Process Economics: Yield, Rate and Cost

Introduction

Industrial chemistry is judged by how reliably it makes useful material, not by yield alone. A route with nearly complete conversion may require an enormous reactor, expensive separation or excessive energy. Another route may sacrifice some single-pass yield yet make product faster and more cheaply through recycle. Understanding these choices begins with three linked quantities: rate of acceptable production, resources consumed, and the equipment required to sustain the process.

Core explanation

Capital cost is paid to build or modify a plant: reactors, compressors, separators, piping, instruments and installation. Operating cost recurs as the plant runs: feedstocks, utilities, catalyst replacement, labour, maintenance, waste treatment and quality losses. A full economic evaluation spreads capital over years and accounts for financing, depreciation, taxes and downtime; a course calculation often uses a simpler annualised capital charge. State the accounting boundary so figures can be compared fairly.

Chemical yield is one component of economics. If 100 mol of A enters and 80 mol of desired P leaves on a one-A-unit basis, the feed-based desired yield is 80%. But the remaining A may be recoverable and recycled, while some of the consumed A may form by-products. A high conversion achieved by very long residence time can lower hourly throughput. A high reaction temperature can speed production but raise energy cost or reduce equilibrium yield for an exothermic reaction. The economically useful operating point balances these effects.

Space-time yield measures productivity of reactor volume: STY = mass of acceptable product produced per unit reactor volume per unit time. Suppose reactor X has volume 10 m³ and makes 500 kg h⁻¹; its STY is 50 kg m⁻³ h⁻¹. Reactor Y has volume 5 m³ and makes 300 kg h⁻¹; its STY is 60 kg m⁻³ h⁻¹. Y uses reactor volume more intensively even though X makes more total product per hour. This metric is useful for comparing reactor productivity but cannot by itself rank whole processes. Y might need larger downstream separators, more expensive catalyst or more energy per kilogram.

An illustrative unit-cost calculation makes the tradeoff clearer. Let a plant's annualised capital charge be 2.0 million currency units and its recurring operating cost be 8.0 million. If it sells 5.0 million kg of on-specification product in that year, the simplified cost is (2.0 + 8.0)/5.0 = 2.0 currency units per kg. If production falls to 4.0 million kg while those costs stay fixed, the apparent unit cost rises to 2.5 per kg. In reality, some operating costs vary with output, so the fixed-versus-variable split should be made explicit before a forecast.

Product quality matters. A plant may produce many kilograms chemically, but off-specification material may need reprocessing or disposal. The denominator of an economic cost per kilogram should normally be saleable output, not raw reactor discharge. Similarly, a process with high atom economy may still consume much electricity, water or solvent; no single metric captures every resource.

Costs are time- and location-dependent. Feedstock prices, electricity tariffs, transport and regulations can change, and a claim that one route is “cheaper” without stated assumptions can quickly become obsolete. The durable reasoning is comparative: enumerate material and energy flows, estimate equipment and running costs on the same basis, then test how the conclusion responds to uncertain inputs. This is why engineers run sensitivity analyses rather than treating one cost estimate as a universal truth.

Step-by-step reasoning

1. Define the product specification, process boundary and time period. 2. Compute feed conversion, desired yield and on-specification output separately. 3. Calculate reactor productivity using mass per reactor volume per operating hour. 4. List capital and recurring operating costs, identifying which are fixed and which vary with throughput. 5. Divide total annualised cost by saleable annual output for a simple unit-cost estimate. 6. Compare routes at the same capacity and quality, then vary uncertain assumptions to test robustness.

Visual explanation

Draw three linked boxes: feed and energy enter a process box, acceptable product and waste leave, and a large equipment box sits beneath. Put a stopwatch beside acceptable output to represent production rate. Space-time yield connects that rate to reactor size; unit cost connects all recurring and annualised equipment charges to acceptable output.

Real-world analogy

Two bakeries can use the same recipe but have different business results. One oven may bake more loaves per hour but use more fuel and need more staff; another may be smaller and efficient yet unable to meet total demand. Counting loaves alone is like counting chemical yield. Economic comparison also needs equipment, operating time, energy and sellable quality.

Real-world example

In ammonia synthesis, a lower temperature favours the exothermic equilibrium but slows reaction. A higher temperature can improve rate yet reduce the equilibrium fraction of ammonia. Product removal and recycle help recover value from unreacted nitrogen and hydrogen, but compressors and separators consume energy and capital. The practical operating point therefore reflects rate, yield and process cost together, as illustrated in OpenStax's equilibrium treatment.

Why?

Why might a lower-yield reactor be economically preferred? If it operates much faster, needs less expensive equipment, or allows cheap recovery of unreacted feed, the cost per kilogram of acceptable product can be lower. Yield is essential information, but it becomes a decision only when combined with throughput and resources.

Common misconception

“Higher space-time yield means lower product cost.” STY includes only product rate and reactor volume. It does not include catalyst price, energy, raw material, separation, maintenance or safety systems. A faster reactor can impose larger costs elsewhere in the plant, so compare the entire process boundary.

Worked example

Reactor X makes 500 kg h⁻¹ acceptable product in 10 m³; reactor Y makes 300 kg h⁻¹ in 5 m³. STY(X) = 500/10 = 50 kg m⁻³ h⁻¹ and STY(Y) = 300/5 = 60 kg m⁻³ h⁻¹. Y is more productive per cubic metre, while X has greater total hourly output. If a customer needs 450 kg h⁻¹, one Y cannot meet demand at the stated rate, whereas one X can. No cost ranking follows until capital and operating costs for each complete process are supplied.

Quick check

1. A 4 m³ reactor produces 240 kg of acceptable product each hour. What is its space-time yield and what cost information is still missing? Answer: STY is 60 kg m⁻³ h⁻¹; feed, energy, capital, separation and other costs are still needed.

Exam focus

Write metric definitions with units and identify the process boundary. Do not confuse yield percentage with hourly production rate, or reactor-volume productivity with whole-plant profitability. For a simple cost-per-kilogram calculation, use on-specification output and include the stated annualised capital charge.

Advanced insight

Economic optimisation is constrained. A process may be barred from operating at the mathematically cheapest point by product quality, equipment pressure ratings, emissions limits or safe operating margins. Sensitivity analysis can show a “switching price” at which one feedstock or route becomes preferable. Thus an optimum is conditional on assumptions and constraints, not an intrinsic property of a chemical reaction.

Summary

Capital cost buys and installs equipment; operating cost pays for continued production. Space-time yield measures acceptable product per reactor volume per time, while unit cost relates annualised spending to saleable output. High conversion, high STY and low cost are related but distinct targets. A defensible comparison uses common boundaries, units, product specifications and sensitivity checks.

Practice questions

1. A reactor makes 900 kg h⁻¹ in 15 m³. Find its space-time yield. Answer: STY = 900/15 = 60 kg m⁻³ h⁻¹. 2. Annualised capital cost is 1.5 million units and operating cost is 6.0 million units for 3.0 million kg saleable output. Find simplified unit cost. Answer: (1.5 + 6.0)/3.0 = 2.5 currency units per kg. 3. Why should rejected off-specification product not be included in the denominator of saleable unit cost? Answer: It cannot be sold as the specified product and may need extra processing or disposal; including it would understate the cost per acceptable kilogram. 4. Give one reason a reactor with lower single-pass conversion might still support a competitive plant. Answer: It may operate faster or permit economical recovery and recycle of unreacted feed, lowering overall cost per saleable product.