0
2.5kviews
A balanced three-phase load connected in delta, drawn a power of 10kW at 440 V at Pf 0.6 lead, find the values of circuit elements and reactive volt-amperes drawn.
1 Answer
2
564views

$P=10\ kW\\ V_L=440\ V\\ pf=0.6\ (lead)$

For a delta - connected load,

(i) Values of circuit elements,

$V_L=V_{ph}=440\ V$

$P=\sqrt 3\ V_L\ I_L\ cos\phi \\ 10\times 10^3=\sqrt 3\times 440\times I_L\times 0.6\\ I_L=21.87\ A$

$I_{ph}=\dfrac{I_L}{\sqrt 3}=\dfrac{21.87}{\sqrt 3}=12.63\ A$

$Z_{ph}=\dfrac{V_{ph}}{I_{ph}}=\dfrac{440}{12.63}=34.84\ \Omega $

$\phi =cos^{-1}(0.6)=53.13^{\circ}\\ R_{ph}=Z_{ph}cos\phi =34.84\times 0.6=20.90\ \Omega \\ X_{ph}=Z_{ph}sin\phi =34.84\times 0.8=27.87\ \Omega $

(ii) Reactive volt- aperes drawn

$Q=\sqrt 3V_LI_Lsin\phi =\sqrt 3\times 440\times 21.87\times 0.8=12.33\ kVAR$

Please log in to add an answer.