Question 9.5: An owner has asked the design team to provide a recommendati......
An owner has asked the design team to provide a recommendation on the most eco nomical among three design options for an electrical transformer. The three options are briefly described here.
(A) Use a premium energy efficiency transformer with an efficiency of 99.7%. The additional cost of this transformer is $10,000 relative to a standard transformer.
(B) Use a medium energy-efficiency transformer with an efficiency of 99.5%. The incremental cost of this transformer relative to a standard transformer is $4,000.
(C) Use a standard energy-efficiency transformer with an efficiency of 99.3%.
Determine the best cost-effective option for the building owner. Assume that the life time of the building electrical system project is 30 years and the discount rate is 5%. The building has a peak load of 750 \ kVA with annual average load factor of 0.876 and power factor of 0.90. Use all the life cycle cost analysis method to determine the best option assuming an electricity cost of $0.12/kWh.
The total cost of operating the transformer is considered for the three options. In this analysis, the salvage value of the transformer and the maintenance fee are assumed to be the same for all three design options and thus are not considered. To present the calculations for LCC analysis, it is recommended to present the results in a tabular format and proceed as shown here:
Therefore, the life cycle cost for option A is the lowest. Thus, it is recommended for the building owner to use a premium efficiency transformer.
This conclusion is different from that obtained by using the simple payback analysis. (Indeed, the payback period for option A—relative to the base case C—is SPP(A)=(\$10,000)/(\$1,400/year)=7.0 \ years, while for option B, SPP(B)=(\$4,000)/(\$718/year)=5.6 \ years.
| Cost Item | Option A | Option B | Option C | |
| Initial investment | ||||
| (a) Incremental cost ($) | 10,000 | 4,000 | 0 | |
| Annual operating costs: | ||||
| (b) Electricity use (kWh/year) | 2,965,396 | 2,971,357 | 2,977,341 | |
| (c) Fuel cost ($) [$0.12 ∗ (b)] | 355,848 | 356,363 | 357,281 | |
| USPW factor | ||||
| d = 5%, N = 30, Equation 9.12
USPW(d,N)=\frac{P}{A}=\frac{(1+d)^N-1}{d(1+d)^N}=\frac{1-(1+d)^{-N}}{d} (9.12) |
15.37 | 15.37 | 15.37 | |
| Present worth ($) | ||||
| (a) + USPW ∗ © | 54,802,449 | 5,485,244 | 5,492,284 | |