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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 |
(Note: EngVert 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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