264 replies to this topic

[PingPong]

Enthalpy calculation is now correct.

[ravindra@096]

So far we have understood that for calculating the duty of Waste Heat Boiler (WHB) we simply need to find the difference between the outlet enthalpy of reformer and inlet enthalpy of WGSR.Since the amount of steam required for the process is known, should we find the temperature difference which can be obtained by using Q=nC_pdT or should we fix the temperature of the inlet and outlet of WHB and calculate the amount of steam which can be generated ?

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

[PingPong]

All heat between reformer effluent and WGSR inlet is converted into HPS in the WHB.

 

Additional HPS will be generated in the convection section of the furnace.

 

All HPS will be superheated in the convection section of the furnace.

 

There is a lot of excess heat available in an SMR unit and one has to make best use of it.

 

 

You could read this topic: https://www.cheresou...ther-condition/

to get a realistic idea of how a real SMR looks like.

Any questions or remarks regarding that SMR topic shall be posted in that topic, not here as it would be confusing to discuss two different design in one topic.

Edited by PingPong, 08 February 2018 - 09:41 AM.

[MurtazaHakim]

This is with reference to your message #99.

 

The WHB duty is around 573 kW. The WGSR effluent enthalpy is 422.6 kW and we have calculated that around 697 kW of energy is required to get the 61.47 kmol/hr of required saturated steam from saturated water at 100 degrees Celsius at atmospheric pressure.So the WGSR effluent enthalpy is not sufficient to produce the required quantity of steam from BFW.

 

The question is what will be the temperature and phase of BFW at the outlet of WGSR effluent-BFW heat exchanger considering the inlet temperature of BFW at 30 degrees Celsius. The BFW temperature after preheat with WGSR effluent is essential in further calculating the temperature increase obtained in WHB section considering the required quantity of steam.

 

Should we fix the entire quantity of steam required in process at once while calculating or should we keep the dT fixed and calculate how much quantity can be generated for a given dT ? 

[PingPong]

The horizontal WHB is located beneath a horizontal steam drum and works by thermosyphon on the shellside (reformer effluent on the tubeside).

The BFW enters the steam drum after preheat against WGSR effluent or in the convection section of the reformer furnace or both.

 

Normally the SMR design maximises HPS production by also generating HPS in the convection section of the furnace. Excess HPS is then exported to other units in the refinery or petrochemical complex in which the SMR is located.

 

If you do not want (or are not allowed) to export HPS then I suggest you calculate back what BFW preheat temperature you need so that WHB HPS production exactly matches that 61.47 kmol/h.

[ravindra@096]

We have performed some calculations and have obtained the following results. It seems that we can produce more than the required steam by utilizing the available heat.

Attached Files

•  heat balance 01.JPG   33KB   0 downloads
•  heat balance 02.JPG   78.86KB   1 downloads

Edited by ravindra@096, 10 February 2018 - 10:43 AM.

[PingPong]

It's not clear to me what you are trying to calculate.

 

If you only want to generate the amount of HPS required for the SMR, being that 61.47 kmol/h, and no excess HPS export, then the BFW preheat temperature is to be roughly 200 to 220 ^oC to absorb the WHB duty. I leave it to you to determine the exact number.

 

The steam drum will have a continuous blowdown (CBD) of say 5% so as to remove salts that are always present in small quantities in the BFW. Let's call the preheated BFW temperature T₁ and the steam drum temperature T₂ then there will be 61.47/0.95 = .... kmol/h BFW supply at T₁ which is converted into 61.47 kmol/h saturated HPS at T₂ and ....kmol/h CBD at T₂ leaving the steam drum.

T₁ can easily be determined from the enthalpy balance of WHB + steam drum.

[MurtazaHakim]

Our objective is to utilize the entire heat available in an optimum manner. After consulting with our supervisor we have decided that we should have export steam if it can be generated with the available energy.

 

 

If you only want to generate the amount of HPS required for the SMR, being that 61.47 kmol/h, and no excess HPS export, then the BFW preheat temperature is to be roughly 200 to 220 ^oC to absorb the WHB duty.

How is preheat temperature calculated ? The available heat between WGSR effluent and condenser inlet for BFW preheat is not sufficient for fulfilling the latent heat for 61.47 kmol/hr at 1 atm. pressure. What is the pressure corresponding to BFW preheat temperature of 200-220 degrees Celsius. What is the phase of BFW at the outlet of WGSR effluent-BFW exchanger.

 

In message#106,we assumed that the available heat of WGSR effluent can be utilized to first heat the BFW from 30 degrees Celsius to 100 degrees Celsius and then to vaporize the BFW at 1 atm. As stated in the spreadsheet,around 90 kW of the available 402 kW goes into raising the temperature of 61.47 kmol/hr from 30 to 100 degrees Celsius and the rest is utilized for vaporization of the part of BFW at 1 atm. The remaining amount of BFW is further vaporized in the WHB.

 

We are not aware of the pressure conditions at the WGSR preheat section as well as WHB.

What are the properties of HPS (temperature and pressure)which is used in the process ?

 

Can we first design a suitable heat exchanger for preheating the BFW and then proceed onto the WHB section since we have decided to have as much steam generated as possible ?

Edited by MurtazaHakim, 12 February 2018 - 04:09 AM.

[PingPong]

Forget for the moment how the BFW will be preheated. There is plenty of heat available for that in the WGRS effluent and convection section of the furnace.

 

In any case the BFW will not be preheated to only 100 ^oC and the BFW certainly will not be vaporized at 1 atm. BFW preheat will be at high pressure.

 

One step at a time: we were doing the heat balance around the WHB, remember.

 

I have mentioned before what the is typical pressure in the steam drum and WHB at which the HPS is generated. Steam table gives you the temperature associated with that.

Our objective is to utilize the entire heat available in an optimum manner. After consulting with our supervisor we have decided that we should have export steam if it can be generated with the available energy.

If the objective is now to maximise energy recovery, and therefor also maximise HPS production, then we do do not need to limit the BFW preheat temperature to 210 - 220 ^oC avoid excess HPS in the WHB, but can assume a BFW preheat temperature of say 15 degrees lower than the operating temperature of the steam drum.

The enthalpy balance around the WHB then allows you to calculate the HPS production in the WHB. Don't forget to take also the CBD into account.

Edited by PingPong, 12 February 2018 - 05:00 AM.

[MurtazaHakim]

From the steam table,the temperature associated with 45 bara pressure is 257.40544 degrees Celsius (calculated by interpolating the value between 256 and 258 degrees Celsius).The BFW temperature after preheat should then be (242.40544) degrees Celsius (15 degrees lower than operating temperature of the steam drum).

 

The enthalpy of (x/0.95) kmol/hr of steam at  242.40544 degrees Celsius is calculated and the enthalpy of x kmol/hr at 374.15 degrees Celsius (max. temperature in saturated steam table taken from our reference) is calculated. The difference between the calculated enthalpies should be equal to the WHB duty of 572.76319 kW. Solving for x, I get 470.191917 kmol/hr which is far to huge amount of steam produced.

 

Is the method adopted for calculating the amount of HPS generated correct ? Please have a look at the attached thumbnails.

 

Attached Files

•  WHB 02.JPG   47.97KB   0 downloads
•  WHB 01.JPG   44.95KB   0 downloads

[PingPong]

Saturated steam from the WHB is not an ideal gas, so you can't calculate its enthalpy in a simple way.

Moreover there is no steam at the WHB inlet. BFW is a liquid and therefor not an ideal gas at all.

 

You need to use the enthalpy data for steam and water from the steamtable.

 

To produce 1 kg/s saturated HPS @ 257 ^oC & 45 bara requires 1.05 kg/s BFW @ 242 ^oC and produces also 0.05 kg/s CBD water @ 257 ^oC.

 

Read their enthalpies (kJ/kg) from the steamtable and you can simply calculate how much kJ/s (kW) it takes to produce 1 kg/s sat'd HPS. And from that how much kg/s sat'd HPS will therefor be produced by the WHB duty.

[ravindra@096]

The WHB duty is 572.76319 kW. The enthalpy of saturated vapor at 257 degrees Celsius is 2797.85 kJ/kg and 1 kg/s of HPS generation requires 2797.85 kW of energy. So (572.76319/2797.85)*1 kg/s HPS will be generated using WHB duty which is equal to 40.94309 kmol/hr of HPS generation.

 

What we understood is that the BFW is a saturated liquid at WHB inlet with temperature 257 degrees Celsius and pressure 45 bara and the WHB duty is utilized in transforming it into HPS.

[PingPong]

Now you disappoint me.

Read again carefully what I wrote:

To produce 1 kg/s saturated HPS @ 257 ^oC & 45 bara requires 1.05 kg/s BFW @ 242 ^oC and produces also 0.05 kg/s CBD water @ 257 ^oC.

 

Read their enthalpies (kJ/kg) from the steamtable and you can simply calculate how much kJ/s (kW) it takes to produce 1 kg/s sat'd HPS.

WHB plus steam drum form a black box for which you need to calculate the enthalpy difference between products (sat'd HPS and CBD) and feed (BFW).

[MurtazaHakim]

We regret the inconvenience caused to you due to our ignorance. We have calculated the HPS generation according to the following.

 

Enthalpy of BFW at 242 ^oC =1047.2 kJ/kg

Enthalpy of 1.05 kg/s of BFW at 242 ^oC = 1047.2*1.05 kJ/s

Enthalpy of 1.05 kg/s of BFW at 242 ^oC = 1099.56 kW .......................... (1)

 

Enthalpy of HPS at 257 ^oC =2797.85 kJ/kg

Enthalpy of 1 kg/s of HPS at 257 ^oC = 2797.85*1 kJ/s

Enthalpy of 1 kg/s of HPS at 257 ^oC = 2797.85 kW ................................ (2)

 

Enthalpy of CBD at 257 ^oC =1120.1 kJ/kg

Enthalpy of 0.05 kg/s of CBD at 257 ^oC = 1120.1*0.05 kJ/s

Enthalpy of 0.05 kg/s of CBD at 257 ^oC = 56.005 kW ............................ (3)

 

Enthalpy of feed = 1099.56 kW

Enthalpy of products (HPS + CBD) = 2797.85 + 56.005 = 2853.855 kW

 

Enthalpy difference = 2853.855 - 1099.56 = 1754.295 kW

 

1754.295 kW generates 1 kg/s HPS and 0.05 kg/s CBD at 257 ^oC

 

572.76319 kW generates (572.76319/1754.295)*1 = 0.3264919 kg/s HPS = (0.3264919*3600/18.015) = 65.24400 kmol/hr HPS

and (0.3264919*0.05) = 0.01632459 kg/s CBD = (0.01632459*3600/18.015) = 3.26219 kmol/hr CBD

 

The values for enthalpy are saturated liquid value for BFW and CBD and saturated vapour value for HPS at their corresponding temperatures.

 

EDIT : The molecular weight of water has been taken with accuracy of three digits and the typing error corrected.

Edited by MurtazaHakim, 15 February 2018 - 05:01 AM.

[PingPong]

Calculated HPS flow is correct.

Edited by PingPong, 14 February 2018 - 11:15 AM.

[MurtazaHakim]

The generated steam is slightly more than the required amount in the process. What should be done after calculation of the steam generated in the WHB section. How to move further in the heat balance ?

[PingPong]

All the HPS is to be superheated to say 385 ^oC (in the convection section of the furnace).

[ravindra@096]

The convection section has flue gas enthalpy of 1060.818 kW at 1000 degrees Celsius from burning tail gas and 389.1797 kW at 1000 degrees Celsius from burning additional fuel gas required for fulfilling radiant coil duty. The total flue gas enthalpy is thus 1449.9977 kW.

The enthalpy of HPS at WHB outlet is 913.47549 kW (excluding the outlet enthalpy of CBD since CBD is taken out from steam drum).

 

The amount of heat available for superheating the 65.244 kmol/hr of HPS then depends on the the outlet temperature and enthalpy of flue gas at stack. We need to determine the flue gas enthalpy at the outlet of convection section so that the difference between the inlet and outlet enthalpies of flue gas provides the available heat for superheating the HPS. What should be the temperature of flue gas at convection section outlet ?

 

Is it sufficient to superheat the HPS to only 385 degrees Celsius since XOT is 550 degrees Celsius ?  

Edited by ravindra@096, 15 February 2018 - 07:05 AM.

[PingPong]

Forget the fluegas for now. Available heat in fluegas is much more than needed for preheat of feeds and superheat of HPS.

 

One step at a time: the HPS is to be superheated to 385 ^oC. Use steam table.

[ravindra@096]

The enthalpy of superheated steam at 385 degrees Celsius and 45 bara pressure is 3169.15 kJ/kg and for 65.244 kmol/hr HPS we get

(3169.15 kJ/kg)*(18.015 kg/kmol)*(65.244 kmol/hr)*(1hr/3600 sec) = 1034.701646 kW .

 

The HPS at the WHB outlet carries 913.4754937 kW, so we need (1034.701646-913.4754937) = 121.2261523 kW additionally to superheat the 65.244 kmol/hr of HPS to 385 degrees Celsius at 45 bara.

 

The source of enthalpy information is NIST steam table.

Edited by ravindra@096, 16 February 2018 - 08:06 AM.

[PingPong]

HPS superheat duty of 121 kW is correct.

 

Most of that superheated HPS will be mixed with the natural gas feed, the excess superheated HPS will be exported.

 

Before it can be mixed with the gas feed the enthalpy of the HPS at 385 ^oC and 45 bara must be recalculated to the same reference as was used for all other gas streams: ideal gas at 25 ^oC, in this case: steam at 25 ^oC and nearly zero pressure. Use steam table for that recalculation.

 

Now you have to decide what the natural gas supply temperature and pressure are at the B.L. of the unit. To avoid a compressor the natural gas supply at B.L. should be not less than 35 bara (15 bar above reformer outlet).

 

The recycle hydrogen is increased in pressure by a small compressor and as a result its temperature will be roughly 100 ^oC before it is mixed with the natural gas.

 

The mixture of natural gas plus recycle hydrogen is preheated to (say) 370 ^oC before entering the HDS reactor.

The ZnO bed effluent is mixed with the required quantity of superheated HPS and further preheated to the previously used reformer inlet temperature.

[ravindra@096]

Enthalpy of steam at 25 ^oC (sat.vapor,3.166 kPa) = 2547.3 kJ/kg

Enthalpy of 1 kmol of steam at 25 ^oC = 2547.3*18.015 = 45889.6095 kJ/kmol

Enthalpy of 65.244 kmol/hr of steam at 25 ^oC = 2994021.682218 kJ/hr = 831.672689505 kW .......................... (1)

 

Enthalpy of HPS at 385 ^oC (45 bara) = 3169.15 kJ/kg

Enthalpy of 1 kmol of HPS at 385 ^oC (45 bara) = 3169.15*18.015 = 57092.23725 kJ/kmol

Enthalpy of 65.244 kmol/hr of HPS at 385 ^oC (45 bara) = 3724925.927139 kJ/hr = 1034.7016464275 kW .......................... (2)

 

Enthalpy difference = 1034.7016464275 - 831.672689505 = 203.0289569225 kW

 

We are yet to confirm the B.L conditions.  

Edited by ravindra@096, 18 February 2018 - 08:47 AM.

[ravindra@096]

There are three values indicated under the enthalpy h in the steam table i.e. sat.Liq., evap , sat. vapor. Which of these values are to be used and how do we decide which to use ?

 

Taking the sat.vapor value of enthalpy for steam at 25 degrees Celsius gives the enthalpy difference of 203.0289569225 kW.

 

The previous calculations are edited 

[PingPong]

There are three values indicated under the enthalpy h in the steam table i.e. sat.Liq., evap , sat. vapor. Which of these values are to be used and how do we decide which to use ?

You need the enthalpy as if it were an ideal gas at 25 ^oC.

The most accurate number you can get for that from a steam table is the enthalpy of saturated steam at 25 ^oC, given in the same column from which before you read the enthalpy of HPS at 385 ^oC & 45 bara.

 

So the specific enthalpy of 385^oC/45bara HPS relative to IG@25C is: 3169 - 2547 = 622 kJ/kg _≈11205 kJ/kmol.

[ravindra@096]

So the edited value of 203.0289 kW for the steam in the process is correct then..... is it ?

How do we move further once the BL conditions are known ?

Why do we find the enthalpy of steam at 25°C and ideal gas state ?