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[abbaschemical]

Dear Experts,
Ammonia in the refrigeration loop undergoes series of cooling in exchangers without interstage stage seperators and finally seperated in a seperator at near -5 C. Here how to calculate the amount or fraction of ammonia condensed when its leaves every exchanger knowing alone the dewpoint and exchanger exit temperatures.
Ammonia in vapor phase exit converter is 155560 Nm3/hr . System pressure of 143 kscg.

[abbaschemical]

Any ideas ?

[abbaschemical]

Thanks Latexman.... but i have asked for syngas....

[gegio1960]

Just to start you need:
- initial gas composition & temperature (other than flowrate & pressure already mentioned)
- pressures at each exchanger outlet (other than temperature already mentioned)

[Atttyub194]

please use commerical software such as hysis or Pro II

[abbaschemical]

To enhance further discussion, i am attaching further details........... ( File attached )


Total Flow (dry gas) inlet of first Exchanger = 725000 Nm3/hr
Inlet gas composition (mole%) : Hydrogen=50.69%, AMMONIA=19.90%, Methane=8.83%, Nitrogen=17.39%, Argon=3.19%
Inlet Pressure in first exchanger is 142 kscg. Assuming Pressure drop in every exchanger is 0.1 Kscg.
How to calculate amount of ammonia condensed in every Heat Exchanger(HE) in the gas ?

Please i not need any calculation idea from software but can someone help me how to calculate manually.

Thank You.

Attached Files

•  HE.png   20.43KB   25 downloads

[abbaschemical]

Dear Kkala, ankur, shivshankar and Art .... expecting your valuable suggestions

[gegio1960]

Dear abbaschemical,
my 2-cents solution with proII-SRK expressed as kmol/h of NH3 in the liquid: 0, 2.93, 3.56, 4.17, 4.72, 5.06
Sorry, I don't perform these calculations by hands from the old good university times, more or less 31 years ago.
But I'm curious to see the results of your hand calculations
Kind regards

[abbaschemical]

Dear Gegio1960,
Thanks for your reply and solution. And i am to working out this manually and definetly i will let you know once i finish it

[abbaschemical]

Dear Kkala, ankur, shivshankar and Art .... expecting your valuable suggestions

[abbaschemical]

waiting for response from everyone....

[abbaschemical]

any ideas ??

[kkala]

1. H2, CH4, N2, Ar have critical temperatures below -80 oC, so no quantity of them shall be liquified at - 8 oC even at 143 kgf/cm2g total pressure. NH3 (crit temp = +132 oC) is liquid at 143 kgf/cm2g and 56 oC; if gases of the mixture were ideal, NH3 partial pressure would be 144*0.199 = 28.7 kgf/cm2a, versus 24.5 kgf/cm2a saturation pressure of pure NH3 at 56 oC (Perry). Difference is attributed to non ideality of gas mixture, all of its components (including NH3) are assumed gaseous at conditions (total P=143 kgf/cm2g, 56 oC). In following notes no ΔP up to separator, no heat loss to ambient, no gases dissolved in liquid NH3 will be considered.
2. Enthalpy of gas mixture entering HE1 (H0) has to be computed, since composition and P of gas mixture is known as well as t=56 oC. Then H1=H0-Q1, where Q1=heat removed from gas mixture in HE1. Q1 could be estimated from the cold stream of HE1.
Note: Hopefully flow rate of cold stream and temperatures in/out are known (even by measurement), though not presently given.
Similarly H2=H1-Q2 (t2=18 oC), H3=H2-Q3 (t3=11 oC), H4=H3-Q4 (t4=3 oC), H5=H4-Q5 (t5=-8 oC) could be estimated; then liquid NH3 of stream 5 will be separated from gas. The question is how to estimate liquid NH3 in a mixture whose enthalpy and temperature is known (P=143 kgf/cm2 g).
3. Let us see few hypothetical simple cases, concerning NH3 condensation in all exchangers (NH3 liquid removed at separator). Temperatures remain same as above, pressure P is different.
α) Inlet to HE1 is pure NH3 (g), no other gases, P=4 kgf/cm2a or higher. All NH3 will be condensed (saturation pressure Ps= 3.2 kgf/cm2a at -8 oC).
β) Inlet gas is NH3 (g) 19.9% and N2 (g) 80.1% (mol/mol), P=10 kgf/cm2a. Assuming no big error for ideal gas, NH3 partial pressure =10*19.9%=1.99 kgf/cm2a <Ps; so no NH3 condensation takes place.
γ) As in 3β, except for P=20 kgf/cm2a. NH3 partial pressure at inlet=20*19.9%=3.98 kgf/cm2a >Ps at -8 oC. In stream 5, part of total NH3 will be liquid and rest gas (of 3.2 kgf/cm2a partial pressure) to occupy the "empty" space along with N2.
Suppose now that enthalpy of whole fluid (stream 5, -8 oC) is known = 340 kJ/kg, based on http://webbook.nist....hemistry/fluid/ '> http://webbook.nist....hemistry/fluid/ (nitrogen) and http://www.engineeri...onia-d_971.html '> http://www.engineeri...onia-d_971.html . Stream 5 contains 0.869 kg N2/kg and 0.131 kg NH3/kg, enthalpies are as follows
N2: 249.46*0.869 = 216.78 kJ/ kg (of stream 5)
NH3: (1435.3x+144.5(1-x))*0.131 kJ/kg (of stream 5), x=NH3 fraction that remains gaseous.
Thus 216.78+(1435.3x+144.5(1-x))*0.131 = 340, so x=0.617; that is 38.3% of total NH3 will be condensed.
Similarly, if enthaly of stream 5 is 290 kJ/kg, x = 0.321, that is 67.9% of total NH3 will be condensed.
In case (γ) pressure, temperature and composition are not adequate to fix extent of NH3 condensation, another parameter like stream 5 enthalpy is needed (depending on Q1, Q2, Q3, Q4, Q5 and H1 - stream 1 does not contain liquid NH3).
4. Even though above indicates a general way, it is difficult to compute enthalpies of gas mixtures (also containing liquid NH3) deviating from ideal gas laws, like in present case. Usually ideal gas enthalpies are calculated for each component (for pressure approaching 0 kgf/cm2 g), then corrections are applied considering each gas at P=143 kgf/cm2 g (Amagat law), as we did as students in a task (1971) (*). But in the present case NH3 would be liquid at P=143 kgf/cm2 g, 56 oC. Any help on this would be welcomed, as well as comments on the whole post.

(*) Look into page 7 of attached "2-Figures and table - Thermodynamics.pdf", from http://www.reviewpe....rmodynamics.pdf '> http://www.reviewpe....rmodynamics.pdf .

Attached Files

•  2 - Figures and table - Thermodynamics.pdf   2.63MB   47 downloads