Consider an opaque horizontal plate that is well insulated on its back side. The irradiation on the plate is 2500 W/m2, of which
500 W/m2 is reflected. The plate is at 227°C and has an emissive power of 1200 W/m2. Air at 127°C flows over the plate with a convection heat transfer coefficient of 15 W/m2 ⋅ K. Determine the emissivity, absorptivity, and radiosity of the plate. What is the net heat transfer rate per unit area?
1 answer:
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Answer:
160N
Explanation:
Moments must be conserved - so.
Answer:
a) the values of the angle α is 45.5°
b) the required magnitude of the vertical force, F is 41 lb
Explanation:
Applying the free equilibrium equation along x-direction
from the diagram
we say
∑Fₓ = 0
Pcosα - 425cos30° = 0
525cosα - 368.06 = 0
cosα = 368.06/525
cosα = 0.701
α = cos⁻¹ (0.701)
α = 45.5°
Also Applying the force equation of motion along y-direction
∑Fₓ = ma
Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)
525sin45.5° + F + 212.5 - 600 = 27.95
374.46 + F + 212.5 - 600 = 27.95
F - 13.04 = 27.95
F = 27.95 + 13.04
F = 40.99 ≈ 41 lb
