## 2880 Kj Of Solar Energy

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Repeat Example 4.2 by considering the system to have a fully mixed storage tank of 100 L and no load. The initial storage tank temperature at the beginning of the day is 40°C and the environmental temperature at the area where the storage tank is located is equal to the ambient air temperature. The tank UA value is 12 W/°C. Calculate the useful energy collected over the day.

### Solution

By using Eq. (5.34), the new storage tank temperature can be considered as the collector inlet. This is correct for the present example but is not very correct in practice because some degree of stratification is unavoidable in the storage tank.

12 X 3600 (Ts 1000 s

The results in this case are shown in Table 5.4. Therefore, the total energy collected over the day = 18350.3 kJ. As can be seen from the results of this example, the collector performance is somewhat lower than those of Example 4.2 because a higher collector inlet s—n

 Time Ta (°C) It (kJ/m2) T (°C) AT/Gt (°C-m2/W) 0 (deg.) K0 Qu (kJ) 6 25 360 40.0 0.150 93.9 0 0 7 26 540 38.6 0.084 80.5 0.394 0.0 8 28 900 37.5 0.038 67.5 0.807 722.2 9 30 1440 38.5 0.021 55.2 0.910 1651.6 10 32 2160 41.9 0.016 44.4 0.952 2734.2 11 34 2880 47.6 0.017 36.4 0.971 3707.9 12 35 3420 55.2 0.021 33.4 0.976 4268.2 13 34 2880 63.2 0.036 36.4 0.971 3078.6 14 32 2160 67.3 0.059 44.4 0.952 1706.1 15 30 1440 67.4 0.094 55.2 0.910 481.5 16 28 900 64.4 0.146 67.5 0.807 0.0 17 26 540 60.5 0.230 80.5 0.394 0 18 25 360 56.8 0.318 93.9 0 0

temperature leads to lower collector efficiency. In this example, too, the use of a spreadsheet program greatly facilitates estimations.

The density of water (and other fluids) drops as its temperature increases. When hot water enters from the collectors and leaves for the load from the top of the tank and cool water flows (cold water returns to the collector and make-up water supply) occur at the bottom, the storage tank will stratify because of the density difference. Additionally, with cool water at the tank bottom, the temperature of water fed to the collector inlet is low, thus the collector performance is enhanced. Moreover, water from the top of the tank, which is at the highest temperature, may meet the heating demand more effectively. The degree of stratification is measured by the temperature difference between the top and bottom of the storage tank and is crucial for the effective operation of a solar system.

There are basically two types of models developed to simulate stratification: the multimode and the plug flow. In the former, the tank modeled is divided into N nodes (or sections) and energy balances are written for each node. This results in a set of N differential equations, which are solved for the temperatures of the N nodes as a function of time. In the latter, segments of liquid of various temperatures are assumed to move through the storage tank in plug-flow and the models keep track of the size, temperature, and position of the segments. Neither of these methods is suitable for hand calculations; however, more details of the plug flow model are given here.

The procedure is presented by Morrison and Braun (1985) and is used in conjunction with TRNSYS thermosiphon model presented in Section 5.1.1.

This model produces the maximum degree of stratification possible. The storage tank is initially represented by three fluid segments. Initially, the change of tank segment temperatures, due to heat loss to the surroundings and conduction between segments, is estimated. The energy input from the collector is determined by considering a constant temperature plug of fluid of volume Vh(= m At/p) entering the tank during the time step At. The plug of fluid entering the tank is placed between existing segments chosen to avoid developing a temperature inversion.

The load flow is considered in terms of another segment of fluid of volume, VL(=mL At/p), and temperature TL, added either to the bottom of the tank or at its appropriate temperature level. Fluid segments are moved up the tank as a result of the addition of the new load flow segment. The net shift of the profile in the tank above the collector return level is equal to the load volume, VL, and that below the collector return is equal to the difference between the collector and load volumes (Vh - VL). After adjusting for the load flow, the auxiliary input is considered, and if sufficient energy is available, segments above the auxiliary input level are heated to the set temperature. According to the situation, the segment containing the auxiliary element is split so that only segments of the tank above the element are heated.

Segments and fractions of segments in the new tank profile that are outside the bounds of the tank are returned to the collector and load. The average temperature of the fluid delivered to the load is given by

where j and a must satisfy i= j-1

The average temperature of fluid returned to the collector is

where n and b must satisfy

The main advantage of this tank model is that small fluid segments are introduced when stratification is developing, while zones of uniform temperature, such as those above the auxiliary heater, are represented by large fluid segments. Additionally, the size of fluid segments used to represent the tank temperature stratification varies with collector flow rate. If the collector flow rate is high, there will be little stratification in the preheat portion of the tank and the algebraic model will produce only a few tank segments. If the collector flow rate is low and the tank is stratified, then small tank segments will be generated. Generally, the number of segments generated in this model is not fixed but depends on many factors, such as the simulation time step, the size of the collector, load flow rates, heat losses, and auxiliary input. To avoid generating an excessive number of segments, adjacent segments are merged if they have a temperature difference of less than 0.5°C.