Currently, I work on hydrocracking of Methyl ester with addition of hydrogen gas into the reactor (mainly Trickle bed or fixed bed reactor). The flow rate of liquid is 128.1mL/min at 1 bar and temperature 85C flowing into the reactor, and hydrogen gas is unknown flow rate. Can somebody help me to calculate hydrogen gas flow rate at 40 bar and temperature at 25C and H2/Oil ratio is 500Nm3/m3 or 500NI/I?
At 25°C ad 40 bars hydrogen behaves almost like an ideal gas. Its compressibility factor is close to unity.Its value of Z at 25 C and 40 bars is about 1.02
So 500 Nm3 of hydrogen at 25 C and 40 bars have a volume 500x(298.15/273.15)x(1.01325/40)x1.02=14.1 m3
Thanks Dr. Soave for your solution.
Would you please explain the value of Z what does it standarding for at 25 C and 40 bars?
Also would you mind if you could show your mathematical calculation , how did you get 1.02?
Once again, thank you for your hints. Its vert helful. I look forward to hearing from you.
Kind regards,
Peter Mponzi
I read 1.02±0.01 from the table given in
http://www.bnl.gov/magnets/staff/gupta/cryogenic-data-handbook/Section3.pdf
page III-E. I could have a more accurate value from the TRC Tables, but they were buried in some drawer. Anyway HYSYS should give a reliable value.
So for a ratio of 500 Nl/l at an oil flow of say 50 ml/min we would need 14.1 x 50 = 705 ml/min H2?
Peter
Dear Dr. Soave,
Apology for my earlier email was very direct. If you have time would you please respond to the equation i have send.
In the meantime, I would appreciate it if you could have a look a general need for the hydrogen flow rate interm of mL/min .
Thank you
Peter
Sorry for misunderstood my question. Please let me rephrase again, this 14.1m3 is the volume flow rate of hydrogen gas at 25 C and 40bars. However, as previously mentioned that 500NI/I or 500Nm3/m3 is the ratio of H2/Oil and let’s presume that oil flow rate is 50mL/min and flowing into the reactor (mainly trickle bed) at 85 C and 40bars. What could be the flow rate (mL/min) of hydrogen gas?
Please take a look calculation if it’s right procedure:
= ratio of 500NI/I at an oil flow of 50mL/min
= 14.1m3*50mL/min= 705mL/min of H2?
Please could you assist?
Thank you,
Peter
if you have 128.1 ml/min of liquid or 0.128l/min, you need:
500Nl/lx0.128Nl/min=64NL/min of H2.
Check how (what unit) your gas flow measurement system operates.
As Dr. Soave explained, you can correct this flow rate to any condition by
PV=z.nRT, where z, the compressibility factor. As n.R = PV/(zT)
Then P1V1/(z1T1) = P2V2/(z2T2).
taking condition 1 as normal standard (1 atm, 273K, z=1) and z2 =1.02 for 40 bar and 298K.
to measure at condition 2, the 64 NL/min= 1,908l/min, IF your instrument is reading the flow rate at this condition. If you are using a rotameter, check at what condition it was calibrated. The manufacturer can help you to correct the value to be read on it.
Dear Dr. Derenzo,
Thank you for great demostration of calculation procedure to be follow. Your explaination has great impact in my current project which is very useful and helpful.
I will come back to you later after achieving good results of my reseach work. I will share with you.
Once again, I do appreciate for your explaination.
Kind regards,
Peter Mponzi
As a design engineer in an engineering team whose work is developing a new gas turbine of 2 MW used for thermal power generation which is fuelled by non-carbon fuels, such as hydrogen and ammonia, your task is to analyse and design the combustion system for providing stable and efficient combustion in the gas turbine.
any idea on how to go about it?
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