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The known conditions are as follows: on the steam side, it is saturated steam at 0.13 Mpa (absolute pressure), with a flow rate of 16.5 t. The temperature at the secondary water inlet is 47°C. What is the minimum amount of water, in tons, required when using a plate heat exchanger to convert all the steam into condensate?
For 130t, what temperature will the secondary water reach, and what will be the temperature of the steam drain?
This post was last edited by wanlirn on 2021-3-18 at 17:14: 550*16.5*1000/(67-47) = 453,750 kg/h. Also, is a plate heat exchanger suitable for using steam?
It’s definitely not as good as shell-and-tube types, but it works fine with low-temperature steam; your calculation is incorrect
The last edit to this post was made by wanlirn on 2021-3-19 at 10:14. Asking about the minimum amount of water required shows that you’re an amateur; there is only a relatively appropriate amount of water that can be used. Besides, if you know how to calculate it, then why ask?
It’s about minimizing water usage, as the client mentioned that the secondary water supply is not sufficient; we want to use as little water as possible. However, the heat exchange efficiency of domestically produced heat exchangers varies, so I’m asking the experts here: to what extent can it achieve optimal performance? Can the secondary water output reach 90%? 95?
To remove water at 90 degrees, approximately 58 tons are needed
How much hydrophobicity have you applied? This is also a huge difference :'(
Everyone says it’s at 90 degrees
550*16.5*1000/(67-47) = 453,750 kg/h. You can replace 67 with any value you prefer; 80, 90, 110 are all options, as long as it’s below the saturation temperature of steam. It’s a simple calculation, right? The higher the temperature, the larger the heat exchange area required. Where you place the steam trap also determines its operating temperature. Also, if you can’t do the calculations, don’t rashly claim that others are wrong. Am I coming across as an amateur just by typing so much?