midterm_soln - J. Buongiorno / Spring 2007 22.313J, 2.59J,...

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J. Buongiorno / Spring 2007 22.313J, 2.59J, 10.536J THERMAL-HYDRAULICS IN POWER TECHNOLOGY OPEN BOOK MID-TERM QUIZ (solutions) 1.5 HOURS Problem 1 (50%) – Bubbly flow of air in a vinegar fermentation tank i) Note that since the vinegar velocity is zero (i.e., the vinegar is stagnant), the air bubble rise velocity coincides with the bubble-liquid relative velocity, v b . For air bubbles of 1-mm equivalent diameter in vinegar we have M 3 × 10 -11 , Eö 0.1 and, thus, from the Re-Eö-M diagram, Re 10 2 . Therefore, from the definition of Re, we get a bubble rise velocity v b 0.1 m/s. ii) The volume of vinegar in the tank is V vin =1.5 m 3 . The tank cross sectional area is A tank = π /4 D 2 1.13 m 2 , where D=1.2 m is the tank diameter. Therefore, the vinegar level prior to air injection is L o =V vin /A tank 1.33 m. Upon air injection the level rises to accommodate the air volume. The total volume of the air-vinegar mixture is V tot =V air +V vin . Since V air = α V tot , one gets V tot =V vin /(1- α ), and thus the new level, L, is: L=V tot /A tank =V vin /[(1- α )A tank ] ( 1 ) where α is the void fraction. According to the drift flux model, the void fraction can be calculated as: vj o v V j C j + = α ( 2 ) where C o =1 and V vj =v b =0.1 m/s, as per the hint in the problem statement. j v and j=j v +j are the air and total superficial velocities, respectively. However, in our case
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midterm_soln - J. Buongiorno / Spring 2007 22.313J, 2.59J,...

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