10.5-(-2.5)])/([10.5+(-2.5)])=2.8145
or ɸ = 70.44o
∴ Power factor, cos ɸ = cos 70.44o = 0.335. (Ans)
(i) For readings of the two wattmeters to be equal,

W1 = W2
or √3 E_L I_L cos⁡〖(〖30〗^o-ɸ)〗=√3 E_L I_L cos⁡〖(〖30〗^o+ɸ)〗
or cos⁡〖(〖30〗^o-ɸ)〗=cos⁡〖(〖30〗^o+ɸ)〗
∴ ɸ=0^o or cos⁡〖ɸ=1〗
i.e., for unity power factor, the readings of two wattmeters are equal. (Ans.)
(ii) As readings of wattmeters are
W_1=√3 E_L I_L cos⁡〖(〖30〗^o-ɸ),and〗
W_2=√3 E_L I_L cos⁡〖(〖30〗^o+ɸ),〗
For reading of one the wattmeters to be zero ɸ must be 60o, which makes W2 = √3 ELIL cos 90o=0

Example 26. A star-connected 3-phase load consists of three similar impedances. When load is
connected to a 3-phase 400 V, 50 Hz supply it takes 30A line current at a power factor of 0.8 lagging.
(i) Calculate the total power taken by the load.
(ii) Calculate the resistance and-reactance of each phase of the load.
(iii) If power is measured by two wattmeters method, then calculate the reading of each wattmeter.

(AMIE Winter, 1996)


Solution. Given: EL = 400 V; f =50 Hz ; IL =Iph = 30 A, cos ɸ =0.8 lagging.
(i) Total power taken by the load, P:
P=√3 E_L I_L cos⁡ɸ
=√3×400×30×0.8=16627 W or 16.627 kW. (Ans)
(ii) Resistance and reactance of each phase of the load, R, X :
E_ph=E_L/√3=400/√3=230.95 V
∴Load impedance,Z=E_ph/I_ph =230.95/30=7.7 Ω
∴Load resistance,R=z cos⁡〖ɸ=7.7×0.8=〗 6.16 Ω. (Ans.)
and,Load resistance,X=z sin⁡〖ɸ=7.7×0.6=〗 4.62 Ω. (Ans.)
(iii) Reading of each wattmeter, W1, W2:
Reading of wattmeter 1, W1 = ELIL cos (30° + ɸ)
Reading of wattmeter 2, W2 = ELIL cos (30° – ɸ)
where ɸ = cos-1 0.8 = 36.9°
∴ W1 = 400 × 30 cos (30° + 36.9°) 7 4708 W or 4.708 kW. (Ans.)
and W2 = 400 × 30 cos (30° – 36.9°) = 11913 W or 11.913 kW. (Ans.)
Example 27. A 3-phase, 3-wire, 415 V system supplies a balanced load of 20 A at a power factor 0.8 lag. The current coil of wattmeter (1) is in phase R and of wattmeter 2 in phase B, Calculate:
(i) The reading on 1 when its voltage coil is across R and Y;
(ii) The reading on 2 when its voltage coil is across B and Y;
(iii) The reading on 1 when its voltage coil is across Y and B.

(AMIE Winter, 1997)


Solution. Given: EL =415 V, IL = 20 A, cos ɸ = 0.8 (lag) or ɸ = cos-1 0.8 =. 36.87° (lag)
(i) The reading on 1 when its voltage coil is across R and Y and current coil is in R phase
= ELIL cos (30° – ɸ) = 415 × 20 × cos (30° – 36.87°)
= 8240 W or 8.24 kW. (Ans.)
(ii) Reading on wattmeter 2 when its voltage coil is across B and Y and current coil is in phase B
= ELIL cos (30° + ɸ) = 415 × 20 cos (30° + 36.87°)
= 3260 W or 3.26 kW. (Ans.)
(iii) When the current coil is in one phase (phasor R) and the voltage coil is across other two phases (Y and R), the wattmeter reads (total kVAR)/√3 and
Total kVAR = √3 (W1-W2)
∴Wattmeter reading= (Total kVAR)/√3=(√3(W_1-W_2))/√3
= W_1-W_2=8240-3260=4980 VAR.(Ans)
Example 28. The ratio of the readings of two wattmeters connected to measure power in a balanced 3 – ɸ, 3 wire load is 5 : 3. The load is known to be inductive with a lagging power factor. Calculate the power factor of the load.
(Punjab University)
Solution. Let the ratio of wattmeter readings be r i.e., W_1/W_2 =r
Now tan⁡〖ɸ=〗 √3 ((W_1-W_2)/(W_1+W_2 ))=√3 (1-W_2/W_1 )/(1+W_2/W_1 )=√3 ((1-r)/(1+r))
and power factor,cos⁡〖ɸ=1/sec⁡ɸ =〗 1/√(1+〖tan〗^2 ɸ)=1/√(1+3((1-r)/(1+r))^2 )
Substituting r= W_1/W_2 =3/5=0.6 in the above expression,we get
cos⁡〖ɸ=1/√(1+3((1-0.6)/(1+0.6))^2 )=0.918 (lag). Ans〗

Example 29. A symmetrical star-connected three-phase load is made of three identical coils having an internal resistance of 12 Ω and inductance of 142 m.H. The power taken by this load when connected to a balanced-3-phase, three wire supply of 400 V, 50 Hz is measured by using the two wattmeters method. The wattmeters used are similar and are of type of 400 V, 5 A, UPF, class 1 accuracy.
(i) Determine expected readings of W1 and W2.
(ii) Compute the total power.

(Gate, 1995)


Solution, Line voltage EL = 400 V
Phase voltage, E_ph=400/√3=231 V
Coil resistance, R_ph=12 Ω
Coil inductive reactance, X_ph=2πfL_ph=2π×50×142×〖10〗^(-3)=44.61 Ω
Coil impedance, Z_ph= √(R_(〖ph〗^2 )+X_(〖ph〗^2 ) )=√(〖12〗^2+〖44.61〗^2 )=46.2 Ω
Coil phase angle, ɸ= 〖cos〗^(-1) R_ph/Z_ph =〖cos〗^(-1) (12/46.2)=〖74.94〗^o (lagging)
Phase current, I_ph= E_ph/Z_ph =231/46.2=5A

(i) Reading of wattmeters W1, W2 :
Readingsof wattmeter, W_1=E_L I_L cos⁡〖(〖30〗^o-ɸ)〗
=400×5.0 cos⁡〖(〖30〗^o-〖74.94〗^o )=1415.7 W. (Ans)〗
∵ I_L=I_ph=5.0A)
Readingsof wattmeter, W_2=E_L I_L cos⁡〖(〖30〗^o+ɸ)〗
=400×5.0 cos⁡〖(〖30〗^o+〖74.24〗^o )=-515.6 W. (Ans)〗

(ii) Total Power:
Total power =W_1+W_2
=1415.7+(-515.6)=900.1 W. (Ans)
Example 30. Two wattmeters are being used to measure power of a balanced load of 30 A at power factor 0.8 lagging being supplied by a 3-phase, 3-wire, 440 V supply. Calculate:
(i) Power consumed.
(ii) Reading of wattmeter No.1
(iii) Reading of wattmeter No.2.
Solution. Line voltage, EL = 440 V
Line current,I_L=30A
Power factor,cos⁡ɸ=0.8
Phase angle, ɸ=〖cos〗^(-1) 0.8=〖36.87〗^o
(i)Power consumed,P=√3 E_L I_L cos⁡ɸ=√3×440×30×0.8
=18290 W or 18.29 kW. (Ans)
(ii)Reading of Wattmeter No.1
=E_L I_L cos (〖30〗^o-ɸ)=440×30 cos (〖30〗^o-〖36.87〗^o )=13105 W. (Ans.)
(iii)Reading of Wattmeter No.2
=E_L I_L cos (〖30〗^o+ɸ)=440×30 cos (〖30〗^o+〖36.87〗^o )
=5185 W (app.) (Ans.)
Example 31. A balanced star-connected load, each having a resistance of 10 Ω and inductive reactance of 20 Ω is connected to a 400 V, 50 Hz supply. The phase sequence is RYB. Two wattmeters connected to read total power have their current coils connected in the red and blue lines respectively. Calculate the reading of each wattmeter.
Draw the circuit and vector diagram.
Solution. Resistance per phase, Rph = 10 Ω
Reactance per phase, X_ph=20 Ω
Line voltage, E_L=400 V
Impedance per phase, Z_ph=√(R_(〖ph〗^2 )+X_(〖ph〗^2 ) )=√(〖10〗^2+〖20〗^2 )=22.36 Ω
Phase voltage, E_ph=E_ph/√3=400/√3=231 V
I_ph (=I_L )=E_ph/Z_ph =231/22.36=10.33 A
Power factor, cos⁡ɸ=R_ph/Z_ph =10/22.36=0.447
∴ ɸ=〖63.43〗^o

Fig-37Fig. 37


Fig-38
Fig. 38


The circuit and vector diagrams are shown in Figs. 37 and 38 respectively.

Reading of wattmeter, W1 = ERY × IR × cos (30° + ɸ) = EL × IL × cos (30° + ɸ)
= 400 × 10.33 cos (30° + 63.43°) = – 247 W (app.), (Ans.)
Reading of wattmeter, W2 = EBY × IB × cos (30° – ɸ) = EL × IL × cos (30° – 63.43°)
= 400 × 10.33 × cos (30° – 63.43°) = 3448 W (app.), (Ans.)

Example 32. Three equal star-connected inductors take 10 kW at power factor 0.8 when connected to a 440 V, 3 ɸ, 3-wire supply. Give the connection diagram of two single-phase wattmeters to measure lilt, power in the circuit. Allocate the reading of each wattmeter. Find the line currents if one of the inductors is short-circuited. Draw the vector diagrams of the currents and voltages under the condition.
Solution. Power, P = 10 kW = 10000 W
Line voltage, EL = 440 V
Power factor, cos ɸ = 0.8 (lag)
∴ ɸ = 36.9° (lag)
The connection diagram of two single-phase wattmeters to measure the power in the circuit is shown in Figs. 39 and 40.
P=√3 E_L I_L cos⁡ɸ
∴Line current,I_L=P/(√3 E_L cos⁡ɸ )=10000/(√3×440×0.8)=16.4 A.

Reading of wattmeter, W1

= ELIL cos (30° – ɸ) = 440 × 16.4 cos (30° – 36.9°)
= 7163.7 W. (Ans.)

Reading of wattmeter, W2

= ELIL cos (30° + ɸ) = 440 × 16.4 cos (30° + 36.9°)
= 2831.1 W. (Ans.)

Phase current, Iph = IL = 16.4 A
Phase impedance,Z_ph=E_ph/I_ph =(440/√3)/16.4=15.49 Ω
Phase resistane,R_ph=Z_ph cos⁡〖ɸ=15.49×0.8=12.39 Ω〗
Phase reactance,X_ph=Z_ph sin⁡〖ɸ=15.49×0.6=9.29 Ω〗
The circuit diagram when one of the inductors is short-circuited is shown in Fig. 39.

Fig.39

Current in line R,I_R=E_RY/Z_ph =440/15.49=28.4 A
Current in line B,I_B=E_BY/Z_ph =440/15.49=28.4 A
The currents in line R and line B are equal in magnitude but differ in phase by 60°, as shown in Fig. 40.

Fig.40

Current in line Y,I_Y=-(I_R+I_(B))
Magnitude of current flowing in line Y,
I_Y=〖2I〗_R cos⁡〖〖30〗^o 〗=2×28.4×cos⁡〖 〖30〗^o 〗=49.19 A. (Ans)
Example 33. A 3-phase, 3-wire balanced load with a lagging power factor is supplied at 440 V (between lines). A single phase wattmeter (scaled in k W) when connected with its current coil in the R-line and voltage coil between Rand Y lines gives a reading of 8 k W. When the same terminals of the voltage coil are switched over to Y-line and B-line, the current coil connections remaining the same, the reading of the wattmeter remains unchanged. Calculate the line current and power factor of the load.

Solution. Line voltage, EL = 440 V
Phase sequence R → Y → B
The current through the wattmeter is IR and potential difference across its pressure coil is ERY• As seen from the ph as or diagram of Fig. 41 the angle between the two is (30° + ɸ).

Fig.41

∴ W1 = ERYIR cos (30° + ɸ) = ELIL cos (30° + ɸ) … (i)
In the second case, current is IR but voltage is EYB. The angle between the two is (90° – ɸ).
∴ W2 = EYIR cos (90° – ɸ) = ELIL cos (90° – ɸ) … (ii)
Since W1 = W2 (given)
ELIL cos (30° + ɸ) = ELIL cos (90° – ɸ)
∴ 30° + ɸ = 90° – ɸ
2ɸ = 60° ∴ ɸ = 30°
∴ Load power factor,
cos ɸ = cos 30° = 0.866 (lag). (Ans.)
Now, W1 = W2 = 8 kW = 8000 W
From (ii) we get 8000 = ELIL cos (90° – ɸ) = 440 × IL cos (90° – 30°)
∴ I_L=8000/(440 ×cos⁡〖〖60〗^o 〗 )=36.36 A
Hence, line current = 36.36 A. (Ans.)

Example 34. A wattmeter indicates 6 kW when its current coil is connected in red phase and its potential coil is connected between the neutral and red phase of a symmetrical 3-phase system supplying a balanced load of 40 A at 440 V. What will be reading of the instrument if the current coil connections remain unchanged, the potential coil be connected between blue and yellow phases?

Solution. When the wattmeter current coil is connected in red phase and its pressure coil is connected between red phase and neutral, the wattmeter reads per phase EphIph cos ɸ
or EphIph cos ɸ = 6 kW = 6000 W
But E_ph=440/√3 V and I_ph=40A (given)
∴Power factor, cos⁡ɸ=6000/(E_ph I_ph )=6000/(440/√3×40)=0.59
and sin⁡〖ɸ=√(1-(0.59)^2 )=0.807〗
When the pressure coil is connected between blue and yellow phases and current coil is connected in red phase, the instrument will read (Total kVAR)/√3 or √3 kVAR per phase.
∴Reading of instrument=√3 E_ph I_ph sin⁡ɸ=√3×440/√3×40×0.807
= 14203 or 14.203 kVAR. (Ans.)

Example 35.A wattmeter reads 5.54 kW when its current coil is connected in R phase and it. voltage coil is connected between the neutral and R phase of a symmetrical3-phase system supplying a balanced load of 30 A at 400 V What will be the reading on the instrument if the connections to the current coil remain unchanged and the voltage coil is connected between B and Y phases ? Take phase sequence RYB. Draw corresponding phasor diagram.

(Punjab University)

Solution. Phase current, Iph = IL = 30 A
Phase voltage,E_ph=E_L/√3=400/√3=231 V
Power supplied per phase, P = reading of wattmeter
= 5.54kW
Load power factor, cos⁡〖ɸ=P/(E_ph.I_ph )〗
=(5.54×〖10〗^3)/(231×30)=0.8
or ɸ = cos-1 0.8 = 36.87°

Fig.42

In the second case when pressure coil is connected across phases Band Y, the phase angle between current in phase Rand voltage across voltage coil is (90° – ɸ) as obvious from the phasor diagram, shown in Fig. 42.

Wattmeter reading    = EYB IR cos (90° – ɸ)

= 400 × 30 × cos (90° – 36.87°)
= 7200 W or 7.2 kW. (Ans.)

Example 36. Two wattmeters A and B give readings as 5000 W and 1000 W respectively during the power measurement of 3 – ɸ, 3 wire balanced load system. (a) Calculate the power and power factor (i) when both meters read direct and (ii) when one of them reads in reverse (b) if the voltage of the circuit is 400 V, what is the value of the capacitance which must be introduced in each phase to cause the whole of the power to appear on A? The frequency of supply is 50 Hz.

(Nagpur University, 1998)

 

Solution. Given : W1 = 5000 W ; W2 = 1000 W ; V = 400 volts; f = 50 Hz.

(a) Power (P) and power factor (cos ɸ):

(i) When both meters read direct:

Total power, P = W1 + W2 = 5000 + 1000 = 6000 W or 6 kW. (Ans.)

We know that, tan⁡ɸ=(√3 (W_1-W_2 ))/(W_1+W_2 )=(√3 (5000-1000))/(5000+1000)=1.1547
ɸ = tan-1 (1.1547) = 49.1°
∴ Power factor, cos ɸ= cos 49.1o = 0.655 (lagging). (Ans.)
(ii) When one of the meters reads in reverse:
Here, W2 = – 1000W
P = W1 + W2 = 5000 + (- 1000) = 4000 W or 4 kW. (Ans.)
Again, tan⁡ɸ=(√3 (W_1-W_2 ))/(W_1+W_2 )=(√3 (5000-(-1000))/(5000-1000)=2.598
ɸ= tan-1 2.598 = 68.95°
∴ cos ɸ = cos 68.95° = 0.359 (lagging). (Ans.)

(b) Capacitance to be introduced in each phase, C :
Reading of wattmeter, B will be zero and whole of the power would appear on wattmeter A when power factor cos ɸ’ = 0.5 or ɸ’ = 60°. It means power factor is to be improved from 0.359 to 0.5.
Load cement per phase,I_ph (=I_ph )=P/(√3 E_L cos⁡ɸ )=4000/(√3×400×0.359)=16.08 A
Load impedance per phase Z_ph=E_ph/I_ph =(400/√3)/16.08=14.36 Ω
Load resistance per phase R_ph=Z_ph cos⁡〖ɸ=14.36〗×0.359=5.156 Ω
Load reactance per phase X_ph=Z_ph sin⁡〖ɸ=14.36〗 sin⁡〖〖cos〗^(-1) (0.359)=13.4 Ω〗
Since there is no change in resistance, therefore,
Reactance per phase in the circuit X’ph = Rph . tan ɸ’ = 5.156 tan 60° = 8.93 Ω
Capacitive reactance introduced in each phase,
Xc =Xph –X’ph = 13.4 – 8.93 = 4.47 Ω
Capacitance introduced in each phase,C=1/(2πfX_L )
=1/(2π×50×4.47)×〖10〗^6 µF=712 µF. (Ans.)
Example 37. A 3-phase, 434 V, 50 Hz supply is connected to a 3-phase, star-connected induction motor and synchronous motor. Impedance of each phase of induction motor is 1.25 + j2.17 Ω. The three-phase synchronous motor is over-excited and it draws a current of 120 A at 0.87 leading power factor. Two wattmeters are connected in usual manner to measure power drawn by the two motors. Calculate:
Reading on each wattmeter,
Combined power factor.

Solution. Line voltage, EL = 434 V
Impedance of each phase of induction motor,
Z1 = 1.25 + j2.17 Ω
Current drawn by the synchronous motor = 120 A at p.f. 0.87 leading
Reading on each wattmeter:
Combined power factor:
Let us assume that the synchronous motor is star-connected. Since it is over-excited it has a leading power factor. The connections and phasor diagrams are shown in Figs. 43 and 44.

Fig.43

Phase voltage in each case =434/√3=250 V
Z1 = 1.25 + j2.17 = 2.5 ∠60°
I_1=250/2.5=100 A
lagging the reference vector ER by 60°
∴ I_1 = 100 ∠-60°=50-j86.6
I_2 = 120∠29.5°=104.4+j59.1
I_R=I_1 +I_2=(50-j86.6)+(104.4+j59.1)
=(154.4-j27.5)=156.8L-10.1°
(i) As shown in Fig. 44, IR lags ER by 10.1°
Similarly, IY lags EY by 10.1°.
As seen from Fig. 43, current through W1 is IR and voltage across it is ERB = ER – EB. As seen, ERB = 434 V lagging by 30°. Phase difference between ERB and IR is 30-10.1 = 19.9°.
Reading of W1 = 434 × 156.8 × cos 19.9°
= 63988 W. (Ans.)
Current IY is also (like IR) the vector sum ofthe line currents draws by the two motors. It is equal to 156.8 A and lags behind its respective phase voltage EY by 10.1°. Current through W2 is IY and voltage across it is EYB = EY – EB. As seen EYB = 43 V.
Phase difference between EYB and IY = 30° + 10.1° = 40. 1° (lag)
∴ Reading of wattmeter, W2 = 434 × 156.8 × cos 40.1° = 52054 W. (Ans.)
(ii) Combined power factor = cos 10.1° = 0.984 (lag). (Ans.)

Example 38. A 3-phase balanced supply with a line voltage of 440 V at a frequency of 50 Hz
supplies a star-connected balance load. Each phase of the load consists of a resistance and a capacitor joined in series and the readings on two wattmeters connected to measure the total power supplied are 810 Wand 2100 W, both positive.

Calculate:
Power factor of the circuit,
Line current, and
Capacitance of each capacitor.

Solution. Reading of wattmeter, W1 = 2100 W
Reading of wattmeter, W2 = 810 W
Line voltage, EL = 440 V
Phase voltage, E_ph=440/√3= 254 V
For a leading power factor, tan⁡ɸ=-(√3 〖(W〗_1-W_2))/((W_1+W_2) )

Fig.45

(i) Power factor of the circuit,
cos⁡ɸ=cos⁡(〖-37.48〗^o )=0.793 (leading). (Ans.)
(ii)Power, P=√3 E_L I_L cos⁡ɸ
(W_1+W_2 )=√3 E_L I_L cos⁡ɸ [∵P=W_1+W_2 ] (2100+810)=√3×440×I_B×0.793
∴ I_L=2910/(√3×440×0.793)=4.81 A
Hence, line current = 4.81A. (Ans.)
(iii) Impedance per phase, Z_ph=E_ph/I_ph =254/4.81 [∵I_ph=I_L=4.81 A] = 52.8 Ω
Z_ph=52.8∠〖-37.48〗^o=41.9-j32.12
∴Capacitive reactance, X_C=32.12 Ω or 1/2πfC=32.12
∴ C=1/(2πf×32.12) F=〖10〗^6/(2π×50×32.12)µF=99.1µF
Hence, capacitance of each capacitor = 99.1 µF. (Ans.)

Example 39. A 3-phase, 3-wire, CRA system supplies a Δ-connected load in which ZAB = 25 ∠90° ZBC = 15 ∠30° ; ZCA = 20 ∠0° ohms. Line voltage is 230 V. Find the wattmeter readings when the two wattmeters method is used with meters in lines A and B.

(A.M.I.E.1998)

Solution. Given: ZAB = 25 ∠90° n , ZBC = 15 ∠30° Ω , ZCA = 20 ∠0° Ω , EL = 230 V.
Wattmeter readings W1 W2 :
Let the voltage EAC be the reference phasor, then
EAC = 230 ∠0°; ECB = 230 ∠-120° V and EBA = 230 ∠120°
Phase current,
I_CA=E_CA/Z_CA =(-E_AC)/Z_CA =(-230∠0^o)/(20∠0^o )
=-11.5∠0^o A or (-11.5+j0)A
Phase current,
I_AB=E_AB/Z_AB =(-E_AB)/Z_AB =(-230∠120^o)/(25∠〖90〗^o )
=-9.2∠30^o A or (-7.967-j4.6)A
Phase current,
I_BC=E_BC/Z_BC =(-E_CB)/Z_BC =(-230∠〖-120〗^o)/(15∠30^o )
=-15.333∠〖-150〗^o
=(13.279+j7.666)A

Fig.46

Now line current, I_A=I_AB-I_CA=(-7.967-j4.6)-(-11.5+j0)
=(3.533-j4.6)=5.8 ∠〖-52.47〗^o
Line current, I_B=I_BC-I_AB=(13.279+j7.666)-(-7.967-j4.6)
=(21.246+j12.266)=24.533 ∠〖30〗^o
Reading of wattmeter W_1=Real part of*E_AC×I_A
=Real part of 230 ∠0^o×5.8 ∠〖-52.47〗^o
=812.64 W. (Ans.) (*means conjugate of)
Reading of wattmeter W_2=Real part of*E_BC×I_B
=-230∠〖-120〗^o×24.533 ∠〖30〗^o=2443.3 W. (Ans.)