A thick-walled tube of stainless steel having a k = 21.63 W/m∙K with dimensions of 0.0254 m ID and 0.0508 m OD is covered with 0
.0254 m thick layer of an insulation (k = 0.2423 W/m∙K). The inside-wall temperature of the pipe is 811 K and the outside surface of the insulation is 310.8 K. For a 0.305 m length of pipe, calculate the heat loss and the temperature at the interface between the metal and the insulation.
1 answer:
Answer:
Q=339.5W
T2=805.3K
Explanation:
Hi!
To solve this problem follow the steps below, the procedure is attached in an image
1. Draw the complete outline of the problem.
2.to find the heat Raise the heat transfer equation for cylinders from the inside of the metal tube, to the outside of the insulation.
3. Once the heat is found, Pose the heat transfer equation for cylinders from the inner part of the metal tube to the outside of the metal tube and solve to find the temperature
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Answer:
W = 112 lb
Explanation:
Given:
- δb = 0.025 in
- E = 29000 ksi (A-36)
- Area A_de = 0.002 in^2
Find:
Compute Weight W attached at C
Solution:
- Use proportion to determine δd:
δd/5 = δb/3
δd = (5/3) * 0.025
δd = 0.0417 in
- Compute εde i.e strain in DE:
εde = δd / Lde
εde = 0.0417 / 3*12
εde = 0.00116
- Compute stress in DE, σde:
σde = E*εde
σde = 29000*0.00116
σde = 33.56 ksi
- Compute the Force F_de:
F_de = σde *A_de
F_de = 33.56*0.002
F_de = 0.0672 kips
- Equilibrium conditions apply:
(M)_a = 0
3*W - 5*F_de = 0
W = (5/3)*F_de
W = (5/3)* 0.0672 = 112 lb
