Answer:
the graph and the answer can be found in the explanation section
Explanation:
Given:
Network rated voltage = 24 kV
Impedance of network = 0.07 + j0.5 Ω/mi, 8 mi
Rn = 0.07 * 8 = 0.56 Ω
Xn = 0.5 * 8 = 4 Ω
If the alternator terminal voltage is equal to network rated voltage will have
Vt = 24 kV/√3 = 13.85 kV/phase
The alternative current is
[tex]I_{a} =\frac{40x10^{6} }{\sqrt{3}*24x10^{3} } =926.2A[/tex]
[tex]X_{s} =0.85\frac{13.85}{926.2} =12.7ohm[/tex]
The impedance Zn is
[tex]\sqrt{0.56^{2}+4^{2} } =4.03ohm[/tex]
The voltage drop is
[tex]I_{a} *Z_{n} =926.2*4.03=3732.58V[/tex]
[tex]r_{dc} =\frac{voltage}{2*current} =\frac{13.85}{2*926.2} =7.476ohm[/tex]
rac = 1.2rdc = 1.2 * 7.476 = 8.97 Ω
The effective armature resistance is
[tex]Z_{s} =\sqrt{R_{a}^{2}+X_{s}^{2} } =\sqrt{8.97^{2}+12.7^{2} } =15.55ohm[/tex]
The induced voltage for leading power factor is
[tex]E_{F} ^{2} =OB^{2} +(BC-CD)^{2}[/tex]
if cosθ = 0.5
[tex]E_{F} =\sqrt{(13850*0.5)^{2}+(\frac{3741}{2}-926.2*12.7)^{2} } =11937.51V[/tex]
if cosθ= 0.6
EF = 12790.8 V
if cosθ = 0.7
EF = 13731.05 V
if cosθ = 0.8
EF = 14741.6 V
if cosθ = 0.9
EF = 15809.02 V
if cosθ = 1
EF = 13975.6 V
The voltage regulation is
[tex]\frac{E_{F}-V_{t} }{V_{t} } *100[/tex]
For each value:
if cosθ = 0.5
voltage regulation = -13.8%
if cosθ = 0.6
voltage regulation = -7.6%
if cosθ = 0.7
voltage regulation = -0.85%
if cosθ = 0.8
voltage regulation = 6.4%
if cosθ = 0.9
voltage regulation = 14%
if cosθ = 1
voltage regulation = 0.9%
the graph is shown in the attached image
for 10% of regulation the power factor is 0.81
