Heat exchanger calculation
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When the water mist temperature drops from 50 degrees Celsius to 40 degrees Celsius and the air flow velocity is 10 meters per second, should the heat exchange area be calculated using 10 meters per second, or should the flow velocity of the water be adjusted to reflect its liquid state?
Calculate the heat exchange area by hand? Should it still be calculated using software like HTRI? What you should determine is the volume, not the flow rate, right?
Pure cooling by temperature reduction can be done at your specified flow rate, while condensation is something different – it’s not just a matter of converting it into a liquid state
Do you need flue gas whitening or steam condensation?
Is there condensation as well as a temperature drop? Should they be calculated separately or combined under the category of condensation?
It’s still the same question as before: The water mist temperature drops from 50 degrees Celsius to 40 degrees Celsius, the air flow rate is 10 meters per second, the concentration of the water mist is 0.5 Kg/m3, and the temperature of the cooling water is 30 degrees Celsius. Calculate the area of the shell-and-tube heat exchanger. The manual calculation approach is as follows: 1. Determine the amounts of gas and liquid of this substance at the two temperatures, that is, the amounts of water and water vapor at 50 degrees Celsius and 40 degrees Celsius; further determine the amount of steam that condenses due to the temperature change. If there is air present, determine its amount as well； 2. Collect parameters of water and water vapor, as well as air (if any), at the qualitative temperature ； 3. Calculate the total heat to be removed ； 4. Calculate the average temperature difference ； 5. Calculate the heat transfer film coefficient ； 6. Calculate the overall heat transfer coefficient ； 7. Calculate the total heat transfer area ； 8. Calculate the resistance loss. Regarding the determination of the material quantity, I believe that at 50 degrees Celsius the density of water vapor is 0.083 Kg/m3; the water in the water mist should be considered as liquid, and the amount of condensation should be calculated based on the changes in the amount of water vapor. I am not sure if there are any omissions or errors in the above information; I would appreciate it if experts could point them out for me.
1. Determine the amounts of gas and liquid for this material at two temperatures, namely the amounts of water and water vapor at 50 degrees Celsius and 40 degrees Celsius; further determine the amount of steam that condenses as the temperature changes, and if there is air, determine the amount of air as well； 2. Collect parameters of water and water vapor, as well as air (if any), at the qualitative temperature ； 3. Calculate the total heat to be removed ； 4. Calculate the average temperature difference ； 5. Calculate the heat transfer film coefficient ； 6. Calculate the overall heat transfer coefficient ； 7. Calculate the total heat transfer area ； 8. Calculate the resistance loss.
There is no problem with the algorithm; the issue lies in the calculation of the overall heat transfer coefficient. Generally, when formulating an equation to relate flow velocity to other parameters, the same basis is used for all calculations, and it’s sufficient to ensure that the equations are consistent with each other. However, in actual production, the calculation results of the correlation equations may be on the high side; this is at least the case for boiling heat transfer.
Thank you; I hope to receive more guidance and have more exchanges.
For the calculation of latent heat and sensible heat, who knows by how much the K value might differ when calculating latent heat compared to when calculating sensible heat (it could be expressed as a multiple)