100 kg of refrigerant-134a at 200 kPa iscontained in a piston-cylinder device whose volume is 12.322 m3. The piston is now moved
1 answer:
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Answer:
Explanation:
Given that
R=8 ft
Width= 10 ft
We know that hydro statics force given as
F=ρ g A X
ρ is the density of fluid
A projected area on vertical plane
X is distance of center mass of projected plane from free surface of water.
Here
X=8/2 ⇒X=4 ft
A=8 x 10=80
So now putting the values
F=ρ g A X
F=62.4(32.14)(80)(4)
Answer:
equivalent stiffness is 136906.78 N/m
damping constant is 718.96 N.s/m
Explanation:
given data
mass = 120 kg
amplitude = 120 N
frequency = 320 r/min
displacement = 5 mm
to find out
equivalent stiffness and damping
solution
we will apply here frequency formula that is
frequency ω = ω(n) √(1-∈ ²) ......................1
here ω(n) is natural frequency i.e = √(k/m)
so from equation 1
320×2π/60 = √(k/120) × √(1-2∈²)
k × ( 1 - 2∈²) = 33.51² ×120
k × ( 1 - 2∈²) = 134752.99 .....................2
and here amplitude ( max ) of displacement is express as
displacement = force / k × ( )
put here value
0.005 = 120/k × ( )
k ×∈ × √(1-2∈²) = 1200 ......................3
so by equation 3 and 2
solve it and we get
∈ = 1.00396
and
∈ = 0.08869
here small value we will consider so
by equation 2 we get
k × ( 1 - 2(0.08869)²) = 134752.99
k = 136906.78 N/m
so equivalent stiffness is 136906.78 N/m
and
damping is express as
damping = 2∈ √(mk)
put here all value
damping = 2(0.08869) √(120×136906.78)
so damping constant is 718.96 N.s/m