Process Design · Study · FE Chemical · FE → PE Prep
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
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-101A/B (pump 101, with installed spare B) or E-204 (exchanger). You are not asked to memorize the whole standard, but you must be able to decode a loop like TT→TIC→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 1957-59=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.
C2=C1(I1I2)
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≈0.6 — the famous six-tenths rule, which says doubling capacity raises cost by only 20.6≈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.
CBCA=(SBSA)n,n≈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 FBM and sums, which captures that exotic-alloy or high-pressure items cost more to install relative to a carbon-steel base.
CFCI=fLCp,delivered,CTCI=CFCI+CWC
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 ±30 to 50% uncertainty and is used to screen ideas. A study or preliminary estimate (±20 to 30%) uses the factored module method on a firm equipment list. A definitive or detailed estimate (±5 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 I2/I1; (2) scale to your capacity with (SA/SB)n, using the tabulated n or 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 updateC2=C1(I1I2)
Escalate a past cost C1 (year-1 index I1) to a present cost C2
Lang factor (installed/fixed-capital cost)CFCI=fLCp,delivered
Module (factored) estimateCTM=∑iCp,iFBM,i
Total capital investmentCTCI=CFCI+CWC
Worked examples
Scale and escalate a heat exchanger
Problem. A 50m2 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 120m2 unit; take CEPCI(2022) =816 and the cost-capacity exponent n=0.60.
Solution. Step 1 — scale to 120m2 at 2015 prices: C=28,000(120/50)0.60=28,000(2.4)0.60
C=28,000(50120)0.60(556.8816)≈$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: fL=5.0 (fixed capital), 6.0
Back out a cost-capacity exponent, then predict
Problem. A vendor quotes two centrifugal pumps from the same family: $9,500 for 100m3/h and $19,800 for 400m3/h
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 I2/I1 moves a price only through time; the exponent (SA/SB)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 30–50% 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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(current index
I2
). CEPCI base
1957-59=100
. Time only — never size.
S is a capacity measure (area, flow, volume, power); n is the cost-capacity exponent. Interpolate within a size class only.
Default when the specific exponent is unknown: cost rises as capacity to the 0.6 power.
Fit n from two cost-capacity data points; slope of the cost line on log-log axes.
Move a base cost across both size and time in one expression.
Each item's purchased cost times its bare-module factor FBM (raises with alloy, pressure); summed for the total module cost.
Fixed capital (built, depreciable) plus working capital (inventory, receivables; typically ∼15% of TCI), recovered at end of life.
. 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=47,346×1.4655=$69,387
.
Final:
≈$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.
(total capital).
Fixed capital:
CFCI=5.0×1.20M=$6.00M
.
Total capital:
CTCI=6.0×1.20M=$7.20M
.
Working capital:
CWC=CTCI−CFCI=7.20−6.00=$1.20M
.
As a fraction of TCI:
1.20/7.20=16.7%
.
Final:
CFCI=$6.00M
,
CTCI=$7.20M
,
CWC=$1.20M
(
16.7%
of TCI — consistent with the
∼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.
(same year). Find the cost-capacity exponent
n
and use it to estimate the cost of a
250m3/h
pump from the same family.
Solution. Exponent from the two points: n=ln(400/100)ln(19,800/9,500)=ln(4)ln(2.084)=1.38630.7345=0.530.
Predict at 250m3/h off the 100m3/h point: C=9,500(250/100)0.530=9,500(2.5)0.530=9,500×1.625=$15,436.
Final: n=0.530, C250≈$15,400. Sanity check: 250m3/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).