A well-insulated rigid tank contains 1.5 kg of a saturated liquid–vapor mixture of water at 200 kPa. Initially, three-quarters o
2 answers:
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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
Where G stands for complex rated power and H signifies the inertia constant of the turbo-generator.
(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
Where M is defined as
(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
Where t is determined from
Consequently,
The change in torque in rpm/s is provided by
The rotor speed in rpm at the culmination of this 10-cycle period is calculated as
Where P indicates the number of poles on the turbo-generator.
Answer:
Volume change percentage is 2.60%
Water level increase is 4.138 mm
Explanation:
Provided data
Water volume V = 500 L
Initial temperature T1 = 20°C
Final temperature T2 = 80°C
Diameter of the vat = 2 m
Objective
We aim to determine percentage change in volume and the rise in water level.
Solution
We will apply the bulk modulus equation, which relates the change in pressure to the change in volume.
It can similarly relate to density changes.
Thus,
E = ................1
And ............2
Here, ρ denotes density. The density at 20°C = 998 kg/m³.
The density at 80°C = 972 kg/m³.
Plugging in these values into equation 2 gives
dV = 0.0130 m³
Therefore, the percentage change in volume will be
dV % = × 100
dV % = × 100
dV % = 2.60 %
Hence, the percentage change in volume is 2.60%
Initial volume v1 = ................3
Final volume v2 = ................4
From equations 3 and 4, subtract v1 from v2.
v2 - v1 =
dV =
Substituting all values yields
0.0130 =
Thus, dl = 0.004138 m.
Consequently, the water level rises by 4.138 mm.
Answer:
Ps=19.62N
Explanation:
A thorough explanation of the answer can be found in the attached files.
Answer:
a) , b)
Explanation:
a) The uniform dresser can be modeled using specific equilibrium equations:
Following some algebraic manipulations, the formulated equation is derived:
b) Similarly, the man can be represented by a set of equilibrium equations:
After some algebraic changes, the expression for the coefficient of static friction comes out as: