Q: Economic life comparison — large vs. small chemical process plants The economic life of a large chemical process plant, compared with that of a small plant, is typically:

Introduction / Context: Economic life is the optimal operating period that minimizes the equivalent annual cost or maximizes economic return before replacement. It is influenced by obsolescence, maintenance growth, and performance decline rather than purely by physical durability. Given Data / Assumptions: Compare large vs. small chemical plants under typical market and technology conditions. Economic life is defined by cost optimum, not the last day equipment can physically run. All else equal (maintenance practices, product market), scale primarily affects capital intensity. Concept / Approach: Larger plants enjoy economies of scale in capital and operating costs but face similar technological obsolescence and market shifts as smaller units. Because economic life is dominated by obsolescence and maintenance escalation, scaling up size does not proportionally extend the optimum replacement time. Textbook rules of thumb therefore state that large plants have an economic life only slightly longer than small plants, not dramatically longer. Step-by-Step Solution: Identify dominant factors: obsolescence, maintenance cost growth, regulatory changes.Recognize that these are largely independent of absolute size.Infer that any extension in economic life with size is modest, not “much more.”Select the option reflecting a small increase. Verification / Alternative check: Practice examples show economic life often clustering in comparable ranges (e.g., 6–12 years) across sizes for similar technologies. Why Other Options Are Wrong: Much more: overstates the effect of scale on obsolescence. Slightly less / Almost equal: do not reflect the small but positive tendency for larger units to remain competitive slightly longer. Common Pitfalls: Equating physical life or design life with economic life. Assuming economies of scale automatically extend useful economic life greatly. Final Answer: Only slightly more

Q: Depreciation methods and interest on investment Which of the listed depreciation methods explicitly accounts for the time value of money (interest on investment) in its annual depreciation calculation?

Introduction / Context: Depreciation schedules allocate cost for accounting; financial evaluation methods incorporate the time value of money. It is essential to distinguish a book method of depreciation from economic analyses that include interest. Given Data / Assumptions: Straight-line and declining-balance are book depreciation methods. Interest on investment refers to time value of money, not interest expense as a separate line item. We are asked which depreciation method itself embeds interest in the calculation of annual depreciation charges. Concept / Approach: Straight-line yields a constant annual charge equal to (P − S)/n without discounting. Declining-balance applies a rate to book value; again, it does not discount future charges to present worth. Methods that explicitly include interest are evaluation tools (e.g., sinking fund planning, present-worth and capital recovery factors), not the two book methods listed. Therefore, neither straight-line nor declining-balance includes interest within the depreciation formula. Step-by-Step Solution: Write SL formula: (P − S)/n → no i (interest) term.Write DB concept: Dep(r) = rate * book_value(r−1) → no discounting.Conclude that neither method embeds interest. Verification / Alternative check: Plant economics texts separate book depreciation schedules from discounted cash flow analyses where interest appears in NPV/IRR or capital recovery factor. Why Other Options Are Wrong: Straight-line or declining-balance alone do not feature an interest rate variable. Choosing both would be incorrect for the same reason. Common Pitfalls: Confusing the use of interest in cash-flow evaluations with its absence in standard book depreciation methods. Final Answer: Neither (a) nor (b)

Q: Working capital in process industries Which of the following is a component of working capital investment for a chemical plant?

Introduction / Context: Working capital is the money tied up in day-to-day operations. Correctly classifying costs as capital (fixed) versus working capital is essential for cash-flow planning and profitability analyses. Given Data / Assumptions: Working capital includes items needed to run the plant: inventories, cash, and receivables. Fixed capital covers long-lived assets such as equipment and utilities plants. Depreciation is a non-cash expense. Concept / Approach: Maintenance and repair inventory (spare parts, consumables) is a current asset required to sustain operations and is part of working capital. Utilities plants and process equipment are part of fixed capital investment. Depreciation is an accounting expense reflecting the allocation of fixed capital cost over time and is not itself an investment in working capital. Step-by-Step Solution: List working capital components: raw material, WIP, finished goods, spares, cash, receivables.Identify which option is a current asset: maintenance and repair inventory (spares).Exclude utilities plants and process equipment (fixed capital) and depreciation (non-cash expense). Verification / Alternative check: Standard cost breakdowns (fixed vs. working capital) place spares within inventories under current assets. Why Other Options Are Wrong: Utilities plants: fixed capital asset. Process equipment: fixed capital. Depreciation: accounting charge, not an asset or cash investment. Common Pitfalls: Counting depreciation as an operational cash outlay (it is a non-cash expense affecting taxes). Misclassifying capital projects (utilities, equipment) as working capital. Final Answer: Maintenance and repair inventory (spares)

Q: Optimum economic pipe diameter for fluid is determined by the

total cost considerations (pumping cost plus fixed cost of the pipe).

Question 1

Taxation concepts for cost estimation: Which type of tax is typically assessed on gross production or sales rather than on net income?

Accepted Answer

Introduction / Context: Cost estimators must distinguish taxes that depend on income versus those tied to production or sales. This affects cash flows, pricing, and profitability analyses. Given Data / Assumptions: Gross earnings here refer to gross sales/production basis, not profits. General definitions applied across jurisdictions. Concept / Approach: Excise taxes are levies on the production, sale, or consumption of specific goods (e.g., fuels, alcohol). They are assessed on units or value sold/produced, not on net income after expenses. Step-by-Step Solution: Identify the base: gross production/sales rather than profit.Map to tax type: excise taxes fit this description.Confirm other taxes target different bases (property value, net income, capital appreciation). Verification / Alternative check: Tax guidelines list excise as transaction/commodity-based; income tax is profit-based after deductions. Why Other Options Are Wrong: Property: based on assessed property value. Income: based on net income. Capital gains: based on gain from asset sales. Common Pitfalls: Equating sales tax and excise; both are gross-basis but excise targets specific goods and may be unit-based. Final Answer: Excise tax

Question 2

Utility cost share in chemical manufacturing: For ordinary chemical process plants, utilities (compressed air, steam, water, power, fuel gases, etc.) typically account for approximately what percentage of total product cost?

Accepted Answer

Introduction / Context: Utility costs are an important component of manufacturing cost but, for many ordinary chemical plants, they are a modest share compared with raw materials and labor/overheads. Given Data / Assumptions: Typical continuous chemical processes (non-energy-intensive specialty products). Normal energy prices and efficient operations. Concept / Approach: Rule-of-thumb cost breakdowns commonly place utilities in the low single-digit percent range of total product cost, unless the process is very energy intensive (e.g., electrolytic or high-temperature furnaces). Step-by-Step Solution: Compare ranges: 1–5% aligns with ordinary plants.Larger ranges (10–45%) are typical only for energy-intensive processes or unusual utility price regimes. Verification / Alternative check: Published cost tables in process design texts show utilities often at a few percent for typical chemical production lines. Why Other Options Are Wrong: 10–45% substantially overstates typical utility shares for ordinary plants. Common Pitfalls: Confusing total manufacturing cost with energy-intensive industries; always consider process specifics. Final Answer: 1% to 5%

Question 3

Profitability measure — simple rate of return: A project costs ₹10,00,000 (total investment) and earns ₹1,50,000 profit each year for 10 years. What is the simple rate of return on investment?

Accepted Answer

Introduction / Context: The simple rate of return (accounting rate of return) is a quick screening metric defined as annual profit divided by total investment, expressed as a percentage. It ignores time value but is frequently used for preliminary comparisons. Given Data / Assumptions: Total investment (cost) = ₹10,00,000. Annual profit = ₹1,50,000 (assumed uniform). Project life = 10 years (not required for the simple rate). Concept / Approach: Simple rate of return = (Annual profit / Total investment) * 100%. Do not discount cash flows for this measure. Step-by-Step Solution: Compute ratio: 1,50,000 / 10,00,000 = 0.15Convert to percent: 0.15 * 100% = 15% Verification / Alternative check: If payback is considered, payback time ≈ 10,00,000 / 1,50,000 ≈ 6.67 years, which is consistent with a 15% per-year profit ratio (but payback is a different metric). Why Other Options Are Wrong: 10% and 1.5% use incorrect denominators or decimal placement; 150% confuses ratio with percent. Common Pitfalls: Using cash inflow instead of profit, or mixing after-tax and before-tax figures without consistency. Final Answer: 15%

Question 4

Working capital proportion: For most chemical plants (excluding strongly seasonal products), the ratio of working capital to total capital investment typically lies in which range?

Accepted Answer

Introduction / Context: Working capital finances inventories, receivables, and operating cash. Its ratio to total capital investment is a standard benchmark in project cost estimation. Given Data / Assumptions: Non-seasonal chemical manufacturing. Normal credit and inventory practices. Concept / Approach: Rules of thumb place working capital commonly around one-tenth to one-fifth of total capital investment for typical chemical plants, reflecting inventory cycles and receivables policies. Step-by-Step Solution: Identify realistic range: 10–20% for most non-seasonal plants.Eliminate extremes: below 2% is too small; 50–60% is atypically high. Verification / Alternative check: Design handbooks show typical working capital estimates of 15% ± for steady, continuous operations. Why Other Options Are Wrong: 0.1–2% underestimates; 50–60% grossly overstates normal needs. Common Pitfalls: Ignoring special cases (seasonal demand, long receivables) which can push the ratio higher. Final Answer: 10% to 20%

Question 5

Plant cost breakdown: Construction expenses are roughly what percentage of the total direct cost of a chemical plant (rule of thumb)?

Accepted Answer

Introduction / Context: Estimating a plant's total erected cost involves apportioning direct costs (equipment, installation, piping, instrumentation, buildings) and indirects. Construction expenses are one component within direct costs. Given Data / Assumptions: Typical chemical process facility. Order-of-magnitude or study estimate stage. Average construction complexity. Concept / Approach: Rule-of-thumb fractions are used when detailed quotes are unavailable. Construction expenses (temporary facilities, field supervision, small tools, etc.) are often on the order of ten percent of total direct cost in many estimating guides. Step-by-Step Solution: Consult standard breakdowns: construction expenses commonly ~10% of total direct cost.Exclude extremes: 2% is too low; 30–50% would imply unusual site conditions or scope. Verification / Alternative check: Cross-check with past projects or estimating handbooks showing construction overheads in the single-digit to low-teens percent range. Why Other Options Are Wrong: 2% understates typical field costs; 30% and 50% are excessive for ordinary projects. Common Pitfalls: Confusing construction expenses (a subset of direct cost) with overall indirect costs or contractor's fee. Final Answer: 10%

Question 6

Cost components and typical percentages: Identify the wrong statement about annual percentages used in chemical plant cost estimates.

Accepted Answer

Introduction / Context: Heuristic percentages are widely used for quick cost appraisals. Recognizing which rules are reasonable and which are unrealistic prevents large estimation errors in feasibility studies. Given Data / Assumptions: Percents refer to typical annualized benchmarks. Values are order-of-magnitude guides, not exact for every plant. Concept / Approach: Typical guides suggest single-digit insurance, low single-digit building depreciation, and equipment depreciation near 10% per year (assuming ~10-year straight-line life). However, R&D as 15% of NSR is far too high for most chemical plants; it is usually much lower (often < 5%, commonly 1–3% for many commodity operations). Step-by-Step Solution: Evaluate each statement against common heuristics.(a) 10% equipment depreciation: reasonable order-of-magnitude.(b) Buildings at ~3%: plausible.(c) Insurance ≈1% of FCI: plausible.(d) R&D = 15% of NSR: unrealistically high for chemical plants. Verification / Alternative check: Industry summaries typically place R&D for process industries in low single digits of sales, much less than 15%. Why Other Options Are Wrong: The other percentages are reasonable heuristics for early estimates; exact values depend on asset lives and insurance markets. Common Pitfalls: Assuming high-tech or pharma R&D percentages apply to commodity chemicals; they do not. Final Answer: Research and development costs are about 15% of net sales realisation (NSR).

Question 7

Plant economics — when are annual depreciation charges non-constant? Annual depreciation cost is not constant under which depreciation calculation method commonly used in chemical engineering economics?

Accepted Answer

Introduction / Context: Depreciation spreads the depreciable value of equipment over its useful life for accounting, taxes, and project evaluation. Different methods alter cash flows and profitability projections. A frequent exam point is identifying which methods yield constant year-to-year charges and which do not. Given Data / Assumptions: We compare common methods: straight-line, sinking fund, present worth (time-value analysis approach), and declining balance. Annual depreciation cost here means the booked depreciation expense each year, not accumulated depreciation. No special salvage timing effects are assumed beyond the usual definitions. Concept / Approach: In the straight-line method, the annual charge equals (P − S)/n, which is constant. In the declining-balance method, a fixed percentage is applied to the book value, so the first year is largest and charges reduce each year. The sinking-fund framework explicitly incorporates interest earnings on the fund and yields increasing annual deposits rather than a constant depreciation expense. Present-worth analysis discounts cash flows for comparison but is not itself an annual depreciation schedule. Of the listed choices, the hallmark non-constant pattern identified in cost accounting questions is the declining balance method. Step-by-Step Solution: State what is constant in straight-line: Dep = (P − S)/n each year (constant).Recall declining balance: Dep(r) = rate * book_value(r−1) → decreases annually.Recognize sinking fund: annual deposit typically escalates with interest assumptions; not a constant charge.Conclude the widely accepted non-constant schedule among standard accounting methods is the declining balance method. Verification / Alternative check: Example tables in plant economics texts show declining-balance charges descending year by year, while straight-line remains flat. Why Other Options Are Wrong: Straight-line method: produces a constant annual charge. Sinking fund method: focuses on building a fund; not the typical depreciation expense schedule used in this context. Present worth method: an evaluation technique, not an annual depreciation formula. Common Pitfalls: Confusing time-value evaluation methods (present worth) with depreciation schedules. Assuming all methods yield constant yearly expenses. Final Answer: Declining balance (diminishing balance) method

Question 8

Economic life comparison — large vs. small chemical process plants The economic life of a large chemical process plant, compared with that of a small plant, is typically:

Accepted Answer

Introduction / Context: Economic life is the optimal operating period that minimizes the equivalent annual cost or maximizes economic return before replacement. It is influenced by obsolescence, maintenance growth, and performance decline rather than purely by physical durability. Given Data / Assumptions: Compare large vs. small chemical plants under typical market and technology conditions. Economic life is defined by cost optimum, not the last day equipment can physically run. All else equal (maintenance practices, product market), scale primarily affects capital intensity. Concept / Approach: Larger plants enjoy economies of scale in capital and operating costs but face similar technological obsolescence and market shifts as smaller units. Because economic life is dominated by obsolescence and maintenance escalation, scaling up size does not proportionally extend the optimum replacement time. Textbook rules of thumb therefore state that large plants have an economic life only slightly longer than small plants, not dramatically longer. Step-by-Step Solution: Identify dominant factors: obsolescence, maintenance cost growth, regulatory changes.Recognize that these are largely independent of absolute size.Infer that any extension in economic life with size is modest, not “much more.”Select the option reflecting a small increase. Verification / Alternative check: Practice examples show economic life often clustering in comparable ranges (e.g., 6–12 years) across sizes for similar technologies. Why Other Options Are Wrong: Much more: overstates the effect of scale on obsolescence. Slightly less / Almost equal: do not reflect the small but positive tendency for larger units to remain competitive slightly longer. Common Pitfalls: Equating physical life or design life with economic life. Assuming economies of scale automatically extend useful economic life greatly. Final Answer: Only slightly more

Question 9

Depreciation methods and interest on investment Which of the listed depreciation methods explicitly accounts for the time value of money (interest on investment) in its annual depreciation calculation?

Accepted Answer

Introduction / Context: Depreciation schedules allocate cost for accounting; financial evaluation methods incorporate the time value of money. It is essential to distinguish a book method of depreciation from economic analyses that include interest. Given Data / Assumptions: Straight-line and declining-balance are book depreciation methods. Interest on investment refers to time value of money, not interest expense as a separate line item. We are asked which depreciation method itself embeds interest in the calculation of annual depreciation charges. Concept / Approach: Straight-line yields a constant annual charge equal to (P − S)/n without discounting. Declining-balance applies a rate to book value; again, it does not discount future charges to present worth. Methods that explicitly include interest are evaluation tools (e.g., sinking fund planning, present-worth and capital recovery factors), not the two book methods listed. Therefore, neither straight-line nor declining-balance includes interest within the depreciation formula. Step-by-Step Solution: Write SL formula: (P − S)/n → no i (interest) term.Write DB concept: Dep(r) = rate * book_value(r−1) → no discounting.Conclude that neither method embeds interest. Verification / Alternative check: Plant economics texts separate book depreciation schedules from discounted cash flow analyses where interest appears in NPV/IRR or capital recovery factor. Why Other Options Are Wrong: Straight-line or declining-balance alone do not feature an interest rate variable. Choosing both would be incorrect for the same reason. Common Pitfalls: Confusing the use of interest in cash-flow evaluations with its absence in standard book depreciation methods. Final Answer: Neither (a) nor (b)

Question 10

Working capital in process industries Which of the following is a component of working capital investment for a chemical plant?

Accepted Answer

Introduction / Context: Working capital is the money tied up in day-to-day operations. Correctly classifying costs as capital (fixed) versus working capital is essential for cash-flow planning and profitability analyses. Given Data / Assumptions: Working capital includes items needed to run the plant: inventories, cash, and receivables. Fixed capital covers long-lived assets such as equipment and utilities plants. Depreciation is a non-cash expense. Concept / Approach: Maintenance and repair inventory (spare parts, consumables) is a current asset required to sustain operations and is part of working capital. Utilities plants and process equipment are part of fixed capital investment. Depreciation is an accounting expense reflecting the allocation of fixed capital cost over time and is not itself an investment in working capital. Step-by-Step Solution: List working capital components: raw material, WIP, finished goods, spares, cash, receivables.Identify which option is a current asset: maintenance and repair inventory (spares).Exclude utilities plants and process equipment (fixed capital) and depreciation (non-cash expense). Verification / Alternative check: Standard cost breakdowns (fixed vs. working capital) place spares within inventories under current assets. Why Other Options Are Wrong: Utilities plants: fixed capital asset. Process equipment: fixed capital. Depreciation: accounting charge, not an asset or cash investment. Common Pitfalls: Counting depreciation as an operational cash outlay (it is a non-cash expense affecting taxes). Misclassifying capital projects (utilities, equipment) as working capital. Final Answer: Maintenance and repair inventory (spares)

Question 11

The amount of compounded interest during 'n' interest periods is

Accepted Answer

p[(1+i)n-1)]

Question 12

Fixed capital investment of a chemical plant is the total amount of money needed to supply the necessary plant and manufacturing facilities plus the working capital for operation of the facilities. Which of the following components of fixed capital …

Accepted Answer

Equipment insulation cost.

Question 13

Optimum economic pipe diameter for fluid is determined by the

Accepted Answer

total cost considerations (pumping cost plus fixed cost of the pipe).

Question 14

A reactor having a salvage value of Rs. 10000 is estimated to have a service life of 10 years. The annual interest rate is 10%. The original cost of the reactor was Rs. 80000. The book value of the reactor after 5 years using sinking fund depreciati…

Accepted Answer

60196

Question 15

The value of a property decreases __________ with time in straight line method of determining depreciation.

Accepted Answer

linearly

Question 16

A present sum of Rs. 100 at the end of one year, with half yearly rate of interest at 10%, will be Rs.

Accepted Answer

121

Question 17

The amount of simple interest during 'n' interest period is (where, i = interest rate based on the length of one interest period, p = principal)

Accepted Answer

p.i.n.

Question 18

Factory manufacturing cost is the sum of the direct production cost

Accepted Answer

fixed charges and plant overhead cost.

Question 19

Total product cost of a chemical plant does not include the __________ cost.

Accepted Answer

market survey

Chemical Engineering Plant Economics Questions