2016 Morning Case 21
Thread Content
At first glance, it seems like this problem is unsolvable; could it be that some information is missing? . . In a 21-liter jacketed reactor, the material at 120°C is continuously added to the bottom of the reactor, while cooling water at 30°C is circulated in the jacket to cool the material. When stirring is stopped, the two fluids flow in parallel, with the outlet temperatures of the material fluid and the cooling water being 90°C and 42°C respectively. If all other conditions in the reactor remain unchanged and stirring is continued to increase the heat transfer coefficient, what will be the outlet temperature of the liquid mixture (°C) in this case? (A) 86.3 (B) 74.1 (C) 67.8 (D) 62.5
During co-current flow, the temperature difference increases by 66.8 k; the temperature difference can only decrease, so D is the only option
As the heat transfer coefficient increases, both the outlet temperature of the liquid material and that of the cooling water rise, or perhaps only one of them rises. But why does the temperature difference decrease? I don’t understand.
:L asked the wrong question. I thought it was asking what the effective temperature difference would become. . .
Can everyone still solve this problem? I asked many people~
Are all four alternative answers to this question correct? My calculation gives 55.7
I don’t know; the wording of some of the questions is incorrect, and everyone has pointed that out to me as well
Could you post the detailed explanation? Let’s discuss it~
The thermal resistance of the dirt on the inside of the tube should be 4*10^(-4)
Even if the dirt coefficient is changed to 10-4, no answer can be obtained.