# Hot Water System Example

The example in this section considers a complete solar water heating system. Although different solar energy systems have different details, the way of han­dling the problems is the same.

Example 12.7

A combined solar and auxiliary energy system is used to meet the same load as in Example 12.5. The total cost of the system to cover 65% of the load (solar fraction) is \$20,000. The owner will pay a down payment of 20% and the rest will be paid over a 20-year period at an interest rate of 7%. Fuel costs are expected to rise at 9% per year. The life of the system is considered to be 20 years, and at the end of this period, the system will be sold for 30% of its origi­nal value. In the first year, the extra maintenance, insurance, and parasitic energy costs are \$120 and the extra property tax is \$300; both are expected to increase by 5% per year. The general market discount rate is 8%. The extra property taxes and interest on the mortgage are deducted from the income tax, which is at a fixed rate of 30%. Find the present worth of the solar savings.

Solution

The values estimated for the various costs and savings for the entire life of the system are shown in Table 12.5. Year zero includes only the down payment, whereas the values for the first year are as given by the problem definition. In the table, the savings are positive and the expenses (or payments) are negative. The down payment is equal to 0.2 X 20,000 = \$4,000. The annual mortgage payment is found from Eq. (12.20):

M 20,000 X 0.8 16,000

Periodic payment = = =

PWF(nL,0, dm) PWF(20, 0, 0.07) 10.594

= \$1,510.29

Table 12.5 Estimated Costs and Savings for the System in Example 12.7

 1 2 3 4 5 6 7 8 Year Extra mortgage payment (\$) Fuel savings (\$) Extra insurance, maintenance, and parasitic cost (\$) Extra property tax (\$) Income tax savings (\$) Solar savings (\$) PW of solar savings (\$) 0 — — — — — -4,000.00 -4,000.00 1 -1,510.29 1,284.70 -120.00 -300.00 426.00 -219.59 -203.32 2 -1,510.29 1,400.32 -126.00 -315.00 422.30 -128.66 -110.31 3 -1,510.29 1,526.35 -132.30 -330.75 418.26 -28.73 -22.80 4 -1,510.29 1,663.72 -138.92 -347.29 413.84 81.07 59.59 5 -1,510.29 1,813.46 -145.86 -364.65 409.01 201.66 137.25 6 -1,510.29 1,976.67 -153.15 -382.88 403.73 334.08 210.52 7 -1,510.29 2,154.57 -160.81 -402.03 397.98 479.42 279.74 8 -1,510.29 2,348.48 -168.85 -422.13 391.71 638.92 345.19 9 -1,510.29 2,559.85 -177.29 -443.24 384.88 813.91 407.16 10 -1,510.29 2,790.23 -186.16 -465.40 377.45 1,005.83 465.89 11 -1,510.29 3,041.35 -195.47 -488.67 369.36 1,216.29 521.64 12 -1,510.29 3,315.07 -205.24 -513.10 360.57 1,447.01 574.63 13 -1,510.29 3,613.43 -215.50 -538.76 351.01 1,699.89 625.05 14 -1,510.29 3,938.64 -226.28 -565.69 340.63 1,977.01 673.10 15 -1,510.29 4,293.12 -237.59 -593.98 329.37 2,280.63 718.95 16 -1,510.29 4,679.50 -249.47 -623.68 317.14 2,613.20 762.77 17 -1,510.29 5,100.65 -261.94 -654.86 303.89 2,977.44 804.71 18 -1,510.29 5,559.71 -275.04 -687.61 289.51 3,376.29 844.91 19 -1,510.29 6,060.08 -288.79 -721.99 273.94 3,812.95 883.51 20 -1,510.29 6,605.49 -303.23 -758.09 257.07 4,290.95 920.62 20 6,000.00 1,287.29 Total present worth of solar savings \$6,186.07

The first year fuel savings is 114.9 GJ X 0.65 = 74.69 GJ. According to Example 12.5, this corresponds to \$1,284.70.

The interest paid for the first year = 16,000 X 0.07 = \$1,120

The principal payment = 1,510.29 — 1,120 = \$390.29 Principal balance = 16,000 — 390.29 = \$15,609.71 Tax savings = 0.3(1,120 + 300) = \$426

The annual solar savings is the sum of the values in columns 2 to 6. These are then brought to a present worth value using the market discount rate of 8%. Year 20 is repeated twice so as to include the resale value of 20,000 X 0.3 = \$6,000. This is a positive value as it is a saving.

The sum of all the values in the last column is the total present worth of the savings of the solar energy system as compared to a fuel-only system. These are the savings the owner would have by installing and operating the solar energy system instead of buying fuel for a conventional system.

As can be understood from the analysis, a supplementary table is required with analysis of the money borrowed (to find tax savings) and cumulative solar savings, as in Table 12.6. The last column is required in the estimation of payback time (see Section 12.2.4).

Table 12.6 Supplementary Table for Example 12.7

 Year Interest paid (\$) Principal payment (\$) Principal balance (\$) Cumulative solar savings (\$) 0 0 0 16,000.00 -4,000.0 1 1,120.00 390.29 15,609.71 -4,203.3 2 1,092.68 417.61 15,192.10 -4,313.6 3 1,063.45 446.84 14,745.26 -4,336.4 4 1,032.17 478.12 14,267.14 -4,276.8 5 998.70 511.59 13,755.55 -4,139.6 6 962.89 547.40 13,208.15 -3,929.1 7 924.57 585.72 12,622.43 -3,649.3 8 883.57 626.72 11,995.71 -3,304.1 9 839.70 670.59 11,325.12 -2,897.0 10 792.76 717.53 10,607.59 -2,431.1 11 742.53 767.76 9,839.84 -1,909.5 12 688.79 821.50 9,018.34 -1,334.8 13 631.28 879.01 8,139.33 -709.8 14 569.75 940.54 7,198.80 -36.7 15 503.92 1,006.37 6,192.42 682.3 16 433.47 1,076.82 5,115.60 1,445.0 17 358.09 1,152.20 3,963.41 2,249.7 18 277.44 1,232.85 2,730.56 3,094.7 19 191.14 1,319.15 1,411.41 3,978.2 20 98.80 1,411.49 0 4,898.8 20 6,186.1

Another way of solving this problem is to carry out separate life cycle anal­yses for the solar and non-solar energy systems. In this case, the total present worth of the solar savings would be obtained by subtracting the total present worth of the two systems. It should be noted, however, that, in this case, equip­ment common to both systems needs to be considered in the analysis, so more information than that given is required.

Updated: August 29, 2015 — 5:07 pm