Particle Swarm Optimization Technique for the Optimal Design of Plate-Type Distillation Column

Abstract

The present study explores the application of non-traditional optimization technique, Particle swarm optimization (PSO), for the optimal design of plate-type distillation column from economic point of view. The optimization procedure involves the selection of the major plate geometric parameters such as hole diameters, ratio of downcomer area to tower area, weir height, fractional hole area in tray, tray spacing, tower diameter, etc. and minimization of total annual cost is considered as design target subjected to operational constraints like flooding, weeping entrainment, quality specifications, etc. The solution space of such type of problem is very complex due to the presence of various constraints and may contain multiple minima. In this chapter, recently proposed PSO method employing feasibility rule for handling constraints is applied for the optimal design of the plate-type distillation column. In contrast to the traditional penalty function method, the rule requires no additional parameters and can guide the swarm to the feasible region quickly. In addition, to avoid the premature convergence, simulated annealing (SA) is applied to the best solution of the swarm to help the algorithm escape from local optima. The presented PSO technique is simple in concept, few in parameters, and easy for implementations. Furthermore, the PSO algorithm explores the good quality solutions quickly, giving the designer more degrees of freedom in the final choice with respect to traditional methods. One case study is presented to demonstrate the effectiveness and accuracy of the proposed algorithm. The PSO approach is able to reduce the total cost of distillation column as compared to cost obtained by commercial simulator.

Appendices

Appendix 1

This section describes the step by step procedure for calculating various constraints given in Table 6.4. The equations are taken from various literatures, namely, Kister [9], Sinnot [18], Wei et al. [20], Douglas [3], McCabe et al. [13], Seader and Ernest [16], Xu and Pless [21], Lieberman [10], etc. As most of the correlations, plots, nomographs, and equations found in literature are in FPS unit, the whole calculations (Eqs. 6.1–6.79) are done in FPS unit and appropriate conversion was made at initial input and final results to convert it to SI unit.

• Step 1: Critical froth velocity (ft/s)

  $$\displaystyle{ U_{\mathrm{dfc}} = \left [\frac{3.077\sigma ^{0.6}(\rho _{L} -\rho _{v})^{0.4}} {\mu _{L}^{0.2}\mu _{L}^{0.8}} \right ]^{0.556} }$$

  (6.15)

• Step 2: Calculate q _ldc—liquid rate to downcomer (gpm)

  $$\displaystyle{ q_{L} = \frac{7.481M_{L}} {60\rho _{L}} }$$

  (6.16)
  $$\displaystyle{ q_{w} = 0.135_{{\ast}qL} }$$

  (6.17)
  $$\displaystyle{ q_{\mathrm{ldc}} = q_{L} - q_{w} }$$

  (6.18)

  where $u_{\mathrm{dfc}} =\mathrm{ critical}$ froth velocity (m/s), σ = surface tension, dynes/cm, ρ _L  = liquid density, kg/m³, ρ _V  = vapor density, kg/m³, μ _L  = liquid viscosity, cP, M _L  = liquid mass flow rate, q _w weep rate (gpm), q _ldc = liquid rate to downcomer (gpm), q _L  = liquid rate down tower (gpm)

• Step 3: Calculate a _dc average downcomer area (ft2)

• Refer to Kister [9], Sinnot [18] for downcomer area calculation for 1 and 2 pass trays. During downcomer area calculation, the following parameters are also evaluated: l _w  = outlet weir length (m), w _dct = Top downcomer (DC) width (m), a _dct = area downcomer at top (m2), l _dcb = bottom downcomer chord length (m), w _dcb = Bottom downcomer (DC) width (m), a _dcb = area downcomer at bottom (m2), l _dcs = bottom downcomer sump chord length (m), w _dcs = downcomer (DC) sump width (m), a _dcs = area of bottom downcomer sump (m2), a _dc = average downcomer area (m2), l _cdct = top center downcomer chord length (m), w _cdct = top center downcomer (DC) width (m), a _cdct = area of center downcomer at top (m2), I _cdcb = bottom center downcomer chord length (m), w _cdcb = bottom center downcomer (DC) width (m), l _cdcs = side center downcomer chord length (m), w _cdcs = side center downcomer (DC) width (m)

• Step 4: Calculate u _bdcv—bubbling area vapor velocity at downcomer critical velocity at downcomer critical velocity

  $$\displaystyle{ u_{\mathrm{bdcvt}} = \left [\left (\frac{u_{\mathrm{dfc}} - \frac{q_{\mathrm{ldc}}} {448a_{\mathrm{dct}}} } {2.49} \right )\left (\frac{\rho _{L} -\rho _{v}} {\rho _{v}} \right )^{0.26}\left (\frac{a_{\mathrm{dct}}} {a_{\mathrm{dc}}} \right )^{0.4}\right ]^{ \frac{1} {0.58} } }$$

  (6.19)
  $$\displaystyle{ u_{\mathrm{bdcvw}} = \left [\left (\frac{u_{\mathrm{dfc}} - \frac{q_{\mathrm{ldc}}} {448a_{\mathrm{dc}}} } {1.379} \right )\left (\frac{\rho _{L} -\rho _{v}} {\rho _{v}} \right )^{0.22}\right )^{ \frac{1} {0.54} } }$$

  (6.20)
  $$\displaystyle{ u_{\mathrm{bdcv}} =\mathrm{ min}(u_{\mathrm{bdcvt}},u_{\mathrm{bdcvw}}) }$$

  (6.21)

  where u _b  = vapor velocity based on bubbling area (m/s), u _bdcvt = u _b at downcomer critical velocity at downcomer entrance (m/s), u _bdcvw = u _b at downcomer critical velocity within downcomer (m/s), u _bdcv = bubbling area vapor velocity at dc critical velocity (m/s)

  Do while flag = True and iteration < 1,000 (Flag = True for first iteration)

• Step 5: Scale q _l for constant 1/v

  $$\displaystyle{ u_{b} = \frac{M_{v}} {3600\rho _{v}a_{b}} }$$

  (6.22)
  $$\displaystyle{ q_{ls} = \frac{q_{l}u_{\mathrm{bdcv}}} {u_{b}} }$$

  (6.23)

  M _V  = vapor mass flow rate, kg/s, q _ls = scaled liquid rate down tower for constant 1/v (gpm)

• Step 6: Calculate Froude numbers

  $$\displaystyle{ F_{r} = \left [\left ( \frac{12\rho _{V }} {p_{L} - p_{V }}\right )\left ( \frac{u_{b}^{2}} {32.18h_{\mathrm{cl}}}\right )\right ] }$$

  (6.24)

• Step 7: Calculate η—vapor/liquid volume ratio on tray

  $$\displaystyle{ \eta = 13.3F_{r}^{0.4}\phi ^{-0.25} }$$

  (6.25)

• Step 8: Calculate α _t —liquid volume fraction on tray

  $$\displaystyle{ \alpha _{t} = \frac{1} {1+\eta } }$$

  (6.26)

• Step 9: Calculate h _f —froth height on tray

  $$\displaystyle{ \frac{h_{\mathrm{cl}}} {\alpha _{t}} }$$

  (6.27)

• Step 10: Calculate c _d —liquid head co-efficient

  $$\displaystyle{ c_{d} = \left \{\begin{array}{@{}l@{\quad }l@{}} 0.61 + 0.08\frac{h_{f}-h_{w}} {h_{w}} \, \frac{h_{f}-h_{w}} {h_{w}} \leq 8.135\quad \\ 1.06\left (1 + \frac{h_{w}} {h_{f}-h_{w}}\right )^{1.5}\,\mbox{ otherwise} \quad \end{array} \right. }$$

  (6.28)

  where c _d  = liquid head coefficient, ϕ = Hole area (fraction of bubbling area)

• Step 11: Calculate h _ct—liq head on tray (m liquid)

  $$\displaystyle{ h_{\mathrm{cl}} =\alpha _{t}h_{w} + 1.02\left (\frac{\alpha _{t}^{0.5}} {c_{d}} \right )^{0.67}\left (\frac{q_{l} - q_{w}} {l_{w}} \right )^{0.45} }$$

  (6.29)

• Step 12: Calculate Δ—pressure drop in liquid

  $$\displaystyle{ \varDelta p = 5.2u_{b}^{2}\left (\frac{1 -\phi ^{2}} {\phi ^{0.2}} \right )\left ( \frac{d_{h}} {\mathrm{deck}_{t}}\right )^{0.2}\left ( \frac{\rho _{v}} {32.185\rho _{L}\phi ^{2}}\right ) + h_{\mathrm{cl}} }$$

  (6.30)

• Step 13: Calculate u _l —horizontal liquid velocity (ft/s)

  $$\displaystyle{ \mathrm{WFP} = \left \{\begin{array}{@{}l@{\quad }l@{}} \frac{l_{w}+\mathrm{max}(l_{\mathrm{dcb}}l_{\mathrm{dcs}})+12d_{t}} {3} \,\mbox{ for pass = 1}\quad \\ \frac{\frac{l_{w}} {2} +\mathrm{max}(l_{\mathrm{cdcb}},l_{\mathrm{cdcs}})} {2} \,\mbox{ for pass = 2A} \quad \\ \frac{l_{\mathrm{cdct}}+\mathrm{max}(l_{\mathrm{dcb}},l_{\mathrm{dcs}})} {2} \,\mbox{ for pass = 2B} \quad \\ \quad \end{array} \right. }$$

  (6.31)
  $$\displaystyle{ a_{h} =\phi a_{b} }$$

  (6.32)
  $$\displaystyle{ u_{t} = \frac{q_{l} - q_{w}} {3.116h_{\mathrm{cl}}\mathrm{WFP}} }$$

  (6.33)

• Step 14: Calculate weep_df—weeping driving force

  $$\displaystyle{ \mathit{weep}_{\mathit{df }} = \frac{h_{cl}} {\alpha _{t}} -\varDelta P }$$

  (6.34)

• Step 15: Calculate q _w —weep rate (gpm)

  $$\displaystyle{ u_{\mathrm{BWP}} = \left (\frac{14.5\phi ^{1.44}} {\rho _{v}^{0.5}} \right )\left (\frac{\rho _{L} -\rho _{v}} {\rho _{v}} \right )^{0.094}\left ( \frac{d_{h}} {\mathrm{deck}_{t}}\right )(\rho _{L}h_{\mathrm{cl}})^{0.144} }$$

  (6.35)
  $$\displaystyle{ q_{W2} = \left \{\begin{array}{@{}l@{\quad }l@{}} \left (94.6a_{h}\alpha _{t}^{2}\right )\left (e^{\left (\frac{-0.043u_{l}d_{h}} {\mathrm{deck}_{t}} \right )}\right )\left (1 -\left [ \frac{u_{b}} {u_{BWP}}\right ]^{0.5}\right )\cdots \quad \\ \cdots \left (64.37\left [\frac{h_{\mathrm{cl}}} {\alpha _{t}} -\varDelta P\right ]^{0.5}\right ),\ \mathrm{weep}_{\mathrm{df}} > 0 \quad \\ 0,\ \mbox{ otherwise} \quad \end{array} \right. }$$

  (6.36)
  $$\displaystyle{ q_{w1} =\mathrm{ max}(q_{w2,0}) }$$

  (6.37)
  $$\displaystyle{ q_{w2} =\mathrm{ max}(q_{w2,0.9q_{l}}) }$$

  (6.38)
  $$\displaystyle{ q_{w} = \frac{q_{w1} + q_{w2}} {2} }$$

  (6.39)
  $$\displaystyle{ q_{w1} = q_{w2} }$$

  (6.40)

• Step 16: Calculate u _bdcv—bubbling area vap velocity at down comer critical velocity

  $$\displaystyle{ q_{\mathrm{ldc}} = q_{\mathrm{ls}} - q_{w} }$$

  (6.41)
  $$\displaystyle{ u_{\mathrm{bdcvt}} = \left [\left (\frac{\left [u_{\mathrm{dfc}} - \frac{q_{\mathrm{ldc}}} {448.8\alpha _{\mathrm{dct}}} \right ]} {2.49} \right )\left (\frac{\left (\rho _{L} -\rho _{V }\right )} {\rho _{V }} \right )^{0.26}\left (\frac{a_{\mathrm{dct}}} {a_{\mathrm{dc}}} \right )^{0.4}\right ]^{ \frac{1} {0.58} } }$$

  (6.42)
  $$\displaystyle{ u_{\mathrm{bdcvw}} = \left [\left (\frac{\left [u_{\mathrm{dfc}} - \frac{q_{\mathrm{ldc}}} {448.8\alpha _{\mathrm{dct}}} \right ]} {1.379} \right )\left (\frac{\left (\rho _{L} -\rho _{V }\right )^{0.22}} {\rho _{V }} \right )\right ]^{ \frac{1} {0.54} } }$$

  (6.43)
  $$\displaystyle{ u_{\mathrm{bdcv2}} =\mathrm{ min}\left (u_{\mathrm{bdcvt}},u_{\mathrm{bdcvw}}\right ) }$$

  (6.44)

  If $\mathrm{abs}(u_{\mathrm{bdcv1}} - u_{\mathrm{bdcv2}}) <\mathrm{ convtol\ Or\ iter}\ \geq \ \mathrm{ maxiter\ Then}$

  $$\displaystyle{ u_{\mathrm{bdcv}} = u_{\mathrm{bdcv2}} }$$

  (6.45)

  Flag = false

  Else

  $$\displaystyle{ u_{\mathrm{bdcv}} = \frac{u_{\mathrm{bdcv1}} + u_{\mathrm{bdcv2}}} {2} }$$

  (6.46)
  $$\displaystyle{ u_{\mathrm{bdcv1}} = u_{\mathrm{bdcv2}} }$$

  (6.47)

  Iter = iter + 1 & loop

• Step 17: Calculate

  $$\displaystyle{ a_{f} = \left \{\begin{array}{@{}l@{\quad }l@{}} a_{t} - 0.5(a_{\mathrm{dct}} + a_{\mathrm{dcb}}),\ \mbox{ for pass =1} \quad \\ a_{t} - 0.5(a_{\mathrm{cdct}} + a_{\mathrm{cdcb}}),\ \mbox{ for pass = 2A}\quad \\ a_{t} - 0.5(a_{\mathrm{sdct}} + a_{\mathrm{sdcb}}),\ \mbox{ for pass = 2B}\quad \\ \quad \end{array} \right. }$$

  (6.48)
  $$\displaystyle{ c_{b} = u_{b}\left [ \frac{p_{V }} {p_{L} - p_{V }}\right ]^{0.5} }$$

  (6.49)
  $$\displaystyle\begin{array}{rcl} c_{bf1} =& & \ 0.9\left [ \frac{p_{V }} {p_{L} - p_{V }}\right ]^{0.04}\left [e^{\left ( \frac{-11} {s_{\mathrm{tray}}} \right )}\right ]\left [\frac{a_{f}} {a_{b}} \right ]^{0.5}\left [e^{ \frac{0.68} {d_{h}+0.73+0.24\left ( \frac{q_{L}} {l_{w}} \right )} }\right ] \\ & & \left [1 - \frac{1} {e^{\phi ^{0.23}\left ( \frac{q_{l}} {l_{w}}+0.25\right )^{0.2} }} \right ]\left [e^{-0.6\left (\frac{s_{s_{\mathrm{traylw}}}} {6q_{L}} \right )^{2} }\right ] {}\end{array}$$

  (6.50)
  $$\displaystyle{ c_{s} = \left [ \frac{M_{V }} {3600\rho _{V }a_{t}}\right ]\left [ \frac{p_{V }} {p_{L} - p_{V }}\right ]^{0.5} }$$

  (6.51)
  $$\displaystyle{ u_{\mathrm{lv}} = \left [ \frac{M_{L}} {3600\rho _{L}a_{t}}\right ] }$$

  (6.52)
  $$\displaystyle{ c_{\mathrm{sp}} = \frac{1.4\left [ \frac{\sigma }{p_{ L}-p_{V }}\right ]^{0.2}\left [ \frac{p_{V }} {p_{L}-p_{V }}\right ]^{-0.5}} {1 + 1.4\left [ \frac{p_{V }} {p_{L}-p_{V }}\right ]^{-0.5}} }$$

  (6.53)
  $$\displaystyle{ c_{\mathrm{ss}} = 0.641c_{\mathrm{sp}}e^{\left [-11.25\left (\frac{u_{\mathrm{lv-0.033}}} {c_{\mathrm{sp}}} \right )^{2}\right ] } }$$

  (6.54)
  $$\displaystyle{ p_{\mathrm{vsl}} = 100 \frac{c_{s}} {c_{\mathrm{ss}}} }$$

  (6.55)

• Step 18: Calculate % jet flood, %DC flood and method

  $$\displaystyle{ c_{\mathrm{afo}} = \left \{\begin{array}{@{}l@{\quad }l@{}} 0.0263S_{\mathrm{tray}} - ( \frac{\rho _{V }} {1560}),\ \mbox{ for $s_{\mathrm{tray}} < 12$} \quad \\ \left (0.58 - \frac{3.29} {s_{\mathrm{tray}}} \right ) -\left (\frac{s_{\mathrm{tray}\rho _{V }-s_{\mathrm{tray}}}} {1560} \right ),\ \mbox{ for }12 \leq S_{\mathrm{tray}} \leq 24 \quad \\ \left (0.58 - \frac{3.29} {s_{\mathrm{tray}}} \right ) -\left ( \frac{s_{\mathrm{tray}\rho _{V }}} {23.88s_{\mathrm{tray}}+987}\right ) + \left (\frac{s_{\mathrm{tray}}} {1560} \right ),\ \mbox{ for } > 24\quad \\ \quad \end{array} \right. }$$

  (6.56)
  $$\displaystyle{ \mbox{ If}\ c_{\mathrm{afo}} > \frac{s_{\mathrm{tray}^{0.65}}} {12} \rho v^{0.167}\ \mbox{ Then}\ c_{\mathrm{ afo}} = \frac{s_{\mathrm{tray}^{0.65}}} {12} \rho v^{0.167} }$$

  (6.57)
  $$\displaystyle{ \mbox{ If}\ c_{\mathrm{afo}}\,=\,0.28\left [1.1\,-\,0.14\rho v\right ]\,+\,0.3\ \mbox{ Then}\ c_{\mathrm{afo}}\,=\,0.28\left [1.1\,-\,0.14\rho v\right ]\,+\,0.3\quad }$$

  (6.58)
  $$\displaystyle{ v_{\mathrm{load}} = c_{b}\ a_{b} }$$

  (6.59)
  $$\displaystyle{ \mathrm{AUD} = \left \{\begin{array}{@{}l@{\quad }l@{}} l_{\mathrm{dcb}}c_{\mathrm{dc}},\ \mbox{ for pass} = 1 \quad \\ 2l_{\mathrm{dcb}}c_{\mathrm{dc}},\ \mbox{ for pass} = \mathrm{2A} \quad \\ 2l_{\mathrm{cdcb}}c_{\mathrm{cdc}},\ \mbox{ for pass} = \mathrm{2B}\quad \\ \quad \end{array} \right.. }$$

  (6.60)
  $$\displaystyle{ \mathrm{FPL} = \left \{\begin{array}{@{}l@{\quad }l@{}} 12d_{t} - w_{\mathrm{dct}} -\mathrm{ max}\left (w_{\mathrm{dcb}},w\mathrm{dcs}\right )\ \mbox{ for pass} = 1 \quad \\ 6d_{t} - w_{\mathrm{dct}} - 0.5\mathrm{max}\left (w_{\mathrm{cdcb}},w\mathrm{cdcs}\right )\ \mbox{ for pass} = \mathrm{2A} \quad \\ 6d_{t} - 0.5w_{\mathrm{dct}} - 0.5\mathrm{max}\left (w_{\mathrm{dcb}},w\mathrm{dcs}\right )\ \mbox{ for pass} = \mathrm{2B}\quad \\ \quad \end{array} \right. }$$

  (6.61)
  $$\displaystyle{ \mathrm{HLUDC} = \left \{\begin{array}{@{}l@{\quad }l@{}} 0.06\left [ \frac{q_{l}} {l\mathrm{dcb}c_{\mathrm{cdc}}} \right ]^{2}\ \mbox{ for pass} = 1 \quad \\ 0.06\left [ \frac{q_{l}} {2l\mathrm{dcb}c_{\mathrm{cdc}}} \right ]^{2}\ \mbox{ for pass} = \mathrm{2A}\quad \\ 0.06\left [ \frac{q_{l}} {2l\mathrm{dcb}c_{\mathrm{cdc}}} \right ]^{2}\ \mbox{ for pass} = \mathrm{2B}\quad \\ \quad \end{array} \right.. }$$

  (6.62)
  $$\displaystyle{ \mathrm{gmpif} = 100\frac{\mathrm{vload} + \frac{q_{l}\mathrm{FPL}} {13000} } {a_{b}\mathrm{caf}_{o}} }$$

  (6.63)
  $$\displaystyle{ \mathrm{FID} = \frac{q_{l}} {a_{\mathrm{dct}}} }$$

  (6.64)
  $$\displaystyle{ \%\mathrm{Downcomerflood} = 100\ \mathrm{max}\left (\frac{\mathrm{FID}} {250}, \frac{\mathrm{FID}} {41\left (\rho _{L} -\rho _{V }\right )}, \frac{\mathrm{FID}} {7.5\left (\rho _{L} -\rho _{V }\right )^{0.5}s_{\mathrm{tray}^{0.5}}} \right ) }$$

  (6.65)
  $$\displaystyle{ \mathrm{weirload} = \frac{q_{l}} {l_{w}} }$$

  (6.66)
  $$\displaystyle{ \mathrm{DCBU} = h_{w} + 0.4 \mathrm{weirdload}^{0.66} + \left (\mathrm{dp} +\mathrm{ HLUDC}\right )\left ( \frac{\rho _{L}} {\rho _{L} -\rho _{V }}\right ) }$$

  (6.67)
  $$\displaystyle{ \%\mathrm{Downcomer\ backup} = 100\frac{\mathrm{DCBU}} {s_{\mathrm{tary}}} }$$

  (6.68)

• Step 19: Calculate final constraints values

  $$\displaystyle{ \mathrm{fmpjfl} = 100 \frac{c_{b}} {c_{\mathrm{bfl}}} }$$

  (6.69)
  $$\displaystyle{ \%\mathrm{jetflood} = \left \{\begin{array}{@{}l@{\quad }l@{}} \mathrm{fmpjf1},\ \mbox{ for vtype = 0} \quad \\ \mathrm{gmpjf },\ \mbox{ for vtype = others}\quad \end{array} \right. }$$

  (6.70)

  where FPL = Flow path length (m), q _ls = scaled liquid rate down tower for constant l/v (gpm), gmpjf = Glitsch method percent jet flood (%), FID = Flow into the downcomer (gpm/ft2), %Downcomer flood = Glitsch method percent downcomer flood (%), HLUDC = Glitsch method head loss under downcomer (m liquid), weir load = Weir loading (gpm/in weir), DCBU = Glitsch method DC backup (m liquid), %Downcomer backup = Glitch method DC backup % of tray spacing (%), %Jet flood = Final percent of jet flood by most appropriate method

  $$\displaystyle{ u_{\mathrm{ldc}} = \frac{M_{L}} {3600d_{\mathrm{dct}}\rho _{L}} }$$

  (6.71)
  $$\displaystyle{ \mathrm{dpdry} =\mathrm{ dp} -\mathrm{ hcl} }$$

  (6.72)
  $$\displaystyle{ \mathrm{dppsi} = \frac{\mathrm{dp}\rho _{L}} {1728} }$$

  (6.73)
  $$\displaystyle{ \mathrm{wfrac} = \frac{qw} {q_{l}} }$$

  (6.74)

• Step 20: Calculate entrainment fraction

  $$\displaystyle{ c_{\mathrm{bfspec}} = c_{\mathrm{bfl}} }$$

  (6.75)
  $$\displaystyle{ c_{\mathrm{bfspec}} = \frac{100c_{b}} {\mathrm{gmpif}},\ \text{ if vtype 0} }$$

  (6.76)
  $$\displaystyle{ c_{\mathrm{bespec}} =\mathrm{ max}(c_{b},0.5c_{\mathrm{bfspec}}) }$$

  (6.77)
  $$\displaystyle{ \mathrm{efrac} =\mathrm{ min}\left (10^{\left [ \frac{c_{\mathrm{bespec}}\ -\ c_{\mathrm{bfspec}}} {\left [ \frac{\rho _{V }} {\rho _{L}-\rho _{V }}\right ]^{-0.2} \frac{1} {31+31 \frac{q_{L}} {l_{w}} }}\right ] },1\right ) }$$

  (6.78)
  $$\displaystyle{ \mathrm{efrac} =\mathrm{ efrac}\frac{M_{V }} {M_{L}},\mathrm{if}\ M_{V } > M_{L} }$$

  (6.79)

  where a _h  = hole area (m²), a _b  = bubbling area (m²), weep_df = weeping driving force, u _BWP = weep point (m/s), deck _t  = deck thickness, m, d _h  = hole diameter, m, q _w  = weep rate (gpm), q _ldc = liquid rate to downcomer (gpm), a _f  = free area (m²), c _b  = capacity factor based on bubbling area (m/s), c _bfl = capacity factor based on bubbling area at const liquid rate jet flood (m/s), u _lv = vertical liquid velocity based on tower area (m/s), c _s  = capacity factor based on tower area (m/s), c _sp = system limit parameter, c _ss = capacity factor at system limit based on tower area (m/s), p _vsl = percent of a vapor system limit at const 1∕v (%), v _load = Glitch method Vload term (m3/s)

Appendix 2

Step 1::

Initialize optimization

Step 1.1:

Initialize algorithm constants, t _max, P

Step 1.2:

Initialize randomly all particle positions x _t ⁱ and velocities v _t ⁱ

Step 1.3:

Evaluate objective function values as f(x _t ⁱ)

Step 1.4:

Assign best positions $p_{t}^{i} = x_{t}^{i}$ with $f(p_{t}^{i}) = f(x_{t}^{i})\ i = 1,\ldots,P$

Step 1.5:

Find $f_{t}^{\mathrm{best}}(p_{t}^{\mathrm{best}}) =\mathrm{ min}\{f(p_{t}^{1}),\ldots,f(p_{t}^{P})\}$ and initialize P _t ^g and $f(p_{t}^{g}) = f_{t}^{\mathrm{best}}(p_{t}^{\mathrm{best}})$

Step 2::

Perform optimization

While ($t \leq t_{\mathrm{max}}$)

Step 2.1:

Update particle positions x _t ⁱ and velocities v _t ⁱ according to Eqs. (7.9) and (7.10) of all P particles.

Step 2.2:

Evaluate objective function values as f(x _t ⁱ)

Step 2.3:

Update particle best position, i.e., p _t+1 ⁱ will be replaced by x _t+1 ⁱ at any of the following scenarios:

• (a) p _t ⁱ is infeasible, but X _t+1 ⁱis feasible.

• (b) Both p _t ⁱ and X _t+1 ⁱ are feasible, but $f(X_{t+1}^{i}) < f(p_{t}^{i})$.

• (c) Both p _t ⁱ and X _t+1 ⁱ are infeasible, but out(X _t+1 ⁱ) < out(p _t ⁱ). where the constraint violation value of an infeasible solution is calculated as follows:

  $$\displaystyle{ \mathrm{out}(x) =\sum _{ j=1}^{N}[\mathrm{max}(g_{ j}(x),0)] }$$

Step 2.4:

Apply SA algorithm to local search for finding gbest. Let p _k ^g denote gbest of the population at generation k and α denote the acceptable probability of a new generated solution.

Step 2.4.1::

In first iteration, let $p_{k}^{g^{'} } = p_{k}^{g}$

Step 2.4.2::

Generate a new solution using the following equation: $x_{k}^{i^{'} } = p_{k}^{g^{'} } +\varepsilon (X_{\mathrm{max}} - X_{\mathrm{min}})N(0,1)$

where $\varepsilon$ is used to control the step size, X _max and X _min denote the upper and lower bounds of the solutions defined by the problem, N(0, 1) denotes a random number normally distributed with mean 0 and variance 1.

Step 2.4.3::

Calculate α according to the following criteria:

$$\displaystyle{ \alpha = \left \{\begin{array}{@{}l@{\quad }l@{}} 1,\ \mbox{ if $x_{k}^{i^{'} }$ is feasible and $p_{k}^{g^{'} }$ is infeasible} \quad \\ 0,\ \mbox{ if $x_{k}^{i^{'} }$ is infeasible and $p_{k}^{g^{'} }$ is feasible } \quad \\ \mathrm{min}(1,e^{\frac{f(p_{k}^{g^{'} })-f(x_{k}^{i^{'} })} {T_{k}} }),\ \mbox{ if $x_{k}^{i^{'} }$ is feasible and $p_{k}^{g^{'} }$ is infeasible} \quad \\ \mathrm{min}(1,e^{[\frac{\mathrm{out}(p_{k}^{g^{'} })-\mathrm{out}(x_{k}^{i^{'} })]} {T_{t}} }),\ \mbox{ if $x_{k}^{i^{'} }$ is infeasible and $p_{k}^{g^{'} }$ is feasible}\quad \\ \quad \end{array} \right. }$$

T _k denotes the temperature at generation k

Step 2.4.4::

If α ≫ U(0, 1) then $p_{k}^{g^{'} } = x_{k}i^{'}$

where U(0,1) represents a random number uniformly distributed in the range of [0, 1].

Step 2.4.5::

Let k = k + 1. If k = L (is a user-defined number of iterations used to stop the local search), stop and output $p_{k}^{g^{'} }$ as the new gbest; else go to Step 2.4.2.

Step 2.5:

Increment iteration count t = t + 1

End while

Step 3::

Report best solution p ^g of the swarm with objective function value f(p ^g)