2017 case, 20 below
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There is a reactor with a diameter D0 of 2.6 m and a tangential height of 3 m, which is mounted on suspension supports on a floorless frame 8 m high. The liquid level inside the container is 2.4 m above the lower tangent line, while the support plane is 2 m away from this lower tangent line. The bottom of the reactor features an elliptical head; what is the wetting area (in m2) of this structure in the event of a fire? (Hint: For containers with oval end caps, the formula for calculating the total external surface area is A=πD0(L+0.3D0)). (A) 16.7 (B) 24.8 (C) 20.7 (D) 41.5 According to the problem statement, the wetted area in the event of a fire should be the external surface area of the cylinder at a distance of 1.5 m from the cylinder’s lower tangent line; thus, A=π×2.6×(1.5+0.3×2.6×0.5)≈15.44. There is no correct answer here – please provide a solution
According to HG/T20570, it should be the approach adopted by the original poster. I can’t get an answer by plugging in the known values from the problem; I don’t know how to solve it.
Could the original poster share the actual exam questions from 2017?
https://bbs.hcbbs.com/thread-2036918-1-1.html See this link: handshake
The total area includes the area of the upper head (flat head), which needs to be deducted.
The L here includes the height of the lower head, so I think one still needs to know how to calculate the height of the head for this problem. Head height approximately = (D/4) + 0.04 = 0.69
Is the container area within 7.5m of the ground surface relatively small? For the external surface area of the container at a distance of 1.5m from the lower head, both ends corresponding to the entire area are elliptical – how can it be calculated using the area formula for circles?
The upper head in the full area should also be oval, right?