A plant processes \( 9000 \) tonnes in a \( 90 \) day season working \( 20 \) hours a day. Its required capacity is
Answer: A
The daily rate is \( 9000/90 = 100 \) tonnes and the hourly rate is \( 100/20 = 5.0 \) tonnes. Sizing on the annual tonnage spread over a full year would give a plant less than a quarter of this and the crop would be lost.
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The plant above has a rated capacity of \( 5.0 \) t/h and operates \( 1800 \) hours a year against a possible \( 8000 \). Its utilisation factor is
Answer: B
Correct answer: 22.5 per cent
\[ \text{Utilisation} = \frac{1800}{8000} = 0.225 = 22.5\% \]The rated capacity of 5 t/h does not enter the calculation - utilisation compares hours used with hours available, and both the actual and rated outputs share the same hourly rate.
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A line has stages rated at \( 6.0 \), \( 4.5 \), \( 7.0 \) and \( 5.5 \) t/h. The line throughput is
Answer: C
Correct answer: 4.5 t/h
A line can pass no more than its slowest stage will take, so the throughput is 4.5 t/h. The capacity above that in the other three stages is simply unusable - capital earning nothing until the bottleneck is uprated.
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Delivered equipment cost is \( 4.2 \) million and a Lang factor of \( 4.7 \) applies. The fixed capital investment is about
Answer: D
Correct answer: 19.7 million
\[ FCI = 4.2 \times 4.7 = 19.7\ \text{million} \]The factor of 4.7 covers everything beyond the equipment itself - installation, piping, instrumentation, electrical work, buildings, engineering and contingency.
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A plant of \( 10 \) t/h capacity cost \( 25 \) million. Using the six-tenths rule, a \( 20 \) t/h plant would cost about
Answer: A
Correct answer: 37.9 million
\[ C_2 = 25 \times \left(\frac{20}{10}\right)^{0.6} = 25 \times 2^{0.6} = 25 \times 1.516 = 37.9\ \text{million} \]Doubling the capacity costs about 52 per cent more, not twice as much - which is the whole argument for building large.
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Fixed costs are \( 8.0 \) million a year, the selling price is \( 4200 \) per tonne and the variable cost is \( 2900 \) per tonne. The breakeven output is
Answer: B
Correct answer: 6154 tonnes
The contribution per tonne is \(4200 - 2900 = 1300\):
\[ Q_{BE} = \frac{8{,}000{,}000}{1300} = 6154\ \text{tonnes} \]Below this output the fixed costs are not covered; above it, every tonne adds 1300 to profit.
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An investment of \( 30 \) million returns a net cash inflow of \( 6.5 \) million a year. The simple payback period is
Answer: C
Correct answer: 4.6 years
\[ \text{Payback} = \frac{30}{6.5} = 4.6\ \text{years} \]Simple payback ignores what happens after that point and the time value of money, so it is a screening measure rather than a decision rule.
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A feed of \( 5000 \) kg/h yields \( 3200 \) kg/h of product and \( 1800 \) kg/h of waste. The process yield is
Answer: D
Correct answer: 64.0 per cent
\[ \text{Yield} = \frac{3200}{5000} \times 100 = 64.0\% \]The waste stream, 1800 kg/h, accounts for the remaining 36 per cent - and the two streams sum to the feed, as a material balance requires.
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Steam demand is \( 2.4 \), \( 1.8 \), \( 0.9 \) and \( 1.5 \) t/h at four consumers, with a diversity factor of \( 0.75 \). The boiler should be rated at about
Answer: A
Correct answer: 4.95 t/h
Sum the demands and apply the diversity factor:
\[ 2.4 + 1.8 + 0.9 + 1.5 = 6.6\ \text{t/h} \]\[ 6.6 \times 0.75 = 4.95\ \text{t/h} \]Sizing on the raw 6.6 t/h would buy a boiler a third larger than the plant will ever draw at one time.
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A plant runs \( 20 \) hours a day for \( 90 \) days at \( 5.0 \) t/h. Its annual throughput is
Answer: B
Correct answer: 9000 tonnes
\[ 5.0 \times 20 \times 90 = 9000\ \text{tonnes} \]Note how short the operating year is - 1800 hours out of a possible 8000 - which is the seasonal plant’s central economic difficulty.
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