Answer to Question #225496 in Electrical Engineering for Zoe

Question

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 (2)
4.2.2 The acceleration of the generator

Answer 1

4.1)

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

Decrease in Mechanic energy is 4MW

4.2.1)

Kinetic energy is given by GH. Where G is the MVA of the 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

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