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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