The document discusses several questions related to evaluating investment options for chemical plants based on their economics. For the first question, four heat exchanger designs to recover lost heat from waste gases are evaluated based on their initial cost, operating costs, heat savings value, and annual savings. Design 4 is determined to have the highest return on incremental investment. The second question evaluates two investment alternatives based on their rate of return, minimum payback period, and capitalized cost to determine which meets the company's investment criteria. The third question analyzes two reactor options based on their initial cost, maintenance costs, salvage value, service time, yield, and profit to determine which meets the ROI condition and maximum 3 year pay

1. Economics of Chemical Plants
Eng. Kareem H. Mokhtar

2. Rules
• ROI=
𝑃𝑟𝑜𝑓𝑖𝑡
𝑇𝑜𝑡𝑎𝑙 𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐼𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡
• ROII=
𝑖𝑛𝑐𝑟𝑒𝑚𝑒𝑛𝑡𝑎𝑙 𝑃𝑟𝑜𝑓𝑖𝑡
𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑛 𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐼𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡
• Minimum Payout Period =
𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑏𝑙𝑒 𝐹𝑖𝑥𝑒𝑑 𝐼𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡
𝑁𝑒𝑡 𝑐𝑎𝑠ℎ 𝑓𝑙𝑜𝑤
• Capitalized cost=
𝐶𝑅 1+𝑖 𝑛
1+𝑖 𝑛−1
+ 𝑉𝑠 +
𝑎𝑛𝑛𝑢𝑎𝑙 𝑐𝑎𝑠ℎ 𝑒𝑥𝑝𝑒𝑛𝑠𝑒𝑠
𝑖
+ 𝑤𝑜𝑟𝑘𝑖𝑛𝑔

3. Question 1
• An existing plant has been operating in such a way that a large
amount of heat is being lost in the waste gases. It has been proposed
to save money by recovering the heat that is now being lost. Four
different heat exchangers have been designed to recover the heat,
and all prices, costs, and savings have been calculated for each of the
designs. The results of these calculations are presented in the
following:

4. Design No. 1 No. 2 No. 3 No. 4
Total initial installed cost 10000 16000 20000 26000
Operating cost $/yr 100 100 100 100
Fixed charges, % of initial cost/yr 20 20 20 20
Value of heat saved $/year 4100 6000 6900 8850

5. • Annual saving= heat saved – operating cost – fixed cost
• For No.1: annual saving = 4100- 100 – (0.2*10000) = 2000$
• For No.2: annual saving = 6000 – 100- (0.2*16000) = 2700$
• For No.3: annual saving = 6900 – 100- (0.2*20000) = 2800$
• For No.4: annual saving = 8850 – 100 – (0.2*26000) = 3550$
Design No. 1 No. 2 No. 3 No. 4
Total initial installed cost 10000 16000 20000 26000
Operating cost $/yr 100 100 100 100
Fixed charges, % of initial cost/yr 20 20 20 20
Value of heat saved $/year 4100 6000 6900 8850

6. • Return on incremental investment can be calculated by dividing the
additional profit from the new method by the increase in cost to get
the new method.
• Return on incremental investment from 1 to 2
• (2700-2000)/ (16000-10000) = 11.7% (accepted)
• From 2 to 3 = 2.5% (refused)
• From 2 to 4 = 850/10000 = 8.5% (refused)

7. Question 2
• A company has two alternative investments which are being
considered Company policies, based on the current economic
situation, dictate a minimum annual return on the original investment
of 13% after taxes must be predicted for any unnecessary investment
with interest on investment not included as a cost. Company policies
also dictate that, where applicable, straight-line depreciation is used,
for time-value of money interpretation, end of year and profit analysis
is used in land value and prestart up costs can be ignored. Given the
following data:

8. Question 2
• Determine and comment which investment, if any, should be made by
alternative –analysis profitability- evaluation methods of:
• (i) Rate of return on initial investment
(ii) Minimum payout period with no interest charge
(iii)Capitalized cost
Investment Inintial cost
$
Working
cost. $
Salvage
Value $
Service life
years
Anuual cash
flow $
Annual
Expenses $
1 360,000 24,000 32,000 7 110,000 60,000
2 450,000 40,000 50,000 8 120,000 44,000

9. Answer - Rate of return on investment
ROII=
𝑃𝑟𝑜𝑓𝑖𝑡
𝑇𝑜𝑡𝑎𝑙 𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐼𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡
Profit= cash flow –depreciation = 110000-[(360000-32000)/7] = 63142.9$
First investment:
63142.9
360000+24000
=16%
Second investment : 14.3%
Investment Inintial cost
$
Working
cost. $
Salvage
Value $
Service life
years
Anuual cash
flow $
Annual
Expenses $
1 360,000 24,000 32,000 7 110,000 60,000
2 450,000 40,000 50,000 8 120,000 44,000

10. Answer - Minimum payout period with no
interest charge
• First alternative:
360000−32000
110000
= 2.98 𝑦𝑒𝑎𝑟𝑠
• Second alternative:
450000−50000
120000
= 3.33
• First option is better
Minimum Payout Period =
𝐷𝑒𝑝𝑟𝑒𝑐𝑖𝑎𝑏𝑙𝑒 𝐹𝑖𝑥𝑒𝑑 𝐼𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡
𝑁𝑒𝑡 𝑐𝑎𝑠ℎ 𝑓𝑙𝑜𝑤
Investment Inintial cost
$
Working
cost. $
Salvage
Value $
Service life
years
Anuual cash
flow $
Annual
Expenses $
1 360,000 24,000 32,000 7 110,000 60,000
2 450,000 40,000 50,000 8 120,000 44,000

11. Answer – Capitalized cost
• 1st: capitalized=
328000 1+0.13 7
1+0.13 7−1
+32000 +
60000
0.13
+ 24000 =1088033.4$
• 2nd: capitalized=
400000 1+0.13 8
1+0.13 8−1
+50000 +
44000
0.13
+ 40000 =1069651.4$
Capitalized cost=
𝐶𝑅 1+1 𝑛
1+𝑖 𝑛−1
+ 𝑉𝑠 +
𝑎𝑛𝑛𝑢𝑎𝑙 𝑐𝑎𝑠ℎ 𝑒𝑥𝑝𝑒𝑛𝑠𝑒𝑠
𝑖
+ 𝑤𝑜𝑟𝑘𝑖𝑛𝑔

12. Question 3
• Two reactors are considered for a company as shown below, for an
effictive annual intrest of 20%, analyze which one of the two reactors
should be bought to fullfill the ROI conditions and maximum payback
time of 3 years
Item Reactor 1 Reactor 2
Initial Cost 25000 15000
Maintenance 2000 4000
Salvage value 3000 -
Sevice Time 4 6
Yield 70% 50%
Feed amount 100 Kg
Pricing of product
300 $

13. Question 3
• Reactor 1:
• Profit = 21000- [2000+((25000-3000)/4)] = 13500$
• ROI = 54%
• Reactor 2:
• Profit = 15000 - [4000+(15000/6)] = 8500$
• ROI = 56%
• Both reactors fulfills the ROI conditions
• ROII: 50% so Reactor one is better in case of return on incremental
investment
• The pay back of both reactors is less than 3 years
• So reactor one is better

14. Question 4
• A plant needs to increase its production by either of two alternatives,
expansion or building a new facility.
• The expansion would cost 130000$, labor cost will increase to
150000$, and regarding the overheads, taxes, depreciation and
insurance will be 60000$ per year.
• Second Alternative will have an initial cost of 200000, labor cost of
120000 and overheads of 70000, the annual insurance and taxes
would be 2% of the initial investment, all other costs except
depreciation is equal. For ROII of minimum 9%, what is the minimum
service life of the new plant to meet this ROII.

15. Question 5
• A plant is producing 20000 t/y of a product. The overall yield is 70% on a
mass basis (kg of product per kg raw materials). The raw material costs
$20/t and the product sell for $38/t. A process modification has been
advised that will increase the yield 80% of the original value. The additional
investment required is $3,000,000, the additional costs are negligible, and
the salvage value is 100,000$
• a) If the same amount of raw material will be used a. Determine ROII and
payback time? b. Is the modification worth making?
• b) If the second investment will increase the raw material consumption by
50% a. Determine ROII and payback time?
b. Is the modification worth making?
• Minimum ROII % is 19%

16. • A] In the case of additional investment
• Production= 20000 * 1.8 = 36000 t/y
A) ROII = profit/ additional investment = [(36000-20000)*38]/ 3000000
= 20 %
• Pay back time = (3000000-100,000)/796571.4 = 3.64 year
• B) Yes it is worth making as ROII > 19%
• B] A) ROII = profit/ additional investment = [(36000-20000)*38]-
[0.5*(36000/0.7)*20]/ 3,000,000= 18.3
• Pay back Time= 2900000/ 757142.8571 = 8.54 years

17. Be prepared for a quiz next time