36 replies to this topic

[invssse]

Dear all,

My question is about relief load calculation for a steam stripper in hydrocracking unit of a grass root refinery.

Background:
(1) steam stripper with condenser;
(2) refer to attached PFD sketch;
(3) Total power failure scenario;
(4) the heat/material balanced data under normal and relief conditions are also attached in speadsheet.
(5)We need to validate if Licensor's data is reliable or not.

Problem:
licensor's data shows the required relief load under Total Power Failure case is 123,000 kg/hr while my calculation gets the result of only 4,900 kg/hr.

Analysis:
The unbalanced heat load method should be used for column relief study.
As you can see from attached spreadsheet, under relief case the bottom product will be flashed to the bubble point at relieving pressure(1.46 MpaG), the total enthalpy increased considerably to 148 M*Kcal/hr from only 117 M*Kcal/hr at normal case. this causes the unbalanced load to be negative.

Question:
Do you think my calculation is correct or not? Can you share with me your comments on this?

Thank you in advance!

Regards,
Invssse

Edited by invssse, 21 November 2012 - 04:39 AM.

[Dacs]

I don't see any PFD, can you attach it again?

[invssse]

the sketch is on the FIRST sheet of the Excel doc. Thanks.

[fallah]

invssse,

I think in the sheet "Total Power Failure" at the box titled "Consumptions" you really meant "Assumptions". Am i right?

Indeed, with total power failure, i.e. while water and air coolers are off, reflux pump filure cannot create flooded condensers.

Please recheck above and let's know the result.

Fallah

[invssse]

Fallah,

Yes, it's "assumption". Sorry for the mistake.
Yes, it's right. but the problem is 20% air cooler duty will be added to the "out" enthalpy that will make the calculated accumulation to be "more" negative.

What I am thinking is that the enthalpy of Bottom flow is very large due to increased temperature under relieving pressure and it takes most unbalanced heat out of the system.

Thanks.

[Dacs]

Oops, the sketch was hidden (need to scroll to the left)

My bad

I'll give my 2 cents once I get off work.

[invssse]

I just updated the sketch to cover a little bit more about the 2 FEEDs and I have a further question.

As can be seen from the sketch, if you look at the feed pressures to the tower, FEED1's pressure is 1.1 MPAG and FEED2 is 1.2 MPAG while the RELIEVING PRESSURE would be 1.43 MPAG. this means these 2 FEEDs' pressure will be lower than relieving pressure and these 2 feeds will stop during relief cases.
However if you look at their upstream pressures before the control valves which are 2.76 and 2.79MPAG respectively.

So my question is that, in this case, these 2 FEEDs will stop or not under relief scenarios? Thanks.

In my previous analysis, I assumed FEED1 and FEED2 will not stop. and the STEAM's pressure is 1.58 MPAG which is higher than relieving pressure so STEAM will definitely continue under relief.

but if FEED1 and FEED2 should assume to stop, then the calculation will be totally different.

Edited by invssse, 15 November 2012 - 05:17 AM.

[Bobby Strain]

Don't overlook the possibility of over-filling due to blocked bottom outlet. This is quite often the controlling case for this column.

Bobby

[CMA010]

If the pressure upstream of the control valves is higher than the relief pressure feeds will continue. This might be at a reduced rate.

I don't understand your unbalanced heat calculation. As most of your condensation is lost and your reflux pump stops your OH product and SW rates will stop. Why do you take credit for OH and SW leaving the system as vapour? Assuming the reflux pump will fail and some condensation will continue, the reflux drum will overfill resulting in a blocked vapour outlet.

Furthermore i do not understand why the pressures / temperatures of you feeds and products are in some cases not equal to relief pressure / temperature but equal to normal pressure / temerature. In my opinion the enthalpy of the feeds should be taken at relief pressure.

Did your licensor also use the unbalanced heat method or did they take the steam rate + vapour from the feeds?

[Dacs]

Here goes my 2 cents:

If we're talking about Total Power Failure (TPF), it's reasonable to assume that you'd have reduced flow on your system, and depending on the operation of your source vessels, you may lose the ability to send the feed altogether (such as the loss of pressure of the source vessels during TPF). That's something worthwhile to look at.

If your bottoms stream serves as a feed to a pump, you can also take note the loss of capacity to send bottoms to its destination.

For the steam, you can take credit for reduced flow (similar with feed).

You will lose the capacity to condense your overhead (due to loss of cooling water) although you may (depending on your company philosophy) be able to take credit for partial cooling in your air cooler due to natural convection.

Looks like your OH is being pumped, you'd lose that when TPF happens.

Your SW on the other hand will continue on.
.
Looking at the whole system:
1. Feed flow may be reduced (or cease to exist)
2. Steam continues on (with reduced flow)
3. Bottoms flow may or may not happen
4. OH will cease to flow
5. SW will still flow
6. Off-gas may or may not continue since:

• You lose the ability to pump the system that will cause overfilling (it may or may not continue depending on the production of condensate minus SW flow), or
• You lose the ability to condense your overhead and continue sending offgas downstream (it may continue)

Keep in mind that flow basis should be based on relieving conditions.

This is not really the unbalanced heat method as prescribed by API per se and I'm bordering something
similar with pinch analysis, but as a part of your study, I'd say it's worthwhile to give this a visit.

If I were doing this, most probably I'd estimate the relief load as roughly around steam flow + vapor portion of the feed flows at relieving condition.

To close, I'd treat the licensor data in the side of being more reliable since this is their technology and they're in the best position to know how their technology works.

My 2 cents

[invssse]

CMA010,
Thanks for your replies.My further comments are as follows.

If the pressure upstream of the control valves is higher than the relief pressure feeds will continue. This might be at a reduced rate.

I don't understand your unbalanced heat calculation. As most of your condensation is lost and your reflux pump stops your OH product and SW rates will stop. Why do you take credit for OH and SW leaving the system as vapour? Assuming the reflux pump will fail and some condensation will continue, the reflux drum will overfill resulting in a blocked vapour outlet.

[Reply]: this is a method used by a world-scale engineering company. the basic assumption of this method is that the Mass Balance is always maintained even during relief. You can see that the flowrate of OH product is added back to form "real accumulation". If as you said the vapor outlet will be blocked, then the final relief rate should be "Steam" rate plus "off-gas" rate which will be 10,348 kg/hr. and that is also far more less than licensor's data.

Let's conduct the calculation as you may mean. the assumptions:
a) FEED1 and FEED2 continues as normal;
b-) Steam continues as normal;
c) Q(air-cooler) =20% normal duty;
d) Q(water-cooler)=0;
e) reflux drum is over-filled and Off-Gas rate=0;
f) OH and SW rate =0;
g) BOTTOM continues at relieving condition.


IN
=========================
Stream Enthalpy(M-Kcal/hr)
-----------------------------------------
FEED1 18.534
FEED2 110.487
STEAM 5.727
-----------------------------------------------------------------
total 134.749


OUT
========================================
Stream Enthalpy(M-Kcal/hr)
---------------------------------------------------------------
OFF-GAS 0
OH 0
SW 0
BOTTOM 148.656
Q(air-cooler) 2.959 (20% normal)
Q(water-cooler) 0
-----------------------------------------------------------------
Total 151.618


the calculated accumulation will also be negative:
Accumulation = (134.749-151.618)/58.7
and the required relief rate: W = W(steam)+W(off-gas)=8387+5433=13,820 kg/hr.

Is this calculation correct?? thanks.

Furthermore i do not understand why the pressures / temperatures of you feeds and products are in some cases not equal to relief pressure / temperature but equal to normal pressure / temerature. In my opinion the enthalpy of the feeds should be taken at relief pressure.
[Reply]: yes, the feeds pressure should be increased to relieving pressure. but the calculated enthalpies change not too much.
the OH and SW's pressure is taken as their partial pressure.

Did your licensor also use the unbalanced heat method or did they take the steam rate + vapour from the feeds?
[Reply]: I don't think they are using latter one. because when I mixed FEEDS with STEAM at relieving pressure, the result vapor rate would be 47,000 kg/hr. and if as you said the steam rate plus vapor from feeds, the result will be even lower.

Edited by invssse, 15 November 2012 - 09:27 PM.

[invssse]

Darcs, Thanks for your comments.

One more thing I would add is the BOTTOM flow will go to an exchanger and a flash drum (Operating Pressure=0.5MPAG) without pump in the flow line.
So I think BOTTOM should assume to continue during TPF.

[Bobby Strain]

Just because a world-scale engineering company uses some method does not mean it is correct. Even the name of the subject procedure is incorrect. Because there is always balanced energy flow in any system. So, if it ain't named right, better be wary.

Bobby

[Adi.putra]

Hi,


Here’s my take on the matter:

Assuming that the source pressure on the Feed and Steam stay the same, i.e. 2.76 Mpag for Feed 1, 2.79 Mpag for Feed 2, and 1.58 Mpag for Steam.

The reduced rate, roughly, will be as follow:

Feed 1 (F1) – 157,739 x ((2.76-1.43) / (2.76-1.04))^(1/2) = 141,193 kg/hr
Feed 2 (F2) – 676,746 x ((2.79-1.43) / (2.79-1.04))^(1/2) = 625,888 kg/hr
Steam (S) – 8,387 x ((1.58-1.43) / (1.58 – 1.04))^(1/2) = 4,420 kg/hr

Before going any further let’s consider the following:

As the air cooler will be on a natural convection mode (20% of normal duty), this will not be sufficient to condense most of the overhead vapor, thus the reflux will be coming back as vapor. This means no liquid coming back to column.

Step 1

Take F1, F2 and S (the reduced ones) and do an H&MB calculation to find out what is the mixture vapor (V0) and liquid (L0) rates and properties.

Step 2

Consider the attached sketch and do another H&MB with these criteria into consideration (simulation would do it)

• Flow through the off-gas line (V3) will be calculated based on the available piping configuration with the column pressure (PRV inlet) at 1.43 Mpag.
• The O.H. flow will be 0 since the level will control valve will shut as the liquid is no longer available.
• The amount of condensate (L2) from the air cooler will be going to the S.W. line. In the case S.W. is capable to deliver more flow than L2 and no liquid level control is available then some vapor will go through here. Calculate this vapor flow (V4), if any, with the same method as V3 calculation.

In this case, the pressure of the column is controlled by the vapor side. The relief rate (V6) shall be calculated as V0 – (V3+V4).

As for the bottom flow, I do not think it will be much of an effect on the relieving rate in this case as it will work based on column pressure. After a certain period, as what the case suggest, the bottom flow will be reduced as an effect of no liquid reflux.

Hope this helps.

Thanks
Adi

Attached Files

•  Step Sketch.pdf   29.4KB   142 downloads

[invssse]

Dear Adi,

thanks for your detailed analysis.

With you method, the flash result shows that V0 will be 32,000 kg/hr. that means the calculated relief rate will be lower than this value. it's still very far from licensor's data of 123,000 kg/hr.

[CMA010]

CMA010,
Thanks for your replies.My further comments are as follows.

If the pressure upstream of the control valves is higher than the relief pressure feeds will continue. This might be at a reduced rate.

I don't understand your unbalanced heat calculation. As most of your condensation is lost and your reflux pump stops your OH product and SW rates will stop. Why do you take credit for OH and SW leaving the system as vapour? Assuming the reflux pump will fail and some condensation will continue, the reflux drum will overfill resulting in a blocked vapour outlet.

[Reply]: this is a method used by a world-scale engineering company. the basic assumption of this method is that the Mass Balance is always maintained even during relief. You can see that the flowrate of OH product is added back to form "real accumulation". If as you said the vapor outlet will be blocked, then the final relief rate should be "Steam" rate plus "off-gas" rate which will be 10,348 kg/hr. and that is also far more less than licensor's data.

Let's conduct the calculation as you may mean. the assumptions:
a) FEED1 and FEED2 continues as normal;
b-) Steam continues as normal;
c) Q(air-cooler) =20% normal duty;
d) Q(water-cooler)=0;
e) reflux drum is over-filled and Off-Gas rate=0;
f) OH and SW rate =0;
g) BOTTOM continues at relieving condition.


IN
=========================
Stream Enthalpy(M-Kcal/hr)
-----------------------------------------
FEED1 18.534
FEED2 110.487
STEAM 5.727
-----------------------------------------------------------------
total 134.749


OUT
========================================
Stream Enthalpy(M-Kcal/hr)
---------------------------------------------------------------
OFF-GAS 0
OH 0
SW 0
BOTTOM 148.656
Q(air-cooler) 2.959 (20% normal)
Q(water-cooler) 0
-----------------------------------------------------------------
Total 151.618


the calculated accumulation will also be negative:
Accumulation = (134.749-151.618)/58.7
and the required relief rate: W = W(steam)+W(off-gas)=8387+5433=13,820 kg/hr.

Is this calculation correct?? thanks.

Furthermore i do not understand why the pressures / temperatures of you feeds and products are in some cases not equal to relief pressure / temperature but equal to normal pressure / temerature. In my opinion the enthalpy of the feeds should be taken at relief pressure.
[Reply]: yes, the feeds pressure should be increased to relieving pressure. but the calculated enthalpies change not too much.
the OH and SW's pressure is taken as their partial pressure.

Did your licensor also use the unbalanced heat method or did they take the steam rate + vapour from the feeds?
[Reply]: I don't think they are using latter one. because when I mixed FEEDS with STEAM at relieving pressure, the result vapor rate would be 47,000 kg/hr. and if as you said the steam rate plus vapor from feeds, the result will be even lower.

I think the formula's are correct, the point i was trying to make was that i don't understand the input. Are you sure the enthalpies of the feeds don't change much? The difference in enthalpy (kg/kcal) might not be big but your flow rates are massive.

Stricktly speaking you bottom temperature can't be higher than the temperature of the feeds while in your calculation it is at least 20 °C higher. This might be the reason that your bottoms product withdraws most of the heat.

Bobby: Personally i have never used the unbalanced heat method for strippers, but it is a widely accepted method within engineering and operating companies.

[Adi.putra]

Dear Invssse,

Do you mind sharing the detailed calculation you did using my method? It's for my reference
I would like know the vapor fraction of F1 and F2 as well as the flow rate figure of the rest, e.g. V0,V1,V3,V4, L2, etc.

Thanks

Adi

Edited by Adi.putra, 16 November 2012 - 04:17 AM.

[invssse]

Hi CMA010,

Thanks for your further comments.

I think the formula's are correct, the point i was trying to make was that i don't understand the input. Are you sure the enthalpies of the feeds don't change much? The difference in enthalpy (kg/kcal) might not be big but your flow rates are massive.

Stricktly speaking you bottom temperature can't be higher than the temperature of the feeds while in your calculation it is at least 20 °C higher. This might be the reason that your bottoms product withdraws most of the heat.

Bobby: Personally i have never used the unbalanced heat method for strippers, but it is a widely accepted method within engineering and operating companies.

For feeds, we normally flash them at relieving pressure adiabatically. By this way the enthalpy will change not so much.

Yes, by normal method, the bottom product at bubble point at normal pressure will be flashed to bubble point at relieving pressure. this makes its temperature is very high and I think this is the problem as you said.

I am thinking that since this is a steam stripping column not a reboiled column, the bottom product should NOT reach bubble point when the pressure increases to relieving pressure.
We just conducted a dynamic simulation of this tower and it's found that when column pressure increased from 1.07 MPAG to 1.33 MPAG when RV lifted, the temperature of bottom can only increase by 4 degC instead of 55 degC by above method!!!
This observation can courage me to do the calculation with an enthalpy of ONLY a little bit higher than NORMAL value for BOTTOM at relieving condition.

Can you agree on this?

As you mention, you personally do not use unbalanced heat load method for stripper. is it because this method is not applicable for stripper or even refining columns with stripping steam instead of reboiler?

There is a paper presented to API mid-year meeting (1978) by M. Sengupta et al where this unbalanced method was used to calculate relief load for CDU. but this is quite old reference.

Thanks.

Edited by invssse, 16 November 2012 - 05:18 AM.

[CMA010]

Hi CMA010,

Thanks for your further comments.

I think the formula's are correct, the point i was trying to make was that i don't understand the input. Are you sure the enthalpies of the feeds don't change much? The difference in enthalpy (kg/kcal) might not be big but your flow rates are massive.

Stricktly speaking you bottom temperature can't be higher than the temperature of the feeds while in your calculation it is at least 20 °C higher. This might be the reason that your bottoms product withdraws most of the heat.

Bobby: Personally i have never used the unbalanced heat method for strippers, but it is a widely accepted method within engineering and operating companies.

For feeds, we normally flash them at relieving pressure adiabatically. By this way the enthalpy will change not so much.

Yes, by normal method, the bottom product at bubble point at normal pressure will be flashed to bubble point at relieving pressure. this makes its temperature is very high and I think this is the problem as you said.

I am thinking that since this is a steam stripping column not a reboiled column, the bottom product should NOT reach bubble point when the pressure increases to relieving pressure.
We just conducted a dynamic simulation of this tower and it's found that when column pressure increased from 1.07 MPAG to 1.33 MPAG when RV lifted, the temperature of bottom can only increase by 4 degC instead of 55 degC by above method!!!
This observation can courage me to do the calculation with an enthalpy of ONLY a little bit higher than NORMAL value for BOTTOM at relieving condition.

Can you agree on this?

As you mention, you personally do not use unbalanced heat load method for stripper. is it because this method is not applicable for stripper or even refining columns with stripping steam instead of reboiler?

There is a paper presented to API mid-year meeting (1978) by M. Sengupta et al where this unbalanced method was used to calculate relief load for CDU. but this is quite old reference.

Thanks.

If the enthalpy of the feeds is taken as the enthalpy upstream of the control valves it is ok.

I agree with you that the temperature of the bottoms product can not increase much, as there is no heat input apart from the feeds. What did your calculation yield with a marginally increased bottoms temperature?

I would not use this method for steam strippers as the unbalanced heat method assumes unlimited quantities of top tray liquid available which are vapourised by the excess heat. Personally I feel that this is not appropriate for steam strippers as vapourisation in the tower is quite limited.

[invssse]

Hi CMA010,

Thanks for your further comments.

I think the formula's are correct, the point i was trying to make was that i don't understand the input. Are you sure the enthalpies of the feeds don't change much? The difference in enthalpy (kg/kcal) might not be big but your flow rates are massive.

Stricktly speaking you bottom temperature can't be higher than the temperature of the feeds while in your calculation it is at least 20 °C higher. This might be the reason that your bottoms product withdraws most of the heat.

Bobby: Personally i have never used the unbalanced heat method for strippers, but it is a widely accepted method within engineering and operating companies.

For feeds, we normally flash them at relieving pressure adiabatically. By this way the enthalpy will change not so much.

Yes, by normal method, the bottom product at bubble point at normal pressure will be flashed to bubble point at relieving pressure. this makes its temperature is very high and I think this is the problem as you said.

I am thinking that since this is a steam stripping column not a reboiled column, the bottom product should NOT reach bubble point when the pressure increases to relieving pressure.
We just conducted a dynamic simulation of this tower and it's found that when column pressure increased from 1.07 MPAG to 1.33 MPAG when RV lifted, the temperature of bottom can only increase by 4 degC instead of 55 degC by above method!!!
This observation can courage me to do the calculation with an enthalpy of ONLY a little bit higher than NORMAL value for BOTTOM at relieving condition.

Can you agree on this?

As you mention, you personally do not use unbalanced heat load method for stripper. is it because this method is not applicable for stripper or even refining columns with stripping steam instead of reboiler?

There is a paper presented to API mid-year meeting (1978) by M. Sengupta et al where this unbalanced method was used to calculate relief load for CDU. but this is quite old reference.

Thanks.

If the enthalpy of the feeds is taken as the enthalpy upstream of the control valves it is ok.

I agree with you that the temperature of the bottoms product can not increase much, as there is no heat input apart from the feeds. What did your calculation yield with a marginally increased bottoms temperature?

I would not use this method for steam strippers as the unbalanced heat method assumes unlimited quantities of top tray liquid available which are vapourised by the excess heat. Personally I feel that this is not appropriate for steam strippers as vapourisation in the tower is quite limited.

I have no idea how the temperature of bottom will be increased even though it's confirmed this is not too much by rigorous dynamic simulation. Can you or anybody else shed some lights on this?

As you mentioned, you think unbalanced heat load method is not appropriate for steam stripper tower, then what method would you apply for this kind of tower to determine the relief load? thanks.

[Dacs]

I will not comment much on the train of discussion above but this is something I can answer:

I have no idea how the temperature of bottom will be increased even though it's confirmed this is not too much by rigorous dynamic simulation. Can you or anybody else shed some lights on this?

I expect to have an increase in temperature as you increase your column pressure pretty much because of the assumption of saturated liquid in your bottom sump.

[invssse]

Thank you all for your contributions to my question. The train is quite long and discussion becomes diverging.
I would just like that you can give me some clear answers or confirmation for the following questions.

(1) Can unbalanced heat load method be used for a column using stripping steam instead of reboiler?
(2) if NOT, can you tell me which method is the industry standard?
(3) If YES, can you tell me if my calculation at #11 is correct? and if bottom product will be saturated liquid at relieving pressure?

I understand normally licensor's relief load data are reliable but I would like to know how this 123000kg/hr comes.

If you like, I can share with you my simulation data of this column based on PROII.

Thank you!

Edited by invssse, 18 November 2012 - 10:01 PM.

[Dacs]

For #1, yes I think it still applies. That method does not preclude the use of other heat source such as stripping steam.

For #2, my opinion is that you have to use your engineering judgment in creating your scenario. As shown by this thread, we differ in our approach on how to obtain the relief load. That said however, as long as your basis are reasonable, I think you can back up your calculations.

For #3, I think its safe to assume that your column is still operating and distillation still takes place inside your column even during relief, so the expectation of having a saturated bottoms is not really far fetched.

Although what I think might happen during TPF is as you lose the ability to send reflux to your column, you'd lose the ability to cool down the vapor (from your feed) that goes up along with your steam from your bottoms, so it's possible that you may get a larger vapor flow than what you'd expect from a system with reflux in place.

Just the same, I'd still estimate that flow at vapor flows from both feeds plus the steam injected in your column.

For what it's worth, I feel that the licensor data is quite big from what I expect.

[invssse]

For #1, yes I think it still applies. That method does not preclude the use of other heat source such as stripping steam.

For #2, my opinion is that you have to use your engineering judgment in creating your scenario. As shown by this thread, we differ in our approach on how to obtain the relief load. That said however, as long as your basis are reasonable, I think you can back up your calculations.

For #3, I think its safe to assume that your column is still operating and distillation still takes place inside your column even during relief, so the expectation of having a saturated bottoms is not really far fetched.

Although what I think might happen during TPF is as you lose the ability to send reflux to your column, you'd lose the ability to cool down the vapor (from your feed) that goes up along with your steam from your bottoms, so it's possible that you may get a larger vapor flow than what you'd expect from a system with reflux in place.

Just the same, I'd still estimate that flow at vapor flows from both feeds plus the steam injected in your column.

For what it's worth, I feel that the licensor data is quite big from what I expect.

Dacs,
Thank you so much for all these helpful clarifications!

[CMA010]

As you mentioned, you think unbalanced heat load method is not appropriate for steam stripper tower, then what method would you apply for this kind of tower to determine the relief load? thanks.

Your system is more complicated and has a bit more going on (two flashing feeds at different locations) so i would not exclude the manual heat balance method.

I think it's going to be very difficult to reproduce the liconsor's data without knowing what method they used and what their input was (flow rates, upstream pressures, mass balance case, etc., etc.). Might be easier to ask them for a detailed calculation.

I don't see how you can have a bottoms temperature significantly higher than the temperature of the feed(s), there is no heat input to the system besides the feeds. So the assumption of a saturated liquid at relief pressure is only possible if the composition changes (more lights).