EPE3701 Assessment 1 (Minor Test 1) Answers Year 2025

Power Engineering Systems

Assessment 1 (Minor Test 1) Answers

Year 2025




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,
, QUESTION 1

The rated values are SRated = 150 MVA

And Prated = 130 MW

The maximum value of the internal voltage is Epu (max) = 2.2

Let the base voltage be Vrated = 11.5 kV

So the maximum internal voltage is

Emax = 2.2 x 11.5 = 25.3 kV

Now the maximum power that can be transferred is

𝐸𝑚𝑎𝑥 × 𝑉𝑟𝑎𝑡𝑒𝑑
𝑃𝑚𝑎𝑥 (𝑝𝑢) =
𝑋𝑠

2.2 × 1
𝑃𝑚𝑎𝑥 (𝑝𝑢) = = 1.692𝑝𝑢
1.3

So the maximum power that can be transferred is

Pmax = 1.692 x 150 = 253.8



1.1

The reactive power delivered by the generator is given by

𝐸×𝑉 𝑉2
𝑄= cos(𝛿) −
𝑋 𝑋

And the reactive power deliverd by the generator is given by

𝐸×𝑉
𝑃= sin(𝛿)
𝑋

The per unit active power is

100𝑀𝑊 2
𝑃𝑝𝑢 = =
150 3