# Hot Water System Optimization Example

When a solar energy system is designed, the engineer seeks to find a solution that gives the maximum life cycle savings of the installation. Such savings rep­resent the money that the user/owner will save because of the use of a solar energy system instead of buying fuel. To find the optimum size system that gives the maximum life cycle savings, various sizes are analyzed economically. When the present values of all future costs are estimated for each of the alter­native systems under consideration, including solar and non-solar options, the system that yields the lowest life cycle cost or the maximum life cycle savings is the most cost effective.

As an example, a graph of life cycle savings against the collector area is shown in Figure 12.1. For this graph, all other parameters except the collec­tor area are kept constant. As can be seen, the life cycle savings start at a nega­tive value for a collector area equal to 0, representing the total value of money required for fuel for a non-solar energy system, and as solar collectors are added to the system, the life cycle savings reach a maximum and then drop. An increase in collector area beyond the maximum point gives lower life cycle savings (than the maximum value), as the bigger initial expenditure required for the solar energy system cannot replace the cost of the fuel saved. It even gives nega­tive life cycle savings for large areas in multiplication of the optimum value, which represent the money lost by the owner in erecting and operating the solar energy system instead of buying the fuel. FiGURE 12.1 Variation of collector area with life cycle savings.

In previous examples, the annual fraction of load met by the solar energy system and the collector area, and thus the solar energy system cost, were given. In the following example, the relationship between the solar fraction, F, and collec­tor area, obtained from thermal performance calculations, are given instead, so the objective is to find the area (system size) that gives the highest life cycle savings.

Example 12.8

If, in Example 12.7, the area-dependent costs are \$250/m2 and the area – independent cost is \$1250 and all other parameters are kept constant, find the optimum area of the solar energy system that maximizes the life cycle savings. From a thermal analysis of the solar process, the relationship of the collector area and solar fraction are as given in Table 12.7.

Table 12.7 Relationship of Collector Area and Solar Fraction in Example 12.8

 Area (m2) Annual solar fraction (F) 0 0 25 0.35 50 0.55 75 0.65 100 0.72 125 0.77 150 0.81

Solution

To solve this problem, the life cycle savings method needs to be applied for each collector area. A spreadsheet calculation of this can very easily be done by changing the collector area and the first year fuel savings, which can be estimated from the annual solar fraction. The complete results are shown in Table 12.8.

Table 12.8 First-Year Fuel and Solar Savings for Example 12.8

 Area (m2) Annual solar fraction (F) Installed cost (\$) First-year fuel savings (\$) Solar savings (\$) 0 0 1,250 0 -5,680.7 25 0.35 7,500 691.8 3,609.9 50 0.55 11,250 1,087.1 6,898.9 75 0.65 20,000 1,284.7 6,186.1 100 0.72 26,250 1,423.1 4,275.0 125 0.77 32,500 1,521.9 1,562.3 150 0.81 38,750 1,600.9 -1,551.2

The life cycle solar savings are plotted against the collector area in Figure 12.2. As can be seen, the maximum occurs at Ac = 60 m2, where the LCS = \$7,013.70. FiGURE 12.2 Life cycle solar savings for each collector area.

12.3.1 Payback Time

The payback time is defined in many ways. Three of the most useful ones are shown in Section 12.1. As indicated, the most common one is the time needed for the cumulative fuel savings to equal to the total initial investment, i. e., it is the time required to get back the money spent to erect the solar energy system from the fuel savings incurred because of the use of the system. This time can be obtained with or without discounting the fuel savings.

NOT DISCOUNTING FUEL SAVINGS

Initially, we consider that the fuel savings are not discounted. The fuel saved in a year (j) is given by

Fuel saved in year j = FLCF1(1 + iF )j_1 (12.23)

where

F = solar fraction, obtained from Eq. (12.1).

CF1 = first year unit energy cost delivered from fuel (like parameters CFA and CFL, it is the product of fuel heating value and heater efficiency) (\$/GJ). iF = fuel inflation rate.

It should be noted that the product FL represents the energy saved because of the use of solar energy. Summing the fuel saved in year (j) over the payback time (nP) and equating to the initial system cost, (Cs), given by Eq. (12.2), gives

np

ЁFLCf1 (1 + iF Г1 = Сх (12.24)

j=1

The summation gives FLCfi [(1 + iF)np – ij  iF

Another way to determine the payback time is to use the PWF values tabu­lated in Appendix 8. Here, the sum of the fuel savings is given by the multi­plication of the first year’s saving, FLCF1 and PWF at zero discount rate. In equation form, this is given by

FLCF1 X PWF(nP, iF, 0) = Cs (12.27)

Therefore, the payback time can be found by interpolation from the tables in Appendix 8, for which, PWF = Cs/FLCF1.

Example 12.9

Find the undiscounted payback time of a solar energy system that covers 63% of an annual load of 185 GJ and costs €15,100. The first-year fuel cost rate is €9.00/GJ and inflates at 9% per year.

Solution    Using Eq. (12.26),

The same result can be obtained from tables of Appendix 8. In this case, PWF = CJFLCF1 = 15,100/(0.63 X 185 X 9) = 14.395. This is very close to the first value (d = 0) of the column for i = 9% for n = 10 (PWF = 14.487). Interpolation can be used to get a more correct answer, but the exact result can be obtained with more accuracy with Eq. (12.26).

DISCOUNTING FUEL SAVINGS

The procedures followed to equate discounted fuel costs to initial investment are similar. For discounted fuel costs, Eq. (12.27) becomes FLCF1 X PWF(nP, iF, d) = Cs

Similarly, the payback time is given by the following.   If iF Ф d,   If iF = d,

Example 12.10

Repeat Example 12.9 with a fuel costs discounted at a rate of 7%.

Solution    Using Eq. (12.29),

Other definitions of payback time are the time required for the annual solar savings to become positive and the time required for the cumulative solar sav­ings to become 0. These can be determined by life cycle analysis. Using the results of Example 12.7, the payback time is

1. The time required for the annual solar savings to become positive = 4 years

2. The time required for the cumulative solar savings to become zero = 15 years