Hydroprocessing Unit material and energy Balance

  • #1
DumpmeAdrenaline
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Homework Statement
The vacuum residue fraction from an Albertan bitumen is hydroprocessed using an ebullated bed technology. The vacuum residue fraction has a density of 1030 kg/m3. The catalyst is a commercial sulfided CoMo/Al2O3 catalyst. The unit is operated at a constant reactor temperature of 428 °C and a total pressure of 10 MPa absolute. The H2-to-feed ratio is maintained at 1000 normal m3/m3 residue feed. The H2 consumption is 1.27 wt% of the residue feed and the average heat of reaction for hydrotreating and cracking expressed in terms of the H2 consumption is 62kJ/mol H2 consumed.

1.1) What is the H2 consumption by reaction in normal m3/m3 residue feed? (Normal conditions:273.15 K, 101.325 kPa).

1.2) Some gaseous products are produced during residue hydroconversion that cannot completely be removed as either liquid phase products, or through scrubbing, i.e. they end up in the H2recycle. For the process considered, these products are methane 1.2 wt%, ethane 0.3 wt%, H2S0.6 wt%, and CO2 0.1 wt% (wt% of the residue feed). If the H2 partial pressure in the H2 recyclemust be at least 8 MPa, what is the purge rate in normal m3/m3 residue feed? (Hint: Non-H2gases in the recycle must be purged at the rate of their formation).

1.3) What is the fresh feed rate of pure H2 in normal m3/m3 residue feed?

1.4) What must the preheater outlet temperature be for the combined feed, i.e. H2 and residue feed? Assume that the average heat capacity for all of the material in the reactor is 2.6 kJ/kg·K at the operating conditions. Consider this question independently of your answers to questions 1.2 and 1.3 and make the following assumption: assume that the ratio of non-H2 gases to H2 in the combined fresh plus recycled H2 feed is 1:6 molar ratio. (Hint: The ebullated bed reactor behaves like an adiabatic continuous stirred tank reactor and the reactions taking place during hydroprocessing are exothermic).
Relevant Equations
Material balance
Ideal gas Law
Energy Balance
1.1) 1 m3 of residue feed has a mass of 1030 kg.
Mass of H2 consumed=(1.27 kg of H2 consumed/100 kg of residue feed)*1030 kg of residue feed=13.081 kg
We can treat H2 as an ideal gas to calculate its volume at normal conditions
101325*V=(13.081/2.016)*1000*8.3145*273.15
V=145.436 m3 normal/m3 residue feed

1.2) To calculate the H2 mass in feed I used the ideal gas
1000*101325=n*8.3145*273.15
n=44614.8 mol H2 =44.6148 kmol
Mass of H2 in feed=44.6148*2.016=89.9434 kg
Mass of Unconverted H2=Mass of H2 in feed-Mass of H2 consumed=89.9434-13.081=76.8624 kg H2

Component in product streamwt% of residue feedm(kg)n (kmol)y (mole fraction)
CH41.212.360.77050.0006
C2H60.33.090.10270.0046
H2S0.66.180.18130.0026
CO20.11.030.023400.0197
H27.462476.862438.12620.9725

We split the product stream into a low pressure purge stream and low pressure recycle stream to avoid the build up of non-condensables that decrease the partial pressure of H2. The composition of all three streams are the same. Given that the partial pressure of H2 in the recycle stream must be 8 MPa I am not sure how this information is enough to compute the purge rate
 
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  • #2
Can you please provide a flow diagram?

You chose as a basis for the calculation 1030 kg residual feed, and you calculated a H2 consumption of 6.49 moles. If the H2 to residue feed ratio is 1000 m^3/m^3, that is a mole ratio of 1000. These two results don't seem consistent to me. What am I missing?
 
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  • #3
Unfortunately, no flow diagram is provided in the problem. I would imagine a simplified diagram of the process to look like the image below. I used 1030 kg as the basis because the questions are asking for the purge and fresh rates per m³ of residue feed, which is equivalent to 1030 kg. However, I am not sure how the volume of 1000 m3 H₂ in normal conditions per m³ of residue feed corresponds to a mole ratio of 1000. The molecular mass of the residue feed to be processed is not given to determine the mole ratio.
 

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  • #4
Maybe that mole ratio is in the feed to the reactor
 
  • #5
How can I deduce that from the problem givens? I think the 1000 m³ of H₂/m³ of residue feed is referring to the rate of the feed to the ebullated reactor. This was the solution posted for the problem. However, I am not sure how the mole fraction of H₂ in the recycle stream is 0.8 when the composition at the splitting point for the three streams (product stream, recycle stream and purge stream) have to be equal to what’s in the table.

1713453278774.png
 
  • #6
DumpmeAdrenaline said:
How can I deduce that from the problem givens? I think the 1000 m³ of H₂/m³ of residue feed is referring to the rate of the feed to the ebullated reactor. This was the solution posted for the problem. However, I am not sure how the mole fraction of H₂ in the recycle stream is 0.8 when the composition at the splitting point for the three streams (product stream, recycle stream and purge stream) have to be equal to what’s in the table.

View attachment 343641
The composition of the recycle steam is exactly the same as the purgee stream. If the partial pressure of H2 is 8 MPa and the total pressure in the system is 10 MPa, the mole fraction of H2 is 0.8.
 
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  • #7
I'm finally understanding the problem statement better. The basis of the calculation is 1 m^3 of residue stream. Ir seems to me everything you've shown in the calculations so far makes sense.
 
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  • #8
Chestermiller said:
The composition of the recycle steam is exactly the same as the purgee stream. If the partial pressure of H2 is 8 MPa and the total pressure in the system is 10 MPa, the mole fraction of H2 is 0.8.
But If the partial pressure of H2 in the recycle stream is 8 MPa I think the total system pressure would not be 10 MPa. According to Daltons law
Partial pressure of H2=Mole fraction of H2*Total pressure in the system
8=0.9725*Total pressure in the system
Total pressure in the system=8/0.9725=8.225 MPa

Similarly, if we want the total pressure in the system to be 10 MPa then the partial pressure of H2 in the recycle stream would be 9.725 MPa
 
  • #9
The mole fraction of H2 in the recycle is not 0.9725.
 
  • #10
The problem states that the non-H₂ components cannot be removed as liquid-phase products from the flash separation unit. Therefore, 1.0779 kmol of non-H₂ components produced during hydroconversion will end up in the gaseous stream. This gaseous stream is split into a low-pressure purge and low-pressure recycle stream. If we use the information that the H₂ partial pressure in the recycle stream is 8MPa, and the total system pressure is 10MPa, this means that the mole fraction of H₂ is 0.8. The mole fractions in the gas stream to be split into the purge stream and recycle stream are the same. The amount of H₂ in the gas stream is given by is x/(1.0799 +x)=0.8 x=4.3116 kmol.

I am not sure how the information provided in the problem is enough to determine how much of the gas stream is split between the recycle and purge.
 
  • #11
DumpmeAdrenaline said:
The problem states that the non-H₂ components cannot be removed as liquid-phase products from the flash separation unit. Therefore, 1.0779 kmol of non-H₂ components produced during hydroconversion will end up in the gaseous stream. This gaseous stream is split into a low-pressure purge and low-pressure recycle stream. If we use the information that the H₂ partial pressure in the recycle stream is 8MPa, and the total system pressure is 10MPa, this means that the mole fraction of H₂ is 0.8. The mole fractions in the gas stream to be split into the purge stream and recycle stream are the same. The amount of H₂ in the gas stream is given by is x/(1.0799 +x)=0.8 x=4.3116 kmol.

I am not sure how the information provided in the problem is enough to determine how much of the gas stream is split between the recycle and purge.
The problem statement, in my judgment, is not precise enough to allow unambiguous solution.
 
  • #12
After further thought, here's my tae on this:

Let M = molar makeup H2 feed to reactor/m^3 residue feed

Let R = molar recycle feed to reactor/m^3 residue feed

Let P = molar purge rate /m^3 residue feed

Then $$M=\frac{13.081}{2.016}+0.8P$$

$$M+0.8R=44.615$$

Does this make sense to you?
 
  • #13
It makes sense to me expressing the makeup in terms of the recycle stream but not for the purge stream.
Shouldnt the Molar H2 makeup to the reactor=Molar H2 in top stream from the the flash separator + Molar H2 in Purge + H2 Consumed

Molar H2 Feed=Molar H2 Make up+ Molar H2 Recycle
Molar H2 Feed=Molar H2 Unconverted+ Molar H2 Consumed
Molar H2 Unconverted =Molar H2 in top stream from the flash separator+ Molar H2 in bottom liquid product
Molar H2 in top stream from the flash separator=Molar H2 Recycled+ Molar H2 Purged
Molar H2 Make up +Molar H2 Recycled=Molar H2 Unconverted+ Molar H2 Consumed
Molar H2 Make up +(Molar H2 in top stream from the flash separator-Molar H2 Purged)=Molar H2 in top stream from the flash separator+ Molar H2 in bottom liquid product +Molar H2 Consumed
Molar H2 Make up=Molar H2 Purged+ Molar H2 in bottom liquid product+ Molar H2 Consumed
=0.8P+(13.081/2.016)+33.8146=0.8P+40.3032=-0.8R+44.615
 
  • #14
DumpmeAdrenaline said:
It makes sense to me expressing the makeup in terms of the recycle stream but not for the purge stream.
I don't think so. I think that the makeup is equal to the converted H2 in the bottom liquid product plus the H2 in the purge stream.
DumpmeAdrenaline said:
Shouldnt the Molar H2 makeup to the reactor=Molar H2 in top stream from the the flash separator + Molar H2 in Purge + H2 Consumed
I don't thnk so. The H2 from the separator is equal to the H2 in the recycle plus the H2 in the Purge. The H2 makeup should be the the H2 leaving in the purge + the converted H2 (leaving with the liquid product).
DumpmeAdrenaline said:
Molar H2 Feed=Molar H2 Make up+ Molar H2 Recycle
Correct
DumpmeAdrenaline said:
Molar H2 Feed=Molar H2 Unconverted+ Molar H2 Consumed
Molar H2 feed = molar H2 consumed + molar H2 in Purge
DumpmeAdrenaline said:
Molar H2 Unconverted =Molar H2 in top stream from the flash separator+ Molar H2 in bottom liquid product
The assumption is that the H2 in the bottom liquid product = H2 converted
DumpmeAdrenaline said:
Molar H2 in top stream from the flash separator=Molar H2 Recycled+ Molar H2 Purged
correct
 
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  • #15
What a careless mistake on my part! The hydrogen must dissolve in the liquid phase to convert the vacuum residue; it makes sense to me now.

But doesn’t this mean that the unconverted H2 ends up in the top stream from the flash separator? The compositions in the table would correspond to the compositions of the top stream, recycled stream, and purged stream yielding the below equations for the makeup of hydrogen in terms of the purge and recyled streams.

M=(13.081/2.016)+0.9725P
M=(89.9434/2.016)-0.9725R
 
  • #16
DumpmeAdrenaline said:
What a careless mistake on my part! The hydrogen must dissolve in the liquid phase to convert the vacuum residue; it makes sense to me now.

But doesn’t this mean that the unconverted H2 ends up in the top stream from the flash separator? The compositions in the table would correspond to the compositions of the top stream, recycled stream, and purged stream yielding the below equations for the makeup of hydrogen in terms of the purge and recyled streams.

M=(13.081/2.016)+0.9725P
M=(89.9434/2.016)-0.9725R
In thess equations, the 0.9724 should be 0.8
 

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