The condenser and separator

The condenser cools the reactor effluent and the separator splits it into a vapour and a liquid. The vapour leaves through two paths, the compressor recycle and the purge; the liquid goes to the stripper. The compressor and its recycle valve belong here too, because their whole job is deciding how much of the separator's vapour goes back around the loop.

Source: teprob.f:473-502 for the equilibrium, teprob.f:585-601 for the flow network and the compressor, teprob.f:674-676 for the condenser duty, and teprob.f:773-777 for the balances.

Vapour-liquid equilibrium

The separator's equilibrium has exactly the shape of the reactor's and shares its code path. The vapour space is VTS less the liquid volume (teprob.f:474); A, B and C get ideal gas partial pressures (teprob.f:481); D through H get Raoult's law with an Antoine vapour pressure (teprob.f:488-490); the composition is \(y_i = p_i / P\) (teprob.f:495); and the vapour holdup comes back out of the ideal gas law at teprob.f:498 and 501.

The one number worth keeping in mind is that PTS floors around 811 mmHg over the whole sampled domain. That matters for the purge clamp below.

The condenser

A smooth saturating function of reactor outlet flow, approaching 0.404655 as the flow grows (teprob.f:674), with the duty taken against the stream 8 temperature rather than the separator's own, and scaled by a disturbance drift factor (teprob.f:675-676):

\[ U A_s = 0.404655 \left(1 - \frac{1}{1 + (F_8/3528.73)^4}\right) \]

\[ Q_s = U A_s \, (T_{ws} - T_{st,8}) \, (1 - 0.25 \, d_{11}) \]

**2 and **4 are integer powers and must not go through pow

gfortran expands an integer exponent into multiplications rather than calling libm, and the shape of that expansion is load-bearing. Measured over 200,000 values with this project's pinned flags:

So it is binary exponentiation, squaring twice, and the two plausible alternatives are each wrong about a third and a half of the time. X**2 is x*x on all 200,000, which is the only thing it could be.

Three ways out

The underflow to the stripper is valve-lagged, and is the simple case (teprob.f:570):

\[ F_{11} = \frac{v_7 R_7}{100} \]

The purge is pressure-driven to atmosphere through valve 6 (teprob.f:585-588):

\[ F_{10} = \frac{v_6 \times 0.151169 \times \sqrt{\max(P_s - 760,\, 0)}}{\overline{M}_{10}} \]

That clamp at zero is the one clamp in the whole flow network that cannot be reached. PTS is a sum of eight partial pressures in a vessel that always holds material, and it floors around 811 mmHg against a threshold of 760, so no trajectory state, no random perturbation and no adversarial boundary takes that branch. It is therefore covered by a unit test at a composition chosen to reach it, rather than by the differential, on the principle that a branch no test enters is indistinguishable from a branch that is wrong.

The recycle goes through the compressor. The operating point on the curve is the pressure ratio between the mixing zone and the separator, clamped at both ends (teprob.f:589-591), and the machine is fixed-speed with a cubic pressure-ratio curve (teprob.f:592-593). The recycle valve then bleeds flow back and the result has a floor (teprob.f:596-599):

\[ \dot m = \max\!\left( F_{\max}\left(1 + \frac{1 - r^3}{1.197}\right) - v_5 \times 53.349 \times \sqrt{\max(P_v - P_s,\, 0)}, \; 10^{-3} \right) \]

with \(r = \mathrm{clamp}(P_v / P_s,\, 1,\, 1.3)\), \(F_{\max} = 280275\) (teprob.f:1170) and the ratio ceiling CPPRMX at 1.3 (teprob.f:1171). The floor at \(10^{-3}\) exists so that the division at teprob.f:600-601 cannot blow up.

PR**3 at teprob.f:593 is an integer power, for the same reason **4 is above, and is written out as three multiplications rather than routed through pow.

The compressor work appears twice: as an enthalpy bump on the recycle stream, and as measurement 20 (teprob.f:594-595, 601, 699):

\[ W = \dot m \, (T_{cs} + 273.15) \times 1.8 \times 10^{-6} \times 1.9872 \times \frac{P_v - P_s}{\overline{M}_9 P_s} \]

Note that the 273.15 on that line is written 273.15D0, double precision, unlike the one in TESUB2 at teprob.f:1411. The original is not consistent about that constant and each occurrence has to be read off its own line.

HST(10) = HST(9) is a snapshot, not an alias

teprob.f:562 copies the separator vapour enthalpy into the purge. Both streams leave the separator vapour space, so at that moment they are the same fluid at the same temperature and the copy is exact.

Then teprob.f:601 adds the compressor work to HST(9). The recycle gains the work; the purge does not, because it was copied first. Reading line 562 as an alias rather than as a copy would give the purge a share of compressor work it never receives, and the two lines are seventy apart, so the ordering is easy to miss. Both energy balances that read stream 9 come after line 601 and therefore see the bumped value.

Balances

One inlet, three outlets, and the condenser duty (teprob.f:762-770 and 773-777):

\[ \frac{dn_i}{dt} = \dot n_{i,8} - \dot n_{i,9} - \dot n_{i,10} - \dot n_{i,11} \]

\[ \frac{dE}{dt} = h_8 F_8 - h_9 F_9 - h_{10} F_{10} - h_{11} F_{11} + Q_s \]

The condenser cooling water wall temperature is YP(38) (teprob.f:791-792).

Variables

Two of the eight shutdown conditions belong to this vessel: separator liquid volume above 12 or below 1 cubic metre (teprob.f:707-708).