Determine the normal stress in a ball, which has an outside diameter of 160 mm and a wall thickness of 3.8 mm, when the ball is
1 answer:
Answer:
<em>The normal stress is 0.7821 MPa</em>
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Explanation:
The external diameter D = 160 mm
The thickness t = 3.8 mm = 3.8 x 10^-3 m
gauge pressure P = 78 kPa = 78 x 10^3 Pa
The maximum shear stress τmax = ?
The external radius of the shell from the external surface R = D/2 = 160/2 = 80 mm
The internal radius of the shell r = R - t
==> 80 - 3.8 = 76.2 mm
Therefore the internal diameter d = 2r = 2 x 76.2 = 152.4 mm
==> d = 152.4 x 10^-3 m
The normal stress σ = =
= 782052.63 Pa
==> σ = <em>0.7821 MPa</em>
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Answer:
a) The temperature on the outer surface of the pipe is approximately 179.97 °C
b) The thickness of the insulation is approximately 0.857 m
Explanation:
We have;
αA = 200 W/(m²·K)
= (T₂ - T₁) × U
= (200 - 180) × 200 = 4,000
For the pipe, we have;
/U= (T₂ - T₁)
∴ 4000× = (180 - T₂)
T₂ ≈ 179.97 °C
The temperature on the outer surface of the pipe, T₂ ≈ 179.97 °C
b) For the insulation, we have;
T₂ - T₃ = 179.97 °C - 40°C ≈ 139.97°C
/U= (T₂ - T₃)
The thickness of the insulation, x ≈ 0.857 m