 what can be done to prevent bridges from collapsing? ( give at least two examples)
1 answer:
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Answer:
Now find the temperature of each surface, we have that the the temperature on the left side of the wall is T∞₁ - Q/h₁A and the temperature on the right side of the wall is T∞₂ + Q/h₂A.
Note: kindly find an attached diagram to the complete question given below.
Sources: The diagram/image was researched and taken from Slader website.
Explanation:
Solution
Let us consider the rate of heat transfer through the plane wall which can be obtained from the relations given below:
Q = T∞₁ -T₁/1/h₁A = T₁ -T₂/L/kA =T₂ -T∞₂/1/h₂A
= T∞₁ - T∞₂/1/h₁A + L/kA + 1/h₂A
Here
The convective heat transfer coefficient on the left side of the wall is h₁, while the convective heat transfer coefficient on the right side of the wall is h₂. the thickness of the wall is L, the thermal conductivity of the wall material is k, and the heat transfer area on one side of the wall is A. Q is refereed to as heat transfer.
Thus
Let us consider the convection heat transfer on the left side of the wall which is given below:
Q = T∞₁ -T₁/1/h₁A
T₁ = T∞₁ - Q/h₁A
Therefore the temperature on the left side of the wall is T∞₁ - Q/h₁A
Now
Let us consider the convection heat transfer on the left side of the wall which is given below:
Q= T₂ -T∞₂/1/h₂A
T₂ = T∞₂ + Q/h₂A
Therefore the temperature on the right side of the wall is T∞₂ + Q/h₂A
Answer:
Not subject to detection
Explanation:
Assuming the value of strain fracture toughness is 77 Mpa
The design stress is half hence
Critical flaw size,
Where Y is dimensionless parameter, is applied stress,
is plane strain fracture toughness,
is critical length of surface crack
The critical length of surface crack is therefore 3.85 mm, which is less than detection apparatus size given as 4 mm
Since the critical flaw size is less than the resolution limit of flaw detection apparatus, the critical flaw for this plate is not subjected to detection.