The amount of BOG from the LPG tank

The amount of BOG multiplied by the heat of vaporization should represent the convective heat transfer between the tank and the surroundings

This post was last edited by arpcd on 11/30/2016 at 10:41. Did the original poster mean that 390 kg/h is actually missing a zero? ? The highest amount of BOG generated by LPG occurs during unloading operations (BOG produced through vaporization due to heat exchange contributes little in other cases). The amount of BOG in such situations depends on the unloading rate; assuming that a 10 dam3 tank is filled in 6 hours, the unloading rate is 10/6 = 1.67 dam3/h. With a density of propane of 1.95 kg/m3, it’s easy to calculate the BOG amount: BOG = 1670 x 1.95 = 3257 kg/h. In reality, the actual BOG amount is slightly lower than this value. For a 10 dam3 tank, the BOG production is generally around 3 to 4 tons per hour. The higher the unloading rate, the greater the amount of BOG generated. . I strongly suspect you missed a zero. . . :Lol, I worked on a BOG compressor project with a capacity of 40 dam3 years ago; the amount of BOG generated was around 10–12 t/h, and the volume of the tank was four times that of the one mentioned by the original poster. Unless the unloading rate is very low. . .

This is my first time doing this, so I don’t quite understand it. I calculated it based on the daily evaporation rate of the tank, considering convection and conduction. The total radiation is calculated as the heat exchange amount divided by the latent heat of vaporization. I’ve also seen people say that it is calculated based on the unloading conditions, but I’m not sure how exactly it’s done. Are there any documents available for reference?

1670*1.95 seems to represent the amount of liquid propane delivered by ship. Why is the amount of gas produced calculated based on this figure? Is it possible that all of that amount gets vaporized?

Ugh~~~~ The so-called unloading condition is easy to understand. Your large tank is currently empty, but the 10 m3 space inside it is filled with propane gas (the pressure under operating conditions is 7–15 kPaG, and the temperature is 20–40°C; you can use either extreme value for calculations). When your LPG ship arrives at the shore and unloading begins, the LPG unloading pump will discharge the gas at a rate of 1670 m3/h, and this pump will be able to fill your large tank in 6 hours. . At this point, the LPG liquid occupies the volume of the gaseous space inside the tank; in other words, it’s like water entering a cup and displacing the air inside it. This is what happens during the unloading process. . . Do you understand? The calculation is extremely simple: unloading rate x gas phase density = BOG. It’s easy to understand – you can figure it out by making a comparison using air and water. The BOG volume is at its maximum at this time! A simple calculation shows that your vaporization amount is less than 400 kg, while the BOG during unloading amounts to several tons. . Go ask your master; don’t make a fool of yourself with things like this. . .

This post was last edited by Xin You Meng Hu on 2016-11-30 at 11:12. I understand how to calculate it after hearing your explanation, but I have another basic question: are low-temperature tanks usually emptied before receiving contents? Or should it be received when only half is left? Also, I saw that the capacity of his unloading arm is listed as 500 m3/h, so I’m wondering if this is the amount of BOG I will have to handle

This post was last edited by arpcd on 2016-11-30 at 11:17. BOG is calculated in order to select the appropriate compressor; the compressor cannot be too small – it must be capable of handling the maximum amount of BOG under operational conditions, with a 15% margin of safety. What you are referring to is a situation during operation that has nothing to do with the calculation of BOG. In fact, it is quite common for the LPG tanks to be emptied before the cargo is unloaded from the ship (and this is why BOG must be calculated on an maximum basis). The design point condition and the operating condition are two different things. . As for the unloading arms you mentioned, there are usually 4 of them: three for unloading and one for returning air. 3 x 500 = 1500, which is pretty close to my estimate of 1670, right? ? ?

That’s the reasoning. Also, as I mentioned earlier, I need to take into account the limitations on the conveying capacity of the unloading arm at the front (there’s a big difference between 500 and the amount calculated based on the volume of the tanks); at such a scale, the impact of heat exchange cannot be ignored. It’s my first time doing this, and I couldn’t find any suitable literature on the subject. Thank you for your guidance.

On his schematic diagram, I see one for unloading and one for returning air. . . . Mark it as 500. I’ll go check again:o

That’s how I understand it: when pairing compressors, one larger and one smaller unit are connected in parallel, with the smaller one used for normal evaporation processes and the larger one used during unloading operations. Otherwise, if only the larger compressor is used, the amount of BOG generated during normal evaporation is so small that the capabilities of that large compressor would not be sufficient, right?