Gyi y2

Assume temperature of water = 25°C; then pw = 997 kg/m ; therefore,

0.048(997)(9.81)(0.94 - 0.05)[J0I048) (0.94 + 0.05) - 2(9.81)(0.05)2(0.94)

P, hydJump

Example 6.4 For the problem in Example 6.3, determine if the power dissipated conforms to the requirement of effective mixing.

Solution: To determine if the power dissipated conforms to the criterion of effective mixing, the G value would be calculated and compared with the values of Table 6.2.

Vjump = 3 y2 W (y1 + y2) = 3(0.94)(0.1)(0.05 + 0.94) = 0.279 m3

FIGURE 6.10 Power dissipation in weir mixers.

From Table 6.2, at to = 5.81 sec, G should range from 1,500 per sec to 4,000 per sec. 2693.43 per sec is within this range and the design conforms to the criteria for effective mixing. Ans

6.4.4 Mixing Power for Weir Mixers

The power dissipation in weir mixers is a result of the conversion of the energy that the water possesses as it drops from the top of the weir to the bottom of the weir. To obtain the dissipation, apply the energy equation between points 1 and 2 in Figure 6.10. This will result in hf = H + Hd (6.34)

H is the head over the weir crest and HD is the drop provided from the weir crest to the surface of the water below. At the points directly below the falling water there is turbulence. As the particles of water reach point 2, however, turbulence ceases and the velocity becomes zero. In other words, the energy at the point of turbulence has been dissipated before reaching point 2. This dissipation is the power dissipation of mixing. Having obtained the friction loss, the power dissipation P is simply

Example 6.5 A suppressed rectangular is used to mix chemical in a wastewater treatment unit. The flow is 0.3 m /s. The length of the weir L and the height P were measured and found to be 2 m and 1 m, respectively. Calculate the power dissipation in the mixer if HD = 2.5 m.

Solution:

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