How to estimate the outlet flow rate of a heat exchanger?
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This post was last edited by 17156614 on 2018-8-24 at 16:56. I would like to ask the experts: For a shell-and-tube heat exchanger, how can seawater at 10°C be used to heat propylene from -47.7°C to -5°C? The flow rate of propylene is 200 m3/h, and the pressure is 25 barg. The calculated heat transfer rate is 3300 kW. In this case, assume that the seawater flow rate is 500 m3/h and the seawater outlet temperature is 4.12°C; if the seawater flow rate is assumed to be 400 m3/h, the seawater outlet temperature is 2.65°C. How should one choose? Is a logarithmic mean temperature difference also required? There is a concept of effective temperature difference in the posts in the forum, but I couldn’t find the specific calculation method
I did a quick calculation: with 300 tons of seawater, the outlet temperature is less than 1°C. To prevent some of the seawater from freezing, it is recommended that the flow rate of seawater be at least 300 tons, and ideally not exceed 500 tons; this way, both the connection point for the seawater and that for propylene can use DN200 pipes. In your operating conditions, the logarithmic mean temperature is around 29 degrees, so it has little impact on the selection process ； Information on the average temperature algorithm can be found on Baidu. For shell-and-tube heat exchangers, seawater should flow on the tube side, using titanium tubes ； Propylene is not very corrosive either; Q345R can likely be used on the shell side, while it’s advisable to line the inside with stainless steel.
This post was last edited by wiseboy on 2018-8-25 23:54. “How do I choose this?” Is it also necessary to calculate the logarithmic mean temperature difference? There is a concept of effective temperature difference in the posts in the forum, but no specific calculation method was found. The complete answer is as follows: 1. It is necessary to recalculate the logarithmic mean temperature difference ； 2. It is also necessary to recalculate the temperature difference correction coefficient; this calculation formula is very complex, and there are diagrams available for reference, but those diagrams are not very accurate ； 3. The effective temperature difference also needs to be recalculated ； 4. Effective temperature difference = logarithmic mean temperature difference × temperature difference correction factor ； 5. The effective temperature difference is the heat transfer temperature difference (driving force) of the heat exchanger, rather than the logarithmic mean temperature difference ； 6. Many people use the logarithmic mean temperature difference to calculate the heat exchanger area, which is incorrect; this leads to calculation errors (resulting in an underestimated area) ranging from 0% to 80%. In other words, sometimes there is no error, while at other times the error is unacceptably large. This is where some of the reasons for failed heat exchanger designs lie. But sadly, they never realize the cause of the design failure and look for reasons elsewhere.