LIFE CYCLE COST ANALYSIS EXAMPLE

     Nitric acid is produced by the gas phase oxidation of nitric oxide by the following reaction:

4NH₃ + 5O₂ --> 4NO + 6H₂O

     A plant used to produce nitric acid could contain several main materials of construction.  Let's consider the following two:

1.  Inconel 600 (78% Ni, 15% Cr, and 7% Fe) will perform fair to good in the conditions

2.  Cr Type Stainless Steel will perform normally excellent in the conditions

From this point on we will refer to the first material as I-600 and the second as SS.  

     Now we make the following assumptions about the nitric acid plant to be constructed:

• Plant produces 50 million gallon NO/year

• Capital Cost of plant with I-600 = $37 million

• Capital Cost of plant with SS = $40 million

• Plant Life = 30 years

• Real interest rate = 8.5 %

• Nitric acid sells for $0.85/gallon (A grade), $0.73/gallon (B grade), $0.65/gallon (C grade)

• The market for nitric acid is considered stable over the life of the plant

     The big difference could be seen in the annual operating costs.  Consider the following; I-600 was previously used in a similar plant and the outcome was as follows:

All I-600 material that came into contact with the nitric acid in the downstream (separation) section of the plant had to be replaced after 10 years of operation.  This plant was designed during a time when the selling price of nitric acid was lower and economics of the time were different so the use of the I-600 material was considered to be a "good decision" and after the material was replaced, the plant actually saved only $100,000 by using the lower grade material.  Assuming that the same scenario were to happen in the new plant (replacement cost of material = $7.5 million, 3 weeks of lost production, 2 days of C grade production, and 1 day of B grade production before A grade production is necessary), what would be the outcome now?

3 weeks of lost production = (50 million gal/yr)/(52 wks/yr)*(3 wks)*($0.85/gal) = $2.45 million

Replacement cost of materials = $7.5 million

2 days of C grade production = (50 million gal/yr)/(52 wks/yr)/(7 days/wk)*(2 days)*(($0.85-$0.65)/gal) = $55,000

1 day of B grade production = (50 million gal/yr)/(52 wks/yr)/(7 days/wk)*(1 day)*(($0.85-$0.73)/gal) = $16,500

TOTAL COST DUE OF REPLACEMENT = $10 MILLION (future dollars)

     Calculating the present value of the money:

PV = ($10 million) (1/(1+0.085))^-10 = $4.4 million

     So what this really means is that 10 years later, you'd actually be adding another $4.4 million to the installed cost of the plant.  With this in mind, another comparison the installed costs shows:

I-600 plant installed cost = $37 million + $4.4 million = $41.4 million

SS plant installed cost = $40 million

     All of the sudden, the cheaper (more attractive) material that was a "good decision" for companies in the past, has become a $1.4 million mistake!  So, if the results shown here were the result of a detailed economic analysis, your options to share with the company executives would be:

1.  Spend $37 million for the plant now and another $10 million in the future for repairs.  Total present cost of the I-600 plant = $41.1 million.

2.  Spend $40 million for the plant now and significantly reduce the risk of  shutdown to replace material and disgruntled customers.

     If your company has $40 million, I suspect that they'd choose option "2"!

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