There are two customers requiring three-phase electrical service, one existing at location A and a new customer at locat

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There Are Two Customers Requiring Three Phase Electrical Service One Existing At Location A And A New Customer At Locat 1 (54.47 KiB) Viewed 11 times

There Are Two Customers Requiring Three Phase Electrical Service One Existing At Location A And A New Customer At Locat 2 (16.65 KiB) Viewed 11 times

There are two customers requiring three-phase electrical service, one existing at location A and a new customer at location B. The load at location A is known to be 110 kVA, and at location B it is contracted to be 280 kVA. Both loads are expected to remain constant indefinitely into the future. Already in service at A are three 100-KVA transformers that were installed some years ago when the load was much greater. Thus, the alternatives are as follows: Alternative A: Install three 100-KVA transformers (new) at B now and replace those at A with three 37.5-kVA transformers only when the existing ones must be retired. Alternative B: Remove the three 100-kVA transformers now at A and relocate them to B. Then install three 37.5-kVA transformers (new) at A. The existing transformers have 10 years of life remaining. Suppose that the before-tax MARR = 8% per year. Use the capitalized worth method to recommend which action to follow. Assume an infinite analysis period with repetitive cycles of replacement and ignore income taxes.
Capital Investment Equipment Installation Property tax Removal cost Market value Useful life (years) Existing and New Transformers Three 37.5-kVA $1,300 $440 3% of capital investment $150 $150 20 Three 100-kVA $2,300 $800 3% of capital investment $125 $125 20
2355N 600 AWNIN 7 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 30 35 40 45 50 Single Payment Discrete Compounding; i = 8% Compound Amount Factor Compound Amount Present Factor To Find F Given P F/P 1.0800 1.1664 1.2597 1.3605 1.4693 1.5869 1.7138 1.8509 1.9990 2.1589 2.3316 2.5182 2.7196 2.9372 3.1722 3.4259 3.7000 3.9960 4.3157 4.6610 5.0338 5.4365 5.8715 6.3412 6.8485 10.0627 14.7853 21.7245 31.9204 46.9016 Worth Factor To Find P Given F P/F 0.9259 0.8573 0.7938 0.7350 0.6806 0.6302 0.5835 0.5403 0.5002 0.4632 0.4289 0.3971 0.3677 0.3405 0.3152 0.2919 0.2703 0.2502 0.2317 0.2145 0.1987 0.1839 0.1703 0.1577 0.1460 0.0994 0.0676 0.0460 0.0313 0.0213 To Find F Given A F/A 1.0000 2.0800 3.2464 4.5061 5.8666 7.3359 8.9228 10.6366 12.4876 14.4866 16.6455 18.9771 21.4953 24.2149 27.1521 30.3243 33.7502 37.4502 41.4463 45.7620 50.4229 55.4568 60.8933 66.7648 73.1059 Uniform Series 113.2832 172.3168 Present Worth Factor To Find P Given A PIA 0.9259 1.7833 2.5771 3.3121 3.9927 4.6229 5.2064 5.7466 6.2469 6.7101 7.1390 7.5361 7.9038 8.2442 8.5595 8.8514 9.1216 9.3719 9.6036 9.8181 10.0168 10.2007 10.3711 10.5288 10.6748 11.2578 11.6546 259.0565 11.9246 386.5056 12.1084 573.7702 12.2335 Sinking Fund Factor To Find A Given F A/F 1.0000 0.4808 0.3080 0.2219 0.1705 0.1363 0.1121 0.0940 0.0801 0.0690 0.0601 0.0527 0.0465 0.0413 0.0368 0.0330 0.0296 0.0267 0.0241 0.0219 0.0198 0.0180 0.0164 0.0150 0.0137 0.0088 0.0058 0.0039 0.0026 0.0017 Capital Recovery Factor To Find A Given P A/P 1.0800 0.5608 0.3880 0.3019 0.2505 0.2163 0.1921 0.1740 0.1601 0.1490 0.1401 0.1327 0.1265 0.1213 0.1168 0.1130 0.1096 0.1067 0.1041 0.1019 0.0998 0.0980 0.0964 0.0950 0.0937 0.0888 0.0858 0.0839 0.0826 0.0817
The CW value for the alternative A is S The CW value for the alternative B is S (Round to the nearest dollar.) (Round to the nearest dollar.)