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I’ve been working on developing software for heat exchanger design recently, and I have a question~ When designing heat exchangers, is K*A a constant? This problem has been bothering me for a long time. Another issue is that it’s difficult to determine the efficiency of the heat exchanger during the calculation process
Your question isn’t quite clear, but whether K*A is a constant depends on how your program is set up. Explain your problem clearly, and maybe everyone can help you.
Reply to 2# sus321: Thank you~ The program is quite simple; the heat exchanger is used to heat air with water, and a finned tube heat exchanger is employed. The efficiency of the heat exchanger and the NTU are determined based on the temperature parameters of water and air, with NTU*Cmin=KA. By setting the ranges of several parameters and using a method of nested loops, the heat exchanger type that meets the pressure drop requirements is identified. Do you think this method is feasible?
I’m not familiar with NTU calculations; I use Q=K*A*T to calculate each model, list the models that meet the composite requirements, and then choose from them.
Reply to 4# sus321: When calculating K, do you start by doing a trial calculation first? First, assume a certain structure and calculate its heat transfer coefficient?
Reply to 5# morning121: Yes,,,,,,,
One more thing: when designing heat exchangers, once the physical property parameters of the cold and hot fluids are determined, is the efficiency then fixed?
This post was last edited by wiseboy on 2011-10-22 at 17:40. K*A can be treated as a constant. The difficulty lies in K and A – what exactly is the value of each? This has been a point of contention for hundreds of years.
Has anyone else written programs before that I could take a look at? My supervisor is forcing me to write programs, and I’m trying to find someone who can help improve the code, but I don’t even know how to start:(
1. The mentor forces one to write programs; obviously, as long as you can manage to meet the requirements for graduation, that’s enough. You can never design practical software just as a game to finish school ; 2. For a fixed structure, KA is a constant. Mathematically, KA has a feasible region, and all KA values within this region are valid.