Balanced or unbalanced load, Figs. 28 and 29 show connection diagrams for star-connected and delta-connected loads respectively. In this method the current coils of the two wattmeters are inserted in any two lines and the pressure (or potential) coil of ach joined to the third line.
Fig. 28. Star-connected load.
Fig. 29. Delta-connected load.
It can be proved that the sum of the instantaneous powers indicated by W1 and W2 gives the instantaneous power absorbed by the three loads L1, L2 and L3. Let us consider a star-connected load (although it can be equally applied to a delta-connected load which can always be replaced by an equivalent star-connected load).
Keeping in mind that it is important to take the direction of the voltage through the circuit as the same as that taken for the current when establishing the readings of the two wattmeters.
Instantaneous current through W_1 = i_R
Instantaneous potential difference across W_1 = e_RB = e_R-e_B
Instantaneous power read by W_1 = i_R (e_R-e_(B))
Instantaneous current through W_2 = i_Y
Instantaneous potential difference across W_2 = e_YB=e_Y-e_B
Instantaneous power read by W_2 = i_Y (e_Y-e_B)
∴ W_1+W_2=i_R (e_R-e_B )+i_Y (e_Y-e_B )=i_R e_R+i_Y e_Y+e_B (i_R+i_Y)
Now according to Kirchhoffs point law
i_R+ i_Y+ i_B=0
∴ i_R+ i_Y=-i_B
∴ W_1+W_2=i_R e_R+i_Y e_Y+i_B e_B=P_1+P_2+P_3
where P1 = power absorbed by load L1, p2 = power absorbed by load L2, and
P3 = power absorbed by load L3.
∴ W1 + W2 = total power absorbed.
This proof is true whether the load is balanced or unbalanced.
In case the load is star-connected, then it should have no neutral connection (i.e., 3-phase, 3-wire connected) and if it has a neutral connection (i.e., 3-phase, 4-wire connected) that it should be exactly balanced so that in each case there is no neutral current iN otherwise Kirchhoff’spoint law will give iR + iY + iB + iN = 0.
In the above derivation we have considered the instantaneous readings. In fact the moving system wattmeter, due to its inertia, cannot quickly follow the variations taking place in cycle, hence it indicates the average power.
∴ Total power W1 + W2
=1/T ∫_0^T▒〖i_R e_RB 〗 dt+ 1/T ∫_0^T▒〖i_Y e_YB 〗 dt
Two Wattmeters Method-Balanced load. The total power consumed by a balanced load can be found by using two wattmeters (Figs. 28 and 29). When load is assumed inductive in Fig.26, the vector diagram for such a balanced star-connected load is shown in Fig. 30.
Let us consider the problem in terms of r.m.s. values (instead of instantaneous values).
Let ER,EY,EB = r.m.s. values of the three phase voltages,
and IR, IY, IB == r.m.s. values of the currents.
Since these voltages and currents are assumed sinusoidal, they can be represented by vectors, the currents lagging behind their respective phase voltages by ɸ.
Fig. 30. Vector diagram-two wattmeters method,
Refer Fig. 30.
Current through wattmeter W1 = IR
Potential difference across pressure coil of wattmeter
W1 = ERB = ER – EB (Vectorially)
The value of ERB is found by compounding ER and EB reversed as shown in Fig. 30. It may be observed that phase difference between ERB and IR = (30° – ɸ),
∴ Reading of wattmeter W1 = ERBIR cos (30° – ɸ) … (i)
Similarly, current through wattmeter W2 = IY
Potential difference across pressure coil of wattmeter
W2= EYB = EY – EBI (Vectorially)
The value of EYB is found by compounding EY and EB reversed as shown in Fig.30. The phase difference between EYB and IY = (30° + ɸ).
∴ Reading of wattmeter W2 = EYBIY cos (30° + ɸ). … (ii)
Since the load is balanced, ERB = EYB = EL (Line voltage)
and, IR = IY = IL (Line current)
∴ W1 = EL IL cos (30° – ɸ)
and W2 = EL IL cos (30° + ɸ)
∴ Total power, P = W1 + W2
= EL cos (30° – ɸ) + EL cos (30° + ɸ)
= EL [cos (30° - ɸ) + cos (30° + ɸ)] = EL [cos 30° cos ɸ + sin 30° sin ɸ + cos 30° cos ɸ - sin 30° sin ɸ] =E_L I_L (2 cos〖〖30〗^2 〗 cosɸ ) 〖=E〗_L I_L×2×√3/2 cosɸ=√3 E_L I_L cosɸ
i.e. P=√3 E_L I_L cosɸ
Hence the sum of the readings of the two wattmeters give the total power consumption in the 3-phase load.
It is worth noting that in the above case the phase sequence of RYB has been assumed, the
readings of the two wattmeters will change if the phase sequence is reversed.
Variations in wattmeter readings. As shown above that for a lagging power factor
W1 = ELIL cos (30° – ɸ)
W2 = ELIL cos (30° + ɸ)
From above it is evident that individual readings of the wattmeters not only depend on the load but also upon its power factor. Let us take up the following cases:
(i) When ɸ = 0
i.e., power factor is unity i.e., load is resistive)
Then W1 = W2 = ELIL cos 30°
The reading of each wattmeter will be equal and opposite i.e., up-scale reading).
(ii) When ɸ = 60°
i.e., power factor = 0.5 (lagging)
Then W2 = ELIL cos (30° + 60°) = 0
Hence, the power is measured by W1 above.
(iii) When 90° >ɸ> 60°
0.5 > p.f. > 0
Then W1 is still positive but reading of W2 is reversed. For a leading p.f., conditions are just the opposite of this. In that case, W1 will read negative because the phase angle between the current and voltage is more than 90°. For getting the total power, the reading of W2 is to be subtracted from that of W1, Under this condition, W2 will read ‘down scale’ i.e., backwards. Hence, to obtain a reading on W2, it is necessary to reverse either it is pressure coil or current coil, usually the former.
All readings taken after reversal of pressure coil are to be taken as negative.
(iv) When ɸ = 90°
(i.e., p.f. = 0 i.e., pure inductive or capacitive load)
Then W1 = ELIL cos (30° – 90°) = ELIL sin 30°
W2 = ELIL cos (30° + 90°) = – ELIL sin 30°
These two readings are equal in magnitude but opposite in sign
∴ W1 + W2=O.
So far we hae considered lagging angles (taken as positive). Now let us discuss how the readings of wattmeters change when the power factor is leading one.
- For ɸ = + 60° (lag) : W2 = 0
- For ɸ = – 60° (lead) : WI = 0
Thus we find that for angles of lead the readings of the two wattmeters are interchanged. Hence, when the power is leading:
W1 = ELIL cos (30° + ɸ)
W2 = ELIL cos (30° – ɸ).
Power Factor-When the load is ‘balanced’. When load is balanced with a lagging power factor and the voltage and currents are sinusoidal:
W1 + W2 = ELIL cos (30° – ɸ) + ELIL cos (30° + ɸ) = √3 ELIL cos ɸ … (i)
Similarly, W1 – W2 = ELIL cos (30° – ɸ) – ELIL cos (30° + ɸ)
= ELIL (2 sin ɸ sin 30°) = ELIL sin ɸ … (ii)
Dividing (ii) by (i), we get
tanɸ=(√3 (W_1-W_2))/((W_1+W_2)) …(8)
For a leading power, this expression becomes
tanɸ=-(√3 (W_1-W_2))/((W_1+W_2)) …(9)
After finding tan ɸ, hence ɸ, the value of power factor cos ɸ can be found (from trigonometrical tables).
One important point which must be kept in mind is that if W2 reading has been taken after reversing the pressure coil i.e., if W2 is negative, then the eqn. (8) becomes
or tanɸ=√3 (W_1 〖 + W〗_2)/(W_1 〖- W〗_2 ) …(9)
The power factor may also be expressed in terms of ratio of the readings of the two wattmeters.
Let= (Smaller reading)/(Larger reading)=W_1/W_2 =α
Then from eqn. (8) above, we have
tanɸ=√(3 [1-(W_2/W_1 )] )/(1+(W_2/W_1 ) )= (√3 (1-α))/((1+α))
We know that, sec2 ɸ = 1 + tan2 ɸ
or 1/(cos^2 ɸ)=1+tan^2 ɸ or cos^2 ɸ=1/(〖1+tan〗^2 ɸ)
cosɸ=1/√(1+tan^2 ɸ)= 1/√(1+ [(√3 (1-α))/(1+α)]^2 )
or = 1/√(1+3 ((1-α)/(1+α))^2 )=(1+α)/√((1+α)^2+3(1-α)^2 )
= (1+α)/(2√(1-α+α^2 ))
i.e., = (1+α)/(2√(1-α+α^2 )) …(11)
Fig. 31. Watt-ratio curve.
Reactive volt amperes (with two wattmeters)
We know that tanɸ=(√3(W_1-W_2))/((W_1+W_2))
As the tangent of the angle of lag between phase current and phase voltage of a circuit is always equal to the ratio of reactive power to the true power (Fig. 32). Hence, in case of a balanced load, the reactive power is given by √3 times the difference of the reading of the two wattmeters used to measure the power of a 3-phase circuit by two wattmeter method.
Mathematical proof is as follows:
√3 (W1 – W2) = √3 [(ELIL cos (30° - ɸ) - ELIL cos (30° + ɸ)] = √3 ELIL [(cos 30° cos ɸ + sin 30° sin ɸ) - (cos 30° cos ɸ - sin 30° sin ɸ)] = √3 ELIL [cos 30° cos ɸ + sin 30° sin ɸ - cos 30° cos ɸ + sin 30° sin ɸ] = √3 ELIL (2 sin 30° sin ɸ) = √3 ELIL sin ɸ
i.e., √3 (W1 – W2) = √3 ELIL sin ɸ