19.2 Process Equipment Selection, Sizing, and Optimization

19.2 Process Equipment Selection, Sizing, and Optimization

The NCEES specification names Process equipment design (e.g., equipment selection, optimization, sizing) under Design. Exam questions in this subtopic are short: pick the right machine for a stated duty, or size one piece of equipment from a standard correlation, or find the economic optimum of a single design variable.

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1. Selection Logic

Pumps

The defining behavioral difference: a centrifugal pump produces a head that falls as flow rises, so closing the discharge simply drives it to shutoff head and it survives (briefly — it will overheat). A positive displacement pump produces whatever pressure it takes to move the fixed displacement, so closing the discharge ruptures the casing or the piping. Every PD pump requires a relief valve on the discharge, upstream of the first block valve.

Compressors

Both types are limited by discharge temperature. For an ideal gas in isentropic compression, (T_2 = T_1 (P_2/P_1)^{(k-1)/k}), so compression ratio is capped by lubricant breakdown and by the material's temperature limit. This is why large ratios are split into multiple stages with intercooling, which also minimizes total work — the optimum stage pressure ratio for (n) stages with perfect intercooling is ((P_{final}/P_{initial})^{1/n}) at each stage.

Heat exchangers

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2. Sizing a Vapor-Liquid Separator

Knockout drums, flash drums, compressor suction scrubbers, and reflux drums are all sized on the same principle: the upward gas velocity must be low enough that entrained droplets settle out rather than being carried over. Equating drag and gravity on a droplet gives the Souders-Brown equation:

$v_{max} = K \sqrt{\frac{\rho_L - \rho_V}{\rho_V}}$

Worked example. A vertical knockout drum with a mesh pad handles (2.0\text{ m}^3\text{/s}) of vapor at (\rho_V = 12\text{ kg/m}^3) with (\rho_L = 700\text{ kg/m}^3). Size the diameter.

$v_{max} = 0.107\sqrt{\frac{700 - 12}{12}} = 0.107\sqrt{57.33} = 0.107(7.572) = 0.810\text{ m/s}$

$A = \frac{\dot{V}}{v_{max}} = \frac{2.0}{0.810} = 2.47\text{ m}^2 \qquad D = \sqrt{\frac{4A}{\pi}} = \sqrt{\frac{4(2.47)}{\pi}} = 1.77\text{ m}$

Round up to a standard (1.8\text{ m}) shell. Height is then set separately by liquid holdup requirements (typically 5-10 minutes between high and low level for surge, more if the drum feeds a pump) plus disengagement space above the inlet nozzle and below the mesh pad.

Note the pressure sensitivity: (v_{max} \propto 1/\sqrt{\rho_V}) and (\rho_V \propto P), so a drum sized at (10\text{ bar}) and then operated at (20\text{ bar}) can handle a larger mass flow but a smaller volumetric one, and the same drum in vacuum service becomes enormous.

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3. Economic Optimization of a Design Variable

Nearly every sizing decision is a contest between capital cost, which rises with size, and operating cost, which falls with size. The optimum is where the total is minimized:

$C_{total}(x) = C_{capital}(x) + C_{operating}(x), \qquad \frac{dC_{total}}{dx} = 0$

The canonical form. When operating cost varies as (1/x) and annualized capital varies linearly with (x):

$C_{total} = \frac{a}{x} + b x \qquad \Longrightarrow \qquad \frac{dC}{dx} = -\frac{a}{x^2} + b = 0 \qquad \Longrightarrow \qquad \boxed{x^* = \sqrt{a/b}}$

Example. Annual heat loss through insulation costs (a/x = 8{,}000/x) ($/\text{yr}) with (x) in inches, and annualized insulation capital is (bx = 500x) ($/\text{yr}). Then (x^* = \sqrt{8{,}000/500} = \sqrt{16} = 4.0\text{ in}), with (C_{min} = 8{,}000/4 + 500(4) = 2{,}000 + 2{,}000 = $4{,}000\text{/yr}).

Notice that at the optimum the two cost terms are exactly equal (($2{,}000) each). That equality is a property of this (1/x)-plus-linear form, and it makes a fast sanity check — but it does not generalize. For (C = a/x^2 + bx) the optimum has capital cost equal to twice the operating cost. Differentiate; do not assume.

The same structure recurs across the discipline:

Optimum reflux ratio deserves its own note because it is shaped differently. As (R \to R_{min}) the required number of stages goes to infinity, so capital cost rises without bound; as (R) increases, stage count falls quickly at first and then flattens while reboiler and condenser duty (and column diameter) keep climbing. The total cost curve is therefore steep on the left and shallow on the right, and the minimum sits at roughly (R_{opt} \approx 1.1) to (1.2 \times R_{min}). Because the curve is flat near the optimum, designers commonly select toward the upper end to buy operating flexibility cheaply.

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4. Heuristics Worth Carrying Into the Exam

These are the first-pass numbers a practicing process engineer uses before any rigorous calculation, and NCEES writes judgment questions against them:

• Liquid line velocity: (1-3\text{ m/s}) (pump discharge), (0.5-1.5\text{ m/s}) (pump suction, to protect NPSH).
• Gas line velocity: (15-30\text{ m/s}).
• Shell-and-tube (U): (\sim 850\text{ W/(m}^2\cdot\text{K)}) water-water, (\sim 300) organic-water, (\sim 30-60) gas-gas.
• Minimum exchanger approach: (\sim 10^\circ\text{C}) with cooling water, (\sim 5^\circ\text{C}) process-process, (1-2^\circ\text{C}) only in plate exchangers.
• Tray spacing (0.45-0.60\text{ m}); overall tray efficiency (50-70%) for hydrocarbons.
• Vessel L/D: (3) for low pressure, (4-5) at elevated pressure (shell cost favors slender vessels as pressure rises).
• Design margin: size equipment for roughly (110-120%) of design rate, and motors for the maximum possible operating point, not the normal one.