6 replies to this topic

[Aznadif]

Dear all, this is my first post; I found this forum very interesting.

 

My request is: In a situation as per drawing 1 , I have to insert and calculate the relief valves.

 

The n° 1 and 2 are for a scenario of external fire, the n° 3 is for thermal expansion.

 

I thought for a normal thermal expansion valve, but my boss, for this last valve, told me He was not sure a normal TRV is sufficient so He asked me to calculate the absorbed heat.

 

In a following post, I will send my reasoning, but in the meantime could someone say as you face front with this problem.

Attached Files

•  Drawing 1.pdf   19.1KB   45 downloads

[shan]

The maximum absorbed heat is the heat needed for the temperature of tube side fluid to reach the temperature of shell side fluid.

[Aznadif]

Here's the calculation that I would do.

 

Hypotheses:

 

a) NH₃ gas in the upper exchanger shell reaches the maximum temperature of C.W. (39°C)

I neglect the thermal resistance of the tubes

c) I consider a linear temperature gradient inside to the tubes (from the periphery to the center)

d) being the tube diameter of reduced entity (20 mm), I do not image large convective movements inside.

 

Based on the above assumptions, I can apply the equations of heat conduction:

 

heat flux density (W/m²) = -lambda * (dT/dx)           and

Power absorbed (W) = flux density * exchange surface

 

lambda is the ammonia conductivity (W/[m*K])

S (m²) is obtained from the exchanger drawings

 

From the hypothesis of the linear gradient, I can write:

 

Power absorbed (W) = -lambda * S* (Ti-Te)/r

 

As told above Te is 39 °C

 

For Ti I have two choices. 1) overestimating, I consider the minimum tube side temperature (-33 °C), 2) I consider an average tube side temperature between Tin (-33 °C) and Tout (15 °C).

 

In case 1, the power absorbed (FI) is 760.4 kW and applying the equation (1) from API 521 (6° Ed., January 2014, page 33) I can calculate the volumetric flow rate to evacuate from the PRV: q=0.633 m³/h (W=431 kg/h)

 

Similarly in he case 2, I can calculate FI=480.4 kW, q=0.400 m³/h (W=273 kg/h)

 

What do you think of what is written above?

Are the assumptions acceptable (for me the weaker one is D))?

What kind of calculation would you have done?

 

Thank you very much in advance to those people who will find time to answer me 

Edited by Aznadif, 06 February 2017 - 10:36 AM.

[SPC]

Dear Aznadif,

 

            Per your PFD details:

 

I will advise to carry out detail scenario analysis as follows for your top gas exchanger  No. 3 PSV as below:

 

1) Closed outlet of tube side fluid ( With inlet flow on)

 

2) Tube rupture case

 

3) Fire case for tube side fluid ( See 10/13 Rule as per API 521)

 

4) Thermal Relief Case ( Liquid filled in entire tube & expansion of total inventory due to Shell side fluid).

 

For this , You can calculate as below :

 

GPM = BH /500 GC

 

Where B= Cubical expansion coefficient per degree F for the liquid at the expected temperature

                   (For your case you can find out B  @Cooling water max. Temperture )

 

G = Specific gravity of liquid

 

C = Specific Heat of trapped fluid in BTU / lb / Degree F( for your case total tube inventory - Consider 10% extra to account for Liq. NH3 piping Volume

 

H= total Heat transfer rate, BTU / hr ( for your case , consider for NH3 upper heat exchanger)

 

Try to calculate Relieving Rate kg/hr for each scenario. Find governing scenario ( giving highest possible relief rate kg/ hr).

 

Remember, if someone tells you to calculate absorbed Heat, meaning is there is expectation to find out  relieving rate under fire scenario. (Refer / Ludwig , Volume1 , 3rd Edition Page No. form 445 to 455)

 

 

Hope this helps you.

 

Regards,

 

SPC

[Aznadif]

SPC, thanks a lot for your reply.

 During the analysis, I already considered the various scenarios.

 

1) closed outlet of tube fluid: it is not a problem because the P(design) of tubes (and also for the shell) is higher than P for pumps shut off.

2) tube rupture: is a little problem because the exchanger is NH3-NH3, the P design is higher than P for pumps shut off and we install a second NH3 heating group as spare (while we repair the exchanger the plant can run with the spare group).

3) Fire scenario: I considered only the case for shell side, the tube external to the exchanger has a little quantity of NH3 but I will rethink about this scenario (thank you).

4) Thermal relief case: In the calculation I reported in my previous post, I used the similar formula for SI. the formula is:

  alfa(v) * FI / (1000*G*C)

 

where alfa(v) is the cubical expansion coefficient per degree C

 

G and C the same you write above (C dimensions are J/[kg*K]) and

 

FI is the total heat transfer rate (espressed in Watt); this one is what I called Power absorbed (maybe not a technical precise name) and it is the quantity my boss want I calculate in a explicit way.

 

My request was if the way I calculate FI in my previous post is correct (almost) or if I am wrong.

 

Again Thank you for this profitable talk.

 

Regards,

Mauro   

[SPC]

Hi Mauro,

 

       Referring your last post statement :" I thought for a normal thermal expansion valve, but my boss, for this last valve, told me He was not sure a normal TRV is sufficient so He asked me to calculate the absorbed heat."

 

I thinks your client is expecting you to carry out fire case ( Heat absorbed).

 

Now, what you are talking is calculation of Heat exchanger duty for Thermal Case 

: Q , Kcal /hr = Total mass of NH3 entering in exchanger (kg/hr)   X  Specific Heat at Avg. Temp. of NH3 X Delta T (For NH3 Side)

 

Consider 20% Design Margin for Heat duty = Q final = Q X 1.2 = ( Kcal /hr X 1000 X 4184 ) /  3600 =  Joules /sec or Watt

 

 

Thanks.

SPC

[Aznadif]

Thanks a lot