Question

Engineering Economics
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A reactor, which will contain corrosive liquid, has been design. If the reactor is made of mild steel, the initial installed cost will be $6000, and the useful life period will be 4 years. Since stainless steel is highly resistant to the corrosive action of the liquids, stainless steel, as the material of construction, has been proposed as an alternative to mild steel, the stainless steel reactor would have an initial installed cost of $16000. The scrape value at the end off the useful life would be zero for either type of reactor, and both could be replaced at a cost equal to the original pri On the basis of equal capitalized cost for both types of reactors, what should be the useful life period for the stainless steel reactor if money is worth 4 percent ?compounded annually 5 نقاط = The capitalized cost-1 $41300.2 O $41400.6 $41323.5 O قم The useful life period for the stainless steel reactor -2 10.4 years o 11.5 years 12.5 years

Answer 1

1. Number of years n=4 yrs Installed Cost of first reactor C1=$ 6000 Interest I=4 %=4/100=0.04 Capitalized cost of First reactor K1=C1+C1/((1+1) ^n-1) K1= $ 6000+ $ 6000 / [(1.04) ^4-1] K1= $6000+ $ 6000/0.16986= $ 6000+ $35323.5 Kl=$ 41323.5 Capitalized cost of First reactor=$ 41323.5 Option A is in-corrected as calculated value is not equal to given value. Option B is in-corrected as calculated value is not equal to given value. Option C is corrected as calculated value is equal to given value. 2. Installed cost of second reactor C2= $16000 Interest I=4 %=4/100=0.04 Number of years n=4 yrs Capitalized cost of second reactors K2=C2+C2/ [(1+1) ^n-1] Since capitalized cost of reactors are equal K2=K1 K2=$ 41323.5 $ 41323.5= C2+C2/((1+1) ^n-1] $ 41323.5= $ 16000+$ 16000/((1+0.04) ^n-1] $ 25323.5= $ 16000/[1.04^n-1]

0.632=1.04^n-1 1.632 =1.04^n Take In on both sides as In X^n=n*In X Ln 1.632/ ln 1.04=n n=12.5 yrs Useful life period of second reactor=12.5 years Option A is incorrect as calculated useful life period is not equal to given value. Option B is incorrect as calculated useful life periodis not equal to given value. Option C is correct as calculated useful life period is equal to given value.