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
![I_{a} =\frac{40x10^{6} }{\sqrt{3}*24x10^{3} } =926.2A](https://tex.z-dn.net/?f=I_%7Ba%7D%20%3D%5Cfrac%7B40x10%5E%7B6%7D%20%7D%7B%5Csqrt%7B3%7D%2A24x10%5E%7B3%7D%20%20%7D%20%3D926.2A)
![X_{s} =0.85\frac{13.85}{926.2} =12.7ohm](https://tex.z-dn.net/?f=X_%7Bs%7D%20%3D0.85%5Cfrac%7B13.85%7D%7B926.2%7D%20%3D12.7ohm)
The impedance Zn is
![\sqrt{0.56^{2}+4^{2} } =4.03ohm](https://tex.z-dn.net/?f=%5Csqrt%7B0.56%5E%7B2%7D%2B4%5E%7B2%7D%20%20%7D%20%3D4.03ohm)
The voltage drop is
![I_{a} *Z_{n} =926.2*4.03=3732.58V](https://tex.z-dn.net/?f=I_%7Ba%7D%20%2AZ_%7Bn%7D%20%3D926.2%2A4.03%3D3732.58V)
![r_{dc} =\frac{voltage}{2*current} =\frac{13.85}{2*926.2} =7.476ohm](https://tex.z-dn.net/?f=r_%7Bdc%7D%20%3D%5Cfrac%7Bvoltage%7D%7B2%2Acurrent%7D%20%3D%5Cfrac%7B13.85%7D%7B2%2A926.2%7D%20%3D7.476ohm)
rac = 1.2rdc = 1.2 * 7.476 = 8.97 Ω
The effective armature resistance is
![Z_{s} =\sqrt{R_{a}^{2}+X_{s}^{2} } =\sqrt{8.97^{2}+12.7^{2} } =15.55ohm](https://tex.z-dn.net/?f=Z_%7Bs%7D%20%3D%5Csqrt%7BR_%7Ba%7D%5E%7B2%7D%2BX_%7Bs%7D%5E%7B2%7D%20%20%20%20%7D%20%3D%5Csqrt%7B8.97%5E%7B2%7D%2B12.7%5E%7B2%7D%20%20%7D%20%3D15.55ohm)
The induced voltage for leading power factor is
![E_{F} ^{2} =OB^{2} +(BC-CD)^{2}](https://tex.z-dn.net/?f=E_%7BF%7D%20%5E%7B2%7D%20%3DOB%5E%7B2%7D%20%2B%28BC-CD%29%5E%7B2%7D)
if cosθ = 0.5
![E_{F} =\sqrt{(13850*0.5)^{2}+(\frac{3741}{2}-926.2*12.7)^{2} } =11937.51V](https://tex.z-dn.net/?f=E_%7BF%7D%20%3D%5Csqrt%7B%2813850%2A0.5%29%5E%7B2%7D%2B%28%5Cfrac%7B3741%7D%7B2%7D-926.2%2A12.7%29%5E%7B2%7D%20%20%20%7D%20%3D11937.51V)
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
![\frac{E_{F}-V_{t} }{V_{t} } *100](https://tex.z-dn.net/?f=%5Cfrac%7BE_%7BF%7D-V_%7Bt%7D%20%20%7D%7BV_%7Bt%7D%20%7D%20%2A100)
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