Calculation of heat transfer rate for air-temperature gasifiers
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This post was last edited by am Little Fairy on 2015-8-17 at 21:51. Liquid nitrogen with a pressure of 1.6 Mpa is vaporized using an air-cooled vaporizer. The gas flow rate is known to be 10 Nm3/h; the design temperature for liquid nitrogen is -196 degrees, while the design temperature for nitrogen is 25 degrees. How should the heat transfer rate and the heat transfer area be calculated? I did some rough calculations: m=10*1.25/810=0.015 kg/h. △Q=m*△H=0.015 kg/h*1351 kcal/kg=20.265 kcal/h. △T=25–196=221 K; 1 K = 7 kcal/(m2·h·°C) (I’m not sure where this value comes from). The heat transfer area S=△Q/(K·△T)=20.265/(7*221)=0.013 m2. This seems completely wrong! ! ! I don’t know where the mistake lies; was the formula chosen incorrectly? Wrong parameter selected? I would be very grateful if you experts could give me some advice! ! !
A phase change has occurred, so your calculation is definitely incorrect; there’s also the issue of latent heat when liquid nitrogen changes from a liquid state to a gas state, which you haven’t taken into account
There are mainly a few points: 1, it should be calculated based on the enthalpy difference before and after the use of liquid nitrogen. For rough calculations, it can be assumed that all are in a saturated state. 2. The temperature difference should be calculated as a logarithmic temperature difference. 3. A heat transfer coefficient of 7 is quite high; generally, a value of 4–5 Kcal is used. But the amount you have is too small, so it’s normal for the calculated area to be small.
Thank you very much for your advice! I’m truly grateful! Due to the relatively poor foundation, I need to ask for further assistance: 1. Where can I find this enthalpy difference? 2. What does it mean that the temperature difference should be calculated as a logarithmic temperature difference? 3. Where can I find the heat transfer coefficient?
1. The enthalpy difference can be found by referring to the T-S diagram of nitrogen in the \"Deep Cryogenics Handbook\", and using the pressure and temperature values to determine the enthalpy values before and after the vaporization of liquid nitrogen on that diagram. 2. Logarithmic temperature difference. . . This is basic knowledge in thermodynamics; you can refer to textbooks on engineering thermodynamics or heat transfer. 3. This heat transfer coefficient is an empirical value; it was mentioned in some books, but I can’t recall the specifics. But generally, 4~5 Kcal is used
Okay, thank you so much! ! !
Thank you! How should this latent heat be taken into consideration?
Should it be understood as: △Q=m*△H+m*i?
When using enthalpy values for calculations, there is no need to consider the specific process. Since enthalpy is a state quantity, there is no need to consider how the gas changes; it is sufficient to know the value of the final state. If heat value calculations are not used and the entire transformation process of the substance is considered, it is necessary to calculate the heat absorption during processes such as vaporization from liquid nitrogen to gaseous nitrogen and regenerative heating. Therefore, in thermodynamic calculations, enthalpy values are generally used as much as possible, as it is simpler and more convenient.
Okay, thank you! How do I read the T-S diagram in the cryogenic handbook? I don’t understand it. Among the projects I’ve been working on recently are air-temperature gasifiers and pressure regulating devices; are there any textbooks or manuals on this topic? I want to learn * a bit.
This post was last edited by arpcd on 2015-8-25 at 14:32. It is recommended that the original poster consult a professional manufacturer for calculations. The reason is simple: your own skills in calculation are too poor. . Take a look at the nitrogen mass flow rate you calculated: m=10*1.25/810=0.015 kg/h. 1.25 is the density of nitrogen, while 810 is the density of liquid nitrogen. What you need to calculate is the mass flow rate – how can these values be used together? ? ? ? What does it become when put together? Have you checked the dimensions of the unit? If 1.25 and 810 are the densities of nitrogen and liquid nitrogen respectively, then 0.015 represents the volumetric flow rate of liquid nitrogen; yet the unit you provided is kg/h... The mass flow rate of nitrogen at 10 Nm3/h is W = 10/22.4*28 = 12.5 kg/h. This is simply high school-level physics knowledge; I won’t go into details about things like enthalpy or latent heat of vaporization. If you really want to calculate it by yourself, there are too many lessons you’ll need to take on your own. . . Be more careful; something will go wrong if you do things this way in construction.