For a nozzle-duct system shown in Fig Q3, the nozzle is designed to produce a Mach number of 2.8 with y = 1.4, The inlet conditi
1 answer:
You might be interested in
Answer:
0
Explanation:
output =transfer function H(s) ×input U(s)
here H(s)=
U(s)= for unit step function
output =H(s)×U(s)
=×
=
taking inverse laplace of output
output=t×
at t=0 putting the value of t=0 in output
output =0
Given Information:
Frequency = f = 60 Hz
Complex rated power = G = 100 MVA
Intertia 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) Find the stored energy in the rotor at synchronous speed.
The stored energy is given by
Where G represents complex rated power and H is the inertia constant of turbo-generator.
(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.
The rotor acceleration is given by
Where M is given by
So, the rotor acceleration is
(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.
The change in torque angle is given by
Where t is given by
So,
The change in torque in rpm/s is given by
The rotor speed in revolutions per minute at the end of this period (10 cycles) is given by
Where P is the number of poles of the turbo-generator.