Afternoon of the second day, 2017, for the Chemical Engineering major – Immersion area of vertical vessels
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Actual exam question: A container is placed on a suspended platform 8 meters high. The diameter of the container is 2.6 m, and its vertical length is 3 m. The liquid level inside the container is 2.4 m above the tangent line of the container’s head; this tangent line is located 2 m away from the platform. What is the wetted area in the event of a fire?
No matter how you calculate this problem, there’s no answer. It’s really strange
Because in the area formula provided by the question setter, L represents the total length, and the heights of the two end caps need to be subtracted
And those questions about energy consumption – are there two calculations whose answers differ from each other???
Then it’s unclear whether to calculate up to a height of 7.5 meters or up to the liquid level height :(
This post was last edited by comic3344 on 2017-9-24 at 22:55. Without that specification, using half of the surface area along with a value of 0.000 and an upward adjustment of 7.5m for the two cores gave the maximum value; it turned out that these two values were identical. . . The liquid level height was provided, but it wasn’t used; it must be incorrect. However, under those circumstances, I had no other choice. Additionally, 3m is the length of the tangent; I think it shouldn’t include the end cap.
The problem is that based on these calculations, I get only 15.42 m2 – there’s no answer, which is awkward. D is 2.6, L is 3m; the distance from the normal liquid level to the lower head is 2.4m, and the distance from the plane of the supports to the lower head is 2m. The height of the frame is 8m. Please ask an expert to help calculate it
This post was last edited by Boroparamita_HSJI on 2017-9-25 07:13 – The container is placed on a suspended platform 8 meters high; its diameter is 2.6 meters and the length of its cylindrical section is 3 meters. The liquid level inside the container is 2.4 meters above the tangent line of the container’s head, with this tangent line being 2 meters away from the platform. I think the distance between the head and the platform is 2 meters. According to the regulations, the area to be considered includes the area of the lower head plus the area of the 1.5-meter-long cylindrical section. The height of the container’s head is 0.25 times the diameter. Therefore, in the formula for the total area, the length L equals 3 + 2*0.25*2.6 = 4.3. After calculating A as 41.47, we subtract 3*3.14*2.6, which represents the area of both heads, and then divide by 2 to get the area of one head, which is 8.5. The wetted area is equal to the area of one head plus 1.5*2.6*3.14, which equals 20.7. I’m not sure if this option exists
You’re absolutely correct; at that time I didn’t know how to calculate the head height