Process Design

7% of exam

Process flow diagrams and piping and instrumentation diagrams, equipment selection, sizing, and scale-up, equipment and facilities cost estimation with cost indices, process design and optimization (sustainability, efficiency, green engineering, inherently safer design), and design standards (regulatory, ASTM, ISO, OSHA).

3 concepts

Exam areas

A. Process flow diagrams and piping and instrumentation diagrams

PFDs, P&IDs and Cost Estimation

Read a PFD versus a P&ID, then turn an equipment list into a capital-cost number with the six-tenths rule, a cost index, and Lang factors.

Process design is communicated through two drawings and priced with a handful of scaling rules, and the FE rewards anyone who is fluent in both. A process flow diagram (PFD) tells you what the plant does — the major equipment, the flow path, and the heat-and-material balance that pins down stream temperatures, pressures, and rates. A piping and instrumentation diagram (P&ID) tells you how it is built and controlled — every line, valve, instrument, and interlock. The cost side is just as systematic: the FE Reference Handbook — Chemical Engineering (Cost Estimation) gives you three tools — a cost index to move a price through time, a capacity exponent to move it across size, and a Lang factor to move from bare equipment to an installed plant. Points are lost here by mixing a PFD detail into a P&ID question, by forgetting to update an old price with an index, or by applying the six-tenths rule to the wrong base. This concept makes each of those automatic.

PFD versus P&ID: what each drawing carries

A PFD is the high-level story of the process. It shows the major equipment as tagged blocks (reactors, columns, exchangers, pumps, drums), the primary process streams with arrows, and a stream table giving flow, composition, temperature, and pressure at each numbered point — it is essentially the material-and-energy balance drawn out. It deliberately omits spare pumps, most valves, line sizes, and instrumentation. A P&ID is the engineering-and-construction drawing: it shows every pipe with its size and spec, every valve, every instrument bubble, control loops, relief devices, and interlocks, but no stream data. Rule of thumb for the exam: if the question is about flows, duties, or what the plant produces, it is a PFD question; if it is about a control loop, a valve, a line spec, or an instrument tag, it is a P&ID question.

Reading instrument and equipment symbols

Instrument bubbles on a P&ID follow ISA-5.1: the first letter is the measured variable (

F

flow,

T

temperature,

P

pressure,

L

level,

A

analysis) and the following letters are the function (

I

indicate,

C

control,

T

transmit,

R

record,

A

alarm). Thus

FIC

is a flow indicating controller and

LAH

is a level-alarm-high. A bubble with no line is field-mounted; a horizontal bar means panel-mounted; a dashed line is an electrical signal and a line with crossing slashes is pneumatic. Equipment carries a tag such as

P\text{-}101A/B

(pump 101, with installed spare

B

) or

E\text{-}204

(exchanger). You are not asked to memorize the whole standard, but you must be able to decode a loop like

TT \to TIC \to TV

as temperature-transmitter feeding a controller that strokes a temperature valve.

Updating a price through time with a cost index

Equipment prices drift with inflation and commodity markets, so a quoted historical cost must be escalated to the year of your estimate using a cost index. The handbook gives the ratio form directly: multiply the old cost by the ratio of the current index to the index in the year of the original quote. The Chemical Engineering Plant Cost Index (CEPCI, base

1957\text{-}59 = 100

) is the usual choice for process plants; the Marshall & Swift index is an alternative. Always escalate before you scale or factor — the index moves a price in time only, never in size.

C_{2} = C_{1}\left(\dfrac{I_{2}}{I_{1}}\right)

Scaling cost with capacity: the six-tenths rule

Equipment cost rises with capacity, but less than proportionally, because a vessel's cost tracks surface (steel) while its duty tracks volume. The handbook states it as a power law: the cost ratio equals the capacity ratio raised to an exponent

n

. When the specific exponent is unknown, the default is

n \approx 0.6

— the famous six-tenths rule, which says doubling capacity raises cost by only

2^{0.6} \approx 1.52

. Real exponents vary (shell-and-tube floating-head exchanger

0.60

, centrifugal fan

0.44

, vacuum drum dryer

0.76

); use the tabulated value when given,

0.6

otherwise. The rule is an interpolation tool — do not extrapolate it across an order of magnitude in size or through a change of equipment type.

\dfrac{C_{A}}{C_{B}} = \left(\dfrac{S_{A}}{S_{B}}\right)^{n}, \qquad n \approx 0.6

From bare equipment to installed plant: Lang factors

The delivered purchased-equipment cost is only a fraction of what a plant costs to build — you must add installation labor, piping, instrumentation, electrical, buildings, service facilities, and the indirect engineering, construction, and contingency. The Lang factor short-cuts all of that with one multiplier on the total delivered equipment cost. The handbook gives factors by plant type: solid-processing plant (

4.0

fixed-capital,

4.7

total-capital), solid-fluid (

4.3

5.0

), and fluid-processing (

5.0

6.0

). A more detailed module method multiplies each item's purchased cost by its own bare-module factor

F_{BM}

and sums, which captures that exotic-alloy or high-pressure items cost more to install relative to a carbon-steel base.

C_{\text{FCI}} = f_{L}\, C_{p,\text{delivered}}, \qquad C_{\text{TCI}} = C_{\text{FCI}} + C_{\text{WC}}

Estimate classes and what they are worth

An estimate is only as good as its definition. An order-of-magnitude (Class 5) estimate from capacity scaling and a Lang factor carries roughly

\pm 30\text{ to }50\%

uncertainty and is used to screen ideas. A study or preliminary estimate (

\pm 20\text{ to }30\%

) uses the factored module method on a firm equipment list. A definitive or detailed estimate (

\pm 5\text{ to }15\%

) waits for quotes and detailed P&IDs. The exam mostly lives in the order-of-magnitude and factored world: you will be handed an equipment cost or capacity and asked to escalate, scale, and factor it — never to produce a contractor-grade bid.

Exam strategy

Do the three cost operations in a fixed order: (1) escalate the base price to your year with the index ratio

I_{2}/I_{1}

; (2) scale to your capacity with

(S_{A}/S_{B})^{n}

, using the tabulated

n

0.6

; (3) factor up to installed or fixed-capital cost with the Lang factor. Mixing the order is harmless when you only escalate-and-scale, but never apply the index to a capacity ratio or vice versa. Confirm the exponent: if a question gives a table value, use it; the bare

0.6

is the fallback. For the drawing questions, sort by content — stream data and major equipment is a PFD, while valves, instrument bubbles, line specs, and control loops are a P&ID. Decode an instrument tag letter by letter: first letter is the variable, the rest are the function.

Key equations

Cost index update

C_{2} = C_{1}\left(\dfrac{I_{2}}{I_{1}}\right)

Escalate a past cost $C_1$ (year-1 index $I_1$ ) to a present cost $C_2$ (current index $I_2$ ). CEPCI base $1957\text{-}59=100$ . Time only — never size.

Capacity scaling (power law)

\dfrac{C_{A}}{C_{B}} = \left(\dfrac{S_{A}}{S_{B}}\right)^{n}

$S$ is a capacity measure (area, flow, volume, power); $n$ is the cost-capacity exponent. Interpolate within a size class only.

Six-tenths rule

C_{A} = C_{B}\left(\dfrac{S_{A}}{S_{B}}\right)^{0.6}

Default when the specific exponent is unknown: cost rises as capacity to the $0.6$ power.

Back-out the exponent

n = \dfrac{\ln\!\left(C_{2}/C_{1}\right)}{\ln\!\left(S_{2}/S_{1}\right)}

Fit $n$ from two cost-capacity data points; slope of the cost line on log-log axes.

Combined escalate-and-scale

C = C_{B}\left(\dfrac{S_{A}}{S_{B}}\right)^{n}\left(\dfrac{I_{2}}{I_{1}}\right)

Move a base cost across both size and time in one expression.

Lang factor (installed/fixed-capital cost)

C_{\text{FCI}} = f_{L}\,C_{p,\text{delivered}}

$f_L$ : fluid plant $5.0$ (fixed) / $6.0$ (total), solid-fluid $4.3/5.0$ , solid $4.0/4.7$ . $C_p$ = total delivered purchased-equipment cost.

Module (factored) estimate

C_{\text{TM}} = \sum_{i} C_{p,i}\,F_{BM,i}

Each item's purchased cost times its bare-module factor $F_{BM}$ (raises with alloy, pressure); summed for the total module cost.

Total capital investment

C_{\text{TCI}} = C_{\text{FCI}} + C_{\text{WC}}

Fixed capital (built, depreciable) plus working capital (inventory, receivables; typically $\sim 15\%$ of TCI), recovered at end of life.

Worked examples

Scale and escalate a heat exchanger

Problem. A $50\ \text{m}^2$ shell-and-tube exchanger (floating head, carbon steel) was purchased in 2015 for $\$28{,}000$ when CEPCI was $556.8$ . Estimate the present (2022) purchased cost of a geometrically similar $120\ \text{m}^2$ unit; take CEPCI(2022) $= 816$ and the cost-capacity exponent $n = 0.60$ .

Solution. Step 1 — scale to

120\ \text{m}^2

at 2015 prices:

C = 28{,}000\,(120/50)^{0.60} = 28{,}000\,(2.4)^{0.60}

. Since

(2.4)^{0.60} = 1.691

C = \$47{,}346

. Step 2 — escalate to 2022: multiply by the index ratio

816/556.8 = 1.4655

, giving

C = 47{,}346 \times 1.4655 = \$69{,}387

. Final:

\approx \boxed{\$69{,}400}

delivered purchased cost. Sanity check: both operations push the price up (bigger and newer), and the scale exponent

<1

means the cost ratio

1.69

is well below the capacity ratio

2.4

— exactly the economy-of-scale the six-tenths rule encodes.

C = 28{,}000\left(\dfrac{120}{50}\right)^{0.60}\left(\dfrac{816}{556.8}\right) \approx \$69{,}400

Lang-factor capital estimate for a fluid plant

Problem. A new continuous, all-fluid process has a total delivered purchased-equipment cost of $\$1.20$ million. Using Lang factors, estimate the fixed capital investment, the total capital investment, and the implied working capital. Comment on whether the working capital is consistent with the usual $15\%$ guideline.

Solution. Fluid-processing Lang factors:

f_L = 5.0

(fixed capital),

6.0

(total capital). Fixed capital:

C_{\text{FCI}} = 5.0 \times 1.20\,\text{M} = \$6.00\,\text{M}

. Total capital:

C_{\text{TCI}} = 6.0 \times 1.20\,\text{M} = \$7.20\,\text{M}

. Working capital:

C_{\text{WC}} = C_{\text{TCI}} - C_{\text{FCI}} = 7.20 - 6.00 = \$1.20\,\text{M}

. As a fraction of TCI:

1.20/7.20 = 16.7\%

. Final:

C_{\text{FCI}} = \boxed{\$6.00\,\text{M}}

C_{\text{TCI}} = \boxed{\$7.20\,\text{M}}

C_{\text{WC}} = \$1.20\,\text{M}

(

16.7\%

of TCI — consistent with the

\sim 15\%

rule of thumb). Sanity check: TCI exceeds FCI, as it must, and the fluid factor exceeds the solid factor because fluid plants are piping- and instrument-intensive.

Back out a cost-capacity exponent, then predict

Problem. A vendor quotes two centrifugal pumps from the same family: $\$9{,}500$ for $100\ \text{m}^3/\text{h}$ and $\$19{,}800$ for $400\ \text{m}^3/\text{h}$ (same year). Find the cost-capacity exponent $n$ and use it to estimate the cost of a $250\ \text{m}^3/\text{h}$ pump from the same family.

Solution. Exponent from the two points:

n = \dfrac{\ln(19{,}800/9{,}500)}{\ln(400/100)} = \dfrac{\ln(2.084)}{\ln(4)} = \dfrac{0.7345}{1.3863} = 0.530

. Predict at

250\ \text{m}^3/\text{h}

off the

100\ \text{m}^3/\text{h}

point:

C = 9{,}500\,(250/100)^{0.530} = 9{,}500\,(2.5)^{0.530} = 9{,}500 \times 1.625 = \$15{,}436

. Final:

n = \boxed{0.530}

C_{250} \approx \boxed{\$15{,}400}

. Sanity check:

250\ \text{m}^3/\text{h}

lies between the two data points, and the predicted price

\$15{,}400

lies between

\$9{,}500

and

\$19{,}800

as it must. The fitted

n = 0.53

is typical of pumps (below the generic

0.6

Common pitfalls

•Asking a PFD question of a P&ID (or vice versa): stream temperatures, duties, and the material balance live on the PFD; valves, line specs, instrument bubbles, and control loops live on the P&ID. Sort the question by its content first.
•Updating size with a cost index or updating time with the capacity exponent. The index $I_2/I_1$ moves a price only through time; the exponent $(S_A/S_B)^n$ moves it only through size. Apply each once, to the right ratio.
•Using the generic $n = 0.6$ when the problem supplies a tabulated exponent. If the handbook gives $0.44$ for a fan or $0.76$ for a vacuum dryer, use it — the six-tenths value is only the fallback.
•Extrapolating the power law across a huge size change or a change of equipment type. It is an interpolation rule within one equipment family and a modest size range, not a universal scaling law.
•Applying the Lang factor to an installed cost or to a single item. The factor multiplies the TOTAL DELIVERED PURCHASED-EQUIPMENT cost of the whole plant to give fixed (or total) capital — not an already-installed price.
•Confusing fixed capital investment with total capital investment: TCI = FCI + working capital. The Lang total-capital factor (e.g., $6.0$ for fluid plants) is larger than the fixed-capital factor (e.g., $5.0$ ) for exactly this reason.
•Forgetting to escalate an old quote at all. A price from 2010 plugged straight into a present-day estimate can be off by $30\text{–}50\%$ purely from index drift.

References

NCEES FE Reference Handbook — Chemical Engineering (Cost Estimation: cost indexes, scaling of equipment costs, capital cost / Lang factors)
NCEES FE Reference Handbook — Engineering Economics (cost indices, capital recovery, working capital)
Peters, Timmerhaus & West, Plant Design and Economics for Chemical Engineers — source of the six-tenths rule, exponent table, and Lang factors
ISA-5.1 Instrumentation Symbols and Identification — instrument tag letters and P&ID symbol conventions

B. Equipment selection

Equipment Selection, Sizing and Scale-Up

Select and size pumps, exchangers, vessels, and columns, scale a unit up by similarity, pick materials for the service, and tell rating from design.

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D. Process design and optimization

Optimization, Green and Inherently Safer Design + Standards

Frame a design as constrained optimization, find the economic optimum where capital trades against operating cost, and apply green and inherently safer design within ASTM/ISO/OSHA standards.

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