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Anastaziya [24]
3 years ago
12

A three-phase wye-connected synchronous generator supplies a network through a transmission line. The network can absorb or deli

ver power while maintaining its terminal voltage constant. The four-pole, 60 Hz generator data rating are 40 MVA and 26 kV with a 0.85 p.u. reactance. The field current of the generator can be adjusted to regulate the excitation (induced) voltage from 0.75 to 1.5 times the rated voltage. The network voltage rating is 24 kV. The transmission line impedance and length are 0.07+j0.5 Ω/mi and 8 mi.
Calculate the generator induced voltage versus power factor, if the network voltage is at the rated value, and the bus absorbs the generator rated power. After this, plot the voltage regulation of the system versus the power factor. Use the induced voltage and network voltage for plotting the voltage regulation. The leading power factor varies from 0.5 to 1. What power factor corresponds to 10% regulation?

Engineering
1 answer:
Amanda [17]3 years ago
8 0

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

X_{s} =0.85\frac{13.85}{926.2} =12.7ohm

The impedance Zn is

\sqrt{0.56^{2}+4^{2}  } =4.03ohm

The voltage drop is

I_{a} *Z_{n} =926.2*4.03=3732.58V

r_{dc} =\frac{voltage}{2*current} =\frac{13.85}{2*926.2} =7.476ohm

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

The induced voltage for leading power factor is

E_{F} ^{2} =OB^{2} +(BC-CD)^{2}

if cosθ = 0.5

E_{F} =\sqrt{(13850*0.5)^{2}+(\frac{3741}{2}-926.2*12.7)^{2}   } =11937.51V

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

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

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Answer:

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