Chemical Engineering Software
MixProps:  Debutanizer Example

Feedstock changes are modifying the operating conditions at a debutanizer.  The distillate (overhead) composition is predicted to be:

Component Flow (lbmol/hr)
Compositions (mol frac)
i-C4 12 0.0256
n-C4 442 0.9444
i-C5 13 0.0278
n-C5 1 0.0022
Totals 468 1.0000

(NoteEngVert can easily convert mol/mass percentages and fractions of streams.)

Predict what the maximum overhead reflux drum pressure and condenser duty (cooling tower water supplied) will be.

The reflux drum pressure is determined by a bubble point pressure calculation.  Cooling water can reasonably be heated to a maximum of 120 F and not cause condenser scaling problems.  Fixing the reflux drum  temperature to 120 F will ensure that the cooling water will never reach this maximum and give the maximum drum pressure to be expected.

NOTE:  To ease data entry, this flash calculation file is stored in the directory where MixProps is installed...

Step 1:  On the Mixprops "Settings" tab, perform a bubble-point pressure calculation with a flux drum temperature of T = 120 F.  Select the PR-78 EOS model.

Step 2:  On the Mixprops "Composition" tab, enter the above distillate composition.

Step 3:  Click on the MixProps "Properties" tab.  Mixprops performs the BUBP calculation (note the liquid phase fraction = 1.0000) and computes all the thermodynamic and transport properties of the system.  Note that compositions and properties are computed for a vapor phase as well.  These compositions and properties are of the first "bubble" of vapor that would form if the reflux drum pressure was lowered or drum temperature raised an infinitesimal amount.

Step 4:  Click on the MixProps "Summary" tab.  Read off the bubble pressure = reflux drum pressure = 69.1 psia.
Note that the liquid enthalpy = -62,823.5 BTU/lbmol.

Step 5:  Click on the MixProps "Settings" tab and perform a dew-point temperature calculation at 69.1 psia.   Note that the vapor enthalpy = -53,814.6 BTU/lbmol.

The condenser duty needed will be:  Q = [-62,823.5 - (-53,814.6)] BTU/lbmol * 468 lbmol/hr = -4.2 MMBTU/hr (heat flow out)

The cooling water supply needed will be:  m = Q/(Cp*dT) = (4.2 MMBtu/hr) / [ (1 Btu/lbmF)(120-85) ] = 120,000 lbm/hr => 240 gpm
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Last Modified:  Sept 30, 2005