Answer to Question #225497 in Electrical Engineering for Zoe

Question #225497
Question 4
4.1 A 600 MVA, 50 Hz turbine-generator has a regulation constant R=0.06 per
unit based on its own rating. If the generator frequency increases by 0.02 Hz
in steady-state, what is the decrease in mechanical power output. Assume
a fixed reference power setting. (5)
4.2 A three-phase, 50 Hz, 100 MVA, 4-pole synchronous generator has an
inertia constant H of 3.5 s and is supplying 0.16 pu real power on a system
base of 500 MVA. The input to the generator is increased to 0.18 pu. Determine:
4.2.1 The kinetic energy stored in the moving parts of the generator
4.2.2 The acceleration of the generator
1
Expert's answer
2021-08-13T02:14:50-0400

4.1)

"\\Delta{f_{p.u}}=\\dfrac{\\Delta{f}}{f_{base}}\\\\\n\\Delta{f_{p.u}}=\\dfrac{0.02}{50}=4\\times10^{-4}per\\ unit\\\\\n\\Delta{p_{mp.u}}=\\Delta{p_{ref}}-\\dfrac{\\Delta{f_{p.u}}}{R}\\\\\n\\Delta{p_{mp.u}}=-\\dfrac{\\Delta{f_{p.u}}}{R}\\\\\n\\Delta{p_{mp.u}}=-\\dfrac{4\\times10^{-4}}{0.06}=-6.67\\times 10^{-3}per\\ unit\\\\\n\\Delta{p_{p.u}}=\\Delta{p_{mp.u}}S_{base}\\\\\n\\Delta{p_{p.u}}=-6.67\\times 10^{-3}\\times600=-4MW"

Decrease in mechanical power output is 4MW.

4.2.1)

Kinetic energy is given by GH.

Where G is the MVA of base machine and H is the inertia constant.

"K.E=GH=100\\times3.5=350MJ"

4.2.2)

Acceleration is "\\dfrac{d^2\\delta}{dt^2}" and is given by

"\\dfrac{d^2\\delta}{dt^2}=\\dfrac{\\pi{f}(P_i-P_e)}{H}\\\\\n\\dfrac{d^2\\delta}{dt^2}=\\dfrac{\\pi{50}(0.18-0.16)}{3.5}=0.898elect.radian\/sec^2\\\\"





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