Answer:
damping natural frequency = 28.76 rad/s
Explanation:
given data
mass = 12 kg
stiffness = 10000 n/m
damping ratio = 0.08
displacement = 8 mm
initial velocity = 1 mm
to find out
damped natural frequency of the system
solution
we first find the natural frequency that is express as
natural frequency ω = ..............1
here k is stiffness and m is mass
so ω =
ω = 28.86 rad/s
so
damping frequency will be
damping frequency = ω × .....................2
here r is damping ration
damping frequency = 28.86 ×
damping natural frequency = 28.76 rad/s
Answer:
<u>Assistants</u><u> </u><u>works alongside and assists the engineers.</u>
Answer:
2062 lbm/h
Explanation:
The air will lose heat and the oil will gain heat.
These heats will be equal in magnitude.
qo = -qa
They will be of different signs because one is entering iits system and the other is exiting.
The heat exchanged by oil is:
qo = Gp * Cpo * (tof - toi)
The heat exchanged by air is:
qa = Ga * Cpa * (taf - tai)
The specific heat capacity of air at constant pressure is:
Cpa = 0.24 BTU/(lbm*F)
Therefore:
Gp * Cpo * (tof - toi) = Ga * Cpa * (taf - tai)
Ga = (Gp * Cpo * (tof - toi)) / (Cpa * (taf - tai))
Ga = (2200 * 0.45 * (150 - 100)) / (0.24 * (300 - 200)) = 2062 lbm/h
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