264 replies to this topic

[PingPong]

Standard reaction enthalpy calculation for multiple simultaneous reactions is the same as for a single reaction:

 

ΔH_R^o = Σ_i (N_i - n_i) * ΔH_f^o_i

 

in which N_i refers to the component quantities in the reaction(s) product

and n_i refers to the component quantities in the feed.

[MurtazaHakim]

The reaction enthalpy at 298.5 K, according to the above mentioned formula has been calculated to be 855.678 KW. Please look into the attached image.

 

Now, how to proceed further in the heat balance ?

Attached Files

•  Reaction Enthalpy @ 298.5 K.JPG   88.08KB   3 downloads

Edited by MurtazaHakim, 30 January 2018 - 09:33 AM.

[PingPong]

Your calculation results are correct.

 

Next step is to calculate how much fuel needs to be fired so as to provide that 1292 kW radiant coil duty.

 

For a cross over temperature (XOT) of 550 ^oC between convection section and radiant coil the flue gas temperature leaving the radiant section will be around 1000 ^oC.

Excess air will usually be in the range of 10 - 15 % (pick your number) when fuel gas is fired.

Assume a temperature for the combustion air. As your furnace will be small I don't think air preheat makes sense, but you may decide differently.

 

Burning fuel gas (in this case tail gas plus natural gas) using air is a chemical reaction so the standard reaction enthalpy can be calculated in the same way as above. Enthalpy of flue gas from burners is enthalpy of fuel gas plus enthalpy of combustion air minus reaction enthalpy. Enthalpy of flue gas at 1000 ^oC you can calculate. Difference between those two shall obviously be that 1292 kW. Again use ideal gas at 25 ^oC as zero point for enthalpy calculations.

 

To determine how much natural gas and combustion air are required to match that 1292 kW should keep you busy for a while as it requires trial and error, or iteration.

 

EDIT: typing error corrected. See minus.

Note that for combustion the reaction enthalpy is negative as it is an exotherm reaction that releases heat.

So minus reaction enthalpy is a positive number.

Edited by PingPong, 02 February 2018 - 05:03 AM.

[MurtazaHakim]

I have understood the following from the above message.

 

Enthalpy of flue gas from burners = enthalpy of fuel gas + enthalpy of combustion air + Std. reaction enthalpy

 

Enthalpy of flue gas from burners - Std. reaction enthalpy = (enthalpy of fuel gas + enthalpy of combustion air) = 1292 KW

 

(1) For computing the enthalpy of flue gas from burners at 1000 degrees Celsius,outlet composition of the flue gas needs to be known.How do we determine the outlet composition of the flue gases ?

(2) Computation of the std. reaction enthalpy too requires the inlet and outlet flow rates of mixture of fuel gas and combustion air respectively. Inlet has (fuel gas + tail gas + combustion air) but what would be the outlet composition ?

Edited by MurtazaHakim, 01 February 2018 - 06:31 AM.

[PingPong]

Enthalpy of flue gas from burners - Std. reaction enthalpy = (enthalpy of fuel gas + enthalpy of combustion air) = 1292 KW

No!

 

Read again what wrote before:

Enthalpy of flue gas from burners is enthalpy of fuel gas plus enthalpy of combustion air minus reaction enthalpy. Enthalpy of flue gas at 1000 ^oC you can calculate. Difference between those two shall obviously be that 1292 kW.

 

Let's do it in multiple steps:

Step 1: do calculation for tail gas fuel only.

Determine how much oxygen is required to completely burn that stream.

Air has 20.95 mol% oxygen and assume rest is nitrogen.

Add also excess air and you have the total fluegas from the tail gas burning.

Calculate standard heat of reaction.

Then calculate enthalpy of flue gas from burners.

And calculate enthalpy of flue gas at 1000 ^oC.

Finally calculate how much heat that releases in the radiant section of the furnace.

That will be less than the required 1292 kW, let's say it is X kW short.

Therefor additional fuel in the form of natural gas is required.

 

Step 2: use the same calculation method for 1 kmol/h of natural gas fuel.

Let's say that that gives Y kW radiant duty.

 

Step 3: now you can easily see how much kmol/h natural gas fuel is required to turn that Y kW into X kW.

 

Step 4: combine all the above to obtain total fuel, total combustion air and total fluegas.

 

 

Concentrate now on Step 1 only.

 

 

EDIT: typing error in my quote corrected. See minus.

Edited by PingPong, 02 February 2018 - 05:07 AM.

[MurtazaHakim]

Kindly have a look at the attached file.The difference between the enthalpy of flue gas at burners and enthalpy of flue gas at 1000 degrees Celsius considering only the tail gas burning is coming out to be approximately 868 kW.

Attached Files

•  Fuel gas requirement.pdf   849.79KB   64 downloads
•  1.JPG   73.85KB   1 downloads
•  2.JPG   57.95KB   0 downloads
•  3.JPG   65.15KB   0 downloads
•  4.JPG   67.01KB   0 downloads

Edited by MurtazaHakim, 03 February 2018 - 07:56 AM.

[PingPong]

Close, but in my opinion not quite right.

 

Fluegas compostion is right, reaction enthalpy is right, but calculated gas enthalpies seems a little off.

 

You now use 30 ^oC for PSA tail gas but that should be 45 ^oC as that is what you used for the condenser/separator upstream the PSA.

 

Note that in message #73 I already questioned your enthalpy formula with the use of τ (tau) in it. The formula does not seem right to me with respect the last term containing D so I suggest you check again. Obviously if wrong it would give a deviation for mixtures with a lot of components in it that have a value for D such as CO₂, O₂ and N₂

 

 

 

EDIT: I now see on the thumbnails that you have corrected the enthalpy formula with respect to the the last term containing D so I expect your gas enthalpy calculations are correct. Small difference with my model will be due to different correlations for C_p in my model compared to the textbook of Smith et al.

 

Smith et al use only three coefficients (either A, B and C, or A, B and D) for component C_p from 298 to 1500 or even 2000 K. That is convenient for manual calculations but not very accurate. And that also has an impact on the calculation of the equilibrium K values for Reforming and WGS as their formulas also use those C_p coefficients.

If you later use a process simulator for this SMR then you will get some different results compared to your manual calculations simply because of this.

Edited by PingPong, 03 February 2018 - 09:25 AM.

[MurtazaHakim]

If we use 45 degrees Celsius for the calculation of enthalpy of fuel gas,can the combustion air temperature still be taken as 30 degrees Celsius ? Will there be no drop in temperature of PSA tail gas due to piping from PSA outlet to reformer furnace burners' inlet ?

 

The formula used is termed Evaluation of the Sensible-Heat Integral in the textbook and has the D term as (D/T₀) multiplied by ((tau-1)/tau) and this is what I have used in my calculations.

 

Taking 45 degrees Celsius as the temperature, the fuel gas enthalpy increases from 1.58 kW to 6.33 kW.

Is the enthalpy of flue gas at 1000 Degrees Celsius correctly calculated ?

Edited by MurtazaHakim, 03 February 2018 - 09:22 AM.

[PingPong]

If we use 45 degrees Celsius for the calculation of enthalpy of fuel gas,can the combustion air temperature still be taken as 30 degrees Celsius ?

The 45 is the result of heat exchange of syngas against cooling water. Anyway it was your choice and it makes sense in the Middle East.

What the average ambient temperature is at your location I don't know.

Will there be no drop in temperature of PSA tail gas due to piping from PSA outlet to reformer furnace burners' inlet ?

Could be, depends on climate if piping were not insulated. But I think that the tail gas piping will be insulated and steam (or electrical) traced to avoid any risk of water condensation in night or winter. All fuel gas piping to furnace burners usually is.

The formula used is termed Evaluation of the Sensible-Heat Integral in the textbook and has the D term as (D/T₀) multiplied by ((tau-1)/tau) and this is what I have used in my calculations.

 

Taking 45 degrees Celsius as the temperature, the fuel gas enthalpy increases from 1.58 kW to 6.33 kW.

Is the enthalpy of flue gas at 1000 Degrees Celsius correctly calculated ?

See my EDIT in my previous post.

Edited by PingPong, 03 February 2018 - 09:42 AM.

[MurtazaHakim]

Is the temperature assumed for combustion air(30°C)correct ? If yes,have we successfully completed step 1 mentioned in your previous post ?

[PingPong]

As I wrote before: I don't know what the ambient air temperature is at your location.

 

If ambient temperature is around 30 ^oC and you do not intend to use air preheat then Step 1 is successfully completed.

[MurtazaHakim]

The natural gas has components other than CH_4. Which combustion reactions are to be considered for burning of C₂₊ since the composition of flue gas depends upon the combustion of constituent gases.

[PingPong]

C2 and heavier will simply burn (react with O₂) completely and produce only CO₂ and H₂O

[MurtazaHakim]

After having exercised the step 1 procedure, we have calculated that 3.623 kmol/hr of natural gas provides 419 kW of heat which when added to 873 kW of heat produced by burning of tail gas, provides the required radiant coil heat of 1292 kW. Kindly have a look into the following thumbnails.

Attached Files

•  01.JPG   69.29KB   0 downloads
•  02.JPG   93.71KB   0 downloads
•  03.JPG   67.19KB   0 downloads
•  04.JPG   66.6KB   0 downloads
•  05.JPG   78.76KB   0 downloads

[PingPong]

Your result is correct.

 

However one small remark: I seem to remember that your wrote in the past that natural gas feed is at 35 ^oC so natural gas fuel should use same temperature. I am not sure what you now used because in the thumbnail you state 318.15 K which is 45 ^oC but the calculated enthalpy of 0.35 kW seems more in line with 35 ^oC. It's of course peanuts on the total enthalpy balance but nevertheless.

[MurtazaHakim]

We have considered the ambient temperature to be around 30 Degrees Celsius.Is it possible to have Natural gas at BL to be at different temperature than the ambient temperature ? Do all the gas (fuel gas + tail gas) need to be at the same temperature at the burners' inlet since we have considered the temp. of combustion air, fuel gas and tail gas to be 30,45 and 45 degrees Celsius respectively.

 

How should we proceed further in the heat balance now ?

[PingPong]

Do all the gas (fuel gas + tail gas) need to be at the same temperature at the burners' inlet since we have considered the temp. of combustion air, fuel gas and tail gas to be 30,45 and 45 degrees Celsius respectively.

No, they can all be different.

Is it possible to have Natural gas at BL to be at different temperature than the ambient temperature ?

Yes, natural gas temperature depends on temperature at supplier and its travel history (distance, above ground, under ground, subsea, et cetera) before it reaches your unit.

 

Any project starts with a Basis Of Design (BOD) which defines what the composition, quantity, temperature and pressure of all feeds are at the battery limit (BL) of the unit. It also defines all requirements and specifications with regards to the products that the unit sends to BL. It also defines ambient conditions (minimum, normal max), design air temperature for aircoolers, cooling water supply and maximum allowed return temperature, conditions of steam systems (T and p) and all other utility systems, and many more things.

How should we proceed further in the heat balance now ?

Next step should be to determine the inlet enthalpy and inlet temperature of the Shift Reactor.

Calculation of inlet enthalpy is straight forward. Calculating the inlet temperature from that requires iteration or trial and error.

 

 

You also need to define (or assume) those kind of BOD things and put them in your final report.

 

For the natural gas feed we need to know both the supply temperature and pressure before we can finish the heat balance. It is up to you to decide, or maybe your supervisor has already, or needs to.

 

The inlet of the SMR requires a certain natural gas inlet pressure (typically 10 - 15 bar higher than the reformer furnace outlet). If the supply pressure is lower than the required inlet pressure then a natural gas compressor is to be included. As a result of that the natural gas temperature also increases due to compressor power input.

 

A sufficient high natural gas supply pressure at BL avoids the compressor.

 

In any case the burner pressure will be much lower than the natural gas supply pressure. As a consequence the pressure of the fuel part of the natural gas will be reduced considerably via a control valve upstream the burners. That results in a temperature drop due to Joule Thomson effect.

 

As the BL conditions of the natural gas are presently not properly defined we can't know what the natural gas fuel temperature at the burners will be.

The impact on the heat balance is peanuts so that is not the problem.

My impression is that your thumnail states that the natural gas fuel temperature is 45 ^oC but the enthalpy of 0.35 kW does not correspond with that as I would expect it to be at least double that value. So there could be a minor error in your calculation method.

Edited by PingPong, 06 February 2018 - 07:05 AM.

[MurtazaHakim]

 

My impression is that your thumnail states that the natural gas fuel temperature is 45 ^oC but the enthalpy of 0.35 kW does not correspond with that as I would expect it to be at least double that value. So there could be a minor error in your calculation method.

 

We have corrected the calculation error and obtained 0.734 kW

 

 

Next step should be to determine the inlet enthalpy and inlet temperature of the Shift Reactor.

Calculation of inlet enthalpy is straight forward. Calculating the inlet temperature from that requires iteration or trial and error.

 

What we understand is that we should first calculate the outlet enthalpy of WGSR at 440 degrees Celsius and then the reaction enthalpy at 25 degrees Celsius (negative due to exothermic nature).The addition of the above two enthalpies should yield the inlet enthalpy which is to be determined by changing the temperature values until we get the value equivalent to the addition of reaction enthalpy and outlet enthalpy. The outlet enthalpy should anyway be greater than the inlet enthalpy since the reaction taking place is exothermic in nature.Please comment whether our understanding is correct or not.

[PingPong]

We have corrected the calculation error and obtained 0.734 kW

Can you post a thumbnail of that?

 

What we understand is that we should first calculate the outlet enthalpy of WGSR at 440 degrees Celsius

Not 440 ^oC (that is the equilibrium temperature for ATE = 20) but use the actual WGSR outlet temperature for enthalpy.

[ravindra@096]

Please go through the following thumbnails. We have attached the thumbnail of fuel gas enthalpy at 45 degrees Celsius.

We have calculated the inlet temperature of WGSR to be around 353 degrees Celsius.

Attached Files

•  WGSR 01.JPG   107.36KB   0 downloads
•  fuel gas @ 45.jpg   81.98KB   0 downloads
•  WGSR 02.JPG   67.42KB   0 downloads
•  WGSR 03.JPG   79.06KB   0 downloads

Edited by ravindra@096, 06 February 2018 - 10:18 AM.

[PingPong]

WGSR inlet at 353 ^oC is correct.

 

With respect to the fuel gas thumbnail: there still is something wrong in the enthalpy calculation for CO₂ and N₂

Moreover I don't understand why you now changed the total amount from 3.623 to 3.589 kmol/h.

 

In all thumbnails you should remove the total number at the bottom of the dH(kJ/kmol) column. It has no meaning as you can't add kJ/kmol for different components.

[MurtazaHakim]

 

With respect to the fuel gas thumbnail: there still is something wrong in the enthalpy calculation for CO₂ and N₂

Moreover I don't understand why you now changed the total amount from 3.623 to 3.589 kmol/h.

Actually we changed the combustion air temperature from 30 to 35 degrees Celsius which in turn changed its enthalpy. We then had to change the natural gas amount (from 3.623 to 3.589) in order to obtain the remaining radiant coil duty.

We again checked the enthalpy of CO₂ and N₂ which comes out to be the same as mentioned in the thumbnail.

 

How to further proceed in the heat balance after determining the inlet temperature of WGSR ?

[ravindra@096]

This is in reference to message #92 regarding the BOD. We had no knowledge of this concept until you enlightened us with it. So far we have only been informed about the natural gas composition. The quantity of natural gas is subject to product (H₂) quantity and fuel gas requirement. As far as the BL conditions are concerned, we do not have any information regarding pressure and temperature at the BL. The assumption of 35 degrees Celsius at BL was arbitrary.

 

What according to you are the suitable parameters for BOD in our case ?

Edited by ravindra@096, 07 February 2018 - 10:12 AM.

[PingPong]

We again checked the enthalpy of CO₂ and N₂ which comes out to be the same as mentioned in the thumbnail.

Then I suggest you check again.

Enthalpy of CO₂ at 318.15 K should be 758. .... kJ/kmol, not your 889. ...

Enthalpy of N₂ at 318.15 K should be 582. .... kJ/kmol, not your 549. ...

 

Before I checked your enthalpy and duty calculations using a process simulator which uses slightly different (better) data than those in Smith et al and therefor I never worried about small differences between your calc and my simulator. It was only until that natural gas fuel enthalpy that was a factor 2 wrong that I looked closer to the dH (kJ/kmol) that you calculated and noticed that most were wrong.

I will not do that again as it takes too much of my time, so I suggest you double check before posting.

 

What according to you are the suitable parameters for BOD in our case ?

I don't know the conditions at the location of your SMR so I suggest you consult your supervisor. Otherwise we'll cross each bridge when we get there.

How to further proceed in the heat balance after determining the inlet temperature of WGSR ?

Calculate now the duty of the Waste Heat Boiler (WHB) between furnace outlet en WGSR inlet and then calculate how much High Pressure Steam (HPS) you can generate with that. For the moment assume that the steam pressure in the WHB is 45 bara (44 barg) and that the Boiler Feed Water (BFW) entering the steam drum is 20 degrees colder than the HPS after preheat against WGSR effluent.

[ravindra@096]

We have corrected the enthalpy now and it comes as you said 758 kJ/kmol for CO₂ and 582 kJ/kmol for N₂. The mistake had taken place due to merging the calculation in excel since the initial 5 components do not have D values whereas the CO₂ and N₂ has D values. Hope this thumbnail is correct. we will try not to repeat such silly mistakes in future.

Attached Files

•  FG enthalpy.JPG   83.05KB   0 downloads