A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the

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A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the

rotor at synchronous speed. (b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses. (c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.

Engineering

1 answer:

Mrrafil [318] · Answer 1 · 2026-02-27T07:45:27+03:00

Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Inertia constant = H = 8 MJ/MVA

Mechanical power = Pmech = 80 MW

Electrical power = Pelec = 50 MW

Number of poles = P = 4

No. of cycles = 10

Required Information:

(a) stored energy =?

(b) rotor acceleration =?

(c) change in torque angle =?

(c) rotor speed =?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

(c) change in torque angle (in elec deg) = 6.75 elec deg

(c) change in torque angle (in rmp/s) = 28.12 rpm/s

(c) rotor speed = 1505.62 rpm

Explanation:

(a) Calculate the rotor's stored energy at synchronous speed.

The stored energy is represented as

$E = G \times H$

Where G stands for complex rated power and H signifies the inertia constant of the turbo-generator.

$E = 100 \times 8 \\\\E = 800 \: MJ$

(b) If we suddenly increase the mechanical input to 80 MW against an electrical load of 50 MW, we shall find the rotor's acceleration while ignoring mechanical and electrical losses.

The formula for rotor acceleration is given by

$P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2}$

Where M is defined as

$M = \frac{E}{180 \times f}$

$M = \frac{800}{180 \times 50}$

$M = 0.0889 \: MJ \cdot s/ elec \: \: deg$

$P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2}$

$30 = 0.0889 \frac{d^2 \delta}{dt^2}$

$\frac{d^2 \delta}{dt^2} = \frac{30}{0.0889}$

$\frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2$

(c) If the acceleration derived in part (b) persists over 10 cycles, we will calculate both the change in torque angle and the rotor speed in revolutions per minute at the end of this duration.

The change in torque angle is expressed as

$\Delta \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2$