The fluid-conditioning components of hydraulic-powered equipment provide fluid that is clean and maintained at an acceptable ope
1 answer:
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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:
Q = 125.538 W
Explanation:
Given data:
D = 30 cm
Temperature degree celcius
Heat coefficient = 12 W/m^2 K
Efficiency 80% = 0.8
Q = 125.538 W