## Mathematical Model

The cross-sectional view of the used mechanical model for the thermal stratifications for numerical analysis is shown in Fig. 1.

Boundary layer equations were used to determine the temperature and velocity distributions in the flow field. The analysis was based on the unsteady, three-dimensional continuity, momentum and energy equations.

The assumptions are used in the present study as:

— The flow is unsteady, turbulent and three-dimensional,

— The thermal conductivity of the tube sheet material does not change with temperature

— The tube material is homogeneous and isotropic.

Three dimensional continuity, momentum and energy equations are solved numerically. The upwind and central difference method used for convections and diffusions, respectively [9].

The hot water temperature is desired to keep constant as long as possible in the upper part of the tank. Therefore, different obstacles are placed into the tank in order to supply higher thermal stratifications. The schematic view of the obstacle type and geometries are shown in figure 2. The details of these geometries are shown in figure 3. Obstacle types and tank matches are

Table 1 Obstacle types and tanks matches

 Obstacle types The obstacle placed tank models. 1 7,8 2 9,10 3 12 4 11 5 12 6 1,2,3,4,5,6

The water temperatures are assumed 333, 320 and 285 for water exit from the tank to usage, water in the tank and water coming from the main lines, respectively.

V.

Figure 1. Cross-sectional view of the used tank model

j________________ ______________________ 1______________ ______________________ L

 1___ . .__ __ 2___ і . 3 _____ 4 11 ‘ .. •■***’*•. .. …………………… • ‘ 1 ‘ n 5 6 9 10 11 12

Figure 2 Obstacle geometries and its assembly shape in the tank.

The dimensions of the obstacles in figure 3 are shown in Table 2. Table 2 The dimensions of the obstacles

 Engel no a s (m) t (m) r1 (m) r2(m) R. i(m) r4(m) l (m) r5 1 — 0.8 0.02 0.96 0.2 — — — — 2 — 0.8 0.2 0.96 0.0 — — — — 3 — — 0.2 — — 1.0 — — — 4 — — 0.2 — — 1.0 0.2 — — 5 — — 0.2 — — — — 0.8 6 20“ — 0.02 — — — 0.8 —