Calculation of the buffering time for hydrogen buffer tanks
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I have a hydrogen buffer tank that is completely sealed. The normal operating pressure is 1 MPaG; the volume of the buffer tank is 100 m3, and the normal gas flow rate is 15,000 m3/h. If there is a sudden failure in the gas supply, how long can the buffer tank provide buffering? That is, how long will it take to fill the buffer tank with these 15,000 m3 of hydrogen?
Pay attention to the gas volumetric flow rate under operating conditions; if it has been converted based on the conditions mentioned above, then the buffering time for this buffer tank is 24 seconds.
Reply to 2# jintchinssen: How was that calculated? Could you explain your method of calculation?
The amount of gas stored in your hydrogen tank can be easily calculated: the volume V is 100 m3, and the pressure P is 1.1 MPa (A) ≈ 11 atm. When the pressure in the tank decreases, the maximum amount of gas that can be released occurs when the tank’s pressure equals atmospheric pressure. Therefore, the volume of gas that can be released, V’, is given by V’ = (P – Po)/Po * V = (11 – 1)/1 * 100 = 1000 Nm3. You, however, require a large amount of gas, namely 15,000 Nm3 per hour. Thus, the buffering time is 1000/15,000 = 0.0667 hours, which is equivalent to 240 seconds. The moderator on the 2nd floor set the time to 0, haha. . Actually, it’s not the moderator’s fault; you didn’t specify whether the gas consumption rate was given in standard cubic meters or in terms of pressure. If it’s in standard cubic meters, that is 15,000 Nm3/h, then the maximum buffer time would be 240 seconds. But if the gas consumption rate is 15,000 m3/h at a pressure of 1 MPaG, then the buffer time would only be 24 seconds. However, the moderator didn’t take this scenario into careful consideration. When the tank releases 100 m3 of gas under these conditions (which is equivalent to 1,000 Nm3, due to the 10-fold difference in pressure), the pressure inside the tank drops to atmospheric level. Yet the subsequent operating conditions require a pressure of 1 MPaG. Therefore, such a 24-second buffer time doesn’t exist, as the gas can’t be released under those conditions; a buffer requires a pressure difference, meaning the pressure used in the subsequent processes must be lower than the pressure in the storage tank. The gas output volume can be calculated using the formula V’ = (P_tank – P_usage) / Po * V. For example, in the case of an instrument air buffer tank, if the user requires a pressure of 0.2 MPaG, then this tank will have 800 Nm3 of gas available for buffering, calculated as (1.1 – 0.3) / 1 * 100. . The principles and formulas are quite simple; the original poster can calculate them by themselves.
Reply to 4# arpcd, thank you! Very detailed.
Hello, is this the standard content? I’d like to take a closer look at it, thank you.
I think these are two separate issues; the original poster didn’t ask clearly enough.