A 15 cm × 15 cm circuit board dissipating 20 W of power uniformly is cooled by air, which approached the circuit board at 20C w
ith a velocity of 6 m/s. Assume the flow to be turbulent, since the electronic components are expected to act as turbulators. Disregard any heat transfer from the back surface of the board, and suppose that radiation heat transfer is negligible. Evaluate thermal properties of the air at a film temperature of 27C. Determine the surface temperature (C) of the electronic components at the following locations i) the leading edge of the board ii) the trailing edge of the board
1 answer:
You might be interested in
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