Answer:
33.429 N-m
Explanation:
Given :
Inclination angle of two shaft, α = 20°
Speed of shaft A, = 1000 rpm
Mass of flywheel, m = 30 kg
Radius of Gyration, k =100 mm
= 0.1 m
Now we know that for maximum velocity,
= 1064.1 rpm
Now we know
Mass of flywheel, m = 30 kg
Radius of Gyration, k =100 mm
= 0.1 m
Therefore moment of inertia of flywheel, I = m.
=30 X
= 0.3 kg-
Now torque on the output shaft
T₂ = I x ω
= 0.3 X 1064.2 rpm
=
= 33.429 N-m
Torque on the Shaft B is 33.429 N-m
Answer:
The initial temperature is 649 K (376 °C).
The final pressure is 0.965 MPa
Explanation:
From the ideal gas equation
PV = nRT
P is the initial pressure of water = 2 MPa = 2×10^6 Pa
V is intial volume = 150 L = 150/1000 = 0.15 m^3
n is the number of moles of water in the container = mass/MW = 1000 g/18 g/mol = 55.6 mol
R is gas constant = 8.314 m^3.Pa/mol.K
T (initial temperature) = PV/nR = (2×10^6 × 0.15)/(55.6 × 8.314) = 649 K = 649 - 273 = 376 °C
From pressure law,
P1/T1 = P2/T2
P2 (final pressure) = P1T2/T1
T2 (final temperature) = 40 °C = 40 + 273 = 313 K
P1 (initial pressure) = 2 MPa
T1 (initial temperature) = 649 K
P2 = 2 × 313/649 = 0.965 MPa
Answer:
1. They have less malfunctions
2. They can be operated from afar
3. Some are self-operated.
Explanation:
Answer:
he must document or remember the order he took it apart so he put it back together
Explanation: