Question
Download Solution PDFTwo wattmeters are used to measure the power in a 3-phase balanced system. What is the power factor of the load when one wattmeter reads twice the other?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
In a two-wattmeter method, for lagging load
The reading of first wattmeter (W1) = VL IL cos (30 + ϕ)
The reading of second wattmeter (W2) = VL IL cos (30 - ϕ)
Total power in the circuit (P) = W1 + W2
Total reactive power in the circuit \(Q=\sqrt{3}\left( {{W}_{1}}-{{W}_{2}} \right)\)
Power factor = cos ϕ
\(\phi =\text{ta}{{\text{n}}^{-1}}\left( \frac{\sqrt{3}\left( {{W}_{1}}-{{W}_{2}} \right)}{{{W}_{1}}+{{W}_{2}}} \right)\)
Calculation:
Given that, W1 = 2 W2
\(\tan \phi = \frac{{\sqrt 3 \left( {2{W_2} - {W_2}} \right)}}{{\left( {2{{\rm{W}}_2} + {W_2}} \right)}} = \frac{1}{{\sqrt 3 }}\)
⇒ ϕ = 30°
Power factor = cos 30° = 0.866
Important Point:
|
p.f. angle (ϕ) |
p.f.(cos ϕ) |
W1 [VLIL cos (30 + ϕ)] |
W2 [VLIL cos (30 - ϕ)] |
W = W1 + W2 [W = √3VLIL cos ϕ] |
Observations |
|
0° |
1 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
√3 VLIL |
W1 = W2 |
|
30° |
0.866 |
\(\frac{{{V_L}I}}{2}\) |
VLIL |
1.5 VL IL |
W2 = 2W1 |
|
60° |
0.5 |
0 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
W1 = 0 |
|
90° |
0 |
\(\frac{{ - {V_L}{I_L}}}{2}\) |
\(\frac{{{V_L}{I_L}}}{2}\) |
0 |
W1 = -ve W2 = +ve |
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