Question

# 1. An agitator is a device, which produces turbulent mixing in a tank (for example, this can be an oscillating grid without a mean flow). The following experiment was carried out to test the mixing efficiency of an agitator in a large tank, i.e., side wall or bottom boundary effects are negligible: 100 g of a 10% NaCl solution (25°C) was released at the water surface and the following concentration measurements were made at a point located 2 m below the release point: a) b) Time, hrs Concentration, ppb 0 0.5 1 1.5 d) 2 3 4 2 130 175 160 127 83 67 Assuming homogeneous, isotropic and stationary turbulence determine the diffusivity. [5 pts] If in actuality the tank is 10 m deep, at what time would the tank bottom exert a significant effect on the concentration value at the measuring point? It is up to you to define "significant effects"! What is the concentration at 5, 10 and 20 hours of the start, respectively? [5 pts] c) If the agitator were not operating (i.e., molecular diffusion), how long would it take for the concentration to reach its maximum value at the distance of 2m? What is the maximum value? The water temperature is 25°C. [5 pts] Assuming that, for some reason, the salt solution is undergoing a first order loss with a half time of 3.5 hours, how would the answers obtained under b) change? [5 pts]/nd) Assuming that, for some reason, the salt solution is undergoing a first order loss with a half time of 3.5 hours, how would the answers obtained under b) change? [5 pts] Note: Do the problem using an analytical solution. If you also do part a and b using a 2D rectangular numerical approach, you will receive 10 pts extra credit. This is much easier to do in Matlab than in Excel even though Excel works great for a few time steps... Solutions: (a) Your solution can vary somewhat depending on your approach, but I found E=2.1 cm2/s works well. (b) The 2 main images at z=20 m and z--20 m do not affect the solution for the 5, 10, and 20 hrs. You would have to wait until close to 1000 hours for the images to make an impact. t (hours) 0 5 10 20 t (sec) 0 18000 36000 72000 (c) Solution: Max concentration Time- (d) t (hours) t (sec) 0 5 10 0 18000 36000 72000 M (mg) 10000 10000 10000 10000 E (cm³/s) 2.1 2.1 M (mg) 10000 10000 10000 10000 2.1 2.1 184.03 ppb 13.00 years E (cm³/s) 2.1 NİNİN Cpeak 0.00E+00 6.11E+01 2.16E+01 7.64E+00 Cpeak 0.00E+00 2.1 6.11E+01 2.1 2.16E+01 2.1 7.64E+00 C(t) ppb 0.00 46.89 18.92 7.19 C(t) ppb 0.00 17.42 2.61 0.14  Fig: 1  Fig: 2