CHE249H1 Lecture Notes - Lecture 8: Embankment Dam

• School: UTSG
• Department: Chemical Engineering and Applied Chemistry
• Course: CHE249H1
• Professor: Yuri Lawryshyn

Document Summary

Pw= $ 300 k k (p / a ,5 ,20)= $ 474 470. 80 earthen dam: Pw= $ 220 k k ( p / a ,5 , 20) $ 35k (p / f ,5 ,8) k (p / f ,5 ,16) Therefore, the concrete reservoir is less expensive and should be chosen. Aw = k ( a / p ,5 ,20) k= . 00 / yr earthen dam: Fc+wood replacement= $ 220k k ( p / f ,5 , 8) k ( p / f ,5 ,16) Aw = . 7215 k ( a / p ,5 ,20) k = $ 41829. 66 / yr. Again ,the concrete reservoir is thebetter choice . *perpetual annuity ( p / a ,i, )= 1 i. Pw= $ 300 k k (p / a ,5 , )= 000. 00 earthen dam: Wood replacement annuity= $ 35k ( a / p ,5 ,8 )= 414. 50 / yr.