264 replies to this topic

[ravindra@096]

We have drawn the composite curve of hot and cold streams using HINT (a heat integration software developed by University of Valladolid Spain) and it appears from the curve that the hot streams have significant heat to heat up all the cold streams. An assumption made while creating the composite curve was that the flue gas stack temp. (T_t) is 200 °C . Assuming the stack temperature of 200 °C results in around 1263 kW (1500-237 kW) of available heat in the convection section of the reformer furnace.The following heat duties remain constant (for fixed flowrate)

1. Heating NG + Rec.H₂  to 370 °C requires 89 kW

2. Heating mixed feeds to 550 °C requires 194 kW 

   

this results in 1263 - (194+89) = 980 kW remaining heat in the convection section. It appears that we require a lot of saturated steam to absorb the 980 kW of available heat. Increasing the quantity of HPS would decrease the WHB steam outlet temp. and BFW temp. at the outlet of WGSR-BFW preheat exchanger. We are confused on how to determine the total HPS we can generate.

Attached Files

•  Composite curves.PNG   9.57KB   3 downloads
•  Hot and Cold streams.PNG   8.02KB   1 downloads

Edited by ravindra@096, 28 February 2018 - 08:37 AM.

[PingPong]

Assuming the stack temperature of 200 °C results in around 1263 kW (1500-237 kW) of available heat in the convection section of the reformer furnace

I doubt whether that 1500 and therefor also that 1263 is correct.

 

There is no point in including the Reformer Effluent in your analysis as you know already that that duty will be only steam generation. So leave it out as it was already solved weeks ago.

 

this results in 1263 - (194+89) = 980 kW remaining heat in the convection section

Not really as you also need 121 kW for HPS superheat so excess heat is less than 980 kW.

 

You can generate and superheat additional HPS using part of that duty.

Required enthalpy difference BFW --> SHPS from steam table.

Make that a separate stream in your pinch analysis and vary its flow until you are satisfied with the result.

How much could be generated is for you to figure out using the composite curves. Note that the flue gas must always be hotter than any cold stream at any point. Use temperature difference with flue gas of say at least 100 ^oC to start with. You may need to slit up the preheat to 370 ^oC in two parts (say first up to 257 ^oC and then from 257 to 370 ^oC) with steam generation (at 257 ^oC) in between to get best heat recovery.

Edited by PingPong, 28 February 2018 - 11:07 AM.

[ravindra@096]

We have modified the composite curve by including the BFW and removing the reformer effluent hot stream, the modified cold curve expands taking more of heat from hot streams. The additional HPS generation and the enthalpy difference associated with it at varying flow rates is calculated and added with the enthalpy required for heating the other cold streams(NG+Rec. H₂,mixed feed, steam superheating) which gives the total required enthalpy for all the cold streams and that total enthalpy is subtracted from the enthalpy of flue gas at 1000 ^oC to give the enthalpy of flue gas leaving the convection section and iteration is done to give the stack temperature at which the enthalpy matches the enthalpy of flue gas leaving the convection section. As we increase the quantity of additional HPS generated and superheated, the stack temperature decreases reaching 150 ^oC for around 56.7 kmol/hr of additional HPS. The thumbnails have been attached herewith.

 

Attached Files

•  modified composite curve.PNG   10.88KB   3 downloads
•  modified streams.PNG   7.77KB   3 downloads
•  Additional HPS.PNG   23.71KB   2 downloads

[PingPong]

I am not familiar with HINT program so I don't know what kind of input options it has for enthalpy data as function of temperature. If that is only m.Cp then you have a problem.

 

When BFW is converted from liquid water at 30 ^oC into SHPS at 385 ^oC than it does not have a constant m.Cp over that temperature range. From 30 to 257 ^oC it has the Cp of liquid water, then it absorbs a lot of vaporization heat at 257 ^oC, and then it is heated as steam from 257 to 385 having the Cp of steam. The graph of H versus T for BFW-->SHPS is not a straight line (with constant m.Cp) but consists of 3 parts with different slopes, with biggest part horizontal (vaporization at constant T). As a consequence the real cold composite curve will have a very different form than you have now.

 

I still don't see why you have 1263 kW cooling duty if flue gas is cooled from 1000 to 200 ^oC, I would expect a lower duty unless you changed the flue gas quantity. Note also that you now state that you aim at a stack temperature of 150 ^oC, not 200 ^oC.

 

Note also that you must include the blowdown in your heat balance.

Edited by PingPong, 03 March 2018 - 09:57 AM.

[ravindra@096]

The enthalpy of flue gas is calculated at 1000 and 150 degrees Celsius and the difference between those enthalpies is calculated as 1282.517 kW (see attached thumbnail). We  have entered the hot and cold streams and split the BFW into one stream from 30 to 257 degrees Celsius another from 257 degrees Celsius sat. liquid to sat. vapor and yet another stream from 257 to 385 degrees Celsius steam superheat. The image of obtained composite curve and added streams is attached herewith.

There is an option of cascade (in HINT) which indicates the cooling duty required. What is that cooling duty ? Is it the enthalpy of flue gas at stack ? We have considered 57.5 kmol/hr of additional HPS generation, the cooling duty at this flowrate is 404.943 kW. When we plug the value of 404.943 kW into the enthalpy table of the flue gas, the corresponding temperature is 320 degrees Celsius. Does it mean that the real stack temperature is 320 and not 150 degrees Celsius and we need to vary the flow rate of additional HPS until we get the cooling duty value equal to the enthalpy of flue gas at 150 degrees Celsius ?

 

The addition of all the enthalpies of cold streams and then subtraction of that enthalpy from the enthalpy of flue gas at 1000 degrees Celsius yields 170.275 kW (for 57.5 kmol/hr flowrate) which corresponds to the flue gas stack temperature of around 152 degrees Celsius.

Which of these values give the stack temp. (the one from the curve or the one calculated from the above said method of addition and subtraction) of flue gas ?

 

 

Note also that you must include the blowdown in your heat balance.

How do we proceed to blowdown addition in heat balance ?

Attached Files

•  FG@1000 C.PNG   17.18KB   2 downloads
•  FG@150 C.PNG   17.41KB   1 downloads
•  streams list.PNG   9.92KB   0 downloads
•  Comp. curve.PNG   10KB   1 downloads
•  Cascade.PNG   17.42KB   1 downloads

Edited by ravindra@096, 06 March 2018 - 08:58 AM.

[PingPong]

The enthalpy of flue gas is calculated at 1000 and 150 degrees Celsius and the difference between those enthalpies is calculated as 1282.517 kW (see attached thumbnail).

Now that you corrected the H at 1000 ^oC from 1500 to 1450 its is correct for a stack T of 150 ^oC.

There is an option of cascade (in HINT) which indicates the cooling duty required. What is that cooling duty ? Is it the enthalpy of flue gas at stack ?

I am not familiar with HINT but I assume that Cascade refers to Cascade Diagram. You can google that and read more about it: https://www.google.c...cascade diagram

 

Your extra HPS generation of 57.5 kmol/h is rougly correct but will decrease a little due to heat lost with extra BD.

How do we proceed to blowdown addition in heat balance ?

Similar as you did for the WHB: BD increases flow of BFW to be preheated to 257 ^oC and then leaves the system as waste water.

 

You did not yet take into account that the WGSR effluent contains a lot of water vapor (steam) that starts condensing when its dew point is reached (about 153 ^oC at 19 bara) and as temperature decreases further more water condenses until at 45 ^oC & 18 bara most is condensed.

Determine the condensation profile and resulting curve of effluent enthalpy (V+L) as function of temperature.

[PingPong]

A few more points:

 

In the table with hot and cold streams you must also include the BFW flow for the WHB preheated up to 242 ^oC that was already calculated weeks ago. Now you only included the extra BFW for the extra HPS for export.

 

In that Cascade window you seem to have entered a Minimum Temperature Difference of only 10 ^oC, which is not correct. To get a result comparable with your graph you must use the same Minimum Temperature Difference value as used for the graph, somewhere between 100 and 130 ^oC is my guess.

Edited by PingPong, 07 March 2018 - 05:59 AM.

[MurtazaHakim]

Meeting after a long time..... Sir, we have revised the calculation of export HPS generation considering the heat lost due to closed blowdown. We are attaching a thumbnail of it, kindly have a look into it. The export HPS is slightly less than 57.5 kmol/hr as you said. The enthalpy difference between flue gas at 1000 degrees Celsius and 150 degrees Celsius is as reported in previous message (1282.51734 kW). Subtracting the duties of process HPS superheat,NG+H2 and mixed feed preheat from 1282.51734, we get 877.69274 kW heat available for export HPS BFW preheat,steam generation and superheat.

 

The final list of hot and cold streams is attached in the thumbnail. Kindly verify the list.

 

How do we generate the condensation profile downstrean the WGSR ?

Attached Files

•  Finallist.PNG   19.01KB   0 downloads
•  ExportHPS.PNG   14.13KB   0 downloads

[PingPong]

I can't really check your numbers as there is not enough explanation about them.

 

In any case you also need to make a flow scheme of the heat exchange between all your hot and cold streams: which streams are exchanged against which and in which order. That is also a check of the feasibility of what you have in mind.

How do we generate the condensation profile downstream the WGSR ?

First determine water dew point at say 19 bara. Then take several temperature points between T_wdpt & 19 bara and condensor outlet at 45 ^oC &18 bara, and at each T&p point determine how much water can be in vapor and consequently how much condensed liquid water is present, and determine enthalpy of V and L of each.

Edited by PingPong, 18 March 2018 - 08:07 AM.

[MurtazaHakim]

I can't really check your numbers as there is not enough explanation about them.

Enthalpy of BFW at 30 ^°C = 125.7 kJ/kg
Enthalpy of 59.63997 kmol/hr of BFW at 30 ^°C = (125.7*18.015*59.63997/3600) kJ/s
Enthalpy of 59.63997 of BFW at 30 ^°C = 37.5149 kW.......................... (1)

Enthalpy of BFW at 257 ^°C = 1120.1 kJ/kg
Enthalpy of 59.63997 kmol/hr of HPS at 257 ^°C = (1120.1*18.015*59.63997/3600) kJ/s
Enthalpy of 59.63997 kmol/hr of HPS at 257 ^°C = 334.2919 kW................................ (2)

Enthalpy of CBD at 257 ^°C = 1120.1 kJ/kg
Enthalpy of (59.63997*0.05) kmol/hr of CBD at 257 ^°C = (1120.1*18.015*59.63997*0.05/3600) kJ/s
Enthalpy of (59.63997*0.05) kmol/hr of CBD at 257 ^°C = 16.7145 kW............................ (3)

Heat of vaporisation at 257 ^°C = 1677.75 kJ/kg
Heat required vaporising (59.63997*0.95) kmol/hr of HPS at 257 ^°C = (1677.75*18.015*59.63997*0.95/3600) kJ/s
Heat required vaporising (59.63997*0.95) kmol/hr of HPS at 257 ^°C = 475.6856 kW.... (4)

Enthalpy of sat. Vapour HPS at 257 ^°C =2797.85 kJ/kg
Enthalpy of (59.63997*0.95) kmol/hr of sat. Vapour HPS at 257 °C = (2797.85*18.015*59.63997*0.95/3600) kJ/s
Enthalpy of (59.63997*0.95) kmol/hr of sat. Vapour HPS at 257 ^°C = 793.26302 kW... (5)

Enthalpy of superheated HPS at 385 ^°C = 3169.15 kJ/kg
Enthalpy of (59.63997*0.95) kmol/hr superheated HPS at 385 ^°C = (3169.15*18.015*59.63997*0.95/3600) kJ/s
Enthalpy of (59.63997*0.95) kmol/hr superheated HPS at 385 ^°C = 898.5362 kW... (6)

Total heat required for (59.63997*0.95) kmol/hr of export HPS generation = heat required to raise 59.63997 kmol/hr of BFW from 30 °C to 257 °C + heat required to vaporise (59.63997*0.95) kmol/hr of sat. Liquid at 257 °C + heat required to superheat (59.63997*0.95) kmol/hr of steam from 257 °C to 385 °C

Total heat required for (59.63997*0.95) kmol/hr of export HPS generation = (2) – (1) + (4) + (6) – (5)

Total heat required for (59.63997*0.95) kmol/hr of export HPS generation =
(334.2919 – 37.5149) + 475.6856 + (898.5362-793.26302)

Total heat required for (59.63997*0.95) kmol/hr of export HPS generation = 877.73578 kW… (7)

Heat required for NG + H₂ preheat from 42 °C to 370 °C = 89 kW............................ (8)

Heat required for mixed feed preheat from 365 °C to 550 °C = 194.5634 kW................. (9)

Heat required for process HPS superheat from 257 °C to 385 °C = 121.23 kW........... (10)

Adding (7) + (8) + (9) + (10) = 1282.52918 kW = Enthalpy difference of flue gas at 1000 °C and 150 °C

The above figure of 59.63997 kmol/hr has been calculated by iterative procedure using solver in MS Excel keeping available heat for Export HPS generation constant at 877.73578 kW.

Edited by MurtazaHakim, 22 March 2018 - 08:01 AM.

[PingPong]

It has been about two weeks since I was into your numbers.

Now I have difficulty to understand or remember the origin of some of them.

 

For example: how did you calculate that 59.63997 kmol/h steam number?

[MurtazaHakim]

Actually it was calculated that the available heat in the convection section (877.73578 kW) could produce 57.5 kmol/hr of export HPS,but later in message # 156 you suggested including closed blowdown which would result in a little heat loss and the export HPS quantity would then decrease a little. We then considered 5% CBD and proceeded further to calculate and that is how we came up with 59.63997 kmol/hr by iteration.

 

95% of 59.63997 is 56.65797 kmol/hr which is the actual quantity of export HPS generated. The calculations in the message #160 explain that 59.63997 kmol/hr is

(1) heated from 30 to 257 degrees Celsius,

     5% of it is taken as CBD in liquid phase at 257 degrees Celsius and

(2) the rest 56.65797 kmol/hr is vaporised and

(3) superheated from 257 to 385 degrees Celsius

 

The heat requirement of the stated tasks (1),(2),(3) is calculated from the steam table and it comes equal to the amount of heat available in the convection section for the stated tasks. The procedure is iterative. By changing the amount of BFW the heat requirements of the stated tasks are calculated until we arrive at a number having heat requirement equal to the heat available in the convection section for the tasks.

The thumbnail in message #158 shows the same thing. The enthalpy calculation in that thumbnail is done according to what is wriiten in message#160. I hope you will understand the procedure used to arrive at that number.

Edited by MurtazaHakim, 20 March 2018 - 12:03 PM.

[PingPong]

So that 59.63997 kmol/h was not steam but BFW, and steam is 95 % of that which is 56.65797 kmol/h.

You seem to have revised post #160 to include that factor 0.95 for steam, but not everywhere. For example: it is still missing in calculation (2).

 

Anyway, 56,7 kmol/h additional HPS generation seems realistic.

Now you need to think about how that is done. Make a flow scheme of the heat exchange between all your hot and cold streams: which streams are exchanged against which and in which order.

[MurtazaHakim]

it is still missing in calculation (2).

Calculation (2) does not include the 0.95 factor because the entire BFW is to be heated to 257 degrees Celsius and not only 95% of that,after which the CBD is drained out at 257 degrees Celsius and the rest is vaporised and superheated consequently. The total heat requirement should therefor include the calculation (2) with the entire BFW and the following calculations with 95% BFW. Is my understanding correct in this regard ?

 

The flow scheme of heat exchange should be such that the WGSR effluent and reformer effluent are to be used to preheat the BFW (meant for generation of process HPS) and vaporisation of process HPS respectively. The heat available due to flue gas is used in the order as follows

1. Preheat NG + recycle H₂ mixture  

2. Superheat process HPS

3. Heat mixed feed 

4. Generate export HPS

What I do not reckon is, how can the target temperature of a cold stream be restricted ? I mean it is possible that one cold stream exceeds the target temperature whereas the temperature of the other remains below target.

Edited by MurtazaHakim, 22 March 2018 - 05:39 AM.

[PingPong]

Calculation (2) does not include the 0.95 factor because the entire BFW is to be heated to 257 degrees Celsius and not only 95% of that

It says HPS in calculation (2), but that should be BFW.

What I do not reckon is, how can the target temperature of a cold stream be restricted ? I mean it is possible that one cold stream exceeds the target temperature whereas the temperature of the other remains below target.

Yes, in the real world everything is possible although the Pinch fundamentalists probably don't like that.

[MurtazaHakim]

We have corrected the typing error in calculation (2). Sir, you have mentioned that for generating the condensation profile of steam downstream the WGSR we need to take several temperature points between steam dew point T_wdpt at 19 bara and 45 ^oC &18 bara and find how much steam has condensed and how much vapor has remained ?  In message #156 you mentioned the dew point at 19 bara is around 153 ^oC. How do we find the dew point at a given pressure ? The steam table indicates the value of 210 ^oC at around 1907.7 kPa. This means that at around 19.07 bara and 210 ^oC the saturated liquid begins to vaporise but how to get a temperature at which a saturated steam begins to condense (reach its dew point) at a given pressure ? Secondly how to determine how much steam has condensed and how much is left as vapor ? 

 

After generating the condensation profile and properly generating the composite curve how do we proceed further ?

 

Kindly give your opinion on the heat exchange flow scheme stated in the previous message having order and combination of hot/cold streams to be exchanged. 

[PingPong]

The steam table indicates the value of 210 ^oC at around 1907.7 kPa. This means that at around 19.07 bara and 210 ^oC the saturated liquid begins to vaporise but how to get a temperature at which a saturated steam begins to condense (reach its dew point) at a given pressure ?

No, that would only be so if the WGSR effluent were pure steam, which it is not. Try again.

Secondly how to determine how much steam has condensed and how much is left as vapor ?

Same method as you used weeks ago to calculate vapor from condensor to PSA unit, except that you can ignore dissolved gases (Henry) in water for intermediate points as those are so small that they do not affect enthalpy.

Kindly give your opinion on the heat exchange flow scheme stated in the previous message having order and combination of hot/cold streams to be exchanged.

That is not a real flow scheme, moreover not correct and not complete.

 

I meant a Process Flow Diagram (PFD) or a Grid Diagram, with temperatures and duties indicated.

Just by hand (pencil on paper), need not be fancy for the moment.

 

Example of a Grid Diagram for a totally different process unit:

 

 

Probably you can do that using an option in HINT.

[ravindra@096]

The dew point calculation has been done by multiplying the total system pressure (19 bara) with the mole fraction of steam in the inlet gas to the condensor. The mole fraction of the steam in the inlet gas to the condensor is calculated as (molar flowrate of steam/Total molar flowrate) in kmol/hr.

 

Mole fraction of steam to the condensor inlet = 31.30407/115.87225 = 0.270160198

Partial pressure of steam to the condensor inlet = 19*0.270160198 = 5.133043762 bara = 513.3043762 kPa

 

The temperature corresponding to 513.3043762 kPa is 152.82501 degrees Celsius (calculated from steam table) which is the dew point for that flowrate.

 

We have created the table depicting the different temperatures between T_wdpt and 45 degrees Celsius with the amount of steam in vapor phase and consequently the amount of water condensed calculated at the respective temperatures.

The formula for calculating the amount of steam in vapor phase = (Amount of dry gas/mole fraction of dry gas)*(mole fraction of steam) [referred from message#54]. The total system pressure is varied for different temperatures since there exists pressure drop across the condensor. The asterisk relates the corresponding total system pressure and the temperature. 

Kindly look into the thumbnail. If the thumbnail calculations are correct then how do we generate the resulting curve of effluent enthalpy for the (V+L) mixture ?

Attached Files

•  condensation profile.PNG   17.69KB   1 downloads

Edited by ravindra@096, 26 March 2018 - 06:48 AM.

[PingPong]

Dewpoint of 152.8 ^oC is correct.

 

Thumbnail seems OK, but normally one would choose say 10 temperatures such that condensation profile is roughly linear. In your table the first condensation is already more than half the total water present. So you better have additional points between 152.8 and 130 ^oC and at least one additional point between 130 and 110 ^oC. And use a gradual pressure profile.

 

Enthalpy of V+L is simply the sum of the enthalpy of the vapor and that of the condensate. Use however same enthalpy basis as for WGSR effluent so adjust liquid water enthalpy to IG@25C basis.

[ravindra@096]

We have calculated the vapour and liquid enthalpy of steam and condensate respectively at different temperatures between the dew point (152.8 C) and 45 C, the obtained values are depicted in the thumbnail attached herewith. Kindly verify the values calculated in the thumbnail and suggest the  necessary corrections required.

Attached Files

•  (V+L) enthalpy.PNG   170.25KB   3 downloads

[PingPong]

It is not clear to me how you calculated those enthalpies.

In any case they cannot be correct.

 

For example: the enthalpy of the vapor at 152.8 ^oC is about 132 kW, not your 37.8 kW.

We have calculated the vapour and liquid enthalpy of steam and condensate respectively at different temperatures between the dew point (152.8 C) and 45 C

Does that mean that you did not take the rest of the vapor into account?

 

You need to calculate the vapor enthalpy in the same way as you did many times before in the past months.

 

The liquid enthalpies cannot be correct either as they should be negative relative to IG@25C.

[ravindra@096]

The calculation of V+L enthalpy has been modified. We have separately calculated the enthalpy of dry gas (excluding steam) at temperatures between 152.8 degrees Celsius and 45 degrees Celsius and added that enthalpy to the enthalpy of steam + enthalpy of condensate at respective temperatures.

 

However, the enthalpy of condensate at temperatures ranging from 152.8 to 45 degrees Celsius is calculated as positive relative to IG at 25 degrees Celsius. We are still a bit skeptical regarding the enthalpy of condensate being negative. Please check our thumbnail.

 

The enthalpy of WGSR effluent at 420 degrees Celsius = 422.6 kW

The enthalpy of WGSR effluent at 153 degrees Celsius = 131.821 kW

Total heat available for exchange before the steam in the mixture reaches dew point = 422.6 - 131.821 = 290.779 kW

 

Total enthalpy of V+L at 152.8 degrees Celsius = 131.6367 kW

Total enthalpy of V+L at 45 degrees Celsius = 20.2232 kW

Total heat available for exchange after the steam in the mixture reaches dew point = 131.6367 - 20.2232 = 111.4135 kW

 

Net heat available = 290.779 + 111.4135 = 402.1925 kW

 

The thumbnails are attached herewith.

 

Attached Files

•  Combined profile.PNG   22.32KB   1 downloads
•  vapor profile.PNG   29.17KB   2 downloads
•  Condensate profile.PNG   29.48KB   1 downloads

[PingPong]

We have separately calculated the enthalpy of dry gas (excluding steam) at temperatures between 152.8 degrees Celsius and 45 degrees Celsius and added that enthalpy to the enthalpy of steam + enthalpy of condensate at respective temperatures.

It is not clear to me what you mean, or what you have done.

 

Nevertheless your vapor enthalpy of 131.6 kW at 152.8 ^oC is about correct.

 

You should not calculate the vapor enthalpy separately for dry gas and steam. Calculate enthalpy for total vapor using the same spreadsheet that you already used many times a month or so ago.

 

The enthalpy of the liquid water condensate at each T you can obtain from the steam balance. However the steam table uses liquid water at triple point (0.01 ^oC) as the zero point for enthalpy, while the vapor from which it is condensed uses IG@25C as the zero point. So you must recalculate the condensate enthalpy to that same basis. It is no different from the recalculation of the HPS enthalpy before mixing it with the natgas+H2 feed that you did a month or so ago.

[ravindra@096]

You should not calculate the vapor enthalpy separately for dry gas and steam. Calculate enthalpy for total vapor using the same spreadsheet that you already used many times a month or so ago.

We calculated the enthalpy for total vapor using the spreadsheet.However, we found no differences in the values obtained previously (separate enthalpy calculations for dry gas and steam) and values obtained now.

 

For calculating the condensate enthalpy, we calculated the specific enthalpy of condensate as the difference between the enthalpy of sat.liquid at temperatures ranging from 152.8 to 45 degrees Celsius and enthalpy of steam as an ideal gas at 25 degrees Celsius.

Example: at 152.8 degrees Celsius,specific enthalpy of condensate = 644.24 - 2547.3 = -1903.06 kJ/kg 

 

Please refer the thumbnail. Is this method correct for evaluation of condensate enthalpy ?

Attached Files

•  Condensate enthalpy.PNG   19.33KB   0 downloads

Edited by ravindra@096, 29 March 2018 - 11:23 AM.

[PingPong]

Example: at 152.8 degrees Celsius,specific enthalpy of condensate = 644.24 - 2547.3 = -1903.06 kJ/kg 

 

Please refer the thumbnail. Is this method correct for evaluation of condensate enthalpy ?

That looks to be correct.